Category Archives: Coding

Bring your firefox 29 to classic look by installing classic Theme


A few days ago, the fine folks at Mozilla released Firefox version 29. The ridiculously high number itself is a result of their attempt to appear as on the ball as Google’s Chrome, which seems to release new versions just for sport. Anyway, this latest Firefox release was the first in a long long while to display any actual change in the user interface, again in an apparent attempt to keep up with Chrome. I’m not a fan of Chrome myself, and have always stayed with Firefox, largely due the Mozilla’s devotion to most of the principles of free and/or open source software. Nothing in particular against Google, but they do seem a bit nosy at times. The new interface for Firefox takes some getting used to. It’s not awful, but some buttons have changed and the tabs look different. Lots of folks seem to hate it. What we’re gonna do here is guide you through the easy process of returning Firefox to its previous user interface in appearance.

It’s pretty simple really. One add-on and you’re in business. Even if you like the new changes, this add-on is handy as it gives you additional things to tweak. The add-on is called Classic Theme Restorer (get it here), and will get your Firefox back to looking like what you’re used to and not like the Chrome wannabe its new appearance implies. Just download it, maybe play with a few settings, and Bob’s your uncle, as they say. I’m not gonna go to the trouble of doing before and after screen-shots, a step by step sequence of what settings to fiddle with, or any of that malarkey, mainly because it’s too much work and you’ll be able to see for yourself what’s changed and what changes back when you install the add-on. It’s really really simple to figure out what to do, and most of it gets done just by installing it. I’ve settled on a blend of features from both new and old, keeping mainly the new curved tabs from the new “Australis” theme, while changing a few things back to the old way. Lots of stuff to play with in the settings here. Have fun.

I’ll take an additional moment here to once more beat the drum for Linux and free software in general, of which Firefox is a great and probably the most popular example. I did a piece a while back on Linux, but to sum up, it’s easy to use, more secure, infinitely configurable, does everything the average user needs to do, is available in an astonishing number of flavors, and best of all, is completely free in (in many cases) every possible sense of the word. Freedom is good. Don’t let the Microsofts and the Apples fool you. Don’t be an insecure sheep. Give it a try. Or at least look into using some of the many great free/open source software products available for all platforms. Some of my favorites include VLC, which is a great player of almost any kind of media you can imagine, the GIMP, which does a lot of what Photoshop does and more, Clementine, an easy to use and lightweight music player/library program (among other things, it handles all 103,000 of my music files), FBReader, which will read almost any kind of e-book format, and of course LibreOffice, a full featured MS Office replacement that is quite compatible for most everyday uses. These are all completely free in every sense of the word, the most important being the freedom thing. The no money thing is nice too.




MySQL Database configuration on linux terminal tutorial

Start the database:

Start the database: /etc/rc.d/init.d/mysqld start
(The script will run mysql_install_db to create a default database in /var/lib/mysql/mysql/ if the mysql init script has never been run before. The install script will not be run again as long as the default database directory exists.)
The database executes as user mysqld and group mysqld.


  • One may manually initialize the database with the command: /usr/bin/mysql_install_db
    Creates system tables in /var/lib/mysql/mysql/
    Only execute the first time MySQL is installed.
  • Databases located in: /var/lib/mysql/
  • Default config file installed by RPM: /etc/my.cnf
    (Ubuntu: /etc/mysql/my.cnf)



Post installation:

  1. Admin user id: root
    Default password: blankThe first task is to assign a password:

    [prompt]$ mysqladmin -u root password 'new-password'

    Note: the following SQL commands will also work:

    mysql> USE mysql;
    mysql> UPDATE user SET Password=PASSWORD('new-password') WHERE user='root';
  2. Create a database: (Creates directory /var/lib/mysql/bedrock)
    [prompt]$ mysqladmin -h localhost -u root -ppassword create bedrock

    (or use SQL command: CREATE DATABASE bedrock;)
    Show all mysql databases: mysqlshow -u root -ppassword


  3. Add tables, data, etc:
    Connect to database and issue the following SQL commands:

    [prompt]$ mysql -h localhost -u root -ppassword
    mysql> show databases;             -- List all databases in MySQL.
    | Database |
    | bedrock  |
    | mysql    |
    | test     |
    mysql> use bedrock;     -- Specify database to connect to. Also refers to path: /var/lib/mysql/bedrock
    mysql> create table employee (Name char(20),Dept char(20),jobTitle char(20));
    mysql> DESCRIBE employee;       -- View table just created. Same as "show columns from employee;"
    | Field    | Type     | Null | Key | Default | Extra |
    | Name     | char(20) | YES  |     | NULL    |       |
    | Dept     | char(20) | YES  |     | NULL    |       |
    | jobTitle | char(20) | YES  |     | NULL    |       |
    3 rows in set (0.03 sec)
    mysql> show tables;
    | Tables_in_bedrock |
    | employee          |
    mysql> INSERT INTO employee VALUES ('Fred Flinstone','Quarry Worker','Rock Digger');
    mysql> INSERT INTO employee VALUES ('Wilma Flinstone','Finance','Analyst');
    mysql> INSERT into employee values ('Barney Rubble','Sales','Neighbor');
    mysql> INSERT INTO employee VALUES ('Betty Rubble','IT','Neighbor');

    Note: Data type used was CHAR. Other data types include:

    • CHAR(M) : Fixed length string. Always stores M characters whether it is holding 2 or 20 characters. Where M can range 1 to 255 characters.
    • VARCHAR(M) : Variable length. Stores only the string. If M is defined to be 200 but the string is 20 characters long, only 20 characters are stored. Slower than CHAR.
    • INT : Ranging from -2147483648 to 2147483647 or unsigned 0 to 4294967295
    • FLOAT(M,N) : FLOAT(4,2) – Four digits total of which 2 are after the decimal. i.e. 12.34 Values are rounded to fit format if they are too large.


  4. Add a user. Use the MySQL SQL console to enter SQL commands. The command mysql with the correct login/password will connect you to the database. The admin tables are stored in the database “mysql”.
    [prompt]$ mysql -h localhost -u root -ppassword
    Welcome to the MySQL monitor.  Commands end with ; or \g.
    Your MySQL connection id is 1 to server version: 3.23.41
    Type 'help;' or '\h' for help. Type '\c' to clear the buffer.
    mysql> USE mysql;
    mysql> SHOW TABLES;
    | Tables_in_mysql |
    | columns_priv    |
    | db              |
    | func            |
    | host            |
    | tables_priv     |
    | user            |
    mysql> INSERT INTO user (Host, User, Password, Select_priv) 
                  VALUES ('', 'Dude1', password('supersecret'), 'Y');
    mysql> FLUSH PRIVILEGES;   -- Required each time one makes a change to the GRANT table
    mysql> GRANT ALL PRIVILEGES ON bedrock.* TO Dude1;
    mysql> FLUSH PRIVILEGES;   -- Required each time one makes a change to the GRANT table
    mysql> quit


    • There is NO space between the -p and the password! You can omit the password and you will be prompted for it.
    • The SQL flush command is equivalent to issuing the command:
      [prompt]$ mysqladmin reload


  5. Test the database:
    mysql> SELECT * from employee;
    | Name            | Dept          | jobTitle    |
    | Fred Flinstone  | Quarry Worker | Rock Digger |
    | Wilma Flinstone | Finance       | Analyst     |
    | Barney Rubble   | Sales         | Neighbor    |
    | Betty Rubble    | IT            | Neighbor    |
    1 row in set (0.00 sec)
    mysql> SELECT name FROM employee WHERE dept='Sales';
    | name          |
    | Barney Rubble |
    1 row in set (0.00 sec)


  6. Quit from the SQL shell:
    [prompt]$ quit


  7. Shutting down the database:
    [prompt]$ mysqladmin -u root -ppassword shutdown       - PREFERRED
    [prompt]$ /etc/rc.d/init.d/mysqld stop
    [prompt]$ service mysqld stop

Documentation in /usr/share/doc/mysql-3.23.41/ (local file)


Users and Security:


Create a database user:

     mysql> CREATE USER david@'localhost' IDENTIFIED BY 'password';


or generate a user by adding them to the user table directly:

     [prompt]$ mysql -h localhost -u root -p 
     mysql> use mysql;
     mysql> INSERT INTO user (Host,User,Password) VALUES('localhost','david',PASSWORD('supersecret'));

Note that the user is defined in the “user” mysql table.


Assign user privileges:

Security and database access is controlled by the GRANT tables. Access to connect to the database and access to process the transaction (table and column access, etc.) are both required. Privileges are searched in the following order:

  1. user table
  2. db and host table
  3. tables_priv
  4. columns_priv

Use the user table to grant connection privileges to database by a user (host, user name and password). Grant database and table access for transaction access. i.e. grant “SELECT”, “UPDATE”, “CREATE”, “DELETE”, “ALTER” etc. permission for database, table, field (columns) or database server access.

Access can be granted by network permissions: GRANT ALL PRIVILEGES on bedrock.* to david@'';
This grants access from nodes – Or the network definitions can reference resolvable names: ‘‘. The host definition of ‘%‘ or ” (null) refers to any host. (..according to the documentation. My experience is that in the mysql.user table use only ‘%’ for “Host” to refer to any host.)

     mysql> GRANT ALL PRIVILEGES on bedrock.* to david@'%';

or (be specific)

     mysql> GRANT SELECT,INSERT,UPDATE,DELETE on bedrock.* to david@'%' identified by 'david';

or (more promiscuous – global privileges rather than database specific)

     mysql> GRANT ALL PRIVILEGES on *.* to david@'%' identified by 'david';

or (be specific by direct assignment in the mysql “Db” table:)

     mysql> use mysql;
     mysql> INSERT INTO db (Host,Db,User,Select_priv,Insert_priv) VALUES('localhost','bedrock','david','Y,'Y');

Note that database specific privileges (eg. Select_priv, Insert_priv, etc) are defined in the “db” mysql table. The mysql “user” table can assign the same (Select_priv, Insert_priv, etc) but global privileges (usually all default to ‘N’).Show privileges: SHOW GRANTS FOR Dude2@'%';

Network security: Use firewall rules (ipchains or iptables) to block internet access to port 3306. (default port used by MySQL)

Note: I have found that when adding access from “anywhere” (‘%’), the MySQL database table ‘user’ requires two entries, ‘localhost’ and ‘%’. Also, it is typically safer to allow more privileges to those with ‘localhost’ access than to users from ‘%’ (“anywhere”).


Passwords and connecting to the databse:

  • Connect: [prompt]$ mysql -h host_name -u user_name -ppassword
  • Using default blank password: [prompt]$ mysql -h localhost -u root -p
    If a password is required, you will be prompted. Note, blank passwords are a security hole which has already lead to one mySQL internet worm. Change any default blank passwords.
  • Delete null/blank users: DELETE FROM user WHERE User = '';
  • Beware of open access permissions from hosts '%': SELECT * FROM db WHERE Host = '%';
  • Change a password:
        [prompt]$ mysqladmin -u root -p password new-password

    You will be prompted to enter the old root password to complete this command.

        [prompt]$ mysqladmin -u root -pold-password password new-password


        mysql> SET PASSWORD FOR root@'localhost' = PASSWORD('supersecret');
        mysql> FLUSH PRIVILEGES;
  • As an added security precaution it is wise to delete any user id not used. i.e. any defaults generated for demonstration purposes.
  • Note that the default port used by MySQL is 3306. This can be protected with firewall rules. See the YoLinux IpTables tutorial.

Debian/Ubuntu upgrades: Note that the Debian/Ubuntu distribution will have an additional file /etc/mysql/debian.conf. This file holds a password for the user “debian-sys-maint” which is used by the install tool dpkg to perform database upgrades. This can also be used in emergencies if you forget the root password. It is also a security hole if the file is available to others.

[Potential Pitfall]: It is very easy to make mistakes which get entered into important tables. If you enter the command twice you may have one incorrect and one correct entry. Look at the table data after a mistake to see what happened in case it needs to be fixed.

mysql> USE mysql;
mysql> SELECT User,Password,Host from user;
| User  | Password         | Host       |
| root  | 99a1544eb571ad63 | localhost  |
|       |                  | localhost  |
| Dude1 | 81a10dba5f6f2144 |            |
| Dude1 |                  |            |
| Dude2 | 92b10dba6f7f3155 | %          |
5 rows in set (0.00 sec)
mysql> DELETE FROM user WHERE User='' AND Host='localhost';
mysql> DELETE FROM user WHERE User='Dude1' AND Password='';
mysql> QUIT

User entries may also be found in the table mysql.db.

mysql> DELETE FROM db WHERE User='Dude3' AND Host='localhost';

[Potential Pitfall]: Any changes (UPDATE) to the user table will require a “FLUSH PRIVILEGES” before the changes will be effective.

mysql> UPDATE user SET Host='%' WHERE User='Dude2';

This will allow a connection with mysql client from any host:
[prompt]$ mysql -u Dude2 -ppassword -h


MySQL root password recovery:

  1. As Linux system root user stop the database process: /etc/init.d/mysql stop
    (or: service mysql stop)
  2. Start MySQL in safe mode and skip the use of the “grant tables”: /usr/bin/mysqld_safe --user=mysql --socket=/var/lib/mysql/mysql.sock --pid-file=/var/run/mysqld/ --datadir=/var/lib/mysql --skip-grant-tables --skip-networking &
  3. Reset the MySQL root password: mysqladmin -u root flush-privileges password newpassword
  4. Stop MySQL running in safe mode: kill `cat /var/run/mysqld/`
  5. Start MySQL: /etc/init.d/mysql start
  6. The new MySQL root password can now be used: mysql -u root -p
    Respond with the password: newpassword


Disabling networking:

If your configuration is a web server interacting with a mySQL database running on the same “localhost” then one may turn off network access to tighten security. Edit shell script:

  • /usr/bin/safe_mysqld (Fedora Core 3)
  • /usr/bin/mysqld_safe (Red Hat Enterprise Linux 4 – MySQL 5.0)
  if test -z "$args"
    $NOHUP_NICENESS $ledir/$MYSQLD $defaults --basedir=$MY_BASEDIR_VERSION \
                      --datadir=$DATADIR $USER_OPTION --pid-file=$pid_file \
                      --skip-networking --skip-locking >> $err_log 2>&1
    eval "$NOHUP_NICENESS $ledir/$MYSQLD $defaults --basedir=$MY_BASEDIR_VERSION \
                      --datadir=$DATADIR $USER_OPTION --pid-file=$pid_file \
                      --skip-networking --skip-locking $args >> $err_log 2>&1"

Add the flag “--skip-networking” marked in bold.

