update

Task/Knapsack-problem-0-1/C/knapsack-problem-0-1.c

@@ -48,8 +48,10 @@ void optimal(int weight, int idx, solution *s)
                 return;
         }
 
-        if (weight < item[idx].weight)
-                return optimal(weight, idx - 1, s);
+        if (weight < item[idx].weight) {
+                optimal(weight, idx - 1, s);
+                return;
+         }
 
         optimal(weight, idx - 1, &v1);
         optimal(weight - item[idx].weight, idx - 1, &v2);
@@ -60,7 +62,7 @@ void optimal(int weight, int idx, solution *s)
         *s = (v1.value >= v2.value) ? v1 : v2;
 }
 
-int main()
+int main(void)
 {
         int i = 0, w = 0;
         solution s = {0, 0};

Task/Knapsack-problem-0-1/Racket/knapsack-problem-0-1-1.rkt

@@ -0,0 +1,56 @@
+#lang racket
+; An ITEM a list of three elements:
+;   a name, a weight, and, a value
+
+; A SOLUTION to a knapsack01 problems is a list of three elements:
+;   the total value, the total weight, and, names of the items to bag
+
+(define (add i s) ; add an item i to the solution s
+  (match-define (list in iw iv) i)
+  (match-define (list v w is) s)
+  (list (+ v iv) (+ w iw) (cons in is)))
+
+(define (knapsack max-weight items)
+  ; return a solution to the knapsack01 problem
+  (define ht (make-hash)) ; (weight number-of-items) -> items
+  (define (get w no-items) (hash-ref ht (list w no-items) #f))
+  (define (update w is x)  (hash-set! ht (list w (length is)) is) x)
+  (define (knapsack1 left items)
+    ; return a solution to the (left, items) problem
+    (cond
+      ; if there are no items, then bag no items:
+      [(empty? items) (list 0 0 '())]
+      ; look up the best solution:
+      [(or (get left (length items))
+           ; the solution haven't been cached, so we
+           ; must compute it and update the cache:
+           (update
+            left items
+            (match items
+              ; let us name the first item
+              [(cons (and (list i w v) x) more)
+               ; if the first item weighs more than the space left,
+               ; we simply find a solution, where it is omitted:
+               (cond [(> w left) (knapsack left more)]
+                     ; there is room for the first item, but
+                     ; we need to choose the best solutions
+                     ; between those with it and that without:
+                     [else
+                      (define without (knapsack left more))
+                      (define value-without (first without))
+                      (define with (knapsack (- left w) more))
+                      (define value-with (+ (first with) v))
+                      ; choose the solutions with greatest value
+                      (if (> value-without value-with)
+                          without
+                          (add x with))])])))]))
+  (knapsack1 max-weight items))
+
+(knapsack 400
+          '((map 9 150) ; 9 is weight, 150 is value
+            (compass 13 35) (water 153 200) (sandwich 50 160)
+            (glucose 15 60) (tin 68 45)(banana 27 60) (apple 39 40)
+            (cheese 23 30) (beer 52 10) (cream 11 70) (camera 32 30)
+            (T-shirt 24 15) (trousers 48 10) (umbrella 73 40)
+            (trousers 42 70) (overclothes 43 75) (notecase 22 80)
+            (glasses 7 20) (towel 18 12) (socks 4 50) (book 30 10)))

Task/Knapsack-problem-0-1/Racket/knapsack-problem-0-1-2.rkt

@@ -0,0 +1 @@
+'(1030 396 (map compass water sandwich glucose banana cream trousers overclothes notecase glasses socks))