Knapsack problem/Unbounded: Difference between revisions

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A traveler gets diverted and has to make an unscheduled stop in what turns out to be Shangri La. Opting to leave, he is allowed to take as much as he likes of the following items, so long as it will fit in his knapsack, and he can carry it.

He knows that he can carry no more than 25 'weights' in total; and that the capacity of his knapsack is 0.25 'cubic lengths'.

Looking just above the bar codes on the items he finds their weights and volumes. He digs out his recent copy of a financial paper and gets the value of each item.

Item	Explanation	Value (each)	weight	Volume (each)
panacea (vials of)	Incredible healing properties	3000	0.3	0.025
ichor (ampules of)	Vampires blood	1800	0.2	0.015
gold (bars)	Shiney shiney	2500	2.0	0.002
Knapsack	For the carrying of	-	<=25	<=0.25

He can only take whole units of any item, but there is much more of any item than he could ever carry

Task

Show how many of each item does he take to maximize the value of items he is carrying away with him.

Note

There are four solutions that maximize the value taken. Only one need be given.

Related tasks

Ada

Translation of: Python

<lang Ada>with Ada.Text_IO;

procedure Knapsack_Unbounded is

  type Bounty is record
     Value  : Natural;
     Weight : Float;
     Volume : Float;
  end record;

  function Min (A, B : Float) return Float is
  begin
     if A < B then
        return A;
     else
        return B;
     end if;
  end Min;

  Panacea : Bounty := (3000,  0.3, 0.025);
  Ichor   : Bounty := (1800,  0.2, 0.015);
  Gold    : Bounty := (2500,  2.0, 0.002);
  Limits  : Bounty := (   0, 25.0, 0.250);
  Best    : Bounty := (   0,  0.0, 0.000);
  Current : Bounty := (   0,  0.0, 0.000);

  Best_Amounts : array (1 .. 3) of Natural := (0, 0, 0);

  Max_Panacea : Natural := Natural (Float'Floor (Min
                             (Limits.Weight / Panacea.Weight,
                              Limits.Volume / Panacea.Volume)));
  Max_Ichor   : Natural := Natural (Float'Floor (Min
                             (Limits.Weight / Ichor.Weight,
                              Limits.Volume / Ichor.Volume)));
  Max_Gold    : Natural := Natural (Float'Floor (Min
                             (Limits.Weight / Gold.Weight,
                              Limits.Volume / Gold.Volume)));

begin

  for Panacea_Count in 0 .. Max_Panacea loop
     for Ichor_Count in 0 .. Max_Ichor loop
        for Gold_Count in 0 .. Max_Gold loop
           Current.Value  := Panacea_Count * Panacea.Value +
                             Ichor_Count * Ichor.Value +
                             Gold_Count * Gold.Value;
           Current.Weight := Float (Panacea_Count) * Panacea.Weight +
                             Float (Ichor_Count) * Ichor.Weight +
                             Float (Gold_Count) * Gold.Weight;
           Current.Volume := Float (Panacea_Count) * Panacea.Volume +
                             Float (Ichor_Count) * Ichor.Volume +
                             Float (Gold_Count) * Gold.Volume;
           if Current.Value  >  Best.Value and
              Current.Weight <= Limits.Weight and
              Current.Volume <= Limits.Volume then
              Best := Current;
              Best_Amounts := (Panacea_Count, Ichor_Count, Gold_Count);
           end if;
        end loop;
     end loop;
  end loop;
  Ada.Text_IO.Put_Line ("Maximum value:" & Natural'Image (Best.Value));
  Ada.Text_IO.Put_Line ("Panacea:" & Natural'Image (Best_Amounts (1)));
  Ada.Text_IO.Put_Line ("Ichor:  " & Natural'Image (Best_Amounts (2)));
  Ada.Text_IO.Put_Line ("Gold:   " & Natural'Image (Best_Amounts (3)));

end Knapsack_Unbounded;</lang>

ALGOL 68

Translation of: Python

<lang algol68>MODE BOUNTY = STRUCT(STRING name, INT value, weight, volume);

[]BOUNTY items = (

              ("panacea", 3000,   3,  25),
              ("ichor",   1800,   2,  15),
              ("gold",    2500,  20,   2)
     );

BOUNTY sack := ("sack", 0, 250, 250);

OP * = ([]INT a,b)INT: ( # dot product operator #

   INT sum := 0;
   FOR i TO UPB a DO sum +:= a[i]*b[i] OD;
   sum

);

OP INIT = (REF[]INT vector)VOID:

   FOR index FROM LWB vector TO UPB vector DO
       vector[index]:=0
   OD;

OP INIT = (REF[,]INT matrix)VOID:

   FOR row index FROM LWB matrix TO UPB matrix DO
       INIT matrix[row index,]
   OD;

PROC total value = ([]INT items count, []BOUNTY items, BOUNTY sack) STRUCT(INT value, weight, volume):(

   ###
   Given the count of each item in the sack return -1 if they can"t be carried or their total value.

   (also return the negative of the weight and the volume so taking the max of a series of return
   values will minimise the weight if values tie, and minimise the volume if values and weights tie).
   ###
   INT weight = items count * weight OF items;
   INT volume = items count * volume OF items;
   IF weight > weight OF sack OR volume > volume OF sack THEN
       (-1, 0, 0)
   ELSE
       ( items count * value OF items, -weight, -volume)
   FI

);

PRIO WRAP = 5; # wrap negative array indices as per python's indexing regime # OP WRAP = (INT index, upb)INT:

 IF index>=0 THEN index ELSE upb + index + 1 FI;

PROC knapsack dp = ([]BOUNTY items, BOUNTY sack)[]INT:(

   ###
   Solves the Knapsack problem, with two sets of weights,
   using a dynamic programming approach
   ###

   # (weight+1) x (volume+1) table #
   # table[w,v] is the maximum value that can be achieved #
   # with a sack of weight w and volume v. #
   # They all start out as 0 (empty sack) #
   [0:weight OF sack, 0:volume OF sack]INT table; INIT table;

   FOR w TO 1 UPB table DO
       FOR v TO 2 UPB table DO
           ### Consider the optimal solution, and consider the "last item" added
           to the sack. Removing this item must produce an optimal solution
           to the subproblem with the sack"s weight and volume reduced by that
           of the item. So we search through all possible "last items": ###
           FOR item index TO UPB items DO
               BOUNTY item := items[item index];
               # Only consider items that would fit: #
               IF w >= weight OF item AND v >= volume OF item THEN
                   # Optimal solution to subproblem + value of item: #
                   INT candidate := table[w-weight OF item,v-volume OF item] + value OF item;
                   IF candidate > table[w,v] THEN
                       table[w,v] := candidate
                   FI
               FI
           OD
       OD
   OD;

   [UPB items]INT result; INIT result;
   INT w := weight OF sack, v := volume OF sack;
   WHILE table[w,v] /= 0 DO
       # Find the last item that was added: #
       INT needle = table[w,v];
       INT item index;
       FOR i TO UPB items WHILE
           item index := i;
           BOUNTY item = items[item index];
           INT candidate = table[w-weight OF item WRAP UPB table, v-volume OF item WRAP 2 UPB table] + value OF item;

WHILE # candidate NE needle DO

         SKIP
       OD;
       # Record it in the result, and remove it: #
       result[item index] +:= 1;
       w -:= weight OF items[item index];
       v -:= volume OF items[item index]
   OD;
   result

);

[]INT max items = knapsack dp(items, sack); STRUCT (INT value, weight, volume) max := total value(max items, items, sack); max := (value OF max, -weight OF max, -volume OF max);

FORMAT d = $zz-d$;

printf(($"The maximum value achievable (by dynamic programming) is "gl$, value OF max)); printf(($" The number of ("n(UPB items-1)(g", ")g") items to achieve this is: ("n(UPB items-1)(f(d)",")f(d)") respectively"l$,

   name OF items, max items));

printf(($" The weight to carry is "f(d)", and the volume used is "f(d)l$,

   weight OF max, volume OF max))</lang>Output:

The maximum value achievable (by dynamic programming) is      +54500
  The number of (panacea, ichor, gold) items to achieve this is: (   9,   0,  11) respectively
  The weight to carry is  247, and the volume used is  247

AutoHotkey

Brute Force. <lang AutoHotkey>Item = Panacea,Ichor,Gold Value = 3000,1800,2500 Weight= 3,2,20 ; *10 Volume= 25,15,2 ; *1000

StringSplit I, Item, `, ; Put input in arrays StringSplit W, Weight,`, StringSplit $, Value, `, StringSplit V, Volume,`,

SetFormat Float, 0.3 W := 250, V := 250, sW:=.1, sV:=.001 ; limits for the total, scale factors p := -1, Wp := -W1, Vp := -V1 ; initial values While (Wp+=W1) <= W && (Vp+=V1) <= V {

  p++, Wi := Wp-W2, Vi := Vp-V2, i := -1
  While (Wi+=W2) <= W && (Vi+=V2) <= V {
     i++, Wg := Wi-W3, Vg := Vi-V3, g := -1
     While (Wg+=W3) <= W && (Vg+=V3) <= V
        If ($ <= Val := p*$1 + i*$2 + ++g*$3)
            t := ($=Val ? t "`n    " : "    ")
          . p "`t   " i "`t   " g "`t  " Wg*sW "`t   " Vg*sV
          , $ := Val
  }

} MsgBox Value = %$%`n`nPanacea`tIchor`tGold`tWeight`tVolume`n%t%</lang>

Bracmat

<lang bracmat>(knapsack=

 ( things
 =   (panacea.3000.3/10.25/1000)
     (ichor.1800.2/10.15/1000)
     (gold.2500.2.2/1000)
 )

& 0:?maxvalue & :?sack & ( add

 =     cumwght
       cumvol
       cumvalue
       cumsack
       name
       wght
       val
       vol
       tings
       n
       ncumwght
       ncumvalue
       ncumvol
   .     !arg
       : ( ?cumwght
         . ?cumvol
         . ?cumvalue
         . ?cumsack
         . (?name.?val.?wght.?vol) ?tings
         )
     & -1:?n
     &   whl
       ' ( 1+!n:?n
         & !cumwght+!n*!wght:~>25:?ncumwght
         & !cumvol+!n*!vol:~>250/1000:?ncumvol
         & !cumvalue+!n*!val:?ncumvalue
         & (   !tings:
             & (   !ncumvalue:>!maxvalue:?maxvalue
                 &     !cumsack
                       ( !n:0&
                       |   ( !cumsack:&Take
                           | Finally
                           )
                           " take "
                           !n
                           " items of "
                           !name
                           ".\n"
                       )
                   : ?sack
               | 
               )
           |   add
             $ ( !ncumwght
               . !ncumvol
               . !ncumvalue
               .   !cumsack
                   ( !n:0&
                   | "Take " !n " items of " !name ".\n"
                   )
               . !tings
               )
           )
         )
 )

& add$(0.0.0..!things) & out$(str$(!sack "The value in the knapsack is " !maxvalue ".")) & );

!knapsack; </lang> Output:

Take 15 items of ichor.
Finally take 11 items of gold.
The value in the knapsack is 54500.

C

figures out the best (highest value) set by brute forcing every possible subset.

<lang c>#include <stdio.h>

include <stdlib.h>

typedef struct {

   char *name;
   double value;
   double weight;
   double volume;

} item_t;

item_t items[] = {

   {"panacea", 3000.0, 0.3, 0.025},
   {"ichor",   1800.0, 0.2, 0.015},
   {"gold",    2500.0, 2.0, 0.002},

};

int n = sizeof (items) / sizeof (item_t); int *count; int *best; double best_value;

void knapsack (int i, double value, double weight, double volume) {

   int j, m1, m2, m;
   if (i == n) {
       if (value > best_value) {
           best_value = value;
           for (j = 0; j < n; j++) {
               best[j] = count[j];
           }
       }
       return;
   }
   m1 = weight / items[i].weight;
   m2 = volume / items[i].volume;
   m = m1 < m2 ? m1 : m2;
   for (count[i] = m; count[i] >= 0; count[i]--) {
       knapsack(
           i + 1,
           value + count[i] * items[i].value,
           weight - count[i] * items[i].weight,
           volume - count[i] * items[i].volume
       );
   }

}

int main () {

   count = malloc(n * sizeof (int));
   best = malloc(n * sizeof (int));
   best_value = 0;
   knapsack(0, 0.0, 25.0, 0.25);
   int i;
   for (i = 0; i < n; i++) {
       printf("%d %s\n", best[i], items[i].name);
   }
   printf("best value: %.0f\n", best_value);
   free(count); free(best);
   return 0;

} </lang>

Output:

9 panacea

0 ichor 11 gold best value: 54500

C_sharp

<lang csharp>/*Knapsack

 This model finds the integer optimal packing of a knapsack

 Nigel_Galloway
 January 29th., 2012

/

using Microsoft.SolverFoundation.Services;

namespace KnapU {

   class Item {
       public string Name {get; set;}
       public int Value {get; set;}
       public double Weight {get; set;}
       public double Volume {get; set;}

       public Item(string name, int value, double weight, double volume) {
           Name = name;
           Value = value;
           Weight = weight;
           Volume = volume;
       }
   }

   class Program
   {
       static void Main(string[] args)
       {
           SolverContext context = SolverContext.GetContext();
           Model model = context.CreateModel();
           Item[] Knapsack = new Item[] {
               new Item("Panacea", 3000, 0.3, 0.025),
               new Item("Ichor", 1800, 0.2, 0.015),
               new Item("Gold", 2500, 2.0, 0.002)
           };
           Set items = new Set(Domain.Any, "items");
           Decision take = new Decision(Domain.IntegerNonnegative, "take", items);
           model.AddDecision(take);
           Parameter value = new Parameter(Domain.IntegerNonnegative, "value", items);
           value.SetBinding(Knapsack, "Value", "Name");
           Parameter weight = new Parameter(Domain.RealNonnegative, "weight", items);
           weight.SetBinding(Knapsack, "Weight", "Name");
           Parameter volume = new Parameter(Domain.RealNonnegative, "volume", items);
           volume.SetBinding(Knapsack, "Volume", "Name");
           model.AddParameters(value, weight, volume);
           model.AddConstraint("knap_weight", Model.Sum(Model.ForEach(items, t => take[t] * weight[t])) <= 25);
           model.AddConstraint("knap_vol", Model.Sum(Model.ForEach(items, t => take[t] * volume[t])) <= 0.25);
           model.AddGoal("knap_value", GoalKind.Maximize, Model.Sum(Model.ForEach(items, t => take[t] * value[t])));
           Solution solution = context.Solve(new SimplexDirective());
           Report report = solution.GetReport();
           System.Console.Write("{0}", report);
       }
   }

}</lang> Produces:

