SEND + MORE = MONEY
SEND + MORE = MONEY 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.
Write a program in your language to solve SEND + MORE = MONEY: A Great Puzzle.
ALGOL 68
This task can be solved without using seven nested loops but then again, it can be solved with them - so why not?.
Uses the observations of the Julia sample (unsuprisingly as this is a translation of the Julia sample).
BEGIN # solve the SEND+MORE=MONEY puzzle - translation of the Julia sample #
INT m = 1;
OP C = ( INT n )CHAR: REPR ( ABS "0" + n ); # convert integer to a digit #
FOR s FROM 8 TO 9 DO
FOR e FROM 0 TO 9 DO
IF e /= m AND e/= s THEN
FOR n FROM 0 TO 9 DO
IF n /= m AND n /= s AND n /= e THEN
FOR d FROM 0 TO 9 DO
IF d /= m AND d /= s AND d /= e AND d /= n THEN
FOR o FROM 0 TO 9 DO
IF o /= m AND o /= s AND o /= e AND o /= n AND o /= d THEN
FOR r FROM 0 TO 9 DO
IF r /= m AND r /= s AND r /= e AND r /= n AND r /= d AND r /= o THEN
FOR y FROM 0 TO 9 DO
IF y /= m AND y /= s AND y /= e AND y /= n AND y /= d AND y /= o AND y /= r THEN
IF ( 1000 * ( s + m ) ) + ( 100 * ( e + o ) ) + ( 10 * ( n + r ) ) + ( d + e )
= ( 10 000 * m ) + ( 1000 * o ) + ( 100 * n ) + ( 10 * e ) + y
THEN
print( ( C s, C e, C n, C d, " + ", C m, C o, C r, C e, " = ", C m, C o, C n, C e, C y
)
)
FI
FI
OD
FI
OD
FI
OD
FI
OD
FI
OD
FI
OD
OD
END- Output:
9567 + 1085 = 10652
BASIC
Rosetta Code problem: https://rosettacode.org/wiki/SEND_%2B_MORE_%3D_MONEY
by Jjuanhdez, 02/2023
BASIC256
m = 1
for s = 8 to 9
for e = 0 to 9
if e <> m and e <> s then
for n = 0 to 9
if n <> m and n <> s and n <> e then
for d = 0 to 9
if d <> m and d <> s and d <> e and d <> n then
for o = 0 to 9
if o <> m and o <> s and o <> e and o <> n and o <> d then
for r = 0 to 9
if r <> m and r <> s and r <> e and r <> n and r <> d and r <> o then
for y = 0 to 9
if y <> m and y <> s and y <> e and y <> n and y <> d and y <> o then
if ((1000*(s+m)) + (100*(e+o)) + (10*(n+r)) + (d+e)) = ((10000* m) + (1000*o) + (100*n) + (10*e) + y) then
print s;e;n;d; " + "; m;o;r;e; " = "; m;o;n;e;y
end if
end if
next y
end if
next r
end if
next o
end if
next d
end if
next n
end if
next e
next s- Output:
Same as FreeBASIC entry.
Gambas
Public Sub Main()
Dim m, s, e, n, d, o, r, y As Byte
m = 1
For s = 8 To 9
For e = 0 To 9
If e <> m And e <> s Then
For n = 0 To 9
If n <> m And n <> s And n <> e Then
For d = 0 To 9
If d <> m And d <> s And d <> e And d <> n Then
For o = 0 To 9
If o <> m And o <> s And o <> e And o <> n And o <> d Then
For r = 0 To 9
If r <> m And r <> s And r <> e And r <> n And r <> d And r <> o Then
For y = 0 To 9
If y <> m And y <> s And y <> e And y <> n And y <> d And y <> o Then
If ((1000*(s+m)) + (100*(e+o)) + (10*(n+r)) + (d+e)) = ((10000* m) + (1000*o) + (100*n) + (10*e) + y) Then
Print s & e & n & d & " + " & m & o & r & e & " = " & m & o & n & e & y
End If
End If
Next
End If
Next
End If
Next
End If
Next
End If
Next
End If
Next
Next
End- Output:
Same as FreeBASIC entry.
Run BASIC
PureBasic
OpenConsole()
m.i = 1
For s.i = 8 To 9
For e.i = 0 To 9
If e <> m And e <> s
For n.i = 0 To 9
If n <> m And n <> s And n <> e
For d.i = 0 To 9
If d <> m And d <> s And d <> e And d <> n
For o.i = 0 To 9
If o <> m And o <> s And o <> e And o <> n And o <> d
For r.i = 0 To 9
If r <> m And r <> s And r <> e And r <> n And r <> d And r <> o
For y.i = 0 To 9
If y <> m And y <> s And y <> e And y <> n And y <> d And y <> o
If ((1000*(s+m))+(100*(e+o))+(10*(n+r))+(d+e)) = ((10000*m)+(1000*o)+(100*n)+(10*e)+y)
PrintN(Str(s)+Str(e)+Str(n)+Str(d)+" + "+Str(m)+Str(o)+Str(r)+Str(e)+" = "+Str(m)+Str(o)+Str(n)+Str(e)+Str(y))
EndIf
EndIf
Next y
EndIf
Next r
EndIf
Next o
EndIf
Next d
EndIf
Next n
EndIf
Next e
Next s
CloseConsole()- Output:
Same as FreeBASIC entry.
