
that may take "a while" as there are infinitely many nontrivial solutions to the single equation x2+y2+z2=g2 (1,2,2,3 would be a simple one of them), and probably (I'm still looking into this) a nonzero fraction (hence also infinitely many) of those also satisfy the three extra conditions.
It really is a challenge that deserves a mathematical approach, not a bruteforce one.





Luc Pattyn wrote: deserves a mathematical approach
True, but it's been about 30 years since I did any of that.
"A Journey of a Thousand Rest Stops Begins with a Single Movement"





You usually don't count building steps, do you, Watson?
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
 Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
 Iain Clarke
[My articles]





This is not a programming question, it is a math problem.
The latter 3 equations define a Euler Brick, all 4 together define a "perfect cuboid", no solution is known so far. See here[^].





Spent many a day on that page.






Hi,
I have to write a program which converts NFA to DFA.I know the algorithm, but I have no idea, how translate it to any C++ / Java code. I would be grateful for advices how to do it (simple pseudocode or something),
Thank you for any sugestion
Paul





I'm puzzled here, you say you know the algorithm, so you have or could write an English description of it; well, that is the pseudocode you want, isn't it? I think what you need most is the courage the get started...
Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles]
I only read formatted code with indentation, so please use PRE tags for code snippets.
I'm not participating in frackin' Q&A, so if you want my opinion, ask away in a real forum (or on my profile page).






the image only shows an example of input and output. it is not an algorithm.
if you know the algorithm, it is expressed either in human language or in a programming language, the former is easier for humans (and possibly ambiguous), the latter is easier for machines.
If you don't have the algorithm, come up with one yourself and/or do your research through google. There's lots of relevant stuff around.
Maybe this CP article is useful: Writing own regular expression parser[^]
Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles]
I only read formatted code with indentation, so please use PRE tags for code snippets.
I'm not participating in frackin' Q&A, so if you want my opinion, ask away in a real forum (or on my profile page).





Hi,
Ok, I done this program, is very stupid, but works
Thanks for all,
Paul





Paul6915 wrote: is very stupid, but works
A program can be very stupid, or it can work. It can't do both. If you've written a program that solves the problem, you have every reason to be proud of it. When someone comes up with a better program to solve the same problem, you have a learning opportunity.
"A Journey of a Thousand Rest Stops Begins with a Single Movement"





Define the DFA "state" of an NDFA as being the list of all states that the NDFA "could" be. Then determine for each possible combination of states, what combinations of states could occur as a result of each possible input character. For example, consider the following NDFA:
1. START STATE. If input is "T", maybe advance to state 2. In any case, maybe stay in state 1.
2. If input is "E", advance to state 3.
3. If input is "S", advance to state 4.
4. If input is "T", advance to state 5.
5. If input is "E", advance to state 6.
6. If input is "R", advance to state 7.
7. ACCEPTING STATE. Stay in state 7 regardless of input.
It's possible that the machine will only be in state 1 (call that DFA state "1").
If it's in DFA state 1 and it gets a "T", it could be in state 1 or 2 (call that DFA state "1/2"); if it gets anything else, it can only be in state 1.
If it's in DFA state "1/2" and it gets a "T", it could be in state 1 or 2; if it gets an "E", it can be in state 1 or 3 ("1/3"). Anything else, it can only be state 1.
If it's in DFA state "1/3" and it gets a "T", it could be in state 1 or 2; if it gets an "S", it can be in state 1 or 4 ("1/4"). Anything else, it can only be state 1.
If it's in DFA state "1/4" and it gets a "T", it could be in state 1, 2, or 5. Anything else, it can only be state 1.
If it's in DFA state "1/2/5" and it gets an "E", it could be in state 1, 3, or 6. Anything else, it can only be state 1.
If it's in DFA state "1/3/6" and it gets an "R", it could be in state 1 or 7 (though state 1 won't really matter); if "S", it can be in state 1 or 4; if "T", state 1 or 2. Anything else, it can only be state 1.
If it's in DFA state "1/7", one could evaluate further state transitions, but since "acceptance" is guaranteed, they're moot.