===Solver Foundation Service Report===
Date: 28/01/2012 17:18:56
Version: Microsoft Solver Foundation 3.0.1.10599 Express Edition
Model Name: DefaultModel
Capabilities Applied: MILP
Solve Time (ms): 210
Total Time (ms): 376
Solve Completion Status: Optimal
Solver Selected: Microsoft.SolverFoundation.Solvers.SimplexSolver
Directives:
Simplex(TimeLimit = -1, MaximumGoalCount = -1, Arithmetic = Default, Pricing = D
efault, IterationLimit = -1, Algorithm = Default, Basis = Default, GetSensitivit
y = False)
Algorithm: Dual
Arithmetic: Double
Variables: 3 -> 3 + 3
Rows: 3 -> 3
Nonzeros: 9
Eliminated Slack Variables: 0
Pricing (double): SteepestEdge
Basis: Current
Pivot Count: 0
Phase 1 Pivots: 0 + 0
Phase 2 Pivots: 0 + 0
Factorings: 3 + 0
Degenerate Pivots: 0 (0.00 %)
Branches: 21
===Solution Details===
Goals:
knap_value: 54500

Decisions:
take(Panacea): 9
take(Ichor): 0
take(Gold): 11

Clojure

<lang lisp>(defstruct item :value :weight :volume)

(defn total [key items quantities]

 (reduce + (map * quantities (map key items))))

(defn max-count [item max-weight max-volume]

 (let [mcw (/ max-weight (:weight item))
       mcv (/ max-volume (:volume item))]
   (min mcw mcv)))</lang>

We have an item struct to contain the data for both contents and the knapsack. The total function returns the sum of a particular attribute across all items times their quantities. Finally, the max-count function returns the most of that item that could fit given the constraints (used as the upper bound on the combination). Now the real work: <lang lisp>(defn knapsacks []

 (let [pan (struct item 3000 0.3 0.025)
       ich (struct item 1800 0.2 0.015)
       gol (struct item 2500 2.0 0.002)
       types [pan ich gol]
       max-w 25.0
       max-v 0.25
       iters #(range (inc (max-count % max-w max-v)))]
   (filter (complement nil?)  
     (pmap 
       #(let [[p i g] %
               w (total :weight types %)
               v (total :volume types %)]
         (if (and (<= w max-w) (<= v max-v))
           (with-meta (struct item (total :value types %) w v) {:p p :i i :g g})))
       (for [p (iters pan)
             i (iters ich)
             g (iters gol)]
         [p i g])))))</lang>

The knapsacks function returns a lazy sequence of all valid knapsacks, with the particular content quantities as metadata. The work of realizing each knapsack is done in parallel via the pmap function. The following then finds the best by value, and prints the result. <lang lisp>(defn best-by-value [ks]

 (reduce #(if (> (:value %1) (:value %2)) %1 %2) ks))

(defn print-knapsack[k]

 (let [ {val :value w :weight  v :volume} k
       {p :p i :i g :g} ^k]
   (println "Maximum value:" (float val))
   (println "Total weight: " (float w))
   (println "Total volume: " (float v))
   (println "Containing:   " p "Panacea," i "Ichor," g "Gold")))</lang>

Calling (print-knapsack (best-by-value (knapsacks))) would result in something like:

Maximum value: 54500
Total weight:  24.7
Total volume:  0.247
Containing:    9 Panacea, 0 Ichor, 11 Gold

Further, we could find all "best" knapsacks rather simply (albeit at the cost of some efficiency): <lang lisp>(defn all-best-by-value [ks]

 (let [b (best-by-value ks)]
   (filter #(= (:value b) (:value %)) ks)))

(defn print-knapsacks [ks]

 (doseq [k ks]
   (print-knapsack k)
   (println)))</lang>

Calling (print-knapsacks (all-best-by-value (knapsacks))) would result in something like:

Maximum value: 54500.0
Total weight:  25.0
Total volume:  0.247
Containing:    0 Panacea, 15 Ichor, 11 Gold

Maximum value: 54500.0
Total weight:  24.9
Total volume:  0.247
Containing:    3 Panacea, 10 Ichor, 11 Gold

Maximum value: 54500.0
Total weight:  24.8
Total volume:  0.247
Containing:    6 Panacea, 5 Ichor, 11 Gold

Maximum value: 54500.0
Total weight:  24.7
Total volume:  0.247
Containing:    9 Panacea, 0 Ichor, 11 Gold

Common Lisp

A dynamic programming O(maxVolume × maxWeight × nItems) solution, where volumes and weights are integral values. <lang lisp>(defun fill-knapsack (items max-volume max-weight)

 "Items is a list of lists of the form (name value weight volume) where weight

and value are integers. max-volume and max-weight, also integers, are the maximum volume and weight of the knapsack. fill-knapsack returns a list of the form (total-value inventory total-volume total-weight) where total-value is the total-value of a knapsack packed with inventory (a list whose elements are elements of items), and total-weight and total-volume are the total weights and volumes of the inventory."

 ;; maxes is a table indexed by volume and weight, where maxes[volume,weight]
 ;; is a list of the form (value inventory used-volume used-weight) where
 ;; inventory is a list of items of maximum value fitting within volume and
 ;; weight, value is the maximum value, and used-volume/used-weight are the
 ;; actual volume/weight of the inventory.
 (let* ((VV (1+ max-volume))
        (WW (1+ max-weight))
        (maxes (make-array (list VV WW))))
   ;; fill in the base cases where volume or weight is 0
   (dotimes (v VV) (setf (aref maxes v 0) (list 0 '() 0 0)))
   (dotimes (w WW) (setf (aref maxes 0 w) (list 0 '() 0 0)))
   ;; populate the rest of the table. The best value for a volume/weight
   ;; combination is the best way of adding an item to any of the inventories
   ;; from [volume-1,weight], [volume,weight-1], or [volume-1,weight-1], or the
   ;; best of these, if no items can be added.
   (do ((v 1 (1+ v))) ((= v VV) (aref maxes max-volume max-weight))
     (do ((w 1 (1+ w))) ((= w WW))
       (let ((options (sort (list (aref maxes v (1- w))
                                  (aref maxes (1- v) w)
                                  (aref maxes (1- v) (1- w)))
                            '> :key 'first)))
         (destructuring-bind (b-value b-items b-volume b-weight) (first options)
           (dolist (option options)
             (destructuring-bind (o-value o-items o-volume o-weight) option
               (dolist (item items)
                 (destructuring-bind (_ i-value i-volume i-weight) item
                   (declare (ignore _))
                   (when (and (<= (+ o-volume i-volume) v)
                              (<= (+ o-weight i-weight) w)
                              (>  (+ o-value  i-value)  b-value))
                     (setf b-value  (+ o-value  i-value)
                           b-volume (+ o-volume i-volume)
                           b-weight (+ o-weight i-weight)
                           b-items  (list* item o-items)))))))
           (setf (aref maxes v w)
                 (list b-value b-items b-volume b-weight))))))))</lang>

Use, having multiplied volumes and weights as to be integral:

> (pprint (fill-knapsack '((panacea 3000  3 25)
                           (ichor   1800  2 15)
                           (gold    2500 20  2))
                         250
                         250))

(54500              ; total-value
 ((ICHOR 1800 2 15) ; 15 ichor
  ...
  (ICHOR 1800 2 15)
  (GOLD 2500 20 2)  ; 11 gold
  ...
  (GOLD 2500 20 2))
 250                ; total volume
 247)               ; total weight

D

Translation of: Python

<lang d>void main() @safe /*@nogc*/ {

   import std.stdio, std.algorithm, std.typecons, std.conv;

   static struct Bounty {
       int value;
       double weight, volume;
   }

   immutable Bounty panacea = {3000,  0.3, 0.025};
   immutable Bounty ichor =   {1800,  0.2, 0.015};
   immutable Bounty gold =    {2500,  2.0, 0.002};
   immutable Bounty sack =    {   0, 25.0, 0.25 };

   immutable maxPanacea = min(sack.weight / panacea.weight,
                              sack.volume / panacea.volume).to!int;
   immutable maxIchor   = min(sack.weight / ichor.weight,
                              sack.volume / ichor.volume).to!int;
   immutable maxGold    = min(sack.weight / gold.weight,
                              sack.volume / gold.volume).to!int;

   Bounty best = {0, 0, 0};
   Tuple!(int, int, int) bestAmounts;

   foreach (immutable nPanacea; 0 .. maxPanacea)
       foreach (immutable nIchor; 0 .. maxIchor)
           foreach (immutable nGold; 0 .. maxGold) {
               immutable Bounty current = {
                   value: nPanacea * panacea.value +
                          nIchor * ichor.value +
                          nGold * gold.value,
                   weight: nPanacea * panacea.weight +
                           nIchor * ichor.weight +
                           nGold * gold.weight,
                   volume: nPanacea * panacea.volume +
                           nIchor * ichor.volume +
                           nGold * gold.volume};

               if (current.value > best.value &&
                   current.weight <= sack.weight &&
                   current.volume <= sack.volume) {
                   best = Bounty(current.value, current.weight, current.volume);
                   bestAmounts = tuple(nPanacea, nIchor, nGold);
               }
           }

   writeln("Maximum value achievable is ", best.value);
   writefln("This is achieved by carrying (one solution) %d" ~
            " panacea, %d ichor and %d gold", bestAmounts[]);
   writefln("The weight to carry is %4.1f and the volume used is %5.3f",
            best.weight, best.volume);

}</lang>

Output:

Maximum value achievable is 54500
This is achieved by carrying (one solution) 0 panacea, 15 ichor and 11 gold
The weight to carry is 25.0 and the volume used is 0.247

Alternative Version

The output is the same. <lang d>void main() {

 import std.stdio, std.algorithm, std.typecons, std.range, std.conv;

 alias Bounty = Tuple!(int,"value", double,"weight", double,"volume");

 immutable panacea = Bounty(3000,  0.3, 0.025);
 immutable ichor =   Bounty(1800,  0.2, 0.015);
 immutable gold =    Bounty(2500,  2.0, 0.002);
 immutable sack =    Bounty(   0, 25.0, 0.25);

 immutable maxPanacea = min(sack.weight / panacea.weight, sack.volume / panacea.volume).to!int;
 immutable maxIchor   = min(sack.weight / ichor.weight,   sack.volume / ichor.volume).to!int;
 immutable maxGold    = min(sack.weight / gold.weight,    sack.volume / gold.volume).to!int;

 immutable best =
   cartesianProduct(maxPanacea.iota, maxIchor.iota, maxGold.iota)
   .map!(t => tuple(Bounty(t[0] * panacea.value  + t[1] * ichor.value  + t[2] * gold.value,
                           t[0] * panacea.weight + t[1] * ichor.weight + t[2] * gold.weight,
                           t[0] * panacea.volume + t[1] * ichor.volume + t[2] * gold.volume), t))
   .filter!(t => t[0].weight <= sack.weight && t[0].volume <= sack.volume)
   .reduce!max;

 writeln("Maximum value achievable is ", best[0].value);
 writefln("This is achieved by carrying (one solution) %d panacea, %d ichor and %d gold", best[1][]);
 writefln("The weight to carry is %4.1f and the volume used is %5.3f", best[0][1..$]);

}</lang>

E

This is a mostly brute-force general solution (the first author of this example does not know dynamic programming); the only optimization is that when considering the last (third) treasure type, it does not bother filling with anything but the maximum amount. <lang e>pragma.enable("accumulator")

/** A data type representing a bunch of stuff (or empty space). */ def makeQuantity(value, weight, volume, counts) {

 def quantity {
   to __printOn(out) { 
     for name => n in counts { out.print(`$n $name  `) }
     out.print(`(val=$value wt=$weight vol=$volume)`)
   }
   to value () { return value  }
   to weight() { return weight }
   to volume() { return volume }
   to counts() { return counts }
   to subtract(other) { return quantity + other * -1 }
   to add(other) {
     return makeQuantity(value  + other.value (),
                         weight + other.weight(),
                         volume + other.volume(),
                         accum counts for name => n in other.counts() { _.with(name, n+counts.fetch(name, fn {0})) })
   }
   to multiply(scalar) {
     return makeQuantity(value  * scalar,
                         weight * scalar,
                         volume * scalar,
                         accum [].asMap() for name => n in counts { _.with(name, n*scalar) })
   }
   /** a.fit(b) the greatest integer k such that a - b * k does not have negative weight or volume. */
   to fit(item) {
     return (weight // item.weight()) \
        .min(volume // item.volume())
   }
 }
 return quantity

}

/** Fill the space with the treasures, returning candidate results as spaceAvailable - the items. */ def fill(spaceAvailable, treasures) {

 if (treasures.size().isZero()) { # nothing to pick
   return [spaceAvailable]
 }
 
 # Pick one treasure type
 def [unit] + otherTreasures := treasures
 
 var results := []
 for count in (0..spaceAvailable.fit(unit)).descending() {
   results += fill(spaceAvailable - unit * count, otherTreasures)
   if (otherTreasures.size().isZero()) {
     break # If there are no further kinds, there is no point in taking less than the most
   }
 }
 return results

}

def chooseBest(emptyKnapsack, treasures) {

 var maxValue := 0
 var best := []
 for result in fill(emptyKnapsack, treasures) {
   def taken := emptyKnapsack - result # invert the backwards result fill() returns
   if (taken.value() > maxValue) {
     best := [taken]
     maxValue := taken.value()
   } else if (taken.value() <=> maxValue) {
     best with= taken
   }
 }
 return best

}

def printBest(emptyKnapsack, treasures) {

 for taken in chooseBest(emptyKnapsack, treasures) { println(`  $taken`) }

}

def panacea := makeQuantity(3000, 0.3, 0.025, ["panacea" => 1]) def ichor := makeQuantity(1800, 0.2, 0.015, ["ichor" => 1]) def gold := makeQuantity(2500, 2.0, 0.002, ["gold" => 1]) def emptyKnapsack \

           := makeQuantity(   0,  25, 0.250, [].asMap())

printBest(emptyKnapsack, [panacea, ichor, gold])</lang>

EchoLisp

Use a cache, and multiply by 10^n to get an integer problem. <lang scheme> (require 'struct) (require 'hash) (require 'sql)

(define H null) ;; cache (define T (make-table (struct goodies (name valeur poids volume )))) (define-syntax-rule (name i) (table-xref T i 0)) (define-syntax-rule (valeur i) (table-xref T i 1)) (define-syntax-rule (poids i) (table-xref T i 2)) (define-syntax-rule (volume i) (table-xref T i 3))

(define goodies

 '(("🍁-panacea"  3000 300 25)
  ("🌵-ichor"    1800 200 15)
  ("⭐️-gold"     2500 2000 2)))

(list->table goodies T)

i = item index, p= remaining weight, v = remaining volume

make an unique hash-key from (i p v)

(define (t-key i p v) (string-append i "|" p "|" v))

retrieve best core for item i returns ( score . quantity)

(define (t-get i p v)

  (if ( < i 0) (cons 0 0)  
     (hash-ref H (t-key i p v )))) ;; may be #f

compute best quantity.score (i), assuming best (i-1 p v) is known

(define (score-qty i p v (q) (score)(smax)(qmax)) (or (t-get i p v) ;; already known (begin

