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Count the coins/0-1

From Rosetta Code
Count the coins/0-1 is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Let say you have some coins in your wallet and you want to have a given sum.

You can use each coin zero or one time.

How many ways can you do it ?

The result should be a number.

For instance the answer is 10 when coins = [1, 2, 3, 4, 5] and sum = 6.

Task

Show the result for the following examples:

  •   coins = [1, 2, 3, 4, 5] and sum = 6
  •   coins = [1, 1, 2, 3, 3, 4, 5] and sum = 6
  •   coins = [1, 2, 3, 4, 5, 5, 5, 5, 15, 15, 10, 10, 10, 10, 25, 100] and sum = 40
Extra
  •   Show the result of the same examples when the order you take the coins doesn't matter. For instance the answer is 3 when coins = [1, 2, 3, 4, 5] and sum = 6.
  •   Show an example of coins you used to reach the given sum and their indices. See Raku for this case.

Go[edit]

Translation of: Wren
package main
 
import "fmt"
 
var cnt = 0  // order unimportant
var cnt2 = 0 // order important
var wdth = 0 // for printing purposes
 
func factorial(n int) int {
    prod := 1
    for i := 2; i <= n; i++ {
        prod *= i
    }
    return prod
}
 
func count(want int, used []int, sum int, have, uindices, rindices []int) {
    if sum == want {
        cnt++
        cnt2 += factorial(len(used))
        if cnt < 11 {
            uindicesStr := fmt.Sprintf("%v", uindices)
            fmt.Printf("  indices %*s => used %v\n", wdth, uindicesStr, used)
        }
    } else if sum < want && len(have) != 0 {
        thisCoin := have[0]
        index := rindices[0]
        rest := have[1:]
        rindices := rindices[1:]
        count(want, append(used, thisCoin), sum+thisCoin, rest,
            append(uindices, index), rindices)
        count(want, used, sum, rest, uindices, rindices)
    }
}
 
func countCoins(want int, coins []int, width int) {
    fmt.Printf("Sum %d from coins %v\n", want, coins)
    cnt = 0
    cnt2 = 0
    wdth = -width
    rindices := make([]int, len(coins))
    for i := range rindices {
        rindices[i] = i
    }
    count(want, []int{}, 0, coins, []int{}, rindices)
    if cnt > 10 {
        fmt.Println("  .......")
        fmt.Println("  (only the first 10 ways generated are shown)")
    }
    fmt.Println("Number of ways - order unimportant :", cnt, "(as above)")
    fmt.Println("Number of ways - order important   :", cnt2, "(all perms of above indices)\n")
}
 
func main() {
    countCoins(6, []int{1, 2, 3, 4, 5}, 7)
    countCoins(6, []int{1, 1, 2, 3, 3, 4, 5}, 9)
    countCoins(40, []int{1, 2, 3, 4, 5, 5, 5, 5, 15, 15, 10, 10, 10, 10, 25, 100}, 20)
}
Output:
Sum 6 from coins [1 2 3 4 5]
  indices [0 1 2] => used [1 2 3]
  indices [0 4]   => used [1 5]
  indices [1 3]   => used [2 4]
Number of ways - order unimportant : 3 (as above)
Number of ways - order important   : 10 (all perms of above indices)

Sum 6 from coins [1 1 2 3 3 4 5]
  indices [0 1 5]   => used [1 1 4]
  indices [0 2 3]   => used [1 2 3]
  indices [0 2 4]   => used [1 2 3]
  indices [0 6]     => used [1 5]
  indices [1 2 3]   => used [1 2 3]
  indices [1 2 4]   => used [1 2 3]
  indices [1 6]     => used [1 5]
  indices [2 5]     => used [2 4]
  indices [3 4]     => used [3 3]
Number of ways - order unimportant : 9 (as above)
Number of ways - order important   : 38 (all perms of above indices)

