# Knapsack problem/Bounded

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

A tourist wants to make a good trip at the weekend with his friends.

They will go to the mountains to see the wonders of nature.   So he needs some items during the trip.   Food, clothing, etc.   He has a good knapsack for carrying the things, but he knows that he can carry only 4 kg weight in his knapsack, because they will make the trip from morning to evening.

He creates a list of what he wants to bring for the trip, but the total weight of all items is too much.   He adds a value to each item.   The value represents how important the thing for the tourist.

The list contains which items are the wanted things for the trip, what is the weight and value of an item, and how many units does he have from each items.

This is the list:

Table of potential knapsack items
item weight (dag) (each) value (each) piece(s)
map 9 150 1
compass 13 35 1
water 153 200 2
sandwich 50 60 2
glucose 15 60 2
tin 68 45 3
banana 27 60 3
apple 39 40 3
cheese 23 30 1
beer 52 10 3
suntan cream 11 70 1
camera 32 30 1
T-shirt 24 15 2
trousers 48 10 2
umbrella 73 40 1
waterproof trousers 42 70 1
waterproof overclothes 43 75 1
note-case 22 80 1
sunglasses 7 20 1
towel 18 12 2
socks 4 50 1
book 30 10 2
knapsack ≤400 dag  ?  ?

The tourist can choose to take any combination of items from the list, and some number of each item is available   (see the column   piece(s)   in the list above).

He may not cut the items, so he can only take whole units of any item.

Show which items does the tourist carry in his knapsack so that their total weight does not exceed 4 kg, and their total value is maximized.

## AutoHotkey

iterative dynamic programming solution

`Item = map,compass,water,sandwich,glucose,tin,banana,apple,cheese,beer,suntan cream      ,camera,tshirt,trousers,umbrella,waterproof trousers,waterproof overclothes,notecase      ,sunglasses,towel,socks,bookWeight= 9,13,153,50,15,68,27,39,23,52,11,32,24,48,73,42,43,22,7,18,4,30Value = 150,35,200,60,60,45,60,40,30,10,70,30,15,10,40,70,75,80,20,12,50,10Bound = 1,1,2,2,2,3,3,3,1,3,1,1,2,2,1,1,1,1,1,2,1,2 StringSplit I, Item,  `, ; Put input in arraysStringSplit W, Weight,`,StringSplit V, Value, `,StringSplit B, Bound, `, W := 400, N := I0, I0 := V0 := W0 := 0 ; W = total weight allowed, maximize total valueLoop %W%   m0_%A_Index% := 0 Loop %N% { ; build achievable value matrix m [ N rows, W columns ]   j := -1+i := A_Index,  m%j%_0 := 0 ; m[i,P] = max value with items 1..i, weight <=P   Loop %W% {                         ; m[i,P] = max_k {m[i-1,P-k*Wi]}      p := A_Index, k := 0, y := m%j%_%p%      While ++k <= B%i% && (r := p - k*W%i%) >= 0         y := y < (c:=m%j%_%r%+k*V%i%) ? c : y      m%i%_%p% := y   }} i := 1+j := N, p := W, s := 0While --i, --j { ; read out solution from value matrix m   If (m%i%_%p% = m%j%_%p%)      Continue   r := p, m := m%i%_%p%, k := 1   While 0 <= (r-=W%i%)  &&  m%j%_%r% != (m-=V%i%)      k++ ; find multiplier   t := k " " I%i% "`n" t, s += k*W%i%, p -= k*W%i%} MsgBox % "Value = " m%N%_%W% "`nWeight = " s "`n`n" t`

## Bracmat

`(knapsack=  ( things  =   (map.9.150.1)      (compass.13.35.1)      (water.153.200.2)      (sandwich.50.60.2)      (glucose.15.60.2)      (tin.68.45.3)      (banana.27.60.3)      (apple.39.40.3)      (cheese.23.30.1)      (beer.52.10.3)      (suntan cream.11.70.1)      (camera.32.30.1)      (T-shirt.24.15.2)      (trousers.48.10.2)      (umbrella.73.40.1)      (waterproof trousers.42.70.1)      (waterproof overclothes.43.75.1)      (note-case.22.80.1)      (sunglasses.7.20.1)      (towel.18.12.2)      (socks.4.50.1)      (book.30.10.2)  )& 0:?maxvalue& :?sack& ( add  =     cumwght        cumvalue        cumsack        name        wght        val        pcs        tings        n        ncumwght        ncumvalue    .     !arg        : ( ?cumwght          . ?cumvalue          . ?cumsack          . (?name.?wght.?val.?pcs) ?tings          )      & -1:?n      &   whl        ' ( 1+!n:~>!pcs:?n          & !cumwght+!n*!wght:~>400:?ncumwght          & !cumvalue+!n*!val:?ncumvalue          & (   !tings:              & (   !ncumvalue:>!maxvalue:?maxvalue                  &     !cumsack                        ( !n:0&                        | (!n.!name)                        )                    : ?sack                |                 )            |   add              \$ ( !ncumwght                . !ncumvalue                .   !cumsack                    (!n:0&|(!n.!name))                . !tings                )            )          )  )& add\$(0.0..!things)& out\$(!maxvalue.!sack)); !knapsack;`
Output:
```  1010
.   (1.map)
(1.compass)
(1.water)
(2.glucose)
(3.banana)
(1.cheese)
(1.suntan cream)
(1.waterproof overclothes)
(1.note-case)
(1.sunglasses)
(1.socks)```

## C

`#include <stdio.h>#include <stdlib.h> typedef struct {    char *name;    int weight;    int value;    int count;} item_t; item_t items[] = {    {"map",                      9,   150,   1},    {"compass",                 13,    35,   1},    {"water",                  153,   200,   2},    {"sandwich",                50,    60,   2},    {"glucose",                 15,    60,   2},    {"tin",                     68,    45,   3},    {"banana",                  27,    60,   3},    {"apple",                   39,    40,   3},    {"cheese",                  23,    30,   1},    {"beer",                    52,    10,   3},    {"suntan cream",            11,    70,   1},    {"camera",                  32,    30,   1},    {"T-shirt",                 24,    15,   2},    {"trousers",                48,    10,   2},    {"umbrella",                73,    40,   1},    {"waterproof trousers",     42,    70,   1},    {"waterproof overclothes",  43,    75,   1},    {"note-case",               22,    80,   1},    {"sunglasses",               7,    20,   1},    {"towel",                   18,    12,   2},    {"socks",                    4,    50,   1},    {"book",                    30,    10,   2},}; int n = sizeof (items) / sizeof (item_t); int *knapsack (int w) {    int i, j, k, v, *mm, **m, *s;    mm = calloc((n + 1) * (w + 1), sizeof (int));    m = malloc((n + 1) * sizeof (int *));    m[0] = mm;    for (i = 1; i <= n; i++) {        m[i] = &mm[i * (w + 1)];        for (j = 0; j <= w; j++) {            m[i][j] = m[i - 1][j];            for (k = 1; k <= items[i - 1].count; k++) {                if (k * items[i - 1].weight > j) {                    break;                }                v = m[i - 1][j - k * items[i - 1].weight] + k * items[i - 1].value;                if (v > m[i][j]) {                    m[i][j] = v;                }            }        }    }    s = calloc(n, sizeof (int));    for (i = n, j = w; i > 0; i--) {        int v = m[i][j];        for (k = 0; v != m[i - 1][j] + k * items[i - 1].value; k++) {            s[i - 1]++;            j -= items[i - 1].weight;        }    }    free(mm);    free(m);    return s;} int main () {    int i, tc = 0, tw = 0, tv = 0, *s;    s = knapsack(400);    for (i = 0; i < n; i++) {        if (s[i]) {            printf("%-22s %5d %5d %5d\n", items[i].name, s[i], s[i] * items[i].weight, s[i] * items[i].value);            tc += s[i];            tw += s[i] * items[i].weight;            tv += s[i] * items[i].value;        }    }    printf("%-22s %5d %5d %5d\n", "count, weight, value:", tc, tw, tv);    return 0;} `
Output:
```map                        1     9   150
compass                    1    13    35
water                      1   153   200
glucose                    2    30   120
banana                     3    81   180
cheese                     1    23    30
suntan cream               1    11    70
waterproof overclothes     1    43    75
note-case                  1    22    80
sunglasses                 1     7    20
socks                      1     4    50
count, weight, value:     14   396  1010```

## C++

C++ DP solution. Initially taken from C but than fixed and refactored.
Remark: The above comment implies there is a bug in the C code, but refers to a much older and very different version Pete Lomax (talk) 19:28, 20 March 2017 (UTC)

`#include <iostream>#include <vector>#include <algorithm>#include <stdexcept>#include <memory>#include <sys/time.h> using std::cout;using std::endl; class StopTimer{public:    StopTimer(): begin_(getUsec()) {}    unsigned long long getTime() const { return getUsec() - begin_; }private:    static unsigned long long getUsec()    {//...you might want to use something else under Windows        timeval tv;        const int res = ::gettimeofday(&tv, 0);        if(res)            return 0;        return tv.tv_usec + 1000000 * tv.tv_sec;    }    unsigned long long begin_;}; struct KnapsackTask{    struct Item    {        std::string name;        unsigned w, v, qty;        Item(): w(), v(), qty() {}        Item(const std::string& iname, unsigned iw, unsigned iv, unsigned iqty):            name(iname), w(iw), v(iv), qty(iqty)        {}    };    typedef std::vector<Item> Items;    struct Solution    {        unsigned v, w;        unsigned long long iterations, usec;        std::vector<unsigned> n;        Solution(): v(), w(), iterations(), usec() {}    };    //...    KnapsackTask(): maxWeight_(), totalWeight_() {}    void add(const Item& item)    {        const unsigned totalItemWeight = item.w * item.qty;        if(const bool invalidItem = !totalItemWeight)            throw std::logic_error("Invalid item: " + item.name);        totalWeight_ += totalItemWeight;        items_.push_back(item);    }    const Items& getItems() const { return items_; }    void setMaxWeight(unsigned maxWeight) { maxWeight_ = maxWeight; }    unsigned getMaxWeight() const { return std::min(totalWeight_, maxWeight_); } private:    unsigned maxWeight_, totalWeight_;    Items items_;}; class BoundedKnapsackRecursiveSolver{public:    typedef KnapsackTask Task;    typedef Task::Item Item;    typedef Task::Items Items;    typedef Task::Solution Solution;     void solve(const Task& task)    {        Impl(task, solution_).solve();    }    const Solution& getSolution() const { return solution_; }private:    class Impl    {        struct Candidate        {            unsigned v, n;            bool visited;            Candidate(): v(), n(), visited(false) {}        };        typedef std::vector<Candidate> Cache;    public:        Impl(const Task& task, Solution& solution):            items_(task.getItems()),            maxWeight_(task.getMaxWeight()),            maxColumnIndex_(task.getItems().size() - 1),            solution_(solution),            cache_(task.getMaxWeight() * task.getItems().size()),            iterations_(0)        {}        void solve()        {            if(const bool nothingToSolve = !maxWeight_ || items_.empty())                return;            StopTimer timer;            Candidate candidate;            solve(candidate, maxWeight_, items_.size() - 1);            convertToSolution(candidate);            solution_.usec = timer.getTime();        }    private:        void solve(Candidate& current, unsigned reminderWeight, const unsigned itemIndex)        {            ++iterations_;             const Item& item(items_[itemIndex]);             if(const bool firstColumn = !itemIndex)            {                const unsigned maxQty = std::min(item.qty, reminderWeight/item.w);                current.v = item.v * maxQty;                current.n = maxQty;                current.visited = true;            }            else            {                const unsigned nextItemIndex = itemIndex - 1;                {                    Candidate& nextItem = cachedItem(reminderWeight, nextItemIndex);                    if(!nextItem.visited)                        solve(nextItem, reminderWeight, nextItemIndex);                    current.visited = true;                    current.v = nextItem.v;                    current.n = 0;                }                if(reminderWeight >= item.w)                {                    for (unsigned numberOfItems = 1; numberOfItems <= item.qty; ++numberOfItems)                    {                        reminderWeight -= item.w;                        Candidate& nextItem = cachedItem(reminderWeight, nextItemIndex);                        if(!nextItem.visited)                            solve(nextItem, reminderWeight, nextItemIndex);                         const unsigned checkValue = nextItem.v + numberOfItems * item.v;                        if ( checkValue > current.v)                        {                            current.v = checkValue;                            current.n = numberOfItems;                        }                        if(!(reminderWeight >= item.w))                            break;                    }                }            }        }        void convertToSolution(const Candidate& candidate)        {            solution_.iterations = iterations_;            solution_.v = candidate.v;            solution_.n.resize(items_.size());             const Candidate* iter = &candidate;            unsigned weight = maxWeight_, itemIndex = items_.size() - 1;            while(true)            {                const unsigned currentWeight = iter->n * items_[itemIndex].w;                solution_.n[itemIndex] = iter->n;                weight -= currentWeight;                if(!itemIndex--)                    break;                iter = &cachedItem(weight, itemIndex);            }            solution_.w = maxWeight_ - weight;        }        Candidate& cachedItem(unsigned weight, unsigned itemIndex)        {            return cache_[weight * maxColumnIndex_ + itemIndex];        }        const Items& items_;        const unsigned maxWeight_;        const unsigned maxColumnIndex_;        Solution& solution_;        Cache cache_;        unsigned long long iterations_;    };    Solution solution_;}; void populateDataset(KnapsackTask& task){    typedef KnapsackTask::Item Item;    task.setMaxWeight( 400 );    task.add(Item("map",9,150,1));    task.add(Item("compass",13,35,1));    task.add(Item("water",153,200,2));    task.add(Item("sandwich",50,60,2));    task.add(Item("glucose",15,60,2));    task.add(Item("tin",68,45,3));    task.add(Item("banana",27,60,3));    task.add(Item("apple",39,40,3));    task.add(Item("cheese",23,30,1));    task.add(Item("beer",52,10,3));    task.add(Item("suntancream",11,70,1));    task.add(Item("camera",32,30,1));    task.add(Item("T-shirt",24,15,2));    task.add(Item("trousers",48,10,2));    task.add(Item("umbrella",73,40,1));    task.add(Item("w-trousers",42,70,1));    task.add(Item("w-overclothes",43,75,1));    task.add(Item("note-case",22,80,1));    task.add(Item("sunglasses",7,20,1));    task.add(Item("towel",18,12,2));    task.add(Item("socks",4,50,1));    task.add(Item("book",30,10,2));} int main(){    KnapsackTask task;    populateDataset(task);     BoundedKnapsackRecursiveSolver solver;    solver.solve(task);    const KnapsackTask::Solution& solution = solver.getSolution();     cout << "Iterations to solve: " << solution.iterations << endl;    cout << "Time to solve: " << solution.usec << " usec" << endl;    cout << "Solution:" << endl;    for (unsigned i = 0; i < solution.n.size(); ++i)    {        if (const bool itemIsNotInKnapsack = !solution.n[i])            continue;        cout << "  " << solution.n[i] << ' ' << task.getItems()[i].name << " ( item weight = " << task.getItems()[i].w << " )" << endl;    }     cout << "Weight: " << solution.w << " Value: " << solution.v << endl;    return 0;}`

## Clojure

We convert the problem to a Knapsack-0/1 problem by replacing (n-max item) vith n-max identical occurences of 1 item. Adapted Knapsack-0/1 problem solution from [1]