		(set! q (min (quotient p (poids i)) (quotient v (volume i)))) ;; max possible q
		(set! smax -Infinity)

( for ((k (1+ q))) ;; try all legal quantities (set! score (+ (first (score-qty (1- i) (- p (* k (poids i))) (- v (* k (volume i))))) (* k (valeur i)))) #:continue (< score smax) (set! smax score) (set! qmax k)) (hash-set H (t-key i p v) (cons smax qmax)))))

compute best scores, starting from last item

(define (task P V)

       (define N (1- (table-count T)))
       (define qty 0) 
       (set! H (make-hash))

(writeln 'total-value (first (score-qty N P V)))

(for/list ((i (in-range N -1 -1))) (set! qty (rest (t-get i P V))) #:continue (= qty 0) (begin0 (cons (name i) (t-get i P V)) (set! P (- P (* (poids i) qty))) (set! V (- V (* (volume i) qty))))))

output

(task 25000 250) total-value 54500

   → (("⭐️-gold" 54500 . 11) ("🌵-ichor" 27000 . 15))

(length (hash-keys H)) ;; # entries in cache

   → 218

</lang>

Eiffel

<lang Eiffel> class KNAPSACK

create make

feature

make do create panacea; panacea := [3000, 0.3, 0.025] create ichor; ichor := [1800, 0.2, 0.015] create gold; gold := [2500, 2.0, 0.002] create sack; sack := [0, 25.0, 0.25] find_solution end

feature {NONE}

panacea: TUPLE [value: INTEGER; weight: REAL_64; volume: REAL_64]

ichor: TUPLE [value: INTEGER; weight: REAL_64; volume: REAL_64]

gold: TUPLE [value: INTEGER; weight: REAL_64; volume: REAL_64]

sack: TUPLE [value: INTEGER; weight: REAL_64; volume: REAL_64]

find_solution -- Solution for unbounded Knapsack Problem. local totalweight, totalvolume: REAL_64 maxpanacea, maxichor, maxvalue, maxgold: INTEGER n: ARRAY [INTEGER] r: TUPLE [value: INTEGER; weight: REAL_64; volume: REAL_64] do maxpanacea := minimum (sack.weight / panacea.weight, sack.volume / panacea.volume).rounded maxichor := minimum (sack.weight / ichor.weight, sack.volume / ichor.volume).rounded maxgold := minimum (sack.weight / gold.weight, sack.volume / gold.volume).rounded create n.make_filled (0, 1, 3) create r across 0 |..| maxpanacea as p loop across 0 |..| maxichor as i loop across 0 |..| maxgold as g loop r.value := g.item * gold.value + i.item * ichor.value + p.item * panacea.value r.weight := g.item * gold.weight + i.item * ichor.weight + p.item * panacea.weight r.volume := g.item * gold.volume + i.item * ichor.volume + p.item * panacea.volume if r.value > maxvalue and r.weight <= sack.weight and r.volume <= sack.volume then maxvalue := r.value totalweight := r.weight totalvolume := r.volume n [1] := p.item n [2] := i.item n [3] := g.item end end end end io.put_string ("Maximum value achievable is " + maxValue.out + ".%N") io.put_string ("This is achieved by carrying " + n [1].out + " panacea, " + n [2].out + " ichor and " + n [3].out + " gold.%N") io.put_string ("The weight is " + totalweight.out + " and the volume is " + totalvolume.truncated_to_real.out + ".") end

minimum (a, b: REAL_64): REAL_64 -- Smaller of 'a' and 'b'. do Result := a if a > b then Result := b end end

end </lang>

Output:

Maximum value achievable is 54500.
This is achieved by carrying 0 panacea, 15 ichor and 11 gold.
The weight is 25 and the volume is 0.247.

Factor

This is a brute force solution. It is general enough to be able to provide solutions for any number of different items. <lang factor>USING: accessors combinators kernel locals math math.order math.vectors sequences sequences.product combinators.short-circuit ; IN: knapsack

CONSTANT: values { 3000 1800 2500 } CONSTANT: weights { 0.3 0.2 2.0 } CONSTANT: volumes { 0.025 0.015 0.002 }

CONSTANT: max-weight 25.0 CONSTANT: max-volume 0.25

TUPLE: bounty amounts value weight volume ;

<bounty> ( items -- bounty )

   [ bounty new ] dip {
       [ >>amounts ]
       [ values v. >>value ]
       [ weights v. >>weight ]
       [ volumes v. >>volume ]
   } cleave ;

valid-bounty? ( bounty -- ? )

   { [ weight>> max-weight <= ]
     [ volume>> max-volume <= ] } 1&& ;

M:: bounty <=> ( a b -- <=> )

   a valid-bounty? [
       b valid-bounty? [
           a b [ value>> ] compare
       ] [ +gt+ ] if
   ] [ b valid-bounty? +lt+ +eq+ ? ] if ;

find-max-amounts ( -- amounts )

   weights volumes [
       [ max-weight swap / ]
       [ max-volume swap / ] bi* min >integer
   ] 2map ;

best-bounty ( -- bounty )

   find-max-amounts [ 1 + iota ] map <product-sequence>
   [ <bounty> ] [ max ] map-reduce ;</lang>

Forth

<lang forth>\ : value ; immediate

weight cell+ ;

volume 2 cells + ;

number 3 cells + ;

\ item value weight volume number create panacea 30 , 3 , 25 , 0 , create ichor 18 , 2 , 15 , 0 , create gold 25 , 20 , 2 , 0 , create sack 0 , 250 , 250 ,

fits? ( item -- ? )

 dup weight @ sack weight @ > if drop false exit then
     volume @ sack volume @ > 0= ;

add ( item -- )

 dup        @        sack        +!
 dup weight @ negate sack weight +!
 dup volume @ negate sack volume +!
 1 swap number +! ;

take ( item -- )

 dup        @ negate sack        +!
 dup weight @        sack weight +!
 dup volume @        sack volume +!
 -1 swap number +! ;

variable max-value variable max-pan variable max-ich variable max-au

.solution

 cr
 max-pan @ . ." Panaceas, "
 max-ich @ . ." Ichors, and "
 max-au  @ . ." Gold for a total value of "
 max-value @ 100 * . ;

check

 sack @ max-value @ <= if exit then
 sack           @ max-value !
 panacea number @ max-pan   !
 ichor   number @ max-ich   !
 gold    number @ max-au    !
 ( .solution ) ;    \ and change <= to < to see all solutions

solve-gold

 gold fits? if gold add  recurse  gold take
 else check then ;

solve-ichor

 ichor fits? if ichor add  recurse  ichor take then
 solve-gold ;

solve-panacea

 panacea fits? if panacea add  recurse  panacea take then
 solve-ichor ;

solve-panacea .solution</lang> Or like this... <lang forth>0 VALUE vials 0 VALUE ampules 0 VALUE bars 0 VALUE bag

250 3 / #250 #25 / MIN 1+ CONSTANT maxvials
250 2/ #250 #15 / MIN 1+ CONSTANT maxampules
250 #20 / #250 2/ MIN 1+ CONSTANT maxbars

RESULTS ( v a b -- k )

3DUP #20 * SWAP 2* + SWAP 3 * + #250 > IF 3DROP -1 EXIT ENDIF 3DUP 2* SWAP #15 * + SWAP #25 * + #250 > IF 3DROP -1 EXIT ENDIF #2500 * SWAP #1800 * + SWAP #3000 * + ;

.SOLUTION ( -- )

CR ." The traveller's knapsack contains " vials DEC. ." vials of panacea, " ampules DEC. ." ampules of ichor, " CR bars DEC. ." bars of gold, a total value of " vials ampules bars RESULTS 0DEC.R ." ." ;

KNAPSACK ( -- )

-1 TO bag maxvials 0 ?DO maxampules 0 ?DO maxbars 0 ?DO

                             K J I RESULTS DUP

bag > IF TO bag K TO vials J TO ampules I TO bars ELSE DROP ENDIF LOOP LOOP LOOP .SOLUTION ;</lang> With the result...

FORTH> knapsack
The traveller's knapsack contains 0 vials of panacea, 15 ampules of ichor, 
11 bars of gold, a total value of 54500. ok

Fortran

Works with: Fortran version 90 and later

A straight forward 'brute force' approach <lang fortran>PROGRAM KNAPSACK

 IMPLICIT NONE

 REAL :: totalWeight, totalVolume
 INTEGER :: maxPanacea, maxIchor, maxGold, maxValue = 0
 INTEGER :: i, j, k
 INTEGER :: n(3)

 TYPE Bounty
   INTEGER :: value
   REAL :: weight
   REAL :: volume
 END TYPE Bounty

 TYPE(Bounty) :: panacea, ichor, gold, sack, current

 panacea = Bounty(3000, 0.3, 0.025)
 ichor   = Bounty(1800, 0.2, 0.015)
 gold    = Bounty(2500, 2.0, 0.002)
 sack    = Bounty(0, 25.0, 0.25)

 maxPanacea = MIN(sack%weight / panacea%weight, sack%volume / panacea%volume)
 maxIchor = MIN(sack%weight / ichor%weight, sack%volume / ichor%volume)
 maxGold = MIN(sack%weight / gold%weight, sack%volume / gold%volume)
 
 DO i = 0, maxPanacea
    DO j = 0, maxIchor
       Do k = 0, maxGold
          current%value = k * gold%value + j * ichor%value + i * panacea%value
          current%weight = k * gold%weight + j * ichor%weight + i * panacea%weight
          current%volume = k * gold%volume + j * ichor%volume + i * panacea%volume       
          IF (current%weight > sack%weight .OR. current%volume > sack%volume) CYCLE
          IF (current%value > maxValue) THEN
             maxValue = current%value
             totalWeight = current%weight
             totalVolume = current%volume
             n(1) = i ; n(2) = j ; n(3) = k
          END IF
       END DO  
    END DO
 END DO

 WRITE(*, "(A,I0)") "Maximum value achievable is ", maxValue
 WRITE(*, "(3(A,I0),A)") "This is achieved by carrying ", n(1), " panacea, ", n(2), " ichor and ", n(3), " gold items"
 WRITE(*, "(A,F4.1,A,F5.3)") "The weight to carry is ", totalWeight, " and the volume used is ", totalVolume

END PROGRAM KNAPSACK</lang> Sample output

Maximum value achievable is 54500
This is achieved by carrying 0 panacea, 15 ichor and 11 gold items
The weight to carry is 25.0 and the volume used is 0.247

Go

Recursive brute-force. <lang go>package main

import "fmt"

type Item struct { Name string Value int Weight, Volume float64 }

type Result struct { Counts []int Sum int }

func min(a, b int) int { if a < b { return a } return b }

func Knapsack(items []Item, weight, volume float64) (best Result) { if len(items) == 0 { return } n := len(items) - 1 maxCount := min(int(weight/items[n].Weight), int(volume/items[n].Volume)) for count := 0; count <= maxCount; count++ { sol := Knapsack(items[:n], weight-float64(count)*items[n].Weight, volume-float64(count)*items[n].Volume) sol.Sum += items[n].Value * count if sol.Sum > best.Sum { sol.Counts = append(sol.Counts, count) best = sol } } return }

func main() { items := []Item{ {"Panacea", 3000, 0.3, 0.025}, {"Ichor", 1800, 0.2, 0.015}, {"Gold", 2500, 2.0, 0.002}, } var sumCount, sumValue int var sumWeight, sumVolume float64

result := Knapsack(items, 25, 0.25)

for i := range result.Counts { fmt.Printf("%-8s x%3d -> Weight: %4.1f Volume: %5.3f Value: %6d\n", items[i].Name, result.Counts[i], items[i].Weight*float64(result.Counts[i]), items[i].Volume*float64(result.Counts[i]), items[i].Value*result.Counts[i])

sumCount += result.Counts[i] sumValue += items[i].Value * result.Counts[i] sumWeight += items[i].Weight * float64(result.Counts[i]) sumVolume += items[i].Volume * float64(result.Counts[i]) }

fmt.Printf("TOTAL (%3d items) Weight: %4.1f Volume: %5.3f Value: %6d\n", sumCount, sumWeight, sumVolume, sumValue) }</lang>

Output:

Panacea  x  9  -> Weight:  2.7  Volume: 0.225  Value:  27000
Ichor    x  0  -> Weight:  0.0  Volume: 0.000  Value:      0
Gold     x 11  -> Weight: 22.0  Volume: 0.022  Value:  27500
TOTAL ( 20 items) Weight: 24.7  Volume: 0.247  Value:  54500

Groovy

Solution: dynamic programming <lang groovy>def totalWeight = { list -> list.collect{ it.item.weight * it.count }.sum() } def totalVolume = { list -> list.collect{ it.item.volume * it.count }.sum() } def totalValue = { list -> list.collect{ it.item.value * it.count }.sum() }

def knapsackUnbounded = { possibleItems, BigDecimal weightMax, BigDecimal volumeMax ->

   def n = possibleItems.size()
   def wm = weightMax.unscaledValue()
   def vm = volumeMax.unscaledValue()
   def m = (0..n).collect{ i -> (0..wm).collect{ w -> (0..vm).collect{ v -> [] } } }
   (1..wm).each { w ->
       (1..vm).each { v ->
           (1..n).each { i ->
               def item = possibleItems[i-1]
               def wi = item.weight.unscaledValue()
               def vi = item.volume.unscaledValue()
               def bi = [w.intdiv(wi),v.intdiv(vi)].min()
               m[i][w][v] = (0..bi).collect{ count ->
                   m[i-1][w - wi * count][v - vi * count] + item:item, count:count
               }.max(totalValue).findAll{ it.count }
           }
       }
   }
   m[n][wm][vm]

}</lang>

Test: <lang groovy>Set solutions = [] items.eachPermutation { itemList ->

   def start = System.currentTimeMillis()
   def packingList = knapsackUnbounded(itemList, 25.0, 0.250)
   def elapsed = System.currentTimeMillis() - start
   
   println "\n  Item Order: ${itemList.collect{ it.name.split()[0] }}"
   println "Elapsed Time: ${elapsed/1000.0} s"
   
   solutions << (packingList as Set)

}

solutions.each { packingList ->

   println "\nTotal Weight: ${totalWeight(packingList)}"
   println "Total Volume: ${totalVolume(packingList)}"
   println " Total Value: ${totalValue(packingList)}"
   packingList.each {
       printf ('  item: %-22s  count:%2d  weight:%4.1f  Volume:%5.3f\n',
               it.item.name, it.count, it.item.weight * it.count, it.item.volume * it.count)
   }

}</lang>

Output:

