RE: [math-fun] Geometry/topology question about planar graphs
-----Original Message----- From: John Conway [mailto:conway@Math.Princeton.EDU] e. However, the "Carpenter's Ruler" problem (are the plane embeddings of a hinged ruler interconvertible in this way, keeping the segment-lengths fixed) is quite difficult - indeed I think it's still unsolved,
This was finally solved, in the affirmative, by Erik Demaine, Robert Connelly, and Guenter Rote. http://theory.lcs.mit.edu/~edemaine/papers/Linkage/
and the corresponding assertion for arbitary graphs is false.
Is there an easy-to-describe counterexample? Andy Latto andy.latto@pobox.com
A semiprime is the product of two primes. On my site, I wondered about the difficulty in identifying semiprimes. I picked googol+37 as an example. Don Reble sent me a note that this is *not* a semiprime.
For example, is 10^100+37 a semiprime?
10^100+37 is 87719765535727771 times a composite number; -- Don Reble Are there any good semiprimality tests? Is there any way of counting the number of factors in a big number other than factoring it? (I know that googolplex+10 has been partially factored into more than 50,000 primes, and one huge composite.) --Ed Pegg Jr
Is there any way of counting the number of factors in a big number other than factoring it?
Well, here's a way to do it without factoring (but it's not fast): The number of distinct prime divisors of n is log_2(d) where d is the number of x in {0,1,...,n-1} such that x^2 mod n = x, that is, d is the number of idempotents in Z/(n). --Edwin
participants (3)
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Andy Latto -
Ed Pegg Jr -
Edwin Clark