Mysql 5.0 configuration: Networking is disabled by default on the default Red Hat and Ubuntu installation.

Red Hat/CentOS: To enable remote database access, add the “bind-address” with the public IP address to the file: /etc/my.cnf. To force local access only without remote access, set the “bind-address” to


Restart the database after making changes.

Ubuntu: To enable remote database access, comment out (or remove) the following line with a “#” in the file: /etc/mysql/my.cnf


bind-address           =


Restart the database after making changes.


A firewall rule can further restrict access to a single server (eg web server at

/sbin/iptables -A INPUT -i eth0 -s -p tcp --destination-port 3306 -j ACCEPT

or LAN only access:

/sbin/iptables -A INPUT -i eth0 -s -p tcp --destination-port 3306 -j ACCEPT


MySQL Admin Commands:


  • Statistics: [prompt]$ mysqladmin version
  • List database environment: [prompt]$ mysqladmin variables
  • Show if database is running: [prompt]$ mysqladmin ping
  • Show databases available:
    [prompt]$ mysqlshow
    | Databases |
    | bedrock   |
    | mysql     |
    | test      |


    mysql> SHOW DATABASES;
  • Delete database: mysql> drop database bedrock;
  • Show list of active threads in server:
    [prompt]$ mysqladmin -h localhost -u root -p processlist
    | Id | User | Host      | db | Command | Time | State | Info             |
    | 15 | root | localhost |    | Query   | 0    |       | show processlist |
  • Delete a database: [prompt]$ mysqladmin drop database-name
  • Execute SQL from Linux command line interface:
    [prompt]$ mysql -h localhost -u root -p -e "select host,db,user from db" mysql
  • Execute SQL command file from Linux command line interface:
    [prompt]$ mysql -h localhost -u root -p database-name < text-file-with-sql-statements.sql
  • Loadtest (benchmark) the system:
    [prompt]$ cd sql-bench
    [prompt]$ run-all-tests
    [prompt]$ mysql -vvf test < ./tests/auto_increment.tst


Sample SQL:

SQL requests are either administrative or data-related. The following are sample SQL segments and are not necessarily pertinent to the previous example:

Create and use a new database named “bedrock”:

mysql> CREATE DATABASE bedrock;   -- Comments follow a double dash
mysql> USE bedrock;

Create and populate table with data:

mysql> CREATE TABLE retired_employee (
                    Name char(20) DEFAULT '' NOT NULL,
                    Dept char(10) DEFAULT '' NOT NULL,
                    JobTitle char(20),
                    UNIQUE name_dept (Name,Dept)
mysql> CREATE UNIQUE index name_dept on employee (name,dept); -- avoids duplicate keys
mysql> INSERT INTO employee VALUES ("Jane Smith","Sales","Customer Rep");
mysql> INSERT INTO employee VALUES ('Jane Smith','Sales','Account Manager');
mysql> INSERT INTO employee VALUES ('Jane Smith','Engineerin','Manager');
mysql> UPDATE employee SET dept='HR' WHERE name='Jane Smith';
mysql> CREATE TABLE pet (
                 name VARCHAR(20), 
                 owner VARCHAR(20), 
                 species VARCHAR(20), 
                 sex CHAR(1), 
                 birth DATE, 
                 death DATE);

Add constraints to a table:

-- Use "auto_increment" integer column:
mysql> ALTER TABLE employee DROP INDEX name_dept;  -- get rid of index

Interrogate an existing database:

mysql> USE bedrock;
mysql> SELECT DATABASE();  -- returns current database. eg. bedrock
mysql> SELECT NOW();
mysql> SELECT USER();
mysql> DESC employee;
mysql> SHOW CREATE TABLE employee;  -- show command used to generate table
mysql> SHOW INDEX FROM employee;
mysql> SELECT DISTINCT dept FROM  bedrock;
mysql> SELECT * FROM  bedrock WHERE Name  LIKE "B%y";  -- "%" match any char: Gives Betty and Barney
mysql> SELECT * FROM  bedrock WHERE Name LIKE "B___y";  -- "_" match space: Gives Betty but not Barney
mysql> SELECT * FROM  bedrock WHERE Name RLIKE "^Betty$";  -- "^" match beginning. "$" to denote end of string
mysql> SELECT COUNT(*) FROM employee;  -- Number of records returned
mysql> SELECT Name, COUNT(*) FROM employee WHERE Name LIKE "B%y";  -- Return Names and number of records returned
mysql> SELECT * FROM pet WHERE species = "snake" OR species = "bird";
mysql> SELECT * FROM pet WHERE species = "dog" AND sex = "f";
mysql> SELECT * FROM pet WHERE birth >= "1998-1-1";
mysql> SELECT name, birth FROM pet ORDER BY birth DESC;
mysql> SELECT * FROM pet WHERE name LIKE "b%";
mysql> SELECT * FROM pet WHERE name REGEXP "^b";
mysql> SELECT species, COUNT(*) FROM pet GROUP BY species;
mysql> SELECT MAX(article) AS article FROM shop;
mysql> SELECT * FROM employee WHERE name LIKE "%Sm%";
mysql> SELECT * FROM employee WHERE name REGEXP "^Ja";

Database cleanup:

mysql> DROP TABLE employee;
mysql> DROP DATABASE bedrock;

See section 3 of MySQL manual for more examples.

Tip: Execute a shell command from the MySQL client interface, use either option:

  • system ls -l
  • \! ls -l

Example: execute the “ls” command to list files from the MySQL client.


Loading Data into the MySQL database:

Loading a SQL file into MySQL:

Import SQL file from MySQL client command line:

  • mysql> source file.sql
  • mysql> \. file.sql

The SQL file may have schema generation statements like CREATE TABLE ... or data load statements like INSERT INTO ... . The statements in the SQL file will be executed as if they were being specified at the MySQL client command line interface.


One may import data into the MySQL database from SQL files or “load” data from CSV or tab delimited files using the LOAD command:

Loading CSV or tab delimeted files into MySQL:

LOAD DATA LOCAL INFILE” vs “LOAD DATA INFILE“: The term “LOCAL” pertains to whether the file is local to the MySQL client. Without the keyword “LOCAL“, the datafile must reside on the same computer as the database server. The location of the client in this case would be irrelevant. The “LOAD DATA INFILE” has many file permission pitfalls and is thus trickey. In fact I have never been sucessful using this method with a user directory.


Load a tab delimited file into the database:

Command: LOAD DATA LOCAL INFILE 'file.dat' INTO TABLE employer;

Input tab delimited file: file.dat

Fred Flinstone  Quarry Worker   Rock Digger
Wilma Flinstone Finance Analyst 
Barney Rubble   Sales   Neighbor
Betty Rubble    IT      Neighbor


  • The number of tab delimeted fields MUST match the number and order of fields in the database.


Load a comma delimited file (CSV) into the database:




  • MS/Windows generated files will have lines terminated by “\r\n”.
  • Linux/Unix generated files will have lines terminated by “\n”.
  • File locations on database server must be absolute path names, relative path or relative to the mysqld process owner’s home directory (typically /var/lib/mysql/). File locations on the client may be fully qualified or relative to the current mysql client directory.
    • Fully qualified: /tmp/TableData.csv
    • Relative to current mysql client directory: ./TableData.csv
      (Verify current directory: mysql> \! pwd)
    • Database process owner home directory: TableData.csv
      (Actual: /var/lib/mysql/TableData.csv)
  • Text strings often are encapsulated by quotes so that the strings may contain a comma without representing a new field.

[Potential Pitfalls]:

  • ERROR 13 (HY000): Can't get stat of '/tmp/TableData.csv' (Errcode: 13)
    The fils is local and you have not specified the “LOCAL” directive.
  • ERROR 29 (HY000): File '/var/lib/mysql/test/TableData.csv' not found (Errcode: 2)
    Error from command LOAD DATA INFILE 'TableData.csv' INTO ... where the file is assumed to be read from the /database-process-home-directory/mysql-database-name/TableData.csv
    (Note: Database name “test” is being used.)
  • ERROR 1045 (28000): Access denied for user 'user1'@'%' (using password: YES)
    ERROR 2 (HY000): File '/tmp/TableData.csv' not found (Errcode: 2)
    Error from command LOAD DATA INFILE '/tmp/TableData.csv' INTO .... This is a common pitfall, trying to load a file located on the database server remotely from a client. Getting the file permissions correct is difficult. Avoid this method. Use the LOAD DATA LOCAL INFILE instead of the LOAD DATA INFILE method (it is so much easier).

Also look at the mysqlimport command.

Dump/Backup/Transfer Database:

The mysqldump command will read the mySQL database and generate a SQL command text file. This allows data to be migrated to other versions of mySQL (i.e. upgrade from typical Red Hat (RH7.x to FC3) mySQL release 3.23.58 to a more advanced mySQL 4.1 or 5.0) or to other SQL databases. SQL command file generated can create tables, insert data, ….


Option Description
Dump all the databases.
Dump the specified databases.
Specify host to connect to.
Specify password. If you do not specify a password, then you will be queried.
Specify user. Defaults to current user logged in.
–opt Same as: –add-drop-table –add-locks –all –extended-insert –quick –lock-tables
–add-drop-table Add a “drop table” SQL statement before each “create” SQL statement.
–add-locks Add “lock” SQL statements around “insert” SQL statements.
Include all mySQL specific SQL “create” options.
Allows utilization of the new, much faster INSERT syntax. Database you are migrating to must support this notation.
Don’t buffer query, dump directly to stdout.
Lock all tables for read.
Display command line options.


  • Dump database to a file:
    • Dump specified database:
      mysqldump --opt database > db-dump-file.sql
    • Dump specified table in database:
      mysqldump --opt database table-name > db-dump-file.sql
    • Dump multiple databases:
      mysqldump --opt --databases database1 database2 database3 > db-dump-file.sql
    • Dump everything:
      mysqldump --opt --all-databases > total-db-dump-file.sql
      mysqldump -u user-id -h host-name --opt --all-databases > total-db-dump-file.sql

    [Potential Pitfall]: If you experience the following error:

    mysqldump: Got error: 1016: Can't open file: 'Database-Name' (errno: 145) when using LOCK TABLES

    Fix with the following command: mysqlcheck -r -u root -p Database-Name

  • Import dumped file:
    mysql database < db-dump-file.sql
  • Export from one database and import to another:
    • Transfer specifed database from one database to another:
      mysqldump --opt database | mysql --host=host-name -C database

Man Page:

Upgrading to 4.1:


Restore MySql Database:

Restore using dump generated by mysqldump above:

  • mysql -h host-name -u user-id -psupersecretpassword < total-db-dump-file.sql
  • mysql database-name -h host-name -u user-id -psupersecretpassword < db-dump-file.sql


System Notes:

[Potential Pitfall]: Ubuntu mysql 5.0 database migration – When migrating the mysql database by copying files from /var/lib/mysql/... and /etc/mysql/... from one system running Ubuntu 6.11 to 8.04, I got nebulous error message in /var/log/syslog. The root cause of the problem was apparmor. If turing off apparmor (/etc/init.d/apparmor stop) allows your database to start and function properly, then go fix your apparmor security rules in /etc/apparmor.d/usr.sbin.mysqld. Also note that you must use the newer script /etc/mysql/debian-start from release 8.04 after copying /etc/mysql/....

Note: Debian and Ubuntu distributions manage mysql package upgrades using a mysql user debian-sys-maint which has its information located in /etc/mysql/debian.cnf. If you ever forget your mysql root password, you can always get back into the mysql database using the user debian-sys-maint and its password held in /etc/mysql/debian.cnf.