  Item Order: [panacea, ichor, gold]
Elapsed Time: 26.883 s

  Item Order: [panacea, gold, ichor]
Elapsed Time: 27.17 s

  Item Order: [ichor, panacea, gold]
Elapsed Time: 25.884 s

  Item Order: [ichor, gold, panacea]
Elapsed Time: 26.126 s

  Item Order: [gold, panacea, ichor]
Elapsed Time: 26.596 s

  Item Order: [gold, ichor, panacea]
Elapsed Time: 26.47 s

Total Weight: 25.0
Total Volume: 0.247
 Total Value: 54500
  item: gold (bars)             count:11  weight:22.0  Volume:0.022
  item: ichor (ampules of)      count:15  weight: 3.0  Volume:0.225

Total Weight: 24.7
Total Volume: 0.247
 Total Value: 54500
  item: gold (bars)             count:11  weight:22.0  Volume:0.022
  item: panacea (vials of)      count: 9  weight: 2.7  Volume:0.225

While this solver can only be used to detect two of the four possible solutions, the other two may be discovered by noting that 5 ampules of ichor and 3 vials of panacea have the same value and the same volume and only differ by 0.1 in weight. Thus the other two solutions can be derived by substitution as follows:

Total Weight: 24.9
Total Volume: 0.247
 Total Value: 54500
  item: gold (bars)             count:11  weight:22.0  Volume:0.022
  item: ichor (ampules of)      count:10  weight: 2.0  Volume:0.150
  item: panacea (vials of)      count: 3  weight: 0.9  Volume:0.075

Total Weight: 24.8
Total Volume: 0.247
 Total Value: 54500
  item: gold (bars)             count:11  weight:22.0  Volume:0.022
  item: ichor (ampules of)      count: 5  weight: 1.0  Volume:0.075
  item: panacea (vials of)      count: 6  weight: 1.8  Volume:0.150

Haskell

This is a brute-force solution: it generates a list of every legal combination of items (options) and then finds the option of greatest value. <lang haskell>import Data.List (maximumBy) import Data.Ord (comparing)

(maxWgt, maxVol) = (25, 0.25) items =

  [Bounty  "panacea"  3000  0.3  0.025,
   Bounty  "ichor"    1800  0.2  0.015,
   Bounty  "gold"     2500  2.0  0.002]

data Bounty = Bounty

  {itemName :: String,
   itemVal :: Int,
   itemWgt, itemVol :: Double}

names = map itemName items vals = map itemVal items wgts = map itemWgt items vols = map itemVol items

dotProduct :: (Num a, Integral b) => [a] -> [b] -> a dotProduct factors = sum . zipWith (*) factors . map fromIntegral

options :: Int options = filter fits $ mapM f items

 where f (Bounty _ _ w v) = [0 .. m]
         where m = floor $ min (maxWgt / w) (maxVol / v)
       fits opt = dotProduct wgts opt <= maxWgt &&
                  dotProduct vols opt <= maxVol

showOpt :: [Int] -> String showOpt opt = concat (zipWith showItem names opt) ++

   "total weight: " ++ show (dotProduct wgts opt) ++
   "\ntotal volume: " ++ show (dotProduct vols opt) ++
   "\ntotal value: " ++ show (dotProduct vals opt) ++ "\n"
 where showItem name num = name ++ ": " ++ show num ++ "\n"

main = putStr $ showOpt $ best options

 where best = maximumBy $ comparing $ dotProduct vals</lang>

Output:

panacea: 9
ichor: 0
gold: 11
total weight: 24.7
total volume: 0.247
total value: 54500

HicEst

<lang hicest>CHARACTER list*1000

NN = ALIAS($Panacea, $Ichor, $Gold, wSack, wPanacea, wIchor, wGold, vSack, vPanacea, vIchor, vGold) NN = (3000, 1800, 2500, 25, 0.3, 0.2, 2.0, 0.25, 0.025, 0.015, 0.002) maxItems = ALIAS(maxPanacea, maxIchor, maxGold) maxItems = ( MIN( wSack/wPanacea, vSack/vPanacea), MIN( wSack/wIchor, vSack/vIchor), MIN( wSack/wGold, vSack/vGold) )

maxValue = 0 DO Panaceas = 0, maxPanacea

 DO Ichors = 0, maxIchor
   DO Golds = 0, maxGold
     weight = Panaceas*wPanacea + Ichors*wIchor + Golds*wGold
     IF( weight <= wSack ) THEN
       volume = Panaceas*vPanacea + Ichors*vIchor + Golds*vGold
       IF( volume <= vSack ) THEN
         value =  Panaceas*$Panacea + Ichors*$Ichor + Golds*$Gold
         IF( value > maxValue ) THEN
           maxValue = value
           ! this restarts the list, removing all previous entries:
           WRITE(Text=list, Name) value, Panaceas, Ichors, Golds, weight, volume, $CR//$LF
         ELSEIF( value == maxValue ) THEN
           WRITE(Text=list, Name, APPend) value, Panaceas, Ichors, Golds, weight, volume, $CR//$LF
         ENDIF
       ENDIF
     ENDIF
   ENDDO
 ENDDO

ENDDO</lang> <lang hicest>value=54500; Panaceas=0; Ichors=15; Golds=11; weight=25; volume=0.247; value=54500; Panaceas=3; Ichors=10; Golds=11; weight=24.9; volume=0.247; value=54500; Panaceas=6; Ichors=5; Golds=11; weight=24.8; volume=0.247; value=54500; Panaceas=9; Ichors=0; Golds=11; weight=24.7; volume=0.247; </lang>

J

Brute force solution. <lang j>mwv=: 25 0.25 prods=: <;. _1 ' panacea: ichor: gold:' hdrs=: <;. _1 ' weight: volume: value:' vls=: 3000 1800 2500 ws=: 0.3 0.2 2.0 vs=: 0.025 0.015 0.002

ip=: +/ .* prtscr=: (1!:2)&2

KS=: 3 : 0

os=. (#:i.@(*/)) mwv >:@<.@<./@:% ws,:vs
bo=.os#~(ws,:vs) mwv&(*./@:>)@ip"_ 1 os
mo=.bo{~{.\: vls ip"1 bo
prtscr &.> prods ([,' ',":@])&.>mo
prtscr &.> hdrs ('total '&,@[,' ',":@])&.> mo ip"1 ws,vs,:vls
LF

)</lang> Example output:

   KS''
panacea: 3
ichor: 10
gold: 11
total weight: 24.9
total volume: 0.247
total value: 54500

Java

With recursion for more than 3 items. <lang java>package hu.pj.alg;

import hu.pj.obj.Item; import java.text.*;

public class UnboundedKnapsack {

   protected Item []  items = {
                              new Item("panacea", 3000,  0.3, 0.025),
                              new Item("ichor"  , 1800,  0.2, 0.015),
                              new Item("gold"   , 2500,  2.0, 0.002)
                              };
   protected final int    n = items.length; // the number of items
   protected Item      sack = new Item("sack"   ,    0, 25.0, 0.250);
   protected Item      best = new Item("best"   ,    0,  0.0, 0.000);
   protected int  []  maxIt = new int [n];  // maximum number of items
   protected int  []    iIt = new int [n];  // current indexes of items
   protected int  [] bestAm = new int [n];  // best amounts

   public UnboundedKnapsack() {
       // initializing:
       for (int i = 0; i < n; i++) {
           maxIt [i] = Math.min(
                          (int)(sack.getWeight() / items[i].getWeight()),
                          (int)(sack.getVolume() / items[i].getVolume())
                       );
       } // for (i)

       // calc the solution:
       calcWithRecursion(0);

       // Print out the solution:
       NumberFormat nf = NumberFormat.getInstance();
       System.out.println("Maximum value achievable is: " + best.getValue());
       System.out.print("This is achieved by carrying (one solution): ");
       for (int i = 0; i < n; i++) {
           System.out.print(bestAm[i] + " " + items[i].getName() + ", ");
       }
       System.out.println();
       System.out.println("The weight to carry is: " + nf.format(best.getWeight()) +
                          "   and the volume used is: " + nf.format(best.getVolume())
                         );

   // calculation the solution with recursion method
   // item : the number of item in the "items" array
   public void calcWithRecursion(int item) {
       for (int i = 0; i <= maxIt[item]; i++) {
           iIt[item] = i;
           if (item < n-1) {
               calcWithRecursion(item+1);
           } else {
               int    currVal = 0;   // current value
               double currWei = 0.0; // current weight
               double currVol = 0.0; // current Volume
               for (int j = 0; j < n; j++) {
                   currVal += iIt[j] * items[j].getValue();
                   currWei += iIt[j] * items[j].getWeight();
                   currVol += iIt[j] * items[j].getVolume();
               }

               if (currVal > best.getValue()
                   &&
                   currWei <= sack.getWeight()
                   &&
                   currVol <= sack.getVolume()
               )
               {
                   best.setValue (currVal);
                   best.setWeight(currWei);
                   best.setVolume(currVol);
                   for (int j = 0; j < n; j++) bestAm[j] = iIt[j];
               } // if (...)
           } // else
       } // for (i)
   } // calcWithRecursion()

   // the main() function:
   public static void main(String[] args) {
       new UnboundedKnapsack();
   } // main()

} // class</lang>

<lang java>package hu.pj.obj;

public class Item {

   protected String name = "";
   protected int value = 0;
   protected double weight = 0;
   protected double volume = 0;

   public Item() {
   }

   public Item(String name, int value, double weight, double volume) {
       setName(name);
       setValue(value);
       setWeight(weight);
       setVolume(volume);
   }

   public int getValue() {
       return value;
   }

   public void setValue(int value) {
       this.value = Math.max(value, 0);
   }

   public double getWeight() {
       return weight;
   }

   public void setWeight(double weight) {
       this.weight = Math.max(weight, 0);
   }

   public double getVolume() {
       return volume;
   }

   public void setVolume(double volume) {
       this.volume = Math.max(volume, 0);
   }

   public String getName() {
       return name;
   }

   public void setName(String name) {
       this.name = name;
   }

} // class</lang>

output:

Maximum value achievable is: 54500
This is achieved by carrying (one solution): 0 panacea, 15 ichor, 11 gold, 
The weight to carry is: 25   and the volume used is: 0,247

JavaScript

Brute force. <lang javascript>var gold = { 'value': 2500, 'weight': 2.0, 'volume': 0.002 },

   panacea = { 'value': 3000, 'weight': 0.3, 'volume': 0.025 },
   ichor = { 'value': 1800, 'weight': 0.2, 'volume': 0.015 },
   
   items = [gold, panacea, ichor],
   knapsack = {'weight': 25, 'volume': 0.25},
   max_val = 0,
   solutions = [],
   g, p, i, item, val;

for (i = 0; i < items.length; i += 1) {

   item = items[i];
   item.max = Math.min(
       Math.floor(knapsack.weight / item.weight),
       Math.floor(knapsack.volume / item.volume)
   );

}

for (g = 0; g <= gold.max; g += 1) {

   for (p = 0; p <= panacea.max; p += 1) {
       for (i = 0; i <= ichor.max; i += 1) {
           if (i * ichor.weight + g * gold.weight + p * panacea.weight > knapsack.weight) {
               continue;
           }
           if (i * ichor.volume + g * gold.volume + p * panacea.volume > knapsack.volume) {
               continue;
           }
           val = i * ichor.value + g * gold.value + p * panacea.value;
           if (val > max_val) {
               solutions = [];
               max_val = val;
           }
           if (val === max_val) {
               solutions.push([g, p, i]);
           }
       }
   }

}

document.write("maximum value: " + max_val + '
'); for (i = 0; i < solutions.length; i += 1) {

   item = solutions[i];
   document.write("(gold: " + item[0] + ", panacea: " + item[1] + ", ichor: " + item[2] + ")
");

}

output:

maximum value: 54500
(gold: 11, panacea: 0, ichor: 15)
(gold: 11, panacea: 3, ichor: 10)
(gold: 11, panacea: 6, ichor: 5)
(gold: 11, panacea: 9, ichor: 0)

</lang>

M4

A brute force solution: <lang M4>divert(-1) define(`set2d',`define(`$1[$2][$3]',`$4')') define(`get2d',`defn(`$1[$2][$3]')') define(`for',

  `ifelse($#,0,``$0,
  `ifelse(eval($2<=$3),1,
  `pushdef(`$1',$2)$4`'popdef(`$1')$0(`$1',incr($2),$3,`$4')')')')

define(`min',

  `define(`ma',eval($1))`'define(`mb',eval($2))`'ifelse(eval(ma<mb),1,ma,mb)')

define(`setv',

  `set2d($1,$2,1,$3)`'set2d($1,$2,2,$4)`'set2d($1,$2,3,$5)`'set2d($1,$2,4,$6)')

dnl name,value (each),weight,volume setv(a,0,`knapsack',0,250,250) setv(a,1,`panacea',3000,3,25) setv(a,2,`ichor',1800,2,15) setv(a,3,`gold',2500,20,2)

define(`mv',0) for(`x',0,min(get2d(a,0,3)/get2d(a,1,3),get2d(a,0,4)/get2d(a,1,4)),

  `for(`y',0,min((get2d(a,0,3)-x*get2d(a,1,3))/get2d(a,2,3),
        (get2d(a,0,4)-x*get2d(a,1,4))/get2d(a,2,4)),
     `

define(`z',min((get2d(a,0,3)-x*get2d(a,1,3)-y*get2d(a,2,3))/get2d(a,3,3),

  (get2d(a,0,4)-x*get2d(a,1,4)-y*get2d(a,2,4))/get2d(a,3,4)))

define(`cv',eval(x*get2d(a,1,2)+y*get2d(a,2,2)+z*get2d(a,3,2))) ifelse(eval(cv>mv),1,

  `define(`mv',cv)`'define(`best',(x,y,z))',
  `ifelse(cv,mv,`define(`best',best (x,y,z))')')
     ')')

divert mv best</lang> Output:

54500 (0,15,11) (3,10,11) (6,5,11) (9,0,11)

Lua

<lang lua>items = { ["panaea"] = { ["value"] = 3000, ["weight"] = 0.3, ["volume"] = 0.025 },

           ["ichor"]  = { ["value"] = 1800, ["weight"] = 0.2, ["volume"] = 0.015 },
           ["gold"]   = { ["value"] = 2500, ["weight"] = 2.0, ["volume"] = 0.002 }
       }

max_weight = 25 max_volume = 0.25

max_num_items = {} for i in pairs( items ) do

  max_num_items[i] = math.floor( math.min( max_weight / items[i].weight, max_volume / items[i].volume ) )

end

best = { ["value"] = 0.0, ["weight"] = 0.0, ["volume"] = 0.0 } best_amounts = {}

for i = 1, max_num_items["panaea"] do

   for j = 1, max_num_items["ichor"] do
       for k = 1, max_num_items["gold"] do
           current = { ["value"]  = i*items["panaea"]["value"] + j*items["ichor"]["value"] + k*items["gold"]["value"],
                       ["weight"] = i*items["panaea"]["weight"] + j*items["ichor"]["weight"] + k*items["gold"]["weight"],
                       ["volume"] = i*items["panaea"]["volume"] + j*items["ichor"]["volume"] + k*items["gold"]["volume"]
                     }
                     
           if current.value > best.value and current.weight <= max_weight and current.volume <= max_volume then
               best = { ["value"] = current.value, ["weight"] = current.weight, ["volume"] = current.volume }
               best_amounts = { ["panaea"] = i, ["ichor"] = j, ["gold"] = k }
           end
       end
   end

end

print( "Maximum value:", best.value ) for k, v in pairs( best_amounts ) do

   print( k, v )

end</lang>

Mathematica

Brute force algorithm: <lang Mathematica>{pva,pwe,pvo}={3000,3/10,1/40}; {iva,iwe,ivo}={1800,2/10,3/200}; {gva,gwe,gvo}={2500,2,2/1000}; wemax=25; vomax=1/4; {pmax,imax,gmax}=Floor/@{Min[vomax/pvo,wemax/pwe],Min[vomax/ivo,wemax/iwe],Min[vomax/gvo,wemax/gwe]};

data=Flatten[Table[{{p,i,g}.{pva,iva,gva},{p,i,g}.{pwe,iwe,gwe},{p,i,g}.{pvo,ivo,gvo},{p,i,g}},{p,0,pmax},{i,0,imax},{g,0,gmax}],2]; data=Select[data,#2<=25&&#3<=1/4&]; First[SplitBy[Sort[data,First[#1]>First[#2]&],First]]</lang> gives back an array of the best solution(s), with each element being value, weight, volume, {number of vials, number of ampules, number of bars}: <lang Mathematica>{{54500,247/10,247/1000,{9,0,11}},{54500,124/5,247/1000,{6,5,11}},{54500,249/10,247/1000,{3,10,11}},{54500,25,247/1000,{0,15,11}}}</lang> if we call the three items by their first letters the best packings are: <lang Mathematica>p:9 i:0 v:11 p:6 i:5 v:11 p:3 i:10 v:11 p:0 i:15 v:11</lang> The volume for all of those is the same, the 'best' solution would be the one that has the least weight: that would the first solution.

