Programming – Page 2 – The Means to Correct

Accidentally skipped over this project earlier and it was mentioned in the poker hands project so I am doubling back to do it. Full description at link. For this post I’m just quoting the descriptions of the subprocedures and tests.

`Card-val`

Write a procedure card-val that takes a single card as its argument and returns the value of that card.

> (card-val 'cq)
2

> (card-val 's7)
0

> (card-val 'ha)
4

> (card-val 'cq)

> (card-val 's7)

> (card-val 'ha)

My answer:

(define (card-val card)
  (cond ((equal? (bf card) 'a) 4)
        ((equal? (bf card) 'k) 3)
        ((equal? (bf card) 'q) 2)
        ((equal? (bf card) 'j) 1)
        (else 0)))

(define (card-val card)

(cond ((equal? (bf card) 'a) 4)

((equal? (bf card) 'k) 3)

((equal? (bf card) 'q) 2)

((equal? (bf card) 'j) 1)

(else 0)))

`High-card-points`

Write a procedure high-card-points that takes a hand as its argument and returns the total number of points from high cards in the hand. (This procedure does not count distribution points.)

> (high-card-points '(sa s10 hq ck c4))
9

> (high-card-points '(sa s10 s7 s6 s2 hq hj h9 ck c4 dk d9 d3))
13

> (high-card-points '(sa s10 hq ck c4))

> (high-card-points '(sa s10 s7 s6 s2 hq hj h9 ck c4 dk d9 d3))

My answer:

(define (high-card-points sent)
  (accumulate + (every card-val sent)))

(define (high-card-points sent)

(accumulate + (every card-val sent)))

`Count-suit`

Write a procedure count-suit that takes a suit and a hand as arguments and returns the number of cards in the hand with the given suit.

> (count-suit 's '(sa s10 hq ck c4))
2

> (count-suit 'c '(sa s10 s7 s6 s2 hq hj h9 ck c4 dk d9 d3))
2

> (count-suit 'd '(h3 d7 sk s3 c10 dq d8 s9 s4 d10 c7 d4 s2))
5

> (count-suit 's '(sa s10 hq ck c4))

> (count-suit 'c '(sa s10 s7 s6 s2 hq hj h9 ck c4 dk d9 d3))

> (count-suit 'd '(h3 d7 sk s3 c10 dq d8 s9 s4 d10 c7 d4 s2))

My answer:

(define (count-suit suit sent)
(appearances suit (accumulate word sent)))

(define (count-suit suit sent)

(appearances suit (accumulate word sent)))

`Suit-counts`

Write a procedure suit-counts that takes a hand as its argument and returns a sentence containing the number of spades, the number of hearts, the number of clubs, and the number of diamonds in the hand.

> (suit-counts '(sa s10 hq ck c4))
(2 1 2 0)

> (suit-counts '(sa s10 s7 s6 s2 hq hj h9 ck c4 dk d9 d3))
(5 3 2 3)

> (suit-counts '(h3 d7 sk s3 c10 dq d8 s9 s4 d10 c7 d4 s2))
(5 1 2 5)

> (suit-counts '(sa s10 hq ck c4))

(2 1 2 0)

> (suit-counts '(sa s10 s7 s6 s2 hq hj h9 ck c4 dk d9 d3))

(5 3 2 3)

> (suit-counts '(h3 d7 sk s3 c10 dq d8 s9 s4 d10 c7 d4 s2))

(5 1 2 5)

My answer:

(define (suit-counts sent)
(se (count-suit 's sent)
    (count-suit 'h sent)
    (count-suit 'c sent)
    (count-suit 'd sent)))

(define (suit-counts sent)

(se (count-suit 's sent)

(count-suit 'h sent)

(count-suit 'c sent)

(count-suit 'd sent)))

`Suit-dist-points`

Write suit-dist-points that takes a number as its argument, interpreting it as the number of cards in a suit. The procedure should return the number of distribution points your hand gets for having that number of cards in a particular suit.

> (suit-dist-points 2)
1

> (suit-dist-points 7)
0

> (suit-dist-points 0)
3

> (suit-dist-points 2)

> (suit-dist-points 7)

> (suit-dist-points 0)

My answer:

(define (suit-dist-points cards_in_suit)
(cond ((equal? cards_in_suit 2) 1)
      ((equal? cards_in_suit 1) 2)
      ((equal? cards_in_suit 0) 3)
      (else 0)))

(define (suit-dist-points cards_in_suit)

(cond ((equal? cards_in_suit 2) 1)

((equal? cards_in_suit 1) 2)

((equal? cards_in_suit 0) 3)

(else 0)))

`Hand-dist-points`

Write hand-dist-points, which takes a hand as its argument and returns the number of distribution points the hand is worth.

