tag:blogger.com,1999:blog-3722233.post4254094867358136250..comments2022-12-07T07:34:35.444-06:00Comments on Computational Complexity: Four answers to the Recip problemLance Fortnowhttp://www.blogger.com/profile/06752030912874378610noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-3722233.post-28357416868131229882013-11-14T04:30:57.037-06:002013-11-14T04:30:57.037-06:00These concepts are well-known, see the entry "...These concepts are well-known, see the entry "Egyptian Fraction" in wikipedia. Trigg in his "Mathematical Quickies" gave as a solution to the problem that every rational can be respresented as an Egyptian fraction: Write it as a sum of unit fractions and eliminate duplicates using 1/n = 1/(n+1) + 1/n(n+1). I think there is a problem remaining: Proof that this process terminates. Is this still a quicky?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-55716300624680421932013-11-12T10:51:47.946-06:002013-11-12T10:51:47.946-06:00This outcome is not a surprise! Approach 1) is hin...This outcome is not a surprise! Approach 1) is hinted at by the given examples for k=3 and k=4. <br />I wonder what would have happened without providing these examples. Anonymousnoreply@blogger.com