Mathprog

<lang mathprog>/*Knapsack

 This model finds the integer optimal packing of a knapsack

 Nigel_Galloway
 January 9th., 2012

/

set Items; param weight{t in Items}; param value{t in Items}; param volume{t in Items};

var take{t in Items}, integer, >=0;

knap_weight : sum{t in Items} take[t] * weight[t] <= 25; knap_vol : sum{t in Items} take[t] * volume[t] <= 0.25;

maximize knap_value: sum{t in Items} take[t] * value[t];

data;

param : Items : weight value volume :=

        panacea          0.3     3000      0.025
        ichor            0.2	  1800	    0.015
        gold		  2.0	  2500      0.002

end;</lang> The solution produced is at Knapsack problem/Unbounded/Mathprog.

Modula-3

Translation of: Fortran

Note that unlike Fortran and C, Modula-3 does not do any hidden casting, which is why FLOAT and FLOOR are used. <lang modula3>MODULE Knapsack EXPORTS Main;

FROM IO IMPORT Put; FROM Fmt IMPORT Int, Real;

TYPE Bounty = RECORD

 value: INTEGER;
 weight, volume: REAL;

END;

VAR totalWeight, totalVolume: REAL;

   maxPanacea, maxIchor, maxGold, maxValue: INTEGER := 0;
   n: ARRAY [1..3] OF INTEGER;
   panacea, ichor, gold, sack, current: Bounty;

BEGIN

 panacea := Bounty{3000, 0.3, 0.025};
 ichor   := Bounty{1800, 0.2, 0.015};
 gold    := Bounty{2500, 2.0, 0.002};
 sack    := Bounty{0, 25.0, 0.25};

 maxPanacea := FLOOR(MIN(sack.weight / panacea.weight, sack.volume / panacea.volume));
 maxIchor := FLOOR(MIN(sack.weight / ichor.weight, sack.volume / ichor.volume));
 maxGold := FLOOR(MIN(sack.weight / gold.weight, sack.volume / gold.volume));

 FOR i := 0 TO maxPanacea DO
   FOR j := 0 TO maxIchor DO
     FOR k := 0 TO maxGold DO
       current.value := k * gold.value + j * ichor.value + i * panacea.value;
       current.weight := FLOAT(k) * gold.weight + FLOAT(j) * ichor.weight + FLOAT(i) * panacea.weight;
       current.volume := FLOAT(k) * gold.volume + FLOAT(j) * ichor.volume + FLOAT(i) * panacea.volume;
       IF current.weight > sack.weight OR current.volume > sack.volume THEN
         EXIT;
       END;
       IF current.value > maxValue THEN
         maxValue := current.value;
         totalWeight := current.weight;
         totalVolume := current.volume;
         n[1] := i; n[2] := j; n[3] := k;
       END;
     END;
   END;
 END;
 Put("Maximum value achievable is " & Int(maxValue) & "\n");
 Put("This is achieved by carrying " & Int(n[1]) & " panacea, " & Int(n[2]) & " ichor and " & Int(n[3]) & " gold items\n");
 Put("The weight of this carry is " & Real(totalWeight) & " and the volume used is " & Real(totalVolume) & "\n");

END Knapsack.</lang> Output:

Maximum value achievable is 54500
This is achieved by carrying 0 panacea, 15 ichor and 11 gold items
The weight of this carry is 25 and the volume used is 0.247

OCaml

This is a brute-force solution: it generates a list of every legal combination of items and then finds the best results: <lang ocaml>type bounty = { name:string; value:int; weight:float; volume:float }

let bounty n d w v = { name = n; value = d; weight = w; volume = v }

let items =

 [ bounty "panacea"  3000  0.3  0.025;
   bounty "ichor"    1800  0.2  0.015;
   bounty "gold"     2500  2.0  0.002; ]

let max_wgt = 25.0 and max_vol = 0.25

let itmax =

 let f it =
   let rec aux n =
     if float n *. it.weight >= max_wgt
     || float n *. it.volume >= max_vol
     then (n)
     else aux (succ n)
   in
   aux 0
 in
 List.map f items

let mklist n m =

 let rec aux i acc =
   if i > m then (List.rev acc)
   else aux (succ i) (i::acc)
 in
 aux n []

let comb_items = List.map (mklist 0) itmax

let combs ll =

 let f hd acc =
   List.concat
     (List.map (fun l -> List.map (fun v -> (v::l)) hd) acc)
 in
 List.fold_right f ll [[]]

let possibles = combs comb_items

let packs =

 let f l =
   let g (v, wgt, vol) n it =
     (v + n * it.value,
      wgt +. float n *. it.weight,
      vol +. float n *. it.volume)
   in
   List.fold_left2 g (0, 0.0, 0.0) l items
 in
 List.map f possibles

let packs = List.combine packs possibles

let results =

 let f (_, wgt, vol) = (wgt <= max_wgt && vol <= max_vol) in
 List.filter (fun v -> f(fst v)) packs

let best_results =

 let max_value = List.fold_left (fun v1 ((v2,_,_),_) -> max v1 v2) 0 results in
 List.filter (fun ((v,_,_),_) -> v = max_value) results

let items_name = List.map (fun it -> it.name) items

let print ((v, wgt, vol), ns) =

 Printf.printf "\
   Maximum value: %d \n \
   Total weight:  %g \n \
   Total volume:  %g \n \
   Containing:    " v wgt vol;
 let f n name = string_of_int n ^ " " ^ name in
 let ss = List.map2 f ns items_name in
 print_endline(String.concat ", " ss);
 print_newline()

let () = List.iter print best_results</lang>

outputs:

Maximum value: 54500 
 Total weight:  24.7 
 Total volume:  0.247 
 Containing:    9 panacea, 0 ichor, 11 gold

Maximum value: 54500 
 Total weight:  24.8 
 Total volume:  0.247 
 Containing:    6 panacea, 5 ichor, 11 gold

Maximum value: 54500 
 Total weight:  24.9 
 Total volume:  0.247 
 Containing:    3 panacea, 10 ichor, 11 gold

Maximum value: 54500 
 Total weight:  25 
 Total volume:  0.247 
 Containing:    0 panacea, 15 ichor, 11 gold

OOCalc

OpenOffice.org Calc has (several) linear solvers. To solve this task, first copy in the table from the task description, then add the extra columns:

Number: (How many chosen, n)
value of n
weight of n
volume of n

Add a TOTALS row to sum the value/weight/volume of n.

The sheet should then look like this:

Open the "Tools->Solver..." menu item and fill in the following items:

Options... (opens a separate popup window, then continue)

OK the solver options window leaving the Solver window open, then select solve to produce in seconds:

Oz

Using constraint propagation and branch and bound search: <lang oz>declare

 proc {Knapsack Sol}
    solution(panacea:P = {FD.decl}
             ichor:  I = {FD.decl}
             gold:   G = {FD.decl} ) = Sol
 in
                                          {Show 0#Sol}
     3 * P +  2 * I + 20 * G =<: 250      {Show 1#Sol}
    25 * P + 15 * I +  2 * G =<: 250      {Show 2#Sol}
    {FD.distribute naive Sol}             {Show d#Sol}
 end

 fun {Value solution(panacea:P ichor:I gold:G)}
    3000 * P + 1800 * I + 2500 * G
 end

 {System.showInfo "Search:"}
 [Best] = {SearchBest Knapsack proc {$ Old New}
                                  {Value Old} <: {Value New}
                               end}

in

 {System.showInfo "\nResult:"}
 {Show Best}
 {System.showInfo "total value: "#{Value Best}}</lang>

If you study the output, you see how the weight and volume equations automagically constrain the domain of the three variables. Afterwards SearchBest only has to evaluate 38 different combinations to find an optimal solution:

Search:
0#solution(gold:_{0#134217726} ichor:_{0#134217726} panacea:_{0#134217726})
1#solution(gold:_{0#12} ichor:_{0#125} panacea:_{0#83})
2#solution(gold:_{0#12} ichor:_{0#16} panacea:_{0#10})
d#solution(gold:0 ichor:0 panacea:0)
d#solution(gold:0 ichor:1 panacea:0)
d#solution(gold:0 ichor:2 panacea:0)
d#solution(gold:0 ichor:3 panacea:0)
d#solution(gold:0 ichor:4 panacea:0)
d#solution(gold:0 ichor:5 panacea:0)
d#solution(gold:0 ichor:6 panacea:0)
d#solution(gold:0 ichor:7 panacea:0)
d#solution(gold:0 ichor:8 panacea:0)
d#solution(gold:0 ichor:9 panacea:0)
d#solution(gold:0 ichor:10 panacea:0)
d#solution(gold:0 ichor:11 panacea:0)
d#solution(gold:0 ichor:12 panacea:0)
d#solution(gold:0 ichor:13 panacea:0)
d#solution(gold:0 ichor:14 panacea:0)
d#solution(gold:0 ichor:15 panacea:0)
d#solution(gold:0 ichor:16 panacea:0)
d#solution(gold:1 ichor:15 panacea:0)
d#solution(gold:1 ichor:16 panacea:0)
d#solution(gold:2 ichor:15 panacea:0)
d#solution(gold:2 ichor:16 panacea:0)
d#solution(gold:3 ichor:15 panacea:0)
d#solution(gold:3 ichor:16 panacea:0)
d#solution(gold:4 ichor:15 panacea:0)
d#solution(gold:4 ichor:16 panacea:0)
d#solution(gold:5 ichor:15 panacea:0)
d#solution(gold:5 ichor:16 panacea:0)
d#solution(gold:6 ichor:15 panacea:0)
d#solution(gold:7 ichor:14 panacea:0)
d#solution(gold:7 ichor:15 panacea:0)
d#solution(gold:8 ichor:14 panacea:0)
d#solution(gold:8 ichor:15 panacea:0)
d#solution(gold:9 ichor:14 panacea:0)
d#solution(gold:9 ichor:15 panacea:0)
d#solution(gold:10 ichor:14 panacea:0)
d#solution(gold:10 ichor:15 panacea:0)
d#solution(gold:11 ichor:14 panacea:0)
d#solution(gold:11 ichor:15 panacea:0)

Result:
solution(gold:11 ichor:15 panacea:0)
total value: 54500

Pascal

With ideas from C, Fortran and Modula-3. <lang pascal>Program Knapsack(output);

uses

 math;

type

 bounty = record
   value: longint;
   weight, volume: real;
 end;

const

 panacea: bounty = (value:3000; weight:  0.3; volume: 0.025);
 ichor:   bounty = (value:1800; weight:  0.2; volume: 0.015);
 gold:    bounty = (value:2500; weight:  2.0; volume: 0.002);
 sack:    bounty = (value:   0; weight: 25.0; volume: 0.25);

var

 totalweight, totalvolume: real;
 maxpanacea, maxichor, maxgold: longint;
 maxvalue: longint = 0;
 n: array [1..3] of longint;
 current: bounty;
 i, j, k: longint;

begin

 maxpanacea := round(min(sack.weight / panacea.weight, sack.volume / panacea.volume));
 maxichor   := round(min(sack.weight / ichor.weight,   sack.volume / ichor.volume));
 maxgold    := round(min(sack.weight / gold.weight,    sack.volume / gold.volume));

 for i := 0 to maxpanacea do
   for j := 0 to maxichor do
     for k := 0 to maxgold do
     begin
       current.value  := k * gold.value  + j * ichor.value  + i * panacea.value;
       current.weight := k * gold.weight + j * ichor.weight + i * panacea.weight;
       current.volume := k * gold.volume + j * ichor.volume + i * panacea.volume;
       if (current.value > maxvalue)      and

(current.weight <= sack.weight) and

          (current.volume <= sack.volume) then

begin

         maxvalue    := current.value;
         totalweight := current.weight;
         totalvolume := current.volume;
         n[1] := i;

n[2] := j; n[3] := k;

       end;
     end;

 writeln ('Maximum value achievable is ', maxValue);
 writeln ('This is achieved by carrying ', n[1], ' panacea, ', n[2], ' ichor and ', n[3], ' gold items');
 writeln ('The weight of this carry is ', totalWeight:6:3, ' and the volume used is ', totalVolume:6:4);

end.</lang> Output:

:> ./Knapsack
Maximum value achievable is 54500
This is achieved by carrying 0 panacea, 15 ichor and 11 gold items
The weight of this carry is 25.000 and the volume used is 0.2470

Perl

Dynamic programming solution. Before you ask, no, it's actually slower for the given data set. See the alternate data set. <lang perl>my (@names, @val, @weight, @vol, $max_vol, $max_weight, $vsc, $wsc);

if (1) { # change 1 to 0 for different data set

       @names  = qw(panacea    icor    gold);
       @val    = qw(3000       1800    2500);
       @weight = qw(3          2       20  );
       @vol    = qw(25         15      2   );
       $max_weight = 250;
       $max_vol = 250;
       $vsc = 1000;
       $wsc = 10;

} else { # with these numbers cache would have been useful

       @names  = qw(panacea    icor    gold    banana  monkey  );
       @val    = qw(17         11      5       3       34      );
       @weight = qw(14         3       2       2       10      );
       @vol    = qw(3          4       2       1       12      );
       $max_weight = 150;
       $max_vol = 100;
       $vsc = $wsc = 1;

}

my @cache; my ($hits, $misses) = (0, 0); sub solu {

       my ($i, $w, $v) = @_;
       return [0, []] if $i < 0;

       if ($cache[$i][$w][$v]) {
               $hits ++;
               return $cache[$i][$w][$v]
       }
       $misses ++;

       my $x = solu($i - 1, $w, $v);

       my ($w1, $v1);
       for (my $t = 1; ; $t++) {
               last if ($w1 = $w - $t * $weight[$i]) < 0;
               last if ($v1 = $v - $t * $vol[$i]) < 0;

               my $y = solu($i - 1, $w1, $v1);

               if ( (my $tmp = $y->[0] + $val[$i] * $t) > $x->[0] ) {
                       $x = [ $tmp, [ @{$y->[1]}, [$i, $t] ] ];
               }
       }

       $cache[$i][$w][$v] = $x

}

my $x = solu($#names, $max_weight, $max_vol); print "Max value $x->[0], with:\n",

       "    Item\tQty\tWeight   Vol    Value\n", '-'x 50, "\n";

my ($wtot, $vtot) = (0, 0); for (@{$x->[1]}) {

       my $i = $_->[0];
       printf  "    $names[$i]:\t% 3d  % 8d% 8g% 8d\n",
               $_->[1],
               $weight[$i] * $_->[1] / $wsc,
               $vol[$i] * $_->[1] / $vsc,
               $val[$i] * $_->[1];

       $wtot += $weight[$i] * $_->[1];
       $vtot += $vol[$i] * $_->[1];

} print "-" x 50, "\n"; printf " Total:\t % 8d% 8g% 8d\n",

       $wtot/$wsc, $vtot/$vsc, $x->[0];

print "\nCache hit: $hits\tmiss: $misses\n";</lang>

Output:

Max value 54500, with:
    Item        Qty     Weight   Vol    Value
--------------------------------------------------
    panacea:      9         2   0.225   27000
    gold:        11        22   0.022   27500
--------------------------------------------------
    Total:                 24   0.247   54500

Cache hit: 0    miss: 218

Cache info is not pertinent to this task, just some info.