> (hand-dist-points '(sa s10 s7 s6 s2 hq hj h9 ck c4 dk d9 d3))
1

> (hand-dist-points '(h3 d7 sk s3 c10 dq d8 s9 s4 d10 c7 d4 s2))
3

> (hand-dist-points '(sa s10 s7 s6 s2 hq hj h9 ck c4 dk d9 d3))

> (hand-dist-points '(h3 d7 sk s3 c10 dq d8 s9 s4 d10 c7 d4 s2))

My answer:

(define (hand-dist-points hand)
  (accumulate + (every suit-dist-points (suit-counts hand))))

(define (hand-dist-points hand)

(accumulate + (every suit-dist-points (suit-counts hand))))

`Bridge-val`

Write a procedure bridge-val that takes a hand as its argument and returns the total number of points that the hand is worth.

> (bridge-val '(sa s10 s7 s6 s2 hq hj h9 ck c4 dk d9 d3))
14

> (bridge-val '(h3 d7 sk s3 c10 dq d8 s9 s4 d10 c7 d4 s2))
8

> (bridge-val '(sa s10 s7 s6 s2 hq hj h9 ck c4 dk d9 d3))

> (bridge-val '(h3 d7 sk s3 c10 dq d8 s9 s4 d10 c7 d4 s2))

My answer:

(define (bridge-val hand)
  (+ (hand-dist-points hand)
     (high-card-points hand)))

(define (bridge-val hand)

(+ (hand-dist-points hand)

(high-card-points hand)))

Full Program

(define (card-val card)
  (cond ((equal? (bf card) 'a) 4)
        ((equal? (bf card) 'k) 3)
        ((equal? (bf card) 'q) 2)
        ((equal? (bf card) 'j) 1)
        (else 0)))

(define (high-card-points hand)
  (accumulate + (every card-val hand)))


(define (count-suit suit sent)
(appearances suit (accumulate word sent)))

(define (suit-counts sent)
(se (count-suit 's sent)
    (count-suit 'h sent)
    (count-suit 'c sent)
    (count-suit 'd sent)))

(define (suit-dist-points cards_in_suit)
(cond ((equal? cards_in_suit 2) 1)
      ((equal? cards_in_suit 1) 2)
      ((equal? cards_in_suit 0) 3)
      (else 0)))

(define (hand-dist-points hand)
  (accumulate + (every suit-dist-points (suit-counts hand))))

(define (bridge-val hand)
  (+ (hand-dist-points hand)
     (high-card-points hand)))

(define (card-val card)

(cond ((equal? (bf card) 'a) 4)

((equal? (bf card) 'k) 3)

((equal? (bf card) 'q) 2)

((equal? (bf card) 'j) 1)

(else 0)))

(define (high-card-points hand)

(accumulate + (every card-val hand)))

(define (count-suit suit sent)

(appearances suit (accumulate word sent)))

(define (suit-counts sent)

(se (count-suit 's sent)

(count-suit 'h sent)

(count-suit 'c sent)

(count-suit 'd sent)))

(define (suit-dist-points cards_in_suit)

(cond ((equal? cards_in_suit 2) 1)

((equal? cards_in_suit 1) 2)

((equal? cards_in_suit 0) 3)

(else 0)))

(define (hand-dist-points hand)

(accumulate + (every suit-dist-points (suit-counts hand))))

(define (bridge-val hand)

(+ (hand-dist-points hand)

(high-card-points hand)))

Other People’s Solutions

I liked AnneB’s suit-counts:

(define (suit-counts hand)
      (every (lambda (suit) (count-suit suit hand)) ‘(s h c d)) )

(define (suit-counts hand)

(every (lambda (suit) (count-suit suit hand)) ‘(s h c d)) )

Category: Programming

Simply Scheme Project – Scoring Bridge Hands

`Card-val`

`High-card-points`

`Count-suit`

`Suit-counts`

`Suit-dist-points`

`Hand-dist-points`

`Bridge-val`

Full Program

Other People’s Solutions