Knapsack problem/Unbounded

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

See also: Knapsack problem/Bounded, Knapsack problem/0-1

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

ItemExplanationValue (each)weightVolume (each)
panacea (vials of)Incredible healing properties30000.30.025
ichor (ampules of)Vampires blood18000.20.015
gold (bars)Shiney shiney25002.00.002
KnapsackFor 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

How many of each item does he take to maximise the value of items he is carrying away with him?

Note:

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


Contents

[edit] Ada

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;

[edit] 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

[edit] AutoHotkey

Brute Force.

Item  = Panacea,Ichor,Gold
Value = 3000,1800,2500
Weight= 3,2,20 ; *10
Volume= 25,15,2 ; *1000
 
StringSplit I, Item, `, ; Put input in arrays
StringSplit W, Weight,`,
StringSplit $, Value, `,
StringSplit V, Volume,`,
 
SetFormat Float, 0.3
W := 250, V := 250, sW:=.1, sV:=.001 ; limits for the total, scale factors
p := -1, Wp := -W1, Vp := -V1 ; initial values
While (Wp+=W1) <= W && (Vp+=V1) <= V {
p++, Wi := Wp-W2, Vi := Vp-V2, i := -1
While (Wi+=W2) <= W && (Vi+=V2) <= V {
i++, Wg := Wi-W3, Vg := Vi-V3, g := -1
While (Wg+=W3) <= W && (Vg+=V3) <= V
If ($ <= Val := p*$1 + i*$2 + ++g*$3)
t := ($=Val ? t "`n " : " ")
. p "`t " i "`t " g "`t " Wg*sW "`t " Vg*sV
, $ := Val
}
}
MsgBox Value = %$%`n`nPanacea`tIchor`tGold`tWeight`tVolume`n%t%

[edit] 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.

[edit] C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
struct {
double val, wgt, vol;
const char * name;
} items[] = { // value in hundreds, volume in thousandths
{30, .3, 25, "panacea"},
{18, .2, 15, "ichor"},
{25, 2., 2, "gold"},
{0,0,0,0}
};
 
/* silly setup for silly task */
int best_cnt[16] = {0}, cnt[16] = {0};
double best_v = 0;
 
void grab_em(int idx, double cap_v, double cap_w, double v)
{
double val;
int t = cap_w / items[idx].wgt;
cnt[idx] = cap_v / items[idx].vol;
 
if (cnt[idx] > t) cnt[idx] = t;
 
while (cnt[idx] >= 0) {
val = v + cnt[idx] * items[idx].val;
if (!items[idx + 1].name) {
if (val > best_v) {
best_v = val;
memcpy(best_cnt, cnt, sizeof(int) * (1 + idx));
}
return;
}
grab_em(idx + 1, cap_v - cnt[idx] * items[idx].vol,
cap_w - cnt[idx] * items[idx].wgt, val);
cnt[idx]--;
}
}
 
int main(void)
{
int i;
 
grab_em(0, 250, 25, 0);
printf("value: %g hundreds\n", best_v);
for (i = 0; items[i].name; i++)
printf("%d %s\n", best_cnt[i], items[i].name);
 
return 0;
}
output
value: 545 hundreds

9 panacea 0 ichor 11 gold

[edit] C#

/*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

[edit] 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

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

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

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

[edit] D

Translation of: Python
void main() {
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.tupleof);
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

[edit] 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..$]);
}

[edit] 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])

[edit] 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.order
math.vectors sequences sequences.product combinators.short-circuit ;
IN: knapsack
 
CONSTANT: values { 3000 1800 2500 }
CONSTANT: weights { 0.3 0.2 2.0 }
CONSTANT: volumes { 0.025 0.015 0.002 }
 
CONSTANT: max-weight 25.0
CONSTANT: max-volume 0.25
 
TUPLE: bounty amounts value weight volume ;
 
: <bounty> ( items -- bounty )
[ bounty new ] dip {
[ >>amounts ]
[ values v. >>value ]
[ weights v. >>weight ]
[ volumes v. >>volume ]
} cleave ;
 
: valid-bounty? ( bounty -- ? )
{ [ weight>> max-weight <= ]
[ volume>> max-volume <= ] } 1&& ;
 
M:: bounty <=> ( a b -- <=> )
a valid-bounty? [
b valid-bounty? [
a b [ value>> ] compare
] [ +gt+ ] if
] [ b valid-bounty? +lt+ +eq+ ? ] if ;
 
