Recursive Programming Brainteaser

I think as Darkwraith said that the only solution is:

526485 + 197485 = 723970

I do not know about the algorithm, but doing it manually took only 10 minutes.

Some hints I used were that:
1. E can be either 0 (no carry over) or 9 (carry over 1) - As adding E and O gave back O.
2. T is even.
3. D and G cannot be 0 as they are the most promonent digit.
4. R cannot be less than 3 as it is D + G which are two different non-zero numbers.

Case 1: E = 0
1. Implies A = 5.
2. There is no carry over from 2nd column implies L < 5.
3. L > 1 as R > 2 and if R = 3, then L = 1, but then D and G can be assigned only 1 and 2 to get 3, which cannot happen. So L can be only 2, 3 or 4.

After a few calculations you can figure out that with E = 0, you cannot find a solution.

Case 2: E = 9
1. Implies A = 4.
2. There will be carryovers from 2nd, 4th and 5th columns.
3. Thus L > 4.
4. As R > 3, so L > 6.
5. So L is either 7 or 8.

After a few calculations you can figure out the numbers as the possibilities will keep on reducing.

I just made a 10 X 10 table with the digits 0 - 9 as rows and different letters as columns. Just kept on crossing the values which a letter cannot take. It is damn easy to do it.

Any comments are welcome. If needed, I can explain any of the above statements or write full explanation.

Quote:

---

Case 2: E = 9
1. Implies A = 4.
2. There will be carryovers from 2nd, 4th and 5th columns.
3. Thus L > 4.
4. As R > 3, so L > 6.
5. So L is either 7 or 8.

---

I see the 2nd and the 4th, but how did you deduce the 5th is a result of a carry-over?


Good going, if you didn't start with the answer and reverse-engineer it (or hear of Steve Ciarcia's Ask Byte) :D

As far as the table goes,
I think this that is the way it was explained by the high-school math-teaching friend that solved it before.

OMG - I had no idea that you just had to GOOGLE the problem.

http://www.google.com/search?sourcei...ERALD%3DROBERT

No, mine is a little bit more sophisticated than that. Its not engineered to one problem. Its engineered to all problems that involve at most 10 symbols where each symbol is distinct.

I was hoping on finding a rule-based way of doing this, but like I said, I did not have enough rules to make it worth while.

[edit] BTW, I did put a note on my post that contained the answer that said that there was only one solution.

I could probably compile it and put the bytecode up here if you want.

I don't need an .exe, but I was considering how I could use the java code. I thought that it didn't get compiled, and that Netscape would run it, but I found out otherwise.

dglienna, this is the explanation you wanted.

In case where I assumes, E = 9 (or wharever is that symbol), there will be a carry from 5th column as well because we know that there was a carry over from 4th column and thus 9 + 1 will itself become 10. So 1 has to carry.


Also it was reverse engineering, but that is what you can do without using the brute force or computational power of computer. I think atleast it is a better way than checking far all possible permutations and finding. This way no one can improve their programming skills just writing some loops and if conditions.

The program using recursion is not very easy but also not something like a challenge.

I also would be really happy to see a general algorithm for any 10 symbol sequence. I think any such algorithm will have to use either the table or the recursion.

In the end I would say that though this was like a high school math teacher, but this is the best and the most methodical way doing it by just pen and paper. You can see the results. I took some 10 minutes. I wonder how many of great programmers among you took less than that in writing code and executing.

Quote:

---

In the end I would say that though this was like a high school math teacher, but this is the best and the most methodical way doing it by just pen and paper. You can see the results. I took some 10 minutes. I wonder how many of great programmers among you took less than that in writing code and executing.

---

I solved for any equation meeting the aforementioned criteria that you could throw at me. You only solved for one. ;)

OIC 654321. I was thinking 123456.
still infinitely easier once you have the answer.

SEND + MORE = MONEY is the only other one that I know.

solved it by hand in about 5 minutes. didn't use a 2, though.
it wasn't anywhere as hard as the first one.

http://www.geocities.com/Athens/Agor...uzzlemenu.html

thanks for the link. i missed that one.

How about trying one that involves multiplication? :)

creating one, or solving one?

Well, if you create one, would that mean that you have to solve it first?

it would not be fair to post a challenge if you don't have the answer already.

[Challenge rule #739]

:D

Then how would you know if your problem had a solution?

you have to define a set of numbers to match each letter, and then use only those letters to use numbers that make an equation, or vice-versa. My puzzle used all of the numbers from 0 to 9. There was a link to a site that contained a lot of puzzles. Take a look there.

:D