SEND + MORE = MONEY: Difference between revisions
Thundergnat (talk | contribs) m Remove double header |
Thundergnat (talk | contribs) m →Fast: Add optimizations and timing |
||
Line 118:
=== Fast ===
Alternately, a version written in 2015 by [http://strangelyconsistent.org/blog/send-more-money-in-perl6 Carl Mäsak]. Not very concise but quite
<syntaxhighlight lang="raku" line>my int $s =
while ++$s <= 9 {
next if $s == 0;
Line 140:
my int $send = $s*1000 + $e*100 + $n*10 + $d;
my int $m =
}
}
Line 189 ⟶ 182:
}
}
}
printf "%.3f elapsed seconds", now - INIT now;</syntaxhighlight>
{{out}}
<pre>SEND + MORE == MONEY
9567 + 1085 == 10652
0.080 elapsed seconds</pre>
=={{header|Ring}}==
|
Revision as of 22:38, 10 February 2023
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
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
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 int $s = 7;
while ++$s <= 9 {
next if $s == 0;
my int $e = -1;
while ++$e <= 9 {
next if $e == $s;
my int $n = -1;
while ++$n <= 9 {
next if $n == $s;
next if $n == $e;
my int $d = -1;
while ++$d <= 9 {
next if $d == $s;
next if $d == $e;
next if $d == $n;
my int $send = $s*1000 + $e*100 + $n*10 + $d;
my int $m = 1;
my int $o = -1;
while ++$o <= 9 {
next if $o == $s;
next if $o == $e;
next if $o == $n;
next if $o == $d;
next if $o == $m;
my int $r = -1;
while ++$r <= 9 {
next if $r == $s;
next if $r == $e;
next if $r == $n;
next if $r == $d;
next if $r == $m;
next if $r == $o;
my int $more = $m*1000 + $o*100 + $r*10 + $e;
my int $y = -1;
while ++$y <= 9 {
next if $y == $s;
next if $y == $e;
next if $y == $n;
next if $y == $d;
next if $y == $m;
next if $y == $o;
next if $y == $r;
my int $money =
$m*10000 + $o*1000 + $n*100 + $e*10 + $y;
next unless $send + $more == $money;
say 'SEND + MORE == MONEY';
say "$send + $more == $money";
}
}
}
}
}
}
}
printf "%.3f elapsed seconds", now - INIT now;
- Output:
SEND + MORE == MONEY 9567 + 1085 == 10652 0.080 elapsed seconds
Ring
t1 = clock() // start
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 "Time: "+ clock() - t1 // end
see "done..." + nl
- Output:
works... SEND = 9567 MORE = 1085 MONEY = 10652 done... Time: 31.9 s
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 sendmore = send + or
var MORE = 1000 * m + 100 * or[0] + 10 * or[1] + send[1]
for (y in digits) {
if (sendmore.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 0.245002 seconds.