LISP is one of the simplest computer languages in terms of syntax and semantics, and also one of the most powerful. It was developed in the mid-1950’s by John McCarthy at M.I.T. as a “LISt Processing language.” It has been historically used for virtually all Artificial Intelligence programs and is often the environment of choice for applications which require a powerful interactive working environment. LISP presents a very different way to think about programming from the “algorithmic” languages, such as Python, C++, and Java.
As its name implies, the basis of LISP is a list. One constructs a list by enumerating elements inside a pair of parentheses. For example, here is a list with four elements (the second element is also a list):
(23 (this is easy) hello 821)
The elements in the list, which are not lists, are called “atoms.” For example, the atoms in the list above are: 23, ‘this, ‘hello, 821, ‘easy, and ‘is. Literals are identified with a single leading quote. Everything in LISP is either an atom or a list (but not both). The only exception is “NIL,” which is both an atom and a list. It can also be written as “()” – a pair of parentheses with nothing inside.
All statements in LISP are function calls with the following syntax: (function arg1 arg2 arg3 … argn). To evaluate a LISP statement, each of the arguments (possibly functions themselves) are evaluated, and then the function is invoked with the arguments. For example, (MULT (ADD 2 3) (ADD 1 4 2)) has a value of 35, since (ADD 2 3) has a value of 5, (ADD 1 4 2) has a value of 7, and (MULT 5 7) has a value of 35. Some functions have an arbitrary number of arguments; others require a fixed number. All statements return a value, which is either an atom or a list.
Basic Functions (SET, SETQ, EVAL, ATOM)
We may assign values to variables using the function SET. For example, the statement (SET ’test 6) would have a value of a 6, and (more importantly, however) would also cause the atom ‘test to be bound to the atom 6. The function SETQ is the same as SET, but it causes LISP to act as if the first argument was quoted. Observe the following examples:
Statement Value Comment (SET ’a ( MULT 2 3)) 6 a is an atom with a vaue of 6 (SET ’a ’(MULT 2 3)) ‘(MULT 2 3) a is a list with 3 elements (SET ’b ’a) ‘a b is an atom with a value of the character a (SET ’c a) ‘(MULT 2 3) c is a list with 3 elements (SETQ EX (ADD 3 (MULT 2 5))) 13 The variable EX has a value of 13 (SETQ VOWELS ’(A E I O U)) ‘(A E I O U) VOWELS is a list of 5 elements
The function EVAL returns the value of its argument, after it has been evaluated. For example, (SETQ z ’(ADD 2 3)) has a value of the list (ADD 2 3); the function (EVAL ’z) has a value of (ADD 2 3); the function (EVAL z) has a value of 5 (but the binding of the atom z has not changed). In this last example, you can think of z being “resolved” twice: once because it is an argument to a function and LISP evaluates all arguments to functions before the function is invoked, and once when the function EVAL is invoked to resolve arguments. The function ATOM can be used to determine whether an item is an atom or a list. It returns either "true", or "NIL" for false. Consider the following examples:
Statement Value Comment (SETQ ’p '(ADD 1 2 3 4)) '(ADD 1 2 3 4) p is a list with 5 elements (ATOM 'p) true The argument to ATOM is the ato p (ATOM p) NIL Because p is not quoted, it is evaluated to the 5-element list. (EVAL p) 10 The argument to EVAL is the value of p; the value of p is 10.
List Functions (CAR, CDR, REVERSE, CONS)
The two most famous LISP functions are CAR and CDR (pronounced: could-er), named after registers of a now long-forgotten IBM machine on which LISP was first developed. The function (CAR x) returns the first item of the list x (and x must be a list or an error will occur); (CDR x) returns the list without its first element (again, x must be a list). The function CONS takes two arguments, of which the second must be a list. It returns a list which is composed by placing the first argument as the first element in the second argument’s list. The function REVERSE returns a list which is its arguments in reverse order.
The CAR and CDR functions are used extensively to grab specific elements of a list or sublist, that there's a shorthand for this: (CADR x) is the same as (CAR (CDR x)), (CAADDAR x) is a shorthand for (CAR (CAR (CDR (CDR (CAR x))))), and so on.
