505 reputation
411
bio website yume3.sourceforge.net
location MT
age 94
visits member for 3 years, 5 months
seen yesterday

Former pictures: Dakota puddles (prairie potholes), 2012-01-30; Yellowstone River icicles, 12 Dec 2011; Black-and-white ducks on YR, ca 6 Jan 2012; Eagle in eastwind snow, on south bank of YR, 15 Jan 2012; Sunlight & shadows on riverbottom, 5 Feb 2012; Duck in park puddle, June 2011; Trumpeter swan pair on YR, 13 Feb 2012; American Avocet on YR, 10 May 2012; Pelicans flying above YR, 24 June 2012, and pelican in YR, 16 May 2012; Rainbow above Livingston Peak, 23 July 2012; Rose/squirrel, 21 Nov 2012; old Bangalore station, Dec 2004; Bighorn sheep, Zion, 8 Nov 2013

Author of programs yume as seen here and qenqote as described here

qenqote automates URL-formatting editing and can minimize the amount of manual editing needed when you include a URL in a stackexchange post.


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
Aug
13
answered Winning on Prize Bonds
Aug
13
awarded  Supporter
Aug
13
awarded  Scholar
Aug
13
accepted Number of $k^p \bmod q$ classes when $q\%p > 1$
Aug
13
comment Number of $k^p \bmod q$ classes when $q\%p > 1$
@Jyrki, I'd like to think I was conversant with the techniques you mentioned, when I took number theory classes long ago. However, I've forgotten quite a lot. The straightforward and basic techniques of André's answer are more helpful to me, while I expect practicing mathematicians will prefer the conciseness of your hints.
Aug
13
awarded  Editor