Sum 40 from coins [1 2 3 4 5 5 5 5 15 15 10 10 10 10 25 100]
  indices [0 1 2 3 4 5 6 7 10] => used [1 2 3 4 5 5 5 5 10]
  indices [0 1 2 3 4 5 6 7 11] => used [1 2 3 4 5 5 5 5 10]
  indices [0 1 2 3 4 5 6 7 12] => used [1 2 3 4 5 5 5 5 10]
  indices [0 1 2 3 4 5 6 7 13] => used [1 2 3 4 5 5 5 5 10]
  indices [0 1 2 3 4 5 6 8]    => used [1 2 3 4 5 5 5 15]
  indices [0 1 2 3 4 5 6 9]    => used [1 2 3 4 5 5 5 15]
  indices [0 1 2 3 4 5 7 8]    => used [1 2 3 4 5 5 5 15]
  indices [0 1 2 3 4 5 7 9]    => used [1 2 3 4 5 5 5 15]
  indices [0 1 2 3 4 5 10 11]  => used [1 2 3 4 5 5 10 10]
  indices [0 1 2 3 4 5 10 12]  => used [1 2 3 4 5 5 10 10]
  .......
  (only the first 10 ways generated are shown)
Number of ways - order unimportant : 464 (as above)
Number of ways - order important   : 3782932 (all perms of above indices)

Julia[edit]

using Combinatorics
 
function coinsum(coins, targetsum; verbose=true)
    println("Coins are $coins, target sum is $targetsum:")
    combos, perms = 0, 0
    for choice in combinations(coins)
        if sum(choice) == targetsum
            combos += 1
            verbose && println("$choice sums to $targetsum")
            for perm in permutations(choice)
                verbose && println("    permutation: $perm")
                perms += 1
            end
        end
    end
    println("$combos combinations, $perms permutations.\n")
end
 
coinsum([1, 2, 3, 4, 5], 6)
coinsum([1, 1, 2, 3, 3, 4, 5], 6)
coinsum([1, 2, 3, 4, 5, 5, 5, 5, 15, 15, 10, 10, 10, 10, 25, 100], 40, verbose=false)
 
Output:
Coins are [1, 2, 3, 4, 5], target sum is 6:
[1, 5] sums to 6
    permutation: [1, 5]
    permutation: [5, 1]
[2, 4] sums to 6
    permutation: [2, 4]
    permutation: [4, 2]
[1, 2, 3] sums to 6
    permutation: [1, 2, 3]
    permutation: [1, 3, 2]
    permutation: [2, 1, 3]
    permutation: [2, 3, 1]
    permutation: [3, 1, 2]
    permutation: [3, 2, 1]
3 combinations, 10 permutations.

Coins are [1, 1, 2, 3, 3, 4, 5], target sum is 6:
[1, 5] sums to 6
    permutation: [1, 5]
    permutation: [5, 1]
[1, 5] sums to 6
    permutation: [1, 5]
    permutation: [5, 1]
[2, 4] sums to 6
    permutation: [2, 4]
    permutation: [4, 2]
[3, 3] sums to 6
    permutation: [3, 3]
    permutation: [3, 3]
[1, 1, 4] sums to 6
    permutation: [1, 1, 4]
    permutation: [1, 4, 1]
    permutation: [1, 1, 4]
    permutation: [1, 4, 1]
    permutation: [4, 1, 1]
    permutation: [4, 1, 1]
[1, 2, 3] sums to 6
    permutation: [1, 2, 3]
    permutation: [1, 3, 2]
    permutation: [2, 1, 3]
    permutation: [2, 3, 1]
    permutation: [3, 1, 2]
    permutation: [3, 2, 1]
[1, 2, 3] sums to 6
    permutation: [1, 2, 3]
    permutation: [1, 3, 2]
    permutation: [2, 1, 3]
    permutation: [2, 3, 1]
    permutation: [3, 1, 2]
    permutation: [3, 2, 1]
[1, 2, 3] sums to 6
    permutation: [1, 2, 3]
    permutation: [1, 3, 2]
    permutation: [2, 1, 3]
    permutation: [2, 3, 1]
    permutation: [3, 1, 2]
    permutation: [3, 2, 1]
[1, 2, 3] sums to 6
    permutation: [1, 2, 3]
    permutation: [1, 3, 2]
    permutation: [2, 1, 3]
    permutation: [2, 3, 1]
    permutation: [3, 1, 2]
    permutation: [3, 2, 1]
9 combinations, 38 permutations.

