how can I find an implemention of traveling salesman problem in java?
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Strictly speaking, the answer to your question is "use\na search engine" :-)\n\nHowever, below are some of the results that came up\nfrom my queries.
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Bruce eckel's thinking in java could be a right place for you to start with\n\nGoogle....\nhttp://www.google.com/search?sourceid=navclient&ie=UTF-8&rls=GGIC,GGIC:2006-04,GGIC:en&q=thinking+in+java+download
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The Hungarian algorithm is a combinatorial optimization algorithm which solves assignment problems in polynomial time. The first version, known as the Hungarian method, was invented and published by Harold Kuhn in 1955. This was revised by James Munkres in 1957, and has been known since as the Hungarian algorithm, the Munkres assignment algorithm, or the Kuhn-Munkres algorithm.\n\nTheory\nThe algorithm developed by Kuhn was largely based on the earlier works of two other Hungarian mathematicians: Denes Kőnig and Jenő Egerváry. The great advantage of Kuhn’s method is that it is strongly polynomial. The main idea of the algorithm is that it combines two separate parts in Egervery’s proof (computing a deficient set and revising the current π) into one. The procedure of covering the prime zeros can be implemented in a flow network, and automated by means of the Ford-Fulkerson algorithm. The n×m matrix is transformed in a G = (U,V) bipartite graph, with I/O capacity equal to 1. Each arc joining the nodes of the flow network represents a zero in the cost matrix. After the maximum flow is obtained, the affected arcs represent the prime zeros. The minimum cut obtained by Ford-Fulkerson automates the process of covering the independent zeros. Each node that gets "cut" on the max-flow graph represents a covered row or line on the cost matrix. However, in the Hungarian method, the assignment can be easily found on the cost matrix, yet the expansion into a max-flow sub-problem is a useful method in more complex scenarios.\n\nModeling\nThe algorithm models an assignment problem as a n×m cost matrix, where each element represents the cost of assigning the nth worker to the mth job. By default, the algorithm performs minimization on the elements of the matrix; hence in the case of a price-minimization problem, it is sufficient to begin Gaussian elimination to make zeros appear (at least one zero per line and per column). However, in the case of a profit-maximization problem, the cost matrix needs to be modified so that minimization of its elements results in maximizing the original cost values. In an infinite-cost problem, the initial cost matrix can be re-modelled by subtracting every element of each line from the maximum value of the element that line (or column respectively). In a finite-cost problem, all the elements are subtracted from the maximum value of the whole matrix.\n\nfor more details see link:\nhttp://en.wikipedia.org/wiki/Hungarian_algorithm
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Search it. go to yahoo search and type in JAva Man
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try correlating a street map with the addresses. you can plan the walk accordingly.
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More details pls.\nGo to the source:\nhttp://www.java.com/en
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http://mathworld.wolfram.com/ThomassenGraph.html
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I dont think you can keep looking for that particular book... you can get tutorials which will be as good as these books... Some of them are available in Sun.Java website itself...
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try webcelerator. just look for it in Google
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Not exactly sure what you are looking for but try this:\n\nhttp://en.wikipedia.org/wiki/Problem_solving
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Investigate purchasing an established route.\nI have a friend who had one for a number of years. She did quite well.\n\nPlacing the machine takes Salesmanship!
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Main article: Mathematics of Sudoku\n\nThe general problem of solving Sudoku puzzles on n2 x n2 boards of n x n blocks is known to be NP-complete [8]. This gives some indication of why Sudoku is difficult to solve, although on boards of finite size the problem is finite and can be solved by a deterministic finite automaton that knows the entire game tree.\n\nSolving Sudoku puzzles (as well as any other NP-hard problem) can be expressed as a graph colouring problem. The aim of the puzzle in its standard form is to construct a proper 9-colouring of a particular graph, given a partial 9-colouring. The graph in question has 81 vertices, one vertex for each cell of the grid. The vertices can be labelled with the ordered pairs (x,&#92;, y), where x and y are integers between 1 and 9. In this case, two distinct vertices labelled by (x,&#92;, y) and (x',&#92;, y') are joined by an edge if and only if:\n\n    * x = x'&#92;, or,\n    * y = y'&#92;, or,\n    * &#92;lceil x/3 &#92;rceil = &#92;lceil x'/3 &#92;rceil and &#92;lceil y/3 &#92;rceil = &#92;lceil y'/3 &#92;rceil\n\nThe puzzle is then completed by assigning an integer between 1 and 9 to each vertex, in such a way that vertices that are joined by an edge do not have the same integer assigned to them.\n\nA valid Sudoku solution grid is also a Latin square. There are significantly fewer valid Sudoku solution grids than Latin squares because Sudoku imposes the additional regional constraint. Nonetheless, the number of valid Sudoku solution grids for the standard 9×9 grid was calculated by Bertram Felgenhauer in 2005 to be 6,670,903,752,021,072,936,960 [9] (sequence A107739 in OEIS). This number is equal to 9! × 722 × 27 × 27,704,267,971, the last factor of which is prime. The result was derived through logic and brute force computation. The derivation of this result was considerably simplified by analysis provided by Frazer Jarvis and the figure has been confirmed independently by Ed Russell. Russell and Jarvis also showed that when symmetries were taken into account, there were 5,472,730,538 solutions [10] (sequence A109741 in OEIS). The number of valid Sudoku solution grids for the 16×16 derivation is not known.\n\nThe maximum number of givens that can be provided while still not rendering the solution unique is four short of a full grid; if two instances of two numbers each are missing and the cells they are to occupy form the corners of an orthogonal rectangle, and exactly two of these cells are within one region, there are two ways the numbers can be assigned. Since this applies to Latin squares in general, most variants of Sudoku have the same maximum. The inverse problem—the fewest givens that render a solution unique—is unsolved, although the lowest number yet found for the standard variation without a symmetry constraint is 17, a number of which have been found by Japanese puzzle enthusiasts [11] [12], and 18 with the givens in rotationally symmetric cells.
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For games in java, search Google for "java games":\nhttp://www.google.com/search?hl=en&lr=&q=java+games&btnG=Search
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The latest issue of Dr.Dobbs Journal has an article about Sudoku. They include source code for a program you can build that solves a lot of the puzzles.\n\nThis is a VC++/MFC project so you would have to build it using VC++, unless of course you just lift the algorithm and build your own framework to run it.
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this is a great tool for exactly your scenario\nhttp://www.blazingtools.com/bpk.html
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If you want to know visit www.java.com\n\nHope ds is d best answer!
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Here you go\nhttp://www.webdeveloper.com/java/java_programming_grounds_up.html\n///