Cheat Engine

nog_lorp · Posted: Mon Oct 20, 2008 7:08 pm Post subject:

For 16 you probably need a digit extraction formula (http://mathworld.wolfram.com/Digit-ExtractionAlgorithm.html) for powers?

For 45, try creating a system of equations:
a(a+1)/2 = b(3b-1)/2
b(3b-1)/2 = c(2c-1)
c(2c-1) = a(a+1)/2
Then try to come up with a single equality that incorporates them all, subtract one side to make it =0 and solve for the roots.

Symbol · Posted: Mon Oct 20, 2008 7:20 pm Post subject:

Yea, I know... 45 was easy, I just used C# (it's the first problem I've used .NET, actually any language other than C++ Razz

), and I accidently typed {1} instead of {0} when printing the result, so it printed the second parameter. Razz

My code was perfectly fine, the first time I ran it, it should've printed the result, but as I said... I made a small mistake.

By the way, thanks for the help, I'll read about the Digit-Extraction Algorithm and see if I understand.

Edit: do you know of any formulas for other number than pi/e?

nog_lorp · Posted: Mon Oct 20, 2008 7:37 pm Post subject:

Spigot Algorithm seems to be a better term for about the same thing, http://en.wikipedia.org/wiki/Spigot_algorithm

Also, follow the link to BBP algorithm and look at "BBP type algorithms". I'm working on it right now too, it is definitely not easy mathematics. There have been algorithms for various LN's.

HalfPrime · Posted: Mon Oct 20, 2008 8:15 pm Post subject:

I'm pretty sure the function for palindromes in base 10 is the same except you'd then have to multiply it by 9 because you have to compute the leading number separately because there's only (possible digits-1) possible digits because we don't want leading 0's. It's done in the binary form, too, but it's just 2-1. and then because base 10 digits take up more bits than binary, to convert the digits, we'd do log10(2^d).
So there's (10^(((log10(2^d))+(log10(2^d))%2)/2-1) )*9 possible base 10 palindromes for d equals binary digits.

nog_lorp · Posted: Mon Oct 20, 2008 9:07 pm Post subject:

Decimal palindromes less than 1mil: 2003

Odecey · Posted: Mon Oct 20, 2008 10:54 pm Post subject:

nog_lorp · Posted: Mon Oct 20, 2008 11:10 pm Post subject:

... I was discussing the concepts involved. There is no published spigot algorithm for 2^x, its just a concept that could be useful.

EDIT: My calculus 3 professor recommends just using an arbitrary precision type to do the problem. He said it "could take months to work out" a spigot algorithm for someone who wasn't an experienced mathematician. Heh.
Finished. For C/C++ you can use GMP. Or use Python of Java or another language that has built in or standard library support for arbitrary precision.