bio | website | andrewdouglasking.com |
---|---|---|
location | Vancouver | |
age | 35 | |
visits | member for | 5 years, 5 months |
seen | Aug 22 at 0:42 | |
stats | profile views | 1,341 |
I am interested in graph theory and combinatorial optimization.
May
29 |
comment |
An example of when nauty, on two different platforms, gives different canonical labels for the same input graph?
Is this comment tongue-in-cheek, or has the implementation somehow been verified using formal methods? (Or do you just mean, if such an example was known it would have been fixed?) |
Apr
25 |
awarded | Notable Question |
Apr
16 |
comment |
A maximum discrepancy hypergraph 2-colouring problem
Because not every vertex is in two sets $S_i$ and $S_j$. In fact, even if no vertex is in two such sets, I don't see how the problem is trivial. |
Apr
16 |
comment |
Defining fuzzy properties of crisp graphs
Agreed. The problem of how to find a good graph clustering is subject to a lot of debate, and is very closely related to this problem. Given a clustering scoring function $f$, and letting $f_o$ be the optimal score of any clustering (vertex partition) of $G$, probably $f_o/f(G)$ would be a good measurement of cliqueness. But in the end, finding the function $f$ is really a matter of taste and context. |
Apr
16 |
comment |
A maximum discrepancy hypergraph 2-colouring problem
Oh yes, you're right. But there may be parallel edges, and you would have to add some fake structure to ensure that every $V_i$ matching has size 8. |
Apr
16 |
answered | Defining fuzzy properties of crisp graphs |
Apr
13 |
comment |
A maximum discrepancy hypergraph 2-colouring problem
Maybe there is some equivalence, but I don't see it. |
Apr
13 |
comment |
A maximum discrepancy hypergraph 2-colouring problem
Note that even if we replace 8 with $2k$ and ask the corresponding question for large $k$, the Local Lemma and random partitioning doesn't work: Prob that $S_i$ fails is $> 1/(2k+1)$, and each $S_i$ can be non-independent with up to $(2k+1)(2)(2k-1)$ other sets. |
Apr
13 |
asked | A maximum discrepancy hypergraph 2-colouring problem |
Apr
9 |
comment |
Area ratio of a minimum bounding rectangle of a convex polygon
Right, in other words that the rectangle actually contains the polygon. |
Apr
8 |
answered | Area ratio of a minimum bounding rectangle of a convex polygon |
Mar
12 |
comment |
Gluing two graphs
I think you need to be more specific about your requirements. When you say there are no multiple edges, do you mean that you ignore multiple edges, or that if you identify $u$ with $a$ and $v$ with $b$, then at most one of $uv$, $ab$ is an edge? I assume you are just ignoring multiple edges. If the two sets are cliques or stable sets, then the operation is clique sum or stable set sum. Otherwise you need to be more specific about what vertex mappings are allowed (i.e. if you are identifying $V_1$ and $V_2$, which bijection you use.) |
Mar
12 |
awarded | Yearling |
Feb
28 |
comment |
Meeting management
I agree that random constructions are a fool's venture; one must pack these partitions into $K_{60}$ with density almost 1/2. Thanks for the insight Kevin. I'm surprised one can go for 11 days under slightly different circumstances. |
Feb
27 |
comment |
Meeting management
PS Jim you posted an answer where a comment would be more appropriate. |
Feb
27 |
comment |
Meeting management
Actually I wrote some code on Friday that attempted to generate a solution randomly. It failed spectacularly. I suspect that this schedule is actually impossible, and maybe some structural analysis could prove it. For example, take "person" $v_1$ and a Monday schedule 1-6, 7-12, etc. For the next four days, $v_1$ will meet with 20 other people, and there are 54 available. $v_1$ can't meet with 5 people from any other group, because they would all have to be on different days, so $v_1$ meets with $\leq 4$ people from each other group... |
Feb
25 |
comment |
Meeting management
For some reason I can't edit the question, but here goes: There are 60 people. On each of five days, the people are partitioned into ten groups of six for day-long meetings. Is it possible to find a set of five partitions such that no two people meet twice? |
Feb
25 |
answered | How to find central vertex in a graph? |
Feb
22 |
comment |
Finding local patterns in a circular list
Sorry, I meant $O(1)$ time, and $(1-\epsilon)n$ nodes. |
Feb
22 |
comment |
Finding local patterns in a circular list
Well, here is a vague set of sufficient conditions: - the pattern always holds somewhere on a circular list - the pattern always holds somewhere on a linear list with property $x$. - given a circular list or a linear list with property $x$, we can find, in $O(\plog n)$ time, a linear sublist with property $x$ containing at most $(1-\epsilon)$ of the nodes. |