First I’m going to implement the filter and accumulate functions in CL (although filter is just another name for the CL library function remove-if-not and accumulate can be trivially modeled with reduce):

(defun filter (predicate sequence)
  (cond ((null sequence) nil)
        ((funcall predicate (car sequence))
          (cons (car sequence)
                (filter predicate (cdr sequence))))
        (t (filter predicate (cdr sequence)))))

(defun accumulate (op initial sequence)
  (if (null sequence)
    (funcall op 
      (car sequence)
      (accumulate op initial (cdr sequence)))))

Exercise 2.33

(defun my-map (p sequence)
    (lambda (x y) (cons (funcall p x) y)) 
    nil sequence))

(defun my-append (seq1 seq2)
  (accumulate #'cons seq2 seq1))

(defun my-length (sequence)
  (accumulate (lambda (x y) (1+ y)) 0 sequence))

Exercise 2.34

We can think of this computation as the accumulation of the current coefficient and x multiplied by next, where next is the same operation for a higher coefficient, until the end is reached (read the Horner formula shown in the book from right to left). This easily translates into the code:

(defun horner-eval (x coeffs)
    (lambda (this-coeff higher-terms)
      (+ (* x higher-terms) this-coeff))

Exercise 2.35

This exercise is an interesting brain-stretcher. The authors show in their template that the last argument to accumulate is a call to map. My solution is different. First of all, it is important to understand that we’ll need a recursive call to count-leaves. This is always the case for tree recursion – accumulate itself only goes over a linear sequence, and we need to delve recursively into the tree. This is a good tip that should get us started.

Now, consider how accumulate works. The lambda it accepts as op has the result of sub-sequence accumulation as its second argument. The first argument is the interesting one, because this is where the real work is done. If x is a leaf (consp returns false), we add 1. If it’s a node with children, we delve recursively (think of the original implementation of count-leaves).

(defun count-leaves (tree)
    (lambda (x y)
      (+  y
          (if (consp x)
            (count-leaves x)

Exercise 2.36

(defun accumulate-n (op init seqs)
  (if (null (car seqs))
    (cons (accumulate op init (mapcar #'car seqs))
          (accumulate-n op init (mapcar #'cdr seqs)))))

Exercise 2.37

(defun dot-product (v w)
  (accumulate #'+ 0 (mapcar #'* v w)))

(defun matrix-*-vector (m v)
    (lambda (row)
      (dot-product row v))

(defun transpose (m)
  (accumulate-n #'cons nil m))

(defun matrix-*-matrix (m n)
  (let ((n-t (transpose n)))
    (mapcar (lambda (row) (matrix-*-vector n-t row)) m)))

Exercise 2.38

First let’s rewrite accumulate as fold-right and add the definition of fold-left:

(defun fold-right (op init seq)
  (if (null seq)
    (funcall op 
      (car seq)
      (fold-right op init (cdr seq)))))

(defun fold-left (op init seq)
  (labels (
    (iter (result rest)
      (if (null rest)
        (iter (funcall op result (car rest))
              (cdr rest)))))
    (iter init seq)))

Note that for simple operations like addition, the two produce equivalent results:

(fold-right #'* 1 '(1 2 3 4 5))
=> 120
(fold-left #'* 1 '(1 2 3 4 5))
=> 120

But for the examples asked about in the exercise:

(fold-right #'/ 1 '(1 2 3))
=> 3/2
(fold-left #'/ 1 '(1 2 3))
=> 1/6

(fold-right #'list nil '(1 2 3))
=> (1 (2 (3 NIL))) 
(fold-left #'list nil '(1 2 3))
=> (((NIL 1) 2) 3)

fold-right and fold-left will produce the same results if op is an associative operation.

Exercise 2.39

(defun reverse-r (seq)
  (fold-right (lambda (x y) (append y (list x))) nil seq))

(defun reverse-l (seq)
  (fold-left (lambda (x y) (cons y x)) nil seq))

Exercise 2.40

Here is the scaffolding for this and the following exercises:

(defun enumerate-interval (low high)
  (if (> low high)
    (cons low (enumerate-interval (1+ low) high))))

(defun flatmap (proc seq)
  (accumulate #'append nil (mapcar proc seq)))

(defun sum (lst)
  (accumulate #'+ 0 lst))

(defun prime-sum? (pair)
  (prime? (sum pair)))

(defun make-pair-sum (pair)
  (list (car pair) (cadr pair) (sum pair)))

The solution for the exercise is:

(defun unique-pairs (n)
    (lambda (i)
      (mapcar (lambda (j) (list i j))
              (enumerate-interval 1 (1- i))))
    (enumerate-interval 1 n)))