`(ns knapsack  (:gen-class)) (def groupeditems [                   ["map", 9, 150, 1]                   ["compass", 13, 35, 1]                   ["water", 153, 200, 3]                   ["sandwich", 50, 60, 2]                   ["glucose", 15, 60, 2]                   ["tin", 68, 45, 3]                   ["banana", 27, 60, 3]                   ["apple", 39, 40, 3]                   ["cheese", 23, 30, 1]                   ["beer", 52, 10, 3]                   ["suntan cream", 11, 70, 1]                   ["camera", 32, 30, 1]                   ["t-shirt", 24, 15, 2]                   ["trousers", 48, 10, 2]                   ["umbrella", 73, 40, 1]                   ["waterproof trousers", 42, 70, 1]                   ["waterproof overclothes", 43, 75, 1]                   ["note-case", 22, 80, 1]                   ["sunglasses", 7, 20, 1]                   ["towel", 18, 12, 2]                   ["socks", 4, 50, 1]                   ["book", 30, 10, 2]                  ]) (defstruct item :name :weight :value) (def items (->> (for [[item wt val n] groupeditems]                   (repeat n [item wt val]))                 (mapcat identity)                 (map #(apply struct item %))                 (vec))) (declare mm) ;forward decl for memoization function(defn m [i w]  (cond    (< i 0) [0 []]    (= w 0) [0 []]    :else    (let [{wi :weight vi :value} (get items i)]      (if (> wi w)        (mm (dec i) w)        (let [[vn sn :as no]  (mm (dec i) w)              [vy sy :as yes] (mm (dec i) (- w wi))]          (if (> (+ vy vi) vn)            [(+ vy vi) (conj sy i)]            no)))))) (def mm (memoize m)) (let [[value indexes] (mm (-> items count dec) 400)      names (map (comp :name items) indexes)]  (println "Items to pack:")  (doseq [[k v] (frequencies names)]    (println (format "%d %s" v k)))  (println "Total value:" value)  (println "Total weight:" (reduce + (map (comp :weight items) indexes)))) `
Output:
```Items to pack:
1 suntan cream
1 map
3 banana
1 water
1 compass
2 glucose
1 note-case
1 waterproof overclothes
1 cheese
1 sunglasses
1 socks
Total value: 1010
Total weight: 396
```

## Common Lisp

`;;; memoize(defmacro mm-set (p v) `(if ,p ,p (setf ,p ,v))) (defun knapsack (max-weight items)  (let ((cache (make-array (list (1+ max-weight) (1+ (length items)))			   :initial-element nil)))     (labels ((knapsack1 (spc items)	(if (not items) (return-from knapsack1 (list 0 0 '())))	 (mm-set (aref cache spc (length items))	   (let* ((i (first items))		  (w (second i))		  (v (third i))		  (x (knapsack1 spc (cdr items))))	     (loop for cnt from 1 to (fourth i) do		   (let ((w (* cnt w)) (v (* cnt v)))		     (if (>= spc w)		       (let ((y (knapsack1 (- spc w) (cdr items))))			 (if (> (+ (first y) v) (first x))			   (setf x (list (+ (first  y) v)					 (+ (second y) w)					 (cons (list (first i) cnt) (third y)))))))))	     x))))       (knapsack1 max-weight items)))) (print  (knapsack 400	    '((map 9 150 1) (compass 13 35 1) (water 153 200 2) (sandwich 50 60 2)              (glucose 15 60 2) (tin 68 45 3) (banana 27 60 3) (apple 39 40 3)	      (cheese 23 30 1) (beer 52 10 3) (cream 11 70 1) (camera 32 30 1)	      (T-shirt 24 15 2) (trousers 48 10 2) (umbrella 73 40 1)	      (trousers 42 70 1) (overclothes 43 75 1) (notecase 22 80 1)	      (glasses 7 20 1) (towel 18 12 2) (socks 4 50 1) (book 30 10 2))))`

## D

Translation of: Python

Solution with memoization.

`import std.stdio, std.typecons, std.functional; immutable struct Item {    string name;    int weight, value, quantity;} immutable Item[] items = [    {"map",           9, 150, 1}, {"compass",      13,  35, 1},    {"water",       153, 200, 3}, {"sandwich",     50,  60, 2},    {"glucose",      15,  60, 2}, {"tin",          68,  45, 3},    {"banana",       27,  60, 3}, {"apple",        39,  40, 3},    {"cheese",       23,  30, 1}, {"beer",         52,  10, 3},    {"suntan cream", 11,  70, 1}, {"camera",       32,  30, 1},    {"t-shirt",      24,  15, 2}, {"trousers",     48,  10, 2},    {"umbrella",     73,  40, 1}, {"w-trousers",   42,  70, 1},    {"w-overcoat",   43,  75, 1}, {"note-case",    22,  80, 1},    {"sunglasses",    7,  20, 1}, {"towel",        18,  12, 2},    {"socks",         4,  50, 1}, {"book",         30,  10, 2}]; Tuple!(int, const(int)[]) chooseItem(in int iWeight, in int idx) nothrow @safe {    alias memoChooseItem = memoize!chooseItem;    if (idx < 0)        return typeof(return)();     int bestV;    const(int)[] bestList;    with (items[idx])        foreach (immutable i; 0 .. quantity + 1) {            immutable wlim = iWeight - i * weight;            if (wlim < 0)                break;             //const (val, taken) = memoChooseItem(wlim, idx - 1);            const val_taken = memoChooseItem(wlim, idx - 1);            if (val_taken[0] + i * value > bestV) {                bestV = val_taken[0] + i * value;                bestList = val_taken[1] ~ i;            }        }     return tuple(bestV, bestList);} void main() {    // const (v, lst) = chooseItem(400, items.length - 1);    const v_lst = chooseItem(400, items.length - 1);     int w;    foreach (immutable i, const cnt; v_lst[1])        if (cnt > 0) {            writeln(cnt, " ", items[i].name);            w += items[i].weight * cnt;        }     writeln("Total weight: ", w, " Value: ", v_lst[0]);}`
Output:
```1 map
1 compass
1 water
2 glucose
3 banana
1 cheese
1 suntan cream
1 w-overcoat
1 note-case
1 sunglasses
1 socks
Total weight: 396 Value: 1010```

## EchoLisp

We convert the problem to a Knapsack-0/1 problem by replacing (n-max item) vith n-max identical occurences of 1 item.

` (lib 'struct)(lib 'sql)(lib 'hash) (define H (make-hash))(define T (make-table (struct goodies (name poids valeur qty))))(define IDX (make-vector 0))  ;; convert to 0/1 PB;; transorm vector [1 2 3 4 (n-max 3) 5 (n-max 2) 6 .. ];; into vector of repeated indices : [1 2 3 4 4 4 5 5 6 ... ] (define (make-01 T)	 (for ((record T) (i (in-naturals)))		 (for ((j (in-range 0 (goodies-qty record))))			 (vector-push IDX i)))	 IDX) (define-syntax-rule (name i) (table-xref T (vector-ref IDX i) 0))(define-syntax-rule (poids i) (table-xref T (vector-ref IDX i) 1))(define-syntax-rule (valeur i) (table-xref T (vector-ref IDX i) 2)) ;;;; code identical to 0/1 problem	;; ;;  make an unique hash-key from (i rest)(define (t-idx i r)  (string-append i "|" r));; retrieve best core for item i, remaining r availbble weight(define (t-get i r)  (or (hash-ref H (t-idx i r)) 0)) ;; compute best score (i), assuming best (i-1 rest) is known(define (score i restant)	 (if (< i 0) 0	 (hash-ref! H (t-idx i restant)		 (if ( >= restant (poids i)) 			 (max 			 (score (1- i) restant) 			 (+ (score (1- i) (- restant (poids i))) (valeur i)))		    (score (1- i) restant)) ;; else not enough	))) ;; compute best scores, starting from last item(define (task W)        (define restant W)        (make-01 T)         (define N (1- (vector-length IDX)))		 (writeln 'total-value (score N W))		 (group		 (for/list  ((i (in-range N -1 -1)))		 #:continue (= (t-get i restant) (t-get (1- i) restant))		(set! restant (- restant (poids i)))		(name i))))  `
Output:
` (define goodies    '((map  9 150 1) (compass 13 35 1)(water 153 200 3) (sandwich 50 60 2)(🍰-glucose 15 60 2)(tin  68 45 3) (🍌-banana  27 60 3)(🍎-apple 39 40 3)(cheese 23 30 1) (beer 52 10 3)(🌞-suntan-cream  11  70 1)(camera 32 30 1) (t-shirt 24 15 2)(trousers 48 10 2)(umbrella 73 40 1) (☔️-trousers  42 70 1)(☔️--overcoat 43 75 1)(note-case 22 80 1) (🌞-sunglasses 7 20 1)(towel 18 12 2)(socks 4 50 1) (book 30 10 2))) (list->table goodies T) (task 400)total-value     1010       → ((socks) (🌞-sunglasses) (note-case) (☔️--overcoat) (🌞-suntan-cream) (cheese) (🍌-banana 🍌-banana 🍌-banana) (🍰-glucose 🍰-glucose) (water) (compass) (map)) (length (hash-keys H))    → 10827 ;; # of entries in cache `

## Go

Solution with caching.

`package main import "fmt" type Item struct {	name               string	weight, value, qty int} var items = []Item{	{"map",			9,	150,	1},	{"compass",		13,	35,	1},	{"water",		153,	200,	2},	{"sandwich",		50,	60,	2},	{"glucose",		15,	60,	2},	{"tin",			68,	45,	3},	{"banana",		27,	60,	3},	{"apple",		39,	40,	3},	{"cheese",		23,	30,	1},	{"beer",		52,	10,	3},	{"suntancream",		11,	70,	1},	{"camera",		32,	30,	1},	{"T-shirt",		24,	15,	2},	{"trousers",		48,	10,	2},	{"umbrella",		73,	40,	1},	{"w-trousers",		42,	70,	1},	{"w-overclothes",	43,	75,	1},	{"note-case",		22,	80,	1},	{"sunglasses",		7,      20,	1},	{"towel",		18,	12,	2},	{"socks",		4,      50,	1},	{"book",		30,	10,	2},} type Chooser struct {	Items []Item	cache map[key]solution} type key struct {	w, p int} type solution struct {	v, w int	qty  []int} func (c Chooser) Choose(limit int) (w, v int, qty []int) {	c.cache = make(map[key]solution)	s := c.rchoose(limit, len(c.Items)-1)	c.cache = nil // allow cache to be garbage collected	return s.v, s.w, s.qty} func (c Chooser) rchoose(limit, pos int) solution {	if pos < 0 || limit <= 0 {		return solution{0, 0, nil}	} 	key := key{limit, pos}	if s, ok := c.cache[key]; ok {		return s	} 	best_i, best := 0, solution{0, 0, nil}	for i := 0; i*items[pos].weight <= limit && i <= items[pos].qty; i++ {		sol := c.rchoose(limit-i*items[pos].weight, pos-1)		sol.v += i * items[pos].value		if sol.v > best.v {			best_i, best = i, sol		}	} 	if best_i > 0 {		// best.qty is used in another cache entry,		// we need to duplicate it before modifying it to		// store as our cache entry.		old := best.qty		best.qty = make([]int, len(items))		copy(best.qty, old)		best.qty[pos] = best_i		best.w += best_i * items[pos].weight	}	c.cache[key] = best	return best} func main() {	v, w, s := Chooser{Items: items}.Choose(400) 	fmt.Println("Taking:")	for i, t := range s {		if t > 0 {			fmt.Printf("  %d of %d %s\n", t, items[i].qty, items[i].name)		}	}	fmt.Printf("Value: %d; weight: %d\n", v, w)}`

(A simple test and benchmark used while making changes to make sure performance wasn't sacrificed is available at /Go_test.)

Output:
```Taking:
1 of 1 map
1 of 1 compass
1 of 2 water
2 of 2 glucose
3 of 3 banana
1 of 1 cheese
1 of 1 suntancream
1 of 1 w-overclothes
1 of 1 note-case
1 of 1 sunglasses
1 of 1 socks
Value: 1010; weight: 396```

## Groovy

Solution: dynamic programming

`def totalWeight = { list -> list.collect{ it.item.weight * it.count }.sum() }def totalValue = { list -> list.collect{ it.item.value * it.count }.sum() } def knapsackBounded = { possibleItems ->    def n = possibleItems.size()    def m = (0..n).collect{ i -> (0..400).collect{ w -> []} }    (1..400).each { w ->        (1..n).each { i ->            def item = possibleItems[i-1]            def wi = item.weight, pi = item.pieces            def bi = [w.intdiv(wi),pi].min()            m[i][w] = (0..bi).collect{ count ->                m[i-1][w - wi * count] + [[item:item, count:count]]            }.max(totalValue).findAll{ it.count }        }    }    m[n][400]}`

Test:

