Knapsack problem/Unbounded
You are encouraged to solve this task according to the task description, using any language you may know.
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.
| 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
How many of each item does he take to maximise the value of items he is carrying away with him?
Note:
- There are four solutions that maximise the value taken. Only one need be given.
ALGOL 68
<lang algol68>MODE BOUNTY = STRUCT(STRING name, INT value, weight, volume);
[]BOUNTY items = (
("panacea", 3000, 3, 25),
("ichor", 1800, 2, 15),
("gold", 2500, 20, 2)
);
BOUNTY sack := ("sack", 0, 250, 250);
OP * = ([]INT a,b)INT: ( # dot product operator #
INT sum := 0; FOR i TO UPB a DO sum +:= a[i]*b[i] OD; sum
);
OP INIT = (REF[]INT vector)VOID:
FOR index FROM LWB vector TO UPB vector DO
vector[index]:=0
OD;
OP INIT = (REF[,]INT matrix)VOID:
FOR row index FROM LWB matrix TO UPB matrix DO
INIT matrix[row index,]
OD;
PROC total value = ([]INT items count, []BOUNTY items, BOUNTY sack) STRUCT(INT value, weight, volume):(
### Given the count of each item in the sack return -1 if they can"t be carried or their total value.
(also return the negative of the weight and the volume so taking the max of a series of return
values will minimise the weight if values tie, and minimise the volume if values and weights tie).
###
INT weight = items count * weight OF items;
INT volume = items count * volume OF items;
IF weight > weight OF sack OR volume > volume OF sack THEN
(-1, 0, 0)
ELSE
( items count * value OF items, -weight, -volume)
FI
);
PRIO WRAP = 5; # wrap negative array indices as per python's indexing regime # OP WRAP = (INT index, upb)INT:
IF index>=0 THEN index ELSE upb + index + 1 FI;
PROC knapsack dp = ([]BOUNTY items, BOUNTY sack)[]INT:(
### Solves the Knapsack problem, with two sets of weights, using a dynamic programming approach ###
# (weight+1) x (volume+1) table # # table[w,v] is the maximum value that can be achieved # # with a sack of weight w and volume v. # # They all start out as 0 (empty sack) # [0:weight OF sack, 0:volume OF sack]INT table; INIT table;
FOR w TO 1 UPB table DO
FOR v TO 2 UPB table DO
### Consider the optimal solution, and consider the "last item" added
to the sack. Removing this item must produce an optimal solution
to the subproblem with the sack"s weight and volume reduced by that
of the item. So we search through all possible "last items": ###
FOR item index TO UPB items DO
BOUNTY item := items[item index];
# Only consider items that would fit: #
IF w >= weight OF item AND v >= volume OF item THEN
# Optimal solution to subproblem + value of item: #
INT candidate := table[w-weight OF item,v-volume OF item] + value OF item;
IF candidate > table[w,v] THEN
table[w,v] := candidate
FI
FI
OD
OD
OD;
[UPB items]INT result; INIT result;
INT w := weight OF sack, v := volume OF sack;
WHILE table[w,v] /= 0 DO
# Find the last item that was added: #
INT needle = table[w,v];
INT item index;
FOR i TO UPB items WHILE
item index := i;
BOUNTY item = items[item index];
INT candidate = table[w-weight OF item WRAP UPB table, v-volume OF item WRAP 2 UPB table] + value OF item;
- WHILE # candidate NE needle DO
SKIP
OD;
# Record it in the result, and remove it: #
result[item index] +:= 1;
w -:= weight OF items[item index];
v -:= volume OF items[item index]
OD;
result
);
[]INT max items = knapsack dp(items, sack); STRUCT (INT value, weight, volume) max := total value(max items, items, sack); max := (value OF max, -weight OF max, -volume OF max);
FORMAT d = $zz-d$;
printf(($"The maximum value achievable (by dynamic programming) is "gl$, value OF max)); printf(($" The number of ("n(UPB items-1)(g", ")g") items to achieve this is: ("n(UPB items-1)(f(d)",")f(d)") respectively"l$,
name OF items, max items));
printf(($" The weight to carry is "f(d)", and the volume used is "f(d)l$,