Coins are [1, 2, 3, 4, 5, 5, 5, 5, 15, 15, 10, 10, 10, 10, 25, 100], target sum is 40:
464 combinations, 3782932 permutations.

MiniZinc[edit]

coins = [1, 2, 3, 4, 5] and sum = 6
 
%Subset sum. Nigel Galloway: January 6th., 2021.
enum Items={a,b,c,d,e};
array[Items] of int: weight=[1,2,3,4,5];
var set of Items: selected;
var int: wSelected=sum(n in selected)(weight[n]);
constraint wSelected=6;
 
Output:
selected = {a, b, c};
----------
selected = {a, e};
----------
selected = {b, d};
----------
==========
%%%mzn-stat: initTime=0
%%%mzn-stat: solveTime=0.001
%%%mzn-stat: solutions=3
%%%mzn-stat: variables=21
%%%mzn-stat: propagators=31
%%%mzn-stat: propagations=236
%%%mzn-stat: nodes=11
%%%mzn-stat: failures=3
%%%mzn-stat: restarts=0
%%%mzn-stat: peakDepth=3
%%%mzn-stat-end
Finished in 172msec
coins = [1, 1, 2, 3, 3, 4, 5] and sum = 6
 
%Subset sum. Nigel Galloway: January 6th., 2021.
enum Items={a,b,c,d,e,f,g};
array[Items] of int: weight=[1,1,2,3,3,4,5];
var set of Items: selected;
var int: wSelected=sum(n in selected)(weight[n]);
constraint wSelected=6;
 
Output:
selected = {a, b, f};
----------
selected = {a, c, d};
----------
selected = {a, c, e};
----------
selected = {a, g};
----------
selected = {b, c, d};
----------
selected = {b, c, e};
----------
selected = {b, g};
----------
selected = {c, f};
----------
selected = {d, e};
----------
==========
%%%mzn-stat: initTime=0
%%%mzn-stat: solveTime=0.001
%%%mzn-stat: solutions=9
%%%mzn-stat: variables=29
%%%mzn-stat: propagators=43
%%%mzn-stat: propagations=820
%%%mzn-stat: nodes=35
%%%mzn-stat: failures=9
%%%mzn-stat: restarts=0
%%%mzn-stat: peakDepth=4
%%%mzn-stat-end
Finished in 187msec
coins = [1, 2, 3, 4, 5, 5, 5, 5, 15, 15, 10, 10, 10, 10, 25, 100] and sum = 40
 
%Subset sum. Nigel Galloway: January 6th., 2021.
enum Items={a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p};
array[Items] of int: weight=[1,2,3,4,5,5,5,5,15,15,10,10,10,10,25,100];
var set of Items: selected;
var int: wSelected=sum(n in selected)(weight[n]);
constraint wSelected=40;
 
Output:
selected = {a, b, c, d, e, f, g, h, k};
----------
selected = {a, b, c, d, e, f, g, h, l};
----------
[ 0 more solutions ]
selected = {a, b, c, d, e, f, g, h, m};
----------
[ 1 more solutions ]
selected = {a, b, c, d, e, f, g, i};
----------
[ 3 more solutions ]
selected = {a, b, c, d, e, f, k, l};
----------
[ 7 more solutions ]
selected = {a, b, c, d, e, g, k, l};
----------
[ 15 more solutions ]
selected = {a, b, c, d, e, j, k};
----------
[ 31 more solutions ]
selected = {a, b, c, d, g, h, l, n};
----------
[ 63 more solutions ]
selected = {a, d, e, h, i, l};
----------
[ 127 more solutions ]
selected = {b, c, e, l, m, n};
----------
[ 206 more solutions ]
selected = {k, l, m, n};
----------
==========
%%%mzn-stat: initTime=0.001
%%%mzn-stat: solveTime=0.011
%%%mzn-stat: solutions=464
%%%mzn-stat: variables=65
%%%mzn-stat: propagators=91
%%%mzn-stat: propagations=122926
%%%mzn-stat: nodes=4203
%%%mzn-stat: failures=1638
%%%mzn-stat: restarts=0
%%%mzn-stat: peakDepth=13
%%%mzn-stat-end
Finished in 207msec