Perl 6

Works with: rakudo version 2015-09-16

Brute force, looked a lot at the Ruby solution.

<lang perl6>class KnapsackItem {

 has $.volume;
 has $.weight;
 has $.value;
 has $.name;
 method new($volume,$weight,$value, $name) {
   self.bless(*, :$volume, :$weight, :$value, :$name)
 }

};

my KnapsackItem $panacea .= new: 0.025, 0.3, 3000, "panacea"; my KnapsackItem $ichor .= new: 0.015, 0.2, 1800, "ichor"; my KnapsackItem $gold .= new: 0.002, 2.0, 2500, "gold"; my KnapsackItem $maximum .= new: 0.25, 25, 0 , "max";

my $max_val = 0; my @solutions; my %max_items;

for $panacea, $ichor, $gold -> $item {

   %max_items{$item.name} = floor [min]
                           $maximum.volume / $item.volume,

$maximum.weight / $item.weight; }

for 0..%max_items<panacea>

      X 0..%max_items<ichor>
          X 0..%max_items<gold>
-> ($p, $i, $g)

{

 next if $panacea.volume * $p + $ichor.volume * $i + $gold.volume * $g > $maximum.volume;
 next if $panacea.weight * $p + $ichor.weight * $i + $gold.weight * $g > $maximum.weight;
 given $panacea.value * $p + $ichor.value * $i + $gold.value * $g {
   if $_ > $max_val { $max_val = $_; @solutions = (); }
   when $max_val    { @solutions.push: $[$p,$i,$g] }
 }

}

say "maximum value is $max_val\npossible solutions:"; say "panacea\tichor\tgold"; .join("\t").say for @solutions;</lang> Output:

maximum value is 54500
possible solutions:
panacea	ichor	gold
0	15	11
3	10	11
6	5	11
9	0	11

PicoLisp

Brute force solution <lang PicoLisp>(de *Items

  ("panacea"  3  25  3000)
  ("ichor"    2  15  1800)
  ("gold"    20   2  2500) )

(de knapsack (Lst W V)

  (when Lst
     (let X (knapsack (cdr Lst) W V)
        (if (and (ge0 (dec 'W (cadar Lst))) (ge0 (dec 'V (caddar Lst))))
           (maxi
              '((L) (sum cadddr L))
              (list
                 X
                 (cons (car Lst) (knapsack (cdr Lst) W V))
                 (cons (car Lst) (knapsack Lst W V)) ) )
           X ) ) ) )

(let K (knapsack *Items 250 250)

  (for (L K  L)
     (let (N 1  X)
        (while (= (setq X (pop 'L)) (car L))
           (inc 'N) )
        (apply tab X (4 2 8 5 5 7) N "x") ) )
  (tab (14 5 5 7) NIL (sum cadr K) (sum caddr K) (sum cadddr K)) )</lang>

Output:

  15 x   ichor    2   15   1800
  11 x    gold   20    2   2500
                250  247  54500

PowerShell

Works with: PowerShell version 3.0

<lang powershell># Define the items to pack $Item = @(

   [pscustomobject]@{ Name = 'panacea'; Unit = 'vials'  ; value = 3000; Weight = 0.3; Volume = 0.025 }
   [pscustomobject]@{ Name = 'ichor'  ; Unit = 'ampules'; value = 1800; Weight = 0.2; Volume = 0.015 }
   [pscustomobject]@{ Name = 'gold'   ; Unit = 'bars'   ; value = 2500; Weight = 2.0; Volume = 0.002 }
   )

Define our maximums

$MaxWeight = 25 $MaxVolume = 0.25

Set our default value to beat

$OptimalValue = 0

Iterate through the possible quantities of item 0, without going over the weight or volume limit

ForEach ( $Qty0 in 0..( [math]::Min( [math]::Truncate( $MaxWeight / $Item[0].Weight ), [math]::Truncate( $MaxVolume / $Item[0].Volume ) ) ) )

   {
   #  Calculate the remaining space
   $RemainingWeight = $MaxWeight - $Qty0 * $Item[0].Weight
   $RemainingVolume = $MaxVolume - $Qty0 * $Item[0].Volume

   #  Iterate through the possible quantities of item 1, without going over the weight or volume limit
   ForEach ( $Qty1 in 0..( [math]::Min( [math]::Truncate( $RemainingWeight / $Item[1].Weight ), [math]::Truncate( $RemainingVolume / $Item[1].Volume ) ) ) )
       {
       #  Calculate the remaining space
       $RemainingWeight2 = $RemainingWeight - $Qty1 * $Item[1].Weight
       $RemainingVolume2 = $RemainingVolume - $Qty1 * $Item[1].Volume

       #  Calculate the maximum quantity of item 2 for the remaining space, without going over the weight or volume limit
       $Qty2 = [math]::Min( [math]::Truncate( $RemainingWeight2 / $Item[2].Weight ), [math]::Truncate( $RemainingVolume2 / $Item[2].Volume ) )

       #  Calculate the total value of the items packed
       $TrialValue =   $Qty0 * $Item[0].Value +
                       $Qty1 * $Item[1].Value +
                       $Qty2 * $Item[2].Value

       #  Describe the trial solution
       $Solution  = "$Qty0 $($Item[0].Unit) of $($Item[0].Name), "
       $Solution += "$Qty1 $($Item[1].Unit) of $($Item[1].Name), and "
       $Solution += "$Qty2 $($Item[2].Unit) of $($Item[2].Name) worth a total of $TrialValue."

       #  If the trial value is higher than previous most valuable trial...
       If ( $TrialValue -gt $OptimalValue )
           {
           #  Set the new number to beat
           $OptimalValue = $TrialValue

           #  Overwrite the previous optimal solution(s) with the trial solution
           $Solutions  = @( $Solution )
           }

       #  Else if the trial value matches the previous most valuable trial...
      ElseIf ( $TrialValue -eq $OptimalValue )
           {
           #  Add the trial solution to the list of optimal solutions
           $Solutions += @( $Solution )
           }
       }
   }

Show the results

$Solutions</lang>

Output:

0 vials of panacea, 15 ampules of ichor, and 11 bars of gold worth a total of 54500.
3 vials of panacea, 10 ampules of ichor, and 11 bars of gold worth a total of 54500.
6 vials of panacea, 5 ampules of ichor, and 11 bars of gold worth a total of 54500.
9 vials of panacea, 0 ampules of ichor, and 11 bars of gold worth a total of 54500.

Prolog

Works with SWI-Prolog and library simplex written by Markus Triska. <lang Prolog>:- use_module(library(simplex)).

% tuples (name, Explantion, Value, weights, volume). knapsack :- L =[( panacea, 'Incredible healing properties', 3000, 0.3, 0.025), ( ichor, 'Vampires blood', 1800, 0.2, 0.015), ( gold , 'Shiney shiney', 2500, 2.0, 0.002)],

gen_state(S0), length(L, N), numlist(1, N, LN),

% to get statistics time((create_constraint_N(LN, L, S0, S1, [], LVa, [], LW, [], LVo), constraint(LW =< 25.0, S1, S2), constraint(LVo =< 0.25, S2, S3), maximize(LVa, S3, S4) )),

sformat(W0, A0, ['Nb']), sformat(W1, A1, ['Items']), sformat(W2, A2, ['Value']), sformat(W3, A33, ['Weigth']), sformat(W4, A44, ['Volume']), format('~w~w~w~w~w~n', [W0, W1,W2,W3,W4]),

print_results(S4, A0, A1, A2, A3, A4, L, LN, 0, 0, 0).

create_constraint_N([], [], S, S, LVa, LVa, LW, LW, LVo, LVo).

create_constraint_N([HN|TN], [(_, _,Va, W, Vo) | TL], S1, SF, LVa, LVaF, LW, LWF, LVo, LVoF) :- constraint(integral(x(HN)), S1, S2), constraint([x(HN)] >= 0, S2, S3), create_constraint_N(TN, TL, S3, SF, [Va * x(HN) | LVa], LVaF, [W * x(HN) | LW], LWF, [Vo * x(HN) | LVo], LVoF).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % compute_lenword([], N, N). compute_lenword([(Name, _, _, _, _)|T], N, NF):- atom_length(Name, L), ( L > N -> N1 = L; N1 = N), compute_lenword(T, N1, NF).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % print_results(_S, A0, A1, A2, A3, A4, [], [], VaM, WM, VoM) :- sformat(W0, A0, [' ']), sformat(W1, A1, [' ']), sformat(W2, A2, [VaM]), sformat(W3, A3, [WM]), sformat(W4, A4, [VoM]), format('~w~w~w~w~w~n', [W0, W1,W2,W3,W4]).

print_results(S, A0, A1, A2, A3, A4, [(Name, _, Va, W, Vo)|T], [N|TN], Va1, W1, Vo1) :- variable_value(S, x(N), X), ( X = 0 -> Va1 = Va2, W1 = W2, Vo1 = Vo2 ; sformat(S0, A0, [X]), sformat(S1, A1, [Name]), Vatemp is X * Va, Wtemp is X * W, Votemp is X * Vo, sformat(S2, A2, [Vatemp]), sformat(S3, A3, [Wtemp]), sformat(S4, A4, [Votemp]), format('~w~w~w~w~w~n', [S0,S1,S2,S3,S4]), Va2 is Va1 + Vatemp, W2 is W1 + Wtemp, Vo2 is Vo1 + Votemp ), print_results(S, A0, A1, A2, A3, A4, T, TN, Va2, W2, Vo2).</lang> Output :

 ?- knapsack.
% 145,319 inferences, 0.078 CPU in 0.079 seconds (99% CPU, 1860083 Lips)
Nb Items       Value    Weigth    Volume
15 ichor       27000      3.00     0.225
11 gold        27500     22.00     0.022
               54500     25.00     0.247
true

PureBasic

Translation of: Fortran

<lang PureBasic>Define.f TotalWeight, TotalVolyme Define.i maxPanacea, maxIchor, maxGold, maxValue Define.i i, j ,k Dim n.i(2)

Enumeration

 #Panacea
 #Ichor
 #Gold
 #Sack
 #Current

EndEnumeration

Structure Bounty

 value.i
 weight.f
 volyme.f

EndStructure

Dim Item.Bounty(4) CopyMemory(?panacea,@Item(#Panacea),SizeOf(Bounty)) CopyMemory(?ichor, @Item(#Ichor), SizeOf(Bounty)) CopyMemory(?gold, @Item(#gold), SizeOf(Bounty)) CopyMemory(?sack, @Item(#Sack), SizeOf(Bounty))

Procedure.f min(a.f, b.f)

 If aItem(#Sack)\weight Or Item(#Current)\volyme>Item(#Sack)\volyme
       Continue
     EndIf
     If Item(#Current)\value>maxValue
       maxValue=Item(#Current)\value
       TotalWeight=Item(#Current)\weight
       TotalVolyme=Item(#Current)\volyme
       n(#Panacea)=i: n(#Ichor)=j: n(#Gold)=k
     EndIf
   Next k
 Next j

Next i
If OpenConsole()
Define txt$ txt$="Maximum value achievable is "+Str(maxValue)+#CRLF$ txt$+"This is achieved by carrying "+Str(n(#Panacea))+" panacea, " txt$+Str(n(#Ichor))+" ichor and "+Str(n(#Gold))+" gold items."+#CRLF$ txt$+"The weight to carry is "+StrF(totalWeight,2) txt$+" and the volume used is "+StrF(TotalVolyme,2) PrintN(txt$) Print(#CRLF$+"Press Enter to quit"): Input()
EndIf
DataSection panacea:
Data.i 3000 Data.f 0.3, 0.025
ichor:
Data.i 1800 Data.f 0.2, 0.015
gold:
Data.i 2500 Data.f 2.0, 0.002
sack:
Data.i 0 Data.f 25.0, 0.25
EndDataSection</lang>
Outputs
Maximum value achievable is 54500 This is achieved by carrying 0 panacea, 15 ichor and 11 gold items The weight to carry is 25.00 and the volume used is 0.25 Press Enter to quit
Python
See Knapsack Problem/Python
R
Brute force method <lang r># Define consts weights <- c(panacea=0.3, ichor=0.2, gold=2.0) volumes <- c(panacea=0.025, ichor=0.015, gold=0.002) values <- c(panacea=3000, ichor=1800, gold=2500) sack.weight <- 25 sack.volume <- 0.25 max.items <- floor(pmin(sack.weight/weights, sack.volume/volumes))
Some utility functions
getTotalValue <- function(n) sum(n*values) getTotalWeight <- function(n) sum(n*weights) getTotalVolume <- function(n) sum(n*volumes) willFitInSack <- function(n) getTotalWeight(n) <= sack.weight && getTotalVolume(n) <= sack.volume
Find all possible combination, then eliminate those that won't fit in the sack
knapsack <- expand.grid(lapply(max.items, function(n) seq.int(0, n))) ok <- apply(knapsack, 1, willFitInSack) knapok <- knapsack[ok,]
Find the solutions with the highest value
vals <- apply(knapok, 1, getTotalValue) knapok[vals == max(vals),]</lang>
panacea ichor gold 2067 9 0 11 2119 6 5 11 2171 3 10 11 2223 0 15 11