weight OF max, volume OF max))</lang>Output:
The maximum value achievable (by dynamic programming) is +54500 The number of (panacea, ichor, gold) items to achieve this is: ( 9, 0, 11) respectively The weight to carry is 247, and the volume used is 247
C
<lang c>#include <stdio.h>
double min(double a, double b) {
return a < b ? a : b;
}
struct Bounty {
int value; double weight, volume;
};
struct Bounty panacea = {3000, 0.3, 0.025},
ichor = {1800, 0.2, 0.015},
gold = {2500, 2.0, 0.002},
sack = { 0, 25.0, 0.25 },
current, best;
- define CALC(V) current.V = npanacea * panacea.V + nichor * ichor.V + ngold * gold.V
int main(void) {
int npanacea, nichor, ngold, max_panacea, max_ichor, max_gold; int best_amounts[3]; best.value = 0; max_panacea = (int)min(sack.weight / panacea.weight, sack.volume / panacea.volume); max_ichor = (int)min(sack.weight / ichor.weight, sack.volume / ichor.volume); max_gold = (int)min(sack.weight / gold.weight, sack.volume / gold.volume);
for (npanacea = 0; npanacea <= max_panacea; npanacea++) {
for (nichor = 0; nichor <= max_ichor; nichor++) {
for (ngold = 0; ngold <= max_gold; ngold++) {
CALC(value);
CALC(weight);
CALC(volume);
if (current.value > best.value && current.weight <= sack.weight &&
current.volume <= sack.volume) {
best.value = current.value;
best.weight = current.weight;
best.volume = current.volume;
best_amounts[0] = npanacea;
best_amounts[1] = nichor;
best_amounts[2] = ngold;
}
}
}
}
printf("Maximum value achievable is %d\n", best.value);
printf("This is achieved by carrying (one solution) %d panacea, %d ichor and %d gold\n",
best_amounts[0], best_amounts[1], best_amounts[2]);
printf("The weight to carry is %4.1f and the volume used is %5.3f\n", best.weight, best.volume);
return 0;
}</lang>
Forth
<lang forth>\ : value ; immediate
- weight cell+ ;
- volume 2 cells + ;
- number 3 cells + ;
\ item value weight volume number create panacea 30 , 3 , 25 , 0 , create ichor 18 , 2 , 15 , 0 , create gold 25 , 20 , 2 , 0 , create sack 0 , 250 , 250 ,
- fits? ( item -- ? )
dup weight @ sack weight @ > if drop false exit then
volume @ sack volume @ > 0= ;
- add ( item -- )
dup @ sack +! dup weight @ negate sack weight +! dup volume @ negate sack volume +! 1 swap number +! ;
- take ( item -- )
dup @ negate sack +! dup weight @ sack weight +! dup volume @ sack volume +! -1 swap number +! ;
variable max-value variable max-pan variable max-ich variable max-au
- .solution
cr max-pan @ . ." Panaceas, " max-ich @ . ." Ichors, and " max-au @ . ." Gold for a total value of " max-value @ 100 * . ;
- check
sack @ max-value @ <= if exit then sack @ max-value ! panacea number @ max-pan ! ichor number @ max-ich ! gold number @ max-au ! ( .solution ) ; \ and change <= to < to see all solutions
- solve-gold
gold fits? if gold add recurse gold take else check then ;
- solve-ichor
ichor fits? if ichor add recurse ichor take then solve-gold ;
- solve-panacea
panacea fits? if panacea add recurse panacea take then solve-ichor ;
solve-panacea .solution</lang> Or like this... <lang forth>0 VALUE vials 0 VALUE ampules 0 VALUE bars 0 VALUE bag
- 250 3 / #250 #25 / MIN 1+ CONSTANT maxvials
- 250 2/ #250 #15 / MIN 1+ CONSTANT maxampules
- 250 #20 / #250 2/ MIN 1+ CONSTANT maxbars
- RESULTS ( v a b -- k )
3DUP #20 * SWAP 2* + SWAP 3 * + #250 > IF 3DROP -1 EXIT ENDIF 3DUP 2* SWAP #15 * + SWAP #25 * + #250 > IF 3DROP -1 EXIT ENDIF #2500 * SWAP #1800 * + SWAP #3000 * + ;
- .SOLUTION ( -- )
CR ." The traveller's knapsack contains " vials DEC. ." vials of panacea, " ampules DEC. ." ampules of ichor, " CR bars DEC. ." bars of gold, a total value of " vials ampules bars RESULTS 0DEC.R ." ." ;
- KNAPSACK ( -- )
-1 TO bag maxvials 0 ?DO maxampules 0 ?DO maxbars 0 ?DO
K J I RESULTS DUP
bag > IF TO bag K TO vials J TO ampules I TO bars ELSE DROP ENDIF LOOP LOOP LOOP .SOLUTION ;</lang> With the result...