Building MySql from source: (on Linux)


  • C compiler: 2.95.2 or later. (Check with the command: rpm -q gcc)

Compile and install: (as root)

  • Downloaded source from
  • Expand tar file: tar xzf mysql-4.1.16.tar.gz
  • cd mysql-4.1.16
  • ./configure --prefix=/opt/mysql --sysconfdir=/opt/etc --localstatedir=/opt/var/mysql --with-unix-socket-path=/opt/tmp/mysql.sock
    (Use the command ./configure --help to see all options.)
    This should create an installation which will not clobber an existing RPM mySQL installation.
  • make
  • make install
  • Create mysql config file: cp support-files/my-medium.cnf /opt/var/my.cnf
  • Create user/group mysql
    • Test if user/group mysql already exists: groups mysql
    • Create group: groupadd mysql
    • Create user: useradd -g mysql -M -r -d /opt/lib/mysql -s /sbin/nologin -c "MySQL Server" mysql
  • chown -R mysql:mysql /opt/var/mysql


  • Install default database: /opt/mysql/bin/mysql_install_db --user=mysql
    Since this command is run as root, specify the --user option to operate command as user mysql.
    Creates help database with SQL script: /opt/mysql/share/mysql/fill_help_tables.sql
  • Start mySQL database: /opt/mysql/bin/mysqld_safe --user=mysql &
  • /opt/mysql/bin/mysqladmin -u root password 'new-password'
  • /opt/mysql/bin/mysqladmin -u root -h yoserver2 password 'new-password'
  • See tutorial above for use and administration.
  • Check defaults: (Defaults from config file: /opt/var/my.cnf)
    • /opt/mysql/bin/my_print_defaults --config-file=my client mysql
    • /opt/mysql/bin/my_print_defaults --config-file=my client mysql mysql_install_db


Commands/Man pages:


  • isamchk – Check and repair of ISAM tables.
  • isamlog – Write info about whats in a nisam log file.
  • msql2mysql
  • my_print_defaults
  • myisamchk
  • myisamlog
  • myisampack
  • mysql – text-based client for mysqld, a SQL-based relational database daemon
  • mysql_config
  • mysql_convert_table_format
  • mysql_find_rows
  • mysql_fix_privilege_tables
  • mysql_install_db
  • mysql_setpermission
  • mysql_zap – a perl script used to kill processes
  • mysqlaccess – Create new users to mysql.
  • mysqlbinlog
  • mysqlbug
  • mysqlcheck
  • mysqld_multi – Used for managing several mysqld processes running in different UNIX sockets and TCP/IP ports.
  • mysqldump – text-based client for dumping or backing up mysql databases , tables and or data.
  • mysqldumpslow
  • mysqlhotcopy
  • mysqlimport
  • mysqlshow – Shows the structure of a mysql database (databases,tables and columns)
  • mysqltest
  • pack_isam
  • perror – used to display a description for a system error code, or an MyISAM/ISAM table handler error code.
  • replace – A utility program to replace changes strings in place in files or on the standard input.
  • resolve_stack_dump
  • resolveip


  • mysqladmin – A utility for performing administrative operations
  • safe_mysqld – The recommended way to start a mysqld daemon on Unix.


Admin GUI Tools:



Interfaces in JAVA

  • The interface is a concept by which we can achieve full abstraction in java.
  • Abstraction is a process of hiding the implementation details and showing only functionality to the user. In other words it shows only important things to the user and hides the internal details
  • By means of interface we are able to achieve multiple inheritance.
  • Interface should be declared with interface keyword and interface name

Ex: interface PrintableInt

  • We can nest Interface inside an Interface.
  • We can have abstract methods in the interface.
  • Interface can extend another Interface.
  • Inside a interface class we can only have abstract methods.
  • Interface takes the method declared as public abstract method by default
  • Interface takes the data members as public,final static by default.

Why do we need Interface

Imagine that you have a project for a Pet animal shop. Here you have an animal class a dog class and a cat class. Now you can create object for Dog and Cat class which means it has a behavior. For example dog it should bark sound and cat should give Meaw sound. But what does animal give. Does this make sense. No so in order to overcome this problem we go for this abstract class where the compiler will not allow us to create object. So in this case Abstract will help

But you have categories as Pet Animal shop, Animal shop and Robot Animal Toy shop. Here if you have animal class and Pet class that have same methods. In your project you want to have multiple inheritance but compiler will not allow you to do that. If compiler allows you to do that, your program will suffer from Death Diamond Problem. In order have multiple inheritance and full abstraction we go for this Interface concept.

Deadly Diamond of Death problem

Consider a example that you have a sub class as ComboDrive now we have got both run method in both super class. By this compiler gets confused which superclass to call. But we have a requirement that we need multiple inheritance. In order to overcome our requirement there is interface concept to help us

A simple program with interface and abstract class

interface Pet


abstract void wow();


abstract class Animal


public static void main(String args[])


System.out.println(“First Interface “);



This is a basic program for just interface syntax

A simple program with interface,multiple inheritance and abstract class

interface PetAnimal
abstract void grow();
interface Animal extends PetAnimal
abstract void grow();
abstract void move();

abstract class Dog implements Animal, PetAnimal
public void grow()
System.out.println(“Ceasar is my pet and it grows”);
public void move()
System.out.println(“Ceasar will walk around and Caesar moves quick”);
class InterSecond extends Dog
public void grow()
System.out.println(“Ceasar is my pet and it grows”);
public void move()
System.out.println(“Ceasar will walk around and Caesar moves quick”);

public static void main(String args[])
InterSecond t=new InterSecond();
System.out.println(“First Interface “);

A simple program with interface,multiple inheritance and abstract class tagged Interface or marker Interface

interface PetAnimal


abstract void grow();


interface Geanny//tagged Interface for serializable, cloneable, Remote



interface Animal extends PetAnimal


int i=10;

void grow();

abstract void move();


abstract class AnimalQualities implements PetAnimal, Geanny, Animal


public void grow()


System.out.println(“Animal will grow “);


public void move()


System.out.println(“Animal moves”);



class Dog extends AnimalQualities


public void grow()


System.out.println(“Ceasar is my pet and it grows”);


public void move()


System.out.println(“Ceasar will walk around and Caesar moves quick”);



class InterThiIntInsideInt


public static void main(String args[])


Dog dogObj=new Dog();



System.out.println(“First Interface “);



Abstract classes in JAVA


We can identify Abstract method or Abstract class by declaration of keyword abstract

Abstract class cannot be instantiated

Object cannot be created for abstract class

Abstract class can be extended

Abstract class

Abstract class cannot be instantiated and has a abstract keyword.

Concrete Class is that what be basically define with class keyword followed by class name and can be extended to a abstract class.

Abstract class can be extended.

Why Abstract class cannot be instantiated?

Imagine that you have a project for a Pet animal shop. Here you have a animal class a dog class and a cat class. Now you can create object for Dog and Cat class which means it has a behavior. For example dog it should bark sound and cat should give Meaw sound. But what does animal give. Does this make sense. No so in order to overcome this problem we go for this abstract class where the compiler will not allow us to create object.

Abstract Method

Abstract method is a method that is declared with abstract keyword and does not have implementation

Abstract method should not have a body and it should end with a semicolon

Abstract methods can be overridden. Most of the cases it will be Over Ridden

Ex : abstract void disp();

Abstract With main method

abstract class AbsMainInAbs


public static void main(String args[])


System.out.println(“Thats Wow”);



Abstract class with Inheritance to concrete class and Abstract method

class AbsBasicTry1 extends AbsMethod


void disp()

{//contents here


public void gearType()


System.out.println(“GearType method and AbsBasicTry1 called”);


void run()


System.out.println(“void run called”);



class AbsBasicTry


public static void main(String args[])


AbsBasicTry1 t=new AbsBasicTry1();





abstract class AbsMethod


void disp()


System.out.println(“this is better”);


abstract void run();

abstract void gearType();


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Conditional functions in PHP


$makefoo = true;

/* We can’t call foo() from here
since it doesn’t exist yet,
but we can call bar() */


if ($makefoo) {
echo “First “;
function foo()
echo “I don’t exist until program execution reaches me.\n”;
echo “Second “;

/* Now we can safely call foo()
since $makefoo evaluated to true */

if ($makefoo) foo();
function bar()
echo “I exist immediately upon program start.\n”;


Check and execute this program and try to modify

Execution steps

First we make the $makefoo as true. and now the function bar(); is called so now this executes the script in bar

Next the execution comes to the if statement on  if ($makefoo) foo(); now this makefoo is true it directly executes the first and second statements . Now the exection comes to the if statement On executing this foo function the condition is true so then the script is and the output comes like this

I exist immediately upon program start. First Second I don’t exist until program execution reaches me.

Note: Also check with foo(); function on top and execute what happens and why it happens

Imperative Programming

Imperative vs Functional Programming

The fragment of LISP we have learned so far is purely functional. We program by defining mathematical functions. Programming this way has the benefit of referential transparency, which means that an experession has the same value and the same behavior, no matter where it is located, and how many times it is invoked. For example, the following function always returns 4 when an input 2 is supplied:

(defun double (x) (* 2 x))

Such programs are easy to develop and debug. The only effect of calling a function is the returned value. Functions become a lot harder to understand and debug when we allow them to produce side effects, that is, affecting subsequent computation in ways other than returning a value. This style of programming, in which side effect is not only permissible but is also the primary means by which we program, is called imperative programming. This is not something new, but is in fact the very kind of programming habits that you have acquired since you learned your first programming language (e.g. C/C++, Pascal, Java). Because of this, we intentionally leave the topic of imperative programming to the end of this series of tutorials. We do so since you already know so much about imperative programming, and since, in my taste, teaching it before teaching functional programming would only reinforce some of the bad habits programmers usually have, and thereby luring you to write C code with LISP, which, of course, make it harder to finish your assignment on time.


Imperative programming is made possible by the notion of program state. The behavior of an expression depends on the exact state a program is in. In turn, the state of a program is defined by, guess what, variables.

Suppose we want to create a global counter to keep track of the times some computational events occur. Common LISP allows us to define global variables as follows.

USER(7): (defparameter *counter* 0 "Counter to keep track of ....")
USER(8): (setf *counter* (1+ *counter*))
USER(9): *counter*
USER(10): (setf *counter* (1+ *counter*))
USER(11): *counter*

We use the defparameter special form to define a global variable *counter*, with zero as its initial value. We also decorate the declaration by a documentation string. We then use the setf special form to assign new values to the variable. The setf special form returns the new value of the variable. Lastly, evaluating the variable gives its current value. Notice that *counter*is evaluated to two different values at different point of time. This is something you will never see if you work with purely functional programs.

Suppose we actually want this counter to reset to zero when it reaches a user defined threshold. We would encapsulate this service into the following definitions:

(defparameter *counter* 0 
  "Counter to keep track of ....")

(defparameter *threshold* 4
  "Counter will be reset to zero when its value reaches this threshold.")

(defun counter-inc ()
  "Increment global counter by one."
  (if (>= (1+ *counter*) *threshold*)
      (setf *counter* 0)
    (setf *counter* (1+ *counter*))))

(defun counter-value ()
  "Return current value of global counter."

(defun counter-threshold ()
  "Return current threshold of global counter."

USER(24): (counter-value)
USER(25): (counter-inc)
USER(26): (counter-inc)
USER(27): (counter-inc)
USER(28): (counter-threshold)
USER(29): (counter-inc)

The function counter-inc can also be defined as follows:

(defun counter-inc ()
  "Increment global counter by one."
  (if (>= (1+ *counter*) *threshold*)
      (setf *counter* 0)
    (incf *counter*)))


With the possibility of state alteration, we also need a new kind of programming construct — sequencing. This allows us to perform multiple state-alterating operations, one after another. For example, we might want to introduce another function to our global counter abstraction: a function to modify the value of *threshold* on the fly:

(defun counter-set-threshold (th)
  "Set counter threshold to TH."
  (setf *threshold* th)
  (if (>= *counter* *threshold*)
      (setf *counter* 0))

The function sets the global value of *threshold* to a new value, and then checks if the current counter value has exceeded the new threshold. If the check succeeds, then the counter is reset to zero. Lastly, the function evaluates and returns the last expression, which is simply the new value of *threshold*. Notice that in this example the body of a function is no longer one expression, but a sequence of expressions. They are evaluated in order, and the value of the last expression is returned as a value of the sequence. The effect is demonstrated as in the following trace:

USER(2): (counter-value)
USER(3): (counter-inc)
USER(4): (counter-inc)
USER(5): (counter-inc)
USER(6): (counter-inc)
USER(7): (counter-inc)
USER(8): (counter-inc)
USER(9): (counter-set-threshold 2)
USER(10): (counter-value)

In fact, forms like let, let*, when and unless all admit sequences as their bodies. The cond form allows sequencing in the body of each alternative. The exception is if, due to its syntax. However, you can always get around by introducing a let form in the alternatives or by using other conditional forms.

Before we move on, notice that we can rewrite the counter-inc function as follows:

(defun counter-inc ()
  "Increment global counter by one."
  (if (>= (1+ *counter*) *threshold*)
      (setf *counter* 0)
    (incf *counter*)))

The (incf x) form is simply a shorthand for (setf x (+ x 1)). In general, the following are useful shorthands when developing imperative programs:

Shorthand Meaning
(incf x delta) (setf x (+ x delta))
(incf x) (incf x 1)
(decf x delta) (setf x (- x delta))
(decf x) (decf x 1)
(push e x) (setf x (cons e x))
(pop x) (let ((e (first x))) (setf x (rest x)) e)


Exercise: Create a global stack abstraction, with interface functions for performing pushing and poping.


The above method for defining an abstraction is not good enough. For one, the global variables are not encapsulated. There is nothing to forbid a careless user to change the value of *counter* in ways violating the invariants of the counter abstraction. To enforce this, we make use of local variables and closures.

As one would expect, the names introduced by a let form turns out not to be a name binding to some immutable values. The names refer to local variables. Such variables follow typical lexical scoping rules, so that LISP always looks up the innermost variable definition when a variable name is to be evaluated.

What is more interesting is that, due to the ability to return functions as values, the local variable has a life span longer than the expression defining it. Consider the following example:

USER(20): (setf inc (let ((counter 0)) 
                      #'(lambda () 
                          (incf counter))))
USER(21): (funcall inc)
USER(22): (funcall inc)
USER(23): (funcall inc)

We assign a value to the global variable inc. That value is obtained by first defining local variable counter using let, and then within the lexical scope of the local variable, a lambda expression is evaluated, thereby creating an annonymous function, which is returned as a value to be assigned to inc. The most interesting part is in the body of that lambda expression — it increments the value of the local variable counter! When the lambda expression is returned, the local variable persists, and is accessible only through the annonymous function.

The thing to remember from this example is that, in other kinds of languages like Pascal and C/C++ the lexical scope of a local variable somehow coincide with its life span. After executation passes beyond the boundary of a lexical scope, all the local variables defined within it cease to exist. This is not true in languages supporting higher-order functions, in particular, expressions that return functions as values. However, that is not to say that lexical scoping does not hold in such languages. In the contrary, lexical scoping is enforced strictly, and therefore the only place from which you can alter the value of counter is, as every true believer of lexical scoping would agree, within the lexical scope of the variable — the lambda expression. As a result, the counter state is effectively encapsulated. The only way to modify it is by going through the annonymous function stored in inc. The technical term to refer to this thing that is stored in inc, this thing that at the same time captures both the definition of a function and the variables referenced within the function body is called a function closure.

