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Jan
24
comment The hardest game of mahjongg
I think you need to address the issue of how to lay out the tiles for arbitrary $n$ before it even makes sense to start thinking about an answer.
Jan
15
comment Maximal tiling without any 3-in-a-rows
@randomra, if the maximum Hamming weight which can be achieved in an $m\times n$ rectangle is $w$ then that gives an upper bound for the density of any infinite tiling of $\frac{w}{mn}$ by the simple mechanism of superimposing an $m \times n$ grid over the infinite tiling.
Jan
9
comment Maximal tiling without any 3-in-a-rows
The existence of (non-tiling) moderately large squares with densities > 1/2 (e.g. 15x15 with 114 bits set) indicates that the case analysis would have to work on larger units than those squares. I very much doubt that the number of cases for this type of approach would be computationally tractable.
Jan
9
comment Maximal tiling without any 3-in-a-rows
I don't understand the argument that cases can be eliminated by the symmetry of swapping 0s and 1s: that's not a symmetry, because 000 is allowed and 111 isn't. I also don't understand what contradiction you see in your first example.
Dec
12
comment If $x^n$ is ivertible in a ring show that $x$ is invertible.
So to be precise, your question is how to show that, given a left inverse of $x$ and a right inverse of $x$, the two must be the same even in a non-commutative ring?
Nov
30
comment Erdős and Szemerédi sums and producs
Note: this looks suspiciously like a certain active contest.
Nov
9
comment Squares with squares
Not only does this not answer the question, but it's not even clear what question it's attempting to answer. You appear to have your own idea of what numbers "of that kind" are, but you haven't defined it, and it's certainly not clear to me why you disqualify e.g. $38^2 = 1444 = 12^2 \times 10 + 2^2$
Nov
9
comment Squares with squares
Fair enough. Updated.
Nov
7
comment Sign table in $2^k$ factorial experiment
Ok, which specific part do you have trouble generalising to the four-factor case?
Oct
31
comment Extended Euclidean Algorithm to find multiplicative inverse of two polynomials
If a polynomial has no remainder when divided by 1, it has no remainder when divided by 2, and vice versa.
Oct
30
comment How many subsets xor to given value?
@user128409235, that's awfully specific. Is this for a programming contest or something similar?
Oct
13
comment A nice form of a given function
@Halbort, I think $\max \{a_i - (a_i \oplus k)\}$ is as simple as it's likely to get.
Oct
1
comment A nice form of a given function
Is that clearer?
Sep
22
comment How to write a double edge in an incidence matrix.
? An edge isn't represented by a column in an adjacency matrix, but by a cell.
Sep
19
comment Definite shape of polyominos
I can see more than one interpretation of the question, so in order to narrow it down: are you asking how to justify the claim that two polyominos are really the same polyomino if there's a way to rotate and reflect one so that it perfectly covers another? Or are you asking how to index them so that when you're counting by hand you can be sure that you don't have duplicates? Or are both of those interpretations wrong and it's something else?
Sep
19
comment What is the probability of having the same (binary) datasets?
On the basis of the information given, you do not have enough information to calculate the probability that the datasets are the same. You can calculate $p(X_1 = 0 \wedge X_2 = 0 \wedge \ldots \wedge X_n = 0)$ (where $X_i$ is 0 if the ith vulnerable is unchanged, and 1 if it changes) given $p(X_1 = 0), p(X_2 = 0), \ldots, p(X_n = 0)$, but the only information you've given us is that the "outcomes are randomly changed" - i.e. that $p(X_1 = 0)$ etc. exist.
Sep
18
comment Is there any winning strategy? 2015 and Game with marbles!!!
A pedantic answer to the question as worded would be "Yes, there is a winning strategy for one of the players". Proof: the game must be finite (taking at most 2015 turns) and cannot end in a draw.
Sep
18
comment Partition of ${1, 2, … , n}$ into subsets with equal sums.
@donbright, the question asks for a proof that any $m, n, k$ satisfying $m \ge n$ and $\frac{n(n+1)}{2} = mk$ has a solution.
Sep
9
comment Find a polynomial with evaluation equal to # of partitions of n into at most k parts
Perhaps instead of $p(n;\le k)=P(n)$ the question statement should say that $p(n;\le k)=\mathrm{round}(P(n))$? E.g. for at most three parts, $p(n;\le 3)$ is the nearest integer to $(n+3)^2/12$
Sep
9
comment How many different sums of parts of a vector
If you plot $S_j$ against $j$ then you get (in general) a zigzag, where each step goes up or down $2$. Now, $S_{n-1} - S_0 = 2 \sum_{i=1}^{n-1}v_i$ so, unless that sum is zero, $S_{n-1}$ and $S_0$ have opposite sign. If you split the line into three by dividing it at the first and last changes of sign (with a special case for sum is zero, sign never changes), you can show that it's only necessary to consider cases where $(\sum_{i=1}^{n-1}v_i) \in \{0,1,2\}$, since any crossing set found with a larger value than those can be "simulated" with a smaller value of the same parity.