FORTH> knapsack The traveller's knapsack contains 0 vials of panacea, 15 ampules of ichor, 11 bars of gold, a total value of 54500. ok
Fortran
A straight forward 'brute force' approach <lang fortran>PROGRAM KNAPSACK
IMPLICIT NONE REAL :: totalWeight, totalVolume INTEGER :: maxPanacea, maxIchor, maxGold, maxValue = 0 INTEGER :: i, j, k INTEGER :: n(3)
TYPE Bounty INTEGER :: value REAL :: weight REAL :: volume END TYPE Bounty
TYPE(Bounty) :: panacea, ichor, gold, sack, current
panacea = Bounty(3000, 0.3, 0.025) ichor = Bounty(1800, 0.2, 0.015) gold = Bounty(2500, 2.0, 0.002) sack = Bounty(0, 25.0, 0.25)
maxPanacea = MIN(sack%weight / panacea%weight, sack%volume / panacea%volume)
maxIchor = MIN(sack%weight / ichor%weight, sack%volume / ichor%volume)
maxGold = MIN(sack%weight / gold%weight, sack%volume / gold%volume)
DO i = 0, maxPanacea
DO j = 0, maxIchor
Do k = 0, maxGold
current%value = k * gold%value + j * ichor%value + i * panacea%value
current%weight = k * gold%weight + j * ichor%weight + i * panacea%weight
current%volume = k * gold%volume + j * ichor%volume + i * panacea%volume
IF (current%weight > sack%weight .OR. current%volume > sack%volume) CYCLE
IF (current%value > maxValue) THEN
maxValue = current%value
totalWeight = current%weight
totalVolume = current%volume
n(1) = i ; n(2) = j ; n(3) = k
END IF
END DO
END DO
END DO
WRITE(*, "(A,I0)") "Maximum value achievable is ", maxValue WRITE(*, "(3(A,I0),A)") "This is achieved by carrying ", n(1), " panacea, ", n(2), " ichor and ", n(3), " gold items" WRITE(*, "(A,F4.1,A,F5.3)") "The weight to carry is ", totalWeight, " and the volume used is ", totalVolume
END PROGRAM KNAPSACK</lang> Sample output
Maximum value achievable is 54500 This is achieved by carrying 0 panacea, 15 ichor and 11 gold items The weight to carry is 25.0 and the volume used is 0.247
Haskell
This is a brute-force solution: it generates a list of every legal combination of items (options) and then finds the option of greatest value.
