Revision of 2.2.3 Sequences as Conventional Interfaces from 30 June 2009 - 8:33pm

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Exercise 2.33


Fill in the missing expressions to complete the following definitions of some basic list-manipulation operations as accumulations:

(define (map p sequence)
  (accumulate (lambda (x y) <??>) nil sequence))
(define (append seq1 seq2)
  (accumulate cons <??> <??>))
(define (length sequence)
  (accumulate <??> 0 sequence))

Exercise 2.34


Evaluating a polynomial in x at a given value of x can be formulated as an accumulation. We evaluate the polynomial

using a well-known algorithm called Horner’s rule, which structures the computation as

In other words, we start with an, multiply by x, add an-1, multiply by x, and so on, until we reach a0.[16] Fill in the following template to produce a procedure that evaluates a polynomial using Horner’s rule. Assume that the coefficients of the polynomial are arranged in a sequence, from a0 through an.

(define (horner-eval x coefficient-sequence)
  (accumulate (lambda (this-coeff higher-terms) <??>)

For example, to compute 1 + 3x + 5x3 + x5 at x = 2 you would evaluate

(horner-eval 2 (list 1 3 0 5 0 1))
[16] According to Knuth (1981), this rule was formulated by W. G. Horner early in the nineteenth century, but the method was actually used by Newton over a hundred years earlier. Horner’s rule evaluates the polynomial using fewer additions and multiplications than does the straightforward method of first computing an xn, then adding an-1xn-1, and so on. In fact, it is possible to prove that any algorithm for evaluating arbitrary polynomials must use at least as many additions and multiplications as does Horner’s rule, and thus Horner’s rule is an optimal algorithm for polynomial evaluation. This was proved (for the number of additions) by A. M. Ostrowski in a 1954 paper that essentially founded the modern study of optimal algorithms. The analogous statement for multiplications was proved by V. Y. Pan in 1966. The book by Borodin and Munro (1975) provides an overview of these and other results about optimal algorithms. [back]

Exercise 2.35


Redefine count-leaves from section 2.2.2 as an accumulation:

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

Exercise 2.36


The procedure accumulate-n is similar to accumulate except that it takes as its third argument a sequence of sequences, which are all assumed to have the same number of elements. It applies the designated accumulation procedure to combine all the first elements of the sequences, all the second elements of the sequences, and so on, and returns a sequence of the results. For instance, if s is a sequence containing four sequences, ((1 2 3) (4 5 6) (7 8 9) (10 11 12)), then the value of (accumulate-n + 0 s) should be the sequence (22 26 30). Fill in the missing expressions in the following definition of accumulate-n:

(define (accumulate-n op init seqs)
  (if (null? (car seqs))
      (cons (accumulate op init <??>)
            (accumulate-n op init <??>))))

Exercise 2.37


Suppose we represent vectors v = (vi) as sequences of numbers, and matrices m = (mij) as sequences of vectors (the rows of the matrix). For example, the matrix

is represented as the sequence ((1 2 3 4) (4 5 6 6) (6 7 8 9)). With this representation, we can use sequence operations to concisely express the basic matrix and vector operations. These operations (which are described in any book on matrix algebra) are the following:

We can define the dot product as[17]

(define (dot-product v w)
  (accumulate + 0 (map * v w)))

Fill in the missing expressions in the following procedures for computing the other matrix operations. (The procedure accumulate-n is defined in exercise 2.36.)

(define (matrix-*-vector m v)
  (map <??> m))
(define (transpose mat)
  (accumulate-n <??> <??> mat))
(define (matrix-*-matrix m n)
  (let ((cols (transpose n)))
    (map <??> m)))
[17] This definition uses the extended version of map described in footnote 12. [back]

Exercise 2.38


The accumulate procedure is also known as fold-right, because it combines the first element of the sequence with the result of combining all the elements to the right. There is also a fold-left, which is similar to fold-right, except that it combines elements working in the opposite direction:

(define (fold-left op initial sequence)
  (define (iter result rest)
    (if (null? rest)
        (iter (op result (car rest))
              (cdr rest))))
  (iter initial sequence))

What are the values of

(fold-right / 1 (list 1 2 3))
(fold-left / 1 (list 1 2 3))
(fold-right list nil (list 1 2 3))
(fold-left list nil (list 1 2 3))

Give a property that op should satisfy to guarantee that fold-right and fold-left will produce the same values for any sequence.

Exercise 2.39


Complete the following definitions of reverse (exercise 2.18) in terms of fold-right and fold-left from exercise 2.38:

(define (reverse sequence)
  (fold-right (lambda (x y) <??>) nil sequence))
(define (reverse sequence)
  (fold-left (lambda (x y) <??>) nil sequence))

Exercise 2.40


Define a procedure unique-pairs that, given an integer n, generates the sequence of pairs (i,j) with 1 ≤ j < in. Use unique-pairs to simplify the definition of prime-sum-pairs given above.

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.

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 possible solution is shown in figure 2.8. 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 kth queen in a position where it does not check any of the queens already on the board. We can formulate this approach recursively: Assume that we have already generated the sequence of all possible ways to place k - 1 queens in the first k - 1 columns of the board. For each of these ways, generate an extended set of positions by placing a queen in each row of the kth column. Now filter these, keeping only the positions for which the queen in the kth column is safe with respect to the other queens. This produces the sequence of all ways to place k queens in the first k columns. By continuing this process, we will produce not only one solution, but all solutions to the puzzle.

We implement this solution as a procedure queens, which returns a sequence of all solutions to the problem of placing n queens on an n× n chessboard. Queens has an internal procedure queen-cols that returns the sequence of all ways to place queens in the first k columns of the board.

(define (queens board-size)
  (define (queen-cols k)  
    (if (= k 0)
        (list empty-board)
         (lambda (positions) (safe? k positions))
          (lambda (rest-of-queens)
            (map (lambda (new-row)
                   (adjoin-position new-row k rest-of-queens))
                 (enumerate-interval 1 board-size)))
          (queen-cols (- k 1))))))
  (queen-cols board-size))

In this procedure rest-of-queens is a way to place k - 1 queens in the first k - 1 columns, and new-row is a proposed row in which to place the queen for the kth column. Complete the program by implementing the representation for sets of board positions, including the procedure adjoin-position, which adjoins a new row-column position to a set of positions, and empty-board, which represents an empty set of positions. You must also write the procedure safe?, which determines for a set of positions, whether the queen in the kth column is safe with respect to the others. (Note that we need only check whether the new queen is safe — the other queens are already guaranteed safe with respect to each other.)

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

 (lambda (new-row)
   (map (lambda (rest-of-queens)
          (adjoin-position new-row k rest-of-queens))
        (queen-cols (- k 1))))
 (enumerate-interval 1 board-size))

Explain why this interchange makes the program run slowly. Estimate how long it will take Louis’s program to solve the eight-queens puzzle, assuming that the program in exercise 2.42 solves the puzzle in time T.


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