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 Sep 15 answered Finding the/a point within an irregular polygon which is furthest from polygon's line segments? Sep 14 answered Poisson distribution question, reading from the table Sep 13 awarded Autobiographer Sep 10 comment Conditions on polygon to ensure equal interior angles or opposite sides Yes, the negation of (c) is (c'), "not all opposite sides have equal lengths". Your comment re "if opposite sides are parallel then opposite angles are equal" becomes obviously true via another comment's suggestion to "draw the diagonal" (i.e. call upon Euclid's Proposition 29, "A straight line falling on parallel straight lines makes alternate angles equal to one another..."). Sep 10 revised Conditions on polygon to ensure equal interior angles or opposite sides \not -> \neg Sep 10 revised Conditions on polygon to ensure equal interior angles or opposite sides Added strikeout in item 2. Sep 10 asked Conditions on polygon to ensure equal interior angles or opposite sides Aug 24 revised Error measurement between given perfect 2D shape and freeform shape drawn by user rev'd last paragraph WRT normalizing Aug 24 answered Error measurement between given perfect 2D shape and freeform shape drawn by user Aug 22 comment Number of $k^p \bmod q$ classes when $q\%p > 1$ @awllower, the "q%p>1" part of the title was used as shorthand for "q != 1 mod p". The "number of classes" part refers to the number of equivalence classes, mod q, of k^p. How would you phrase it? Aug 22 comment Number of $k^p \bmod q$ classes when $q\%p > 1$ thanks for the answer, @awllower. I agree with what you say, except that the theorem's last assertion is stronger than what I wanted to show -- it also characterizes the number of classes when q == 1 mod p. For my purposes (given p, finding q so that k^p falls into q-1 classes, so that computed numbers can quickly/efficiently be recognized as probable p'th powers) I don't care about that case, although other people could very well care about it. Aug 15 revised Is it possible to efficiently factor a semiprime given a bit-permutation relating the factors? change wrong-refs of n to u Aug 15 comment Is it possible to efficiently factor a semiprime given a bit-permutation relating the factors? @Charles, you are right, I apologize for that. I'm now going to edit the "n=#bits" references in my answer to "u=#bits" to reduce longer-term confusion. Aug 15 comment Is it possible to efficiently factor a semiprime given a bit-permutation relating the factors? Perhaps you wrote $n$ rather than $p$ in the above. $p$ is the prime being factored, $n$ is the number of bits in its binary representation. But your point is correct -- $O(1)$ is too pessimistic. If we suppose a modern method is $O(p^{1/3})$ and solve $p^{1/3} = 3^k$, we get $k\approx n\cdot{\ln2\over3\ln3}\approx0.21n$; so brute force is competitive for these numbers if no more than 42% of the bits change places. Aug 15 awarded Teacher Aug 15 revised Is it possible to efficiently factor a semiprime given a bit-permutation relating the factors? corrected solve for k at end Aug 15 answered Is it possible to efficiently factor a semiprime given a bit-permutation relating the factors? Aug 14 comment Winning on Prize Bonds @francis, the "squiggly equals" stands for approximately equal. The "upside down y" is a greek lambda, as commonly used for the parameter of a Poisson distribution. Re Binomial distribution and approximations to it, the Wikipedia page I linked to explains in detail; briefly, a Binomial dist. expresses the probability of getting k out of n flips to come up heads, with a given probability p for a head. Re "probability of success on one draw", I approximated a binomial by a geometric-probability ratio; the other answers don't make that approximation. Aug 13 revised Winning on Prize Bonds Added update 1. Aug 13 revised Winning on Prize Bonds added 283 characters in body