SEND + MORE = MONEY: Difference between revisions
m
→{{header|Phix}}: minor tidy
(SEND + MORE = MONEY in PureBasic) |
m (→{{header|Phix}}: minor tidy) |
||
| (17 intermediate revisions by 10 users not shown) | |||
Line 93:
{{out}}
<pre>Same as FreeBASIC entry.</pre>
==={{header|Gambas}}===
{{trans|FreeBASIC}}
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>Same as FreeBASIC entry.</pre>
==={{header|Run BASIC}}===
{{works with|Just BASIC}}
{{works with|Liberty BASIC}}
<syntaxhighlight lang="lb"></syntaxhighlight>
==={{header|PureBasic}}===
Line 201 ⟶ 247:
{{out}}
<pre>Same as FreeBASIC entry.</pre>
=={{header|C}}==
{{trans|Julia}}
<syntaxhighlight lang="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;
}</syntaxhighlight>
{{out}}
<pre>
9567 + 1085 = 10652
</pre>
=={{header|FreeBASIC}}==
Line 320 ⟶ 406:
Took 1.149804ms.
</pre>
=={{header|J}}==
'''Tacit Solution'''
<syntaxhighlight lang="j">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.</syntaxhighlight>
Example use:
<syntaxhighlight lang="j"> solve ''
9567 + 1085 = 10652, random tries 248241
solve ''
9567 + 1085 = 10652, random tries 246504
solve ''
9567 + 1085 = 10652, random tries 3291556</syntaxhighlight>
The code is tacit and fixed (in other words, it is point-free):
<syntaxhighlight lang="j"> _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:</syntaxhighlight>
=={{header|jq}}==
{{works with|jq}}
'''Works with gojq, the Go implementation of jq, and with fq'''
Straight out of the [https://en.wikipedia.org/wiki/Jq_(programming_language) wikipedia] page, except for {} instead of [] in the last line of the def:
<syntaxhighlight lang=jq>
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
</syntaxhighlight>
{{output}}
<pre>
{"s":9,"e":5,"n":6,"d":7,"m":1,"o":0,"r":8,"y":2}
</pre>
Line 357 ⟶ 510:
end
</syntaxhighlight>{{out}} 9567 + 1085 == 10652
=={{header|Nim}}==
{{trans|Julia}}
<syntaxhighlight lang="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}"
</syntaxhighlight>
{{out}}
<pre>9567 + 1085 = 10652
</pre>
=={{header|Pascal}}==
==={{header|Free Pascal}}===
Line 612 ⟶ 793:
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</pre>
=={{header|Perl}}==
{{trans|Raku}}
=== Exhaustive ===
<syntaxhighlight lang="perl" line>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();</syntaxhighlight>
{{out}}
<pre>SEND + MORE == MONEY
9567 + 1085 == 10652</pre>
=== Fine-tuned ===
<syntaxhighlight lang="perl" line>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";
}
}
}
}
}
}
}
</syntaxhighlight>
{{out}}
<pre>SEND + MORE == MONEY
9567 + 1085 == 10652</pre>
=={{header|Phix}}==
<!--(phixonline)-->
atom t0 = time()
constant mp = new_dict() -- keys 'A'..'Z', values 0..9
sequence front, -- 1 if wordstart, else 0, for (0|1)..9
multh, -- multiplier hash, see below
used
-- mp ends up holding the (first) acceptable solution.
-- Letters which start any word cannot be 0 (in the rules).
-- In SEND+MORE=MONEY, 'E' is 100 + 1 - 10 = 91 from the
-- three places E occurs in the puzzle, stored in multh.
-- Hence sum(letter_values*multh)==0 means it is solved.
