Impact
~385k
people reached
- 0 posts edited
- 1 helpful flag
- 110 votes cast
Oct
24 |
awarded | Good Answer |
Oct
21 |
awarded | Yearling |
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). |
Sep
12 |
awarded | Good Answer |
Aug
2 |
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! |
Jun
23 |
comment |
Bounds on the probability of k-of-n events in terms of bounds on single and pairwise probabilities
It sounds like you could really use something akin to positive/negative association such as $P(A_i\cap A_j)\leq \geq P(A_i)P(A_j)$. Do you happen to know if these hold on some large subset of index pairs? |
Jun
22 |
revised |
Semicircle law universality elsewhere
added 4 characters in body |
Jun
22 |
comment |
Why is this distribution exponential?
Try working this out with just two points. |
Jun
22 |
comment |
Why is this distribution exponential?
Basically by rescaling the interval as a function of $n$ the waiting time between successive points becomes a poisson process. This is because the probability of seeing the next point is proportional to the interval you are looking at. |
Jun
19 |
comment |
Semicircle law universality elsewhere
@jon bannon: I'm not intimately familiar with it but I basically lumped it with random matrix theory. If this is wrong of me I would love to see a note on this. |
Jun
19 |
revised |
Semicircle law universality elsewhere
deleted 1 character in body |
Jun
19 |
asked | Semicircle law universality elsewhere |