\n
I’ll try to break this down to essentials. But first let me commend you for tackling a problem that’s “known” to be enormously hard. No progress can be made without risk taking.<\/p>\n
You are approaching TSP in terms of a recursive expression for S(a, b, I), the length of a shortest path from city a to city b, a \\ne b, passing through each city in the unordered set I exactly once. <\/p>\n
With S in hand, the TSP is easy to solve. For the set of cities C, find<\/p>\n
min( D(b, a) + S(a, b, C\\a\\b) ) over all pairs a, b drawn from C where a \\ne b\n<\/code><\/pre>\nHere D(x, y) = D(y, x) is the distance from city x to y and C\\a\\b is C with a and b removed.<\/p>\n
The recursive expression you propose for S is<\/p>\n
S(a, b, I) = min( D(a, p) + S(p, q, I\\p\\q) + D(q, b) ) \n over all pairs p, q drawn from I where p \\ne q , \n<\/code><\/pre>\nThe base cases are where I has zero or one element(s). These are pretty obvious.<\/p>\n
You are proposing to cache values of S(a, b, I) so that no such computation is ever repeated. (This is called memoizing by the way.) <\/p>\n
So what is the cost of this computation, or equivalently the size of the cache? We can write a recurrence for it, where the parameter n = |I| is the number of cities in the intermediate set:<\/p>\n
C(n) = M(n, 2) C(n - 2) = ( n(n-1)\/2 ) C(n - 2)\nC(0) = C(1) = 1\n<\/code><\/pre>\nHere M(n, m) is the combination of n things taken m at a time, n! \/ (m! (n-m)!)<\/p>\n
We can solve this. For even n:<\/p>\n
C(n) = n! \/ 2^(n\/2)\n<\/code><\/pre>\nI’ll let you work out the odd case. <\/p>\n
For the tour among m cities, we’d need to repeat this for all city pairs and corresponding intermediate sets:<\/p>\n
(m(m-1)\/2) C(m-2) = m! \/ 2^(m\/2-2)\n<\/code><\/pre>\nSo your method does avoid an exponential amount of work with respect to the na\u00efve algorithm of generating all possible tours, but the factorial still dominates: this function is super-exponential. <\/p>\n
NB on your other “stopping criteria:” Above is the cost of computing all possible values of S(a,b,I) exactly once. To get a poly time algorithm, you will have to come up with a scheme for skipping a super-exponential number (a,b,I) of triples entirely. It’s unlikely you can do this, but don’t let this dampen your enthusiasm.<\/p>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
1<\/span> <\/p><\/div>\n<\/div>\n[ad_2]<\/p>\n
solved Java: Traveling Salesman – Found polynomial algorithm <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] I’ll try to break this down to essentials. But first let me commend you for tackling a problem that’s “known” to be enormously hard. No progress can be made without risk taking. You are approaching TSP in terms of a recursive expression for S(a, b, I), the length of a shortest path from city … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[771,323,4372,3898,2421],"class_list":["post-17706","post","type-post","status-publish","format-standard","hentry","category-solved","tag-graph","tag-java","tag-np-complete","tag-np-hard","tag-traveling-salesman"],"yoast_head":"\n[Solved] Java: Traveling Salesman - Found polynomial algorithm - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-java-traveling-salesman-found-polynomial-algorithm\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Java: Traveling Salesman - Found polynomial algorithm - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] I’ll try to break this down to essentials. 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