6.090 IAP '05 - Homework 4 Solutions

Problem 1:

1.
(define foo
  (lambda (x)
    (+ x 5)))

((lambda (x) x) 1)

((lambda (foo bar) (if foo bar #f))
 (= x 1) 7)

(define weird
  (lambda (x y z)
    (lambda (foo)
      (+ x y z foo))))

2.
(define x 5)
unspec

(define (y) (+ 7 7))
unspec

(let ((x 3))
  (+ x x))
6

(let ((x (y))
      (y 7))
  (if (> x 3)
      7
      y)
7

(let ((mit 12))
  (let ((is (+ mit 1)))
    (let ((hard (- is 7)))
      (+ mit is hard))))
31

Problem 2:

1.
The (fact 10) evaluation becomes more indented as the computation
proceeds.  At the end, it stops indenting more and sharply unindents
as is computes a bunch of *n 's.

The (remainder 30 3) evaluation stays at the same indentation the
whole time.  The end doesn't look any different.

(fact 10) is a recursive process, so it keeps deferring the *n
operation.  Each time it has to remember to come back and do a *n it
indents another 2 spaces.  At the end, it reaches the bottom of the
recursion and is able to do all the *n's.  (remainder 30 3) is an
iterative process.  It doesn't have any deferred operations and thus
stays at the same indentation level the whole time.

2.
slow-add1 is a recursive process.  Deferred op (inc ...)
slow-add2 is an iterative process.  No deferred ops

3.

(define (quotient x y)
  (quotient-helper x y 0))

(define (quotient-helper x y answer)
  (if (< x y)
      answer
      (quotient-helper (- x y) y (+ 1 answer))))

Problem 3:

1.
+-+-+   +-+-+
|.|.--->|.|\|
+|+-+   +|+-+
 |       |
 V       V
+-+-+    2
|.|\|
+|+-+
 |   
 V   
 1   

+-+-+         +-+-+
|.|.--------->|.|\|
+|+-+         +|+-+
 |             |
 V             V
+-+-+   +-+-+  3
|.|.--->|.|\|
+|+-+   +|+-+
 |       |
 V       V
+-+-+    2
|.|\|
+|+-+
 |   
 V   
 1   


+-+-+
|\|\|
+-+-+


+-+-+   +-+-+   +-+-+
|.|.--->|.|.--->|.|\|
+|+-+   +|+-+   +|+-+
 |       |       |
 V       V       V
 3       2       1

2.
(list 7)

(list "this" "is" "yummy")

(list (list nil))

(list (list "apples" 3) (list "oranges 2))

3.

(define (sum-list lst)
  (if (null? lst)
      0
      (+ (car lst) (sum-list (cdr lst)))))

4.

(define (seven-on-the-end lst)
  (if (null? lst)
      (list 7)
      (cons (car lst) (seven-on-the-end (cdr lst)))))