True BASIC
LET m = 1
FOR s = 8 TO 9
FOR e = 0 TO 9
IF e <> m AND e <> s THEN
FOR n = 0 TO 9
IF n <> m AND n <> s AND n <> e THEN
FOR d = 0 TO 9
IF d <> m AND d <> s AND d <> e AND d <> n THEN
FOR o = 0 TO 9
IF o <> m AND o <> s AND o <> e AND o <> n AND o <> d THEN
FOR r = 0 TO 9
IF r <> m AND r <> s AND r <> e AND r <> n AND r <> d AND r <> o THEN
FOR y = 0 TO 9
IF y <> m AND y <> s AND y <> e AND y <> n AND y <> d AND y <> o THEN
IF ((1000*(s+m))+(100*(e+o))+(10*(n+r))+(d+e)) = ((10000*m)+(1000*o)+(100*n)+(10*e)+y) THEN
PRINT STR$(s); STR$(e); STR$(n); STR$(d); " + ";
PRINT STR$(m); STR$(o); STR$(r); STR$(e); " = ";
PRINT STR$(m); STR$(o); STR$(n); STR$(e); STR$(y)
END IF
END IF
NEXT y
END IF
NEXT r
END IF
NEXT o
END IF
NEXT d
END IF
NEXT n
END IF
NEXT e
NEXT s
END
- Output:
Same as FreeBASIC entry.
Yabasic
m = 1
for s = 8 to 9
for e = 0 to 9
if e <> m and e <> s then
for n = 0 to 9
if n <> m and n <> s and n <> e then
for d = 0 to 9
if d <> m and d <> s and d <> e and d <> n then
for o = 0 to 9
if o <> m and o <> s and o <> e and o <> n and o <> d then
for r = 0 to 9
if r <> m and r <> s and r <> e and r <> n and r <> d and r <> o then
for y = 0 to 9
if y <> m and y <> s and y <> e and y <> n and y <> d and y <> o then
if ((1000*(s+m)) + (100*(e+o)) + (10*(n+r)) + (d+e)) = ((10000* m) + (1000*o) + (100*n) + (10*e) + y) ? str$(s), str$(e), str$(n), str$(d), " + ", str$(m), str$(o), str$(r), str$(e), " = ", str$(m), str$(o), str$(n), str$(e), str$(y)
fi
next y
fi
next r
fi
next o
fi
next d
fi
next n
fi
next e
next s
end- Output:
Same as FreeBASIC entry.
C
#include <stdio.h>
int main() {
int m = 1, s, e, n, d, o, r, y, sum1, sum2;
const char *f = "%d%d%d%d + %d%d%d%d = %d%d%d%d%d\n";
for (s = 8; s < 10; ++s) {
for (e = 0; e < 10; ++e) {
if (e == m || e == s) continue;
for (n = 0; n < 10; ++n) {
if (n == m || n == s || n == e) continue;
for (d = 0; d < 10; ++d) {
if (d == m || d == s || d == e || d == n) continue;
for (o = 0; o < 10; ++o) {
if (o == m || o == s || o == e || o == n || o == d) continue;
for (r = 0; r < 10; ++r) {
if (r == m || r == s || r == e || r == n || r == d || r == o) continue;
for (y = 0; y < 10; ++y) {
if (y == m || y == s || y == e || y == n || y == d || y == o) continue;
sum1 = 1000*s + 100*e + 10*n + d + 1000*m + 100*o + 10*r + e;
sum2 = 10000*m + 1000*o + 100*n + 10*e + y;
if (sum1 == sum2) {
printf(f, s, e, n, d, m, o, r, e, m, o, n, e, y);
}
}
}
}
}
}
}
}
return 0;
}
- Output:
9567 + 1085 = 10652
FreeBASIC
Dim As Byte m = 1, s, e, n, d, o, r, y
For s = 8 To 9
For e = 0 To 9
If e <> m And e <> s Then
For n = 0 To 9
If n <> m And n <> s And n <> e Then
For d = 0 To 9
If d <> m And d <> s And d <> e And d <> n Then
For o = 0 To 9
If o <> m And o <> s And o <> e And o <> n And o <> d Then
For r = 0 To 9
If r <> m And r <> s And r <> e And r <> n And r <> d And r <> o Then
For y = 0 To 9
If y <> m And y <> s And y <> e And y <> n And y <> d And y <> o Then
If ((1000*(s+m)) + (100*(e+o)) + (10*(n+r)) + (d+e)) = _
((10000* m) + (1000*o) + (100*n) + (10*e) + y) Then
Print s & e & n & d & " + " & _
m & o & r & e & " = " & m & o & n & e & y
End If
End If
Next y
End If
Next r
End If
Next o
End If
Next d
End If
Next n
End If
Next e
Next s
Sleep- Output:
9567 + 1085 = 10652
Go
package main
import (
"fmt"
"time"
)
func contains(a []int, v int) bool {
for i := 0; i < len(a); i++ {
if a[i] == v {
return true
}
}
return false
}
func main() {
start := time.Now()
var sends [][4]int
var ors [][2]int
m := 1
digits := []int{0, 2, 3, 4, 5, 6, 7, 8, 9}
for s := 8; s <= 9; s++ {
for _, e := range digits {
if e == s {
continue
}
for _, n := range digits {
if n == s || n == e {
continue
}
for _, d := range digits {
if d == s || d == e || d == n {
continue
}
sends = append(sends, [4]int{s, e, n, d})
}
}
}
}
for _, o := range digits {
for _, r := range digits {
if r != o {
ors = append(ors, [2]int{o, r})
}
}
}
fmt.Println("Solution(s):")
for _, send := range sends {
SEND := 1000*send[0] + 100*send[1] + 10*send[2] + send[3]
for _, or := range ors {
send2 := send[:]
or2 := or[:]
if contains(send2, or[0]) || contains(send2, or[1]) {
continue
}
MORE := 1000*m + 100*or[0] + 10*or[1] + send[1]
for _, y := range digits {
if contains(send2, y) || contains(or2, y) {
continue
}
MONEY := 10000*m + 1000*or[0] + 100*send[2] + 10*send[1] + y
if SEND+MORE == MONEY {
fmt.Printf("%d + %d = %d\n", SEND, MORE, MONEY)
}
}
}
}
fmt.Printf("\nTook %s.\n", time.Since(start))
}
- Output:
Solution(s): 9567 + 1085 = 10652 Took 1.149804ms.