: find-max-amounts ( -- amounts )
weights volumes [
[ max-weight swap / ]
[ max-volume swap / ] bi* min >integer
] 2map ;
 
: best-bounty ( -- bounty )
find-max-amounts [ 1 + iota ] map <product-sequence>
[ <bounty> ] [ max ] map-reduce ;

[edit] Forth

\ : value ; immediate
: weight cell+ ;
: volume 2 cells + ;
: number 3 cells + ;
 
\ item value weight volume number
create panacea 30 , 3 , 25 , 0 ,
create ichor 18 , 2 , 15 , 0 ,
create gold 25 , 20 , 2 , 0 ,
create sack 0 , 250 , 250 ,
 
: fits? ( item -- ? )
dup weight @ sack weight @ > if drop false exit then
volume @ sack volume @ > 0= ;
 
: add ( item -- )
dup @ sack +!
dup weight @ negate sack weight +!
dup volume @ negate sack volume +!
1 swap number +! ;
 
: take ( item -- )
dup @ negate sack +!
dup weight @ sack weight +!
dup volume @ sack volume +!
-1 swap number +! ;
 
variable max-value
variable max-pan
variable max-ich
variable max-au
 
: .solution
cr
max-pan @ . ." Panaceas, "
max-ich @ . ." Ichors, and "
max-au @ . ." Gold for a total value of "
max-value @ 100 * . ;
 
: check
sack @ max-value @ <= if exit then
sack @ max-value !
panacea number @ max-pan  !
ichor number @ max-ich  !
gold number @ max-au  !
( .solution ) ; \ and change <= to < to see all solutions
 
: solve-gold
gold fits? if gold add recurse gold take
else check then ;
 
: solve-ichor
ichor fits? if ichor add recurse ichor take then
solve-gold ;
 
: solve-panacea
panacea fits? if panacea add recurse panacea take then
solve-ichor ;
 
solve-panacea .solution

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

[edit] 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

[edit] 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

[edit] 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

[edit] Haskell

This is a brute-force solution: it generates a list of every legal combination of items (options) and then finds the option of greatest value.

import Data.List (maximumBy)
import Data.Ord (comparing)
 
(maxWgt, maxVol) = (25, 0.25)
items =
[Bounty "panacea" 3000 0.3 0.025,
Bounty "ichor" 1800 0.2 0.015,
Bounty "gold" 2500 2.0 0.002]
 
data Bounty = Bounty
{itemName :: String,
itemVal :: Int,
itemWgt, itemVol :: Double}
 
names = map itemName items
vals = map itemVal items
wgts = map itemWgt items
vols = map itemVol items
 
dotProduct :: (Num a, Integral b) => [a] -> [b] -> a
dotProduct factors = sum . zipWith (*) factors . map fromIntegral
 
options :: [[Int]]
options = filter fits $ mapM f items
where f (Bounty _ _ w v) = [0 .. m]
where m = floor $ min (maxWgt / w) (maxVol / v)
fits opt = dotProduct wgts opt <= maxWgt &&
dotProduct vols opt <= maxVol
 
showOpt :: [Int] -> String
showOpt opt = concat (zipWith showItem names opt) ++
"total weight: " ++ show (dotProduct wgts opt) ++
"\ntotal volume: " ++ show (dotProduct vols opt) ++
"\ntotal value: " ++ show (dotProduct vals opt) ++ "\n"
where showItem name num = name ++ ": " ++ show num ++ "\n"
 
main = putStr $ showOpt $ best options
where best = maximumBy $ comparing $ dotProduct vals

Output:

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

[edit] HicEst

CHARACTER list*1000
 
NN = ALIAS($Panacea, $Ichor, $Gold, wSack, wPanacea, wIchor, wGold, vSack, vPanacea, vIchor, vGold)
NN = (3000, 1800, 2500, 25, 0.3, 0.2, 2.0, 0.25, 0.025, 0.015, 0.002)
maxItems = ALIAS(maxPanacea, maxIchor, maxGold)
maxItems = ( MIN( wSack/wPanacea, vSack/vPanacea), MIN( wSack/wIchor, vSack/vIchor), MIN( wSack/wGold, vSack/vGold) )
 
maxValue = 0
DO Panaceas = 0, maxPanacea
DO Ichors = 0, maxIchor
DO Golds = 0, maxGold
weight = Panaceas*wPanacea + Ichors*wIchor + Golds*wGold
IF( weight <= wSack ) THEN
volume = Panaceas*vPanacea + Ichors*vIchor + Golds*vGold
IF( volume <= vSack ) THEN
value = Panaceas*$Panacea + Ichors*$Ichor + Golds*$Gold
IF( value > maxValue ) THEN
maxValue = value
! this restarts the list, removing all previous entries:
WRITE(Text=list, Name) value, Panaceas, Ichors, Golds, weight, volume, $CR//$LF
ELSEIF( value == maxValue ) THEN
WRITE(Text=list, Name, APPend) value, Panaceas, Ichors, Golds, weight, volume, $CR//$LF
ENDIF
ENDIF
ENDIF
ENDDO
ENDDO
ENDDO
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;

