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15h
comment A combinatorial game theory problem
What do you mean by "vicinal squares"? Does that cover diagonally adjacent, orthogonally adjacent, or both? And why do you call this a combinatorial game theory problem? It seems to be a single combinatorial question.
17h
answered Generalized Dyck words with alphabet of size $k$
2d
awarded  Constituent
Dec
17
revised Maximum absolute value of polynomial coefficients
added 657 characters in body
Dec
10
comment How to solve this tough recurrence relation?
Cross-post on MO
Dec
9
awarded  Caucus
Dec
6
revised Why are asymptotically one half of the integer compositions gap-free?
Replace tautology with definition of gap-free
Nov
29
reviewed Close Cardinality of set difference
Nov
27
answered Explicit formula for Nth string of Gray Code.
Nov
27
comment Give the generator polynomial of a binary cyclic [9, 2] code.
FWIW I checked the factors of $x^9-1$ over $GF(2)$ and they're correct.
Nov
25
comment Secret Santa Perfect Loop problem
Why is $A\rightarrow B\rightarrow C\rightarrow D\rightarrow A$ the only perfect loop? What's wrong with e.g. $A\rightarrow C\rightarrow B\rightarrow D\rightarrow A$?
Nov
17
reviewed Leave Open Primes of some particular form.
Nov
17
revised Primes of some particular form.
Parameterise the reference to Dirichlet's theorem; make the question explicit
Nov
15
reviewed Close A finite semi-group with cancellation laws is a group .
Nov
15
reviewed Close How many pure trees with a fixed number of nodes exist?
Nov
15
comment Upper bound for the widest matrix with no two subsets of columns with the same vector sum
Related
Nov
15
comment Upper bound for the widest matrix with no two subsets of columns with the same vector sum
It's not actually the case that "any column cannot be the linear combination of any 2 other columns" - if it were then the upper bound would be 1. The point is that general linear combination allows multiples other than 0 or 1, whereas property X does not.
Nov
12
comment Counting all possible legal board states in Quoridor
Sketch of an approach: augment the previous approach by a set of variables which track whether the cells above and below a row of intersections are reachable from the top row and from the bottom row, and another which tracks totals. This will increase the number of nodes in the graph by a factor of on the order of tens of thousands, but the number of edges won't increase by nearly such a big factor, so it should still be a feasible calculation.
Nov
11
reviewed Leave Closed The dog language BOW
Nov
8
comment Counting all possible board positions in Quoridor
@lameK, if you're happy with the answer then there's a tick-in-a-circle button which you can use to mark it as accepted. That takes it out of the "Unanswered questions" list, and will be appreciated by people looking for genuinely unanswered questions. As for tougher Quoridor questions, in principle I'm interested but I always have a few things on the go and I can't guarantee a quick response. If you want to discuss by e-mail then I have a catch-all for anything to cheddarmonk.org.