Discussion 9: Scheme, Scheme Lists

disc09.pdf

Switch to Pensieve:

Everyone: Go to pensieve.co, log in with your @berkeley.edu email, and enter your group number as the room number (which was in the email that assigned you to this discussion). As long as you all enter the same number (any number), you'll all be using a shared document.

Once you're on Pensieve, you don't need to return to this page; Pensieve has all the same content (but more features). If for some reason Penseive doesn't work, return to this page and continue with the discussion.

Attendance

Fill out this discussion attendance form with the unique number you receive from your TA. As soon as you get your number, fill out the form, selecting arrival (not departure -- that's later).

Getting Started

If there are fewer than 3 people in your group, feel free to merge your group with another group in the room.

Everybody say your name, and then figure out who most recently pet a dog. (Feel free to share dog photos. Even cat photos are acceptable.)

Scheme

An if expression has 3 subexpressions:

The first one, called the predicate, is always evaluated to choose between the next two.
The second one, called the consequent, is evaluated if the predicate's value is true.
The third one, called the alternative, is evaluated if the predicate's value is false. Only #f is a false value in Scheme.

The value of the second or third expression (whichever gets evaluated) is the value of the whole if expression.

Q1: Perfect Fit

Definition: A perfect square is k*k for some integer k.

Implement fit, which takes non-negative integers total and n. It returns whether there are n different positive perfect squares that sum to total.

Important: Don't use the Scheme interpreter to tell you whether you've implemented it correctly. Discuss! On the final exam, you won't have an interpreter.

Your Answer

Solution

    ; Return whether there are n perfect squares with no repeats that sum to total

    (define (fit total n)
        (define (f total n k)
            (if (and (= n 0) (= total 0))
                #t
            (if (< total (* k k))
                #f
            (or (f total n (+ k 1)) (f (- total (* k k)) (- n 1) (+ k 1)))
            )))
        (f total n 1))

    (expect (fit 10 2) #t)  ; 1*1 + 3*3
    (expect (fit 9 1)  #t)  ; 3*3
    (expect (fit 9 2)  #f)  ;
    (expect (fit 9 3)  #f)  ; 1*1 + 2*2 + 2*2 doesn't count because of repeated 2*2
    (expect (fit 25 1)  #t) ; 5*5
    (expect (fit 25 2)  #t) ; 3*3 + 4*4

Use the (or _ _) special form to combine two recursive calls: one that uses k*k in the sum and one that does not. The first should subtract k*k from total and subtract 1 from n; the other should leaves total and n unchanged.

Discussion Time: As a group, come up with one sentence describing how your implementation makes sure that all n positive perfect squares are different (no repeats). If you're not sure why, talk to a TA!

Scheme Lists & Quotation

Scheme lists are linked lists. Lightning review:

nil and () are the same thing: the empty list.
(cons first rest) constructs a linked list with first as its first element. and rest as the rest of the list, which should always be a list.
(car s) returns the first element of the list s.
(cdr s) returns the rest of the list s.
(list ...) takes n arguments and returns a list of length n with those arguments as elements.
(append ...) takes n lists as arguments and returns a list of all of the elements of those lists.
(draw s) draws the linked list structure of a list s. It only works on code.cs61a.org/scheme. Try it now with something like (draw (cons 1 nil)).

Quoting an expression leaves it unevaluated. Examples:

'four and (quote four) both evaluate to the symbol four.
'(2 3 4) and (quote (2 3 4)) both evaluate to a list containing three elements: 2, 3, and 4.
'(2 3 four) and (quote (2 3 four)) evaluate to a list containing 2, 3, and the symbol four.

Here's an important difference between list and quotation:

scm> (list 2 (+ 3 4))
(2 7)
scm> '(2 (+ 3 4))
(2 (+ 3 4))

Q2: Nested Lists

Create the nested list depicted below three different ways: using list, quote, and cons.

First, describe the list together: "It looks like there are four elements, and the first element is ..." If you get stuck, look at the hint below. (But try to describe it yourself first!)

A four-element list in which the first element is a list containing both a and b, the second element is c, the third element is d, and the fourth element is a list containing just e.

Next, use calls to list to construct this list. If you run this code and then (draw with-list) in code.cs61a.org, the draw procedure will draw what you've built.

Your Answer

Solution

    (define with-list
        (list
            (list 'a 'b) 'c 'd (list 'e)
        )
    )
    ; (draw with-list)  ; Uncomment this line to draw with-list

Every call to list creates a list, and there are three different lists in this diagram: a list containing a and b: (list 'a 'b), a list containing e: (list 'e), and the whole list of four elements: (list _ 'c 'd _). Try to put these expressions together.

Now, use quote to construct this list.

Your Answer

Solution

    (define with-quote
        '(
            (a b) c d (e)
        )

    )
    ; (draw with-quote)  ; Uncomment this line to draw with-quote

One quoted expression is enough, but it needs to match the structure of the linked list using Scheme notation. So, your task is to figure out how this list would be displayed in Scheme.

The nested list drawn above is a four-element list with lists as its first and last elements: ((a b) c d (e)). Quoting that expression will create the list.

Now, use cons to construct this list. Don't use list. You can use first in your answer.

(define first
    (cons 'a (cons 'b nil)))

Your Answer

Solution

    (define with-cons
        (cons
            first (cons 'c (cons 'd (cons (cons 'e nil) nil)))
        )
    )
    ; (draw with-cons)  ; Uncomment this line to draw with-cons

The provided first is the first element of the result, so the answer takes the form:

first ____

You can either fill in the blank with a quoted three-element list:

'(___ ___ ___)
   c   d  (e)

or with nested calls to cons:

(cons ___ (cons ___ (cons ___ nil)))
       c         d        (e)

Q3: Pair Up

Implement pair-up, which takes a list s. It returns a list of lists that together contain all of the elements of s in order. Each list in the result should have 2 elements. The last one can have up to 3.

Look at the examples together to make sure everyone understands what this procedure does.

Your Answer

Solution

    ;;; Return a list of pairs containing the elements of s.
    ;;;
    ;;; scm> (pair-up '(3 4 5 6 7 8))
    ;;; ((3 4) (5 6) (7 8))
    ;;; scm> (pair-up '(3 4 5 6 7 8 9))
    ;;; ((3 4) (5 6) (7 8 9))
    (define (pair-up s)
        (if (<= (length s) 3)
            (list s)
            (cons (list (car s) (car (cdr s))) (pair-up (cdr (cdr s))))
        ))

    (expect (pair-up '(3 4 5 6 7 8)) ((3 4) (5 6) (7 8)) )
    (expect (pair-up '(3 4 5 6 7 8 9)) ((3 4) (5 6) (7 8 9)) )

pair-up takes a list (of numbers) and returns a list of lists, so when (length s) is less than or equal to 3, return a list containing the list s. For example, (pair-up (list 2 3 4)) should return ((2 3 4)).

Use (cons _ (pair-up _)) to create the result, where the first argument to cons is a list with two elements: the (car s) and the (car (cdr s)). The argument to pair-up is everything after the first two elements.

Discussion: What's the longest list s for which (pair-up (pair-up s)) will return a list with only one element? (Don't just guess and check; discuss!)

Document the Occasion

Let your TA know you're done so that you can each get a departure number, and fill out the attendance form again (this time selecting departure instead of arrival). If your TA isn't in the room, go find them next door.