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Ed H


Member


Joined: 30 Nov 2007
Posts: 138

Posted: 30 Sep 2008 12:31:51 am    Post subject:

Write a program that takes three inputs, a base B, an exponent E, and a modulus M, and returns the value of B^E (mod M).

One restraint is that the program cannot actually multiply B E times; it must be fast enough for semi-practical use.


Last edited by Guest on 14 Oct 2008 10:03:15 pm; edited 1 time in total
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Weregoose
Authentic INTJ


Super Elite (Last Title)


Joined: 25 Nov 2004
Posts: 3976

Posted: 30 Sep 2008 02:40:08 am    Post subject:

67 bytes:
Spoiler wrote:
1→A
For(I,1,[font="verdana"]E
‾5+log(2E)/log(2
If int(2fPart(E/2^I
AB-Mint(AB/M→A
B²-Mint(B²/M→B
End
A
I have a hunch that this can be improved still, but it was necessary to keep updating B separately to keep it from overflowing...

Last edited by Guest on 30 Jul 2010 04:26:14 am; edited 1 time in total
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Ed H


Member


Joined: 30 Nov 2007
Posts: 138

Posted: 01 Oct 2008 10:44:28 pm    Post subject:

I got 56 bytes using the same algorithm. However, I used a while loop instead of a for loop. This is what I had:
Quote:
1→R
While E
If fPart(.5E
BR-Mint(BR/M→R
int(.5E→E
B²-Mint(B²/M→B
End
R
Other than the loop, our code is almost identical.

Edit: to give credit where it's due, I found the algorithm on the Wikipedia entry for modular exponentiation.


Last edited by Guest on 30 Jul 2010 04:22:28 am; edited 1 time in total
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Weregoose
Authentic INTJ


Super Elite (Last Title)


Joined: 25 Nov 2004
Posts: 3976

Posted: 05 Oct 2008 03:41:18 pm    Post subject:

Do you mind if I display that on my website? Once I can figure out a safe range of inputs...

Last edited by Guest on 05 Oct 2008 03:42:52 pm; edited 1 time in total
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Ed H


Member


Joined: 30 Nov 2007
Posts: 138

Posted: 05 Oct 2008 11:07:59 pm    Post subject:

I mind not at all.

I find that you're most likely to lose precision when the modulus is very high - I know E is good up to 2E13, and B is probably good up to around 10^7.
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