What if we want to define multiple interface functions for the encapsulated counter? Simple, just return all of them:

USER(34): (setf list-of-funcs (let ((counter 0))
                                (list #'(lambda ()
                                          (incf counter))
                                      #'(lambda ()
                                          (setf counter 0)))))
USER(35): (setf inc (first list-of-funcs))
USER(36): (setf reset (second list-of-funcs))
USER(37): (funcall inc)
USER(38): (funcall inc)
USER(39): (funcall inc)
USER(40): (funcall reset)
USER(41): (funcall inc)


Exercise: Define an encapsulated global stack.

Poor-Man’s Object

Having only one instance of this encapsulated counter abstraction is not good enough. Imagine cases when we need multiple instances of such counters, each having its own state. Well, the answer is simple, we just evaluate the function-returning let form everytime we need a new instance of counter. Since, a new local variable will be created everytime we evaluate the let form, the function closure that is returned each time will be associated with a different incarnation of the local variable counter. To automate this process, of course we define a constructor for such closures:

(defun make-counter ()
  "Create a new instance of a counter object."
  (let ((counter 0))
    (list #'(lambda ()
	      (incf counter))
	  #'(lambda ()
	      (setf counter 0)))))

Notice how this allows us to maintain independently encapsulated states:

USER(44): (setf c1 (make-counter))
USER(44): (setf c2 (make-counter))
USER(45): (funcall (first c1))
USER(46): (funcall (first c1))
USER(47): (funcall (second c1))
USER(48): (funcall (first c2))
USER(49): (funcall (first c2))
USER(50): (funcall (first c2))
USER(51): (funcall (first c1))
USER(52): (funcall (first c1))
USER(53): (funcall (second c2))
USER(54): (funcall (first c1))

To make invoking counter interface functions easier, we can define the following shorthands:

(defun counter-increment (counter)
  "Increment counter."
  (funcall (first counter)))

(defun counter-reset (counter)
  "Reset counter."
  (funcall (second counter)))

And now we have an all-rounded counter abstraction.

USER(56): (counter-increment c1)
USER(57): (counter-reset c1)

The moral of this store is that, function closures are encapsulated states. They are a poor-man’s version of objects, which, after all, are nothing but encapsulated states. (Yes, I am aware that Common LISP has a Common LISP Object System (CLOS), and there is no point of using closures in this manner if all we want is simply object orientation. But No, I want you to understand what higher-order functions buy you, and how they serve as a building block for other advanced programming language constructs. Lastly, I don’t want you to spend time struggling to learn yet another object system.)


Exercise: Implement a constructor for your encapsulated stack abstraction. Define appropriate shorthands for convenient invocation of interface functions.

Iterative Constructs

To round off this tutorial, we discuss something that you know so much about — iterations. We start with something very simple:

USER(1): (dotimes (i 10) (print i))


The (dotimes (i n) body) form executes body N times. In addition, it defines a local variable i, which receives an integer value ranging from 0 to n-1. The body of the form could be a sequence of expressions. The form returns NIL. The (print x) form prints the LISP object xto the console.

A similar construct is defined for iterating through a list:

USER(2): (dolist (i '(a b c d e)) (print i))


The most general looping construct is the do form:

(defun fibonacci (n)
  "Compute the N'th Fibonacci number."
  (do ((f1 0 f2)
       (f2 1 (+ f1 f2))
       (i  0 (1+ i)))
      ((= i n) f2)
      ; empty body

The first list following the do keyword is a list of local variables and their initializations and update forms. Each member of this list has the format (var init-form update-form). Within this loop are defined three local variables f1, f2 and i. The three initialization forms 0, 1 and 0 are evaluated first, and then assigned to the three locals simultaneously. In each subsequent iteration, the three update forms f2, (+ f1 f2) and (1+ i) will be evaluated first, and then the values are assigned to the three local variables simultaneously. The second list following the do keyword (i.e. ((= i n) f2)) has the general format of (exit-condition return-form). The exit condition (= i n) is checked after the initialization is done, or after the update is performed in subsequent iterations. If the test succeeds, then the return form is evaluated, and its value is returned. Otherwise, the body of the doform, which in this case is empty, will be executed with the updated value of the local variables.

Indefinite looping can be achieved using the (loop body) form. One may exit from such loop using the return-from form:

(defun fib (n)
  "Compute the N'th Fibonacci number."
  (let ((f1 0)
	(f2 1)
	(i  0))
     (if (= i n)
	 (return-from fib f2))
     ; empty body
     (psetf f1 f2
	    f2 (+ f1 f2)
	    i  (1+ i)))))

The fib function has the same meaning as the fibonacci function coded in terms of do. The psetf is a variation of setf that implements “parallel” assignment.   

Data Abstraction

Binary Trees

Suppose we want to create a new kind of recursive data type, our familiar binary trees. The first thing we have to do is to define the data type in terms of its constructors, selectors and recognizers. In the case of binary trees, we have the following:

  1. Constructors: We have two kinds of binary trees, leaves and nodes. Accordingly, we need a constructor for each kind:
    • (make-bin-tree-leaf E): A leaf is a composite object with one component, the element E.
    • (make-bin-tree-node E B1 B2): A node consists of three components, an element E, a left subtree B1 and a right subtree B2. Each of B1 and B2 is a binary tree.

    Notice that the definition of binary tree is inherently recursive (as in the case of nodes). Larger binary trees can be composed from smaller ones.

  2. Selectors: We need to define a selector for each component of each kind of binary tree.
    • (bin-tree-leaf-element L): Retrieve the element of a leaf L.
    • (bin-tree-node-element N): Retrieve the element of a node N.
    • (bin-tree-node-left N): Retrieve the left subtree of a node N.
    • (bin-tree-node-right N): Retrieve the right subtree of a node N.
  3. Recognizers: We define one recognizer for each kind of binary tree.
    • (bin-tree-leaf-p B): Test if a given binary tree B is a leaf.
    • (bin-tree-node-p B): Test if a given binary tree B is a node.

Notice that we have not written a line of code yet, and still we are able to write down the function signature of all the constructors, selectors and recognizers. The process is more or less mechanical:

  1. Define a constructor for each variant of the recursive data type. The parameters for a constructor defines the components of a composite object.
  2. For each parameter of each constructor, define a selector to retrieve the corresponding component.
  3. For each constructor, define a corresponding recognizer.

The next question is how we are to represent a binary tree as a LISP object. Of course, a list is the first thing that comes to our mind:

  • We represent an leaf with element E by a singleton list containing E (i.e. (list E)).
  • A node with element E, left subtree B1 and right subtree B2 is represented as a list containing the three components (i.e. (list E B1 B2)).

Fixing the representation, we can thus implement the recursive data type functions:

;; Binary Trees

;; Constructors for binary trees

(defun make-bin-tree-leaf (E)
  "Create a leaf."
  (list E))

(defun make-bin-tree-node (E B1 B2)
  "Create a node with element K, left subtree B1 and right subtree B2."
  (list E B1 B2))

;; Selectors for binary trees

(defun bin-tree-leaf-element (L)
  "Retrieve the element of a leaf L."
  (first L))

(defun bin-tree-node-element (N)
  "Retrieve the element of a node N."
  (first N))

(defun bin-tree-node-left (N)
  "Retrieve the left subtree of a node N."
  (second N))

(defun bin-tree-node-right (N)
  "Retrieve the right subtree of a node N."
  (third N))

;; Recognizers for binary trees

(defun bin-tree-leaf-p (B)
  "Test if binary tree B is a leaf."
  (and (listp B) (= (list-length B) 1)))

(defun bin-tree-node-p (B)
  "Test if binary tree B is a node."
  (and (listp B) (= (list-length B) 3)))

The representation scheme works out like the following:

USER(5): (make-bin-tree-node '*
                             (make-bin-tree-node '+
                                                 (make-bin-tree-leaf 2)
                                                 (make-bin-tree-leaf 3))
                             (make-bin-tree-node '-
                                                 (make-bin-tree-leaf 7)
                                                 (make-bin-tree-leaf 8)))
(* (+ (2) (3)) (- (7) (8)))

The expression above is a binary tree node with element * and two subtrees. The left subtree is itself a binary tree node with + as its element and leaves as its subtress. The right subtree is also a binary tree node with -as its element and leaves as its subtrees. All the leaves are decorated by numeric components.

           / \
          /   \
         /     \
        +       -
       / \     / \
      2   3   7   8

Searching Binary Trees

As discussed in previous tutorials, having recursive data structures defined in the way we did streamlines the process of formulating structural recursions. We review this concept in the following examples.

Suppose we treat binary trees as containers. An expression E is a member of a binary tree B if:

  1. B is a leaf and its element is E.
  2. B is a node and either its element is E or E is a member of one of its subtrees.

For example, the definition asserts that the members of (* (+ (2) (3)) (- (7) (8))) are *, +, 2, 3, -, 7 and 8. Such a definition can be directly implemented by our recursive data type functions:

(defun bin-tree-member-p (B E)
  "Test if E is an element in binary tree B."
  (if (bin-tree-leaf-p B)
      (equal E (bin-tree-leaf-element B))
    (or (equal E (bin-tree-node-element B))
        (bin-tree-member-p (bin-tree-node-left B) E)
	(bin-tree-member-p (bin-tree-node-right B) E))))

The function can be made more readable by using the letform:

(defun bin-tree-member-p (B E)
  "Test if E is an element in binary tree B."
  (if (bin-tree-leaf-p B)
      (equal E (bin-tree-leaf-element B))
	((elmt  (bin-tree-node-element B))
	 (left  (bin-tree-node-left    B))
	 (right (bin-tree-node-right   B)))
      (or (equal E elmt)
	  (bin-tree-member-p left E)
	  (bin-tree-member-p right E)))))

Tracing the execution of bin-tree-member-p, we get:

USER(14): (trace bin-tree-member-p)
USER(15): (bin-tree-member-p '(+ (* (2) (3)) (- (7) (8))) 7) 
 0: (BIN-TREE-MEMBER-P (+ (* (2) (3)) (- (7) (8))) 7)
   1: (BIN-TREE-MEMBER-P (* (2) (3)) 7)
     2: (BIN-TREE-MEMBER-P (2) 7)
     2: returned NIL
     2: (BIN-TREE-MEMBER-P (3) 7)
     2: returned NIL
   1: returned NIL
   1: (BIN-TREE-MEMBER-P (- (7) (8)) 7)
     2: (BIN-TREE-MEMBER-P (7) 7)
     2: returned T
   1: returned T
 0: returned T


Exercise: Let size(B) be the number of members in a binary tree B. Give a recursive definition of size(B), and then implement a LISP function (bin-tree-size B) that returns size(B).

Traversing Binary Trees

Let us write a function that will reverse a tree in the sense that the left and right subtrees of every node are swapped:

(defun binary-tree-reverse (B)
  "Reverse binary tree B."
  (if (bin-tree-leaf-p B)
	((elmt  (bin-tree-node-element B))
	 (left  (bin-tree-node-left    B))
	 (right (bin-tree-node-right   B)))
      (make-bin-tree-node elmt
		          (binary-tree-reverse right)
		          (binary-tree-reverse left)))))

The correctness of the above implementation can be articulated as follows. Given a binary tree B and an object E, either the binary tree is a leaf or it is a node:

  • Case 1: B is a leaf.
    Then the reversal of B is simply B itself.
  • Case 2: B is a node.
    Then B has three components, namely, an element elmt, a left subtree left and a right subtree right. The reversal of B is a node with element elmt, left subtree the reversal of right, and right subtree the reversal of left.

The following shows us how the recursion unfolds:

USER(21): (trace bin-tree-reverse)
USER(22): (bin-tree-reverse '(* (+ (2) (3)) (- (7) (8))))
 0: (BIN-TREE-REVERSE (* (+ (2) (3)) (- (7) (8))))
   1: (BIN-TREE-REVERSE (- (7) (8)))
     2: (BIN-TREE-REVERSE (8))
     2: returned (8)
     2: (BIN-TREE-REVERSE (7))
     2: returned (7)
   1: returned (- (8) (7))
   1: (BIN-TREE-REVERSE (+ (2) (3)))
     2: (BIN-TREE-REVERSE (3))
     2: returned (3)
     2: (BIN-TREE-REVERSE (2))
     2: returned (2)
   1: returned (+ (3) (2))
 0: returned (* (- (8) (7)) (+ (3) (2)))
(* (- (8) (7)) (+ (3) (2)))

The resulting expression represents the following tree:

           / \
          /   \
         /     \
        -       +
       / \     / \
      8   7   3   2

Let us implement a function that will extract the members of a given binary tree, and put them into a list in preorder.