Using Dynamic Programming
<lang r> Data_<-structure(list(item = c("Panacea", "Ichor", "Gold"), value = c(3000, 1800, 2500), weight = c(3, 2, 20), volume = c(25, 15, 2)), .Names = c("item", "value", "weight", "volume"), row.names = c(NA, 3L), class = "data.frame")
knapsack_volume<-function(Data, W, Volume, full_K) {
# Data must have the colums with names: item, value, weight and volume. K<-list() # hightest values K_item<-list() # itens that reach the hightest value K<-rep(0,W+1) # The position '0' K_item<-rep(,W+1) # The position '0' for(w in 1:W) { temp_w<-0 temp_item<- temp_value<-0 for(i in 1:dim(Data)[1]) # each row { wi<-Data$weight[i] # item i vi<- Data$value[i] item<-Data$item[i] volume_i<-Data$volume[i] if(wi<=w & volume_i <= Volume) { back<- full_KVolume-volume_i+1[w-wi+1] temp_wi<-vi + back
if(temp_w < temp_wi) { temp_value<-temp_wi temp_w<-temp_wi temp_item <- item } } } Kw+1<-temp_value K_itemw+1<-temp_item } return(list(K=K,Item=K_item)) }

Un_knapsack<-function(Data,W,V) { K<-list();K_item<-list() K1<-rep(0,W+1) #the line 0 K_item1<-rep(, W+1) #the line 0 for(v in 1:V) { best_volum_v<-knapsack_volume(Data, W, v, K) Kv+1<-best_volum_v$K K_itemv+1<-best_volum_v$Item }
return(list(K=data.frame(K),Item=data.frame(K_item,stringsAsFactors=F))) }
retrieve_info<-function(knapsack, Data) { W<-dim(knapsack$K)[1] itens<-c() col<-dim(knapsack$K)[2] selected_item<-knapsack$Item[W,col] while(selected_item!=) { selected_item<-knapsack$Item[W,col] if(selected_item!=) { selected_item_value<-Data[Data$item == selected_item,] W <- W - selected_item_value$weight itens<-c(itens,selected_item) col <- col - selected_item_value$volume } } return(itens) }
main_knapsack<-function(Data, W, Volume) { knapsack_result<-Un_knapsack(Data,W,Volume) items<-table(retrieve_info(knapsack_result, Data)) K<-knapsack_result$K[W+1, Volume+1] cat(paste('The Total profit is: ', K, '\n')) cat(paste('You must carry:', names(items), '(x',items, ') \n')) }
main_knapsack(Data_, 250, 250) </lang> <lang r> Output: The Total profit is: 54500 You must carry: Gold (x 11 ) You must carry: Panacea (x 9 ) </lang>
Racket
<lang racket>
lang racket
(struct item (name explanation value weight volume) #:prefab)
(define items
(list (item "panacea (vials of)" "Incredible healing properties" 3000 0.3 0.025) (item "ichor (ampules of)" "Vampires blood" 1800 0.2 0.015) (item "gold (bars)" "Shiney shiney" 2500 2.0 0.002)))
(define (fill-sack items volume-left weight-left sack sack-value)
(match items ['() (values (list sack) sack-value)] [(cons (and (item _ _ item-val weight volume) item) items) (define max-q-wgt (floor (/ weight-left weight))) (define max-q-vol (floor (/ volume-left volume))) (for/fold ([best (list sack)] [best-val sack-value]) ([n (exact-round (add1 (min max-q-vol max-q-wgt)))]) (define-values [best* best-val*] (fill-sack items (- volume-left (* n volume)) (- weight-left (* n weight)) (cons (cons n item) sack) (+ sack-value (* n item-val)))) (cond [(> best-val* best-val) (values best* best-val*)] [(= best-val* best-val) (values (append best best*) best-val*)] [else (values best best-val)]))]))
(define (display-sack sack total)
(for ([sk sack]) (define qty (car sk)) (define name (item-name (cdr sk))) (if (zero? qty) (printf "Leave ~a\n" name) (printf "Take ~a ~a\n" qty name))) (printf "GRAND TOTAL: ~a\n\n" total))
(call-with-values (λ() (fill-sack items 0.25 25 '() 0))
(λ(sacks total) (for ([s sacks]) (display-sack s total))))
</lang>

Output:
Take 11 gold (bars) Take 15 ichor (ampules of) Leave panacea (vials of) GRAND TOTAL: 54500 Take 11 gold (bars) Take 10 ichor (ampules of) Take 3 panacea (vials of) GRAND TOTAL: 54500 Take 11 gold (bars) Take 5 ichor (ampules of) Take 6 panacea (vials of) GRAND TOTAL: 54500 Take 11 gold (bars) Leave ichor (ampules of) Take 9 panacea (vials of) GRAND TOTAL: 54500
REXX
Translation of: Fortran
displays 1st solution
<lang rexx>/*REXX program solves the knapsack/unbounded problem: highest value, weight, and volume.*/
/* value weight volume */
maxPanacea=0 /* ═══════ ══════ ══════ */ maxIchor =0; panacea.$ = 3000 ; panacea.w = 0.3 ; panacea.v = 0.025 maxGold =0; ichor.$ = 1800 ; ichor.w = 0.2 ; ichor.v = 0.015 max$ =0; gold.$ = 2500 ; gold.w = 2 ; gold.v = 0.002 now. =0; sack.$ = 0 ; sack.w = 25 ; sack.v = 0.25
maxPanacea= min(sack.w / panacea.w, sack.v / panacea.v) maxIchor = min(sack.w / ichor.w, sack.v / ichor.v) maxGold = min(sack.w / gold.w, sack.v / gold.v)
do p=0 to maxPanacea do i=0 to maxIchor do g=0 to maxGold now.$= g * gold.$ + i * ichor.$ + p * panacea.$ now.w= g * gold.w + i * ichor.w + p * panacea.w now.v= g * gold.v + i * ichor.v + p * panacea.v if now.w > sack.w | now.v > sack.v then iterate if now.$ > max$ then do; maxP=p; maxI=i; maxG=g max$=now.$; maxW=now.w; maxV=now.v end end /*g (gold) */ end /*i (ichor) */ end /*p (panacea)*/
Ctot = maxP + maxI + maxG; L = length(Ctot) + 1 say ' panacea in sack:' right(maxP, L) say ' ichors in sack:' right(maxI, L) say ' gold items in sack:' right(maxG, L) say '════════════════════' copies("═", L) say 'carrying a total of:' right(cTot, L)
say left(, 40) "total value: " max$ / 1 say left(, 40) "total weight: " maxW / 1 say left(, 40) "total volume: " maxV / 1 /*stick a fork in it, we're all done. */</lang>
output
panacea in sack: 0 ichors in sack: 15 gold items in sack: 11 ════════════════════ ═══ carrying a total of: 26 total value: 54500 total weight: 25 total volume: 0.247
displays all solutions
<lang rexx>/*REXX program solves the knapsack/unbounded problem: highest value, weight, and volume.*/
maxPanacea=0 /* value weight volume */ maxIchor =0 /* ═══════ ═══════ ══════ */ maxGold =0; panacea.$ = 3000 ; panacea.w = 0.3 ; panacea.v = 0.025 max$ =0; ichor.$ = 1800 ; ichor.w = 0.2 ; ichor.v = 0.015 now. =0; gold.$ = 2500 ; gold.w = 2 ; gold.v = 0.002
=0; sack.$ = 0 ; sack.w = 25 ; sack.v = 0.25
L =0 maxPanacea= min(sack.w / panacea.w, sack.v / panacea.v) maxIchor = min(sack.w / ichor.w, sack.v / ichor.v) maxGold = min(sack.w / gold.w, sack.v / gold.v)
do p=0 to maxPanacea do i=0 to maxIchor do g=0 to maxGold now.$ = g * gold.$ + i * ichor.$ + p * panacea.$ now.w = g * gold.w + i * ichor.w + p * panacea.w now.v = g * gold.v + i * ichor.v + p * panacea.v if now.w > sack.w | now.v > sack.v then iterate i if now.$ > max$ then do; #=0; max$=now.$; end if now.$ = max$ then do; #=#+1; maxP.#=p; maxI.#=i; maxG.#=g max$.#=now.$; maxW.#=now.w; maxV.#=now.v L=max(L, length(p + i + g) ) end end /*g (gold) */ end /*i (ichor) */ end /*p (panacea)*/ L=L + 1 do j=1 for #; say; say copies('▒', 70) "solution" j say ' panacea in sack:' right(maxP.j, L) say ' ichors in sack:' right(maxI.j, L) say ' gold items in sack:' right(maxG.j, L) say '════════════════════' copies("═", L) say 'carrying a total of:' right(maxP.j + maxI.j + maxG.j, L) say left(, 40) "total value: " max$.j / 1 say left(, 40) "total weight: " maxW.j / 1 say left(, 40) "total volume: " maxV.j / 1 end /*j*/ /*stick a fork in it, we're all done. */</lang>
output
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ solution 1 panacea in sack: 0 ichors in sack: 15 gold items in sack: 11 ════════════════════ ═══ carrying a total of: 26 total value: 54500 total weight: 25 total volume: 0.247 ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ solution 2 panacea in sack: 3 ichors in sack: 10 gold items in sack: 11 ════════════════════ ═══ carrying a total of: 24 total value: 54500 total weight: 24.9 total volume: 0.247 ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ solution 3 panacea in sack: 6 ichors in sack: 5 gold items in sack: 11 ════════════════════ ═══ carrying a total of: 22 total value: 54500 total weight: 24.8 total volume: 0.247 ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ solution 4 panacea in sack: 9 ichors in sack: 0 gold items in sack: 11 ════════════════════ ═══ carrying a total of: 20 total value: 54500 total weight: 24.7 total volume: 0.247
Ruby

Brute force method,

Translation of: Tcl
<lang ruby>KnapsackItem = Struct.new(:volume, :weight, :value) panacea = KnapsackItem.new(0.025, 0.3, 3000) ichor = KnapsackItem.new(0.015, 0.2, 1800) gold = KnapsackItem.new(0.002, 2.0, 2500) maximum = KnapsackItem.new(0.25, 25, 0)
max_items = {} for item in [panacea, ichor, gold]
max_items[item] = [(maximum.volume/item.volume).to_i, (maximum.weight/item.weight).to_i].min
end
maxval = 0 solutions = []
0.upto(max_items[ichor]) do |i|
0.upto(max_items[panacea]) do |p| 0.upto(max_items[gold]) do |g| break if i*ichor.weight + p*panacea.weight + g*gold.weight > maximum.weight break if i*ichor.volume + p*panacea.volume + g*gold.volume > maximum.volume val = i*ichor.value + p*panacea.value + g*gold.value if val > maxval maxval = val solutions = i, p, g elsif val == maxval solutions << [i, p, g] end end end
end
puts "The maximal solution has value #{maxval}" solutions.each do |i, p, g|
printf " ichor=%2d, panacea=%2d, gold=%2d -- weight:%.1f, volume=%.3f\n", i, p, g, i*ichor.weight + p*panacea.weight + g*gold.weight, i*ichor.volume + p*panacea.volume + g*gold.volume
end</lang>

Output:
The maximal solution has value 54500 ichor= 0, panacea= 9, gold=11 -- weight:24.7, volume=0.247 ichor= 5, panacea= 6, gold=11 -- weight:24.8, volume=0.247 ichor=10, panacea= 3, gold=11 -- weight:24.9, volume=0.247 ichor=15, panacea= 0, gold=11 -- weight:25.0, volume=0.247
SAS
This is yet another brute force solution. <lang SAS>data one;
wtpanacea=0.3; wtichor=0.2; wtgold=2.0; volpanacea=0.025; volichor=0.015; volgold=0.002; valpanacea=3000; valichor=1800; valgold=2500; maxwt=25; maxvol=0.25;
/* we can prune the possible selections */ maxpanacea = floor(min(maxwt/wtpanacea, maxvol/volpanacea)); maxichor = floor(min(maxwt/wtichor, maxvol/volichor)); maxgold = floor(min(maxwt/wtgold, maxvol/volgold)); do i1 = 0 to maxpanacea; do i2 = 0 to maxichor; do i3 = 0 to maxgold; panacea = i1; ichor=i2; gold=i3; output; end; end; end;
run; data one; set one;
vals = valpanacea*panacea + valichor*ichor + valgold*gold; totalweight = wtpanacea*panacea + wtichor*ichor + wtgold*gold; totalvolume = volpanacea*panacea + volichor*ichor + volgold*gold; if (totalweight le maxwt) and (totalvolume le maxvol);
run; proc sort data=one;
by descending vals;
run; proc print data=one (obs=4);
var panacea ichor gold vals;
run;</lang> Output:
Obs panacea ichor gold vals 1 0 15 11 54500 2 3 10 11 54500 3 6 5 11 54500 4 9 0 11 54500
Use SAS/OR: <lang sas>/* create SAS data set */ data mydata;
input Item $1-19 Value weight Volume; datalines;
panacea (vials of) 3000 0.3 0.025 ichor (ampules of) 1800 0.2 0.015 gold (bars) 2500 2.0 0.002