[edit] J

Brute force solution.

mwv=: 25 0.25
prods=: <;. _1 ' panacea: ichor: gold:'
hdrs=: <;. _1 ' weight: volume: value:'
vls=: 3000 1800 2500
ws=: 0.3 0.2 2.0
vs=: 0.025 0.015 0.002
 
ip=: +/ .*
prtscr=: (1!:2)&2
 
KS=: 3 : 0
os=. (#:i.@(*/)) mwv >:@<.@<./@:% ws,:vs
bo=.os#~(ws,:vs) mwv&(*./@:>)@ip"_ 1 os
mo=.bo{~{.\: vls ip"1 bo
prtscr &.> prods ([,' ',":@])&.>mo
prtscr &.> hdrs ('total '&,@[,' ',":@])&.> mo ip"1 ws,vs,:vls
LF
)

Example output:

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

[edit] 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

[edit] 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>

[edit] 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,volume
setv(a,0,`knapsack',0,250,250)
setv(a,1,`panacea',3000,3,25)
setv(a,2,`ichor',1800,2,15)
setv(a,3,`gold',2500,20,2)
 
define(`mv',0)
for(`x',0,min(get2d(a,0,3)/get2d(a,1,3),get2d(a,0,4)/get2d(a,1,4)),
`for(`y',0,min((get2d(a,0,3)-x*get2d(a,1,3))/get2d(a,2,3),
(get2d(a,0,4)-x*get2d(a,1,4))/get2d(a,2,4)),
`
define(`z',min((get2d(a,0,3)-x*get2d(a,1,3)-y*get2d(a,2,3))/get2d(a,3,3),
(get2d(a,0,4)-x*get2d(a,1,4)-y*get2d(a,2,4))/get2d(a,3,4)))
define(`cv',eval(x*get2d(a,1,2)+y*get2d(a,2,2)+z*get2d(a,3,2)))
ifelse(eval(cv>mv),1,
`define(`mv',cv)`'define(`best',(x,y,z))',
`ifelse(cv,mv,`define(`best',best (x,y,z))')')
')')
divert
mv best

Output:

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

[edit] Lua

items = {   ["panaea"] = { ["value"] = 3000, ["weight"] = 0.3, ["volume"] = 0.025 },
["ichor"] = { ["value"] = 1800, ["weight"] = 0.2, ["volume"] = 0.015 },
["gold"] = { ["value"] = 2500, ["weight"] = 2.0, ["volume"] = 0.002 }
}
 
max_weight = 25
max_volume = 0.25
 
max_num_items = {}
for i in pairs( items ) do
max_num_items[i] = math.floor( math.min( max_weight / items[i].weight, max_volume / items[i].volume ) )
end
 
best = { ["value"] = 0.0, ["weight"] = 0.0, ["volume"] = 0.0 }
best_amounts = {}
 
for i = 1, max_num_items["panaea"] do
for j = 1, max_num_items["ichor"] do
for k = 1, max_num_items["gold"] do
current = { ["value"] = i*items["panaea"]["value"] + j*items["ichor"]["value"] + k*items["gold"]["value"],
["weight"] = i*items["panaea"]["weight"] + j*items["ichor"]["weight"] + k*items["gold"]["weight"],
["volume"] = i*items["panaea"]["volume"] + j*items["ichor"]["volume"] + k*items["gold"]["volume"]
}
 
if current.value > best.value and current.weight <= max_weight and current.volume <= max_volume then
best = { ["value"] = current.value, ["weight"] = current.weight, ["volume"] = current.volume }
best_amounts = { ["panaea"] = i, ["ichor"] = j, ["gold"] = k }
end
end
end
end
 
print( "Maximum value:", best.value )
for k, v in pairs( best_amounts ) do
print( k, v )
end

[edit] 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:11
p:6 i:5 v:11
p:3 i:10 v:11
p: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.

[edit] 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.