Nim[edit]

Translation of: Go

Using a solver object rather than global variables. 

import math, sequtils, strutils
 
type Solver = object
  want: Positive
  count1: Natural
  count2: Natural
  width: Natural
 
proc count(solver: var Solver; sum: int; used, have, uindices, rindices: seq[int]) =
  if sum == solver.want:
    inc solver.count1
    inc solver.count2, fac(used.len)
    if solver.count1 < 11:
      let uindiceStr = ($uindices.join(" ")).alignLeft(solver.width)
      echo "  indices $1  →  used $2".format(uindiceStr, used.join(" "))
  elif sum < solver.want and have.len != 0:
    let thisCoin = have[0]
    let index = rindices[0]
    let rest = have[1..^1]
    let rindices = rindices[1..^1]
    solver.count(sum + thisCoin, used & thisCoin, rest, uindices & index, rindices)
    solver.count(sum, used, rest, uindices, rindices)
 
proc countCoins(want: int; coins: openArray[int]; width: int) =
  echo "Sum $# from coins $#".format(want, coins.join(" "))
  var solver = Solver(want: want, width: width)
  var rindices = toSeq(0..coins.high)
  solver.count(0, newSeq[int](), @coins, newSeq[int](), rindices)
  if solver.count1 > 10:
    echo "  ......."
    echo "  (only the first 10 ways generated are shown)"
  echo "Number of ways – order unimportant : ", solver.count1, " (as above)"
  echo "Number of ways – order important   : ", solver.count2, " (all perms of above indices)\n"
 
when isMainModule:
  countCoins(6, [1, 2, 3, 4, 5], 5)
  countCoins(6, [1, 1, 2, 3, 3, 4, 5], 7)
  countCoins(40, [1, 2, 3, 4, 5, 5, 5, 5, 15, 15, 10, 10, 10, 10, 25, 100], 18)
Output:
Sum 6 from coins 1 2 3 4 5
  indices 0 1 2  →  used 1 2 3
  indices 0 4    →  used 1 5
  indices 1 3    →  used 2 4
Number of ways – order unimportant : 3 (as above)
Number of ways – order important   : 10 (all perms of above indices)

Sum 6 from coins 1 1 2 3 3 4 5
  indices 0 1 5    →  used 1 1 4
  indices 0 2 3    →  used 1 2 3
  indices 0 2 4    →  used 1 2 3
  indices 0 6      →  used 1 5
  indices 1 2 3    →  used 1 2 3
  indices 1 2 4    →  used 1 2 3
  indices 1 6      →  used 1 5
  indices 2 5      →  used 2 4
  indices 3 4      →  used 3 3
Number of ways – order unimportant : 9 (as above)
Number of ways – order important   : 38 (all perms of above indices)

Sum 40 from coins 1 2 3 4 5 5 5 5 15 15 10 10 10 10 25 100
  indices 0 1 2 3 4 5 6 7 10  →  used 1 2 3 4 5 5 5 5 10
  indices 0 1 2 3 4 5 6 7 11  →  used 1 2 3 4 5 5 5 5 10
  indices 0 1 2 3 4 5 6 7 12  →  used 1 2 3 4 5 5 5 5 10
  indices 0 1 2 3 4 5 6 7 13  →  used 1 2 3 4 5 5 5 5 10
  indices 0 1 2 3 4 5 6 8     →  used 1 2 3 4 5 5 5 15
  indices 0 1 2 3 4 5 6 9     →  used 1 2 3 4 5 5 5 15
  indices 0 1 2 3 4 5 7 8     →  used 1 2 3 4 5 5 5 15
  indices 0 1 2 3 4 5 7 9     →  used 1 2 3 4 5 5 5 15
  indices 0 1 2 3 4 5 10 11   →  used 1 2 3 4 5 5 10 10
  indices 0 1 2 3 4 5 10 12   →  used 1 2 3 4 5 5 10 10
  .......
  (only the first 10 ways generated are shown)
Number of ways – order unimportant : 464 (as above)
Number of ways – order important   : 3782932 (all perms of above indices)