<lang haskell>import Data.List (maximumBy) import Data.Ord (comparing)
(maxWgt, maxVol) = (25, 0.25) items =
[Bounty "panacea" 3000 0.3 0.025, Bounty "ichor" 1800 0.2 0.015, Bounty "gold" 2500 2.0 0.002]
data Bounty = Bounty
{itemName :: String,
itemVal :: Int,
itemWgt, itemVol :: Double}
names = map itemName items vals = map itemVal items wgts = map itemWgt items vols = map itemVol items
dotProduct :: (Num a, Integral b) => [a] -> [b] -> a dotProduct factors = sum . zipWith (*) factors . map fromIntegral
options :: Int options = filter fits $ mapM f items
where f (Bounty _ _ w v) = [0 .. m]
where m = floor $ min (maxWgt / w) (maxVol / v)
fits opt = dotProduct wgts opt <= maxWgt &&
dotProduct vols opt <= maxVol
showOpt :: [Int] -> String showOpt opt = concat (zipWith showItem names opt) ++
"total weight: " ++ show (dotProduct wgts opt) ++ "\ntotal volume: " ++ show (dotProduct vols opt) ++ "\ntotal value: " ++ show (dotProduct vals opt) ++ "\n" where showItem name num = name ++ ": " ++ show num ++ "\n"
main = putStr $ showOpt $ best options
where best = maximumBy $ comparing $ dotProduct vals</lang>
Output:
panacea: 9 ichor: 0 gold: 11 total weight: 24.7 total volume: 0.247 total value: 54500
J
Brute force solution. <lang j> mwv=: 25 0.25 prods=: <;. _1 ' panacea: ichor: gold:' hdrs=: <;. _1 ' weight: volume: value:' vls=: 3000 1800 2500 ws=: 0.3 0.2 2.0 vs=: 0.025 0.015 0.002
ip=: +/ .* prtscr=: (1!:2)&2
KS=: 3 : 0
os=. (#:i.@(*/)) mwv >:@<.@<./@:% ws,:vs
bo=.os#~(ws,:vs) mwv&(*./@:>)@ip"_ 1 os
mo=.bo{~{.\: vls ip"1 bo
prtscr &.> prods ([,' ',":@])&.>mo
prtscr &.> hdrs ('total '&,@[,' ',":@])&.> mo ip"1 ws,vs,:vls
LF
) </lang> Example output:
KS'' panacea: 3 ichor: 10 gold: 11 total weight: 24.9 total volume: 0.247 total value: 54500
Modula-3
Note that unlike Fortran and C, Modula-3 does not do any hidden casting, which is why FLOAT and FLOOR are used. <lang modula3> MODULE Knapsack EXPORTS Main;
FROM IO IMPORT Put; FROM Fmt IMPORT Int, Real;
TYPE Bounty = RECORD
value: INTEGER; weight, volume: REAL;
END;
VAR totalWeight, totalVolume: REAL;
maxPanacea, maxIchor, maxGold, maxValue: INTEGER := 0; n: ARRAY [1..3] OF INTEGER; panacea, ichor, gold, sack, current: Bounty;
BEGIN
panacea := Bounty{3000, 0.3, 0.025};
ichor := Bounty{1800, 0.2, 0.015};
gold := Bounty{2500, 2.0, 0.002};
sack := Bounty{0, 25.0, 0.25};
maxPanacea := FLOOR(MIN(sack.weight / panacea.weight, sack.volume / panacea.volume)); maxIchor := FLOOR(MIN(sack.weight / ichor.weight, sack.volume / ichor.volume)); maxGold := FLOOR(MIN(sack.weight / gold.weight, sack.volume / gold.volume));
FOR i := 0 TO maxPanacea DO
FOR j := 0 TO maxIchor DO
FOR k := 0 TO maxGold DO
current.value := k * gold.value + j * ichor.value + i * panacea.value;
current.weight := FLOAT(k) * gold.weight + FLOAT(j) * ichor.weight + FLOAT(i) * panacea.weight;
current.volume := FLOAT(k) * gold.volume + FLOAT(j) * ichor.volume + FLOAT(i) * panacea.volume;
IF current.weight > sack.weight OR current.volume > sack.volume THEN
EXIT;
END;
IF current.value > maxValue THEN
maxValue := current.value;
totalWeight := current.weight;
totalVolume := current.volume;
n[1] := i; n[2] := j; n[3] := k;
END;
END;
END;
END;
Put("Maximum value achievable is " & Int(maxValue) & "\n");
Put("This is achieved by carrying " & Int(n[1]) & " panacea, " & Int(n[2]) & " ichor and " & Int(n[3]) & " gold items\n");
Put("The weight of this carry is " & Real(totalWeight) & " and the volume used is " & Real(totalVolume) & "\n");
END Knapsack. </lang>
Output:
Maximum value achievable is 54500 This is achieved by carrying 0 panacea, 15 ichor and 11 gold items The weight of this carry is 25 and the volume used is 0.247
Python
Simple Solution
<lang python> class Bounty:
def __init__(self, value, weight, volume):
self.value, self.weight, self.volume = value, weight, volume
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) best = Bounty( 0, 0, 0) current = Bounty( 0, 0, 0)
