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thornahawk
μολών λαβέ
Active Member
Joined: 27 Mar 2005
Posts: 569
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 Posted: 19 Nov 2007 05:13:15 am Post subject: |
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You are given two lists in L1 and L2, possibly of different lengths, each representing a list of polynomial coefficients. (With regards to coefficient ordering, use whichever of "constant term first" or "coefficient of largest-degree term first" is more convenient).
The challenge is to write the shortest program that generates the Sylvester matrix corresponding to those two polynomials.
thornahawk |
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angel14995
Member
Joined: 13 Oct 2007
Posts: 181
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| Posted: 19 Nov 2007 08:56:38 pm Post subject: |
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I've got it, it's not pretty, but I got it
Spoiler wrote: :SortA(L1
:SortA(L2
:-1+dim(L1 /->/ M
:-1+dim(L2 /->/ N
:{(M+N),(M+N)} /->/ dim([A]
:Fill(0,[A]
:For(A,1,N
:A /->/ C
:For(B,C,M+C
:L1(B-C+1 /->/ [A](A,B
:End
:End
:For(A,1,M
:A /->/ C
:For(B,C,N+C
:L2(B-C+1 /->/ [A](A+N,B
:End
:End
:[A]
And it's 143 bytes to boot! |
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Weregoose
Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004
Posts: 3976
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| Posted: 20 Nov 2007 06:10:39 am Post subject: |
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angel14995, your program doesn't work for all inputs (notably small ones). I'm pretty sure you can go back and optimize it a bit, too...
Here's mine ([s]102[/s] 101 bytes): Spoiler wrote: [s]1+max(dim(L1),dim(L2→dim(L2[/s] dim(L1)-2+dim(L2→dim(L2
For(X,0,1
Repeat Ans(dim(Ans
dim(L2→dim(L1
List►matr(L1,[B]
[B]
If X or not(L1(1
augment([A],[B]
Ans→[A]
augment({0},L1→L1
End
L2→L1
End
[A][font="arial"]τ In this program, problems arise if the first or last coefficients are explicitly defined to be zero. This also destroys the original lists. :(
[EDIT] – Fix made in the first line.
P.S., removing the transpose instruction would offer another matrix having the same determinant and would also save one more byte, bringing the total to 100. :)
[EDIT×2]
Don't tell me that no one spotted the other error that I just now fixed after three days. This program should work now.
Last edited by Guest on 04 Sep 2010 08:33:34 pm; edited 1 time in total |
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angel14995
Member
Joined: 13 Oct 2007
Posts: 181
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| Posted: 20 Nov 2007 06:54:52 am Post subject: |
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Weregoose wrote: angel14995, your program doesn't work for all inputs (notably small ones).
Unless I am reading it wrong, or I am just being plain stupid, here is how I interpreted it:
Spoiler wrote: In the list, the value of m and n are one less than the total number of values in the list. Since M+N gives us the width and height of the matrix, then they are two less than the dimensions of L1 and L2 added. If the dimensions are 1 and 2, then it would be a 1*1 matrix. If it is 2 and 2, it would be a 2*2 matrix. If it is 3*2, it is a 3*3 matrix. I don't know if the outputs for anything lower than 1*2 is accurate, but then I'll fix it.
And I know there are optimizations, I just need to find them  |
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Weregoose
Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004
Posts: 3976
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| Posted: 20 Nov 2007 06:56:05 am Post subject: |
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Oh wow, mine is the one that screws up, not yours. I misread the articles on Sylvester Matrices. So, I'll attempt to fix mine now, sorry.  |
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angel14995
Member
Joined: 13 Oct 2007
Posts: 181
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| Posted: 20 Nov 2007 06:59:24 am Post subject: |
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| I win in finishing it first, you'll win it getting it uber small. And misreading the article. Lol |
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Weregoose
Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004
Posts: 3976
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| Posted: 20 Nov 2007 07:03:13 am Post subject: |
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Okay, my fix allowed me to save more space. :biggrin:
It is now roughly 4:00 AM here, and I'm about to go to bed. (Excuses, excuses...)
Last edited by Guest on 20 Nov 2007 07:04:15 am; edited 1 time in total |
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JoostinOnline
Active Member
Joined: 22 Aug 2007
Posts: 559
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| Posted: 20 Nov 2007 11:57:38 am Post subject: |
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I don't know who Spoiler is, but he sure is smart...
Last edited by Guest on 20 Nov 2007 01:09:29 pm; edited 1 time in total |
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