/* call OPTMODEL procedure in SAS/OR */ proc optmodel;
/* declare sets and parameters, and read input data */ set <str> ITEMS; num value {ITEMS}; num weight {ITEMS}; num volume {ITEMS}; read data mydata into ITEMS=[item] value weight volume;
/* declare variables, objective, and constraints */ var NumSelected {ITEMS} >= 0 integer; max TotalValue = sum {i in ITEMS} value[i] * NumSelected[i]; con WeightCon: sum {i in ITEMS} weight[i] * NumSelected[i] <= 25; con VolumeCon: sum {i in ITEMS} volume[i] * NumSelected[i] <= 0.25;
/* call mixed integer linear programming (MILP) solver */ solve;
/* print optimal solution */ print TotalValue; print NumSelected;
/* to get all optimal solutions, call CLP solver instead */ solve with CLP / findallsolns;
/* print all optimal solutions */ print TotalValue; for {s in 1.._NSOL_} print {i in ITEMS} NumSelected[i].sol[s];
quit;</lang>
MILP solver output:
TotalValue 54500 [1] NumSelected gold (bars) 11 ichor (ampules of) 0 panacea (vials of) 9
CLP solver output:
TotalValue 54500 [1] gold (bars) 11 ichor (ampules of) 15 panacea (vials of) 0 [1] gold (bars) 11 ichor (ampules of) 10 panacea (vials of) 3 [1] gold (bars) 11 ichor (ampules of) 5 panacea (vials of) 6 [1] gold (bars) 11 ichor (ampules of) 0 panacea (vials of) 9
Scala
<lang Scala>import scala.annotation.tailrec
case class Item(name: String, value: Int, weight: Double, volume: Double)
val items = List(
Item("panacea", 3000, 0.3, 0.025), Item("ichor", 1800, 0.2, 0.015), Item("gold", 2500, 2.0, 0.002))
val (maxWeight, maxVolume) = (25, 0.25)
def show(is: List[Item]) =
(items.map(_.name) zip items.map(i => is.count(_ == i))).map { case (i, c) => s"$i: $c" }.mkString(", ")
case class Knapsack(items: List[Item]) {
def value = items.foldLeft(0)(_ + _.value) def weight = items.foldLeft(0.0)(_ + _.weight) def volume = items.foldLeft(0.0)(_ + _.volume) def isFull = !((weight <= maxWeight) && (volume <= maxVolume)) override def toString = s"[${show(items)} | value: $value, weight: $weight, volume: $volume]"
}
def fill(knapsack: Knapsack): List[Knapsack] =
items.map(i => Knapsack(i :: knapsack.items))
//cause brute force def distinct(list: List[Knapsack]) =
list.map(k => Knapsack(k.items.sortBy(_.name))).distinct
@tailrec def f(notPacked: List[Knapsack], packed: List[Knapsack]): List[Knapsack] =
notPacked match { case Nil => packed.sortBy(_.value).takeRight(4) case _ => val notFull = distinct(notPacked.flatMap(fill)).filterNot(_.isFull) f(notFull, notPacked ::: packed) }
f(items.map(i => Knapsack(List(i))), Nil).foreach(println)</lang> Output:
panacea: 0, ichor: 15, gold: 11 | value: 54500, weight: 24.99999999999999, volume: 0.2470000000000001] [panacea: 3, ichor: 10, gold: 11 | value: 54500, weight: 24.899999999999995, volume: 0.24700000000000003] [panacea: 6, ichor: 5, gold: 11 | value: 54500, weight: 24.8, volume: 0.24699999999999997] [panacea: 9, ichor: 0, gold: 11 | value: 54500, weight: 24.700000000000006, volume: 0.24699999999999997]
Seed7
<lang seed7>$ include "seed7_05.s7i";
include "float.s7i";
const type: bounty is new struct
var integer: value is 0; var float: weight is 0.0; var float: volume is 0.0; end struct;
const func bounty: bounty (in integer: value, in float: weight, in float: volume) is func
result var bounty: bountyVal is bounty.value; begin bountyVal.value := value; bountyVal.weight := weight; bountyVal.volume := volume; end func;
const proc: main is func
local const bounty: panacea is bounty(3000, 0.3, 0.025); const bounty: ichor is bounty(1800, 0.2, 0.015); const bounty: gold is bounty(2500, 2.0, 0.002); const bounty: sack is bounty(0, 25.0, 0.25); const integer: maxPanacea is trunc(min(sack.weight / panacea.weight, sack.volume / panacea.volume)); const integer: maxIchor is trunc(min(sack.weight / ichor.weight, sack.volume / ichor.volume)); const integer: maxGold is trunc(min(sack.weight / gold.weight, sack.volume / gold.volume)); var bounty: current is bounty.value; var bounty: best is bounty.value; var array integer: bestAmounts is 3 times 0; var integer: numPanacea is 0; var integer: numIchor is 0; var integer: numGold is 0; begin for numPanacea range 0 to maxPanacea do for numIchor range 0 to maxIchor do for numGold range 0 to maxGold do current.value := numGold * gold.value + numIchor * ichor.value + numPanacea * panacea.value; current.weight := flt(numGold) * gold.weight + flt(numIchor) * ichor.weight + flt(numPanacea) * panacea.weight; current.volume := flt(numGold) * gold.volume + flt(numIchor) * ichor.volume + flt(numPanacea) * panacea.volume; if current.value > best.value and current.weight <= sack.weight and current.volume <= sack.volume then best := current; bestAmounts := [] (numPanacea, numIchor, numGold); end if; end for; end for; end for; writeln("Maximum value achievable is " <& best.value); writeln("This is achieved by carrying " <& bestAmounts[1] <& " panacea, " <& bestAmounts[2] <& " ichor and " <& bestAmounts[3] <& " gold items"); writeln("The weight of this carry is " <& best.weight <& " and the volume used is " <& best.volume digits 4); end func;</lang>
Output:
Maximum value achievable is 54500 This is achieved by carrying 0 panacea, 15 ichor and 11 gold items The weight of this carry is 25.0 and the volume used is 0.2470
Sidef
Translation of: Perl
<lang ruby>struct KnapsackItem {
Number volume, Number weight, Number value, String name,
}
var items = [
KnapsackItem(25, 3, 3000, "panacea") KnapsackItem(15, 2, 1800, "ichor" ) KnapsackItem( 2, 20, 2500, "gold" )
]
var (
max_weight = 250, max_vol = 250, vsc = 1000, wsc = 10
)
func solve(i, w, v) is cached {
return [0, []] if i.is_neg;
var x = solve(i.dec, w, v);
var (w1, v1); Inf.times { |t| var item = items[i]; break if ((w1 = (w - t*item.weight)).is_neg) break if ((v1 = (v - t*item.volume)).is_neg)
var y = solve(i.dec, w1, v1); if ((var tmp = (y[0] + t*item.value)) > x[0]) { x = [tmp, [y[1]..., [i, t]]]; } }
return x
}
var x = solve(items.end, max_weight, max_vol)
print <<"EOT" Max value #{x[0]}, with:
Item Qty Weight Vol Value
{"-" * 50}
EOT
var (wtot=0, vtot=0); x[1].each { |s|
var item = items[s[0]]; " #{item.name}:\t% 3d % 8d% 8g% 8d\n".printf( s[1], item.weight * s[1] / wsc, item.volume * s[1] / vsc, item.value * s[1] ); wtot += (item.weight * s[1]); vtot += (item.volume * s[1]);
}
print <<"EOT"
{"-" * 50}
Total:\t #{"%8d%8g%8d" % (wtot/wsc, vtot/vsc, x[0])}
EOT</lang>

Output:
Max value 54500, with: Item Qty Weight Vol Value -------------------------------------------------- panacea: 9 2 0.225 27000 gold: 11 22 0.022 27500 -------------------------------------------------- Total: 24 0.247 54500
Tcl
The following code uses brute force, but that's tolerable as long as it takes just a split second to find all 4 solutions. The use of arrays makes the quote quite legible: <lang Tcl>#!/usr/bin/env tclsh proc main argv {
array set value {panacea 3000 ichor 1800 gold 2500} array set weight {panacea 0.3 ichor 0.2 gold 2.0 max 25} array set volume {panacea 0.025 ichor 0.015 gold 0.002 max 0.25}
foreach i {panacea ichor gold} { set max($i) [expr {min(int($volume(max)/$volume($i)), int($weight(max)/$weight($i)))}] } set maxval 0 for {set i 0} {$i < $max(ichor)} {incr i} { for {set p 0} {$p < $max(panacea)} {incr p} { for {set g 0} {$g < $max(gold)} {incr g} { if {$i*$weight(ichor) + $p*$weight(panacea) + $g*$weight(gold) > $weight(max)} continue if {$i*$volume(ichor) + $p*$volume(panacea) + $g*$volume(gold) > $volume(max)} continue set val [expr {$i*$value(ichor)+$p*$value(panacea)+$g*$value(gold)}] if {$val == $maxval} { lappend best [list i $i p $p g $g] } elseif {$val > $maxval} { set maxval $val set best [list [list i $i p $p g $g]] } } } } puts "maxval: $maxval, best: $best"
} main $argv</lang>
$ time tclsh85 /Tcl/knapsack.tcl maxval: 54500, best: {i 0 p 9 g 11} {i 5 p 6 g 11} {i 10 p 3 g 11} {i 15 p 0 g 11} real 0m0.188s user 0m0.015s sys 0m0.015s
Ursala
The algorithm is to enumerate all packings with up to the maximum of each item, filter them by the volume and weight restrictions, partition the remaining packings by value, and search for the maximum value class. <lang Ursala>#import nat
import flo
vol = iprod/<0.025,0.015,0.002>+ float* val = iprod/<3000.,1800.,2500.>+ float* wgt = iprod/<0.3,0.2,2.0>+ float*
packings = ~&lrlrNCCPCS ~&K0=> iota* <11,17,13>
solutions = fleq$^rS&hl |=&l ^(val,~&)* (fleq\25.+ wgt)*~ (fleq\0.25+ vol)*~ packings
cast %nmL
human_readable = ~&p/*<'panacea','ichor','gold'> solutions</lang> output:
< <'panacea': 0,'ichor': 15,'gold': 11>, <'panacea': 3,'ichor': 10,'gold': 11>, <'panacea': 6,'ichor': 5,'gold': 11>, <'panacea': 9,'ichor': 0,'gold': 11>>
Visual Basic
See: Knapsack Problem/Visual Basic
The above Link contains a longer version (which perhaps runs a bit faster), whilst the one below is focussing more on expressing/solving the problem in less lines of code. <lang vb>Function Min(E1, E2): Min = IIf(E1 < E2, E1, E2): End Function 'small Helper-Function
Sub Main() Const Value = 0, Weight = 1, Volume = 2, PC = 3, IC = 4, GC = 5 Dim P&, I&, G&, A&, M, Cur(Value To Volume) Dim S As New Collection: S.Add Array(0) '<- init Solutions-Coll.
Const SackW = 25, SackV = 0.25 Dim Panacea: Panacea = Array(3000, 0.3, 0.025) Dim Ichor: Ichor = Array(1800, 0.2, 0.015) Dim Gold: Gold = Array(2500, 2, 0.002)
For P = 0 To Int(Min(SackW / Panacea(Weight), SackV / Panacea(Volume))) For I = 0 To Int(Min(SackW / Ichor(Weight), SackV / Ichor(Volume))) For G = 0 To Int(Min(SackW / Gold(Weight), SackV / Gold(Volume))) For A = Value To Volume: Cur(A) = G * Gold(A) + I * Ichor(A) + P * Panacea(A): Next If Cur(Value) >= S(1)(Value) And Cur(Weight) <= SackW And Cur(Volume) <= SackV Then _ S.Add Array(Cur(Value), Cur(Weight), Cur(Volume), P, I, G), , 1 Next G, I, P Debug.Print "Value", "Weight", "Volume", "PanaceaCount", "IchorCount", "GoldCount" For Each M In S '<- enumerate the Attributes of the Maxima If M(Value) = S(1)(Value) Then Debug.Print M(Value), M(Weight), M(Volume), M(PC), M(IC), M(GC) Next
End Sub</lang>

Output:

Value Weight Volume PanaceaCount IchorCount GoldCount 54500 24.7 0.247 9 0 11 54500 24.8 0.247 6 5 11 54500 24.9 0.247 3 10 11 54500 25 0.247 0 15 11
zkl
Translation of: D
<lang zkl>panacea:=T(3000, 0.3, 0.025); // (value,weight,volume) ichor :=T(1800, 0.2, 0.015); gold :=T(2500, 2.0, 0.002); sack :=T( 0, 25.0, 0.250); const VAL=0, W=1, VOL=2;
maxes:=T(panacea,ichor,gold)
.apply('wrap(t){ (sack[W]/t[W]).min(sack[VOL]/t[VOL]).toInt().walker() });
best:=Utils.Helpers.cprod3(maxes.xplode())
.apply('wrap(t){ T(T(panacea[VAL]*t[0] + ichor[VAL]*t[1] + gold[VAL]*t[2], panacea[W] *t[0] + ichor[W] *t[1] + gold[W] *t[2], panacea[VOL]*t[0] + ichor[VOL]*t[1] + gold[VOL]*t[2]), t) }) .filter('wrap(t){ t[0][W]<=sack[W] and t[0][VOL]<=sack[VOL] }) .reduce(fcn(a,b){ a[0][VAL] > b[0][VAL] and a or b });
println("Maximum value achievable is %,d".fmt(best[0][VAL])); println(("This is achieved by carrying (one solution):"
" %d panacea, %d ichor and %d gold").fmt(best[1].xplode()));
println("The weight to carry is %4.1f and the volume used is %5.3f"
.fmt(best[0][1,*].xplode()));</lang>
cprod3 is the Cartesian product of three lists or iterators.

Output:
Maximum value achievable is 54,500 This is achieved by carrying (one solution): 9 panacea, 0 ichor and 11 gold The weight to carry is 24.7 and the volume used is 0.247

@@ Line 1: / Line 1: @@
 {{task|Classic CS problems and programs}}
-See also: [[Knapsack problem/Bounded]], [[Knapsack problem/0-1]]
-A traveller gets diverted  and has to make an unscheduled stop in what turns out to be Shangri La. Opting to leave, he is allowed to take as much as he likes of the following items, so long as it will fit in his knapsack, and he can carry it.
+A traveler gets diverted and has to make an unscheduled stop in what turns out to be Shangri La. &nbsp; Opting to leave, he is allowed to take as much as he likes of the following items, so long as it will fit in his knapsack, and he can carry it.
-He knows that he can carry no more than 25 'weights'  in total; and that the capacity of his knapsack is 0.25 'cubic lengths'.
+He knows that he can carry no more than &nbsp; 25 &nbsp; 'weights' in total; &nbsp; and that the capacity of his knapsack is &nbsp; 0.25 &nbsp; 'cubic lengths'.
-Looking just above the bar codes on the items he finds their weights and volumes. He digs out his recent copy of a  financial paper and gets the value of each item.
+Looking just above the bar codes on the items he finds their weights and volumes. &nbsp; He digs out his recent copy of a  financial paper and gets the value of each item.
 <table
  style="text-align: left; width: 80%;" border="4"
@@ Line 46: / Line 46: @@
  nowrap="nowrap" valign="middle">&lt;=25</td><td
  style="background-color: rgb(255, 204, 255);" align="left"
- nowrap="nowrap" valign="middle">&lt;=0.25&nbsp;</td></tr></table>
+ nowrap="nowrap" valign="middle">&lt;=0.25&nbsp;</td></tr>
+</table>
+<br>
 He can only take whole units of any item, but there is much more of any item than he could ever carry
-'''How many of each item does he take to maximise the value of items he is carrying away with him?'''
+;Task:
-Note:
+Show how many of each item does he take to maximize the value of items he is carrying away with him.
-# There are four solutions that maximise the value taken. Only one ''need'' be given.
+;Note:
+* &nbsp; There are four solutions that maximize the value taken. &nbsp; Only one ''need'' be given.
 <!-- All solutions
@@ Line 65: / Line 71: @@
 # (9, 0, 11) also minimizes weight and volume within the limits of calculation
 -->
+;Related tasks:
+* &nbsp; [[Knapsack problem/Bounded]]
+* &nbsp; [[Knapsack problem/0-1]]
+<br><br>
 =={{header|Ada}}==