uniform distribution for a class of k-paradoxical oriented graphs
Post on 06-Jan-2016
32 Views
Preview:
DESCRIPTION
TRANSCRIPT
Uniform distribution for a class of k-paradoxical oriented graphs
Joint work with undergraduate students J. C. Schroeder and D. J. Pleshinger (Ohio Northern University, 2012)
With many thanks to the Miami University Fall Conference (2012), where the chosen topic was “Statistics in Sports”. Not being a statistician and definitely wanting to attend and present something, I had to find a suitable topic on short notice. I thought that distribution properties of the dominating sets in the Paley tournaments might not be too far from the conference theme (although some may beg to differ ). Eventually two Ohio Northern University seniors who like to visualize and draw nice pictures of graphs joined me in this investigation, and presented at the 2013 Joint Meetings.
Objects of interest: oriented graphs (no 2 – cycles, no loops, no multiple edges).
(" ")x y x dominates y
Notation:
(" ") if for all x S x dominates S x z z S
( ) { | } (note that ( ) )D S x x S D S S
: ( ) whenever S is a set of vertices with | | .k paradoxical D S S k
1 – paradoxical
2 – paradoxical
Featured in the paper "On a problem in graph theory“ By P. Erdős (The Mathematical Gazette (1963) 47: 220–223)
1
2 mod 7
4
x x
x x
x x
Paley tournaments
A Paley tournament on p = 11 verticesSource:
www.ams.jhu.edu/~leslie/paley.eps
2
: F where 3 (mod 4) ( ) has
: where and 1 ( 1) 2p
p
vertices pG p
edges x x t x F t p
Paley tournament on p = 23 vertices
Existence of k – paradoxical tournaments:If k is fixed, then large enough n, most tournaments on n vertices are k –paradoxical.
1 random tournament with vertices, | | , Pr 1
2kT n S k x A S x S
1Probability that nobody deafeats all teams from : Pr ( for every ) = 1
2
n k
kS x S x A S
is not paradoxical: for some , there is no team in defeating all teams from .kT k S P A A S S
: | | and S S k x A S x S
Probability of a randomly selected tournament T being paradoxicalnot k
exp
1Pr 1 0 ( fixed, )
2k
n k
kS P A
poly
nx A S x S k n
k
Erdős (1963): Non-constructive, probabilistic proof.
Sketch of proof
Graham and Spencer (1971) used Paley tournaments G(p) and Weil estimates to provide explicit examples of k – paradoxical tournaments. In , if with , then ( ) for an absolute implied constant.
2p k
pG p S F S k D S O k p
Proof - preliminaries
: multiplicative character of order
: 1 , where of degree , not a perfect power
q
q
x F q
F rWeil estimates P x d q
P x F x d r th
1 2 1 2
1 2 1 2
odd prime, 2, 1 mod ; : multiplicative character of order ; 1;
2, ,..., exp ,0 1 ; , ,..., for ;
; ; , ,..., ; , ,..., : # |
p
t r t p i j
t t p
p r p r F r t
k iU k r d d d F d d i j
r
N N p d d d x F x
for 1, 2,...,i id i t
APPLICATION OF WEIL's ESTIMATES TO DISTRIBUTION OF POWER RESIDUES
THEOREM 1: , with an absolute implied constant.t
pN O t p
r
2
A MORE EXPLICIT BOUND:
( ) is paradoxical if 4kG p k p k
1 2, ,...,
and 1 for 1,...,
k
ip
S a a a
a xx S x F S i k
p
1 2( ) # 1 for 1, 2,..., ; ; , ,..., ; 1, 1,..., 1| ip k
x aD S x F i k N p a a a
p p
( ) with an absolute implied constant2k
pD S O k p
Proof
Or and 1 for 1,..., , since 3 mod 4ip
x ax F S i k p
p
With the notation from THEOREM 1,
( ) if is fixed and is large.2k
pD S k p
Success of the Paley construction (in providing an explicit example for an indirect, probabilistic result)
explained (after the fact) by the - quasi random character of the sequence of quadratic residue .
Chung and Graham (1992, - , Journal Of Combinatorial Theory A)N
s
Quasi random subsets of
1 2 3 1Historical: Harold Davenport in 1930's considered the 1 sequences of interest: , , ,...,
p
p p p p
Global symmetry: L/R symmetric if p is of the form 4k+1, anti-symmetric if p is of the form 4k+3.
Cumulative sums, p = 17489
1( ) behaves as a quasi random subset of of density
2 2pk k
pD S F
Cumulative sums, p = 17491
Distribution of the dominating set ( ) in Paley tournaments ( )D S G p
Intuitive understanding: if , and | , then we expect
2 2
p p
k k
S F S k I x F p x p
I pD S I I
Method of proof: similar with the classical Graham and Spencer proof, only that instead of
Weil estimates we will use estimates for .incomplete character sums with polynomial arguments
1
THEOREM 2 : log N H
x N
Burgess estimates P x K p p
: multiplicative character of order , of degree , not a perfect power
holds uniformly for and , where depends on the degree of ( ) only.
q qF r P x F x d r th
N H K P x
0 5
6
: G(103) with 0,5,6
11,27,45,48,53,75,80,90,94,95,101
EXAMPLE D
3
10311
2D S
Uniform distribution Paley Tournaments
1,..., pI N N H F
3 mod 4p
THEOREM 2
The number of elements of the dominating set in the interval is given by
log , where the implied constant de2k
Uniform distribution for the dominating sets in Paley tournaments
D S I
HD S I O p p
pends on only.k
1
1 11 , where
2 2
ki
k kx I S i
x aD S I A B
p
1 1
1 , and 1k k
i i
x I x I Si i
x a x aA B
p p
1 2, ,..., kS a a a
1 2
1 21 1 ...
