University of Virginia, Department of Computer Science CS200: Computer Science, Spring 2004

Notes: Wednesday 11 February 2004

• Wednesday, 11 February: Problem Set 3. Please get started on this early (i.e., already!).
• Friday, 13 Feburary: Friday's class will be in the Tuttle Lounge. Please go there at 2:00 instead of the normal classroom.
• Monday, 16 Feburary: Problem Set 4
• Monday, 23 February: Exam 1 (will be handed out on Friday, 20 February)
• Before 10 March: Read rest of GEB part I (Chapters 2-4 and 6-9, in addition to Chapters 1 and 5 you have already read).
Pegboard Puzzle

```;;; A board is a pair of the number of rows and the empty squares
(define (make-board rows holes) (cons rows holes))
(define (board-holes board) (cdr board))
(define (board-rows board) (car board))

;;; make-position creates an row col coordinate that represents a position on the board
;;; e.g.             1,1
;;;               2,1   2,2
;;;            3,1   3,2   3,3
(define (make-position row col) (cons row col))
(define (get-row posn) (car posn))
(define (get-col posn) (cdr posn))

(define (same-position pos1 pos2)
(and (= (get-row pos1) (get-row pos2))
(= (get-col pos1) (get-col pos2))))

;;; on-board? takes a board and a position and returns true if it is
;;; contained in the board.
(define (on-board? board posn)
(and (>= (get-row posn) 1)  (>= (get-col posn) 1)
(<= (get-row posn) (board-rows board))
(<= (get-col posn) (get-row posn))))

;;; There are rows + (rows - 1) + ... + 1 squares (holes or pegs)
(define (board-squares board) (count-squares (board-rows board)))
(define (count-squares nrows)
(if (= nrows 1) 1 (+ nrows (count-squares (- nrows 1)))))

;;; peg? returns true if the position on board has a peg in it, and false if it doesn't
(define (peg? board posn)
(contains (lambda (pos) (same-position posn pos)) (board-holes board)))

;;; remove-peg evaluates to the board you get by removing a peg at posn from
;;; the passed board
(define (remove-peg board posn)
(make-board (board-rows board) (cons posn (board-holes board))))

;;; add-peg evaluates to the board you get by adding a peg at posn to board
(define (add-peg board posn)
(make-board (board-rows board) (remove-hole (board-holes board) posn)))

(define (remove-hole lst posn)
(filter (lambda (pos) (not (same-position pos posn))) lst))

;;; Here's a different way to define filter
(define (filter f lst)
(insertl (lambda (el rest) (if (f el) (cons el rest) rest))
lst
null))

(define (contains f lst) (> (length (filter f lst)) 0))

;;; move creates a list of three posn, a start (the posn that the jumping
;;; peg starts from), a jump (the posn that is being jumped over), and end
;;; (the posn that the peg will end up in)
(define (make-move start jump end) (list start jump end))
(define (get-start move) (first move))
(define (get-jump move) (second move))
(define (get-end move) (third move))

;;; execute-move evaluates to the board after making move move on board.
(define (execute-move board move)
(add-peg (remove-peg (remove-peg board (get-start move))
(get-jump move))
(get-end move)))

;;; generate-moves evaluates to all possible moves that move a peg into
;;; the position empty, even if they are not contained on the board.
(define (generate-moves empty)
(map (lambda (hops)
(let ((hop1 (car hops)) (hop2 (cdr hops)))
(make-move (make-position (+ (get-row empty) (car hop1)) (+ (get-col empty) (cdr hop1)))
(make-position (+ (get-row empty) (car hop2)) (+ (get-col empty) (cdr hop2)))
empty)))
(list
(cons (cons 2 0) (cons 1 0))         ;; right of empty, hopping left
(cons (cons -2 0) (cons -1 0))       ;; left of empty, hopping right
(cons (cons 0 2) (cons 0 1))         ;; below, hopping up
(cons (cons 0 -2) (cons 0 -1))       ;; above, hopping down
(cons (cons 2 2) (cons 1 1))         ;; above right, hopping down-left
(cons (cons -2 2) (cons -1 1))       ;; above left, hopping down-right
(cons (cons 2 -2) (cons 1 -1))       ;; below right, hopping up-left
(cons (cons -2 -2) (cons -1 -1)))))) ;; below left, hopping up-right

(define (all-possible-moves board)
(apply append (map generate-moves (board-holes holes))))

;;; legal-moves filters the moves on a board to produce only those that are valid.

(define (legal-move? move)
;; A move is valid if:
;;    o the start and end positions are on the board
;;    o there is a peg at the start position
;;    o there is a peg at the jump position
;;    o there is not a peg at the end position
(and (on-board? board (get-start move))
(on-board? board (get-end move))
(peg? board (get-start move))
(peg? board (get-jump move))
(not (peg? board (get-end move)))))

(define (legal-moves board)
(filter legal-move? (all-possible-moves board)))

(define (is-winning-position? board)
;; A board is a winning position if only one hole contains a peg
(= (length (board-holes board)) (- (board-squares board) 1)))

(define (find-first-winner board moves)
(if (null? moves)
(if (is-winning-position? board)
null ;; Found a winning game, no moves needed to win (eval to null)
#f)  ;; A losing position, no more moves, but too many pegs.
;;; See if the first move is a winner
(let ((result (solve-pegboard (execute-move board (car moves)))))
(if result ;; anything other than #f is a winner (null is not #f)
(cons (car moves) result) ;; found a winner, this is the first move
(find-first-winner board (cdr moves))))))

;;; solve-pegboard evaluates to:
;;;    #f if the board is a losing position (there is no sequence of moves to win from here)
;;;    or a list of moves to win from this position
;;;
;;; NOTE: null is a winning result!  It means the board has one peg in it right now and
;;;    no moves are required to win.
(define (solve-pegboard board)
(find-first-winner board (legal-moves board)))
```