`def items = [         [name:"map",                    weight:  9, value:150, pieces:1],        [name:"compass",                weight: 13, value: 35, pieces:1],        [name:"water",                  weight:153, value:200, pieces:2],        [name:"sandwich",               weight: 50, value: 60, pieces:2],        [name:"glucose",                weight: 15, value: 60, pieces:2],        [name:"tin",                    weight: 68, value: 45, pieces:3],        [name:"banana",                 weight: 27, value: 60, pieces:3],        [name:"apple",                  weight: 39, value: 40, pieces:3],        [name:"cheese",                 weight: 23, value: 30, pieces:1],        [name:"beer",                   weight: 52, value: 10, pieces:3],        [name:"suntan cream",           weight: 11, value: 70, pieces:1],        [name:"camera",                 weight: 32, value: 30, pieces:1],        [name:"t-shirt",                weight: 24, value: 15, pieces:2],        [name:"trousers",               weight: 48, value: 10, pieces:2],        [name:"umbrella",               weight: 73, value: 40, pieces:1],        [name:"waterproof trousers",    weight: 42, value: 70, pieces:1],        [name:"waterproof overclothes", weight: 43, value: 75, pieces:1],        [name:"note-case",              weight: 22, value: 80, pieces:1],        [name:"sunglasses",             weight:  7, value: 20, pieces:1],        [name:"towel",                  weight: 18, value: 12, pieces:2],        [name:"socks",                  weight:  4, value: 50, pieces:1],        [name:"book",                   weight: 30, value: 10, pieces:2],] def start = System.currentTimeMillis()def packingList = knapsackBounded(items)def elapsed = System.currentTimeMillis() - start println "Elapsed Time: \${elapsed/1000.0} s"println "Total Weight: \${totalWeight(packingList)}"println " Total Value: \${totalValue(packingList)}"packingList.each {    printf ('  item: %-22s  weight:%4d  value:%4d  count:%2d\n',            it.item.name, it.item.weight, it.item.value, it.count)}`
Output:
```Elapsed Time: 0.603 s
Total Weight: 396
Total Value: 1010
item: map                     weight:   9  value: 150  count: 1
item: compass                 weight:  13  value:  35  count: 1
item: water                   weight: 153  value: 200  count: 1
item: glucose                 weight:  15  value:  60  count: 2
item: banana                  weight:  27  value:  60  count: 3
item: cheese                  weight:  23  value:  30  count: 1
item: suntan cream            weight:  11  value:  70  count: 1
item: waterproof overclothes  weight:  43  value:  75  count: 1
item: note-case               weight:  22  value:  80  count: 1
item: sunglasses              weight:   7  value:  20  count: 1
item: socks                   weight:   4  value:  50  count: 1```

Directly lifted from 1-0 problem:

`inv = 	[("map",9,150,1), ("compass",13,35,1), ("water",153,200,2), ("sandwich",50,60,2),	("glucose",15,60,2), ("tin",68,45,3), ("banana",27,60,3), ("apple",39,40,3),	("cheese",23,30,1), ("beer",52,10,3), ("cream",11,70,1), ("camera",32,30,1),	-- what to do if we end up taking one trouser?	("tshirt",24,15,2), ("trousers",48,10,2), ("umbrella",73,40,1), ("wtrousers",42,70,1),	("woverclothes",43,75,1), ("notecase",22,80,1), ("sunglasses",7,20,1), ("towel",18,12,2),	("socks",4,50,1), ("book",30,10,2)] knapsack = foldr addItem (repeat (0,[])) where	addItem (name,w,v,c) old = foldr inc old [1..c] where		inc i list = left ++ zipWith max right new where			(left, right) = splitAt (w * i) list			new = map (\(val,itms)->(val + v * i, (name,i):itms)) old main = print \$ (knapsack inv) !! 400`
Output:
```(1010,[("socks",1),("sunglasses",1),("notecase",1),("woverclothes",1),("cream",1),("cheese",1),("banana",3),("glucose",2),("water",1),("compass",1),("map",1)])
```

The above uses merging lists for cache. It's faster, and maybe easier to understand when some constant-time lookup structure is used for cache (same output):

`import Data.Array -- snipped the item list; use the one from aboveknapsack items cap = (solu items) ! cap where	solu = foldr f (listArray (0,cap) (repeat (0,[])))	f (name,w,v,cnt) ss = listArray (0,cap) \$ map optimal [0..] where		optimal ww = maximum \$ (ss!ww):[prepend (v*i,(name,i)) (ss!(ww - i*w))					| i <- [1..cnt], i*w < ww]		prepend (x,n) (y,s) = (x+y,n:s) main = do print \$ knapsack inv 400`

## J

Brute force solution:

`'names numbers'=:|:".;._2]0 :0  'map';                      9       150        1  'compass';                 13        35        1  'water';                  153       200        2  'sandwich';                50        60        2  'glucose';                 15        60        2  'tin';                     68        45        3  'banana';                  27        60        3  'apple';                   39        40        3  'cheese';                  23        30        1  'beer';                    52        10        3  'suntan cream';            11        70        1  'camera';                  32        30        1  'T-shirt';                 24        15        2  'trousers';                48        10        2  'umbrella';                73        40        1  'waterproof trousers';     42        70        1  'waterproof overclothes';  43        75        1  'note-case';               22        80        1  'sunglasses';               7        20        1  'towel';                   18        12        2  'socks';                    4        50        1  'book';                    30        10        2) 'weights values pieces'=:|:numbersdecode=: (pieces+1)&#: pickBest=:4 :0  NB. given a list of options, return the best option(s)  n=. decode y  weight=. n+/ .*weights  value=.  (x >: weight) * n+/ .*values  (value = >./value)#y) bestCombo=:3 :0   limit=. */pieces+1   i=. 0   step=. 1e6   best=. ''   while.i<limit do.      best=. 400 pickBest best,(#~ limit&>)i+i.step      i=. i+step   end.   best)    bestCombo''978832641`

`   decode 9788326411 1 1 0 2 0 3 0 1 0 1 0 0 0 0 0 1 1 1 0 1 0    (0<decode 978832641) # (":,.decode 978832641),.' ',.names1 map                   1 compass               1 water                 2 glucose               3 banana                1 cheese                1 suntan cream          1 waterproof overclothes1 note-case             1 sunglasses            1 socks                    weights +/ .* decode 978832641396   values +/ .* decode 9788326411010`

Dynamic programming solution (faster):

`dyn=:3 :0  m=. 0\$~1+400,+/pieces NB. maximum value cache  b=. m                 NB. best choice cache  opts=.+/\0,pieces     NB. distinct item counts before each piece  P=. */\1+0,pieces   NB. distinct possibilities before each piece  for_w.1+i.400 do.    for_j.i.#pieces do.      n=. i.1+j{pieces         NB. possible counts for this piece      W=. n*j{weights          NB. how much they weigh      s=. w>:W                 NB. permissible options      v=. s*n*j{values         NB. consequent values      base=. j{opts            NB. base index for these options      I=. <"1 w,.n+base        NB. consequent indices      i0=. <w,base             NB. status quo index      iN=. <"1 (w-s*W),.base   NB. predecessor indices      M=. >./\(m{~i0)>.v+m{~iN NB. consequent maximum values      C=. (n*j{P)+b{~iN        NB. unique encoding for each option      B=. >./\(b{~i0)>. C * 2 ~:/\ 0,M NB. best options, so far      m=. M I} m       NB. update with newly computed maxima      b=. B I} b       NB. same for best choice    end.  end.  |.(1+|.pieces)#:{:{:b)    dyn''1 1 1 0 2 0 3 0 1 0 1 0 0 0 0 0 1 1 1 0 1 0`

Note: the brute force approach would return multiple "best answers" if more than one combination of choices would satisfy the "best" constraint. The dynamic approach arbitrarily picks one of those choices. That said, with this particular choice of item weights and values, this is an irrelevant distinction.

## Java

General dynamic solution after wikipedia. The solution extends the method of Knapsack problem/0-1#Java .

`package hu.pj.alg.test; import hu.pj.alg.BoundedKnapsack;import hu.pj.obj.Item;import java.util.*;import java.text.*; public class BoundedKnapsackForTourists {    public BoundedKnapsackForTourists() {        BoundedKnapsack bok = new BoundedKnapsack(400); // 400 dkg = 400 dag = 4 kg         // making the list of items that you want to bring        bok.add("map", 9, 150, 1);        bok.add("compass", 13, 35, 1);        bok.add("water", 153, 200, 3);        bok.add("sandwich", 50, 60, 2);        bok.add("glucose", 15, 60, 2);        bok.add("tin", 68, 45, 3);        bok.add("banana", 27, 60, 3);        bok.add("apple", 39, 40, 3);        bok.add("cheese", 23, 30, 1);        bok.add("beer", 52, 10, 3);        bok.add("suntan cream", 11, 70, 1);        bok.add("camera", 32, 30, 1);        bok.add("t-shirt", 24, 15, 2);        bok.add("trousers", 48, 10, 2);        bok.add("umbrella", 73, 40, 1);        bok.add("waterproof trousers", 42, 70, 1);        bok.add("waterproof overclothes", 43, 75, 1);        bok.add("note-case", 22, 80, 1);        bok.add("sunglasses", 7, 20, 1);        bok.add("towel", 18, 12, 2);        bok.add("socks", 4, 50, 1);        bok.add("book", 30, 10, 2);         // calculate the solution:        List<Item> itemList = bok.calcSolution();         // write out the solution in the standard output        if (bok.isCalculated()) {            NumberFormat nf  = NumberFormat.getInstance();             System.out.println(                "Maximal weight           = " +                nf.format(bok.getMaxWeight() / 100.0) + " kg"            );            System.out.println(                "Total weight of solution = " +                nf.format(bok.getSolutionWeight() / 100.0) + " kg"            );            System.out.println(                "Total value              = " +                bok.getProfit()            );            System.out.println();            System.out.println(                "You can carry te following materials " +                "in the knapsack:"            );            for (Item item : itemList) {                if (item.getInKnapsack() > 0) {                    System.out.format(                        "%1\$-10s %2\$-23s %3\$-3s %4\$-5s %5\$-15s \n",                        item.getInKnapsack() + " unit(s) ",                        item.getName(),                        item.getInKnapsack() * item.getWeight(), "dag  ",                        "(value = " + item.getInKnapsack() * item.getValue() + ")"                    );                }            }        } else {            System.out.println(                "The problem is not solved. " +                "Maybe you gave wrong data."            );        }     }     public static void main(String[] args) {        new BoundedKnapsackForTourists();    }} // class`
`package hu.pj.alg; import hu.pj.obj.Item;import java.util.*; public class BoundedKnapsack extends ZeroOneKnapsack {    public BoundedKnapsack() {}     public BoundedKnapsack(int _maxWeight) {        setMaxWeight(_maxWeight);    }     public BoundedKnapsack(List<Item> _itemList) {        setItemList(_itemList);    }     public BoundedKnapsack(List<Item> _itemList, int _maxWeight) {        setItemList(_itemList);        setMaxWeight(_maxWeight);    }     @Override    public List<Item> calcSolution() {        int n = itemList.size();         // add items to the list, if bounding > 1        for (int i = 0; i < n; i++) {            Item item = itemList.get(i);            if (item.getBounding() > 1) {                for (int j = 1; j < item.getBounding(); j++) {                    add(item.getName(), item.getWeight(), item.getValue());                }            }        }         super.calcSolution();         // delete the added items, and increase the original items        while (itemList.size() > n) {            Item lastItem = itemList.get(itemList.size() - 1);            if (lastItem.getInKnapsack() == 1) {                for (int i = 0; i < n; i++) {                    Item iH = itemList.get(i);                    if (lastItem.getName().equals(iH.getName())) {                        iH.setInKnapsack(1 + iH.getInKnapsack());                        break;                    }                }            }            itemList.remove(itemList.size() - 1);        }         return itemList;    }     // add an item to the item list    public void add(String name, int weight, int value, int bounding) {        if (name.equals(""))            name = "" + (itemList.size() + 1);        itemList.add(new Item(name, weight, value, bounding));        setInitialStateForCalculation();    }} // class`
`package hu.pj.alg; import hu.pj.obj.Item;import java.util.*; public class ZeroOneKnapsack {    protected List<Item> itemList  = new ArrayList<Item>();    protected int maxWeight        = 0;    protected int solutionWeight   = 0;    protected int profit           = 0;    protected boolean calculated   = false;     public ZeroOneKnapsack() {}     public ZeroOneKnapsack(int _maxWeight) {        setMaxWeight(_maxWeight);    }     public ZeroOneKnapsack(List<Item> _itemList) {        setItemList(_itemList);    }     public ZeroOneKnapsack(List<Item> _itemList, int _maxWeight) {        setItemList(_itemList);        setMaxWeight(_maxWeight);    }     // calculte the solution of 0-1 knapsack problem with dynamic method:    public List<Item> calcSolution() {        int n = itemList.size();         setInitialStateForCalculation();        if (n > 0  &&  maxWeight > 0) {            List< List<Integer> > c = new ArrayList< List<Integer> >();            List<Integer> curr = new ArrayList<Integer>();             c.add(curr);            for (int j = 0; j <= maxWeight; j++)                curr.add(0);            for (int i = 1; i <= n; i++) {                List<Integer> prev = curr;                c.add(curr = new ArrayList<Integer>());                for (int j = 0; j <= maxWeight; j++) {                    if (j > 0) {                        int wH = itemList.get(i-1).getWeight();                        curr.add(                            (wH > j)                            ?                            prev.get(j)                            :                            Math.max(                                prev.get(j),                                itemList.get(i-1).getValue() + prev.get(j-wH)                            )                        );                    } else {                        curr.add(0);                    }                } // for (j...)            } // for (i...)            profit = curr.get(maxWeight);             for (int i = n, j = maxWeight; i > 0  &&  j >= 0; i--) {                int tempI   = c.get(i).get(j);                int tempI_1 = c.get(i-1).get(j);                if (                    (i == 0  &&  tempI > 0)                    ||                    (i > 0  &&  tempI != tempI_1)                )                {                    Item iH = itemList.get(i-1);                    int  wH = iH.getWeight();                    iH.setInKnapsack(1);                    j -= wH;                    solutionWeight += wH;                }            } // for()            calculated = true;        } // if()        return itemList;    }     // add an item to the item list    public void add(String name, int weight, int value) {        if (name.equals(""))            name = "" + (itemList.size() + 1);        itemList.add(new Item(name, weight, value));        setInitialStateForCalculation();    }     // add an item to the item list    public void add(int weight, int value) {        add("", weight, value); // the name will be "itemList.size() + 1"!    }     // remove an item from the item list    public void remove(String name) {        for (Iterator<Item> it = itemList.iterator(); it.hasNext(); ) {            if (name.equals(it.next().getName())) {                it.remove();            }        }        setInitialStateForCalculation();    }     // remove all items from the item list    public void removeAllItems() {        itemList.clear();        setInitialStateForCalculation();    }     public int getProfit() {        if (!calculated)            calcSolution();        return profit;    }     public int getSolutionWeight() {return solutionWeight;}    public boolean isCalculated() {return calculated;}    public int getMaxWeight() {return maxWeight;}     public void setMaxWeight(int _maxWeight) {        maxWeight = Math.max(_maxWeight, 0);    }     public void setItemList(List<Item> _itemList) {        if (_itemList != null) {            itemList = _itemList;            for (Item item : _itemList) {                item.checkMembers();            }        }    }     // set the member with name "inKnapsack" by all items:    private void setInKnapsackByAll(int inKnapsack) {        for (Item item : itemList)            if (inKnapsack > 0)                item.setInKnapsack(1);            else                item.setInKnapsack(0);    }     // set the data members of class in the state of starting the calculation:    protected void setInitialStateForCalculation() {        setInKnapsackByAll(0);        calculated     = false;        profit         = 0;        solutionWeight = 0;    }} // class`
`package hu.pj.obj; public class Item {    protected String name    = "";    protected int weight     = 0;    protected int value      = 0;    protected int bounding   = 1; // the maximal limit of item's pieces    protected int inKnapsack = 0; // the pieces of item in solution     public Item() {}     public Item(Item item) {        setName(item.name);        setWeight(item.weight);        setValue(item.value);        setBounding(item.bounding);    }     public Item(int _weight, int _value) {        setWeight(_weight);        setValue(_value);    }     public Item(int _weight, int _value, int _bounding) {        setWeight(_weight);        setValue(_value);        setBounding(_bounding);    }     public Item(String _name, int _weight, int _value) {        setName(_name);        setWeight(_weight);        setValue(_value);    }     public Item(String _name, int _weight, int _value, int _bounding) {        setName(_name);        setWeight(_weight);        setValue(_value);        setBounding(_bounding);    }     public void setName(String _name) {name = _name;}    public void setWeight(int _weight) {weight = Math.max(_weight, 0);}    public void setValue(int _value) {value = Math.max(_value, 0);}     public void setInKnapsack(int _inKnapsack) {        inKnapsack = Math.min(getBounding(), Math.max(_inKnapsack, 0));    }     public void setBounding(int _bounding) {        bounding = Math.max(_bounding, 0);        if (bounding == 0)            inKnapsack = 0;    }     public void checkMembers() {        setWeight(weight);        setValue(value);        setBounding(bounding);        setInKnapsack(inKnapsack);    }     public String getName() {return name;}    public int getWeight() {return weight;}    public int getValue() {return value;}    public int getInKnapsack() {return inKnapsack;}    public int getBounding() {return bounding;}} // class`
Output:
```Maximal weight           = 4 kg
Total weight of solution = 3,96 kg
Total value              = 1010

