# Knapsack problem/Unbounded

Knapsack problem/Unbounded
You are encouraged to solve this task according to the task description, using any language you may know.

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

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.

## 360 Assembly

Translation of: Visual Basic

The program uses two ASSIST macros (XDECO,XPRNT) to keep the code as short as possible.

`*        Knapsack problem/Unbounded   04/02/2017KNAPSACK CSECT         USING  KNAPSACK,R13       base register         B      72(R15)            skip savearea         DC     17F'0'             savearea         STM    R14,R12,12(R13)    prolog         ST     R13,4(R15)         " <-         ST     R15,8(R13)         " ->         LR     R13,R15            " addressability         MVC    S,=F'0'            s(1,kva)=0;         LA     R11,0              ns=0         LA     R1,KW              kw         SLA    R1,2               *4         L      R2,PANACEA-4(R1)   panacea(kw)         L      R4,SACKW           sackw         SRDA   R4,32              ~         DR     R4,R2              sackw/panacea(kw)         ST     R5,XP              xp=sackw/panacea(kw)         LA     R1,KV              kv         SLA    R1,2               *4         L      R2,PANACEA-4(R1)   panacea(kv)         L      R4,SACKV           sackv         SRDA   R4,32              ~         DR     R4,R2              r5=sackv/panacea(kv)         C      R5,XP              if r5<xp         BNL    EMINXP         ST     R5,XP     xp=min(sackw/panacea(kw),sackv/panacea(kv))EMINXP   LA     R1,KW              kw         SLA    R1,2               *4         L      R2,ICHOR-4(R1)     ichor(kw)         L      R4,SACKW           sackw         SRDA   R4,32              ~         DR     R4,R2              sackw/ichor(kw)         ST     R5,XI              xi=sackw/ichor(kw)         LA     R1,KV              kv         SLA    R1,2               *4         L      R2,ICHOR-4(R1)     ichor(kv)         L      R4,SACKV           sackv         SRDA   R4,32              ~         DR     R4,R2              r5=sackv/ichor(kv)         C      R5,XI              if r5<xi         BNL    EMINXI         ST     R5,XI     xi=min(sackw/ichor(kw),sackv/ichor(kv))EMINXI   LA     R1,KW              kw         SLA    R1,2               *4         L      R2,GOLD-4(R1)      gold(kw)         L      R4,SACKW           sackw         SRDA   R4,32              ~         DR     R4,R2              sackw/gold(kw)         ST     R5,XG              xg=sackw/gold(kw)         LA     R1,KV              kv         SLA    R1,2               *4         L      R2,GOLD-4(R1)      gold(kv)         L      R4,SACKV           sackv         SRDA   R4,32              ~         DR     R4,R2              r5=sackv/gold(kv)         C      R5,XG              if r5<xg         BNL    EMINXG         ST     R5,XG     xg=min(sackw/gold(kw),sackv/gold(kv))EMINXG   SR     R10,R10            ip=0LOOPIP   C      R10,XP             do ip=0 to xp         BH     ELOOPIP         SR     R9,R9              ii=0LOOPII   C      R9,XI              do ii=0 to xi         BH     ELOOPII         SR     R8,R8              ig=0LOOPIG   C      R8,XG              do ig=0 to xg         BH     ELOOPIG         LA     R7,KVA             m=kvaLOOPM    C      R7,=A(KV)          do m=kva to kv         BH     ELOOPM         LR     R1,R7              m         SLA    R1,2               *4         LR     R5,R8              ig         M      R4,GOLD-4(R1)      *gold(m)         LR     R2,R5              r2=ig*gold(m)         LR     R5,R9              ii         M      R4,ICHOR-4(R1)     *ichor(m)         AR     R2,R5              r2=ig*gold(m)+ii*ichor(m)         LR     R5,R10             ip         M      R4,PANACEA-4(R1)   *panacea(m)         AR     R2,R5              r2=r2+ip*panacea(m)         ST     R2,CUR-4(R1)       cur(m)=r2         LA     R7,1(R7)           m=m+1         B      LOOPMELOOPM   LA     R1,KVA             kva         SLA    R1,2               *4         L      R2,CUR-4(R1)       cur(kva)         C      R2,S-4(R1)         if cur(kva)>=s(1,kva)         BL     ENDIF         LA     R1,KW              kw         SLA    R1,2               *4         L      