Comments on: Euler Problem 188

By: Michael

Michael — Fri, 26 Mar 2010 17:33:00 +0000

Hi Christian -

I’m an engineer (for 3 more months) and hobbyist programmer solving math problems, not a mathematician solving programming ones so I can’t really venture a creditable mathematical opinion. There are those who hang around DDoE who can (i.e fzz), I’m just not one of them. When you have the right answer, you can check into the Euler problem and read the insights of other solvers, some of whom are math-types. All that said, my knee quivers that if the general case is bad, the understanding is bad.

In the case of Euler #188, the loop can be too large or too small by a significant number relative to 1855, and still give the right answer. And while Euler is fond of problems that support a millisec-quick answer, I think in this case it’s happenstance. I lucked up with a bad algorithm giving a good result.

…mrt

By: Christian

Christian — Thu, 25 Mar 2010 18:26:00 +0000

I've spent some time thinking about the reason for the correct behaviour of the code. I'm not sure if it's a 100% correct, so I'd be glad to hear your opinion:
If a and n are coprime there is an integer e called the order of a modulo n with the property that x=y mod e implies a^x=a^y mod n. In case a=1777 and n=10^8, I think the order of a mod n is 2^4*5^7 or maybe 2^5*5^7. In any case it is a divisor of 10^8.
In the general situation, this must not necessarily hold (e.g. consider n=5 and a=2)

By: Michael

Michael — Wed, 24 Mar 2010 14:31:00 +0000

The reason the original code provided the right answer is that the loop settled upon it in an amazingly few number of runs. I found this by dumping the answers into a spreadsheet.

...mrt

By: Michael

Michael — Tue, 23 Mar 2010 19:09:00 +0000

Hi Christian -

There is a logic bug in the above code when m < b, and since nothing depends on i, there is no need to count backwards. I was influenced wrongly by having to go right-to-left. Thanks, I'd guess ;-) for making me relook at that code. I never considered that case.

For Euler's case, this works:

a = 1777
Answer = a
For i = 1 to 1855-1
Answer = Pow(a, Answer, 10 ^ 8)
Next i

For your case:

a = 12
Answer = a
For i = 1 to 2-1
Answer = Pow(a,answer,10)
Next i

I can't satisfy myself with an articulation of the logic bug, so I'd invite your comments.

...mrt

By: Michael

Michael — Mon, 22 Mar 2010 22:21:00 +0000

Christian -

My apology, I saw that right after I posted. There’s something screwy using mod 10. If I use mod 100, it seems right.

I’ll think about it and post back. On the positive side, that code does check in the right answer

…mrt

By: Christian

Christian — Mon, 22 Mar 2010 22:09:00 +0000

My point is that 12^12 is 12^^2

By: Michael

Michael — Mon, 22 Mar 2010 21:52:00 +0000

Christian -

It's not 12^12. It's hyper-exponentiation. It's 12^^12, or in this case it's not 1777^1885, it's 1777^^1885.

See the Wiki article referenced above.

...mrt

By: Christian

Christian — Mon, 22 Mar 2010 21:30:00 +0000

Let me explain, what irritated me. Have a look at the code

Answer=1
For i = 2 To 1 Step -1
Answer = Pow(12, Answer, 10 )
Next i

After the first iteration, we have Answer=Pow(12, 1, 10 )=2, and so also after the second iteration, we have Answer=Pow(12, 2, 10 )=4.

On the other hand we have MOD(12^12,10)=6.

By: Michael

Michael — Mon, 22 Mar 2010 15:49:00 +0000

Christian -

In spreadsheet terms, it uses MOD(a*b,n) = MOD(MOD(a,n)*MOD(b,n),n) which is valid.

...mrt

By: Christian

Christian — Wed, 17 Mar 2010 18:44:00 +0000

I’m not sure, if modular theory is used correctly in this example… When writing

For i = 1855 To 1 Step -1
Answer = Pow(a, Answer, 10 ^
Next i

you seem to use the relation a^b = a^(b mod n) mod n; but this relation is false in general. Please correct me, if I’m wrong…