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first of all thanks for replying... i've examined the laplace expansion. But in high order matrices finding the minor matrices or the cofactors is hard to code for me and i think, i can turn the square matrice to the triangle square matrice. so i just multiply the a[j,j] s then i find the determinant of it.. by the way.. my code is just turn the square matrice to a triangle square matrice. thanks for comments...
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Hi,
I do not know if your code works or not, but if I were you I would use the definition of determinant to compute it.
The hardest part is to code a function to get permutations.
Regards
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thanks for replying my message. my code is working for me for all inputs and it works only square matrices you know... and yeah you are very right... and by the way if you want to try it you can use excel microsoft vb.. thanks for advance...
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But ... determinants are only defined for a square matrix... isn't then the range what you want to compute ?
Regards
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sure i am... i need only square matrices... n unkowns n linear equations. best regards....
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Hey all,
I'm not sure this is the right forum, as I'm not sure what I'm looking for is an algorithm. 
I have an IP address along with a port.
My goal is to determine which running application is using this port.
Googling taught me that this process is called Port-To-Application Mapping (aka PAM),
but I couldn't find any code which does that...
The best thing I found was a program which knows how to do it.
Anyone knows any PAM algorithms?
Thanks in advance,
Shy.
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anyone know step by step mathematical processes for image convolution using laplacian? plz explain...
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Hi,
If I remember well ...
If your image M is considered as a x*y matrix (ie gray scaled) of pixels and a point P is then described as P(i,j) where 0 <= i < n and 0 <= j < n, and the laplacian is considered as a 3*3 matrix L defined as [[a,b,a],[b,c,b],[a,b,a]], the discrete convolution gives you another n*m matrix R where R(x, y) = Sum(i=0, i < 3, Sum(j=0, j < 3, L(i,j)*P(x + i - 1, y + j -1))) (0 outbounds)
Regards
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anyone can help find out any source code for prediction using bayesian belief network algorithm pls...A.S.A.P
respect,relax,respon
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i am looking for bayes network free source code for my preliminary analysis research...
respect,relax,respon
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anyone know about this technique...amswering me pls???
respect,relax,respon
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is there an algorithm(c++/c#) or existing database, which can be used to turn English words from tense or plural form to their original form?
thx~
http://chinajuanbob.spaces.live.com/
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acording to MS, you shouldn't try to do this because while it's (barely) possible to do it for a single language, internationalizing will require writing a set of special rules for every language supported. Instead write your text in a plurality nuetral fashion. Instead of "1 cat, 2 dogs" use "cats: 1, dogs: 2".
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You wrote that? That's cool. 
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Holy crap. My head hurts. I give you a 5 for being smarter than I. 
---
Shawn Poulson
spoulson@explodingcoder.com
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Im having a problem with a A* path finding algorithm I made. Its able to find the path but it will rarely make diagonal moves when it should.
Strart
X
X
X X X End
I think I got the math right, its the algorithm that isnt working right. Iv googled around and found some good articals but most fail to explain how to impliment it. Some come with source code but is too complex and bloated for me to grasp. Im close to giving up on using the internet to help solve my problem.
static int Sqrt(int x) { if (x<0) throw new ArgumentOutOfRangeException(); int temp, y=0, b=0x8000, bshft=15, v=x; do { if (v>=(temp=(y<<1)+b<<bshft--)) {="" y+="b;" v-="temp;" }="" while="" ((b="">>=1)>0); return y; 
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Henize wrote: it will rarely make diagonal moves when it should
Are diagonal cells considered as neighbours in your algorithm ? If no, then that is your problem 
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pathfinding isn't simple. FI you don't believe me, just look at teh number of RTS games that have had totally fubar systems over the years.
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The first step is to find out why the A* chose one path over another. For instance if you compare distance, diagnal is longer on a per-case basis, the lookahead may not be evaluating far enough.
A* and D* both use a series of evaluations to reduce testing everything, find out what it is eliminating and why. Walk the through the algorithm and at test cases output evaluations either to your debugger or even with good old print statements. You are trying to understand why it is making the choices it is.
Once you know why the diagnal path is elliminated then the algorithm can be adjusted.
_________________________
Asu no koto o ieba, tenjo de nezumi ga warau.
Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)
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Your right, the look ahead is not looking far enough ahead. That seems to be my problem, I dont know how I could implement that. The A* algorithm seemed strait forward at first glance but it has eluded me. I have came so close to getting it working 3 times now(3 differnent implimintations) but all have failed.
static int Sqrt(int x) { if (x<0) throw new ArgumentOutOfRangeException(); int temp, y=0, b=0x8000, bshft=15, v=x; do { if (v>=(temp=(y<<1)+b<<bshft--)) {="" y+="b;" v-="temp;" }="" while="" ((b="">>=1)>0); return y;
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Thanks but that impimitation is too different than the one im using. The code Im using is barrowed from an artical from codeproject but Iv heavily modified it and improved it. However there is a flaw that will take some time to figure out. The problem is this...
S = start
X = unwalkable
E = end
. = current position
When it makes the next move it will go strait right because it costs less
X
S.X E
X
the next move would be to go up or down by 1, however I need it to calculate the total cost of those two moves and compare it to just moving diagonally in the first place. I cant seem to figure out how to modify it to beable to do that. Once it makes the move it will set its parent to the previous and if it finds a better path then it would have to go through and change the parents. I may need to redesign the whole thing from scratch.
static int Sqrt(int x) { if (x<0) throw new ArgumentOutOfRangeException(); int temp, y=0, b=0x8000, bshft=15, v=x; do { if (v>=(temp=(y<<1)+b<<bshft--)) {="" y+="b;" v-="temp;" }="" while="" ((b="">>=1)>0); return y;
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