Pascal[edit]

first count all combinations by brute force.Order is important.Modified nQueens.
Next adding weigths by index.
Using Phix wording 'silly'. 

program Coins0_1;
{$IFDEF FPC}
   {$MODE DELPHI}
   {$OPTIMIZATION ON,ALL}
{$ELSE}
  {$Apptype console}
{$ENDIF}
 
uses
  sysutils;// TDatetime
const
  coins1 :array[0..4] of byte = (1, 2, 3, 4, 5);
  coins2 :array[0..6] of byte = (1, 1, 2, 3, 3, 4, 5);
  coins3 :array[0..15] of byte = (1,2,3,4,5,5,5,5,15,15,10,10,10,10,25,100);
  nmax = High(Coins3);
type
{$IFNDEF FPC}
  NativeInt = Int32;
{$ENDIF}
  tFreeCol = array[0..nmax] of Int32;
var
  FreeIdx,
  IdxWeight : tFreeCol;
  n,
  gblCount : nativeUInt;
 
procedure AddNextWeight(Row,sum:nativeInt);
//order is important
var
  i,Col,Weight : nativeInt;
begin
  IF row <= n then
  begin
    For i := row to n do
    begin
      Col := FreeIdx[i];
      Weight:= IdxWeight[col];
      IF Sum+Weight <= 0 then
      Begin
        Sum +=Weight;
        If Sum = 0 then
        Begin
          Sum -=Weight;
          inc(gblCount);
        end
        else
        begin
          FreeIdx[i] := FreeIdx[Row];
          FreeIdx[Row] := Col;
 
          AddNextWeight(Row+1,sum);
          //Undo
          Sum -=Weight;
          FreeIdx[Row] := FreeIdx[i];
          FreeIdx[i] := Col;
        end;
      end;
    end;
  end;
end;
 
procedure CheckBinary(n,MaxIdx,Sum:NativeInt);
//order is not important
Begin
  if sum = 0 then
    inc(gblcount);
  If (sum < 0) AND (n <= MaxIdx) then
  Begin
    //test next sum 
    CheckBinary(n+1,MaxIdx,Sum);// add nothing
    CheckBinary(n+1,MaxIdx,Sum+IdxWeight[n]);//or the actual index
  end;
end;
 
procedure CheckAll(i,MaxSum:NativeInt);
Begin
  n := i;
  gblCount := 0;
  AddNextWeight(0,-MaxSum);
  Write(MaxSum:6,gblCount:12);
  gblCount := 0;
  CheckBinary(0,i,-MaxSum);
  WriteLn(gblCount:12);
end;
 
var
  i: nativeInt;
 
begin
  writeln('sum':6,'very silly':12,'silly':12);
  For i := 0 to High(coins1) do
  Begin
    FreeIdx[i] := i;
    IdxWeight[i] := coins1[i];
  end;
  CheckAll(High(coins1),6);
 
  For i := 0 to High(coins2) do
  Begin
    FreeIdx[i] := i;
    IdxWeight[i] := coins2[i];
  end;
  CheckAll(High(coins2),6);
 
  For i := 0 to High(coins3) do
  Begin
    FreeIdx[i] := i;
    IdxWeight[i] := coins3[i];
  end;
  CheckAll(High(coins3),40);
end.
 