You can carry te following materials in the knapsack:
1 unit(s)  map                     9   dag   (value = 150)
1 unit(s)  compass                 13  dag   (value = 35)
1 unit(s)  water                   153 dag   (value = 200)
2 unit(s)  glucose                 30  dag   (value = 120)
3 unit(s)  banana                  81  dag   (value = 180)
1 unit(s)  cheese                  23  dag   (value = 30)
1 unit(s)  suntan cream            11  dag   (value = 70)
1 unit(s)  waterproof overclothes  43  dag   (value = 75)
1 unit(s)  note-case               22  dag   (value = 80)
1 unit(s)  sunglasses              7   dag   (value = 20)
1 unit(s)  socks                   4   dag   (value = 50)
```

## JavaScript

Based on the (dynamic) J implementation. Expressed as an htm page:

`<html><head><title></title></head><body></body></html> <script type="text/javascript">var data= [  {name: 'map',                    weight:  9, value:150, pieces:1},  {name: 'compass',                weight: 13, value: 35, pieces:1},  {name: 'water',                  weight:153, value:200, pieces:2},  {name: 'sandwich',               weight: 50, value: 60, pieces:2},  {name: 'glucose',                weight: 15, value: 60, pieces:2},  {name: 'tin',                    weight: 68, value: 45, pieces:3},  {name: 'banana',                 weight: 27, value: 60, pieces:3},  {name: 'apple',                  weight: 39, value: 40, pieces:3},  {name: 'cheese',                 weight: 23, value: 30, pieces:1},  {name: 'beer',                   weight: 52, value: 10, pieces:3},  {name: 'suntan, cream',          weight: 11, value: 70, pieces:1},  {name: 'camera',                 weight: 32, value: 30, pieces:1},  {name: 'T-shirt',                weight: 24, value: 15, pieces:2},  {name: 'trousers',               weight: 48, value: 10, pieces:2},  {name: 'umbrella',               weight: 73, value: 40, pieces:1},  {name: 'waterproof, trousers',   weight: 42, value: 70, pieces:1},  {name: 'waterproof, overclothes',weight: 43, value: 75, pieces:1},  {name: 'note-case',              weight: 22, value: 80, pieces:1},  {name: 'sunglasses',             weight:  7, value: 20, pieces:1},  {name: 'towel',                  weight: 18, value: 12, pieces:2},  {name: 'socks',                  weight:  4, value: 50, pieces:1},  {name: 'book',                   weight: 30, value: 10, pieces:2}]; function findBestPack() {	var m= [[0]]; // maximum pack value found so far	var b= [[0]]; // best combination found so far	var opts= [0]; // item index for 0 of item 0 	var P= [1]; // item encoding for 0 of item 0	var choose= 0;	for (var j= 0; j<data.length; j++) {		opts[j+1]= opts[j]+data[j].pieces; // item index for 0 of item j+1		P[j+1]= P[j]*(1+data[j].pieces); // item encoding for 0 of item j+1	}	for (var j= 0; j<opts[data.length]; j++) {		m[0][j+1]= b[0][j+1]= 0; // best values and combos for empty pack: nothing	}	for (var w=1; w<=400; w++) {		m[w]= [0];		b[w]= [0];		for (var j=0; j<data.length; j++) {			var N= data[j].pieces; // how many of these can we have?			var base= opts[j]; // what is the item index for 0 of these?			for (var n= 1; n<=N; n++) {				var W= n*data[j].weight; // how much do these items weigh?				var s= w>=W ?1 :0; // can we carry this many?				var v= s*n*data[j].value; // how much are they worth?				var I= base+n; // what is the item number for this many?				var wN= w-s*W; // how much other stuff can we be carrying?				var C= n*P[j] + b[wN][base]; // encoded combination				m[w][I]= Math.max(m[w][I-1], v+m[wN][base]); // best value				choose= b[w][I]= m[w][I]>m[w][I-1] ?C :b[w][I-1];			}		}	}	var best= [];	for (var j= data.length-1; j>=0; j--) {		best[j]= Math.floor(choose/P[j]);		choose-= best[j]*P[j];	}	var out='<table><tr><td><b>Count</b></td><td><b>Item</b></td><th>unit weight</th><th>unit value</th>';	var wgt= 0;	var val= 0;	for (var i= 0; i<best.length; i++) {		if (0==best[i]) continue;		out+='</tr><tr><td>'+best[i]+'</td><td>'+data[i].name+'</td><td>'+data[i].weight+'</td><td>'+data[i].value+'</td>'		wgt+= best[i]*data[i].weight;		val+= best[i]*data[i].value;	}	out+= '</tr></table><br/>Total weight: '+wgt;	out+= '<br/>Total value: '+val;	document.body.innerHTML= out;}findBestPack();</script>`

This will generate (translating html to mediawiki markup):

 unit weight unit value Count Item 1 map 9 150 1 compass 13 35 1 water 153 200 2 glucose 15 60 3 banana 27 60 1 cheese 23 30 1 suntan, cream 11 70 1 waterproof, overclothes 43 75 1 note-case 22 80 1 sunglasses 7 20 1 socks 4 50

Total weight: 396
Total value: 1010

## Julia

Works with: Julia version 0.6

The solution uses Julia's MathProgBase. Most of the work is done with this package's `mixintprog` function.

Type and Function:

`using MathProgBase, Cbc struct KPDSupply{T<:Integer}    item::String    weight::T    value::T    quant::TendBase.show(io::IO, kdps::KPDSupply) = print(io, kdps.quant, " ", kdps.item, " (\$(kdps.weight) kg, \$(kdps.value) €)") function solve(gear::Vector{KPDSupply{T}}, capacity::Integer) where T<:Integer    w = getfield.(gear, :weight)    v = getfield.(gear, :value)    q = getfield.(gear, :quant)    sol = mixintprog(-v, w', '<', capacity, :Int, 0, q, CbcSolver())    sol.status == :Optimal || error("this problem could not be solved")     if all(q .== 1) # simpler case        return gear[sol.sol == 1.0]    else        pack = similar(gear, 0)        s = round.(Int, sol.sol)        for (i, g) in enumerate(gear)            iszero(s[i]) && continue            push!(pack, KPDSupply(g.item, g.weight, g.value, s[i]))        end        return pack    endend`

Main:

`gear = [KPDSupply("map", 9, 150, 1),        KPDSupply("compass", 13, 35, 1),        KPDSupply("water", 153, 200, 2),        KPDSupply("sandwich", 50, 60, 2),        KPDSupply("glucose", 15, 60, 2),        KPDSupply("tin", 68, 45, 3),        KPDSupply("banana", 27, 60, 3),        KPDSupply("apple", 39, 40, 3),        KPDSupply("cheese", 23, 30, 1),        KPDSupply("beer", 52, 10, 3),        KPDSupply("suntan cream", 11, 70, 1),        KPDSupply("camera", 32, 30, 1),        KPDSupply("T-shirt", 24, 15, 2),        KPDSupply("trousers", 48, 10, 2),        KPDSupply("umbrella", 73, 40, 1),        KPDSupply("waterproof trousers", 42, 70, 1),        KPDSupply("waterproof overclothes", 43, 75, 1),        KPDSupply("note-case", 22, 80, 1),        KPDSupply("sunglasses", 7, 20, 1),        KPDSupply("towel", 18, 12, 2),        KPDSupply("socks", 4, 50, 1),        KPDSupply("book", 30, 10, 2)] pack = solve(gear, 400)println("The hiker should pack: \n - ", join(pack, "\n - "))println("\nPacked weight: ", sum(getfield.(pack, :weight)), " kg")println("Packed value: ", sum(getfield.(pack, :value)), " €")`
Output:
```The hiker should pack:
- 1 map (9 kg, 150 €)
- 1 compass (13 kg, 35 €)
- 1 water (153 kg, 200 €)
- 2 glucose (15 kg, 60 €)
- 3 banana (27 kg, 60 €)
- 1 cheese (23 kg, 30 €)
- 1 suntan cream (11 kg, 70 €)
- 1 waterproof overclothes (43 kg, 75 €)
- 1 note-case (22 kg, 80 €)
- 1 sunglasses (7 kg, 20 €)
- 1 socks (4 kg, 50 €)

Packed weight: 327 kg
Packed value: 830 €```

## Kotlin

Translation of: C
`// version 1.1.2 data class Item(val name: String, val weight: Int, val value: Int, val count: Int) val items = listOf(    Item("map", 9, 150, 1),    Item("compass", 13, 35, 1),    Item("water", 153, 200, 2),    Item("sandwich", 50, 60, 2),    Item("glucose", 15, 60, 2),    Item("tin", 68, 45, 3),    Item("banana", 27, 60, 3),    Item("apple", 39, 40, 3),    Item("cheese", 23, 30, 1),    Item("beer", 52, 10, 3),    Item("suntan cream", 11, 70, 1),    Item("camera", 32, 30, 1),    Item("T-shirt", 24, 15, 2),    Item("trousers", 48, 10, 2),    Item("umbrella", 73, 40, 1),    Item("waterproof trousers", 42, 70, 1),    Item("waterproof overclothes", 43, 75, 1),    Item("note-case", 22, 80, 1),    Item("sunglasses", 7, 20, 1),    Item("towel", 18, 12, 2),    Item("socks", 4, 50, 1),    Item("book", 30, 10, 2)) val n = items.size const val MAX_WEIGHT = 400 fun knapsack(w: Int): IntArray {    val m  = Array(n + 1) { IntArray(w + 1) }    for (i in 1..n) {        for (j in 0..w) {            m[i][j] = m[i - 1][j]            for (k in 1..items[i - 1].count) {                if (k * items[i - 1].weight > j) break                val v = m[i - 1][j - k * items[i - 1].weight] + k * items[i - 1].value                if (v > m[i][j]) m[i][j] = v            }        }    }    val s = IntArray(n)    var j = w    for (i in n downTo 1) {        val v = m[i][j]        var k = 0        while (v != m[i - 1][j] + k * items[i - 1].value) {            s[i - 1]++            j -= items[i - 1].weight            k++        }    }    return s} fun main(args: Array<String>) {   val s = knapsack(MAX_WEIGHT)   println("Item Chosen             Weight Value  Number")   println("---------------------   ------ -----  ------")   var itemCount = 0   var sumWeight = 0   var sumValue  = 0   var sumNumber = 0   for (i in 0 until n) {       if (s[i] == 0) continue       itemCount++       val name   = items[i].name       val number = s[i]       val weight = items[i].weight * number       val value  = items[i].value  * number       sumNumber += number       sumWeight += weight       sumValue  += value       println("\${name.padEnd(22)}    \${"%3d".format(weight)}   \${"%4d".format(value)}    \${"%2d".format(number)}")   }   println("---------------------   ------ -----  ------")   println("Items chosen \$itemCount           \${"%3d".format(sumWeight)}   \${"%4d".format(sumValue)}    \${"%2d".format(sumNumber)}")}`
Output:
```Item Chosen             Weight Value  Number
---------------------   ------ -----  ------
map                         9    150     1
compass                    13     35     1
water                     153    200     1
glucose                    30    120     2
banana                     81    180     3
cheese                     23     30     1
suntan cream               11     70     1
waterproof overclothes     43     75     1
note-case                  22     80     1
sunglasses                  7     20     1
socks                       4     50     1
---------------------   ------ -----  ------
Items chosen 11           396   1010    14
```

## Mathematica

`[email protected]{#[[;; , 1]],     LinearProgramming[-#[[;; , 3]], -{#[[;; , 2]]}, -{400},     {0, #[[4]]} & /@ #, Integers]} &@{{"map", 9, 150, 1},  {"compass", 13, 35, 1},  {"water", 153, 200, 2},  {"sandwich", 50, 60, 2},  {"glucose", 15, 60, 2},  {"tin", 68, 45, 3},  {"banana", 27, 60, 3},  {"apple", 39, 40, 3},  {"cheese", 23, 30, 1},  {"beer", 52, 10, 3},  {"suntan cream", 11, 70, 1},  {"camera", 32, 30, 1},  {"T-shirt", 24, 15, 2},  {"trousers", 48, 10, 2},  {"umbrella", 73, 40, 1},  {"waterproof trousers", 42, 70, 1},  {"waterproof overclothes", 43, 75, 1},  {"note-case", 22, 80, 1},  {"sunglasses", 7, 20, 1},  {"towel", 18, 12, 2},  {"socks", 4, 50, 1},  {"book", 30, 10, 2}}`
Output:
```{{"map", 1}, {"compass", 1}, {"water", 1}, {"sandwich",
0}, {"glucose", 2}, {"tin", 0}, {"banana", 3}, {"apple",
0}, {"cheese", 1}, {"beer", 0}, {"suntan cream", 1}, {"camera",
0}, {"T-shirt", 0}, {"trousers", 0}, {"umbrella",
0}, {"waterproof trousers", 0}, {"waterproof overclothes",
1}, {"note-case", 1}, {"sunglasses", 1}, {"towel", 0}, {"socks",
1}, {"book", 0}}```

## 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 quantity{t in Items}; var take{t in Items}, integer, >=0, <=quantity[t]; knap_weight : sum{t in Items} take[t] * weight[t] <= 400; maximize knap_value: sum{t in Items} take[t] * value[t]; data; param : Items          : weight   value   quantity :=         map		  9	   150        1         compass          13	   35	      1         water		  153	   200        2         sandwich	  50	   60	      2         glucose	  15	   60	      2         tin		  68	   45	      3         banana		  27	   60	      3         apple		  39	   40	      3         cheese		  23	   30	      1         beer		  52	   10	      3         suntancream	  11	   70	      1         camera		  32	   30	      1         T-shirt	  24	   15	      2         trousers	  48	   10	      2         umbrella	  73	   40	      1         w-trousers	  42	   70	      1         w-overclothes	  43	   75	      1         note-case	  22	   80	      1         sunglasses	  7        20	      1         towel		  18	   12	      2         socks		  4        50	      1         book		  30	   10	      2; end;`

The solution produced using glpk is here: Knapsack problem/Bounded/Mathprog

lpsolve may also be used. The result may be found here: File:Knap_objective.png

The constraints may be found here: File:Knap_constraint.png

## 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)
• weight of n
• value of n

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

The sheet should then look like this:

table pre solving

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

• Target Cell: \$H\$27
• Optimise result to: maximum
• By Changing cells: \$F\$5:\$F\$26
• Limiting conditions:
• \$F\$5:\$F\$26 <= \$E\$5:\$E\$26
• \$G\$27 <= 400
• Options... (opens a separate popup window, then continue)
• Solver engine: OpenOffice.org Linear Solver
• Settings:
• Assume variables as integer: True
• Assume variables as non-negative: True
• Epsilon level: 0
• Limit branch-and-bound depth: True
• Solving time limit (seconds): 100

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

Table solved

## OxygenBasic

` type KnapSackItem string name,sys dag,value,tag KnapSackItem it[100] sys dmax=400sys items=22 it=> "map",	                  9, 150, 0,"compass",               13,  35, 0,"water",                153, 200, 0,"sandwich",              50, 160, 0,"glucose",               15,  60, 0,"tin",                   68,  45, 0,"banana",                27,  60, 0,"apple",                 39,  40, 0,"cheese",                23,  30, 0,"beer",                  52,  10, 0,"suntan cream",          11,  70, 0,"camera",                32,  30, 0,"T-shirt",               24,  15, 0,"trousers",              48,  10, 0,"umbrella",              73,  40, 0,"waterproof trousers",   42,  70, 0,"waterproof overclothes",43,  75, 0,"note-case",             22,  80, 0,"sunglasses",             7,  20, 0,"towel",                 18,  12, 0,"socks",                  4,  50, 0,"book",                  30,  10, 0 tot=0for i=1 to items  tot+=it(i).dagnextxs=tot-dmax 'REMOVE LOWEST PRIORITY ITEMS TILL XS<=0 cr=chr(13)+chr(10)tab=chr(9)pr="remove: " crc=0 do  v=1e9  w=0  k=0  '  'FIND NEXT LEAST VALUE ITEM  '  for i=1 to items    if it[i].tag=0      'w=it[i].value               'TEST PRIORITY ONLY      w=1000*it[i].value/it[i].dag 'TEST PRIORIT/WEIGHT VALUE      if w<v then v=w : k=i    end if  next  '  'LOG AND REMOVE FROM LIST  '  if k    xs-=it[k].dag 'deduct from excess weight    it[k].tag=1    pr+=it(k).name tab it(k).dag tab it(k).value cr    if xs<=0 then exit do 'Weight within dmax  end if  c++  if c>=items then exit doend do'pr+=cr "Knapsack contents: " cr'for i=1 to items  if it(i).tag=0    pr+=it(i).name tab it(i).dag tab it(i).value cr  end ifnext 'TRY FITTING IN LOWER PRIORITY ITEMS av=-xs for i=1 to items  if it[i].tag    if av-it[i].dag > 0 then      pr+="Can include: " it(i).name tab it(i).dag tab it(i).value cr      av-=it[i].dag    end if  end ifnextpr+=cr "Weight: " dmax+xs'putfile "s.txt",prprint pr'Knapsack contents: 'map             	9	150'compass         	13	35'water           	153	200'sandwich        	50	160'glucose         	15	60'banana          	27	60'suntan cream     	11	70'waterproof trousers	42	70'waterproof overclothes	43	75'note-case        	22	80'sunglasses       	7	20'socks           	4	50''Weight: 396 `

## Oz

Using constraint programming.

`declare  %% maps items to tuples of  %% Weight(hectogram), Value and available Pieces  Problem = knapsack('map':9#150#1                     'compass':13#35#1                     'water':153#200#2                     'sandwich':50#60#2                     'glucose':15#60#2                     'tin':68#45#3                     'banana':27#60#3                      'apple':39#40#3                     'cheese':23#30#1                     'beer':52#10#3                     'suntan cream':11#70#1                     'camera':32#30#1                     't-shirt':24#15#2                     'trousers':48#10#2                     'umbrella':73#40#1                     'waterproof trousers':42#70#1                     'waterproof overclothes':43#75#1                      'note-case':22#80#1                     'sunglasses':7#20#1                     'towel':18#12#2                     'socks':4#50#1                     'book':30#10#2                    )   %% item -> Weight  Weights = {Record.map Problem fun {\$ X} X.1 end}  %% item -> Value  Values =  {Record.map Problem fun {\$ X} X.2 end}   proc {Knapsack Solution}     %% a solution maps items to finite domain variables     %% whose maximum values depend on the item type     Solution = {Record.map Problem fun {\$ _#_#Max} {FD.int 0#Max} end}     %% no more than 400 hectograms     {FD.sumC Weights Solution '=<:' 400}      %% search through valid solutions     {FD.distribute naive Solution}  end   proc {PropagateLargerValue Old New}     %% propagate that new solutions must yield a higher value     %% than previously found solutions (essential for performance)     {FD.sumC Values New '>:' {Value Old}}   end   fun {Value Candidate}     {Record.foldL {Record.zip Candidate Values Number.'*'} Number.'+' 0}  end   fun {Weight Candidate}     {Record.foldL {Record.zip Candidate Weights Number.'*'} Number.'+' 0}  end   [Best] = {SearchBest Knapsack PropagateLargerValue}in  {System.showInfo "Items: "}  {Record.forAllInd Best   proc {\$ I V}      if V > 0 then	 {System.showInfo I#": "#V}      end   end  }  {System.printInfo "\n"}  {System.showInfo "total value: "#{Value Best}}  {System.showInfo "total weight: "#{Weight Best}}`
Output:
```Items:
banana: 3
cheese: 1
compass: 1
glucose: 2
map: 1
note-case: 1
socks: 1
sunglasses: 1
suntan cream: 1
water: 1
waterproof overclothes: 1

