Zack Wolske
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Registered User
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May 7 |
awarded | ● Disciplined |
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Apr 30 |
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Verifying the correctness of a Sudoku solution @Ralph: how would one verify anything other than a complete set of $9$ elements? You can check that subsets of them do not contain repeats, but knowing that can only verify a full array after you check at least $3$ subsets (when the subsets are 8/9), which seems to be a lot more work. |
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Apr 30 |
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Verifying the correctness of a Sudoku solution I think you can do $1$ better for s, assuming I have the right construction in mind which works except on a corner set (flip pairs of elements along each side of the rectangle, so that rows keep the same elements on horizontal flips, and the number of vertical flips is even, and vice versa for columns). This just requires an even number of squares be chosen in each row and column, which you do with a corner set, taking $2$, $2$, and $0$. You can make the same construction with $2$ squares in each row and column (like the complement of a minimal set of squares that meets all corner sets). |
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Apr 18 |
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When integer polynomials take integer values, does their GCD also take integer values? However it's defined (monic or not), surely you must have $P/D$ and $Q/D$ polynomials with integer coefficients. What do you take for $2x$ and $2x^2+x$? Either it's a counterexample, or you aren't requiring the quotients to be in $\mathbb{Z}[x]$. |
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Apr 15 |
awarded | ● Nice Answer |
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Apr 10 |
revised |
partly obscured Rubik’s cube updated link |
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Apr 5 |
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Is there a deep reason for the fecundity of involutions? Many plants (think of flowers and fruit) and simple animals (think of starfish, jellyfish and anemones) exhibit 3-, 5- or 6-fold symmetry, and often higher. Flowers have been "making useful use of automorphisms of order three" for ages. Jellyfish, possibly the most successful animal to ever exist on the planet, have such striking and high order radial symmetry that they are the biologist's go-to example for demonstrating this phenomenon in nature. |
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Mar 30 |
answered | Enumerating 0-1 finite boxes without null rays. |
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Mar 30 |
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Enumerating 0-1 finite boxes without null rays. To get your formula for $M(n,2,2)$ via inclusion-exclusion, stack $n$ levels of $2 \times 2$ matrices, then use $M(2,2)=7$ to overcount the number of ways to fill each level. This includes $4 \cdot 2^n$ cases where one of the four $n$-lines is all $0$s, and that overcounts the $2$ cases when two lines are all $0$s. The same inclusion/exclusion idea gives $M(n,3,2) = 25^n -6(8^n+3^n) +6(3\cdot2^n +1)$, and the asymptotics $M(m,n,p) \sim M(m,n)^p$. |
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Mar 29 |
accepted | Series defined by a fixed-point functional equation |
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Mar 10 |
revised |
Series defined by a fixed-point functional equation Added general method at the end |
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Mar 10 |
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Series defined by a fixed-point functional equation Yes, and it seems to me to be much easier when all of your polynomials $P_i$ are homogeneous. I'll add it to the answer, because there is a bit of enumeration and subscripts don't look so nice in comments. |
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Mar 7 |
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Series defined by a fixed-point functional equation A typo in the code, $t+1/2$ instead of $t/2+1$, made the last function give the wrong values. It's now corrected, and agrees with "the number of maximal balanced binary trees" from your paper in Theoretical Computer Science (Feb. 2012, 420). |
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Mar 7 |
revised |
Series defined by a fixed-point functional equation deleted 588 characters in body |
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Mar 7 |
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How to call covers not covering anything else? Perhaps those 120 people also had not heard of this "standard" term. I would suggest that if you use it, you include the definition, otherwise your reader might grow tired of looking for the term, just like you did. Searching for exhaustion sets brings up as many references to weightlifting as it does to set theory. |
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Mar 7 |
answered | Series defined by a fixed-point functional equation |
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Mar 2 |
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Sane bound on number of moves for Maker-Breaker game on $\mathbb R^2$ for $\{0,1,2,3,4\}$ @François: Is any version of gomoku on a hexagonal board known to be a first player win? The Erdős-Selfridge result says it is with those two advantages on a large enough board, but is it true without those? |
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Feb 14 |
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Semimagic Squares and Partitions Why not do this by induction on $s$? It's true if every entry is $1$, and if they are not all $1$, then you can find a set of $n$ entries, one from each row and column, which are all greater than $1$ (consider the labels of Latin squares and use pigeonhole principle). Assign the same single element to each of those, and you've reduced $s$ by $1$. |
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Feb 7 |
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To what extent should a pure mathematician care about the meaning of things? What is the physical meaning of a derivative, to you? It could be the rate of change of something, if your function is measured against a time variable. It could be the slope of a tangent line, if your function is a graph of one variable. It could be the best linear hyperplane approximation, if your function has a vector of inputs. The nice part about using the derivative abstractly is that it could have any of those physical meanings, or many others, but it doesn't have to. You are free to use any reasonable physical interpretation, or none at all. |
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Feb 7 |
answered | Diophantine question |
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Feb 7 |
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Diophantine question There are still quite a lot besides those. For example, take $x=2$, and any $z$. Then the equation simplifies to $n_z((y-1)^2 -1)=q^2$, which is Pell's equation if $n_z$ is not a square. Choose $z$ so that $6(z^2-z)$ is not a square (another Pell equation, but this time we want non-solutions) and you'll get infinitely many more solutions. I think the same tricks will work with any $x$ value: complete squares, get a Pell-type equation for $z$ in terms of $x$ and choose a non-solution $n_z$, then write the Pell equation for $y$ in terms of that, and solve. |
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Jan 21 |
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Sequence inequality No. Take any integer $m>C$, with sequences $a_n=0$ for $n\neq m$, $a_m=m$; $b_n=1$ if $n \leq m$, and $0$ otherwise. |
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Jan 4 |
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Can the difference of two distinct Fibonacci numbers be a square infinitely often? @joro: No, I mean that the squares 9 and 49 do not divide it. Since it can be factored as a Fibonacci number times a Lucas number, and 3 and 7 both divide it, and 3 and 7 so not divide the Lucas number, then both of their squares (i.e. 9 or 49) must divide the Fibonacci part for it to be a square. I've edited now to clarify. |
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Jan 4 |
revised |
Can the difference of two distinct Fibonacci numbers be a square infinitely often? added 58 characters in body |
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Dec 18 |
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Would this hold for any 2 by 3 matrix? May be better to ask for some help on math stack exchange. You'll also want to tell them which matrix norm you're using, what your base field is, and why you care about this. |
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Dec 17 |
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Useless math that became useful This sort of appendix seems contrary to the nature of mathematics. The argument isn't countered by providing a list of other ideas that people might have said were useless. Instead, why not focus on the education aspects? According to the Wikipedia article, the search has led a few computer programmers into what is ostensibly number theory, and may have introduced many young people to a fundamental idea behind proofs - even if you haven't found a palindrome by $10^9, there might still be one. Sounds a lot like Skewes' number, also probably called useless. |
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Dec 2 |
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There are n horses. At a time only k horse can run in the single race. How many minimum races are required to find the top m fastest horses? @David: This is even weaker than partial sorting, since we don't need to know the exact rank of elements $1$ to $m$, just the set of elements in those positions. |
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Nov 21 |
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why Skolemization? @Mariano: Can we call the practice of rebranding those processes Suárez-Alvarezization? |