J
Tacit Solution
SEND=. 10 #. 0 1 2 3&{
MORE=. 10 #. 4 5 6 1&{
MONEY=. 10 #. 4 5 2 1 7&{
M=. 4&{
entry=. 0&{::
try=. 1&{::
sample=. (10 ?~ 8:) ; 1 + try NB. counting tries to avoid a premature convergence
good=. (not=. -.) (o=.@:) (0 = M) (and=. *.) (SEND + MORE) = MONEY
answer=. (": o SEND , ' + ' , ": o MORE , ' = ' , ": o MONEY) o entry
tries=. ', random tries ' , ": o try
while=. ^: (^:_)
solve=. (answer , tries) o (sample while (not o good o entry)) o ( 0 ;~ i.) o 8: f.
Example use:
solve ''
9567 + 1085 = 10652, random tries 248241
solve ''
9567 + 1085 = 10652, random tries 246504
solve ''
9567 + 1085 = 10652, random tries 3291556
The code is tacit and fixed (in other words, it is point-free):
_80 [\ (5!:5)<'solve'
((":@:(10 (#.) 0 1 2 3&({ )) , ' + ' , ":@:(10 (#.) 4 5 6 1&({ )) , ' = ' , ":@:
(10 (#.) 4 5 2 1 7&({ )))@:(0&({::)) , ', random tries ' , ":@:(1&({::)))@:(((10
?~ 8:) ; 1 + 1&({::))^:(-.@:(-.@:(0 = 4&({ )) *. ((10 (#.) 0 1 2 3&({ )) + 10 (
#.) 4 5 6 1&({ )) = 10 (#.) 4 5 2 1 7&({ ))@:(0&({::)))^:_)@:(0 ;~ i.)@:8:
jq
Works with gojq, the Go implementation of jq, and with fq
Straight out of the wikipedia page, except for {} instead of [] in the last line of the def:
def send_more_money:
def choose(m;n;used): ([range(m;n+1)] - used)[];
def num(a;b;c;d): 1000*a + 100*b + 10*c + d;
def num(a;b;c;d;e): 10*num(a;b;c;d) + e;
first(
1 as $m
| 0 as $o
| choose(8;9;[]) as $s
| choose(2;9;[$s]) as $e
| choose(2;9;[$s,$e]) as $n
| choose(2;9;[$s,$e,$n]) as $d
| choose(2;9;[$s,$e,$n,$d]) as $r
| choose(2;9;[$s,$e,$n,$d,$r]) as $y
| select(num($s;$e;$n;$d) + num($m;$o;$r;$e) == num($m;$o;$n;$e;$y))
| {$s,$e,$n,$d,$m,$o,$r,$e,$m,$o,$n,$e,$y} );
send_more_money- Output:
{"s":9,"e":5,"n":6,"d":7,"m":1,"o":0,"r":8,"y":2}
Julia
A hoary old task, solved with pencil before electricity was a thing.
Since the M in Money is the result of carry in base 10 of two single digits it is a 1 (we exclude 0 here though that would work, but then MONEY would be spelled ONEY).
In addition, the S plus 1 then needs to result in a carry, so S is 8 or 9, depending on whether there is a carry into that column. Pencil and paper can continue, but from here the computer is likely quicker.
let
m = 1
for s in 8:9
for e in 0:9
e in [m, s] && continue
for n in 0:9
n in [m, s, e] && continue
for d in 0:9
d in [m, s, e, n] && continue
for o in 0:9
o in [m, s, e, n, d] && continue
for r in 0:9
r in [m, s, e, n, d, o] && continue
for y in 0:9
y in [m, s, e, n, d, o] && continue
if 1000s + 100e + 10n + d + 1000m + 100o + 10r + e ==
10000m + 1000o + 100n + 10e + y
println("$s$e$n$d + $m$o$r$e == $m$o$n$e$y")
end
end
end
end
end
end
end
end
end
- Output:
9567 + 1085 == 10652
Nim
import std/strformat
let m = 1
for s in 8..9:
for e in 0..9:
if e in [m, s]: continue
for n in 0..9:
if n in [m, s, e]: continue
for d in 0..9:
if d in [m, s, e, n]: continue
for o in 0..9:
if o in [m, s, e, n, d]: continue
for r in 0..9:
if r in [m, s, e, n, d, o]: continue
for y in 0..9:
if y in [m, s, e, n, d, o]: continue
if 1000 * s + 100 * e + 10 * n + d + 1000 * m + 100 * o + 10 * r + e ==
10000 * m + 1000 * o + 100 * n + 10 * e + y:
echo &"{s}{e}{n}{d} + {m}{o}{r}{e} = {m}{o}{n}{e}{y}"
- Output:
9567 + 1085 = 10652
Pascal
Free Pascal
simple brute force. Permutation stolen by nQueens.