total value: 1010
total weight: 396
```

Takes about 3 seconds on a slow netbook.

## Perl

Recursive solution with caching.

`#!/usr/bin/perl use strict; my \$raw = <<'TABLE';map     	9       150     1compass 	13      35      1water   	153     200     2sandwich        50      60      2glucose 	15      60      2tin     	68      45      3banana  	27      60      3apple  		39      40      3cheese  	23      30      1beer    	52      10      1suntancream     11      70      1camera  	32      30      1T-shirt 	24      15      2trousers        48      10      2umbrella        73      40      1w_trousers     	42      70      1w_overcoat  	43      75      1note-case       22      80      1sunglasses      7       20      1towel   	18      12      2socks   	4       50      1book    	30      10      2TABLE my @items;for (split "\n", \$raw) {        my @x = split /\s+/;	push @items, {		name	=> \$x[0],		weight	=> \$x[1],		value	=> \$x[2],		quant	=> \$x[3],	}} my \$max_weight = 400; my %cache;sub pick {	my (\$weight, \$pos) = @_;	if (\$pos < 0 or \$weight <= 0) {		return 0, 0, []	} 	@{ \$cache{\$weight, \$pos} //= [do{	# odd construct: for caching		my \$item = \$items[\$pos];		my (\$bv, \$bi, \$bw, \$bp) = (0, 0, 0, []); 		for my \$i (0 .. \$item->{quant}) {			last if \$i * \$item->{weight} > \$weight;			my (\$v, \$w, \$p) = pick(\$weight - \$i * \$item->{weight}, \$pos - 1);			next if (\$v += \$i * \$item->{value}) <= \$bv; 			(\$bv, \$bi, \$bw, \$bp) = (\$v, \$i, \$w, \$p);		} 		my @picked = ( @\$bp, \$bi );		\$bv, \$bw + \$bi * \$item->{weight}, \@picked	}]}} my (\$v, \$w, \$p) = pick(\$max_weight, \$#items);for (0 .. \$#\$p) {	if (\$p->[\$_] > 0) {		print "\$p->[\$_] of \$items[\$_]{name}\n";	}}print "Value: \$v; Weight: \$w\n";`
Output:
```1 of map
1 of compass
1 of water
2 of glucose
3 of banana
1 of cheese
1 of suntancream
1 of w_overcoat
1 of note-case
1 of sunglasses
1 of socks
Value: 1010; Weight: 396```

## Perl 6

Works with: Rakudo version 2017.01
`my class KnapsackItem { has \$.name; has \$.weight; has \$.unit; } multi sub pokem ([],           \$,  \$v = 0) { \$v }multi sub pokem ([\$,  *@],     0,  \$v = 0) { \$v }multi sub pokem ([\$i, *@rest], \$w, \$v = 0) {  my \$key = "{[email protected]} \$w \$v";  (state %cache){\$key} or do {    my @skip = pokem @rest, \$w, \$v;    if \$w >= \$i.weight { # next one fits      my @put = pokem @rest, \$w - \$i.weight, \$v + \$i.unit;      return (%cache{\$key} = |@put, \$i.name).list if @put[0] > @skip[0];    }    return (%cache{\$key} = |@skip).list;  }} my \$MAX_WEIGHT = 400;my @table = flat map -> \$name,  \$weight,  \$unit,     \$count {     KnapsackItem.new( :\$name, :\$weight, :\$unit ) xx \$count;},        'map',                         9,      150,    1,        'compass',                     13,     35,     1,        'water',                       153,    200,    2,        'sandwich',                    50,     60,     2,        'glucose',                     15,     60,     2,        'tin',                         68,     45,     3,        'banana',                      27,     60,     3,        'apple',                       39,     40,     3,        'cheese',                      23,     30,     1,        'beer',                        52,     10,     3,        'suntan cream',                11,     70,     1,        'camera',                      32,     30,     1,        'T-shirt',                     24,     15,     2,        'trousers',                    48,     10,     2,        'umbrella',                    73,     40,     1,        'waterproof trousers',         42,     70,     1,        'waterproof overclothes',      43,     75,     1,        'note-case',                   22,     80,     1,        'sunglasses',                  7,      20,     1,        'towel',                       18,     12,     2,        'socks',                       4,      50,     1,        'book',                        30,     10,     2        ; my (\$value, @result) = pokem @table, \$MAX_WEIGHT; (my %hash){\$_}++ for @result; say "Value = \$value";say "Tourist put in the bag:";say "  # ITEM";for %hash.sort -> \$item {  say "  {\$item.value} {\$item.key}";}`
Output:
```Value = 1010
Tourist put in the bag:
# ITEM
3 banana
1 cheese
1 compass
2 glucose
1 map
1 note-case
1 socks
1 sunglasses
1 suntan cream
1 water
1 waterproof overclothes```

## Phix

### Very dumb and very slow brute force version

Of no practical use, except for comparison against improvements.

`atom t0 = time() constant goodies = {-- item                     weight value pieces{"map",                     9,      150,    1},{"compass",                 13,     35,     1},{"sandwich",                50,     60,     2},{"glucose",                 15,     60,     2},{"tin",                     68,     45,     3},{"banana",                  27,     60,     3},{"apple",                   39,     40,     3},{"cheese",                  23,     30,     1},{"beer",                    52,     10,     3},{"suntan cream",            11,     70,     1},{"water",                   153,    200,    2},{"camera",                  32,     30,     1},{"T-shirt",                 24,     15,     2},{"trousers",                48,     10,     2},{"umbrella",                73,     40,     1},{"waterproof trousers",     42,     70,     1},{"waterproof overclothes",  43,     75,     1},{"note-case",               22,     80,     1},{"sunglasses",              7,      20,     1},{"towel",                   18,     12,     2},{"socks",                   4,      50,     1},{"book",                    30,     10,     2}} function knapsack(integer max_weight, integer at)    integer best_points = 0, points    sequence best_choices = {}, choices    atom act_weight = 0, sub_weight    if at>=1 then        integer {?,witem,pitem,imax} = goodies[at]        for i=0 to imax do            integer wlim = max_weight-i*witem            if wlim<0 then exit end if            {points,sub_weight,choices} = knapsack(wlim, at-1)            points += i*pitem            if points>best_points then                best_points = points                best_choices = choices&i                act_weight = sub_weight+i*witem            end if        end for    end if    return {best_points, act_weight, best_choices}end function sequence res = knapsack(400, length(goodies))   -- {points,act_weight,choices} atom weight = 0, witematom points = 0, pitemstring idescfor i=1 to length(goodies) do    integer c = res[3][i]    if c then        {idesc,witem,pitem} = goodies[i]        printf(1,"%d %s\n",{c,idesc})        weight += c*witem        points += c*pitem    end ifend forif points!=res[1] then ?9/0 end if  -- sanity checkprintf(1,"Value %d, weight %g [%3.2fs]\n",{points,weight,time()-t0})`
Output:
```1 map
1 compass
2 glucose
3 banana
1 cheese
1 suntan cream
1 water
1 waterproof overclothes
1 note-case
1 sunglasses
1 socks
Value 1010, weight 396 [17.53s]
```

### Dynamic Programming version

Much faster but limited to integer weights

Translation of: C
`sequence items = {                  {"map",                    9,   150,   1},                  {"compass",               13,    35,   1},                  {"water",                153,   200,   2},                  {"sandwich",              50,    60,   2},                  {"glucose",               15,    60,   2},                  {"tin",                   68,    45,   3},                  {"banana",                27,    60,   3},                  {"apple",                 39,    40,   3},                  {"cheese",                23,    30,   1},                  {"beer",                  52,    10,   3},                  {"suntan cream",          11,    70,   1},                  {"camera",                32,    30,   1},                  {"T-shirt",               24,    15,   2},                  {"trousers",              48,    10,   2},                  {"umbrella",              73,    40,   1},                  {"waterproof trousers",   42,    70,   1},                  {"waterproof overclothes",43,    75,   1},                  {"note-case",             22,    80,   1},                  {"sunglasses",             7,    20,   1},                  {"towel",                 18,    12,   2},                  {"socks",                  4,    50,   1},                  {"book",                  30,    10,   2},                 }; sequence {names,weights,points,counts} = columnize(items) constant n = length(items) function knapsack(int w)int v-- m is the achievable points matrix:-- Note that Phix uses 1-based indexes, so m[1][1] --  actually holds points for 0 items of weight 0,--  and m[n+1][w+1] is for n items at weight w.seq m = repeat(repeat(0,w+1),n+1)    for i=1 to n do        for j=1 to w+1 do       -- (0 to w really)            m[i+1][j] = m[i][j]            for k=1 to counts[i] do                if k*weights[i]>j-1 then                    exit                end if                v = m[i][j-k*weights[i]]+k*points[i]                if v>m[i+1][j] then                    m[i+1][j] = v                end if            end for        end for    end for    seq s = repeat(0,n)    int j = w+1                 -- (w -> 0 really)    for i=n+1 to 2 by -1 do     -- (n to 1 really)        v = m[i][j]        int k = 0        while v!=m[i-1][j]+k*points[i-1] do            s[i-1] += 1            j -= weights[i-1]            k += 1        end while    end for    return send function int tc = 0, tw = 0, tv = 0seq s = knapsack(400)for i=1 to n do    int si = s[i]    if si then        printf(1,"%-22s %5d %5d %5d\n", {names[i], si, si*weights[i], si*points[i]})        tc += si        tw += si*weights[i]        tv += si*points[i]    end ifend forprintf(1,"%-22s %5d %5d %5d\n", {"count, weight, points:", tc, tw, tv})`
Output:
```map                        1     9   150
compass                    1    13    35
water                      1   153   200
glucose                    2    30   120
banana                     3    81   180
cheese                     1    23    30
suntan cream               1    11    70
waterproof overclothes     1    43    75
note-case                  1    22    80
sunglasses                 1     7    20
socks                      1     4    50
count, weight, points:    14   396  1010
```

### Range cache version

The main problem with the dynamic programming solution is that it is only practical for integer weights. You could multiply by 1000 and truncate to get an approximation to the nearest 0.001kg, but the memory use would obviously increase dramatically. A naive cache could also suffer similarly, if it retained duplicate solutions for w=15.783, 15.784, 15.785, and everything in between. Using a (bespoke) range cache solves this problem, although the simplistic version given could probably be improved with a binary search or similar. It also significantly reduces the workload; for instance if you find a solution for 170 that actually weighs 150 then any subsequent query in 150..170 requires zero work, unlike the naive cache and dp solutions.

`---- demo\rosetta\knapsackB.exw-- ==========================--atom t0 = time() enum HI,PTS,ACTW,SOLNsequence range_cache = {} integer cache_entries = 0 procedure add_range(integer at, atom weight, atom actual_weight, atom points, sequence soln)    if actual_weight>weight then ?9/0 end if    for i=length(range_cache)+1 to at do -- (while too small do)        if i=at then            range_cache = append(range_cache,{{weight,points,actual_weight,soln}})            cache_entries += 1            return        end if        range_cache = append(range_cache,{})    end for    for i=1 to length(range_cache[at]) do        sequence rcati = range_cache[at][i]        if weight=rcati[ACTW] then            if rcati[PTS..SOLN]!={points,actual_weight,soln} then ?9/0 end if            return        elsif weight<rcati[ACTW] then            -- (we cannot extend an existing range down, since it starts at            --  the actual weight, that must also be the minimum weight...)            if soln=rcati[SOLN] then ?9/0 end if            -- insert a new range            range_cache[at][i..i-1] = {{weight,points,actual_weight,soln}}            cache_entries += 1            return        elsif soln=rcati[SOLN] then            if rcati[PTS..SOLN]!={points,actual_weight,soln} then ?9/0 end if            if weight>rcati[HI] then        -- extend existing range up                rcati = {}                range_cache[at][i][HI] = weight            end if            return        elsif weight<=rcati[HI] then            ?9/0        -- duplicate solution?? (or discard as below)--          return                          -- (discard)        end if    end for    range_cache[at] = append(range_cache[at],{weight,points,actual_weight,soln})    cache_entries += 1end procedure function in_range(integer at, atom weight)    if at<=length(range_cache) then        for i=1 to length(range_cache[at]) do            sequence rcati = range_cache[at][i]            if weight<=rcati[HI] then                if weight>=rcati[ACTW] then                    return rcati[PTS..SOLN] -- {pts,act_weight,soln}                end if                exit            end if        end for    end if    return {}   -- (no suitable cache entry found)end function constant goodies = {-- item                     weight value pieces{"map",                     9,      150,    1},{"compass",                 13,     35,     1},{"sandwich",                50,     60,     2},{"glucose",                 15,     60,     2},{"tin",                     68,     45,     3},{"banana",                  27,     60,     3},{"apple",                   39,     40,     3},{"cheese",                  23,     30,     1},{"beer",                    52,     10,     3},{"suntan cream",            11,     70,     1},{"water",                   153,    200,    2},{"camera",                  32,     30,     1},{"T-shirt",                 24,     15,     2},{"trousers",                48,     10,     2},{"umbrella",                73,     40,     1},{"waterproof trousers",     42,     70,     1},{"waterproof overclothes",  43,     75,     1},{"note-case",               22,     80,     1},{"sunglasses",              7,      20,     1},{"towel",                   18,     12,     2},{"socks",                   4,      50,     1},{"book",                    30,     10,     2}} integer cache_hits = 0integer cache_misses = 0 function knapsack(integer max_weight, integer at)    integer best_points = 0, points    sequence best_choices = {}, choices    atom act_weight = 0, sub_weight    if at>=1 then        sequence soln = in_range(at,max_weight)        if length(soln) then            cache_hits += 1            return soln        end if        cache_misses += 1        integer {?,witem,pitem,imax} = goodies[at]        best_choices = repeat(0,at)        for i=0 to imax do            integer wlim = max_weight-i*witem            if wlim<0 then exit end if            {points,sub_weight,choices} = knapsack(wlim, at-1)            points += i*pitem            if points>best_points then                best_points = points                best_choices = choices&i                act_weight = sub_weight+i*witem            end if        end for        add_range(at,max_weight,act_weight,best_points,best_choices)    end if    return {best_points, act_weight, best_choices}end function sequence res = knapsack(400, length(goodies))   -- {points,act_weight,choices} atom weight = 0, witematom points = 0, pitemstring idescfor i=1 to length(goodies) do    integer c = res[3][i]    if c then        {idesc,witem,pitem} = goodies[i]        printf(1,"%d %s\n",{c,idesc})        weight += c*witem        points += c*pitem    end ifend forif points!=res[1] then ?9/0 end if  -- sanity checkif weight!=res[2] then ?9/0 end if  -- sanity checkprintf(1,"Value %d, weight %g [%3.2fs]\n",{points,weight,time()-t0}) printf(1,"cache_entries:%d, hits:%d, misses:%d\n",{cache_entries,cache_hits,cache_misses})`
Output:
```1 map
1 compass
2 glucose
3 banana
1 cheese
1 suntan cream
1 water
1 waterproof overclothes
1 note-case
1 sunglasses
1 socks
Value 1010, weight 396 [0.00s]
cache_entries:409, hits:1226, misses:882
```

The distributed version contains additional comments and extra code for comparing this against a naive cache and no cache (CACHE_RANGE shown above).
CACHE_SIMPLE: (as above but ending)

```Value 1010, weight 396 [0.08s]
cache_entries:5549, hits:7707, misses:5549
```

Even on this simple integer-only case, range cache reduces cache size better than 10-fold and effort 6-fold.
We also know the dp solution matrix has 9223 entries, admittedly each being much smaller than a cache entry.
And finally CACHE_NONE (the dumb version): (as above but ending)

```Value 1010, weight 396 [18.14s]
cache_entries:0, hits:0, misses:33615741
```

## PicoLisp

`(de *Items   ("map" 9 150 1)                     ("compass" 13 35 1)   ("water" 153 200 3)                 ("sandwich" 50 60 2)   ("glucose" 15 60 2)                 ("tin" 68 45 3)   ("banana" 27 60 3)                  ("apple" 39 40 3)   ("cheese" 23 30 1)                  ("beer" 52 10 3)   ("suntan cream" 11 70 1)            ("camera" 32 30 1)   ("t-shirt" 24 15 2)                 ("trousers" 48 10 2)   ("umbrella" 73 40 1)                ("waterproof trousers" 42 70 1)   ("waterproof overclothes" 43 75 1)  ("note-case" 22 80 1)   ("sunglasses" 7 20 1)               ("towel" 18 12 2)   ("socks" 4 50 1)                    ("book" 30 10 2) ) # Dynamic programming solution(de knapsack (Lst W)   (when Lst      (cache '*KnapCache (cons W Lst)         (let X (knapsack (cdr Lst) W)            (if (ge0 (- W (cadar Lst)))               (let Y (cons (car Lst) (knapsack (cdr Lst) @))                  (if (> (sum caddr X) (sum caddr Y)) X Y) )               X ) ) ) ) ) (let K   (knapsack      (mapcan                                   # Expand multiple items         '((X) (need (cadddr X) NIL X))         *Items )      400 )   (for I K      (apply tab I (3 -24 6 6) NIL) )   (tab (27 6 6) NIL (sum cadr K) (sum caddr K)) )`
Output:
```   map                          9   150
compass                     13    35
water                      153   200
glucose                     15    60
glucose                     15    60
banana                      27    60
banana                      27    60
banana                      27    60
cheese                      23    30
suntan cream                11    70
waterproof overclothes      43    75
note-case                   22    80
sunglasses                   7    20
socks                        4    50
396  1010```

## Prolog

Works with: SWI Prolog

### Library clpfd

Library: clpfd

Library clpfd is written by Markus Triska. Takes about 3 seconds to compute the best solution.