best_amounts = (0, 0, 0)
max_panacea = int(min(sack.weight // panacea.weight, sack.volume // panacea.volume)) max_ichor = int(min(sack.weight // ichor.weight, sack.volume // ichor.volume)) max_gold = int(min(sack.weight // gold.weight, sack.volume // gold.volume))
for npanacea in xrange(max_panacea):
for nichor in xrange(max_ichor):
for ngold in xrange(max_gold):
current.value = npanacea * panacea.value + nichor * ichor.value + ngold * gold.value
current.weight = npanacea * panacea.weight + nichor * ichor.weight + ngold * gold.weight
current.volume = npanacea * panacea.volume + nichor * ichor.volume + ngold * gold.volume
if current.value > best.value and current.weight <= sack.weight and \
current.volume <= sack.volume:
best = Bounty(current.value, current.weight, current.volume)
best_amounts = (npanacea, nichor, ngold)
print "Maximum value achievable is", best.value print "This is achieved by carrying (one solution) %d panacea, %d ichor and %d gold" % \
(best_amounts[0], best_amounts[1], best_amounts[2])
print "The weight to carry is %4.1f and the volume used is %5.3f" % (best.weight, best.volume) </lang>
General Solution
Requires Python V.2.6+ <lang python>from itertools import product, izip from collections import namedtuple
Bounty = namedtuple('Bounty', 'name value weight volume')
sack = Bounty('sack', 0, 25.0, 0.25)
items = [Bounty('panacea', 3000, 0.3, 0.025),
Bounty('ichor', 1800, 0.2, 0.015),
Bounty('gold', 2500, 2.0, 0.002)]
def tot_value(items_count):
"""
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).
"""
global items, sack
weight = sum(n * item.weight for n, item in izip(items_count, items))
volume = sum(n * item.volume for n, item in izip(items_count, items))
if weight <= sack.weight and volume <= sack.volume:
return sum(n * item.value for n, item in izip(items_count, items)), -weight, -volume
else:
return -1, 0, 0
def knapsack():
global items, sack # find max of any one item max1 = [min(int(sack.weight // item.weight), int(sack.volume // item.volume)) for item in items]
# Try all combinations of bounty items from 0 up to max1 return max(product(*[xrange(n + 1) for n in max1]), key=tot_value)
max_items = knapsack()
maxvalue, max_weight, max_volume = tot_value(max_items)
max_weight = -max_weight
max_volume = -max_volume
print "The maximum value achievable (by exhaustive search) is %g." % maxvalue item_names = ", ".join(item.name for item in items) print " The number of %s items to achieve this is: %s, respectively." % (item_names, max_items) print " The weight to carry is %.3g, and the volume used is %.3g." % (max_weight, max_volume) </lang>
Sample output
The maximum value achievable (by exhaustive search) is 54500 The number of panacea, ichor, gold items to achieve this is: (9, 0, 11), respectively The weight to carry is 24.7, and the volume used is 0.247
General Dynamic Programming solution
A dynamic programming approach using a 2-dimensional table (One dimension for weight and one for volume). Because this approach requires that all weights and volumes be integer, I multiplied the weights and volumes by enough to make them integer. This algorithm takes O(w*v) space and O(w*v*n) time, where w = weight of sack, v = volume of sack, n = number of types of items. To solve this specific problem it's much slower than the brute force solution.
<lang python>from itertools import product, izip from collections import namedtuple
Bounty = namedtuple('Bounty', 'name value weight volume')
- "namedtuple" is only available in Python 2.6+; for earlier versions use this instead:
- class Bounty:
- def __init__(self, name, value, weight, volume):
- self.name = name
- self.value = value
- self.weight = weight
- self.volume = volume
sack = Bounty('sack', 0, 250, 250)
items = [Bounty('panacea', 3000, 3, 25),
Bounty('ichor', 1800, 2, 15),
Bounty('gold', 2500, 20, 2)]
def tot_value(items_count, items, sack):
""" 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).
