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 wothout 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
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.