`:- use_module(library(clpfd)). % tuples (name, weights, value, nb pieces).knapsack :-	L = [(   map, 	        9, 	150, 	1),	     (   compass, 	13, 	35, 	1),	     (   water, 	153, 	200, 	2),	     (   sandwich, 	50, 	60, 	2),	     (   glucose, 	15, 	60, 	2),	     (   tin, 	68, 	45, 	3),	     (   banana, 	27, 	60, 	3),	     (   apple, 	39, 	40, 	3),	     (   cheese, 	23, 	30, 	1),	     (   beer, 	52, 	10, 	3),	     (   'suntan cream', 	11, 	70, 	1),	     (   camera, 	32, 	30, 	1),	     (   'T-shirt', 	24, 	15, 	2),	     (   trousers, 	48, 	10, 	2),	     (   umbrella, 	73, 	40, 	1),	     (   'waterproof trousers', 	42, 	70, 	1),	     (   'waterproof overclothes', 	43, 	75, 	1),	     (   'note-case', 	22, 	80, 	1),	     (   sunglasses, 	7, 	20, 	1),	     (   towel, 	18, 	12, 	2),	     (   socks, 	4, 	50, 	1),	     (   book, 	30, 	10, 	2)], 	% Takes is the list of the numbers of each items	% these numbers are between 0 and the 4th value of the tuples of the items        maplist(collect, L, Ws, Vs, Takes),        scalar_product(Ws, Takes, #=<, 400),        scalar_product(Vs, Takes, #=, VM), 	% to have statistics on the resolution of the problem.	time(labeling([max(VM), down], Takes)),        scalar_product(Ws, Takes, #=, WM), 	%% displayinf of the results.	compute_lenword(L, 0, Len),	sformat(A1, '~~w~~t~~~w|', [Len]),	sformat(A2, '~~t~~w~~~w|', [4]),	sformat(A3, '~~t~~w~~~w|', [5]),	print_results(A1,A2,A3, L, Takes, WM, VM). collect((_, W, V, N), W, V, Take) :-	Take in 0..N. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%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(A1,A2,A3, [], [], WM, WR) :-	sformat(W0, '~w ', [' ']),	sformat(W1, A1, [' ']),	sformat(W2, A2, [WM]),	sformat(W3, A3, [WR]),	format('~w~w~w~w~n', [W0,W1,W2,W3]).  print_results(A1,A2,A3, [_H|T], [0|TR], WM, VM) :-	!,	print_results(A1,A2,A3, T, TR, WM, VM). print_results(A1, A2, A3, [(Name, W, V, _)|T], [N|TR], WM, VM) :-	sformat(W0, '~w ', [N]),	sformat(W1, A1, [Name]),	sformat(W2, A2, [W]),	sformat(W3, A3, [V]),	format('~w~w~w~w~n', [W0,W1,W2,W3]),	print_results(A1, A2, A3, T, TR, WM, VM).`
Output:
``` ?- knapsack.
% 21,154,932 inferences, 3.187 CPU in 3.186 seconds (100% CPU, 6638515 Lips)
1 map                      9  150
1 compass                 13   35
1 water                  153  200
2 glucose                 15   60
3 banana                  27   60
1 cheese                  23   30
1 suntan cream            11   70
1 waterproof overclothes  43   75
1 note-case               22   80
1 sunglasses               7   20
1 socks                    4   50
396 1010
true```

### Library simplex

Library simplex is written by Markus Triska. Takes about 10 seconds to compute the best solution.

`:- use_module(library(simplex)).% tuples (name, weights, value, pieces).knapsack :-	L = [(map, 	9, 	150, 	1),	     (   compass, 	13, 	35, 	1),	     (   water, 	153, 	200, 	2),	     (   sandwich, 	50, 	60, 	2),	     (   glucose, 	15, 	60, 	2),	     (   tin, 	68, 	45, 	3),	     (   banana, 	27, 	60, 	3),	     (   apple, 	39, 	40, 	3),	     (   cheese, 	23, 	30, 	1),	     (   beer, 	52, 	10, 	3),	     (   'suntan cream', 	11, 	70, 	1),	     (   camera, 	32, 	30, 	1),	     (   'T-shirt', 	24, 	15, 	2),	     (   trousers, 	48, 	10, 	2),	     (   umbrella, 	73, 	40, 	1),	     (   'waterproof trousers', 	42, 	70, 	1),	     (   'waterproof overclothes', 	43, 	75, 	1),	     (   'note-case', 	22, 	80, 	1),	     (   sunglasses, 	7, 	20, 	1),	     (   towel, 	18, 	12, 	2),	     (   socks, 	4, 	50, 	1),	     (   book, 	30, 	10, 	2)], 	 gen_state(S0),	 length(L, N),	 numlist(1, N, LN),	 time(( create_constraint_N(LN, L, S0, S1),		maplist(create_constraint_WV, LN, L, LW, LV),		constraint(LW =< 400, S1, S2),		maximize(LV, S2, S3)	      )),	compute_lenword(L, 0, Len),	sformat(A1, '~~w~~t~~~w|', [Len]),	sformat(A2, '~~t~~w~~~w|', [4]),	sformat(A3, '~~t~~w~~~w|', [5]),	print_results(S3, A1,A2,A3, L, LN, 0, 0).  create_constraint_N([], [], S, S). create_constraint_N([HN|TN], [(_, _, _, Nb) | TL], S1, SF) :-	constraint(integral(x(HN)), S1, S2),	constraint([x(HN)] =< Nb, S2, S3),	constraint([x(HN)] >= 0, S3, S4),	create_constraint_N(TN, TL, S4, SF). create_constraint_WV(N, (_, W, V, _), W * x(N), V * x(N)). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%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, A1, A2, A3, [], [], WM, VM) :-	sformat(W0, '~w ', [' ']),	sformat(W1, A1, [' ']),	sformat(W2, A2, [WM]),	sformat(W3, A3, [VM]),	format('~w~w~w~w~n', [W0, W1,W2,W3]).  print_results(S, A1, A2, A3, [(Name, W, V,_)|T], [N|TN], W1, V1) :-	variable_value(S, x(N), X),	(   X = 0 -> W1 = W2, V1 = V2	;   sformat(S0, '~w ', [X]),	    sformat(S1, A1, [Name]),	    sformat(S2, A2, [W]),	    sformat(S3, A3, [V]),	    format('~w~w~w~w~n', [S0, S1,S2,S3]),	    W2 is W1 + X * W,	    V2 is V1 + X * V),	print_results(S, A1, A2, A3, T, TN, W2, V2).`

## Python

Not as dumb a search over all possible combinations under the maximum allowed weight:

`from itertools import groupbyfrom collections import namedtuplefrom pprint import pprint as pp def anyvalidcomb(items,  val=0, wt=0):    ' All combinations below the maxwt '    if not items:        yield [], val, wt    else:        this, *items = items            # car, cdr        for n in range(this.number+1):            v = val + n*this.value            w = wt  + n*this.weight            if w > maxwt:                break            for c in anyvalidcomb(items, v, w):                yield [this]*n + c[COMB], c[VAL], c[WT] maxwt = 400COMB, VAL, WT = range(3)Item  = namedtuple('Items', 'name weight value number')items = [ Item(*x) for x in          (            ("map", 9, 150, 1),            ("compass", 13, 35, 1),            ("water", 153, 200, 3),            ("sandwich", 50, 60, 2),            ("glucose", 15, 60, 2),            ("tin", 68, 45, 3),            ("banana", 27, 60, 3),            ("apple", 39, 40, 3),            ("cheese", 23, 30, 1),            ("beer", 52, 10, 3),            ("suntan cream", 11, 70, 1),            ("camera", 32, 30, 1),            ("t-shirt", 24, 15, 2),            ("trousers", 48, 10, 2),            ("umbrella", 73, 40, 1),            ("waterproof trousers", 42, 70, 1),            ("waterproof overclothes", 43, 75, 1),            ("note-case", 22, 80, 1),            ("sunglasses", 7, 20, 1),            ("towel", 18, 12, 2),            ("socks", 4, 50, 1),            ("book", 30, 10, 2),           ) ]   bagged = max( anyvalidcomb(items), key=lambda c: (c[VAL], -c[WT])) # max val or min wt if values equalprint("Bagged the following %i items\n  " % len(bagged[COMB]) +      '\n  '.join('%i off: %s' % (len(list(grp)), item.name)                  for item,grp in groupby(sorted(bagged[COMB]))))print("for a total value of %i and a total weight of %i" % bagged[1:])`
Output:
```Bagged the following 14 items
3 off: banana
1 off: cheese
1 off: compass
2 off: glucose
1 off: map
1 off: note-case
1 off: socks
1 off: sunglasses
1 off: suntan cream
1 off: water
1 off: waterproof overclothes
for a total value of 1010 and a total weight of 396```

### Dynamic programming solution

This is much faster. It expands the multiple possible instances of an item into individual instances then applies the zero-one dynamic knapsack solution:

`from itertools import groupby try:    xrangeexcept:    xrange = range maxwt = 400 groupeditems = (            ("map", 9, 150, 1),            ("compass", 13, 35, 1),            ("water", 153, 200, 3),            ("sandwich", 50, 60, 2),            ("glucose", 15, 60, 2),            ("tin", 68, 45, 3),            ("banana", 27, 60, 3),            ("apple", 39, 40, 3),            ("cheese", 23, 30, 1),            ("beer", 52, 10, 3),            ("suntan cream", 11, 70, 1),            ("camera", 32, 30, 1),            ("t-shirt", 24, 15, 2),            ("trousers", 48, 10, 2),            ("umbrella", 73, 40, 1),            ("waterproof trousers", 42, 70, 1),            ("waterproof overclothes", 43, 75, 1),            ("note-case", 22, 80, 1),            ("sunglasses", 7, 20, 1),            ("towel", 18, 12, 2),            ("socks", 4, 50, 1),            ("book", 30, 10, 2),           )items = sum( ([(item, wt, val)]*n for item, wt, val,n in groupeditems), []) def knapsack01_dp(items, limit):    table = [[0 for w in range(limit + 1)] for j in xrange(len(items) + 1)]     for j in xrange(1, len(items) + 1):        item, wt, val = items[j-1]        for w in xrange(1, limit + 1):            if wt > w:                table[j][w] = table[j-1][w]            else:                table[j][w] = max(table[j-1][w],                                  table[j-1][w-wt] + val)     result = []    w = limit    for j in range(len(items), 0, -1):        was_added = table[j][w] != table[j-1][w]         if was_added:            item, wt, val = items[j-1]            result.append(items[j-1])            w -= wt     return result  bagged = knapsack01_dp(items, maxwt)print("Bagged the following %i items\n  " % len(bagged) +      '\n  '.join('%i off: %s' % (len(list(grp)), item[0])                  for item,grp in groupby(sorted(bagged))))print("for a total value of %i and a total weight of %i" % (    sum(item[2] for item in bagged), sum(item[1] for item in bagged)))`

### Non-zero-one solution

`items = (	("map",		9,	150,	1),	("compass",	13,	35,	1),	("water",	153,	200,	3),	("sandwich",	50,	60,	2),	("glucose",	15,	60,	2),	("tin",		68,	45,	3),	("banana",	27,	60,	3),	("apple",	39,	40,	3),	("cheese",	23,	30,	1),	("beer",	52,	10,	3),	("suntan cream",11,	70,	1),	("camera",	32,	30,	1),	("t-shirt",	24,	15,	2),	("trousers",	48,	10,	2),	("umbrella",	73,	40,	1),	("w-trousers",	42,	70,	1),	("w-overcoat",	43,	75,	1),	("note-case",	22,	80,	1),	("sunglasses",	7,	20,	1),	("towel",	18,	12,	2),	("socks",	4,	50,	1),	("book",	30,	10,	2),) #cache: could just use memoize module, but explicit caching is clearerdef choose_item(weight, idx, cache):    if idx < 0: return 0, []     k = (weight, idx)    if k in cache: return cache[k]     name, w, v, qty = items[idx]    best_v, best_list = 0, []     for i in range(0, qty + 1):        wlim = weight - i * w        if wlim < 0: break         val, taken = choose_item(wlim, idx - 1, cache)        if val + i * v > best_v:            best_v = val + i * v            best_list = taken[:]            best_list.append(i)     cache[k] = [best_v, best_list]    return best_v, best_list  v, lst = choose_item(400, len(items) - 1, {})w = 0for i, cnt in enumerate(lst):    if cnt > 0:        print cnt, items[i][0]        w = w + items[i][1] * cnt print "Total weight:", w, "Value:", v`

## R

`library(tidyverse)library(rvest) task_html= read_html("http://rosettacode.org/wiki/Knapsack_problem/Bounded")task_table= html_nodes(html, "table")[[1]] %>%  html_table(table, header= T, trim= T) %>%  set_names(c("items", "weight", "value", "pieces")) %>%  filter(items != "knapsack") %>%  mutate(weight= as.numeric(weight),         value= as.numeric(value),         pieces= as.numeric(pieces))`

Solution of the task using genetic algorithm.

`library(rgenoud) fitness= function(x= rep(1, nrow(task_table))){  total_value= sum(task_table\$value * x)  total_weight= sum(task_table\$weight * x)  ifelse(total_weight <= 400, total_value, 400-total_weight)} allowed= matrix(c(rep(0, nrow(task_table)), task_table\$pieces), ncol = 2)set.seed(42)evolution= genoud(fn= fitness,                   nvars= nrow(allowed),                   max= TRUE,                  pop.size= 10000,                  data.type.int= TRUE,                   Domains= allowed) cat("Value: ", evolution\$value, "\n")cat("Weight:", sum(task_table\$weight * evolution\$par), "dag", "\n")data.frame(item= task_table\$items, pieces= as.integer(solution)) %>%  filter(solution> 0)`
Output:
```Value:  1010
Weight: 396 dag
item pieces
1                     map      1
2                 compass      1
3                sandwich      1
4                 glucose      2
5                  banana      3
6                   apple      2
7                  cheese      1
8            suntan cream      1
9  waterproof overclothes      1
10              note-case      1
11             sunglasses      1
12                  towel      1
13                  socks      1
```

## Racket

The algorithm is nearly a direct translation of Haskell's array-based implementation. However, the data is taken from the webpage itself.

