Reputation
5,300
Top tag
Next privilege 10,000 Rep.
Access moderator tools
Badges
1 14 28
Newest
 Yearling
Impact
~113k people reached

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
answered Definite shape of polyominos
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
reviewed Close Normal Distribution Statistics
Sep
19
reviewed Close What is the probability of having the same (binary) datasets?
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
19
reviewed Close Mental Math Visual Retention
Sep
19
reviewed Close Finding Supremum and Infimum of a given set. Is my solution correct?
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.
Sep
4
answered Integer Tetrahedra
Aug
31
comment Would this proof strategy work for proving the lonely runner conjecture?
@DavidvonRudisill, following r.e.s.' reasoning, we can consider the general case for three runners of speeds $0$, $a$, $b$. Then wlog the equispacing would have $at = \frac{3\alpha + 1}{3}$ and $bt = \frac{3\beta + 2}{3}$ with $\alpha, \beta \in \mathbb{N}$. But then $\frac{a}{b} = \frac{3\alpha + 1}{3\beta + 2} \in \mathbb{Q}$, so your strategy can at best work for a subset of runner speeds which has measure $0$ in the space of all possible runner speeds. This is (related to) the point James Hines was trying to make with the comment on irrational speeds.
Aug
23
comment How many legal states of chess exists?
web.archive.org/web/20140601124247/http://homepages.cwi.nl/…
Aug
10
comment Complexity of subset-generation algorithm
At a high level, $2^n$. At a low level, it depends on implementation details. There are probably some questions about it on the sister site stackoverflow.com
Aug
9
revised A combinatorial identity
OP pointed out in comments that multiplying top and bottom by 2^i i! allows eliminating the double-factorial and cancels more terms
Aug
9
comment A combinatorial identity
1. Not a typo: the $!!$ is a double factorial. 2. You're quite right. I've edited.
Aug
9
answered A combinatorial identity
Aug
9
comment Complexity of subset-generation algorithm
It's not very clear to me what most of the operations are, or what the purpose of k is (it seems to be unused). If what you want is a way to iterate through a powerset efficiently then you should look into Gray codes.