program SymbolToDigit;
{$IFDEF FPC}{$MODE DELPHI}{$Optimization ON,All}{$ENDIF}
{$IFDEF Windows}{$APPTYPE CONSOLE}{$ENDIF}
uses
sysutils;// TDatetime
const
nmax = 9;
maxLen = 7;
type
tFreeDgt = array[0..nmax+1] of Int32;
tSymbWord = String[maxLen];
tDgtWord = record
DW_DgtsIdx: array[1..maxLen] of UInt8;
DW_maxIdx: Uint8;
end;
tDgtFront = record
DW_DgtsIdx: array[1..nmax+1] of UInt8;
DW_maxIdx: Uint8;
end;
tInUse = set of 0..nmax+1;
const
{
maxIDx = 2;
cSumWords : array[0..maxIDx] of tSymbWord =('SEND','MORE','MONEY');
}
{
maxIDx = 4;
cSumWords : array[0..maxIDx] of tSymbWord =('ABRA','CADABRA','ABRA','CADABRA','HOUDINI');
}
//MANYOTHERS=M2A7N6Y4O1T9H5E0R8S3
maxIDx = 41;
cSumWords : array[0..maxIDx] of tSymbWord =(
'SO','MANY','MORE','MEN','SEEM','TO','SAY','THAT',
'THEY','MAY','SOON','TRY','TO','STAY','AT','HOME',
'SO','AS','TO','SEE','OR','HEAR','THE','SAME','ONE',
'MAN','TRY','TO','MEET','THE','TEAM','ON','THE',
'MOON','AS','HE','HAS','AT','THE','OTHER','TEN',
'TESTS');
var
{$ALIGN 32}
DigitSample,
DigitSampleSolution : tFreeDgt;
SymbInUse : array[0..10] of char;
Words :array[0..maxIDx] of tSymbWord;
DgtWords : array[0..maxIDx] of tDgtWord;
DgtFrontWords :tDgtFront;
SymbInUseCount,gblCount : Uint32;
fullStop: boolean;
ch : char;
procedure OneSol(idx:int32;const DS:tFreeDgt);
var
i,symbolIdx : Int32;
begin
For i := maxlen downto 1 do
begin
symbolIdx := DgtWords[idx].DW_DgtsIdx[i];
if symbolIdx = 0 then
write(' ')
else
write(DS[symbolIdx]);
end;
writeln(cSumWords[idx]:maxLen+2);
end;
procedure RevString(var s:tSymbWord);
var
i,j: NativeInt;
begin
i := 1;
j := Length(s);
while j>i do
begin
ch:= s[i];s[i]:= s[j];s[j] := ch;
inc(i);dec(j);
end;
end;
procedure GetSymbols;
var
//CHR(ORD('A')-1) = '@' is placeholder for no Symbol
SymbToIdx : array['@'..'Z'] of byte;
FrontSymbols :tInUse;
i,j : Int32;
Begin
fillchar(SymbToIdx,SizeOf(SymbToIdx),#255);
SymbToIdx['@'] := 0;
SymbInUseCount := 1;//['@'] is always zero
For i := 0 to maxIDx do
begin
Words[i] := cSumWords[i];
j := length(Words[i]);
//position of highest symbol
DgtWords[i].DW_maxIdx := j;
// extend by '@' aka zero
RevString(Words[i]);
setlength(Words[i],maxlen);
For j := j+1 to maxLen do
Words[i][j] := Low(SymbToIdx);
end;
// find all symbols
for j := 1 to High(tSymbWord) do
Begin
For i := 0 to maxIdx do
begin
ch := Words[i][j];
if SymbToIdx[ch] = 255 then
begin
SymbToIdx[ch] := SymbInUseCount;
SymbInUse[SymbInUseCount] := ch;
inc(SymbInUseCount);
end;
end;
end;
dec(SymbInUseCount);
For i := 1 to SymbInUseCount do
write(SymbInUse[i]);
writeln(SymbInUseCount:4,' symbols');
//get index for every symbol in word
For i := 0 to maxIdx do
with DgtWords[i] do
for j := 1 to High(tSymbWord) do
DW_DgtsIdx[j]:= SymbToIdx[Words[i][j]];
//find all first symbols
FrontSymbols := [];
For i := 0 to maxIDx do
with DgtWords[i] do
include(FrontSymbols,DW_DgtsIdx[DW_maxIdx]);
j := 1;
For i := 0 to nmax+1 do
if i in FrontSymbols then
Begin
DgtFrontWords.DW_DgtsIdx[j] := i;
inc(j);
end;
DgtFrontWords.DW_maxIdx := j-1;
end;
function AddWords(const DS:tFreeDgt):boolean;