The following examples illustrate the use of CAR, CDR, REVERSE, and CONS:
Statement Value (REVERSE ’(8 7 6)) (6 7 8) (REVERSE ’(1 (2 3 4) 5 6)) (6 5 (2 3 4) 1) (CAR ’(This is a list)) This (CAR (CDR ’(This is a list))) Is (SETQ x (CDR (CDR ’(a b c d)))) (c d) (CDR ’(This is a list )) (is a list) (CAR ’(hi)) Hi (CDR ’(one)) NIL or () (CAR ’((1) (2 3) (4 5 6))) (1) (CONS 32 ’(22 12)) (32 22 12) (CONS ’first ’(last)) (first last) (CONS ’(one) ’(two three)) ((one) two three) (CDR (CAR ’((red white) blue))) (white) (CONS (CAR ’(red green blue)) (CDR ’(brown tan gold))) (red tan gold) (CAR ’hi) Error (CAR ()) Error (CAR ’(ADD 2 2)) ADD (CONS 78 NIL) (78) (CDR ’((red green) (blue white)) ((blue white)) (SETQ fruit ’(apple orange)) (apple orange) (SETQ more (CONS ’grape fruit)) (grape apple orange) (CDR (CDR (REVERSE more))) (grape) (SETQ colors ’(red green yellow)) (red green yellow) (SETQ area (MULT 2 3)) 6 (SETQ A (CONS area ’(CDR colors))) (6 CDR colors) (SETQ B (CONS ’area (CDR colors))) (area green yellow) (SETQ C (REVERSE colors)) (yellow green red) (CAR (CDR C)) Green (CONS ’pink (CONS ’orange (CONS C NIL))) (pink orange (yellow green red))
As you have probably deduced, the function ADD simply summed its arguments. We’ll also be using the following arithmetic functions:
Function Result (ADD x1 x2 …) sum of all arguments (MULT x1 x2 …) product of all arguments (SUB a b) a-b (DIV a b) a/b (SQUARE a) a*a (EXP a n) an (EQ a b) true if a and b are equal, NIL otherwise (POS a) true if a is positive, NIL otherwise (NEG a) true if a is negative, NIL otherwise
Some examples of these functions are as follows:
Statement Value (ADD (EXP 2 3) (SUB 4 1) (DIV 54 4)) 24.5 (SUB (MULT 3 2) (SUB 12 (ADD 2 2))) -2 (ADD (SQUARE 3) (SQUARE 4)) 25
LISP also allows us to create our own functions using the DEF function. (We will sometimes use DEFUN rather than DEF, as it is a bit more standard terminology.) For example,
(DEF SECOND (parms) (CAR (CDR parms)))
defines a new function called SECOND which operates on a single parameter named “parms”. SECOND will take the CDR of the parameter and then the CAR of that result. So, for example: (SECOND ’(a b c d e)) would first CDR the list (yielding (b c d e)) and then CAR the result. So the value would be the single character “b”. Consider the following program fragment:
- (SETQ X ’(a c s l))
- (DEF WHAT(parms) (CONS parms (REVERSE (CDR parms))))
- (DEF SECOND(parms) (CONS (CAR (CDR parms)) NIL))
The following chart illustrates the use of the user-defined functions WHAT and SECOND:
Statement Value (WHAT X) ((a c s l) l s c) (SECOND X) (c) (SECOND (WHAT X)) (l) (WHAT (SECOND X)) ((c))
There are many online LISP interpreters available on the Internet. The one that ACSL uses for testing its programs is CLISP that is accessible from JDoodle. This interpreter is nice because it is quite peppy in running programs, and functions are not case sensitive. So,
(CAR (CDR x)) is legal as is
(car (cdr x)) One drawback of this interpreter is the print function changes lowercase input into uppercase.
Questions from this topic will typically present a line of LISP code or a short sequence of statements and ask what is the value of the (final) statement.
Sample Problem 1
Evaluate the following expression.
(EXP (MULT 2 (SUB 5 (DIV (ADD 5 3 4) 2)) 3) 3)
(EXP (MULT 2 (SUB 5 (DIV (ADD 5 3 4) 2)) 3) 3) (EXP (MULT 2 (SUB 5 (DIV 12 2)) 3) 3) (EXP (MULT 2 (SUB 5 6) 3) 3) (EXP (MULT 2-1 3) 3) (EXP –6 3) -216
Sample Problem 2
(CDR ’((2 (3))(4 (5 6) 7)))
Solution: The CDR function takes the first element of its parameter (which is assumed to be a list) and returns the list with the first element removed. The first element of the list is (2 (3)) and the list without this element is ((4 (5 6) 7)), a list with one element.
Sample Problem 3
Consider the following program fragment:
(SETQ X ’(RI VA FL CA TX)) (CAR (CDR (REVERSE X)))
What is the value of the CAR expression?
Solution: The first statement binds variable X to the list ‘(RI VA FL CA TX). The REVERSE of this list is ‘(TX CA FL VA RI) whose CDR is ‘(CA FL VA RI). The CAR of this list is just the atom CA (without the quote).
Sample Problem 4
Given the function definitions for HY and FY as follows:
(DEFUN HY(PARM) (REVERSE (CDR PARM))) (DEFUN FY(PARM) (CAR (HY (CDR PARM))))
What is the value of the following?
(FY ’(DO RE (MI FA) SO))
Solution: To evaluate (FY ’(DO RE (MI FA) SO)), we must evaluate
(CAR (HY (CDR ’(DO RE (MI FA) SO)))).
Thus, HY is invoked with
PARM = ’(RE (MI FA) SO), and we evaluate (REVERSE (CDR ’(RE (MI FA) SO)))
This has a value of (SO (MI FA)) which is returned to FY. FY now takes the CAR of this which is SO (without the quotes).
The following YouTube videos show ACSL students and advisors working out some ACSL problems that have appeared in previous contests. Some of the videos contain ads; ACSL is not responsible for the ads and does not receive compensation in any form for those ads.
|LISP very gentle intro (CalculusNguyenify)
A very gentle introduction to the LISP category, covering the LISP syntax and the basic operators, SETQ, ADD, SUB, MULT, and DIV.
|LISP Basics (for ACSL) (Tangerine Code)
Explains the LISP operators CAR, CDR, and REVERSE, in the context of solving an ACSL All-Star Contest problem.
|LISP Basics (for ACSL) (Tangerine Code)
Completes the problem that was started in the above video.