Did you try a search?[^]
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Walt Fair, Jr., P. E.
Comport Computing
Specializing in Technical Engineering Software





Hey does anyone know of an algorithm to find the point that resides inside a polygon that is farthest (on average) to any other point in the polygon? The polygons don't have to be regular and not all points can see the other points (IE you may not be able to draw a line between every point without intersecting the outer edge of the polygon)





Look up how to calculate the centroid of an arbitrary area. This comes out of the more general study of moments of area.
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My understanding of centroid means that if your polygon was in the shape of the letter C the centroid would be somewhere outside the polygon (in the middle of the c that is). I could be wrong though.





Yes, certainly with concave polygons the centroid may but doesn't necessarily lie outside.
If I'm understanding your problem this time, you need to formulate it by taking the double integral over the area of the polygon of the distance to each point and an arbitrary point (x,y). Personally, I'd use distance squared and get rid of the pesky square root. Then once you've established the equation of the distance squared over the area, you take the partial derivatives with respect to x and y and find the maximum. The rub probably comes when you have to restrict (x, y) to lie "inside" the polygon. Off the top of my head, that part isn't clear. I'm on the road and tired so that part is left up to the student.
The wonderful thing about the Darwin Awards is that everyone wins, especially the members of the audience.





Thanks Tim, this has put me on the right track. I am developing an algorithm that will do just this, I hadn't even considered that it was a maximization problem until now. I'll get back to you with my results.





I'd like to hear, especially the constrait to the body of the polygon part. It's easy to say but, I think, mathematically it might me more difficult to express. At least generally.
I made it to the airport and, hopefully, tonight I'll be home.
The wonderful thing about the Darwin Awards is that everyone wins, especially the members of the audience.





This is a very interesting question. I don't know if there is a known algorithm for that (but I guess there must be).
Just a thought: maybe you can start working on a basic idea like drawing a line from each vertex to the middle point of each side (excluding the two sides on which the vertex stands). Then examine the intersection points of these lines (excluding those that lie outside the polygon) and find the one which is farthest on average from all vertexes.
Not sure if this is the right way to go, but I hope I gave you some idea to work on.
Let me know if you find the solution, I'll be very interested.
2+2=5 for very large amounts of 2
(always loved that one hehe!)





To calculate the centroid of a polygon:
Source: http://en.wikipedia.org/wiki/Centroid[^] and http://local.wasp.uwa.edu.au/~pbourke/geometry/polyarea/[^]
PointF centroid(PointF[] p) {
float cx = 0;
float cy = 0;
float a = 0;
float ca;
for (int i = 0; i < p.Length; i++) {
ca = x[i] * y[(i + 1) % p.Length]  x[(i + 1) % p.Length] * y[i];
cx = cx + (x[i] + x[(i + 1) % p.Length]) * ca;
cy = cy + (y[i] + y[(i + 1) % p.Length]) * ca;
a = a + ca / 2;
}
cx = cx / (6 * a);
cy = cy / (6 * a);
return new PointF(cx , cy);
}
By the way, I am not sure if what you need is a centroid. Hope this helps, anyway.





I have raw metal sheets of various rectangular sizes (x1*y1, x2*y2,
... xn*yn).I want select best fit sheet which fullfill my requirement for getting new sheets of X*Y it also has some tolerence.
e.g
I have metals sheets like 200*100,300*200,250*500,50*100,100*100
and I Require 3 pieces of 100*100 with 3mm tolerence.so how do I find the best fit soluion and how do I implement this logic.
Thanks in advance.





I am guessing that the effort involved in implementing would be extensive and difficult unless you are a major gemotry geek. After a little searching it seems some of the relevant keywords that lead to academic articles on the topic are "Nesting", and "No Fit Polygon". Since nesting could get you more bird articles you might want to include "sheet metal" or "CAD" in your search. I know it isn't an answer, but it should help your search.