[edit] 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

[edit] 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

[edit] 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:

table pre solving

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

solver menu
  • Options... (opens a separate popup window, then continue)
solver popup options menu

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

Table solved

[edit] 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

[edit] 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

[edit] 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.

[edit] Perl 6

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

[edit] 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

[edit] 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 

[edit] PureBasic

Translation of: Fortran
Define.f TotalWeight, TotalVolyme
Define.i maxPanacea, maxIchor, maxGold, maxValue
Define.i i, j ,k
Dim n.i(2)
 
Enumeration
#Panacea
#Ichor
#Gold
#Sack
#Current
EndEnumeration
 
Structure Bounty
value.i
weight.f
volyme.f
EndStructure
 
Dim Item.Bounty(4)
CopyMemory(?panacea,@Item(#Panacea),SizeOf(Bounty))
CopyMemory(?ichor, @Item(#Ichor), SizeOf(Bounty))
CopyMemory(?gold, @Item(#gold), SizeOf(Bounty))
CopyMemory(?sack, @Item(#Sack), SizeOf(Bounty))
 
Procedure.f min(a.f, b.f)
If a<b
ProcedureReturn a
Else
ProcedureReturn b
EndIf
EndProcedure
 
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 j
Next i
 
If OpenConsole()
Define txt$
txt$="Maximum value achievable is "+Str(maxValue)+#CRLF$
txt$+"This is achieved by carrying "+Str(n(#Panacea))+" panacea, "
txt$+Str(n(#Ichor))+" ichor and "+Str(n(#Gold))+" gold items."+#CRLF$
txt$+"The weight to carry is "+StrF(totalWeight,2)
txt$+" and the volume used is "+StrF(TotalVolyme,2)
PrintN(txt$)
 
Print(#CRLF$+"Press Enter to quit"): Input()
EndIf
 
DataSection
panacea:
Data.i 3000
Data.f 0.3, 0.025
ichor:
Data.i 1800
Data.f 0.2, 0.015
gold:
Data.i 2500
Data.f 2.0, 0.002
sack:
Data.i 0
Data.f 25.0, 0.25
EndDataSection

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

[edit] Python

See Knapsack Problem/Python

[edit] R

Brute force method

# Define consts
weights <- c(panacea=0.3, ichor=0.2, gold=2.0)
volumes <- c(panacea=0.025, ichor=0.015, gold=0.002)
values <- c(panacea=3000, ichor=1800, gold=2500)
sack.weight <- 25
sack.volume <- 0.25
max.items <- floor(pmin(sack.weight/weights, sack.volume/volumes))
 
# Some utility functions
getTotalValue <- function(n) sum(n*values)
getTotalWeight <- function(n) sum(n*weights)
getTotalVolume <- function(n) sum(n*volumes)
willFitInSack <- function(n) getTotalWeight(n) <= sack.weight && getTotalVolume(n) <= sack.volume
 
# Find all possible combination, then eliminate those that won't fit in the sack
knapsack <- expand.grid(lapply(max.items, function(n) seq.int(0, n)))
ok <- apply(knapsack, 1, willFitInSack)
knapok <- knapsack[ok,]
 
# Find the solutions with the highest value
vals <- apply(knapok, 1, getTotalValue)
knapok[vals == max(vals),]
     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 )
 

[edit] 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

[edit] REXX

(Any resemblence to the Fortran code is 110% coincidental.)

/*REXX program solves a  knapsack/unbounded  problem.                   */
maxPanacea=0
maxIchor =0
maxGold =0
max$ =0
current. =0
 
/* value weight volume */
/* ═══════ ═══════ ══════ */
panacea.$= 3000 ; panacea.w= 0.3 ; panacea.v= 0.025
ichor.$= 1800 ; ichor.w= 0.2 ; ichor.v= 0.015
gold.$= 2500 ; gold.w= 2  ; gold.v= 0.002
sack.$= 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
current.$=g*gold.$ + i*ichor.$ + p*panacea.$
current.w=g*gold.w + i*ichor.w + p*panacea.w
current.v=g*gold.v + i*ichor.v + p*panacea.v
if current.w>sack.w | current.v>sack.v then iterate
if current.$>max$ then do
max$ = current.$
totalW = current.w
totalV = current.v
maxP=p; maxI=i; maxG=g
end
end /*g (gold) */
end /*i (ichor) */
end /*p (panacea)*/
 
cTot=maxP+maxI+maxG
L=length(cTot)+1
say ' panacea in sack:' right(maxP,L)
say ' ichors in sack:' right(maxI,L)
say ' gold items in sack:' right(maxG,L)
say '════════════════════' copies('═',L)
say 'carrying a total of:' right(cTot,L)
say left('',40) 'total value: ' max$/1
say left('',40) 'total weight: ' totalW/1
say left('',40) 'total volume: ' totalV/1
/*stick a fork in it, we're 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 