var
col,row,
sum,carry : NativeUInt;
begin
// check for zero in first symbols of words
with DgtFrontWords do
For col := DW_maxIdx downto 1 do
begin
if DS[DW_DgtsIdx[col]] = 0 then
EXIT(false);
end;
carry := 0;
For col := 1 to maxLen do
Begin
sum := carry;
carry := 0;
// add one column
For row := maxIdx-1 downto 0 do
sum := sum+DS[DgtWords[row].DW_DgtsIdx[col]];
if sum > 9 then
begin
carry := sum DIV 10;
sum := sum - 10 * carry;
end;
//digit of sum
if sum <> DS[DgtWords[maxIDx].DW_DgtsIdx[col]] then
EXIT(false);
end;
If Carry = 0 then
DigitSampleSolution := DS;
EXIT(true);
end;
procedure NextPermute(Row:nativeInt;var DS:tFreeDgt);
var
i,Col : nativeInt;
begin
if fullStop then EXIT;
IF row <= 10 then
begin
NextPermute(Row+1,DS);
For i := row+1 to 10 do
begin
//swap
Col := DS[i];
DS[i] := DS[Row];
DS[Row] := Col;
NextPermute(Row+1,DS);
//Undo swap
DS[Row] := DS[i];
DS[i] := Col;
end
end
else
begin
fullStop := AddWords(DS);
inc(gblCount);
end
end;
var
T1,T0: TDateTime;
i,j : Uint32;
begin
DigitSample[0] := 0;
For i := 0 to nmax do
DigitSample[i+1] := i;
GetSymbols;
t0 := time;
gblCount := 0;
fullStop := false;
NextPermute(1,DigitSample);
t1:= time;
IF maxIDx < 10 then
For i := 0 to High(DgtWords)do
OneSol(i,DigitSampleSolution);
writeln;
For i := 1 to SymbInUseCount do
begin
j := DigitSampleSolution[i];
write(SymbInUse[i],'=',j,' ');
end;
writeln;
WriteLn(gblCount,' checks ',FormatDateTime(' NN:SS.ZZZ',T1-t0),' secs');
end.
- @TIO.RUN:
DEYNROSM 8 symbols 9567 SEND 1085 MORE 10652 MONEY D=7 E=5 Y=2 N=6 R=8 O=0 S=9 M=1 2704147 checks 00:00.043 secs //shorthened 'SO','MANY','MORE','MEN','SEEM','TO','SAY', OYENMTSRAH 10 symbols O=1 Y=4 E=0 N=6 M=2 T=9 S=3 R=8 A=7 H=5 496179 checks 00:00.013 secs
Perl
Exhaustive
use v5.36;
use enum <D E M N O R S Y>;
use Algorithm::Combinatorics <combinations permutations>;
sub solve {
for my $p (map { permutations $_ } combinations [0..9], 8) {
return $p if @$p[M] > 0 and join('',@$p[S,E,N,D])+join('',@$p[M,O,R,E]) == join('',@$p[M,O,N,E,Y]);
}
}
printf "SEND + MORE == MONEY\n%d + %d == %d", join('',@$_[S,E,N,D]), join('',@$_[M,O,R,E]), join '',@$_[M,O,N,E,Y]) for solve();
- Output:
SEND + MORE == MONEY 9567 + 1085 == 10652
Fine-tuned
use v5.36;
my $s = 7;
while (++$s <= 9) {
my $e = -1;
while (++$e <= 9) {
next if $e == $s;
my $n = -1;
while (++$n <= 9) {
next if grep { $n == $_ } $s,$e;
my $d = -1;
while (++$d <= 9) {
next if grep { $d == $_ } $s,$e,$n;
my $send = $s*10**3 + $e*10**2 + $n*10 + $d;
my ($m, $o) = (1, -1);
while (++$o <= 9) {
next if grep { $o == $_ } $s,$e,$n,$d,$m;
my $r = -1;
while (++$r <= 9) {
next if grep { $r == $_ } $s,$e,$n,$d,$m,$o;
my $more = $m*10**3 + $o*10**2 + $r*10 + $e;
my $y = -1;
while (++$y <= 9) {
next if grep { $y == $_ } $s,$e,$n,$d,$m,$o,$r;
my $money = $m*10**4 + $o*10**3 + $n*10**2 + $e*10 + $y;
next unless $send + $more == $money;
say "SEND + MORE == MONEY\n$send + $more == $money";
}
}
}
}
}
}
}
- Output:
SEND + MORE == MONEY 9567 + 1085 == 10652
Phix
Mighta gota bit carried away here...