` #lang racket(require net/url html xml xml/path) (struct item (name mass value count) #:transparent) ;this section is to convert the web page on the problem into the data for the problem;i don't got time to manually type tables, nevermind that this took longer(define (group-n n l)  (let group-n ([l l] [acc '()])    (if (null? l)        (reverse acc)        (let-values ([(takes drops) (split-at l n)])          (group-n drops (cons takes acc))))))(define (atom? x) (not (or (pair? x) (null? x))));modified from little schemer...finds nested list where regular member would've returned non-#f(define (member* x t)  (cond [(null? t) #f]        [(atom? (car t)) (or (and (equal? (car t) x) t)                             (member* x (cdr t)))]        [else (or (member* x (car t))                  (member* x (cdr t)))]))(define (addr->xexpr f) (compose xml->xexpr f read-html-as-xml get-pure-port string->url))(define (read-page) ((addr->xexpr cadr) "http://rosettacode.org/wiki/Knapsack_problem/Bounded"))(define (get-xml-table xe) (member* 'table xe))(define (xml-table->item-list xe-table)  ;all html table datas  (let* ([strs (se-path*/list '(td) xe-table)]         ;4 columns per row         [rows (group-n 4 strs)]         ;the last two rows belong to the knapsack entry which we don't want         [rows (take rows (- (length rows) 2))])    (for/list ([r rows])      (let ([r (map string-trim r)])        ;convert the weight, value, and pieces columns to numbers and structify it        (apply item (car r) (map string->number (cdr r)))))));top-level function to take a string representing a URL and gives the table I didn't want to type(define addr->item-list (compose xml-table->item-list get-xml-table read-page)) ;stores best solution(struct solution (value choices) #:transparent) ;finds best solution in a list of solutions(define (best sol-list)  (let best ([l (cdr sol-list)] [bst (car sol-list)])    (match l      ['() bst]      [(cons s l) (if (> (solution-value s) (solution-value bst))                      (best l s)                      (best l bst))]))) ;stores the choices leading to best solution...item name and # of that item taken(struct choice (name count) #:transparent) ;algorithm is derived from Haskell's array-based example;returns vector of solutions for every natural number capacity up to the input max(define (solution-vector capacity)  ;find best value and items that gave it, trying all items  ;first we set up a vector for the best solution found for every capacity  ;the best solution found at the beginning has 0 value with no items  (for/fold ([solutions (make-vector (add1 capacity) (solution 0 '()))])            ([i (addr->item-list)])    ;the new solutions aren't accumulated until after processing all the way through a particular    ;capacity, lest capacity c allow capacity c+1 to reuse an item that had been exhausted, i.e.    ;we have to move on to the next item before we save ANY solutions found for that item, so they    ;must be saved "in parallel" by overwriting the entire vector at once    (for/vector ([target-cap (range 0 (add1 capacity))])      (match-let ([(item name mass value count) i])        ;find best solution for item out of every number that can be taken of that item        ;we'll call an "item and count of item" a choice        (best (for/list ([n (range 0 (add1 count))])                ;ensure mass of this choice is not greater than target capacity                (let ([mass-choice (* n mass)])                  (if (> mass-choice target-cap)                      (solution 0 '())                      ;subtract from target capacity the amount taken up by this choice                      ;use it to get the best solution for THAT capacity                      (let ([remaining-cap-solution (vector-ref solutions (- target-cap mass-choice))])                        ;the new solution found adds the value of this choice to the above value and                        ;adds the choice itself to the list of choices                        (solution (+ (* n value) (solution-value remaining-cap-solution))                                  (cons (choice name n) (solution-choices remaining-cap-solution)))))))))))) ;top level function to convert solution vector into single best answer(define (knapsack capacity)  ;the best solution is not necessarily the one that uses max capacity,  ;e.g. if max is 5 and you have items of mass 4 and 5, and the 4 is worth more  (match-let ([(solution value choices) (best (vector->list (solution-vector capacity)))])    (solution value (filter (compose positive? choice-count) choices)))) `
Output:
```(solution
1010
(list
(choice "socks" 1)
(choice "sunglasses" 1)
(choice "note-case" 1)
(choice "waterproof overclothes" 1)
(choice "suntan cream" 1)
(choice "cheese" 1)
(choice "banana" 3)
(choice "glucose" 2)
(choice "water" 1)
(choice "compass" 1)
(choice "map" 1)))
```

## REXX

Originally, the combination generator/checker subroutine   findBest   was recursive and made the program solution generic.

However, a recursive solution also made the solution much more slower, so the combination generator/checker was
"unrolled" and converted into discrete combination checks (based on the number of items).   The unused combinatorial
checks were discarded and only the pertinent code was retained.

`/*REXX program solves a  knapsack problem  (22 items + repeats, with weight restriction.*/call @gen                                        /*generate items and initializations.  */call @sort                                       /*sort items by decreasing their weight*/call build                                       /*build a list of choices (objects).   */call showOBJ                                     /*display the list of choices (objects)*/call findBest                                    /*examine and find the possible choices*/call showBest                                    /*display best choice  (weight, value).*/exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/@gen: @.=;        @.1  = 'map                       9  150'                  @.2  = 'compass                  13   35'                  @.3  = 'water                   153  200   2'                  @.4  = 'sandwich                 50   60   2'                  @.5  = 'glucose                  15   60   2'                  @.6  = 'tin                      68   45   3'                  @.7  = 'banana                   27   60   3'                  @.8  = 'apple                    39   40   3'                  @.9  = 'cheese                   23   30'                  @.10 = 'beer                     52   10   3'                  @.11 = 'suntan_cream             11   70'                  @.12 = 'camera                   32   30'                  @.13 = 'T-shirt                  24   15   2'                  @.14 = 'trousers                 48   10   2'                  @.15 = 'umbrella                 73   40'                  @.16 = 'waterproof_trousers      42   70'                  @.17 = 'waterproof_overclothes   43   75'                  @.18 = 'note-case                22   80'                  @.19 = 'sunglasses                7   20'                  @.20 = 'towel                    18   12   2'                  @.21 = 'socks                     4   50'                  @.22 = 'book                     30   10   2'    highQ = 0                                    /*maximum quantity specified (if any). */     maxL = length('knapsack items')             /* "     "    width for the table names*/     maxW = length('weight')                     /* "     "      "    "    "   weights. */     maxV = length('value')                      /* "     "      "    "    "   values.  */     maxQ = length('pieces')                     /* "     "      "    "    "   quantity.*/     maxWeight=400                               /*the maximum weight for the knapsack. */     items= 0;   i.=;    w.=0;    v.=0;   q.=0   /*initialize some stuff and things.    */        Tw= 0;   Tv=0;   Tq=0;    m=maxWeight    /*     "     more   "    "     "       */     say;  say 'maximum weight allowed for a knapsack: '   commas(maxWeight);     say     return/*──────────────────────────────────────────────────────────────────────────────────────*/@sort:        do j=1  while @.j\==''             /*process each choice and sort the item*/              @.j=space(@.j);   _wt=word(@.j, 2) /*choose first item (arbitrary).       */                  do k=j+1  while @.k\==''       /*find a possible heavier item.        */                  ?wt=word(@.k, 2)                  if ?wt>_wt  then do;  [email protected].k;  @.[email protected].j;  @.j=_;  _wt=?wt;  end   /*swap*/                  end   /*k*/              end       /*j*/                    /* [↑]  minimizes the # of combinations*/      obj=j-1                                    /*adjust for the   DO   loop index.    */      return/*──────────────────────────────────────────────────────────────────────────────────────*/build:        do j=1  for obj                    /*build a list of choices (objects).   */              parse var  @.j  item  w  v  q  .   /*parse the original choice for table. */              if w>maxWeight  then iterate       /*Is the weight > maximum?  Then ignore*/              Tw=Tw+w;  Tv=Tv+v;   Tq=Tq+1       /*add the totals up  (for alignment).  */              maxL=max(maxL, length(item))       /*find the maximum width for an item.  */              if q==''  then q=1              highQ=max(highQ, q)              items=items+1                      /*bump the item counter.               */              i.items=item;  w.items=w;  v.items=v;  q.items=q                  do k=2  to q  ;  items=items+1 /*bump the item counter  (each piece). */                  i.items=item;  w.items=w;  v.items=v;  q.items=q                                Tw=Tw+w;    Tv=Tv+v;    Tq=Tq+1                  end   /*k*/              end       /*j*/      maxW = max(maxW, length( commas(Tw) ) )    /*find the maximum width for weight.   */      maxV = max(maxV, length( commas(Tv) ) )    /*  "   "     "      "    "  value.    */      maxQ = max(maxQ, length( commas(Tq) ) )    /*  "   "     "      "    "  quantity. */      maxL = maxL + maxL %4 + 4                  /*extend the width of name for table.  */      return                                     /* [↑]    %  is REXX integer division. *//*──────────────────────────────────────────────────────────────────────────────────────*/commas: procedure;  parse arg _;   n=_'.9';   #=123456789;    b=verify(n, #, "M");   x=','        e=verify(n, #'0', , verify(n, #"0.", 'M') ) - 4     /* [↓]  add commas to number*/           do j=e  to b  by -3;    _=insert(x, _, j);  end  /*j*/;                return _/*──────────────────────────────────────────────────────────────────────────────────────*/hdr:  parse arg _item_, _;         if highq\==1  then _=center('pieces', maxq)      call show center(_item_,  maxL), center('weight', maxW), center('value',  maxV), ,                center(_     ,  maxQ);                          call hdr2;        return/*──────────────────────────────────────────────────────────────────────────────────────*/hdr2: _=maxQ;   x='═';          if highq==1  then _=0      call show copies(x, maxL), copies(x, maxW), copies(x, maxV), copies(x, _);  return/*──────────────────────────────────────────────────────────────────────────────────────*/j?:   parse arg _,?;  \$=value('Z'_);    do k=1  for _;  ?=? value('J'k);  end;    return/*──────────────────────────────────────────────────────────────────────────────────────*/show: parse arg _item, _weight, _value, _quant      say translate(left(_item, maxL,'─'), ,'_')  right(commas(_weight), maxW),          right(commas(_value ), maxV)            right(commas(_quant ), maxQ);   return/*──────────────────────────────────────────────────────────────────────────────────────*/showOBJ: call hdr 'item';     do j=1  for obj             /*show the formatted choices. */                              parse var  @.j  item weight value q .                              if highq==1  then  q=                                           else  if q==''  then q=1                              call show  item, weight, value, q                              end   /*j*/         say;    say  'number of   unique   named items: '    obj                 say  'number of items (including reps): '  items;    say;        return/*──────────────────────────────────────────────────────────────────────────────────────*/showBest:                     do words(?);  ?=strip(space(?), "L", 0);  end   /*words(?)*/         bestC=?;   bestW=0;   bestV=\$;   highQ=0;   totP=words(bestC);   say         call hdr 'best choice'                              do j=1  to totP         /*J  is modified within  DO  loop.*/                              _=word(bestC, j);     _w=w._;     _v=v._;     q=1                              if _==0  then iterate                                  do k=j+1  to totP;  __=word(bestC, k)  /*get a choice.*/                                  if i._\==i.__  then leave              /*not equal ?  */                                  j=j+1;       w._=w._+_w;       v._=v._+_v;         q=q+1                                  end   /*k*/                              call show  i._,  w._,  v._,  q;            bestW=bestw+w._                              end       /*j*/         call hdr2;    say         call show 'best weight'   ,     bestW        /*show a nicely formatted winnerW.*/         call show 'best value'    , ,   bestV        /*  "  "    "       "     winnerV.*/         call show 'knapsack items', , , totP         /*  "  "    "       "     pieces. */         return/*─────────────────────────────────────────────────────────────────────────────────────────────────────────*/findBest:      h=items;      \$=0 do j1 =0  for h+1;                                       w1=    w.j1 ; z1=    v.j1 ;if  z1>\$ then call j?  1 do j2 =j1 +(j1 >0) to h;if w.j2 +w1 >m then iterate  j1; w2=w1 +w.j2 ; z2=z1 +v.j2 ;if  z2>\$ then call j?  2 do j3 =j2 +(j2 >0) to h;if w.j3 +w2 >m then iterate  j2; w3=w2 +w.j3 ; z3=z2 +v.j3 ;if  z3>\$ then call j?  3 do j4 =j3 +(j3 >0) to h;if w.j4 +w3 >m then iterate  j3; w4=w3 +w.j4 ; z4=z3 +v.j4 ;if  z4>\$ then call j?  4 do j5 =j4 +(j4 >0) to h;if w.j5 +w4 >m then iterate  j4; w5=w4 +w.j5 ; z5=z4 +v.j5 ;if  z5>\$ then call j?  5 do j6 =j5 +(j5 >0) to h;if w.j6 +w5 >m then iterate  j5; w6=w5 +w.j6 ; z6=z5 +v.j6 ;if  z6>\$ then call j?  6 do j7 =j6 +(j6 >0) to h;if w.j7 +w6 >m then iterate  j6; w7=w6 +w.j7 ; z7=z6 +v.j7 ;if  z7>\$ then call j?  7 do j8 =j7 +(j7 >0) to h;if w.j8 +w7 >m then iterate  j7; w8=w7 +w.j8 ; z8=z7 +v.j8 ;if  z8>\$ then call j?  8 do j9 =j8 +(j8 >0) to h;if w.j9 +w8 >m then iterate  j8; w9=w8 +w.j9 ; z9=z8 +v.j9 ;if  z9>\$ then call j?  9 do j10=j9 +(j9 >0) to h;if w.j10+w9 >m then iterate  j9;w10=w9 +w.j10;z10=z9 +v.j10;if z10>\$ then call j? 10 do j11=j10+(j10>0) to h;if w.j11+w10>m then iterate j10;w11=w10+w.j11;z11=z10+v.j11;if z11>\$ then call j? 11 do j12=j11+(j11>0) to h;if w.j12+w11>m then iterate j11;w12=w11+w.j12;z12=z11+v.j12;if z12>\$ then call j? 12 do j13=j12+(j12>0) to h;if w.j13+w12>m then iterate j12;w13=w12+w.j13;z13=z12+v.j13;if z13>\$ then call j? 13 do j14=j13+(j13>0) to h;if w.j14+w13>m then iterate j13;w14=w13+w.j14;z14=z13+v.j14;if z14>\$ then call j? 14 do j15=j14+(j14>0) to h;if w.j15+w14>m then iterate j14;w15=w14+w.j15;z15=z14+v.j15;if z15>\$ then call j? 15 do j16=j15+(j15>0) to h;if w.j16+w15>m then iterate j15;w16=w15+w.j16;z16=z15+v.j16;if z16>\$ then call j? 16 do j17=j16+(j16>0) to h;if w.j17+w16>m then iterate j16;w17=w16+w.j17;z17=z16+v.j17;if z17>\$ then call j? 17 do j18=j17+(j17>0) to h;if w.j18+w17>m then iterate j17;w18=w17+w.j18;z18=z17+v.j18;if z18>\$ then call j? 18 do j19=j18+(j18>0) to h;if w.j19+w18>m then iterate j18;w19=w18+w.j19;z19=z18+v.j19;if z19>\$ then call j? 19 do j20=j19+(j19>0) to h;if w.j20+w19>m then iterate j19;w20=w19+w.j20;z20=z19+v.j20;if z20>\$ then call j? 20 do j21=j20+(j20>0) to h;if w.j21+w20>m then iterate j20;w21=w20+w.j21;z21=z20+v.j21;if z21>\$ then call j? 21 do j22=j21+(j21>0) to h;if w.j22+w21>m then iterate j21;w22=w21+w.j22;z22=z21+v.j22;if z22>\$ then call j? 22 do j23=j22+(j22>0) to h;if w.j23+w22>m then iterate j22;w23=w22+w.j23;z23=z22+v.j23;if z23>\$ then call j? 23 do j24=j23+(j23>0) to h;if w.j24+w23>m then iterate j23;w24=w23+w.j24;z24=z23+v.j24;if z24>\$ then call j? 24 do j25=j24+(j24>0) to h;if w.j25+w24>m then iterate j24;w25=w24+w.j25;z25=z24+v.j25;if z25>\$ then call j? 25 do j26=j25+(j25>0) to h;if w.j26+w25>m then iterate j25;w26=w25+w.j26;z26=z25+v.j26;if z26>\$ then call j? 26 do j27=j26+(j26>0) to h;if w.j27+w26>m then iterate j26;w27=w26+w.j27;z27=z26+v.j27;if z27>\$ then call j? 27 do j28=j27+(j27>0) to h;if w.j28+w27>m then iterate j27;w28=w27+w.j28;z28=z27+v.j28;if z28>\$ then call j? 28 do j29=j28+(j28>0) to h;if w.j29+w28>m then iterate j28;w29=w28+w.j29;z29=z28+v.j29;if z29>\$ then call j? 29 do j30=j29+(j29>0) to h;if w.j30+w29>m then iterate j29;w30=w29+w.j30;z30=z29+v.j30;if z30>\$ then call j? 30 do j31=j30+(j30>0) to h;if w.j31+w30>m then iterate j30;w31=w30+w.j31;z31=z30+v.j31;if z31>\$ then call j? 31 do j32=j31+(j31>0) to h;if w.j32+w31>m then iterate j31;w32=w31+w.j32;z32=z31+v.j32;if z32>\$ then call j? 32 do j33=j32+(j32>0) to h;if w.j33+w32>m then iterate j32;w33=w32+w.j33;z33=z32+v.j33;if z33>\$ then call j? 33 do j34=j33+(j33>0) to h;if w.j34+w33>m then iterate j33;w34=w33+w.j34;z34=z33+v.j34;if z34>\$ then call j? 34 do j35=j34+(j34>0) to h;if w.j35+w34>m then iterate j34;w35=w34+w.j35;z35=z34+v.j35;if z35>\$ then call j? 35 do j36=j35+(j35>0) to h;if w.j36+w35>m then iterate j35;w36=w35+w.j36;z36=z35+v.j36;if z36>\$ then call j? 36 do j37=j36+(j36>0) to h;if w.j37+w36>m then iterate j36;w37=w36+w.j37;z37=z36+v.j37;if z37>\$ then call j? 37 end;end;end;end;end;end;end;end;end;end;end;end;end;end;end;end;end;end;end;end;end;end;end;end;end;end;end     end;end;end;end;end;end;end;end;end;end;        return      /* [↑]  there is one END for each DO loop.*/`