[edit] 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].min
end
 
maxval = 0
solutions = []
 
0.upto(max_items[ichor]) do |i|
0.upto(max_items[panacea]) do |p|
0.upto(max_items[gold]) do |g|
next if i*ichor.weight + p*panacea.weight + g*gold.weight > maximum.weight
next if i*ichor.volume + p*panacea.volume + g*gold.volume > maximum.volume
val = i*ichor.value + p*panacea.value + g*gold.value
if val > maxval
maxval = val
solutions = [[i, p, g]]
elsif val == maxval
solutions << [i, p, g]
end
end
end
end
 
puts "The maximal solution has value #{maxval}"
solutions.each do |i, p, g|
printf " ichor=%2d, panacea=%2d, gold=%2d -- weight:%.1f, volume=%.3f\n",
i, p, g,
i*ichor.weight + p*panacea.weight + g*gold.weight,
i*ichor.volume + p*panacea.volume + g*gold.volume
end
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

[edit] 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

[edit] 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

[edit] 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 tclsh
proc main argv {
array set value {panacea 3000 ichor 1800 gold 2500}
array set weight {panacea 0.3 ichor 0.2 gold 2.0 max 25}
array set volume {panacea 0.025 ichor 0.015 gold 0.002 max 0.25}
 
foreach i {panacea ichor gold} {
set max($i) [expr {min(int($volume(max)/$volume($i)),
int($weight(max)/$weight($i)))}]
}
set maxval 0
for {set i 0} {$i < $max(ichor)} {incr i} {
for {set p 0} {$p < $max(panacea)} {incr p} {
for {set g 0} {$g < $max(gold)} {incr g} {
if {$i*$weight(ichor) + $p*$weight(panacea) + $g*$weight(gold)
> $weight(max)} continue
if {$i*$volume(ichor) + $p*$volume(panacea) + $g*$volume(gold)
> $volume(max)} continue
set val [expr {$i*$value(ichor)+$p*$value(panacea)+$g*$value(gold)}]
if {$val == $maxval} {
lappend best [list i $i p $p g $g]
} elseif {$val > $maxval} {
set maxval $val
set best [list [list i $i p $p g $g]]
}
}
}
}
puts "maxval: $maxval, best: $best"
}
main $argv
$ 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

[edit] 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>>

[edit] Visual Basic

See: Knapsack Problem/Visual Basic

The above Link contains a longer version (which perhaps runs a bit faster), whilst the one below is focussing more on expressing/solving the problem in less lines of code.

Function Min(E1, E2): Min = IIf(E1 < E2, E1, E2): End Function 'small Helper-Function

Sub Main()
Const Value = 0, Weight = 1, Volume = 2, PC = 3, IC = 4, GC = 5
Dim P&, I&, G&, A&, M, Cur(Value To Volume)
Dim S As New Collection: S.Add Array(0) '<- init Solutions-Coll.

Const SackW = 25, SackV = 0.25
Dim Panacea: Panacea = Array(3000, 0.3, 0.025)
Dim Ichor: Ichor = Array(1800, 0.2, 0.015)
Dim Gold: Gold = Array(2500, 2, 0.002)
 
For P = 0 To Int(Min(SackW / Panacea(Weight), SackV / Panacea(Volume)))
For I = 0 To Int(Min(SackW / Ichor(Weight), SackV / Ichor(Volume)))
For G = 0 To Int(Min(SackW / Gold(Weight), SackV / Gold(Volume)))
For A = Value To Volume: Cur(A) = G * Gold(A) + I * Ichor(A) + P * Panacea(A): Next
If Cur(Value) >= S(1)(Value) And Cur(Weight) <= SackW And Cur(Volume) <= SackV Then _
S.Add Array(Cur(Value), Cur(Weight), Cur(Volume), P, I, G), , 1
Next G, I, P
 
Debug.Print "Value", "Weight", "Volume", "PanaceaCount", "IchorCount", "GoldCount"
For Each M In S '<- enumerate the Attributes of the Maxima
If M(Value) = S(1)(Value) Then Debug.Print M(Value), M(Weight), M(Volume), M(PC), M(IC), M(GC)
Next
End Sub
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 
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