with javascript_semantics sequence answer procedure solve(sequence sums,solve_order,firsts,correct_to, integer l,avail,done=0) done += 1 integer d = solve_order[done] for ad=firsts[done] to 9 do integer m = power(2,ad) if and_bits(avail,m) then answer[d] = ad integer ct = correct_to[done], carry = 0, bOK = true for i=1 to ct do carry += sum(extract(answer,sums[i][1])) if remainder(carry,10)!=answer[sums[i][2]] then bOK = false exit end if carry = floor(carry/10) end for if bOK then if ct=length(sums) then if carry=0 then ?answer end if else solve(sums,solve_order,firsts,correct_to,l,avail-m,done) end if end if answer[d] = -1 end if end for end procedure procedure cryptarithm(string s) integer eq = find('=',s) sequence ans = trim(s[eq+1..$]), words = apply(split(s[1..eq-1],'+'),trim), res = join(unique(join(words,"")&ans,"STABLE"),"") integer l = length(res) for i,w in words do words[i] = apply(true,find,{w,{res}}) end for ans = apply(true,find,{ans,{res}}) sequence sums = {}, solve_order = {} for i=1 to length(ans) do sequence set = {} for w in words do if length(w)>=i then integer wi = w[-i] set &= wi if not find(wi,solve_order) then solve_order &= wi end if end if end for integer ai = ans[-i] if not find(ai,solve_order) then solve_order &= ai end if sums = append(sums,{set,ai}) end for assert(length(solve_order)=l) sequence firsts = repeat(0,l), correct_to = repeat(0,l) for f in unique(join(apply(true,extract,{words,{{1}}}),{})&ans[1]) do firsts[find(f,solve_order)] = 1 end for integer mm = 0 for i,fs in apply(sums,flatten) do integer m = max(apply(true,find,{fs,{solve_order}})) mm = max(m,mm) correct_to[mm] = i end for assert(correct_to[$]=length(sums)) answer = repeat(-1,l) printf(1,"%s - %s is ",{s,res}) solve(sums,solve_order,firsts,correct_to,l,0b1_111_111_111) end procedure constant tests = { `SEND+MORE=MONEY`, `TO+GO=OUT`, `SEND + A + TAD + MORE = MONEY`, `ABRA + CADABRA + ABRA + CADABRA = HOUDINI`, `I + GUESS + THE + TRUTH = HURTS`, `THATS + THE + THEORY = ANYWAY`, -- tad slow, 5.8s vs 0.2s for all above: `SO+MANY+MORE+MEN+SEEM+TO+SAY+THAT+ THEY+MAY+SOON+TRY+TO+STAY+AT+HOME+ SO+AS+TO+SEE+OR+HEAR+THE+SAME+ONE+ MAN+TRY+TO+MEET+THE+TEAM+ON+THE+ MOON+AS+HE+HAS+AT+THE+OTHER+TEN =TESTS`} papply(tests,cryptarithm)
- Output:
SEND+MORE=MONEY - SENDMORY is {9,5,6,7,1,0,8,2}
TO+GO=OUT - TOGU is {2,1,8,0}
SEND + A + TAD + MORE = MONEY - SENDATMORY is {9,2,8,3,7,4,1,0,6,5}
ABRA + CADABRA + ABRA + CADABRA = HOUDINI - ABRCDHOUIN is {7,4,5,1,9,3,6,0,8,2}
I + GUESS + THE + TRUTH = HURTS - IGUESTHR is {5,2,6,8,1,4,7,9}
THATS + THE + THEORY = ANYWAY - THASEORYNW is {8,6,9,7,3,2,4,1,5,0}
SO+MANY+MORE+MEN+SEEM+TO+SAY+THAT+
THEY+MAY+SOON+TRY+TO+STAY+AT+HOME+
SO+AS+TO+SEE+OR+HEAR+THE+SAME+ONE+
MAN+TRY+TO+MEET+THE+TEAM+ON+THE+
MOON+AS+HE+HAS+AT+THE+OTHER+TEN
=TESTS - SOMANYRETH is {3,1,2,7,6,4,8,0,9,5}
Translation of Raku
Quite a bit slower than the above
enum S, E, N, D, M, O, R, Y function check(sequence p) if p[M]!=0 and sum(sq_mul(extract(p,{S,E,N,D}),{1000,100,10,1}))+ sum(sq_mul(extract(p,{M,O,R,E}),{1000,100,10,1}))= sum(sq_mul(extract(p,{M,O,N,E,Y}),{10000,1000,100,10,1})) then printf(1," %d%d%d%d\n",extract(p,{S,E,N,D})) printf(1," + %d%d%d%d\n",extract(p,{M,O,R,E})) printf(1,"= %d%d%d%d%d\n",extract(p,{M,O,N,E,Y})) return false end if return true -- continue end function {} = permutes(tagset(9,0),check,8)
- Output:
9567 + 1085 = 10652
Python
# SEND + MORE = MONEY by xing216
m = 1
for s in range(8,10):
for e in range(10):
if e in [m, s]: continue
for n in range(10):
if n in [m, s, e]: continue
for d in range(10):
if d in [m, s, e, n]: continue
for o in range(10):
if o in [m, s, e, n, d]: continue
for r in range(10):
if r in [m, s, e, n, d, o]: continue
for y in range(10):
if y in [m, s, e, n, d, o]: continue
if 1000 * s + 100 * e + 10 * n + d + 1000 * m + 100 * o + 10 * r + e == \
10000 * m + 1000 * o + 100 * n + 10 * e + y:
print(f"{s}{e}{n}{d} + {m}{o}{r}{e} = {m}{o}{n}{e}{y}")
- Output:
9567 + 1085 = 10652
Raku
Idiomatic
enum <D E M N O R S Y>;
sub find_solution ( ) {
for ('0'..'9').combinations(8) -> @c {
.return with @c.permutations.first: -> @p {
@p[M] !== 0 and
@p[ S,E,N,D].join
+ @p[ M,O,R,E].join
== @p[M,O,N,E,Y].join
}
}
}
my @s = find_solution();
say " {@s[ S,E,N,D].join}";
say " + {@s[ M,O,R,E].join}";
say "== { @s[M,O,N,E,Y].join}";
- Output:
9567 + 1085 == 10652
Fast
Alternately, a version written in 2015 by Carl Mäsak. Not very concise but quite speedy. Applying the observation that M must be 1 and S must be either 8 or 9 gets the runtime under a tenth of a second.
