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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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which article and was the original working before you modified it?
_________________________
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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Jeffry J. Brickley wrote: which article and was the original working before you modified it?
http://www.codeproject.com/cs/algorithms/seunghyopback.asp[^]
It worked similerly to mine except all nodes has the same gcost so it would move diagonaly much more than it should. I modified it and cleaned it up the best I could so I could put it to good use and I really want to get it working correctly so I can move on. Here is my implimintation.
http://www.codeproject.com/useritems/PathFinding_AI.asp[^]
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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Can anyone suggest me an algorithm or a method to find "Whether a Point is exist on a Bezier Curve or not"?
ie. I have x & y and I want to check whether that coordinate exist on the bezier curve or not.
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Try the following link (from Google 'bezier curves')
http://www.moshplant.com/direct-or/bezier/math.html
John P.
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PremalathaP wrote: Can anyone suggest me an algorithm or a method to find "Whether a Point is exist on a Bezier Curve or not"?
Try to search on the PopCap developer forum, I remember there were lots of discussions about Bezier Curve. (obviously lots of people are trying to clone Zuma  )
http://developer.popcap.com/forums.php[^]
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Are you using vb.net if so I have something that might help. If your not using vb you may be able to convert it to c or whatever flvor you like, I think the code I originally got my example from was java and I rewrote and expanded on it for vb.net.
The program draws a curve, then you click on the curve, if the x,y appears on the curve the curve is redrawn showing where you clicked and splitting the curve.
It's not commented but you should be able to figure it out.
If interested let me know and I can send the code, it's app. 68kb and I did not want to clutter up the board.
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Funny![^]
There are II kinds of people in the world, those who understand binary and those who understand Roman numerals. Web - Blog - RSS - Math
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I am getting "Page Not Found" Error.
Regards,
Arun Kumar.A
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It is gone now.
"Patriotism is your conviction that this country is superior to all other countries because you were born in it." - George Bernard Shaw
Web - Blog - RSS - Math - LinkedIn - BM
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