(defun bin-tree-preorder (B)
  "Create a list containing keys of B in preorder."
  (if (bin-tree-leaf-p B)
      (list (bin-tree-leaf-element B))
	((elmt  (bin-tree-node-element B))
	 (left  (bin-tree-node-left    B))
	 (right (bin-tree-node-right   B)))
      (cons elmt
	    (append (bin-tree-preorder left)
		    (bin-tree-preorder right))))))

Tracing the execution of the function, we obtain the following:

USER(13): (trace bin-tree-preorder)
USER(14): (bin-tree-preorder '(* (+ (2) (3)) (- (7) (8))))
 0: (BIN-TREE-PREORDER (* (+ (2) (3)) (- (7) (8))))
   1: (BIN-TREE-PREORDER (+ (2) (3)))
     2: (BIN-TREE-PREORDER (2))
     2: returned (2)
     2: (BIN-TREE-PREORDER (3))
     2: returned (3)
   1: returned (+ 2 3)
   1: (BIN-TREE-PREORDER (- (7) (8)))
     2: (BIN-TREE-PREORDER (7))
     2: returned (7)
     2: (BIN-TREE-PREORDER (8))
     2: returned (8)
   1: returned (- 7 8)
 0: returned (* + 2 3 - 7 8)
(* + 2 3 - 7 8)

As we have discussed before, the append call in the code above is a source of inefficiency that can be obtimized away:

(defun fast-bin-tree-preorder (B)
  "A tail-recursive version of bin-tree-preorder."
  (preorder-aux B nil))

(defun preorder-aux (B A)
  "Append A to the end of the list containing elements of B in preorder."
  (if (bin-tree-leaf-p B)
      (cons (bin-tree-leaf-element B) A)
	((elmt  (bin-tree-node-element B))
	 (left  (bin-tree-node-left    B))
	 (right (bin-tree-node-right   B)))
      (cons elmt
	    (preorder-aux left
			  (preorder-aux right A))))))

An execution trace of the implementation is the following:

USER(15): (trace fast-bin-tree-preorder preorder-aux)          
USER(16): (fast-bin-tree-preorder '(* (+ (2) (3)) (- (7) (8))))
 0: (FAST-BIN-TREE-PREORDER (* (+ (2) (3)) (- (7) (8))))
   1: (PREORDER-AUX (* (+ (2) (3)) (- (7) (8))) NIL)
     2: (PREORDER-AUX (- (7) (8)) NIL)
       3: (PREORDER-AUX (8) NIL)
       3: returned (8)
       3: (PREORDER-AUX (7) (8))
       3: returned (7 8)
     2: returned (- 7 8)
     2: (PREORDER-AUX (+ (2) (3)) (- 7 8))
       3: (PREORDER-AUX (3) (- 7 8))
       3: returned (3 - 7 8)
       3: (PREORDER-AUX (2) (3 - 7 8))
       3: returned (2 3 - 7 8)
     2: returned (+ 2 3 - 7 8)
   1: returned (* + 2 3 - 7 8)
 0: returned (* + 2 3 - 7 8)
(* + 2 3 - 7 8)

Exercise: Implement a function that will create a list containing members of a given binary tree in postorder. Implement also a tail-recursive version of the same function.

Exercise: Repeat the last exercise with inorder.

Abstract Data Types

Abstract data types are blackboxes. They are defined in terms of their external interfaces, and not their implementation. For example, a set abstraction offers the following operations:

  • (make-empty-set) creates an empty set.
  • (set-insert S E) returns a set containing all members of set S plus an additional member E.
  • (set-remove S E) returns a set containing all members of set S except for E.
  • (set-member-p S E) returns true if E is a member of set S.
  • (set-empty-p S) returns true if set S is empty.

To implement an abstract data type, we need to decide on a representation. Let us represent a set by a list with no repeated members.

(defun make-empty-set ()
  "Creates an empty set."

(defun set-insert (S E)
  "Return a set containing all the members of set S plus the element E."
  (adjoin E S :test #'equal))

(defun set-remove (S E)
  "Return a set containing all the members of set S except for element E."
  (remove E S :test #'equal))

(defun set-member-p (S E)
  "Return non-NIL if set S contains element E."
  (member E S :test #'equal))

(defun set-empty-p (S)
  "Return true if set S is empty."
  (null S))


Exercise: Look up the definition of adjoin, remove and member from CLTL2. In particular, find out how the :test keyword is used to specify the equality test function to be used by the three functions. What will happen if we omit the :test keyword and the subsequent #'equal when invoking the three functions?

Notice that we have implemented an abstract data type (sets) using a more fundamental recursive data structure (lists) with additional computational constraints (no repetition) imposed by the interface functions.

Binary Search Trees

Another way of implementing the same set abstraction is to use the more efficient binary search tree (BST). Binary search trees are basically binary trees with the following additional computational constraints:

  • All the members in the left subtree of a tree node is no greater than the element of the node.
  • All the members in the right subtree of a tree node is greater than the element of the node.
  • All the leaf members are distinct.

Again, we are implementing an abstract data type (sets) by a more fundamental recursive data structure (binary trees) with additional computational constraints. In particular, we use the leaves of a binary tree to store the member of a set, and the tree nodes for providing indexing information that improves search performance. for example, a BST representing the set {1 2 3 4} would look like:

           / \
          /   \
         /     \
        1       3
       / \     / \
      1   2   3   4

An empty BST is represented by NIL, while a nonempty BST is represented by a binary tree. We begin with the constructor and recognizer for empty BST.

(defun make-empty-BST ()
  "Create an empty BST."

(defun BST-empty-p (B)
  "Check if BST B is empty."
  (null B))

Given the additional computational constraints, membership test can be implemented as follows:

(defun BST-member-p (B E)
  "Check if E is a member of BST B."
  (if (BST-empty-p B)
    (BST-nonempty-member-p B E)))

(defun BST-nonempty-member-p (B E)
  "Check if E is a member of nonempty BST B."
  (if (bin-tree-leaf-p B)
      (= E (bin-tree-leaf-element B))
    (if (<= E (bin-tree-node-element B))
	(BST-nonempty-member-p (bin-tree-node-left B) E)
      (BST-nonempty-member-p (bin-tree-node-right B) E))))

Notice that we handle the degenerate case of searching an empty BST separately, and apply the well-known recursive search algorithm only on nonempty BST.

USER(16): (trace BST-member-p BST-nonempty-member-p)
USER(17): (BST-member-p '(2 (1 (1) (2)) (3 (3) (4))) 3)
 0: (BST-MEMBER-P (2 (1 (1) (2)) (3 (3) (4))) 3)
   1: (BST-NONEMPTY-MEMBER-P (2 (1 (1) (2)) (3 (3) (4))) 3)
     2: (BST-NONEMPTY-MEMBER-P (3 (3) (4)) 3)
       3: (BST-NONEMPTY-MEMBER-P (3) 3)
       3: returned T
     2: returned T
   1: returned T
 0: returned T

Insertion is handled by the following family of functions:

(defun BST-insert (B E)
  "Insert E into BST B."
  (if (BST-empty-p B)
      (make-bin-tree-leaf E)
    (BST-nonempty-insert B E)))

(defun BST-nonempty-insert (B E)
  "Insert E into nonempty BST B."
  (if (bin-tree-leaf-p B)
      (BST-leaf-insert B E)
    (let ((elmt  (bin-tree-node-element B))
	  (left  (bin-tree-node-left    B))
	  (right (bin-tree-node-right   B)))
      (if (<= E (bin-tree-node-element B))
	  (make-bin-tree-node elmt
			      (BST-nonempty-insert (bin-tree-node-left B) E)
	(make-bin-tree-node elmt
			    (BST-nonempty-insert (bin-tree-node-right B) E))))))

(defun BST-leaf-insert (L E)
  "Insert element E to a BST with only one leaf."
  (let ((elmt (bin-tree-leaf-element L)))
    (if (= E elmt)
      (if (< E elmt)
	  (make-bin-tree-node E
			      (make-bin-tree-leaf E)
			      (make-bin-tree-leaf elmt))
	(make-bin-tree-node elmt
			    (make-bin-tree-leaf elmt)
			    (make-bin-tree-leaf E))))))

As before, recursive insertion to nonempty BST is handled outside of the general entry point of BST insertion. Traversing down the index nodes, the recursive algorithm eventually arrives at a leaf. In case the element is not already in the tree, the leaf is turned into a node with leaf subtrees holding the inserted element and the element of the original leaf. For example, if we insert 2.5 into the tree represented by (2 (1 (1) (2)) (3 (3) (4))), the effect is the following:

            2                      2
           / \                    / \
          /   \                  /   \
         /     \       ==>      /     \
        1       3              1       3
       / \     / \            / \     / \
      1   2   3   4          1   2  2.5  4
                                    / \
                                  2.5  3

A trace of the insertion operation is given below:

USER(22): (trace BST-insert BST-nonempty-insert BST-leaf-insert)
USER(23): (BST-insert '(2 (1 (1) (2)) (3 (3) (4))) 2.5)
 0: (BST-INSERT (2 (1 (1) (2)) (3 (3) (4))) 2.5)
   1: (BST-NONEMPTY-INSERT (2 (1 (1) (2)) (3 (3) (4))) 2.5)
     2: (BST-NONEMPTY-INSERT (3 (3) (4)) 2.5)
       3: (BST-NONEMPTY-INSERT (3) 2.5)
         4: (BST-LEAF-INSERT (3) 2.5)
         4: returned (2.5 (2.5) (3))
       3: returned (2.5 (2.5) (3))
     2: returned (3 (2.5 (2.5) (3)) (4))
   1: returned (2 (1 (1) (2)) (3 (2.5 (2.5) (3)) (4)))
 0: returned (2 (1 (1) (2)) (3 (2.5 (2.5) (3)) (4)))
(2 (1 (1) (2)) (3 (2.5 (2.5) (3)) (4)))

Removal of elements is handled by the following family of functions:

(defun BST-remove (B E)
  "Remove E from BST B."
  (if (BST-empty-p B)
    (if (bin-tree-leaf-p B)
	(BST-leaf-remove B E)
      (BST-node-remove B E))))

(defun BST-leaf-remove (L E)
  "Remove E from BST leaf L."
  (if (= E (bin-tree-leaf-element L))

(defun BST-node-remove (N E)
  "Remove E from BST node N."
      ((elmt  (bin-tree-node-element N))
       (left  (bin-tree-node-left    N))
       (right (bin-tree-node-right   N)))
    (if (<= E elmt)
	(if (bin-tree-leaf-p left)
	    (if (= E (bin-tree-leaf-element left))
	  (make-bin-tree-node elmt (BST-node-remove left E) right))
      (if (bin-tree-leaf-p right)
	  (if (= E (bin-tree-leaf-element right))
	(make-bin-tree-node elmt left (BST-node-remove right E))))))

This time, removal from empty BST’s and BST’s with a single leaf are both degenerate cases. The recursive removal algorithm deals with BST nodes. Traversing down the index nodes, the recursive algorithm searches for the parent node of the leaf to be removed. In case it is found, the sibling of the leaf to be removed replaces its parent node. For example, the effect of removing 2 from the BST represented by (2 (1 (1) (2)) (3 (3) (4)))is depicted as follows:

            2                      2
           / \                    / \
          /   \                  /   \
         /     \       ==>      /     \
        1       3              1       4
       / \     / \            / \     
      1   2   3   4          1   2  

A trace of the deletion operation is given below:

USER(4): (trace BST-remove BST-node-remove)
USER(5): (BST-remove '(2 (1 (1) (2)) (3 (3) (4))) 3)
 0: (BST-REMOVE (2 (1 (1) (2)) (3 (3) (4))) 3)
   1: (BST-NODE-REMOVE (2 (1 (1) (2)) (3 (3) (4))) 3)
     2: (BST-NODE-REMOVE (3 (3) (4)) 3)
     2: returned (4)
   1: returned (2 (1 (1) (2)) (4))
 0: returned (2 (1 (1) (2)) (4))
(2 (1 (1) (2)) (4))


Exercise: A set can be implemented as a sorted list, which is a list storing distinct members in ascending order. Implement the sorted list abstraction.


We demonstrate how one can perform symbolic computation using LISP. To begin with, we define a new recursive data type for polynomials, which is defined recursively as follows:

  • If num is a number, then (make-constant num) is a polynomial;
  • If sym is a symbol, then (make-variable sym) is a polynomial;
  • If poly1 and poly2are polynomials, then the following are also polynomials:
    • (make-sum poly1 poly2)
    • (make-product poly1 poly2)
  • If poly is a polynomial and num is a number, then (make-power poly num) is a polynomial.

One can represent polynomials in the most standard way:

;; Constructors for polynomials

(defun make-constant (num)

(defun make-variable (sym)

(defun make-sum (poly1 poly2)
  (list '+ poly1 poly2))

(defun make-product (poly1 poly2)
  (list '* poly1 poly2))

(defun make-power (poly num)
  (list '** poly num))

For example, (make-power (make-sum (make-variable 'x) (make-constant 1)) 2) is represented by the LISP form (** (+ x 1) 2), which denotes the polynomail (x + 1)2in our usual notation.

We then define a recognizer for each constructor:

;; Recognizers for polynomials

(defun constant-p (poly)
  (numberp poly))

(defun variable-p (poly)
  (symbolp poly))

(defun sum-p (poly)
  (and (listp poly) (eq (first poly) '+)))

(defun product-p (poly)
  (and (listp poly) (eq (first poly) '*)))

(defun power-p (poly)
  (and (listp poly) (eq (first poly) '**)))

We then need to define selectors for the composite polynomials. We define a selector for each component of each composite constructor.

;; Selectors for polynomials

(defun constant-numeric (const)

(defun variable-symbol (var)

(defun sum-arg1 (sum)
  (second sum))

(defun sum-arg2 (sum)
  (third sum))

(defun product-arg1 (prod)
  (second prod))

(defun product-arg2 (prod)
  (third prod))

(defun power-base (pow)
  (second pow))

(defun power-exponent (pow)
  (third pow))

One may ask why we define so many trivial looking functions for carrying out the same task (sum-arg1 and product-arg1have exactly the same implementation). The reason is that we may end up changing the representation in the future, and there is no guarantee that sums and products will be represented similarly in the future. Also, programs written like this tends to be self-commenting.

Now that we have a completely defined polynomial data type, let us do something interesting with it. Let us define a function that carries out symbolic differentiation. In particular, we want a function (d poly x) which returns the derivative of polynomial poly with respect to variable x. Let us review our first-year differential calculus:

  • The derivative (dC / dx) of a constant C is zero.
  • The derivative (dy/dx) of a variable y is 1 if the x = y. Otherwise, we leave the derivative unevaluated. We represent unevaluated derivatives using the following functions
    ;; Unevaluated derivative
    (defun make-derivative (poly x)
        (list 'd poly x))
    (defun derivative-p (poly x)
      (and (listp poly) (eq (first poly) 'd)))
  • The derivative (d(F+G)/dx) of a sum (F+G) is (dF/dx) + (dG/dx).
  • The derivative (d(F*G)/dx) of a product (F*G) is F*(dG/dx) + G*(dF/dx).
  • The derivative (d(FN)/dx) of a power FN is N * FN-1 * (dF/dx).