my $s = 7;
while ++$s ≤ 9 {
my $e = -1;
while ++$e ≤ 9 {
next if $e == $s;
my $n = -1;
while ++$n ≤ 9 {
next if $n == $s|$e;
my $d = -1;
while ++$d ≤ 9 {
next if $d == $s|$e|$n;
my $send = $s×10³ + $e×10² + $n×10 + $d;
my ($m, $o) = 1, -1;
while ++$o ≤ 9 {
next if $o == $s|$e|$n|$d|$m;
my $r = -1;
while ++$r ≤ 9 {
next if $r == $s|$e|$n|$d|$m|$o;
my $more = $m×10³ + $o×10² + $r×10 + $e;
my $y = -1;
while ++$y ≤ 9 {
next if $y == $s|$e|$n|$d|$m|$o|$r;
my $money = $m×10⁴ + $o×10³ + $n×10² + $e×10 + $y;
next unless $send + $more == $money;
say 'SEND + MORE == MONEY' ~ "\n$send + $more == $money";
}
}
}
}
}
}
}
printf "%.3f elapsed seconds", now - INIT now;
- Output:
SEND + MORE == MONEY 9567 + 1085 == 10652 0.080 elapsed seconds
Ring
// Bert Mariani 2023-02-09 | A Monte Carlo method to solve the encryted message | SEND + MORE = MONEY
t1 = clock() //
See "Start Clock: "+ t1 +nl
counter = 1
aSendory = [["s","-"],["e","-"],["n","-"],["d","-"],["o","-"],["r","-"],["y","-"]]
aRandom = List(10) // 0-9
for j = 1 to 100000000
aRandom = GenRandomUniq() // 5 2 0 8 7 1 6 4 3 9
for i = 1 to 7
if aRandom[1] != 1 // m = 1
aSendory[i][2] = aRandom[1]
del(aRandom,1) // Shorten list, remove value entry picked
else
del(aRandom,1)
i--
ok
next
if (TrySolution(aSendory)) break else counter++ ok // True=1 = Solution Found
next
See "End Clock.: "+ (clock() - t1) +nl
See "Count cycles: "+ counter +nl
Func GenRandomUniq()
throwLimit = 10 // 0-9, Ring does 1-10
aList = 1:throwLimit
aOut = []
while len(aOut) != throwLimit
nSize = len(aList)
if nSize > 0
nIndex = random(nSize) // Random pointer into list
if nIndex = 0 nIndex=1 ok // Ignore 0, Ring Index at 1-10
aOut + (aList[nIndex] -1) // -1 fix value 0-9, Ring +1 Extract list entry content
del(aList,nIndex) // Shorten list, remove value entry picked
else
aOut + aList[1]
aList = []
ok
end
return aOut
Func TrySolution(aTry)
s1 = ( aTry[1][2]) * 1000 // send
e1 = ( aTry[2][2]) * 100
n1 = ( aTry[3][2]) * 10
d1 = ( aTry[4][2]) * 1
nbr1 = s1 + e1 + n1 + d1
m1 = 1 * 1000 // more
o1 = ( aTry[5][2]) * 100
r1 = ( aTry[6][2]) * 10
e1 = ( aTry[2][2]) * 1
nbr2 = m1 + o1 + r1 + e1
m1 = 1 * 10000 // money
o1 = ( aTry[5][2]) * 1000
n1 = ( aTry[3][2]) * 100
e1 = ( aTry[2][2]) * 10
y1 = ( aTry[7][2]) * 1
nbr3 = m1 + o1 +n1 + e1 + y1
nbr4 = nbr1 + nbr2
if (nbr3 = nbr4)
See "Solved: SEND: "+ nbr1 +" MORE: "+ nbr2 +" MONEY: "+ nbr3 +" Check "+ nbr4 +nl
return (nbr3 = nbr4 ) // True
ok
return False- Output:
// Output // Start Clock: 32 // Solved: SEND: 9567 MORE: 1085 MONEY: 10652 Check 10652 // End Clock.: 3792 // Count cycles: 28316
original
// Author: Gal Zsolt 2023-02-08
see "works..." + nl + nl
aListSend = []
aListMore = []
for s = 0 to 9
for e1 = 0 to 9
for n = 0 to 9
for d = 0 to 9
bool = s!=e1 and s!=n and s!=d and e1!=n and e1!=d and n!=d
if bool
sendmore = s*1000+e1*100+n*10+d
add(aListSend,sendmore)
add(aListMore,sendmore)
ok
next
next
next
next
for ind1 = len(aListSend) to 1 step -1
for ind2 = 1 to len(aListMore)
strSend = string(aListSend[ind1])
strMore = string(aListMore[ind2])
m = substr(strMore,1,1)
o = substr(strMore,2,1)
r = substr(strMore,3,1)
e2 = substr(strMore,4,1)
bool1 = substr(strSend,m)
bool2 = substr(strSend,o)
bool3 = substr(strSend,r)
if substr(strSend,2,1) = substr(strMore,4,1)
bool4 = 0
else
bool4 = 1
ok
boolSendMore = bool1 + bool2 + bool3 + bool4
if boolSendMore < 1
if substr(strSend,2,1) = substr(strMore,4,1)
for y = 0 to 9
strMoney1 = substr(strMore,1,1) + substr(strMore,2,1) + substr(strSend,3,1)
strMoney2 = substr(strMore,4,1) + string(y)
strMoney = strMoney1 + strMoney2
numMoney = number(strMoney)
numSend = number(strSend)
numMore = number(strMore)
y1 = substr(strMoney,5,1)
ySend = substr(strSend,y1)
yMore = substr(strMore,y1)
yCheck = ySend + yMore
r = substr(strMore,3,1)
rCheck = substr(strSend,r)
if (numSend + numMore = numMoney) and yCheck < 1 and rCheck < 1
see "SEND = "+strSend+" MORE = "+strMore+" MONEY = "+strMoney+nl+nl
exit 3
ok
next
ok
ok
next
next
see "done..." + nl- Output:
works... SEND = 9567 MORE = 1085 MONEY = 10652 done...