output

```maximum weight allowed for a knapsack:  400

item               weight value pieces
═══════════════════════════════ ══════ ═════ ══════
water──────────────────────────    153   200      2
umbrella───────────────────────     73    40      1
tin────────────────────────────     68    45      3
beer───────────────────────────     52    10      3
sandwich───────────────────────     50    60      2
trousers───────────────────────     48    10      2
waterproof overclothes─────────     43    75      1
waterproof trousers────────────     42    70      1
apple──────────────────────────     39    40      3
camera─────────────────────────     32    30      1
book───────────────────────────     30    10      2
banana─────────────────────────     27    60      3
T-shirt────────────────────────     24    15      2
cheese─────────────────────────     23    30      1
note-case──────────────────────     22    80      1
towel──────────────────────────     18    12      2
glucose────────────────────────     15    60      2
compass────────────────────────     13    35      1
suntan cream───────────────────     11    70      1
map────────────────────────────      9   150      1
sunglasses─────────────────────      7    20      1
socks──────────────────────────      4    50      1

number of items (with repetitions):  37

best choice           weight value pieces
═══════════════════════════════ ══════ ═════ ══════
water──────────────────────────    153   200      1
waterproof overclothes─────────     43    75      1
banana─────────────────────────     81   180      3
cheese─────────────────────────     23    30      1
note-case──────────────────────     22    80      1
glucose────────────────────────     30   120      2
compass────────────────────────     13    35      1
suntan cream───────────────────     11    70      1
map────────────────────────────      9   150      1
sunglasses─────────────────────      7    20      1
socks──────────────────────────      4    50      1
═══════════════════════════════ ══════ ═════ ══════

best weight────────────────────    396
best value─────────────────────        1,010
knapsack items─────────────────                  14
```

## Ruby

Translation of: Python
` # Item struct to represent each item in the problemStruct.new('Item', :name, :weight, :value, :count) \$items = [  Struct::Item.new('map', 9, 150, 1),  Struct::Item.new('compass', 13, 35, 1),  Struct::Item.new('water', 153, 200, 3),  Struct::Item.new('sandwich', 50, 60, 2),  Struct::Item.new('glucose', 15, 60, 2),  Struct::Item.new('tin', 68, 45, 3),  Struct::Item.new('banana', 27, 60, 3),  Struct::Item.new('apple', 39, 40, 3),  Struct::Item.new('cheese', 23, 30, 1),  Struct::Item.new('beer', 52, 10, 3),  Struct::Item.new('suntan cream', 11, 70, 1),  Struct::Item.new('camera', 32, 30, 1),  Struct::Item.new('t-shirt', 24, 15, 2),  Struct::Item.new('trousers', 48, 10, 2),  Struct::Item.new('umbrella', 73, 40, 1),  Struct::Item.new('w-trousers', 42, 70, 1),  Struct::Item.new('w-overcoat', 43, 75, 1),  Struct::Item.new('note-case', 22, 80, 1),  Struct::Item.new('sunglasses', 7, 20, 1),  Struct::Item.new('towel', 18, 12, 2),  Struct::Item.new('socks', 4, 50, 1),  Struct::Item.new('book', 30, 10, 2)] def choose_item(weight, id, cache)  return 0, [] if id < 0   k = [weight, id]  return cache[k] unless cache[k].nil?  value = \$items[id].value  best_v = 0  best_list = []  (\$items[id].count+1).times do |i|    wlim = weight - i * \$items[id].weight    break if wlim < 0    val, taken = choose_item(wlim, id - 1, cache)    if val + i * value > best_v      best_v = val + i * value      best_list = taken + [i]    end  end  cache[k] = [best_v, best_list]  return [best_v, best_list]end val, list = choose_item(400, \$items.length - 1, {})w = 0list.each_with_index do |cnt, i|  if cnt > 0    print "#{cnt} #{\$items[i].name}\n"    w += \$items[i][1] * cnt  endend p "Total weight: #{w}, Value: #{val}" `

## SAS

Use MILP solver in SAS/OR:

`/* create SAS data set */data mydata;   input item \$1-23 weight value pieces;   datalines;map                      9  150  1  compass                 13   35  1  water                  153  200  2  sandwich                50   60  2  glucose                 15   60  2  tin                     68   45  3  banana                  27   60  3  apple                   39   40  3  cheese                  23   30  1  beer                    52   10  3  suntan cream            11   70  1  camera                  32   30  1  T-shirt                 24   15  2  trousers                48   10  2  umbrella                73   40  1  waterproof trousers     42   70  1  waterproof overclothes  43   75  1  note-case               22   80  1  sunglasses               7   20  1  towel                   18   12  2  socks                    4   50  1  book                    30   10  2  ; /* call OPTMODEL procedure in SAS/OR */proc optmodel;   /* declare sets and parameters, and read input data */   set <str> ITEMS;   num weight {ITEMS};   num value {ITEMS};   num pieces {ITEMS};   read data mydata into ITEMS=[item] weight value pieces;    /* declare variables, objective, and constraints */   var NumSelected {i in ITEMS} >= 0 <= pieces[i] integer;   max TotalValue = sum {i in ITEMS} value[i] * NumSelected[i];   con WeightCon:      sum {i in ITEMS} weight[i] * NumSelected[i] <= 400;    /* call mixed integer linear programming (MILP) solver */   solve;    /* print optimal solution */   print TotalValue;   print {i in ITEMS: NumSelected[i].sol > 0.5} NumSelected;quit;`

Output:

```TotalValue
1010

[1] NumSelected
banana 3
cheese 1
compass 1
glucose 2
map 1
note-case 1
socks 1
sunglasses 1
suntan cream 1
water 1
waterproof overclothes 1
```

## Sidef

Translation of: Perl
`var raw = <<'TABLE'map           9     150      1compass      13      35      1water       153     200      2sandwich     50      60      2glucose      15      60      2tin          68      45      3banana       27      60      3apple        39      40      3cheese       23      30      1beer         52      10      1suntancream  11      70      1camera       32      30      1T-shirt      24      15      2trousers     48      10      2umbrella     73      40      1w_trousers   42      70      1w_overcoat   43      75      1note-case    22      80      1sunglasses    7      20      1towel        18      12      2socks         4      50      1book         30      10      2TABLE struct KnapsackItem {    String name,    Number weight,    Number value,    Number quant,} var items = []raw.each_line{ |row|    var fields = row.words;    items << KnapsackItem(          name: fields[0],        weight: fields[1].to_n,         value: fields[2].to_n,         quant: fields[3].to_n,    )} func pick(weight, pos) is cached {     if (pos.is_neg || weight.is_neg || weight.is_zero) {        return (0, 0, [])    }     var (bv=0, bi=0, bw=0, bp=[])    var item = items[pos];     for i in range(0, item.quant) {        break if (i*item.weight > weight)        var (v, w, p) = pick(weight - i*item.weight, pos.dec)        next if ((v += i*item.value) <= bv)        (bv, bi, bw, bp) = (v, i, w, p)    }     (bv, bw + bi*item.weight, [bp..., bi])} var (v, w, p) = pick(400, items.end)p.range.each { |i|    say "#{p[i]} of #{items[i].name}" if p[i].is_pos}say "Value: #{v}; Weight: #{w}"`
Output:
```1 of map
1 of compass
1 of water
2 of glucose
3 of banana
1 of cheese
1 of suntancream
1 of w_overcoat
1 of note-case
1 of sunglasses
1 of socks
Value: 1010; Weight: 396
```

## Tcl

Classic dumb search algorithm:

`# The list of items to consider, as list of listsset items {    {map			9	150	1}    {compass			13	35	1}    {water			153	200	2}    {sandwich			50	60	2}    {glucose			15	60	2}    {tin			68	45	3}    {banana			27	60	3}    {apple			39	40	3}    {cheese			23	30	1}    {beer			52	10	3}    {{suntan cream}		11	70	1}    {camera			32	30	1}    {t-shirt			24	15	2}    {trousers			48	10	2}    {umbrella			73	40	1}    {{waterproof trousers}	42	70	1}    {{waterproof overclothes}	43	75	1}    {note-case			22	80	1}    {sunglasses			7	20	1}    {towel			18	12	2}    {socks			4	50	1}    {book			30	10	2}} # Simple extraction functionsproc countednames {chosen} {    set names {}    foreach item \$chosen {	lappend names [lindex \$item 3] [lindex \$item 0]    }    return \$names}proc weight {chosen} {    set weight 0    foreach item \$chosen {	incr weight [expr {[lindex \$item 3] * [lindex \$item 1]}]    }    return \$weight}proc value {chosen} {    set value 0    foreach item \$chosen {	incr value [expr {[lindex \$item 3] * [lindex \$item 2]}]    }    return \$value} # Recursive function for searching over all possible choices of itemsproc knapsackSearch {items {chosen {}}} {    # If we've gone over the weight limit, stop now    if {[weight \$chosen] > 400} {	return    }    # If we've considered all of the items (i.e., leaf in search tree)    # then see if we've got a new best choice.    if {[llength \$items] == 0} {	global best max	set v [value \$chosen]	if {\$v > \$max} {	    set max \$v	    set best \$chosen	}	return    }    # Branch, so recurse for chosing the current item or not    set this [lindex \$items 0]    set rest [lrange \$items 1 end]    knapsackSearch \$rest \$chosen    for {set i 1} {\$i<=[lindex \$this 3]} {incr i} {	knapsackSearch \$rest \	    [concat \$chosen [list [lreplace \$this end end \$i]]]    }} # Initialize a few global variablesset best {}set max 0# Do the brute-force searchknapsackSearch \$items# Pretty-print the resultsputs "Best filling has weight of [expr {[weight \$best]/100.0}]kg and score [value \$best]"puts "Best items:"foreach {count item} [countednames \$best] {    puts "\t\$count * \$item"}`
Output:
```Best filling has weight of 3.96kg and score 1010
Best items:
1 * map
1 * compass
1 * water
2 * glucose
3 * banana
1 * cheese
1 * suntan cream
1 * waterproof overclothes
1 * note-case
1 * sunglasses
1 * socks
```

## Ursala

Instead of an ad-hoc solution, we can convert this task to a mixed integer programming problem instance and solve it with the lpsolve library, which is callable in Ursala.

`#import std#import flo#import lin items = # name: (weight,value,limit) <   'map': (9,150,1),   'compass': (13,35,1),   'water': (153,200,3),   'sandwich': (50,60,2),   'glucose': (15,60,2),   'tin': (68,45,3),   'banana': (27,60,3),   'apple': (39,40,3),   'cheese': (23,30,1),   'beer': (52,10,3),   'suntan cream': (11,70,1),   'camera': (32,30,1),   't-shirt': (24,15,2),   'trousers': (48,10,2),   'umbrella': (73,40,1),   'waterproof trousers': (42,70,1),   'waterproof overclothes': (43,75,1),   'note-case': (22,80,1),   'sunglasses': (7,20,1),   'towel': (18,12,2),   'socks': (4,50,1),   'book': (30,10,2)> system = # convert the item list to mixed integer programming problem specification linear_system\$[   integers: ~&nS,   upper_bounds: * ^|/~& [email protected],   lower_bounds: @nS ~&\*0.+ :/'(slack)',   costs: * ^|/~& negative+ [email protected],   equations: ~&iNC\400.+ :/(1.,'(slack)')+ * ^|rlX/~& [email protected]] format = @t --^*p\pad` @nS @mS printf/*'%0.0f ' #show+ main =  format solution system items`

The `linear_system\$[`...`]` function takes the item list as an argument and returns a data structure with the following fields, which is passed to the `solution` function, which calls the `lpsolve` routines.

• `integers` declares that all variables in the list take integer values
• `upper_bounds` associates an upper bound for each variable directly as given
• `lower_bounds` are zero for all variables in the table, and for an additional slack variable, which is not required to be an integer and won't appear in the solution
• `costs` are also taken from the table and negated because their value is to be maximized rather than minimized as in the standard formulation
• `equations` specifies the single constraint that the total weights of the selected items in the selected quantities plus the slack equal 400

The `format` function converts the output list to a readable form.

Output:
```3 banana
1 cheese
1 compass
2 glucose
1 map
1 note-case
1 socks
1 sunglasses
1 suntan cream
1 water
1 waterproof overclothes```

## zkl

`fcn addItem(old,[(nm,w,v,c)]){  // old:list:(cost of:,(name,#,...))   [1..c].reduce(fcn(list,i,nm,w,v,old){      wi,left,right:=w*i,list[0,wi],list[wi,*];      new:=old.apply('wrap([(val,itms)]){ T(val + v*i,itms.append(nm,i)) });      left.extend(right.zipWith(  // inc	    fcn([(v1,_)]a,[(v2,_)]b){ v1>v2 and a or b  },new));   },old,nm,w,v,old);}//--> new list`
`items:=T( // item: (name,weight,value,#)   T("apple",      39, 40,3),T("banana",   27,60,3),   T("beer",       52, 10,3),T("book",     30,10,2),T("camera",      32, 30,1),   T("cheese",     23, 30,1),T("compass",  13,35,1),T("glucose",     15, 60,2),   T("map",         9,150,1),T("note-case",22,80,1),T("sandwich",    50, 60,2),   T("socks",       4, 50,1),T("sunglasses",7,20,1),T("suntan cream",11, 70,1),   T("t-shirt",    24, 15,2),T("tin",      68,45,3),T("towel",       18, 12,2),   T("trousers",   48, 10,2),T("umbrella", 73,40,1),T("water",      153,200,2),   T("overclothes",43, 75,1),T("waterproof trousers",42,70,1) );weight:='wrap(knapsack){ // knapsack is (cost, (nm,#,nm,#...))   knapsack[1].pump(List,Void.Read, // read nm,#, first read is implicit      'wrap(nm,n){ items[items.filter1n(fcn(it,nm){ it[0]==nm },nm)][1]*n })      .sum(0)};`
`const MAX_WEIGHT=400;knapsack:=items.reduce(addItem,(MAX_WEIGHT).pump(List,T(0,T).copy))[-1];knapsack.toString(*).println(weight(knapsack));`
```L(1010,L("banana",3,"cheese",1,"compass",1,"glucose",2,"map",1,"note-case",1,"socks",1,"sunglasses",1,"suntan cream",1,"water",1,"overclothes",1))396