Output:
   sum  very silly       silly
     6          10           3
     6          38           9
    40     3782932         464

// real	0m0,080s ( fpc 3.2.0 -O3 -Xs AMD 2200G 3.7 Ghz )

Perl[edit]

#!/usr/bin/perl
 
use strict; # https://rosettacode.org/wiki/Count_the_coins/0-1
use warnings;
 
countcoins( 6, [1, 2, 3, 4, 5] );
countcoins( 6, [1, 1, 2, 3, 3, 4, 5] );
countcoins( 40, [1, 2, 3, 4, 5, 5, 5, 5, 15, 15, 10, 10, 10, 10, 25, 100] );
 
my $count;
 
sub countcoins
  {
  my ($want, $coins) = @_;
  print "\nsum $want coins @$coins\n";
  $count = 0;
  count($want, [], 0, $coins);
  print "Number of ways: $count\n";
  }
 
sub count
  {
  my ($want, $used, $sum, $have) = @_;
  if( $sum == $want ) { $count++ }
  elsif( $sum > $want or @$have == 0 ) {}
  else
    {
    my ($thiscoin, @rest) = @$have;
    count( $want, [@$used, $thiscoin], $sum + $thiscoin, \@rest);
    count( $want, $used, $sum, \@rest);
    }
  }
Output:
sum 6 coins 1 2 3 4 5
Number of ways: 3

sum 6 coins 1 1 2 3 3 4 5
Number of ways: 9

sum 40 coins 1 2 3 4 5 5 5 5 15 15 10 10 10 10 25 100
Number of ways: 464

Phix[edit]

sane way[edit]

In the second test, this counts [1,2,3] as one way to get 6, not counting the other [1,2,3] or the other [1,2,3] or the other [1,2,3]. 

with javascript_semantics
function choices(sequence coins, integer tgt, cdx=1)
    integer count = 0
    if tgt=0 then
        count += 1
    elsif tgt>0 and cdx<=length(coins) then
        object ci = coins[cdx]
        integer {c,n} = iff(sequence(ci)?ci:{ci,1})
        for j=0 to n do
            count += choices(coins,tgt-j*c,cdx+1)
        end for
    end if
    return count
end function
 
constant tests = {{{1,2,3,4,5},6},
                  {{{1,2},2,{3,2},4,5},6},
                  {{1,2,3,4,{5,4},{15,2},{10,4},25,100},40}}
printf(1,"%V\n",{apply(false,choices,tests)})
Output:
{3,5,33}

silly way[edit]

In the second test, this counts [1,2,3] as four ways to get 6, since there are two 1s and two 3s... ("order unimportant") 

with javascript_semantics
function silly(sequence coins, integer tgt, cdx=1)
    integer count = 0
    if tgt=0 then
        count += 1
    elsif tgt>0 and cdx<=length(coins) then
        count += silly(coins,tgt-coins[cdx],cdx+1)
        count += silly(coins,tgt,cdx+1)
    end if
    return count
end function
 
constant tests = {{{1,2,3,4,5},6},
                  {{1,1,2,3,3,4,5},6},
                  {{1,2,3,4,5,5,5,5,15,15,10,10,10,10,25,100},40}}
printf(1,"%V\n",{apply(false,silly,tests)})
Output:
{3,9,464}

very silly way[edit]

In the second test, this counts [1,2,3] as 24 ways to get 6, from 2 1s, 2 3s, and 6 ways to order them ("order important") 

with javascript_semantics
function very_silly(sequence coins, integer tgt, cdx=1, taken=0)
    integer count = 0
    if tgt=0 then
        count += factorial(taken)
    elsif tgt>0 and cdx<=length(coins) then
        count += very_silly(coins,tgt-coins[cdx],cdx+1,taken+1)
        count += very_silly(coins,tgt,cdx+1,taken)
    end if
    return count
end function
 
constant tests = {{{1,2,3,4,5},6},
                  {{1,1,2,3,3,4,5},6},
                  {{1,2,3,4,5,5,5,5,15,15,10,10,10,10,25,100},40}}
printf(1,"%V\n",{apply(false,very_silly,tests)})
Output:
{10,38,3782932}

Raku[edit]

First part is combinations filtered on a certain property. Second part (extra credit) is permutations of those combinations.