Ruby
Solving for the string "SEND + 1ORE == 1ONEY" using 'tr' , which translates characters to other characters. The resulting string is brutally evalled.
str = "SEND + 1ORE == 1ONEY"
digits = [0,2,3,4,5,6,7,8,9] # 1 is absent
uniq_chars = str.delete("^A-Z").chars.uniq.join
res = digits.permutation(uniq_chars.size).detect do |perm|
num_str = str.tr(uniq_chars, perm.join)
next if num_str.match?(/\b0/) #no words can start with 0
eval num_str
end
puts str.tr(uniq_chars, res.join)
- Output:
9567 + 1085 == 10652
Vala
void main() {
int m = 1, s, e, n, d, o, r, y, sum1, sum2;
string f = "%d%d%d%d + %d%d%d%d = %d%d%d%d%d\n";
for (s = 8; s < 10; ++s) {
for (e = 0; e < 10; ++e) {
if (e == m || e == s) continue;
for (n = 0; n < 10; ++n) {
if (n == m || n == s || n == e) continue;
for (d = 0; d < 10; ++d) {
if (d == m || d == s || d == e || d == n) continue;
for (o = 0; o < 10; ++o) {
if (o == m || o == s || o == e || o == n || o == d) continue;
for (r = 0; r < 10; ++r) {
if (r == m || r == s || r == e || r == n || r == d || r == o) continue;
for (y = 0; y < 10; ++y) {
if (y == m || y == s || y == e || y == n || y == d || y == o) continue;
sum1 = 1000*s + 100*e + 10*n + d + 1000*m + 100*o + 10*r + e;
sum2 = 10000*m + 1000*o + 100*n + 10*e + y;
if (sum1 == sum2) {
print(f, s, e, n, d, m, o, r, e, m, o, n, e, y);
}
}
}
}
}
}
}
}
}
- Output:
9567 + 1085 = 10652
Wren
Clearly M = 1 and S must be 8 or 9. Brute force can be used to solve for the other letters.
var start = System.clock
var sends = []
var ors = []
var m = 1
var digits = (0..9).toList
digits.remove(m)
for (s in 8..9) {
for (e in digits) {
if (e == s) continue
for (n in digits) {
if (n == s || n == e) continue
for (d in digits) {
if (d == s || d == e || d == n) continue
sends.add([s, e, n, d])
}
}
}
}
for (o in digits) {
for (r in digits) {
if (r == o) continue
ors.add([o, r])
}
}
System.print("Solution(s):")
for (send in sends) {
var SEND = 1000 * send[0] + 100 * send[1] + 10 * send[2] + send[3]
for (or in ors) {
if (send.contains(or[0]) || send.contains(or[1])) continue
var MORE = 1000 * m + 100 * or[0] + 10 * or[1] + send[1]
for (y in digits) {
if (send.contains(y) || or.contains(y)) continue
var MONEY = 10000 * m + 1000 * or[0] + 100 * send[2] + 10 * send[1] + y
if (SEND + MORE == MONEY) {
System.print("%(SEND) + %(MORE) = %(MONEY)")
}
}
}
}
System.print("\nTook %(System.clock - start) seconds.")
- Output:
Solution(s): 9567 + 1085 = 10652 Took 0.051735 seconds.
XPL0
include xpllib; \for Print
def M = 1;
int S, E, N, D, O, R, Y;
begin \ Solve the SEND+MORE=MONEY puzzle - Translation of the Algol 68 sample
for S:= 8 to 9 do
for E:= 0 to 9 do
if E # M and E # S then
for N:= 0 to 9 do
if N # M and N # S and N # E then
for D:= 0 to 9 do
if D # M and D # S and D # E and D # N then
for O:= 0 to 9 do
if O # M and O # S and O # E and O # N and O # D then
for R:= 0 to 9 do
if R # M and R # S and R # E and R # N and R # D and R # O then
for Y:= 0 to 9 do
if Y # M and Y # S and Y # E and Y # N and Y # D and Y # O and Y # R then
if 1000*(S+M) + 100*(E+O) + 10*(N+R) + D + E = 10_000*M + 1000*O + 100*N + 10*E + Y then
Print("%d%d%d%d + %d%d%d%d = %d%d%d%d%d\n", S, E, N, D, M, O, R, E, M, O, N, E, Y);
end- Output:
9567 + 1085 = 10652