R2,CUR-4(R1)       cur(kw)         C      R2,SACKW           if cur(kw)<=sackw         BH     ENDIF         LA     R1,KV              kv         SLA    R1,2               *4         L      R2,CUR-4(R1)       cur(kv)         C      R2,SACKV           if cur(kv)<=sackv         BH     ENDIF         LR     R6,R11             j=nsLOOPJ    C      R6,=F'1'           do j=ns to 1 by -1         BL     ELOOPJ         LR     R1,R6              j         MH     R1,=H'24'          *24         LA     R2,S(R1)           s(j+1,1)         LA     R3,S-24(R1)        s(j,1)         MVC    0(24,R2),0(R3)     s(j+1,*)=s(j,*)         BCTR   R6,0               j=j-1         B      LOOPJELOOPJ   LA     R1,KVA             kva         SLA    R1,2               *4         L      R2,CUR-4(R1)       cur(kva)         ST     R2,S-4(R1)         s(1,kva)=cur(kva)         LA     R1,KW              kw         SLA    R1,2               *4         L      R2,CUR-4(R1)       cur(kw)         ST     R2,S-4(R1)         s(1,kw)=cur(kw)         LA     R1,KV              kv         SLA    R1,2               *4         L      R2,CUR-4(R1)       cur(kv)         ST     R2,S-4(R1)         s(1,kv)=cur(kv)         LA     R1,KP              kp         SLA    R1,2               *4         ST     R10,S-4(R1)        s(1,kp)=ip         LA     R1,KI              ki         SLA    R1,2               *4         ST     R9,S-4(R1)         s(1,ki)=ii         LA     R1,KG              kg         SLA    R1,2               *4         ST     R8,S-4(R1)         s(1,kg)=ig         L      R2,S               r2=s(1,1)         C      R2,S+24            if s(1,1)>s(2,1)         BNH    ELSE         LA     R11,1              ns=1         B      ENDIFELSE     LA     R11,1(R11)         ns+1ENDIF    LA     R8,1(R8)           ig=ig+1         B      LOOPIGELOOPIG  LA     R9,1(R9)           ii=ii+1         B      LOOPIIELOOPII  LA     R10,1(R10)         ip=ip+1         B      LOOPIPELOOPIP  XPRNT  TITLE,72         LA     R6,1               j=1         LA     R3,S-4             [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */LOOPJP   CR     R6,R11             do j=1 to ns         BH     ELOOPJP         LA     R3,4(R3)           ++         L      R1,0(R3)           s(j,kva)         XDECO  R1,PG              edit         LA     R3,4(R3)           ++         L      R1,0(R3)           s(j,kw)         XDECO  R1,PG+12           edit         LA     R3,4(R3)           ++         L      R1,0(R3)           s(j,kv)         XDECO  R1,PG+24           edit         MVC    PG+20(2),PG+21     shift         MVI    PG+22,C'.'         decimal point         LA     R3,4(R3)           ++         L      R1,0(R3)           s(j,kp)         XDECO  R1,PG+36           edit         MVC    PG+31(2),=C'0.'    decimal point         LA     R3,4(R3)           ++         L      R1,0(R3)           s(j,ki)         XDECO  R1,PG+48           edit         LA     R3,4(R3)           ++         L      R1,0(R3)           s(j,kg)         XDECO  R1,PG+60           edit         XPRNT  PG,L'PG            print buffer         LA     R6,1(R6)           j=j+1         B      LOOPJPELOOPJP  L      R13,4(0,R13)       epilog          LM     R14,R12,12(R13)    " restore         XR     R15,R15            " rc=0         BR     R14                exitKVA      EQU    1KW       EQU    2KV       EQU    3KP       EQU    4KI       EQU    5KG       EQU    6SACKW    DC     F'250'SACKV    DC     F'250'PANACEA  DC     F'3000',F'3',F'25'ICHOR    DC     F'1800',F'2',F'15'GOLD     DC     F'2500',F'20',F'2'XP       DS     FXI       DS     FXG       DS     FCUR      DS     3FS        DS     60FTITLE    DC     CL36'       Value      Weight      Volume'         DC     CL36'     Panacea       Ichor        Gold'PG       DS     CL72         YREGS         END    KNAPSACK`
Output:
```       Value      Weight      Volume     Panacea       Ichor        Gold
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       0.247           0          15          11
```

Translation of: Python
`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;`

## ALGOL 68

Translation of: Python
`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))`
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.