This is pretty much duplicating other tasks, in process if not wording. Even though I am adding a solution, my vote would be for deletion as it doesn't really add anything to the other tasks; Combinations, Permutations, Subset sum problem and to a large extent 4-rings or 4-squares puzzle. 

for <1 2 3 4 5>, 6
   ,<1 1 2 3 3 4 5>, 6
   ,<1 2 3 4 5 5 5 5 15 15 10 10 10 10 25 100>, 40
  -> @items, $sum {
 
    put "\n\nHow many combinations of [{ @items.join: ', ' }] sum to $sum?";
 
    given ^@items .combinations.grep: { @items[$_].sum == $sum } {
        .&display;
        display .race.map( { Slip(.permutations) } ), '';
    }
}
 
sub display ($list, $un = 'un') {
    put "\nOrder {$un}important:\nCount: { +$list }\nIndices" ~ ( +$list > 10 ?? ' (10 random examples):' !! ':' );
    put $list.pick(10).sort».join(', ').join: "\n"
}
Output:
How many combinations of [1, 2, 3, 4, 5] sum to 6?

Order unimportant:
Count: 3
Indices:
0, 1, 2
0, 4
1, 3

Order important:
Count: 10
Indices:
0, 1, 2
0, 2, 1
0, 4
1, 0, 2
1, 2, 0
1, 3
2, 0, 1
2, 1, 0
3, 1
4, 0


How many combinations of [1, 1, 2, 3, 3, 4, 5] sum to 6?

Order unimportant:
Count: 9
Indices:
0, 1, 5
0, 2, 3
0, 2, 4
0, 6
1, 2, 3
1, 2, 4
1, 6
2, 5
3, 4

Order important:
Count: 38
Indices (10 random examples):
0, 4, 2
1, 2, 3
1, 2, 4
1, 5, 0
1, 6
2, 1, 3
2, 1, 4
2, 4, 0
3, 2, 0
6, 0


How many combinations of [1, 2, 3, 4, 5, 5, 5, 5, 15, 15, 10, 10, 10, 10, 25, 100] sum to 40?

Order unimportant:
Count: 464
Indices (10 random examples):
0, 1, 2, 3, 5, 7, 10, 11
0, 1, 2, 3, 5, 8, 12
0, 1, 2, 3, 6, 7, 11, 13
0, 3, 5, 7, 9, 13
0, 3, 9, 10, 11
1, 2, 5, 7, 9, 13
4, 5, 10, 12, 13
5, 6, 7, 9, 10
5, 6, 10, 11, 12
5, 8, 10, 12

Order important:
Count: 3782932
Indices (10 random examples):
0, 11, 3, 4, 7, 5, 6, 1, 2
1, 10, 5, 4, 6, 2, 0, 3, 7
2, 7, 13, 4, 1, 3, 5, 6, 0
2, 12, 4, 13, 10, 1
3, 0, 5, 4, 7, 13, 6, 2, 1
5, 7, 9, 4, 0, 1, 2, 3
6, 2, 7, 11, 0, 3, 5, 1, 4
10, 0, 12, 6, 5, 3, 4
13, 0, 1, 5, 7, 3, 2, 12
13, 6, 10, 1, 4, 3, 2, 0

Wren[edit]

Translation of: Perl

Well, after some huffing and puffing, the house is still standing so I thought I'd have a go at it. Based on the Perl algorithm but modified to deal with the extra credit. 