"""
weight = sum(n * item.weight for n, item in izip(items_count, items))
volume = sum(n * item.volume for n, item in izip(items_count, items))
if weight <= sack.weight and volume <= sack.volume:
return sum(n * item.value for n, item in izip(items_count, items)), -weight, -volume
else:
return -1, 0, 0
def knapsack_dp(items, sack):
""" 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) table = [[0] * (sack.volume + 1) for i in xrange(sack.weight + 1)]
for w in xrange(sack.weight + 1):
for v in xrange(sack.volume + 1):
# 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 in items:
# Only consider items that would fit:
if w >= item.weight and v >= item.volume:
table[w][v] = max(table[w][v],
# Optimal solution to subproblem + value of item:
table[w - item.weight][v - item.volume] + item.value)
# Backtrack through matrix to re-construct optimum:
result = [0] * len(items)
w = sack.weight
v = sack.volume
while table[w][v]:
# Find the last item that was added:
aux = [table[w-item.weight][v-item.volume] + item.value for item in items]
i = aux.index(table[w][v])
# Record it in the result, and remove it:
result[i] += 1
w -= items[i].weight
v -= items[i].volume
return result
max_items = knapsack_dp(items, sack)
max_value, max_weight, max_volume = tot_value(max_items, items, sack)
max_weight = -max_weight
max_volume = -max_volume
print "The maximum value achievable (by exhaustive search) is %g." % max_value item_names = ", ".join(item.name for item in items) print " The number of %s items to achieve this is: %s, respectively." % (item_names, max_items) print " The weight to carry is %.3g, and the volume used is %.3g." % (max_weight, max_volume)</lang>
Sample 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
Tcl
The following code uses brute force, but that's tolerable as long as it takes just a split second to find all 4 solutions. The use of arrays makes the quote quite legible: <lang Tcl>
- !/usr/bin/env tclsh
proc main argv {
array set value {panacea 3000 ichor 1800 gold 2500}
array set weight {panacea 0.3 ichor 0.2 gold 2.0 max 25}
array set volume {panacea 0.025 ichor 0.015 gold 0.002 max 0.25}
foreach i {panacea ichor gold} {
set max($i) [expr {min(int($volume(max)/$volume($i)),
int($weight(max)/$weight($i)))}]
}
set maxval 0
for {set i 0} {$i < $max(ichor)} {incr i} {
for {set p 0} {$p < $max(panacea)} {incr p} {
for {set g 0} {$g < $max(gold)} {incr g} {
if {$i*$weight(ichor) + $p*$weight(panacea) + $g*$weight(gold)
> $weight(max)} continue
if {$i*$volume(ichor) + $p*$volume(panacea) + $g*$volume(gold)
> $volume(max)} continue
set val [expr {$i*$value(ichor)+$p*$value(panacea)+$g*$value(gold)}]
if {$val == $maxval} {
lappend best [list i $i p $p g $g]
} elseif {$val > $maxval} {
set maxval $val
set best [list [list i $i p $p g $g]]
}
}
}
}
puts "maxval: $maxval, best: $best"
} main $argv
$ 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 </lang>
Visual Basic
<lang vb>Option Explicit
Type TreasureType
Name As String Units As String Value As Currency weight As Single Volume As Single
End Type
Type SolutionType
Desc As String Value As Currency
End Type
Type KnapsackType
Contents() As Integer CapacityWeight As Single CapacityVolume As Single
End Type
Dim Treasures() As TreasureType
Public Sub Main()
SetupTreasureShangriLa Debug.Print CalcKnapsack(25, 0.25)
End Sub