-- Obviously used stops us using digits more than once.
function solve_rec(string uniq, int i, atom s)
-- Aside: integer s is fine on 64-bit, but reaches
-- a high of 13,304,757,742 & crashes on 32-bit.
if i > length(uniq) then return s==0 end if
integer chdx = uniq[i]-'A'+1
for v=front[chdx] to 9 do
if not used[v+1] then
used[v+1] = true
if solve_rec(uniq,i+1,s+v*multh[chdx]) then
setd(uniq[i],v,mp)
return true
end if
used[v+1]
end if
end for
return false
end function
function solve(string puzzle)
destroy_dict(mp,true) -- empty, but keep
used = repeat(false,10) -- nb [1..10] for 0..9
multh = repeat(0,26) -- see above
front = repeat(0,26) -- 1 if 1st in any word
string uniq = ""
sequence words = split_any(puzzle," +=\n")
for iw,word in words do
front[word[1]-'A'+1] = 1
integer l = length(word),
m = iff(iw=length(words)?-1:+1)
for i,ch in word do
multh[ch-'A'+1] += m*power(10,l-i)
if not find(ch,uniq) then
uniq &= ch
end if
end for
end for
if not solve_rec(uniq,1,0) then
return "no solution"
end if
for i,ch in puzzle do
if ch>='A' and ch<='Z' then
puzzle[i] = getd(ch,mp)+'0'
end if
end for
return puzzle
end function
constant tests = {
"SEND + MORE == MONEY",
"I + BB == ILL",
"A == B",
"ACA + DD == BD",
"A + A + A + A + A + A + A + A + A + A + A + B == BCC",
"AS + A == MOM",
"NO + NO + TOO == LATE",
"HE + SEES + THE == LIGHT",
"AND + A + STRONG + OFFENSE + AS + A + GOOD == DEFENSE",
"SIX + SEVEN + SEVEN = TWENTY",
"THIS+A+FIRE+THEREFORE+FOR+ALL+HISTORIES+I+TELL+A+TALE+THAT+"&
"FALSIFIES+ITS+TITLE+TIS+A+LIE+THE+TALE+OF+THE+LAST+FIRE+"&
"HORSES+LATE+AFTER+THE+FIRST+FATHERS+FORESEE+THE+HORRORS+THE+"&
"LAST+FREE+TROLL+TERRIFIES+THE+HORSES+OF+FIRE+THE+TROLL+RESTS+"&
"AT+THE+HOLE+OF+LOSSES+IT+IS+THERE+THAT+SHE+STORES+ROLES+OF+"&
"LEATHERS+AFTER+SHE+SATISFIES+HER+HATE+OFF+THOSE+FEARS+A+TASTE+"&
"RISES+AS+SHE+HEARS+THE+LEAST+FAR+HORSE+THOSE+FAST+HORSES+THAT+"&
"FIRST+HEAR+THE+TROLL+FLEE+OFF+TO+THE+FOREST+THE+HORSES+THAT+"&
"ALERTS+RAISE+THE+STARES+OF+THE+OTHERS+AS+THE+TROLL+ASSAILS+AT+"&
"THE+TOTAL+SHIFT+HER+TEETH+TEAR+HOOF+OFF+TORSO+AS+THE+LAST+HORSE"&
"+FORFEITS+ITS+LIFE+THE+FIRST+FATHERS+HEAR+OF+THE+HORRORS+THEIR+"&
"FEARS+THAT+THE+FIRES+FOR+THEIR+FEASTS+ARREST+AS+THE+FIRST+FATHERS"&
"+RESETTLE+THE+LAST+OF+THE+FIRE+HORSES+THE+LAST+TROLL+HARASSES+"&
"THE+FOREST+HEART+FREE+AT+LAST+OF+THE+LAST+TROLL+ALL+OFFER+THEIR+"&
"FIRE+HEAT+TO+THE+ASSISTERS+FAR+OFF+THE+TROLL+FASTS+ITS+LIFE+"&
"SHORTER+AS+STARS+RISE+THE+HORSES+REST+SAFE+AFTER+ALL+SHARE+HOT+"&
"FISH+AS+THEIR+AFFILIATES+TAILOR+A+ROOFS+FOR+THEIR+SAFE == FORTRESSES",
"TO + GO = OUT",
"SEND + A + TAD + MORE = MONEY",
"ABRA + CADABRA + ABRA + CADABRA = HOUDINI",
"I + GUESS + THE + TRUTH = HURTS",
"THATS + THE + THEORY = ANYWAY",
`SO + MANY + MORE + MEN + SEEM + TO + SAY + THAT +
THEY + MAY + SOON + TRY + TO + STAY + AT + HOME +
