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### SICP Exercise 2.43: Eight queens: interchange the order of the nested mappings

2021-11-02

Exercise 2.43: Louis Reasoner is having a terrible time doing exercise 2.42. His queens procedure seems to work, but it runs extremely slowly. (Louis never does manage to wait long enough for it to solve even the 6× 6 case.) When Louis asks Eva Lu Ator for help, she points out that he has interchanged the order of the nested mappings in the flatmap, writing it as

### SICP Exercise 2.42: Eight queens puzzle

2021-10-29

Exercise 2.42: The “eight-queens puzzle” asks how to place eight queens on a chessboard so that no queen is in check from any other (i.e., no two queens are in the same row, column, or diagonal).

One way to solve the puzzle is to work across the board, placing a queen in each column. Once we have placed

`k - 1`

queens, we must place the`k`

^{th}queen in a position where it does not check any of the queens already on the board.

### SICP Exercise 2.41: Triple sum

2021-10-18

Exercise 2.41: Write a procedure to find all ordered triples of distinct positive integers

`i`

,`j`

, and`k`

less than or equal to a given integer`n`

that sum to a given integer`s`

.

`unique-triples`

can be written easily base on `unique-pairs`

in 2.40:

1(define (unique-triples n) 2 (flatmap 3 (lambda (i) 4 (flatmap 5 (lambda (j) 6 (map (lambda (k) (list i j k)) 7 (enumerate-interval 1 (- j 1)))) 8 (enumerate-interval 1 (- i 1)))) 9 (enumerate-interval 1 n)))

### SICP Exercise 2.35: Counting leaves of a tree

2021-10-14

Exercise 2.35: Redefine count-leaves from section 2.2.2 as an accumulation:

1(define (count-leaves t) 2 (accumulate <??> <??> (map <??> <??>)))

The `count-leaves`

procedure from section 2.2.2:

1(define (count-leaves x) 2 (cond ((null? x) 0) 3 ((not (pair? x)) 1) 4 (else (+ (count-leaves (car x)) 5 (count-leaves (cdr x))))))

### SICP Exercise 2.27: Reversing nested lists

2021-10-12

Exercise 2.27: Modify your reverse procedure of exercise 2.18 to produce a deep-reverse procedure that takes a list as argument and returns as its value the list with its elements reversed and with all sublists deep-reversed as well. For example,

1(define x (list (list 1 2) (list 3 4))) 2 3x 4((1 2) (3 4)) 5 6(reverse x) 7((3 4) (1 2)) 8 9(deep-reverse x) 10((4 3) (2 1))

First, look at my `reverse`

procedure:

1#lang racket/base 2(require racket/trace) 3 4(define (reverse items) 5 (iter items null)) 6 7(define (iter remaining result) 8 (trace iter) 9 (if (null? remaining) 10 result 11 (iter (cdr remaining) (cons (car remaining) result)))) 12 13(trace reverse) 14(reverse (list (list 1 2) (list 3 4)))

### SICP Exercise 1.25: A simpler expmod?

2021-09-30

Exercise 1.25: Alyssa P. Hacker complains that we went to a lot of extra work in writing

`expmod`

. After all, she says, since we already know how to compute exponentials, we could have simply written:1(define (expmod base exp m) 2 (remainder (fast-expt base exp) m))Is she correct? Would this procedure serve as well for our fast prime tester? Explain.

First, look at the original algorithm:

1 (define (expmod base exp m) 2 (cond ((= exp 0) 1) 3 ((even? exp) 4 (remainder 5 (square (expmod base (/ exp 2) m)) 6 m)) 7 (else 8 (remainder 9 (* base (expmod base (- exp 1) m)) 10 m))))

### SICP Exercise 1.16: Iterative Exponentiation

2021-09-29

I am reading SICP.

Section 1.2.4 talks about the problem of computing the exponential of a given number.

The authors start with a recursive procedure:

1#lang sicp 2 3(define (expt b n) 4 (if (= n 0) 5 1 6 (* b (expt b (- n 1)))))

This requires O(n) steps and O(n) space.

then an iterative procedure:

1#lang sicp 2 3(define (expt b n) 4 (expt-iter b n 1)) 5 6(define (expt-iter b counter product) 7 (if (= counter 0) 8 product 9 (expt-iter b 10 (- counter 1) 11 (* b product))))

### SICP Exercise 1.14: orders of growth of count-change

2021-09-28

Exercise 1.14: Draw the tree illustrating the process generated by the

`count-change`

procedure of section 1.2.2 in making change for 11 cents. What are the orders of growth of the space and number of steps used by this process as the amount to be changed increases?

The `count-change`

procedure:

1(trace-define (count-change amount) 2 (cc amount 5)) 3 4(trace-define (cc amount kinds-of-coins) 5 (cond ((= amount 0) 1) 6 ((or (< amount 0) (= kinds-of-coins 0)) 0) 7 (else (+ (cc amount 8 (- kinds-of-coins 1)) 9 (cc (- amount 10 (first-denomination kinds-of-coins)) 11 kinds-of-coins))))) 12 13(define (first-denomination kinds-of-coins) 14 (cond ((= kinds-of-coins 1) 1) 15 ((= kinds-of-coins 2) 5) 16 ((= kinds-of-coins 3) 10) 17 ((= kinds-of-coins 4) 25) 18 ((= kinds-of-coins 5) 50)))