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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 2 years, 9 months |
| seen | 5 hours ago | |
| stats | profile views | 157 |
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Oct 21 |
comment |
Why are very large prime numbers important in cryptography? Nice answer, Arturo, but you still miss a key point (which I'm sure, you know, but didn't mention): The discrete logarithm problem is only difficult if $p-1$ has a big prime factor (Pohlig–Hellman algorithm). Like factorization is only difficult if the given number has at least two big prime factors (Lenstra's elliptic curve factorization finds quickly "small" prime factors). |
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Oct 21 |
comment |
Why are very large prime numbers important in cryptography? I doubt that factoring a 256-bit RSA modulus (being the product of two 128-bit primes) takes longer than a minute on a current PC using the best known factorization algorithm. Even for home use I'd recommend to use at least 256-bit primes. By the way, about a year ago a team around Arjen Lenstra factored the 768-bit RSA challenge. |
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Oct 6 |
comment |
Why is this coin-flipping probability problem unsolved? @Joseph: What do you mean with the expected future maximum ration? |
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Oct 6 |
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Why is this coin-flipping probability problem unsolved? @Joseph: Thanks to the law of the iterated logarithm you are guaranteed \$50. But what will your strategy be for continuing or stopping after having drawn $k$ after $n$ throws? The expectation depends on your strategy, which has to be a function of $k$ and $n$. Your graph hides the fact that you have to stop without knowing the future. |
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Oct 6 |
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Why is this coin-flipping probability problem unsolved? Would you go on with a game when winning in 70% of the cases a little money and loosing a lot in the remaining 30% of the cases? |
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Oct 5 |
comment |
Grouping numbers that are “close to each other” @whuber: I agree. That's why I posted it just as a comment. It's maybe "the" most trivial approach (which might be anyway enough for jeffp). @jeffp: Even if you won't implement the methods in the answer of whuber, I'd recommend to accept his answer (or whatever better answer will be posted), as it provides a better solution in general. |
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Oct 5 |
comment |
Grouping numbers that are “close to each other” The differences belonging to your example are 0.6, 0.5, 0.2, 11.3, 2, 33, 8, 1. The 4 biggest values are 33 (= 49-16), 11.3 (= 14-2.7) and 8 (= 57-49) giving your result. |
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Oct 5 |
comment |
Grouping numbers that are “close to each other” You are just looking for the $N$ maximal values of $x_{i+1}-x_i$ after sorting your set $S = \{x_1, \dots, x_n\}$. I doubt this has some name. |
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Oct 5 |
comment |
fixed-point free elements in transitive permutation groups Yes, as the group is transitive, $N$ either doesn't fix any element or it fixes all elements (since the set of fixed points of $N$ is union of orbits of $G$). |
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Oct 5 |
comment |
fixed-point free elements in transitive permutation groups As $P$ is a $p$-group, it has a nontrivial center $Z(P) \ne 1$. Take an element $1 \ne z \in Z(P)$ (central element = element of the center). |
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Oct 4 |
comment |
fixed-point free elements in transitive permutation groups SPOILER (solution of step 2): Let $G$ acting on a set $\Omega$ have a normal subgroup $N$. Then $G$ acts on the set of fixed points of $N$: If $x$ is fixed by $N$, given $n\in N$, $g\in G$, as $N$ normal, there is an $n' \in N$ such that $n'g = gn$. Now $x^g = x^{n'g} = x^{g n} shows that $n$ fixes also $x^g$. |
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Oct 4 |
comment |
fixed-point free elements in transitive permutation groups SPOILER (solution of step 1): For a $p$-Sylow $P$ of $G$ the one-point stabilizer $P_x = P \cap G_x$ has index $p^k$ in $P$, hence $P$ is transitive. |
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Oct 4 |
answered | fixed-point free elements in transitive permutation groups |
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Aug 26 |
awarded | Scholar |
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Aug 26 |
accepted | Estimates involving sums with binomials |
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Aug 26 |
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Estimates involving sums with binomials @J.Mangaldan: Thanks for your interest. In the moment I'm content with Qiaochu Yuan's answer. |
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Aug 26 |
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Estimates involving sums with binomials @J.Mangaldan: Sorry for the late response; I found a different approach calculating the probabilities. I doubt the "bigger sum" would help as each summand has yet another coefficient. But if you are curious: The next bigger term has $(a+1)$ in the place of $a$, $b$ might be greater and the sum is starting at $(c+2)$. |
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Aug 25 |
awarded | Commentator |
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Aug 25 |
comment |
Estimates involving sums with binomials @Moron: This probability is only one term in a bigger sum, and it's too complicated to get into details about the variables. BigOh won't do. |
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Aug 25 |
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Estimates involving sums with binomials @Moron: I'm not sure about constraints on a, b and c yet, since the probabilities are supposed to help me finding parameters for some algorithm. A closed form would be nice for the running time analysis of the algorithm, but maybe I can live without it. |