The above calculus can be encoded in LISP as follows:

;; ;; Differentiation function ;; (defun d (poly x) (cond ((constant-p poly) 0) ((variable-p poly) (if (equal poly x) 1 (make-derivative poly x))) ((sum-p poly) (make-sum (d (sum-arg1 poly) x) (d (sum-arg2 poly) x))) ((product-p poly) (make-sum (make-product (product-arg1 poly) (d (product-arg2 poly) x)) (make-product (product-arg2 poly) (d (product-arg1 poly) x)))) ((power-p poly) (make-product (make-product (power-exponent poly) (make-power (power-base poly) (1- (power-exponent poly)))) (d (power-base poly) x))))) 

Test driving the differentiation function we get:

USER(11): (d '(+ x y) 'x) (+ 1 (D Y X)) USER(12): (d '(* (+ x 1) (+ x 1)) 'x) (+ (* (+ X 1) (+ 1 0)) (* (+ X 1) (+ 1 0))) USER(13): (d '(** (+ x 1) 2) 'x) (* (* 2 (** (+ X 1) 1)) (+ 1 0)) 

The result is correct but very clumsy. We would like to simplify the result a bit using the following rewriting rules:

  • E + 0 = E
  • 0 + E = E
  • E * 0 = 0
  • 0 * E = 0
  • E * 1 = E
  • 1 * E = E
  • E0 = 1
  • E1 = E

This can be done by defining a simplification framework, in which we can implement such rules:

;; ;; Simplification function ;; (defun simplify (poly) "Simplify polynomial POLY." (cond ((constant-p poly) poly) ((variable-p poly) poly) ((sum-p poly) (let ((arg1 (simplify (sum-arg1 poly))) (arg2 (simplify (sum-arg2 poly)))) (make-simplified-sum arg1 arg2))) ((product-p poly) (let ((arg1 (simplify (product-arg1 poly))) (arg2 (simplify (product-arg2 poly)))) (make-simplified-product arg1 arg2))) ((power-p poly) (let ((base (simplify (power-base poly))) (exponent (simplify (power-exponent poly)))) (make-simplified-power base exponent))) ((derivative-p poly) poly))) 

The simplify function decomposes a composite polynomial into its components, apply simplification recursively to the components, and then invoke the type-specific simplification rules (i.e. make-simplified-sum, make-simplified-product, make-simplified-power) based on the type of the polynomial being processed.

The simplification rules are encoded in LISP as follows:

(defun make-simplified-sum (arg1 arg2) "Given simplified polynomials ARG1 and ARG2, construct a simplified sum of ARG1 and ARG2." (cond ((and (constant-p arg1) (zerop arg1)) arg2) ((and (constant-p arg2) (zerop arg2)) arg1) (t (make-sum arg1 arg2)))) (defun make-simplified-product (arg1 arg2) "Given simplified polynomials ARG1 and ARG2, construct a simplified product of ARG1 and ARG2." (cond ((and (constant-p arg1) (zerop arg1)) (make-constant 0)) ((and (constant-p arg2) (zerop arg2)) (make-constant 0)) ((and (constant-p arg1) (= arg1 1)) arg2) ((and (constant-p arg2) (= arg2 1)) arg1) (t (make-product arg1 arg2)))) (defun make-simplified-power (base exponent) "Given simplified polynomials BASE and EXPONENT, construct a simplified power with base BASE and exponent EXPONENT." (cond ((and (constant-p exponent) (= exponent 1)) base) ((and (constant-p exponent) (zerop exponent)) (make-constant 1)) (t (make-power base exponent)))) 

Let us see how all these pay off:

USER(14): (simplify (d '(* (+ x 1) (+ x 1)) 'x)) (+ (+ X 1) (+ X 1)) USER(15): (simplify (d '(** (+ x 1) 2) 'x)) (* 2 (+ X 1)) 

Comparing to the original results we saw before, this is a lot more reasonable.


Exercise: Extend the symbolic polynomial framework in the following ways:

  • Define a new type of polynomial — difference. If poly1 and poly2 are polynomials, then (make-difference poly1 poly2) is also a polynomial. Implement the constructor, recognizer and selectors for this type of polynomial.
  • The derivative (d(F-G)/dx) of a difference (F-G) is (dF/dx) – (dG/dx). Extend the differentiation function to incorporate this.
  • Implement the following simplification rule:
    • E – 0 = E


Exercise: Extend the symbolic polynomial framework in the following ways:

  • Define a new type of polynomial — negation. If poly1 is a polynomial, then (make-negation poly) is also a polynomial. Implement the constructor, recognizer and selectors for this type of polynomial.
  • The derivative (d(-F)/dx) of a negation -F is -(dF/dx). Extend the differentiation function to incorporate this.
  • Implement the following simplification rules:
    • -0 = 0
    • -(-E) = E


Exercise: The simplification rules we have seen so far share a common feature: the right hand sides do not involve any new polynomial constructor. For example, -(-E) is simply E. However, some of the most useful simplification rules are those involving constructors on the right hand sides:

  • 0 – E = -E
  • E1 + (-E2) = E1 – E2
  • (-E1) + E2 = E2 – E1
  • E1 – (-E2) = E1 + E2
  • E * (-1) = -E
  • (-1) * E = -E

Within the type-specific simplification functions, if we naively apply the regular constructors to build the expressions on the right hand sides, then we run into the risk of constructing polynomials that are not fully simplified. For example, -x and -1 are both fully simplified, but if we now construct their product (-1) * (-x), the last simplification rule above says that we can rewrite the product into -(-x), which needs further simplification. One naive solution is to blindly apply full simplification to the newly constructed polynomials, but this is obviously an overkill. What then is an efficient and yet correct implementation of the above simplification rules?


Exercise: If all the components of a composite polynomial are constants, then we can actually perform further simplification. For example, (+ 1 1) should be simplified to 2. Extend the simplification framework to incorporate this.

Tower of Hanoi

The Tower of Hanoi problem is a classical toy problem in Artificial Intelligence: There are N disks D1, D2, …, Dn, of graduated sizes and three pegs 1, 2, and 3. Initially all the disks are stacked on peg 1, with D1, the smallest, on top and Dn, the largest, at the bottom. The problem is to transfer the stack to peg 3 given that only one disk can be moved at a time and that no disk may be placed on top of a smaller one. [Pearl 1984]

We call peg 1 the “from” peg, peg 3 the “to” peg. Peg 2 is a actually a buffer to facilitate movement of disks, and we call it an “auxiliary” peg. We can move N disks from the “from” peg to the “to” peg using the following recursive scheme.

  1. Ignoring the largest disk at the “from” peg, treat the remaining disks as a Tower of Hanoi problem with N-1 disks. Recursively move the top N-1 disks from the “from” peg to the “auxiliary” peg, using the “to” peg as a buffer.
  2. Now that the N-1 smaller disks are in the “auxiliary” peg, we move the largest disk to the “to” peg.
  3. Ignoring the largest disk again, treat the remaining disks as a Tower of Hanoi problem with N-1 disks. Recursively move the N-1 disks from the “auxiliary” peg to the “to” peg, using the “from” peg as a buffer.

To code this solution in LISP, we need to define some data structure. First, we represent a disk by a number, so that Di is represented by i. Second, we represent a stack of disks by a tower, which is nothing but a list of numbers, with the first element representing the top disk. We define the usual constructors and selectors for the tower data type.

;; ;; A tower is a list of numbers ;; (defun make-empty-tower () "Create tower with no disk." nil) (defun tower-push (tower disk) "Create tower by stacking DISK on top of TOWER." (cons disk tower)) (defun tower-top (tower) "Get the top disk of TOWER." (first tower)) (defun tower-pop (tower) "Remove the top disk of TOWER." (rest tower)) 

Third, we define the hanoi data type to represent a Tower of Hanoi configuration. In particular, a hanoi configuration is a list of three towers. The elementary constructors and selectors are given below:

;; ;; Hanoi configuration ;; (defun make-hanoi (from-tower aux-tower to-tower) "Create a Hanoi configuration from three towers." (list from-tower aux-tower to-tower)) (defun hanoi-tower (hanoi i) "Select the I'th tower of a Hanoi construction." (nth (1- i) hanoi)) 

Working with towers within a Hanoi configuration is tedious. We therefore define some shortcut to capture recurring operations:

;; ;; Utilities ;; (defun hanoi-tower-update (hanoi i tower) "Replace the I'th tower in the HANOI configuration by tower TOWER." (cond ((= i 1) (make-hanoi tower (second hanoi) (third hanoi))) ((= i 2) (make-hanoi (first hanoi) tower (third hanoi))) ((= i 3) (make-hanoi (first hanoi) (second hanoi) tower)))) (defun hanoi-tower-top (hanoi i) "Return the top disk of the I'th tower in the HANOI configuration." (tower-top (hanoi-tower hanoi i))) (defun hanoi-tower-pop (hanoi i) "Pop the top disk of the I'th tower in the HANOI configuration." (hanoi-tower-update hanoi i (tower-pop (hanoi-tower hanoi i)))) (defun hanoi-tower-push (hanoi i disk) "Push DISK into the I'th tower of the HANOI configuration." (hanoi-tower-update hanoi i (tower-push (hanoi-tower hanoi i) disk))) 

The fundamental operator we can perform on a Hanoi configuration is to move a top disk from one peg to another:

;; ;; Operator: move top disk from one tower to another ;; (defun move-disk (from to hanoi) "Move the top disk from peg FROM to peg TO in configuration HANOI." (let ((disk (hanoi-tower-top hanoi from)) (intermediate-hanoi (hanoi-tower-pop hanoi from))) (hanoi-tower-push intermediate-hanoi to disk))) 

We are now ready to capture the logic of our recursive solution into the following code:

;; ;; Subgoal: moving a tower from one peg to another ;; (defun move-tower (N from aux to hanoi) "In the HANOI configuration, move the top N disks from peg FROM to peg TO using peg AUX as an auxiliary peg." (if (= N 1) (move-disk from to hanoi) (move-tower (- N 1) aux from to (move-disk from to (move-tower (- N 1) from to aux hanoi))))) 

We use the driver function solve-hanoi to start up the recursion:

;; ;; Driver function ;; (defun solve-hanoi (N) "Solve the Tower of Hanoi problem." (move-tower N 1 2 3 (make-hanoi (make-complete-tower N) nil nil))) (defun make-complete-tower (N) "Create a tower of N disks." (make-complete-tower-aux N (make-empty-tower))) (defun make-complete-tower-aux (N A) "Push a complete tower of N disks on top of tower A." (if (zerop N) A (make-complete-tower-aux (1- N) (tower-push A N)))) 

To solve a Tower of Hanoi problem with 3 disks, we call (solve-hanoi 3):

USER(50): (solve-hanoi 3) (NIL NIL (1 2 3)) 

All we get back is the final configuration, which is not as interesting as knowing the sequence of moves taken by the algorithm. So we trace usage of the move-disk operator:

USER(51): (trace move-disk) (MOVE-DISK) USER(52): (solve-hanoi 3) 0: (MOVE-DISK 1 3 ((1 2 3) NIL NIL)) 0: returned ((2 3) NIL (1)) 0: (MOVE-DISK 1 2 ((2 3) NIL (1))) 0: returned ((3) (2) (1)) 0: (MOVE-DISK 3 2 ((3) (2) (1))) 0: returned ((3) (1 2) NIL) 0: (MOVE-DISK 1 3 ((3) (1 2) NIL)) 0: returned (NIL (1 2) (3)) 0: (MOVE-DISK 2 1 (NIL (1 2) (3))) 0: returned ((1) (2) (3)) 0: (MOVE-DISK 2 3 ((1) (2) (3))) 0: returned ((1) NIL (2 3)) 0: (MOVE-DISK 1 3 ((1) NIL (2 3))) 0: returned (NIL NIL (1 2 3)) (NIL NIL (1 2 3)) 

From the trace we can actually read off the sequence of operator applications necessary for one to achieve the solution configuration. This is good, but not good enough. We want to know why each move is being taken. So we trace also the high-level subgoals:

USER(53): (trace move-tower) (MOVE-TOWER) USER(54): (solve-hanoi 3) 0: (MOVE-TOWER 3 1 2 3 ((1 2 3) NIL NIL)) 1: (MOVE-TOWER 2 1 3 2 ((1 2 3) NIL NIL)) 2: (MOVE-TOWER 1 1 2 3 ((1 2 3) NIL NIL)) 3: (MOVE-DISK 1 3 ((1 2 3) NIL NIL)) 3: returned ((2 3) NIL (1)) 2: returned ((2 3) NIL (1)) 2: (MOVE-DISK 1 2 ((2 3) NIL (1))) 2: returned ((3) (2) (1)) 2: (MOVE-TOWER 1 3 1 2 ((3) (2) (1))) 3: (MOVE-DISK 3 2 ((3) (2) (1))) 3: returned ((3) (1 2) NIL) 2: returned ((3) (1 2) NIL) 1: returned ((3) (1 2) NIL) 1: (MOVE-DISK 1 3 ((3) (1 2) NIL)) 1: returned (NIL (1 2) (3)) 1: (MOVE-TOWER 2 2 1 3 (NIL (1 2) (3))) 2: (MOVE-TOWER 1 2 3 1 (NIL (1 2) (3))) 3: (MOVE-DISK 2 1 (NIL (1 2) (3))) 3: returned ((1) (2) (3)) 2: returned ((1) (2) (3)) 2: (MOVE-DISK 2 3 ((1) (2) (3))) 2: returned ((1) NIL (2 3)) 2: (MOVE-TOWER 1 1 2 3 ((1) NIL (2 3))) 3: (MOVE-DISK 1 3 ((1) NIL (2 3))) 3: returned (NIL NIL (1 2 3)) 2: returned (NIL NIL (1 2 3)) 1: returned (NIL NIL (1 2 3)) 0: returned (NIL NIL (1 2 3)) (NIL NIL (1 2 3)) 

