To earn full credit the work must be done before the end of the lab and should contain a list of at least 3 notes. Each note is a short paragraph with one or two sentences and a new LISP function (or link to a file containing the function). The sentences should say what the function does and what you learned by writing it.
Let me know by calling me across to your workstation when done.
:set showmatch
Here are some simple LISP expressions/commands. Load the LISP interpreter and input each in turn. Try to predict what each will return as a value before inputting it:
()
(+ 1 2)
(1 + 2)
(* (+ 1 2) (+ 3 4))
(+ 1 2 3 4)
(A B C)
'(A B C)
A
'A
(A)If you get one wrong... you may need to go back to [ 11.html ] again.
( define (a) 4321)
(a)
a
'a
(a 1)Copy and paste the above XLISP commands into XLISP in a terminal window.
Then define and test a new function called answer that returns the value 42.
(DEFINE (square x) (* x x))Here is how you test it...
(square 3)
(square 4)
(square 5)
Here is how you can use it:
(square (+ 1 2 3))
(+ (square 3) (square 4) )
(+ 3 (square 3) )Test the above!
Here is how XLISP can list it:
squareXLISP does not let you edit a function however!
Do not leave LISP until you complete the next two steps.
Here is a function for the cube:
(DEFINE (cube x) (* x (square x)))Test it.
Define a function called fourth that returns the fourth power of a number. Use the fact that the fourth power on n is the square of the square of n:
Test it. And save it...
Here is a definition of a function with two arguments in LISP:
(DEFINE (pythagoras x y) (+ (square x) (square y)))
(pythagoras 3 4)Some common errors
(pythagoras 4)
(pythagoras 1 3 4)
(pythagoras )In our XLISP the value of the function name is an expression defining a function:
pythagoras
Here is a function that return the larger of two expressions:
(define (max2 a b) (if (> a b) a b))
(max2 1 17)
(max2 17 1)
(max2 (max2 3 5) (max2 4 1))
(max2 (square 3) (cube 2))
Define a function called min2 that returns the smallest of two arguments. Test it and save it....
(define (binroot target lo hi error )
(let (( mid (/ (+ lo hi) 2.0))) ; this saves the time to recalculate mid
(if (<= (- hi lo) error)
mid
(if (< (square mid) target)
(binroot target mid hi error)
(binroot target lo mid error)
)
)
)
)Here is how I tested it:
(binroot 50 0 100 0.05)Test it further and trace it.
The above algorithm will find roots of any monotonic increasing function. Modify it to find cube roots and fourth roots of positive numbers.
(define (power x n)
(cond
((= n 0) 1)
((= n 1) x)
(T (* x (power x (- n 1))))
)
)Here are two test cases
(power 2 3)
(power 3 2)Test with the trace function....
However this is not a very fast way to calculate powers. There
is another one based on these facts:
Net
Can you figure out how to speed up the original power function?
(define (factorial n)
(if (= n 0) 1 (* n ( factorial ( - n 1 ) ))
)
)
Here you print out the definition of factorial:
factorialHere you input expressions that apply factorial to simple numbers:
(factorial 0)
(factorial 3)
(factorial 5)
Trace how it works:
(trace factorial)
(factorial 0)
(factorial 3)
(factorial 5)
(untrace factorial)
Below you can use factorial is more complicated ways:
(factorial (factorial 3))
(setq n 3)
(factorial n)
(setq f (factorial n))
f
(setq f (factorial f))
f
The problem with this factorial is that n! quite a small n is too large to be represented as an integer. Worse n! when n<0 is infinitely small! As a result we should define a safe factorial that returns a string "error" when n<0 or n is too big.
Program and test this one.
. . . . . . . . . ( end of section Optional experiments if you have time) <<Contents | End>>
. . . . . . . . . ( end of section CS320 Lab 12 LISP Laboratory Number 2) <<Contents | End>>