sci.math.research Google Group

Re: Request regarding counting algorithms in graph theory

2010-03-19T19:30:02Z

In article <hnqlgq$5b...@news.acm.uiuc.ed u>,
I don't know; however, I believe that the following paper by Goodall and
Noble ("Counting cocircuits and convex two-colorings is #P-complete")
shows that your problem is #P-complete.
[link]
Perhaps their reduction will allow you to apply algorithms for other

Re: Request regarding counting algorithms in graph theory

2010-03-19T16:30:01Z

int partitions(G):
V=vertices(G)
if 1<=|V|: return 0
select x in V
return count({x}, neighbors(G, x), {}, G)
int count(yes, maybe, no, G):
if 0==|maybe|:
if G[vertices(G)-yes] connected : return 1
else : return 0
select x in maybe
return count(yes, maybe-{x}, no + {x}, G) +

Re: Request regarding counting algorithms in graph theory

2010-03-19T16:30:01Z

In article <hntke9$gf...@news.acm.uiuc.ed u>,
No, it is specifically for situations where you don't want to do that.
The largest solution to a problem of this kind that I have computed is
the number of chiral stellations of the deltoidal icositerahedron:
44,688,470,607,269,500. My algorithm took several hours. I am looking

Re: Request regarding counting algorithms in graph theory

2010-03-18T16:30:01Z

Does the algorithm involve actually generating the sets?
If so, the obvious thing to do is check each set of remaining
nodes to see if they induce a connected subgraph.

Re: 3^n - 2^n and relatives

2010-03-18T16:30:01Z

A couple of years ago I posted some discussion here which were
kindly considered by some posters here. It occured, that the
ideas, with which I connected my own 1-cycle-attempt in the
collatz-problem with the approximation of |S*log2 - N*log3| were
essentially the same as that of Ray Steiner (and later John

Re: Request regarding counting algorithms in graph theory

2010-03-18T16:30:01Z

In article <hnqlgq$5b...@news.acm.uiuc.ed u>,
I misunderstood Knuth's example: it counts sets of edges which connect
the nodes of the graph.
But I think that an algorithm could be devised to count the partitions
of the nodes of the graph into two sets, each of which is connected by
arcs of the graph. I think the problems like this can be solved by means

Is there a not-easy-solvable PDE with probabilistic representation with known (or approximable) probability density?

2010-03-17T13:30:01Z

It is well-known that solution of many equations could be represented
as expectation of some functional of random processes (e.g. Feynman�
Kac formula and etc.)
So the solution is the expectation which could be calculated by Monte-
Carlo method.
But if the probabilistic solution is the expectation of the random

Request regarding counting algorithms in graph theory

2010-03-17T13:30:02Z

Donald Knuth, in _The Art of Computer Programming_ Section 7.1.4 [1],
describes an algorithm for counting the connected sets of nodes in a
non-directed graph: see Exercise 55.
I am interested in a related problem, namely to count the partitions of
such a graph into two connected components. Are there any published

Call for papers XA2010

2010-03-15T13:40:50Z

XA2010
"EUROPEAN CONFERENCE ON
COMPUTER SCIENCES & APPLICATIONS"
3rd Edition
Timi ?oara, Rom?nia, September 24-25, 2010
AIM
The 3rd European Conference on Computer Sciences and Applications
intends to stimulate the research activity and to establish
interactions between Romanian and foreign researchers, teachers, B.

Nine papers published by Geometry & Topology Publications

2010-03-14T02:24:30Z

Seven papers have been published by Algebraic & Geometric Topology
(1) Algebraic & Geometric Topology 10 (2010) 525-530
Faithfulness of a functor of Quillen
by William G Dwyer, Andrei Radulescu-Banu and Sebastian Thomas
URL: [link]
DOI: 10.2140/agt.2010.10.525

Re: decomposing rationals

2010-03-12T22:06:54Z

If a and b are rational,
this becomes an integer linear programming problem in two variables.
Such things are solvable in polynomial time.
I don't remember details.
I think that you can solve the linear relaxation
and plug away with Gomory fractional cuts.
Selecting the approximations could be tricky.

Re: minimum height of a 0-1 simplex

2010-03-12T19:13:15Z

The above is roughly 1/n, but there are worse examples:
e_j for j in 1..k
k
SUM e_j + e_L for L in k+1..n
j=1
k n
The vertices satisfy SUM x_j + (1-k) SUM x_j = 1
j=1 j=k+1
For k=2n/3, the distance to zero is roughly sqrt(27/(4n**3)).

Call for Papers Reminder (extended): The World Congress on Engineering WCE 2010

2010-03-12T10:42:22Z

CFP: The World Congress on Engineering WCE 2010
WCE 2010: London, U.K., 30 June - 2 July, 2010
[link]
Draft Paper Submission Deadline (extended): 18 March, 2010
The WCE 2010 is organized by International Association of Engineers
(IAENG), a non-profit international association for the engineers and

Re: minimum height of a 0-1 simplex

2010-03-10T19:48:09Z

That is what I get, also.
The base satisfies the linear equation
n-1
SUM x[j] + (2-n)*x[n] = 1
j=1
Lower bound, anyone?

Re: minimum height of a 0-1 simplex

2010-03-10T12:37:27Z

I'm not sure what this means,
but I think that it is not true even in three dimension.
that a minimal-height-defining line segment of a
tetrahedron lies within the tetrahedron.
Consider a tetrahedron with the following vertices:
(0, 0, 0), (3, 2, 0), (3, -2, 0) and (4, 0, 1).
Its minimal-height-defining segment is (4, 0, 1)_(4, 0, 0).