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Last Active:: Jun 05 2012 11:25 AM
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In Topic: Making Cryptarithm Solver - Improving It

Posted 24 May 2012

Hmm, I wrote this up, but never used the 'defs'. This doesn't work and only outputs the values of send, more and money.

for s in range(1,10):
	for e in range(0,10):
		if s!=e:
			for n in range(0,10):
				if n not in [s,e]:
					for d in range(0,10):
						if d not in [s,e,n]:
							print (s,e,n,d)
for m in range(1,10): 
	for o in range(0,10): 
		if m!=o: 
			for r in range(0,10): 
				if r not in [m,o]: 
					for e in range(0,10): 
						if e not in [m,o,r]: 
							print (m,o,r,e)
for m in range(1,10):
	for o in range(0,10):
		if m!=o:
			for n in range(0,10):
				if n not in [m,o]:
					for e in range(0,10):
						if e not in [m,o,n]:
							for y in range(0,10):
								if y not in [m,o,n,e]:
									print (m,o,n,e,y)
send = s*1000 + e*100 + n*10 + d
more = m*1000 + o*100 + r*10 + e
money = m*10000 + o*1000 + n*100 + e*10 + y
if send + more == money:
	print "  %1d%1d%1d%1d" % (s, e, n, d)
	print " +%1d%1d%1d%1d" % (m, o, r, y)
	print "-------"
	print " %1d%1d%1d%1d%1d" % (m, o, n, e, y)

In Topic: Making Cryptarithm Solver - Improving It

Posted 22 May 2012

Hmm, I don't quite understand those yet, maybe I should go and learn them. Though, this...

def process():
	for s in range(1,10):
		for e in range(0,10):
			if s!=e:
				for n in range(0,10):
					if n not in [s,e]:
						for d in range(0,10):
							if d not in [s,e,n]:
								def doSend(s,e,n,d)

def doSendMore(s,e,n,d,m,o,r):					
	print (m,o,r,e)
	
def doSend(s,e,n,d):
	print (s,e,n,d)
	for m in range(1,10): 
		for o in range(0,10): 
			if m!=o: 
				for r in range(0,10): 
					if r not in [m,o]:
						doSendMore(s,e,n,d,m,o,r)

I believe I understand this a bit and can probably try and keep going. Would I need another def to figure out if 'send + more = money'?

In Topic: Making Cryptarithm Solver - Improving It

Posted 22 May 2012

Can't figure out how to edit posts, so I got this together.

for s in range(1,10):
	for e in range(0,10):
		if s!=e:
			for n in range(0,10):
				if n not in [s,e]:
					for d in range(0,10):
						if d not in [s,e,n]:
							print (s,e,n,d)
for m in range(1,10): 
	for o in range(0,10): 
		if m!=o: 
			for r in range(0,10): 
				if r not in [m,o]: 
					for e in range(0,10): 
						if e not in [m,o,r]: 
							print (m,o,r,e)
for m in range(1,10):
	for o in range(0,10):
		if m!=o:
			for n in range(0,10):
				if n not in [m,o]:
					for e in range(0,10):
						if e not in [m,o,n]:
							for y in range(0,10):
								if y not in [m,o,n,e]:
									print (m,o,n,e,y)
word1 = s,e,n,d
word2 = m,o,r,e
word3 = m,o,n,e,y

if word1 + word2 == word3:
    print word1
    print word2
    print word3

Currently I'm working on trying to get everything from declaring the word1, word2, etc. and down.

In Topic: Making Cryptarithm Solver - Improving It

Posted 22 May 2012
Sorry, it was my birthday this weekend and we went camping, anyways, may I ask what the "v" in your code means? Or is it something that I just didn't learn?

I was thinking of grabbing all the values using the first code you showed me above, and then checking using an if statement whether send + more = money.

Using your code, I made the second word as well.
```
for m in range(1,10): 
	for o in range(0,10): 
		if m!=o: 
			for r in range(0,10): 
				if r not in [m,o]: 
					for e in range(0,10): 
						if e not in [m,o,r]: 
							print (m,o,r,e)
```
In Topic: Making Cryptarithm Solver - Improving It

Posted 16 May 2012
baavgai, on 16 May 2012 - 06:11 AM, said:
It looks like you're doing this:
```
for s in range(1,10):
	for e in range(0,10):
		if e not in [s]:
			for n in range(0,10):
				if n not in [s,e]:
			...
```
If so, then a one liner...
Spoiler

[ (s,e,n,d,m,o,r,y) for y in range(10) if y not in (s,e,n,d,m,o,r) and ((s*1000 + e*100 + n*10 + d) + (m*1000 + o*100 + r*10 + e)) == (m*10000 + o*1000 + n*100 + e*10 + y) for r in range(10) if r not in (s,e,n,d,m,o) for o in range(10) if o not in (s,e,n,d,m) for m in range(10) if m not in (s,e,n,d) for d in range(10) if d not in (s,e,n) for n in range(10) if n not in (s,e) for e in range(0,10) if e!=s for s in range(10) ]
It's not a complete one liner, because at that point I get zero results. I was rather bummed. I looked over the logic.

Ultimately, I need this to be true: s*1000 + e*100 + n*10 + d + m*1000 + o*100 + r*10 + e = m*10000 + o*1000 + n*100 + e*10 + y

Which means, this must be true: s*1000 + e*91 + n*90 + d - m*9000 - o*900 - r*10 - y = 0

Just looking at that, if d!=y, then I don't think it can ever be true. Nor do I believe there is a set within the given range where it can be true, regardless of limits.

So, the shortest form of the given problem in python is pass.
Okay, I was just trying to follow what you showed me in your earlier post, where did I go wrong?

I guess you could solve for the first word at first using...
```
for s in range(1,10):
	    if s!=e:
```
and keep using that for all the letters in 'SEND'. Or am I thinking the wrong way?

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