`Item  = Panacea,Ichor,GoldValue = 3000,1800,2500Weight= 3,2,20                         ; *10Volume= 25,15,2                        ; *1000 StringSplit I, Item,  `,               ; Put input in arraysStringSplit W, Weight,`,StringSplit \$, Value, `,StringSplit V, Volume,`, SetFormat Float, 0.3W := 250, V := 250, sW:=.1, sV:=.001   ; limits for the total, scale factorsp := -1, Wp := -W1, Vp := -V1          ; initial valuesWhile (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%`

## 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; `

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.

`#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;} `
Output:
```9 panacea
0 ichor
11 gold
best value: 54500

```

## C_sharp

`/*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);        }    }}`

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

`(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)))`

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:

`(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])))))`

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.

`(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")))`

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):

`(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)))`

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.

`(defun fill-knapsack (items max-volume max-weight)  "Items is a list of lists of the form (name value weight volume) where weightand value are integers. max-volume and max-weight, also integers, are themaximum volume and weight of the knapsack. fill-knapsack returns a list of theform (total-value inventory total-volume total-weight) where total-value is thetotal-value of a knapsack packed with inventory (a list whose elements areelements of items), and total-weight and total-volume are the total weights andvolumes 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))))))))`

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
`void main() @safe [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */*/ {    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);}`
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.

`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..\$]);}`

## 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.

`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])`

## EchoLisp

Use a cache, and multiply by 10^n to get an integer problem.