MAN + TRY + TO + MEET + THE + TEAM + ON + THE +
MOON + AS + HE + HAS + AT + THE + OTHER + TEN =TESTS`,
}
for t in tests do
printf(1,"%s\n%s\n\n",{shorten(t,""),shorten(solve(t),"")})
end for
?elapsed(time()-t0)
</syntaxhighlight>
{{out}}
<pre>
SEND + MORE ==
9567 + 1085 == 10652
I + BB == ILL
1 + 99 == 100
A == B
no solution
ACA + DD == BD
no solution
A + A + A + A + A + ...A + A + A + B == BCC
9 + 9 + 9 + 9 + 9 + ...9 + 9 + 9 + 1 == 100
AS + A == MOM
92 + 9 == 101
NO + NO + TOO == LATE
74 + 74 + 944 == 1092
HE + SEES + THE == LIGHT
54 + 9449 + 754 == 10257
AND + A + STRONG + O... A + GOOD == DEFENSE
503 + 5 + 691208 + 2... 5 + 8223 == 3474064
SIX + SEVEN + SEVEN = TWENTY
650 + 68782 + 68782 = 138214
THIS+A+FIRE+THEREFOR...R+SAFE == FORTRESSES
9874+1+5730+98030563...3+4150 == 5639304404
TO + GO = OUT
21 + 81 = 102
SEND + A + TAD + MORE = MONEY
9283 + 7 + 473 + 1062 = 10825
ABRA + CADABRA + ABRA + CADABRA = HOUDINI
7457 + 1797457 + 7457 + 1797457 = 3609828
I + GUESS + THE + TRUTH = HURTS
5 + 26811 + 478 + 49647 = 76941
THATS + THE + THEORY = ANYWAY
86987 + 863 + 863241 = 951091
SO + MANY + MORE + M...+ OTHER + TEN =TESTS
31 + 2764 + 2180 + 2...+ 19508 + 906 =90393
"4.8s"
</pre>
Line 757 ⟶ 1,041:
+ 1085
= 10652
</pre>
=={{header|Python}}==
{{trans|Nim}}
<syntaxhighlight lang="python3">
# 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}")
</syntaxhighlight>
{{out}}
<pre>
9567 + 1085 = 10652
</pre>
Line 790 ⟶ 1,101:
=== Fast ===
Alternately, a version written in 2015 by [http://strangelyconsistent.org/blog/send-more-money-in-perl6 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.
<syntaxhighlight lang="raku" line>my
while ++$s
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;
next if $
while ++$
next if $
my $y = -1;
next if $y == $s|$e|$n|$d|$m|$o|$r;
my
next unless $send + $more == $money;
say 'SEND + MORE == MONEY' ~ "\n$send + $more == $money";
}
}
Line 940 ⟶ 1,225:
===original===
<syntaxhighlight lang="ring">
// Author: Gal Zsolt 2023-02-08
see "works..." + nl + nl
aListSend = []
Line 1,001 ⟶ 1,286:
next
next
see "done..." + nl
</syntaxhighlight>
Line 1,009 ⟶ 1,293:
SEND = 9567 MORE = 1085 MONEY = 10652
done...
</pre>
=={{header|Ruby}}==
Solving for the string "SEND + 1ORE == 1ONEY" using 'tr' , which translates characters to other characters. The resulting string is brutally evalled.
<syntaxhighlight lang="ruby">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)
</syntaxhighlight>
{{out}}
<pre>9567 + 1085 == 10652
</pre>
=={{header|Vala}}==
{{trans|C}}
<syntaxhighlight lang="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);
}
}
}
}
}
}
}
}
}</syntaxhighlight>
{{out}}
<pre>
9567 + 1085 = 10652
</pre>
=={{header|Wren}}==
Clearly M = 1 and S must be 8 or 9. Brute force can be used to solve for the other letters.
<syntaxhighlight lang="
var sends = []
var ors = []
| |||