(defun prime-sum-pairs (n)
    (filter #'prime-sum? (unique-pairs n))))

Exercise 2.41

(defun unique-triples (n)
  "Unique triples of numbers <= n" 
    (lambda (i)
        (lambda (j)
            (lambda (k) (list i j k))
            (enumerate-interval 1 (1- j))))
        (enumerate-interval 1 (1- i))))
    (enumerate-interval 1 n)))

(defun triples-sum-s (s n)
  "Triples of numbers <= n summing to s" 
    (lambda (triple) 
      (= (sum triple) s))
    (unique-triples n)))

Exercise 2.42

A single position is defined by a row and a column:

(defun make-position (row col)
  (cons row col))

(defun position-row (pos)
  (car pos))

(defun position-col (pos)
  (cdr pos))

(defun positions-equal (a b)
  (equal a b))

A set of positions is just a list of position objects:

(defvar empty-board '())

(defun adjoin-position (row col positions)
  (append positions (list (make-position row col))))

Note how adjoin-position is implemented: in order to append the new position to the list’s end, I use append. But append expects lists as its argument, so the new position is wrapped in a call to list. There must be a more elegant way to do this :-)

And this is the implementation of safe? and its helper function attacks?:

(defun attacks? (a b)
  "Both a and b are positions. This function
  checks if a queen in position a attacks the
  queen in position b." 
  (let ((a-row (position-row a))
        (a-col (position-col a))
        (b-row (position-row b))
        (b-col (position-col b)))
      ((= a-row b-row) t) ; row attack
      ((= a-col b-col) t) ; column attack
      ((= (abs (- a-col b-col)) ; diagonal attack
          (abs (- a-row b-row))) t)
      (t nil))))

(defun safe? (k positions)
  "Is the queen in the kth column safe with
  respect to the queens in columns 1..k-1?" 
  (let ((kth-pos (nth (1- k) positions)))
    (if (null (find-if 
                (lambda (pos)
                  (and  (not (positions-equal kth-pos pos))
                        (attacks? kth-pos pos)))

To complete the picture, this is queens translated to CL:

(defun queens (board-size)  
    (queen-cols board-size board-size))

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

I tested this implementation and it works correctly for boards with size < 8. However, for board size 8 and higher, CLISP reports a stack overflow. When I traced queen-cols to see what the problem is, it appeared that the lists returned by queen-cols are very large. Since they are probably returned by value on the stack, this is what causes the overflow. I tried to increase CLISP’s memory consumption parameter, but it doesn’t help. It is apparently possible to do some more serious tweaking with CLISP’s stack size allocation, but it’s far from trivial. Seems like an annoying limitation in CLISP (imagine that this code ran without problems in the environment the authors used 25 years ago!).

When I tried it on SBCL running on Linux (Ubuntu 7), it worked without problems and generated correct solutions for board size 8.

Exercise 2.43

Here is Louis Reasoner’s version of queens:

(defun louis-queens (board-size)
  (louis-queen-cols board-size board-size))

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

Since the recursive call to queen-cols is a costly operation in terms of time, Louis’s implementation looks suspicious just for the fact that it places it inside another loop1. Let’s analyze it a little further:

The “normal” implementation takes the result of queen-cols for one column less, and for each position in the list attaches all the possible placements of the new queen. These new sets of positions are then filtered by safe?.

Louis’s implementation, on the other hand, does it in a different order, which is crucial. For each new possible position, it re-generates the result of queen-cols for one column less, and attaches the new position to it. Note that the call to louis-queen-cols doesn’t need the “loop variable” new-row, and hence it’s quite clear that placing it in the inner loop is needless2.

Runtime analysis: queen-cols calls itself only once per execution. Therefore, for some board size N, queen-cols is called N+1 times (the call with N=0 returns immediately, though).

louis-queen-cols, however, calls itself board-size times per execution. But this continues recursively, because each louis-queen-cols called calls itself board-size times too. So the call tree is N-nary, each node has N children. The tree is N levels deep, so the total amount of nodes (calls) in it is O(N^N). To find the ratio between it and the original queen-cols, we divide: O(N^N) / N+1 = O(N^N). Therefore, asymptotically louis-queen-cols is O(N^N) times slower.

1 Think of filter, flatmap and mapcar as loops, since they are sequence functions.

2 To be completely fair, neither does the call to enumerate-interval in the inner loop of queen-cols use the loop variable. We can actually precompute the list returned by (enumerate-interval 1 board-size) and use it in each iteration. However, since enumerate-interval is very fast, the speed gain won’t be too significant.


comments powered by Disqus