` (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  `

## 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 `
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.
```

## Elixir

Brute Force:

`defmodule Item do  defstruct volume: 0.0, weight: 0.0, value: 0  def new(volume, weight, value) do    %__MODULE__{volume: volume, weight: weight, value: value}  endend defmodule Knapsack do  def solve_unbounded(items, maximum) do    {max_volume, max_weight} = {maximum.volume, maximum.weight}    max_items = Enum.map(items, fn {name,item} ->      {name, trunc(min(max_volume / item.volume, max_weight / item.weight))}    end)    Enum.map(max_items, fn {name,max} -> for i <- 0..max, do: {name,i} end)    |> product    |> total(items)    |> Enum.filter(fn {_kw, {volume,weight,_}} -> volume <= max_volume and                                                  weight <= max_weight end)    |> Enum.group_by(fn {_kw, {_,_,value}} -> value end)    |> Enum.max    |> print  end   defp product([x]), do: x  defp product([a,b]), do: for x <- a, y <- b, do: [x,y]  defp product([h|t]), do: for x <- h, y <- product(t), do: [x | y]   defp total(lists, items) do    Enum.map(lists, fn kwlist ->      total = Enum.reduce(kwlist, {0,0,0}, fn {name,n},{volume,weight,value} ->        {volume + n * items[name].volume,         weight + n * items[name].weight,         value  + n * items[name].value}      end)      {kwlist, total}    end)  end   defp print({max_value, data}) do    IO.puts "Maximum value achievable is #{max_value}\tvolume  weight  value"    Enum.each(data, fn {kw,{volume,weight,value}} ->      :io.format "~s =>\t~6.3f, ~5.1f, ~6w~n", [(inspect kw), volume, weight, value]    end)  endend items = %{panacea: Item.new(0.025, 0.3, 3000),           ichor:   Item.new(0.015, 0.2, 1800),          gold:    Item.new(0.002, 2.0, 2500) }maximum = Item.new(0.25, 25, 0)Knapsack.solve_unbounded(items, maximum)`
Output:
```Maximum value achievable is 54500       volume  weight  value
[gold: 11, ichor: 0, panacea: 9] =>      0.247,  24.7,  54500
[gold: 11, ichor: 5, panacea: 6] =>      0.247,  24.8,  54500
[gold: 11, ichor: 10, panacea: 3] =>     0.247,  24.9,  54500
[gold: 11, ichor: 15, panacea: 0] =>     0.247,  25.0,  54500
```

## Factor

This is a brute force solution. It is general enough to be able to provide solutions for any number of different items.

`USING: accessors combinators kernel locals math math.ordermath.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 25.0CONSTANT: 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 ;`

## Forth

`\ : value ; immediate: weight cell+ ;: volume 2 cells + ; : number 3 cells + ; \      item value weight volume numbercreate 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-valuevariable max-panvariable max-ichvariable 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`

Or like this...

`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 ;`

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

`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`

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.

`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)}`

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

`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]}`

Test:

`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)    }}`

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```

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.

`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 itemsvals = map itemVal itemswgts = map itemWgt itemsvols = map itemVol items dotProduct :: (Num a, Integral b) => [a] -> [b] -> adotProduct 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] -> StringshowOpt 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`

Output:

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

## 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 = 0DO 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  ENDDOENDDO`
`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; `

## J

Brute force solution.

`mwv=: 25 0.25prods=: <;. _1 ' panacea: ichor: gold:'hdrs=: <;. _1 ' weight: volume: value:'vls=: 3000 1800 2500ws=: 0.3 0.2 2.0vs=: 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)`

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.

`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`
`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`

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.

`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 + '<br>');for (i = 0; i < solutions.length; i += 1) {    item = solutions[i];    document.write("(gold: " + item[0] + ", panacea: " + item[1] + ", ichor: " + item[2] + ")<br>");} output:<pre>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)</pre>`

## Kotlin

Translation of: C

Recursive brute force approach:

`// version 1.1.2 data class Item(val name: String, val value: Double, val weight: Double, val volume: Double) val items = listOf(    Item("panacea", 3000.0, 0.3, 0.025),    Item("ichor", 1800.0, 0.2, 0.015),    Item("gold", 2500.0, 2.0, 0.002)) val n = items.sizeval count = IntArray(n)val best  = IntArray(n)var bestValue = 0.0 const val MAX_WEIGHT = 25.0const val MAX_VOLUME = 0.25 fun knapsack(i: Int, value: Double, weight: Double, volume: Double) {    if (i == n) {        if (value > bestValue) {            bestValue = value            for (j in 0 until n) best[j] = count[j]        }        return    }    val m1 = Math.floor(weight / items[i].weight).toInt()    val m2 = Math.floor(volume / items[i].volume).toInt()    val m  = minOf(m1, m2)    count[i] = m    while (count[i] >= 0) {        knapsack(            i + 1,            value  + count[i] * items[i].value,            weight - count[i] * items[i].weight,            volume - count[i] * items[i].volume        )        count[i]--    }} fun main(args: Array<String>) {    knapsack(0, 0.0, MAX_WEIGHT, MAX_VOLUME)    println("Item Chosen  Number Value  Weight  Volume")    println("-----------  ------ -----  ------  ------")    var itemCount = 0    var sumNumber = 0    var sumWeight = 0.0    var sumVolume = 0.0    for (i in 0 until n) {        if (best[i] == 0) continue        itemCount++        val name   = items[i].name        val number = best[i]        val value  = items[i].value  * number        val weight = items[i].weight * number        val volume = items[i].volume * number        sumNumber += number        sumWeight += weight        sumVolume += volume        print("\${name.padEnd(11)}   \${"%2d".format(number)}    \${"%5.0f".format(value)}   \${"%4.1f".format(weight)}")        println("    \${"%4.2f".format(volume)}")    }    println("-----------  ------ -----  ------  ------")    print("\${itemCount} items       \${"%2d".format(sumNumber)}    \${"%5.0f".format(bestValue)}   \${"%4.1f".format(sumWeight)}")    println("    \${"%4.2f".format(sumVolume)}")}`
