{"id":28704,"date":"2023-01-02T02:25:13","date_gmt":"2023-01-01T20:55:13","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-write-an-algorithm-to-efficiently-find-all-i-and-j-for-any-given-n-such-that-nij\/"},"modified":"2023-01-02T02:25:13","modified_gmt":"2023-01-01T20:55:13","slug":"solved-write-an-algorithm-to-efficiently-find-all-i-and-j-for-any-given-n-such-that-nij","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-write-an-algorithm-to-efficiently-find-all-i-and-j-for-any-given-n-such-that-nij\/","title":{"rendered":"[Solved] Write an algorithm to efficiently find all i and j for any given N such that N=i^j"},"content":{"rendered":"

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Given that i^j=N<\/code>, you can solve the equation for j<\/code> by taking the log<\/code> of both sides:
j log(i) = log(N)<\/code> or j = log(N) \/ log(i)<\/code>. So the algorithm becomes<\/p>\n

for i=2 to N\n{\n    j = log(N) \/ log(i)\n    if((Power(i,j)==N)\n       print(i,j)\n}\n<\/code><\/pre>\n
Note that due to rounding errors with floating point calculations, you might want to check j-1<\/code> and j+1<\/code> as well, but even so, this is an O(n)<\/em> solution.<\/p>\n
Also, you need to skip i=1<\/code> since log(1) = 0<\/code> and that would result in a divide-by-zero error. In other words, N=1<\/code> needs to be treated as a special case. Or not allowed, since the solution for N=1<\/code> is i=1<\/code> and j=any value<\/code>.<\/p>\n
As M Oehm pointed out in the comments, an even better solution is to iterate over j<\/code>, and compute i<\/code> with pow(n,1.0\/j)<\/code>. That reduces the time complexity to O(logN)<\/em>, since the maximum value for j<\/code> is log2(N)<\/code>.<\/p>\n<\/p><\/div>\n
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solved Write an algorithm to efficiently find all i and j for any given N such that N=i^j <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Given that i^j=N, you can solve the equation for j by taking the log of both sides:j log(i) = log(N) or j = log(N) \/ log(i). So the algorithm becomes for i=2 to N { j = log(N) \/ log(i) if((Power(i,j)==N) print(i,j) } Note that due to rounding errors with floating point calculations, you … 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":[457],"class_list":["post-28704","post","type-post","status-publish","format-standard","hentry","category-solved","tag-algorithm"],"yoast_head":"\n[Solved] Write an algorithm to efficiently find all i and j for any given N such that N=i^j - 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-write-an-algorithm-to-efficiently-find-all-i-and-j-for-any-given-n-such-that-nij\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Write an algorithm to efficiently find all i and j for any given N such that N=i^j - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Given that i^j=N, you can solve the equation for j by taking the log of both sides:j log(i) = log(N) or j = log(N) \/ log(i). 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