import "/fmt" for Fmt
import "/math" for Int
 
var cnt  = 0 // order unimportant
var cnt2 = 0 // order important
var wdth = 0 // for printing purposes
 
var count // recursive
count = Fn.new { |want, used, sum, have, uindices, rindices|
    if (sum == want) {
        cnt = cnt + 1
        cnt2 = cnt2 + Int.factorial(used.count)
        if (cnt < 11) Fmt.print("  indices $*n => used $n", wdth, uindices, used)
    } else if (sum < want && !have.isEmpty) {
        var thisCoin = have[0]
        var index = rindices[0]
        var rest = have.skip(1).toList
        var rindices = rindices.skip(1).toList
        count.call(want, used + [thisCoin], sum + thisCoin, rest, uindices + [index], rindices)
        count.call(want, used, sum, rest, uindices, rindices)
    }
}
 
var countCoins = Fn.new { |want, coins, width|
    System.print("Sum %(want) from coins %(coins)")
    cnt  = 0
    cnt2 = 0
    wdth = -width
    count.call(want, [], 0, coins, [], (0...coins.count).toList)
    if (cnt > 10) {
        System.print("  .......")
        System.print("  (only the first 10 ways generated are shown)")
    }
    System.print("Number of ways - order unimportant : %(cnt) (as above)")
    System.print("Number of ways - order important   : %(cnt2) (all perms of above indices)\n")
}
 
countCoins.call(6, [1, 2, 3, 4, 5], 9)
countCoins.call(6, [1, 1, 2, 3, 3, 4, 5], 9)
countCoins.call(40, [1, 2, 3, 4, 5, 5, 5, 5, 15, 15, 10, 10, 10, 10, 25, 100], 28)
Output:
Sum 6 from coins [1, 2, 3, 4, 5]
  indices [0, 1, 2] => used [1, 2, 3]
  indices [0, 4]    => used [1, 5]
  indices [1, 3]    => used [2, 4]
Number of ways - order unimportant : 3 (as above)
Number of ways - order important   : 10 (all perms of above indices)

Sum 6 from coins [1, 1, 2, 3, 3, 4, 5]
  indices [0, 1, 5] => used [1, 1, 4]
  indices [0, 2, 3] => used [1, 2, 3]
  indices [0, 2, 4] => used [1, 2, 3]
  indices [0, 6]    => used [1, 5]
  indices [1, 2, 3] => used [1, 2, 3]
  indices [1, 2, 4] => used [1, 2, 3]
  indices [1, 6]    => used [1, 5]
  indices [2, 5]    => used [2, 4]
  indices [3, 4]    => used [3, 3]
Number of ways - order unimportant : 9 (as above)
Number of ways - order important   : 38 (all perms of above indices)

Sum 40 from coins [1, 2, 3, 4, 5, 5, 5, 5, 15, 15, 10, 10, 10, 10, 25, 100]
  indices [0, 1, 2, 3, 4, 5, 6, 7, 10] => used [1, 2, 3, 4, 5, 5, 5, 5, 10]
  indices [0, 1, 2, 3, 4, 5, 6, 7, 11] => used [1, 2, 3, 4, 5, 5, 5, 5, 10]
  indices [0, 1, 2, 3, 4, 5, 6, 7, 12] => used [1, 2, 3, 4, 5, 5, 5, 5, 10]
  indices [0, 1, 2, 3, 4, 5, 6, 7, 13] => used [1, 2, 3, 4, 5, 5, 5, 5, 10]
  indices [0, 1, 2, 3, 4, 5, 6, 8]     => used [1, 2, 3, 4, 5, 5, 5, 15]
  indices [0, 1, 2, 3, 4, 5, 6, 9]     => used [1, 2, 3, 4, 5, 5, 5, 15]
  indices [0, 1, 2, 3, 4, 5, 7, 8]     => used [1, 2, 3, 4, 5, 5, 5, 15]
  indices [0, 1, 2, 3, 4, 5, 7, 9]     => used [1, 2, 3, 4, 5, 5, 5, 15]
  indices [0, 1, 2, 3, 4, 5, 10, 11]   => used [1, 2, 3, 4, 5, 5, 10, 10]
  indices [0, 1, 2, 3, 4, 5, 10, 12]   => used [1, 2, 3, 4, 5, 5, 10, 10]
  .......
  (only the first 10 ways generated are shown)
Number of ways - order unimportant : 464 (as above)
Number of ways - order important   : 3782932 (all perms of above indices)
Retrieved from "https://rosettacode.org/mw/index.php?title=Count_the_coins/0-1&oldid=334212"
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