Output:
```Item Chosen  Number Value  Weight  Volume
-----------  ------ -----  ------  ------
panacea        9    27000    2.7    0.23
gold          11    27500   22.0    0.02
-----------  ------ -----  ------  ------
2 items       20    54500   24.7    0.25
```

## M4

A brute force solution:

`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,volumesetv(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))')')      ')')divertmv best`

Output:

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

## 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 = 25max_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    endend print( "Maximum value:", best.value )for k, v in pairs( best_amounts ) do    print( k, v )end`

## Mathematica

Brute force algorithm:

`{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]]`

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}:

`{{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}}}`

if we call the three items by their first letters the best packings are:

`p:9 i:0 v:11p:6 i:5 v:11p:3 i:10 v:11p:0 i:15 v:11`

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

`/*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;`

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.

`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.`

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:

`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`

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:

`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}}`

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.

`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.`

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.

`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";`
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 2016.08

Brute force, looked a lot at the Ruby solution.

`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;`

Output:

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

## Phix

For each goodie, fill yer boots, then (except for the last) recursively try with fewer and fewer.
Increase profit and decrease weight/volume to pick largest profit for least weight and space.

`---- demo\rosetta\knapsack.exw-- =========================--integer attempts = 0function knapsack(sequence res, goodies, atom profit, weight, volume, at=1, sequence chosen={})    atom {pitem,witem,vitem} = goodies[at][2]    integer n = min(floor(weight/witem),floor(volume/vitem))    chosen &= n    profit += n*pitem   -- increase profit    weight -= n*witem   -- decrease weight left    volume -= n*vitem   -- decrease space left    if at=length(goodies) then        attempts += 1        if length(res)=0        or res<{profit,weight,volume} then            res = {profit,weight,volume,chosen}        end if    else        while n>=0 do            res = knapsack(res,goodies,profit,weight,volume,at+1,chosen)            n -= 1            chosen[\$] = n            profit -= pitem            weight += witem            volume += vitem        end while    end if    return resend function constant goodies = {-- item           profit weight volume{"panacea",      {3000,   0.3, 0.025}},{"ichor",        {1800,   0.2, 0.015}},{"shiney shiney",{2500,   2.0, 0.002}}} sequence res -- {profit,(weight left),(space left),{counts}}res = knapsack({},goodies,0,25,0.25)integer {p,i,g} = res[4]sequence {d,pwv} = columnize(goodies),         {?,w,v} = columnize(pwv)atom weight = sum(sq_mul(res[4],w)),     volume = sum(sq_mul(res[4],v))printf(1,"Profit %d: %d %s, %d %s, %d %s\n",        {res[1],p,d[1],i,d[2],g,d[3]})printf(1," [weight:%g, volume:%g, %d attempts]\n",        {weight,volume,attempts})`
Output:
```Profit 54500: 9 panacea, 0 ichor, 11 shiney shiney
[weight:24.7, volume:0.247, 98 attempts]
```

You get the same result whatever order the goodies are in, but with a different number of attempts, gold/ichor/panacea being the highest at 204.

## PicoLisp

Brute force solution

`(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)) )`

Output:

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

## PowerShell

Works with: PowerShell version 3.0
`#  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 limitForEach ( \$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`
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.

`:- 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)	      )), 	% we display the results	compute_lenword(L, 0, Len),	sformat(A0, '~~w~~t~~~w|', [3]),	sformat(A1, '~~w~~t~~~w|', [Len]),	sformat(A2, '~~t~~w~~~w|', [10]),	sformat(A3, '~~t~~2f~~~w|', [10]),	sformat(A4, '~~t~~3f~~~w|', [10]),	sformat(A33, '~~t~~w~~~w|', [10]),	sformat(A44, '~~t~~w~~~w|', [10]), 	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).`

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
`Define.f TotalWeight, TotalVolymeDefine.i maxPanacea, maxIchor, maxGold, maxValueDefine.i i, j ,kDim n.i(2) Enumeration   #Panacea  #Ichor  #Gold  #Sack  #CurrentEndEnumeration Structure Bounty  value.i  weight.f  volyme.fEndStructure 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 a<b    ProcedureReturn a  Else    ProcedureReturn b  EndIfEndProcedure maxPanacea=min(Item(#Sack)\weight/Item(#Panacea)\weight,Item(#Sack)\volyme/Item(#Panacea)\volyme)maxIchor  =min(Item(#Sack)\weight/Item(#Ichor)\weight,  Item(#Sack)\volyme/Item(#Ichor)\volyme)maxGold   =min(Item(#Sack)\weight/Item(#Gold)\weight,   Item(#Sack)\volyme/Item(#Gold)\volyme) For i=0 To maxPanacea   For j=0 To maxIchor    For k=0 To maxGold      Item(#Current)\value=k*Item(#Gold)\value  +j*item(#Ichor)\value +i*item(#Panacea)\value      Item(#Current)\weight=k*Item(#Gold)\weight+j*Item(#Ichor)\weight+i*Item(#Panacea)\weight      Item(#Current)\volyme=k*Item(#Gold)\volyme+j*Item(#Ichor)\volyme+i*Item(#Panacea)\volyme      If Item(#Current)\weight>Item(#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 jNext 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 DataSectionpanacea:  Data.i 3000  Data.f 0.3, 0.025ichor:  Data.i 1800  Data.f 0.2, 0.015gold:  Data.i 2500  Data.f 2.0, 0.002sack:  Data.i  0  Data.f  25.0, 0.25EndDataSection`

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
```