The trace gives us information as to what subgoals each operator application is trying to establish. For example, the top level subgoals are the following:

 0: (MOVE-TOWER 3 1 2 3 ((1 2 3) NIL NIL)) 1: (MOVE-TOWER 2 1 3 2 ((1 2 3) NIL NIL)) ... 1: returned ((3) (1 2) NIL) 1: (MOVE-DISK 1 3 ((3) (1 2) NIL)) 1: returned (NIL (1 2) (3)) 1: (MOVE-TOWER 2 2 1 3 (NIL (1 2) (3))) ... 1: returned (NIL NIL (1 2 3)) 0: returned (NIL NIL (1 2 3)) 

They translate directly to the following: In order to move a tower of 3 disks from peg 1 to peg 3 using peg 2 as a buffer (i.e. (MOVE-TOWER 3 1 2 3 ((1 2 3) NIL NIL))) we do the following:

  1. 1: (MOVE-TOWER 2 1 3 2 ((1 2 3) NIL NIL))
    Move a tower of 2 disks from peg 1 to peg 2 using peg 3 as a buffer. The result of the move is the following:
    1: returned ((3) (1 2) NIL)
  2. 1: (MOVE-DISK 1 3 ((3) (1 2) NIL))
    Move a top disk from peg 1 to peg 3. The result of this move is:
    1: returned (NIL (1 2) (3))
  3. 1: (MOVE-TOWER 2 2 1 3 (NIL (1 2) (3)))
    Move a tower of 2 disks from peg 2 to peg 3 using peg 1 as a buffer, yielding the following configuration:
    1: returned (NIL NIL (1 2 3))

Advanced Functional Programming in LISP

Auxiliary Functions and Accumulator Variables

We define the reversal of a list L to be a list containing exactly the elements of L in reversed order. The Common LISP built-in function (reverse L) returns the reversal of L:

USER(1): (reverse '(1 2 3 4))
(4 3 2 1)
USER(2): (reverse '(1 (a b) (c d) 4))
(4 (c d) (a b) 1)
USER(3): (reverse nil)

Implementing a version of reverse is not difficult, but finding an efficient implementation is not as trivial. To review what we learned in the last tutorial, let us begin with a naive implementation of reverse:

(defun slow-list-reverse (L)
  "Create a new list containing the elements of L in reversed order."
  (if (null L)
    (list-append (slow-list-reverse (rest L)) 
                 (list (first L)))))

Notice that this linearly recursive implementation calls the function list-append we implemented in the the last tutorial. Notice also the new function list, which returns a list containing its arguments. For example, (list 1 2) returns the list (1 2). In general, (list x1 x2 ... xn) is just a shorthand for (cons x1 (cons x2 ... (cons xn nil) ... )). So, the expression (list (first L)) in the listing above returns a singleton list containing the first element of L.

So, why does (slow-list-reverse L) returns the reversal of L? The list L is constructed either by nil or by cons:

  • Case 1: L is nil.
    The reversal of L is simply nil.
  • Case 2: L is constructed from a call to cons.
    Then L has two components: (first L) and (rest L). If we append (first L) to the end of the reversal of (rest L), then we obtain the reversal of L. Of course, we could make use of list-append to do this. However, list-append expects two list arguments, so we need to construct a singleton list containing (first L) before we pass it as a second argument to list-append.

Let us trace the execution of the function to see how the recursive calls unfold:

USER(3): (trace slow-list-reverse)
USER(4): (slow-list-reverse '(1 2 3 4))
 0: (SLOW-LIST-REVERSE (1 2 3 4))
   1: (SLOW-LIST-REVERSE (2 3 4))
     2: (SLOW-LIST-REVERSE (3 4))
       3: (SLOW-LIST-REVERSE (4))
         4: returned NIL
       3: returned (4)
     2: returned (4 3)
   1: returned (4 3 2)
 0: returned (4 3 2 1)
(4 3 2 1)

Everything looks fine, until we trace also the unfolding of list-append:

USER(9): (trace list-append)
USER(10): (slow-list-reverse '(1 2 3 4))
 0: (SLOW-LIST-REVERSE (1 2 3 4))
   1: (SLOW-LIST-REVERSE (2 3 4))
     2: (SLOW-LIST-REVERSE (3 4))
       3: (SLOW-LIST-REVERSE (4))
         4: returned NIL
         4: (LIST-APPEND NIL (4))
         4: returned (4)
       3: returned (4)
       3: (LIST-APPEND (4) (3))
         4: (LIST-APPEND NIL (3))
         4: returned (3)
       3: returned (4 3)
     2: returned (4 3)
     2: (LIST-APPEND (4 3) (2))
       3: (LIST-APPEND (3) (2))
         4: (LIST-APPEND NIL (2))
         4: returned (2)
       3: returned (3 2)
     2: returned (4 3 2)
   1: returned (4 3 2)
   1: (LIST-APPEND (4 3 2) (1))
     2: (LIST-APPEND (3 2) (1))
       3: (LIST-APPEND (2) (1))
         4: (LIST-APPEND NIL (1))
         4: returned (1)
       3: returned (2 1)
     2: returned (3 2 1)
   1: returned (4 3 2 1)
 0: returned (4 3 2 1)
(4 3 2 1)

What we see here is revealing: given a list of N element, slow-list-reverse makes O(N) recursive calls, with each level of recursionl involving a call to the linear-time function list-append. The result is that slow-list-reverse is an O(N2)function.

We can in fact build a much more efficient version of reverse using auxiliary functions and accumulator variables:

(defun list-reverse (L)
  "Create a new list containing the elements of L in reversed order."
  (list-reverse-aux L nil))

(defun list-reverse-aux (L A)
  "Append list A to the reversal of list L."
  (if (null L)
    (list-reverse-aux (rest L) (cons (first L) A))))

The function list-reverse-aux is an auxiliary function (or a helper function). It does not perform any useful function by itself, but the driver function list-reverse could use it as a tool when building a reversal. Specifically, (list-reverse-aux L A) returns a new list obtained by appending list A to the reversal of list L. By passing nil as A to list-reverse-aux, the driver function list-reverse obtains the reversal of L.

Let us articulate why (list-reverse-aux L A) correctly appends A to the reversal of list L. Again, we know that either L is nil or it is constructed by cons:

  • Case 1: L is nil.
    The reversal of L is simply nil. The result of appending A to the end of an empty list is simply A itself.
  • Case 2: L is constructed by cons.
    Now L is composed of two parts: (first L) and (rest L). Observe that (first L) is the last element in the reversal of L. If we are to append A to the end of the reversal of L, then (first L) will come immediately before the elements of A. Observing the above, we recognize that we obtain the desired result by recursively appending (cons (first L) A) to the reversal of (rest L).

Tracing both list-reverse and list-reverse-aux, we get the following:

USER(17): (trace list-reverse list-reverse-aux)
USER(18): (list-reverse '(1 2 3 4))
 0: (LIST-REVERSE (1 2 3 4))
   1: (LIST-REVERSE-AUX (1 2 3 4) NIL)
     2: (LIST-REVERSE-AUX (2 3 4) (1))
       3: (LIST-REVERSE-AUX (3 4) (2 1))
         4: (LIST-REVERSE-AUX (4) (3 2 1))
           5: (LIST-REVERSE-AUX NIL (4 3 2 1))
           5: returned (4 3 2 1)
         4: returned (4 3 2 1)
       3: returned (4 3 2 1)
     2: returned (4 3 2 1)
   1: returned (4 3 2 1)
 0: returned (4 3 2 1)
(4 3 2 1)

For each recursive call to list-reverse-aux, notice how the first element of L is “peeled off”, and is then “accumulated” in A. Because of this observation, we call the variable A an accumulator variable.

Factorial Revisited

To better understand how auxiliary functions and accumulator variables are used, let us revisit the problem of computing factorials. The following is an alternative implementation of the factorial function:

(defun fast-factorial (N)
  "A tail-recursive version of factorial."
  (fast-factorial-aux N 1))

(defun fast-factorial-aux (N A)
  "Multiply A by the factorial of N."
  (if (= N 1)
    (fast-factorial-aux (- N 1) (* N A))))

Let us defer the explanation of why the function is named “fast-factorial“, and treat it as just another way to implement factorial. Notice the structural similarity between this pair of functions and those for computing list reversal. The auxiliary function (fast-factorial-aux N A) computes the product of A and the N‘th factorial. The driver function computes N! by calling fast-factorial-aux with A set to 1. Now, the correctness of the auxiliary function (i.e. that (fast-factorial-aux N A) indeed returns the product of N! and A) can be established as follows. Nis either one or larger than one.

  • Case 1: N = 1
    The product of A and 1! is simply A * 1! = A.
  • Case 2: N > 1
    Since N! = N * (N-1)!, we then have N! * A = (N-1)! * (N * A), thus justifying our implementation.

Tracing both fast-factorial and fast-factorial-aux, we get the following:

USER(3): (trace fast-factorial fast-factorial-aux)  
USER(4): (fast-factorial 4)
     2: (FAST-FACTORIAL-AUX 3 4)
       3: (FAST-FACTORIAL-AUX 2 12)
         4: (FAST-FACTORIAL-AUX 1 24)
         4: returned 24
       3: returned 24
     2: returned 24
   1: returned 24
 0: returned 24

If we compare the structure of fast-factorial with list-reverse, we notice certain patterns underlying the use of accumulator variables in auxiliary functions:

  1. An auxiliary function generalizes the functionality of the driver function by promising to compute the function of interest and also combine the result with the value of the accumulator variable. In the case of list-reverse-aux, our original interest were computing list reversals, but then the auxiliary function computes a more general concept, namely, that of appending an auxiliary list to some list reversal. In the case of fast-factorial-aux, our original interest were computing factorials, but the auxiliary function computes a more general value, namely, the product of some auxiliary number with a factorial.
  2. At each level of recursion, the auxiliary function reduces the problem into a smaller subproblem, and accumulating intermediate results in the accumulator variable. In the case of list-reverse-aux, recursion is applied to the sublist (rest L), while (first L) is cons‘ed with A. In the case of fast-factorial, recursion is applied to (N – 1)!, while N is multiplied with A.
  3. The driver function initiates the recursion by providing an initial value for the auxiliary variable. In the case of computing list reversals, list-reverse initializes A to nil. In the case of computing factorials, fast-factorial initializes A to 1.

Now that you understand how fast-factorial works, we explain where the adjective “fast” come from …

Tail Recursions

Recursive functions are usually easier to reason about. Notice how we articulate the correctness of recursive functions in this and the previous tutorial. However, some naive programmers complain that recursive functions are slow when compared to their iterative counter parts. For example, consider the original implementation of factorialwe saw in the previous tutorial:

(defun factorial (N)
  "Compute the factorial of N."
  (if (= N 1)
    (* N (factorial (- N 1)))))

It is fair to point out that, as recursion unfolds, stack frames will have to be set up, function arguments will have to be pushed into the stack, so on and so forth, resulting in unnecessary runtime overhead not experienced by the iterative counterpart of the above factorialfunction:

int factorial(int N) {
  int A = 1;
  while (N != 1) {
    A = A * N;
    N = N - 1;
  return A;

Because of this and other excuses, programmers conclude that they could write off recursive implementations …

Modern compilers for functional programming languages usually implement tail-recursive call optimizations which automatically translate a certain kind of linear recursion into efficient iterations. A linear recursive function is tail-recursive if the result of each recursive call is returned right away as the value of the function. Let’s examine the implementation of fast-factorial again:

(defun fast-factorial (N)
  "A tail-recursive version of factorial."
  (fast-factorial-aux N 1))

(defun fast-factorial-aux (N A)
  "Multiply A by the factorial of N."
  (if (= N 1)
    (fast-factorial-aux (- N 1) (* N A))))

Notice that, in fast-factorial-aux, there is no work left to be done after the recursive call (fast-factorial-aux (- N 1) (* N A)). Consequently, the compiler will not create new stack frame or push arguments, but instead simply bind (- N 1) to N and (* N A) to A, and jump to the beginning of the function. Such optimization effectively renders fast-factorial as efficient as its iterative counterpart. Notice also the striking structural similarity between the two.

When you implement linearly recursive functions, you are encouraged to restructure it as a tail recursion after you have fully debugged your implementation. Doing so allows the compiler to optimize away stack management code. However, you should do so only after you get the prototype function correctly implemented. Notice that the technique of accumulator variables can be used even when we are not transforming code to tail recursions. For some problems, the use of accumulator variables offers the most natural solutions.

Exercise: Recall that the N‘th triangular number is defined to be 1 + 2 + 3 + … + N. Give a tail-recursive implementation of the function (fast-triangular N) which returns the N‘th triangular number.

Exercise: Give a tail-recursive implementation of the function (fast-power B E) that raises B to the power E (assuming that both B and E are non-negative integers).

Exercise: Give a tail-recursive implementation of the function (fast-list-length L), which returns the length of a given list L.

Functions as First-Class Objects

A data type is first-class in a programming language when you can pass instances of the data type as function arguments or return them as function values. We are used to treating numeric and Boolean values as first-class data types in languages like Pascal and C. However, we might not be familiar to the notion that functions could be treated as first-class objects, that is, functions can be passed as function arguments and returned as function values. This unique feature of Common LISP and other functional languages makes it easy to build very powerful abstractions. In the remaining of this tutorial, we will look at what passing functional arguments buys us. In the fourth tutorial, when we talk about imperative programming in LISP, we will return to the topic of returning functional values.