Public Sub SetupTreasureShangriLa()
ReDim Treasures(3) As TreasureType
With Treasures(1)
.Name = "panacea"
.Units = "vials"
.Value = 3000
.weight = 0.3
.Volume = 0.025
End With
With Treasures(2)
.Name = "ichor"
.Units = "ampules"
.Value = 1800
.weight = 0.2
.Volume = 0.015
End With
With Treasures(3)
.Name = "gold"
.Units = "bars"
.Value = 2500
.weight = 2
.Volume = 0.002
End With
End Sub
Public Function CalcKnapsack(ByVal sCapacityWeight As Single, ByVal sCapacityVolume As Single) As String Dim Knapsack As KnapsackType Dim Solution As SolutionType
Knapsack.CapacityVolume = sCapacityVolume Knapsack.CapacityWeight = sCapacityWeight ReDim Knapsack.Contents(UBound(Treasures)) As Integer Call Stuff(Knapsack, Solution, 1) Debug.Print "Maximum value: " & Solution.Value Debug.Print "Ideal Packing(s): " & vbCrLf & Solution.Desc
End Function
Private Sub Stuff(ByRef Knapsack As KnapsackType, ByRef Solution As SolutionType, ByVal nDepth As Integer) Dim nI As Integer Dim curVal As Currency Dim sWeightRemaining As Single Dim sVolumeRemaining As Single Dim nJ As Integer
sWeightRemaining = CalcWeightRemaining(Knapsack) sVolumeRemaining = CalcvolumeRemaining(Knapsack)
With Treasures(nDepth)
If nDepth = UBound(Treasures) Then
Knapsack.Contents(nDepth) = Min(Fix(sWeightRemaining / .weight), Fix(sVolumeRemaining / .Volume))
curVal = CalcValue(Knapsack)
If curVal > Solution.Value Then
Solution.Value = curVal
Solution.Desc = BuildDesc(Knapsack)
ElseIf curVal = Solution.Value Then
Solution.Desc = Solution.Desc & vbCrLf & "or" & vbCrLf & vbCrLf & BuildDesc(Knapsack)
End If
Else
For nI = 0 To Min(Fix(sWeightRemaining / .weight), Fix(sVolumeRemaining / .Volume))
Knapsack.Contents(nDepth) = nI
For nJ = nDepth + 1 To UBound(Treasures)
Knapsack.Contents(nJ) = 0
Next nJ
Call Stuff(Knapsack, nDepth + 1)
Next nI
End If
End With
End Sub
Private Function CalcValue(ByRef Knapsack As KnapsackType) As Currency Dim curTmp As Currency Dim nI As Integer
For nI = 1 To UBound(Treasures)
curTmp = curTmp + (Treasures(nI).Value * Knapsack.Contents(nI))
Next nI
CalcValue = curTmp
End Function
Private Function Min(ByVal vA As Variant, ByVal vB As Variant) As Variant
If vA < vB Then
Min = vA
Else
Min = vB
End If
End Function
Private Function CalcWeightRemaining(ByRef Knapsack As KnapsackType) As Single Dim sTmp As Single Dim nI As Integer
For nI = 1 To UBound(Treasures)
sTmp = sTmp + (Treasures(nI).weight * Knapsack.Contents(nI))
Next nI
CalcWeightRemaining = Knapsack.CapacityWeight - sTmp
End Function
Private Function CalcvolumeRemaining(ByRef Knapsack As KnapsackType) As Single Dim sTmp As Single Dim nI As Integer
For nI = 1 To UBound(Treasures)
sTmp = sTmp + (Treasures(nI).Volume * Knapsack.Contents(nI))
Next nI
CalcvolumeRemaining = Knapsack.CapacityVolume - sTmp
End Function
Private Function BuildDesc(ByRef Knapsack As KnapsackType) As String Dim cTmp As String Dim nI As Integer
For nI = 1 To UBound(Treasures)
cTmp = cTmp & Knapsack.Contents(nI) & " " & Treasures(nI).Units & " of " & Treasures(nI).Name & vbCrLf
Next nI
BuildDesc = cTmp
End Function</lang>
Output:
Maximum value: 54500 Ideal Packing(s): 0 vials of panacea 15 ampules of ichor 11 bars of gold or 3 vials of panacea 10 ampules of ichor 11 bars of gold or 6 vials of panacea 5 ampules of ichor 11 bars of gold or 9 vials of panacea 0 ampules of ichor 11 bars of gold