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comment Fast Upper Triangular Matrix Exponentiation
@Federico Poloni: unfortunately you also need the resulting pair of matrices to commute which isn't the case with that decomposition
Dec
12
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revised Fast Upper Triangular Matrix Exponentiation
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asked Fast Upper Triangular Matrix Exponentiation
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Sep
15
comment “Most Similar Vector Problem” on an Integer Lattice?
@BerkU.: it should be the global best solution because the global solution must be located at one of the shaded box corners (you will have to check in the negative direction as well).
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Aug
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answered “Most Similar Vector Problem” on an Integer Lattice?
Aug
2
comment “Most Similar Vector Problem” on an Integer Lattice?
Curiously, why isn't this solvable by zig-zagging the lattice? For example if $n=2$, of the four points on the unit square, pick the closest to your vector. Lets say this point is $(1,1)$, in the positive quadrant. Look one step up and right and determine which location is closer to $u$. Move to that point, and repeat. Collect all distances this way, and pick the minimum one. You might have to repeat this with the initial best point in the opposite quadrant to check the other direction. It seems like this algorithm is $O(Mn)$. Or just find all unit boxes $u$ intersects and poll each vertex.
Jul
11
comment Which subgroup order of the symmetric group is the most frequent?
@EricWofsey: It seems like a difficult question actually, see the top answer here: math.stackexchange.com/questions/76176/…
Jul
9
comment Another name for coin-flipping polynomials
They are homogenous polynomials for each $n$ in variables $x,y=(1-x)$ I'm not sure what else you can say without making some kind of restrictions.
Jul
9
comment Another name for coin-flipping polynomials
A general singleton has probability $x^k(1-x)^{n-k}$ in the coin space. Any event is thus $\sum_{k\in E}x^k(1-x)^{n-k}$, where $E$ is an index multiset set of the event, which is the form of the most general coin polynomial.
Jul
5
comment Distribution of trivial subset sums
If you add $L+1$ to the interval, then you're looking at the number of ways to sum to $L+1$ from $[1,L]$. You necessarily need all your chosen numbers to be less than $L-1$, which occurs with probability $\binom{L-1}{n}$, after which you are evaluating $p(L-1)$: the number of partitions of $L-1$ with distinct parts, which equals the number of partitions with odd parts by Euler's theorem. I'm guessing the growth rate is known via circle methods or other analytic combinatorics, probably something like $O(\exp(c\sqrt{n})/n)$.
Jun
25
comment Logarithmic integral, $π(x)$ and $x/(\ln x)$
Not yet. See the discussion here: en.wikipedia.org/wiki/Skewes'_number#More_recent_estimates
Jun
25
comment Extrapolation between longest increasing and longest alternating subsequences
A trivial observation: there's some strange parity here. The Tracy widom distribution is biased toward one side, whereas the Gaussian is symmetric about the mean. This is evident here since for small $m$, we basically have reflection symmetry. So I would expect the transition to occur when $m\gg n/2$.
Jun
23
comment Semicircle law universality elsewhere
thanks for all your answers!