## R

Brute force method

`# Define constsweights <- 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 <- 25sack.volume <- 0.25max.items <- floor(pmin(sack.weight/weights, sack.volume/volumes)) # Some utility functionsgetTotalValue <- 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 sackknapsack <- 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 valuevals <- apply(knapok, 1, getTotalValue)knapok[vals == max(vals),]`
```     panacea ichor gold
2067       9     0   11
2119       6     5   11
2171       3    10   11
2223       0    15   11
```

Using Dynamic Programming

` 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_K[[Volume-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				}				}		}	K[[w+1]]<-temp_value	K_item[[w+1]]<-temp_item	}return(list(K=K,Item=K_item))}  Un_knapsack<-function(Data,W,V){	K<-list();K_item<-list()	K[[1]]<-rep(0,W+1) #the line 0	K_item[[1]]<-rep('', W+1) #the line 0	for(v in 1:V)	{		best_volum_v<-knapsack_volume(Data, W, v, K)		K[[v+1]]<-best_volum_v\$K		K_item[[v+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) `
` Output:The Total profit is:  54500 You must carry: Gold (x 11 ) You must carry: Panacea (x 9 ) `

## 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)))) `
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

`/*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 = 0.025maxGold   =0;      ichor.\$ = 1800  ;     ichor.w =  0.2 ;     ichor.v = 0.015max\$      =0;       gold.\$ = 2500  ;      gold.w =  2   ;      gold.v = 0.002now.      =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) + 1say '    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. */`

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

`/*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 = 0.025max\$      =0;       ichor.\$ = 1800  ;    ichor.w =  0.2  ;    ichor.v = 0.015now.      =0;        gold.\$ = 2500  ;     gold.w =  2    ;     gold.v = 0.002#         =0;        sack.\$ =    0  ;     sack.w = 25    ;     sack.v = 0.25L         =0maxPanacea= 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. */`

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
`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].minend maxval = 0solutions = [] 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  endend 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`
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.

`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;`

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:

`/* create SAS data set */data mydata;   input Item \$1-19 Value weight Volume;   datalines;panacea (vials of) 3000 0.3 0.025ichor (ampules of) 1800 0.2 0.015gold (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;`

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

`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 forcedef distinct(list: List[Knapsack]) =  list.map(k => Knapsack(k.items.sortBy(_.name))).distinct @tailrecdef 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)`

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

`\$ 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;`

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
`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`
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:

`#!/usr/bin/env tclshproc 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`
```\$ 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.

`#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`

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

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.

`Function Min(E1, E2): Min = IIf(E1 < E2, E1, E2): End Function 'small Helper-FunctionSub Main()Const Value = 0, Weight = 1, Volume = 2, PC = 3, IC = 4, GC = 5Dim 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 = 0.25Dim 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)  NextEnd Sub`
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
`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()));`

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
```