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Ed H
Member
Joined: 30 Nov 2007
Posts: 138
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 Posted: 19 May 2008 11:23:10 pm Post subject: |
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Well, here's the rules: make the smallest program that returns the GCD of a set of positive integers. The input list is from either Ans or L1, and the output must be a number, not a list.
I've got it down to 46 but I know there must be a better solution. |
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Weregoose
Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004
Posts: 3976
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| Posted: 19 May 2008 11:48:21 pm Post subject: |
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Sixteen bytes:
Spoiler wrote: min(gcd(Ans,min(Ans
Last edited by Guest on 02 Aug 2010 02:04:14 am; edited 1 time in total |
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Ed H
Member
Joined: 30 Nov 2007
Posts: 138
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| Posted: 19 May 2008 11:51:27 pm Post subject: |
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Weregoose wrote: Sixteen bytes:
Spoiler wrote: min(gcd(Ans,min(Ans
Oh a, I was looking too much into the programming aspect and ignoring the math.
Last edited by Guest on 02 Aug 2010 02:04:34 am; edited 1 time in total |
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DarkerLine
ceci n'est pas une |
Super Elite (Last Title)
Joined: 04 Nov 2003
Posts: 8328
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| Posted: 20 May 2008 11:12:24 am Post subject: |
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Weregoose wrote: Sixteen bytes:
Spoiler wrote: min(gcd(Ans,min(Ans
[post="123859"]<{POST_SNAPBACK}>[/post]
That doesn't work: try that on {6,10,15}. The GCD should be 1, but your code returns 2.
Instead, this works (25 bytes):
Spoiler wrote: While stdDev(Ans
gcd(Ans,min(Ans
End
min(Ans
It's by no means the fastest, though, but GCD is thankfully a fairly fast operation, so even doing it O(n2) times is not a problem. Also, it doesn't work for 1-element lists. Anyway, the next routine, only at 34 bytes, should be close to optimal in the number of times it calls GCD as well:
Spoiler wrote: min(L1
For(I,1,dim(L1
gcd(Ans,L1(I
End
Ans
Last edited by Guest on 02 Aug 2010 02:03:58 am; edited 1 time in total |
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