Frequently, we need to apply a transformation multiple times on the same data object. For example, we could define the following transformation:

(defun double (x)
  "Multiple X by 2."
  (* 2 x))

We could compute 24 by applying the doubletransformation 4 times on 1:

USER(10): (double (double (double (double 1))))

Not only is this clumsy, but it also fails to express the very idea that the same transformation is applied multiple times. It would be nice if we can repeat applying a given transformation F on X for N times by simply writing (repeat-transformation F N X). The following function does just that:

(defun repeat-transformation (F N X)
  "Repeat applying function F on object X for N times."
  (if (zerop N)
    (repeat-transformation F (1- N) (funcall F X))))

The definition follows the standard tail recursive pattern. Notice the form (funcall F X). Given a function F and objects X1 X2Xn, the form (funcall F X1 X2 ... Xn) invoke the function F with arguments X1, X2, …, Xn. The variable N is a counter keeping track of the remaining number of times we need to apply function F to the accumulator variable X.

To pass a the function double as an argument to repeat-transformation, we need to annotate the function name double by a closure constructor, as in the following:

USER(11): (repeat-transformation (function double) 4 1)

There is nothing magical going on, the closure constructor is just a syntax for telling Common LISP that what follows is a function rather than a local variable name. Had we not included the annotation, Common LISP will treat the name double as a variable name, and then report an error since the name doubleis not defined.

To see how the evaluation arrives at the result 16, we could, as usual, trace the execution:

USER(12): (trace repeat-transformation)
USER(13): (repeat-transformation #'double 4 1)
         4: (REPEAT-TRANSFORMATION # 0 16)
         4: returned 16
       3: returned 16
     2: returned 16
   1: returned 16
 0: returned 16

Higher-Order Functions

Notice that exponentiation is not the only use of the repeat-transformation function. Let’s say we want to build a list containing 10 occurrences of the symbol blah. We can do so with the help of repeat-transformation:

USER(30): (defun prepend-blah (L) (cons 'blah L))
USER(31): (repeat-transformation (function prepend-blah) 10 nil)

Suppose we want to fetch the 7’th element of the list (a b c d e f g h i j). Of course, we could use the built in function seventh to do the job, but for the fun of it, we could also achieve what we want in the following way:

USER(32): (first (repeat-transformation (function rest) 6 '(a b c d e f g h i j)))

Basically, we apply rest six times before apply first to get the seventh element. In fact, we could have defined the function list-nth(see previous tutorial) in the following way:

(defun list-nth (N L)
  (first (repeat-transformation (function rest) N L)))

(list-nthnumbers the member of a list from zero onwards.)

As you can see, functions that accepts other functions as arguments are very powerful abstractions. You can encapsulate generic algorithms in such a function, and parameterize their behavior by passing in different function arguments. We call a function that has functional parameters (or return a function as its value) a higher-order function.

One last point before we move on. The closure constructor function is used very often when working with higher-order functions. Common LISP therefore provide another equivalent syntax to reduce typing. When we want Common LISP to interpret a name F as a function, instead of typing (function F), we can also type the shorthand #'F. The prefix #' is nothing but an alternative syntax for the closure constructor. For example, we could enter the following:

USER(33): (repeat-transformation #'double 4 1)
USER(34): (repeat-transformation #'prepend-blah 10 nil)
USER(35): (first (repeat-transformation #'rest 6 '(a b c d e f g h i j)))

Lambda Expressions

Some of the functions, like prepend-blah for example, serves no other purpose but to instantiate the generic algorithm repeat-transformation. It would be tedious if we need to define it as a global function using defun before we pass it into repeat-transformation. Fortunately, LISP provides a mechanism to help us define functions “in place”:

USER(36): (repeat-transformation #'(lambda (L) (cons 'blah L)) 10 nil)

The first argument (lambda (L) (cons 'blah L)) is a lambda expression. It designates an anonymous function (nameless function) with one parameter L, and it returns as a function value (cons 'blah L). We prefix the lambda expression with the closure constructor #' since we want Common LISP to interpret the argument as a function rather than a call to a function named lambda.

Similarly, we could have computed powers as follows:

USER(36): (repeat-transformation #'(lambda (x) (* 2 x)) 4 1)

Exercise: Define a function (apply-func-list L X) so that, given a list L of functions and an object X, apply-func-list applies the functions in L to Xin reversed order. For example, the following expression

(apply-func-list (list #'double #'list-length #'rest) '(1 2 3))

is equivalent to

(double (list-length (rest '(1 2 3))))

Exercise: Use apply-func-listto compute the following:

  1. 10 times the fourth element of the list (10 20 30 40 50),
  2. the third element of the second element in the list ((1 2) (3 4 5) (6)),
  3. the difference between 10 and the length of (a b c d e f),
  4. a list containing a list containing the symbol blah.

Iterating Through a List

We have seen how we could iterate through a list using linear recursion. We have also seen how such recursion can be optimized if structured in a tail-recursive form. However, many of the the list processing functions look striking similar. Consider the following examples:

(defun double-list-elements (L)
  "Given a list L of numbers, return a list containing the elements of L multiplied by 2."
  (if (null L)
    (cons (double (first L)) (double-list-elements (rest L)))))

(defun reverse-list-elements (L)
  "Given a list L of lists, return a list containing the reversal of L's members."
  (if (null L)
    (cons (reverse (first L)) (reverse-list-elements (rest L)))))

We could come up with infinitely many more examples of this sort. Having to write a new function everytime we want to iterate through a list is both time-consuming and error-prone. The solution is to capture the generic logic of list iteration in higher-order functions, and specialize their behaviors by passing in functional arguments.

If we look at the definitions of double-list-elements and reverse-list-elements, we see that they apply a certain function to the first element of their input, then recursively invoke themselves to process the rest of the input list, and lastly combine the result of the two function calls using cons. We could capture this logic into the following function:

(defun mapfirst (F L)
  "Apply function F to every element of list L, and return a list containing the results."
  (if (null L)
    (cons (funcall F (first L)) (mapfirst F (rest L)))))

The functions double-list-elements and reverse-list-elements can be replaced by the following:

USER(18): (mapfirst #'double '(1 2 3 4))
(2 4 6 8)
USER(19): (mapfirst #'reverse '((1 2 3) (a b c) (4 5 6) (d e f)))
((3 2 1) (C B A) (6 5 4) (F E D))

Of course, you could also pass lambda abstractions as arguments:

USER(20): (mapfirst #'(lambda (x) (* x x)) '(1 2 3 4))
(1 4 9 16)

In fact, the higher-order function is so useful that Common LISP defines a function mapcar that does exactly what mapfirst is intended for:

USER(22): (mapcar #'butlast '((1 2 3) (a b c) (4 5 6) (d e f)))
((1 2) (A B) (4 5) (D E))

The reason why it is called mapcar is that the function first was called car in some older dialects of LISP (and rest was called cdr in those dialects; Common LISP still supports car and cdr but we strongly advice you to stick with the more readable first and rest). We suggest you to consider using mapcarwhenever you are tempted to write your own list-iterating functions.

The function mapcar is an example of generic iterators, which capture the generic logic of iterating through a list. If we look at what we do the most when we iterate through a list, we find that the following kinds of iterations occurs most frequently in our LISP programs:

  1. Transformation iteration: transforming a list by systematically applying a monadic function to the elements of the list.
  2. Search iteration: searching for a list member that satisfies a given condition.
  3. Filter iteration: screening out all members that does not satisfy a given condition.

As we have already seen, mapcar implements the generic algorithm for performing transformation iteration. In the following, we will look at the analogous of mapcarfor the remaining iteration categories.

Search Iteration

Let us begin by writing a function that returns an even element in a list of numbers:

(defun find-even (L)
  "Given a list L of numbers, return the leftmost even member."
  (if (null L)
    (if (evenp (first L))
        (first L)
      (find-even (rest L)))))

Exercise: Implement a function that, when given a list L of lists, return a non-empty member of L.

We notice that the essential logic of searching can be extracted out into the following definition:

(defun list-find-if (P L)
  "Find the leftmost element of list L that satisfies predicate P."
  (if (null L)
    (if (funcall P (first L))
        (first L)
      (list-find-if P (rest L)))))

The function list-find-if examines the elements of L one by one, and return the first one that satisfies predicate P. The function can be used for locating even or non-nilmembers in a list:

USER(34): (list-find-if #'evenp '(1 3 5 8 11 12))
USER(35): (list-find-if #'(lambda (X) (not (null X))) '(nil nil (1 2 3) (4 5)))
(1 2 3)

Common LISP defines a built-in function find-if which is a more general version of list-find-if. It can be used just like list-find-if:

USER(37): (find-if #'evenp '(1 3 5 8 11 12))
USER(38): (find-if #'(lambda (X) (not (null X))) '(nil nil (1 2 3) (4 5)))
(1 2 3)

Exercise: Use find-if to define a function that searches among a list of lists for a member that has length at least 3.

Exercise: Use find-if to define a function that searches among a list of lists for a member that contains an even number of elements.

Exercise: Use find-if to define a function that searches among a list of numbers for a member that is divisible by three.

Filter Iteration

Given a list of lists, suppose we want to screen out all the member lists with length less than three. We could do so by the following function:

(defun remove-short-lists (L)
  "Remove all members of L that has length less than three."
  (if (null L)
    (if (< (list-length (first L)) 3)
        (remove-short-lists (rest L))
      (cons (first L) (remove-short-lists (rest L))))))

To articulate the correctness of this implementation, consider the following. The list L is either nil or constructed by cons.

  • Case 1: L is nil.
    Removing short lists from an empty list simply results in an empty list.
  • Case 2: L is constructed by cons.
    L has two components: (first L) and (rest L). We have two cases: either (first L)has fewer than 3 members or it has at least 3 members.

    • Case 2.1: (first L) has fewer than three elements.
      Since (first L) is short, and will not appear in the result of removing short lists from L, the latter is equivalent to the result of removing short lists from (rest L).
    • Case 2.2: (first L) has at least three elements.
      Since (first L) is not short, and will appear in the result of removing short lists from L, the latter is equivalent to adding (first L) to the result of removing short lists from (rest L).

A typical execution trace is the following:

USER(17): (remove-short-lists '((1 2 3) (1 2) nil (1 2 3 4)))
 0: (REMOVE-SHORT-LISTS ((1 2 3) (1 2) NIL (1 2 3 4)))
   1: (REMOVE-SHORT-LISTS ((1 2) NIL (1 2 3 4)))
     2: (REMOVE-SHORT-LISTS (NIL (1 2 3 4)))
       3: (REMOVE-SHORT-LISTS ((1 2 3 4)))
         4: returned NIL
       3: returned ((1 2 3 4))
     2: returned ((1 2 3 4))
   1: returned ((1 2 3 4))
 0: returned ((1 2 3) (1 2 3 4))
((1 2 3) (1 2 3 4))

Alternatively, we could have removed short lists using Common LISP’s built-in function remove-if:

USER(19): (remove-if #'(lambda (X) (< (list-length X) 3)) '((1 2 3) (1 2) nil (1 2 3 4)))
((1 2 3) (1 2 3 4))

The function (remove-if P L) constructs a new version of list L that contains only members not satisfying predicate P. For example, we can remove all even members from the list (3 6 8 9 10 13 15 18) by the following:

USER(21): (remove-if #'(lambda (X) (zerop (rem x 2))) '(3 6 8 9 10 13 15 18))
(3 9 13 15)

Without remove-if, we would end up having to implement a function like the following:

(defun remove-even (L)
  "Remove all members of L that is an even number."
  (if (null L)
    (if (zerop (rem (first L) 2))
        (remove-even (rest L))
      (cons (first L) (remove-even (rest L))))))

Exercise: Demonstrate the correctness of remove-even using arguments you have seen in this tutorial.

Exercise: Observe the recurring pattern in remove-short-lists and remove-even, and implement your own version of remove-if.

We could actually implement list-intersection using remove-if and lambda abstraction:

(defun list-intersection (L1 L2)
  "Compute the intersection of lists L1 and L2."
  (remove-if #'(lambda (X) (not (member X L2))) L1))

In the definition above, the lambda abstraction evaluates to a predicate that returns true if its argument is not a member of L2. Therefore, the remove-if expression removes all elements of L1 that is not a member of L2. This precisely gives us the intersection of L1 and L2.

Exercise: Use remove-if and lambda abstractions to implement list-difference.

Exercise: Look up the functionality of remove-if-not in CTRL2. Re-implement list-intersection using remove-if-not and lambda abstraction.

Functions Returning Multiple Values

The functions we have seen so far return a single value, but some LISP functions return two or more values. For example, if two arguments number and divisor are passed to the floor function, it returns two values, the quotient q and the remainder r so that number = divisor * q + r.

USER(11): (floor 17 5)
USER(12): (floor -17 5)

Regular binding constructs like let and let*only allows us to catch the first returned value of a multiple-valued function, as the following example illustrates:

USER(13): (let ((x (floor 17 5))) x)

One can use the multiple-value-bindto receive the results of a multiple-valued function:

USER(14): (multiple-value-bind (x y) (floor 17 5) 
            (+ x y))

In the above expression, (x y) is the list of variables binding to the returned values, (floor 17 5) is the expression generating multiple results. Binding the two values of (floor 17 5) to x and y, LISP then evaluates the expression (+ x y).

One may also write LISP functions that return multiple values:

(defun order (a b)
  "Return two values: (min a b) and (max a b)."
  (if (>= a b)
      (values b a)
    (values a b)))

The valuesspecial form returns its arguments as multiple values.

For more information about LISP constructs for handling multiple values, consult section 7.10 of CLTL2.

Exercise: Implement a tail-recursive function (list-min-max L) that returns two values: the minimum and maximum members of a list L of numbers.