Can the Travelling-Salesman Problem be solved in Polynomial Time?
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right.  so, yeah, you're not going to solve the P & NP problem on yahoo answers.\n\nand to give you the proper answer to your question:\nfor all we know there could be, but we aren't even certain of this possibility.  it's conjectured that all NP-complete problems are not solvable in polynomial time, but whether this is a provable conjecture is unknown.
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U can use either Kruskal's algorithm or prims algorithm. More information about these algorithm u can find on internet.\n\nPrims algorithm is based on adjacency matrix.
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Close, but no cigar...\n\nABCDE FGHIJ KLMNO\nCJDMG OEKHB INALF\nGFBKL NAOJC DHIEM\nEKNCI LMGAH JOFBD\n\nConstraint: Fifteen riders yields (15 * 14 / 2) = 105 unordered pairs.  A single five rider heat yields (5 * 4 / 2) = 10 unordered pairs.  Over twelve heats that makes 120 unordered pairs -- a little bit more than the 105 that we want to cover.  So we want to have each pair of riders be together in at least one heat, with no pair of riders being together in more than two heats.\n\nResult: Not quite -- 21 "head-to-head" matchups do not take place.  But no pair of riders are together more than twice.\n\nAK BI BM BN CF CH CL\nDK DL DN EF EG EJ EL\nFM GN GO HN IO JK JL\n\nConstraint: There are four sets and five positions.  So we want to have each rider in a different position in each heat.\n\nResult: Constraint is met!\n\nConstraint: There are three heats per set and fours sets. So we want to have each rider be the first seat of a set, the second heat of a set, and the third heat of a set (obviously the fourth set will double up on one of the heats).\n\nResult: Not quite -- 5 riders miss one of the first/second/third heats in a set.\n\nC:310 D:202 G:220 H:031 O:022\n\nAlternative:\n\nABCDE FGHIJ KLMNO\nJDLGM HOEBK INFCA\nLFOAH DEGKN MIBJC\nEHNMI OCJFD GKALB\n\nThe same on the first two constraints.  On the third constraint there are now 7 riders (up from 5) missing one of the first/second/third heats in a set.  But they are all 2-2-0 with none of the 3-1-0's of the first schedule.\n\nAJ AM BF BN CG CH CK\nCL DH DI EF EJ EL FK\nFM GO IK IL IO JK JN\n\nA:202 D:220 E:220 H:220 K:022 L:202 M:202
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Optimisation problems are normally done using partial derivatives, however you're problem is discreet so it will be handled differently. First make a map of the network lets say with coloured lines to represent the individual paths from the beginning to the end of the manufacturing process and nodes where they intersect, i.e. nodes that represent each machine. Then there are multiple ways to run the algorithm. For the case of minimum number search through the nodes to see if there is one that bundles all the lines through it if not start looking for pairs that would have all the lines going through one or the other of the nodes and so on until you find a solution. Unless visually it is a simple graph you will need a computer to do the processing.\nThe problem with a cost assigned is more difficult. In this case each node has a value associated with it. In this case generate an ordering from cheapest to dearest of the possible costs of selecting any number of nodes. The test each of these in succession until you find one that corresponds to having all the lines pass through at least one of the selected nodes associated with that particular cost.\nIn knowing details of the way in which the graph would look or the cost for certain locations it may be possible to modify the algorithm so it runs quicker, but this should do it for the information given.
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Yes, this is a difficult one.\nI'll keep trying, but don't hold your breath.\nAll I can say so far is that I could find\nonly one solution : (x, y) = (7,1).\nThere do not seem to be any more\nsolutions with x >= 8, and this is what\nneeds to be proved. So good luck with\nfinding someone who can do this.
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Organizing the different legs of a network
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I'm not sure if I understood correctly but check out the pigeonhole principle
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there is a problem in maths that is named "graphes" and the funtion of these routers is to use this solution, u can find it in math books!!!
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If there are no other constraints, the roles can be assigned in \n15*14*13*12*11*10 ways. The first role can be filled in 15 ways and the second in 14 ways and so on. This is also called a permutation. See http://mathworld.wolfram.com/Permutation.html
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There used to be a lot of them, but with computers these days they are being solved by brute force. Examples include Fermat's Last Theorem and the Four Color problem in topology. You can look those up to see what they are, and you will probably find sites listing other unsolved mathematical puzzles.
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My suspicion upon seeing this question is that it can be solved using mathematical induction.  After all, couldn't we simply 'build' such a (maximally connected) graph starting with n=2 vertices and 'grow' it from there?\n\nRather than take on the proof myself -- mostly because I am unsure of the definition of 'components' -- I leave it as an exercise for you.  However, it bears noting that the formula shown is the commonly accepted "sum of the first n integers" formula used in almost every text on induction.\n\nGood luck!
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solve three body problems
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Fundamental Theorem of Linear Programming:\n\nIf the constraints define a feasible, bounded convex polyhedron, then the maxima and minima are attained on the boundary points of that polyhedron.\n\nFor an example, check out http://people.brunel.ac.uk/~mastjjb/jeb/or/morelp.html
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For the ones who answered before me, P is a problem that is solvable within polynomial time and NP is a problem that has nondeterministic polynomial time.\n\nP=NP is an open problem. It has not been solved.\n\nHowever, it is known that if one NP is P, all NP is P. To prove that P=NP, all you need to do is that one particular NP is P. To disprove that P=NP, you need to prove that all NP is not P.
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There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
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plano concave\nconvexo concave\nconvergent
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An algorithm is a way of solving a problem, kind of like an equation or a formula. It would provide a framework for plugging in variables and give you an answer. \nThe traveling salesman problem is scenario in which a salesman has to visit a number of cities all once. He wants to do this in such a way that he minimizes the total distance traveled. At a first glance, it seems like an easy problem - and it is when there are only a few cities. But it quickly gets so complicated that even the best computers can't come up with the shortest path. In relation to an algorithm, it would be very beneficial to find one equation that always shows us the shortest distance. If you can find it, there's a $1 million dollar prize... Hope this helps! Also, check out the website I've included as a source for more info.