1
The main term satisfies
...1 1 = log log
(the implied constant depends on only), while the 'small term' satisfies 2 .
k
t
ki i it
x I t i i i k
k
A
x a x a x aA I O p p H O p p
p
k B B
Thus log , where the implied constant depends on only.2k
HD S I O p p k
Proof
log , where the implied constant depends on only.2k
HD S I O p p k
1Corollary: If is fixed and log for some 0 then
1, 2,...,2k
k H p p
H HD S I D S N N N H D S
p
exactly what one should expect under the hypothesis of a random tournament
paradoxical oriented graphs with a (relatively) small number of edgesk
From Dirichelet's Theorem: there are infinitely many primes
of the form 4 2 1 (note that gcd(4 ,2 1) 1)
p
p mq q q q
1DEFINITION. Let 1, 0 fixed. Let 1, odd, with .k q q
q
Observation: since is odd, any such prime satisfies 3 mod 4 , so is defined.q p G p
2Let : a multiplicative character of order exactly 2 .p qF U q
Define the oriented graph as follows
Vertex set
Edges: where and a nonzero power of order 2 .
q
p
p p
G p
F
x x t x F t F q
=1.x y iff y x
Dominating sets for G : | 1 for all q q pp D S x F y x y S
3 mod 4 , and any nonzero power of order 2 is also a nonzero perfect squarep q
subgraph of the Paley tournament qG p G p
1REGULARITY: there are 2 1 emerging edges per vertex
2
(and the same # of incoming edges)
pm
q
Small (relatively) number of edges:
1=
2q
E G pp pE G p E G p
q q
3:(43).
It has 43 vertices, out degree 2 1 7 each.
Example: the oriented regular graph
3, 4 2 1 43.
G
m
q m p mq q
3(127) fragment
127 vertices, out degree 2 1 21 each.
Shown: vertices emerging from 4, 37, 77, 98
Example:
3, 10, 4 2 1 127.
G
m
q m mq q
THEOREM 3: If , are fixed, and is odd, then for all large enough primes in the arithmetic progression
4 2 1, the oriented graphs are paradoxical. Moreover, for every integer interval q
k q q p
p mq q G p k
1, 2,..., modulo , we have log where the implied 2
constant depends on and only.
q k
pI N N N H p D S I O p p
q
k q
Main uniform distribution result for the oriented graphs qG p
Therefore the oriented graphs are paradoxical for all large enough 2 1 mod 4 qG p k p q q
Indeed log 0 if is large enough and log for some 0.2
k
pO p p p H p p
q
PROOF: similar to that of THEOREM 2 only that Burgess' character sum estimates
are used with a character of order 2 instead of a quadratic character.q
2 1
1 211
1,if 1KEY: if , ,..., , the quantity 1 is
0, else 2
qkj q
p p k ikji
x D Sx F S F a a a a x
q
COROLLARY: For every positive integer and every 0 there exists a paradoxical subgraph of a Paley tournament,
with a number of edges less than a fraction of out of the number of edges of the un
k k
derlying tournament.
NOTE: are quasirandom subsets of in the sense described by F.R.K. Chung and R. Graham
since the incomplete exponential sums with nontrivial additive characters evaluate as .q
q p
x D S
D S
x o p
A closer look: fragment of a 2-paradoxical oriented graph with 67 vertices. Outgoing edges from x to x+1, x+9, x+14, x+15, x+22, x+24, x+25, x+40, x+59, x+62, x+64 for any x modulo 67. For better visibility, only the outgoing edges from vertices 0,1,…,19 are shown, with the ones emerging from vertex 0 marked in red. This is a subgraph of the Paley tournament G(67) , with one-third the number of edges.
PART II: ALTERNATIVE METHODOLOGY, BEYOND POWER RESIDUES
Any good (pseudo )random tournament construction is bound to provide similar examples of paradoxical tournaments. k
The quasi random behavior of power residues well known ( . . Chung and Graham, - ,
Journal of Combinatorial Theory 61(1992)). Not surprisingly, the Graham and Spencer constructNe g Quasi random subsets of
ion works.
OUR ALTERNATIVE : uses one of the animating ideas of the undergraduate research program
at ONU: the greatest prime factor sequences ('GPF sequences')
0 1 10
0 1 1 2 2
GPF sequence of order : A prime sequence with , ,..., given
and ... for ( , not all zero)
n kn
n n n k n k j
k q q q q
q gpf a a x a x a x n k a
THE GPF CONJECTURE: Every GPF sequence is .ultimately periodic
1 Proved for 1 in the special case ( ) with | .n nk q gpf aq b a b
1 2 Proved for 2 in the special case ( )
(' ', G. Back and M. Caragiu, Fibonacci Quarterly, 2010).
Unique limit cycle 7,3,5,2
n n nk q gpf q q
GPF Fibonacci
1 Higher dimensional analogues investigated ( ' ' applies componentwise).
are prime vectors. A a nonnegative square matrix. UP proved in some special cases.
Computational evidence: U
n n
n
Q gpf AQ B gpf
Q
P appears to hold true in general.
1123 1n nq P q
0I. 658545674551q 0II. 6599q
LC: 587, 2777, 1109, 59, 191, 691, 467, 373, 37, 569, 17497, 31649, 1259, 7039, 2243
LC: 419, 353, 167, 10271, 631667, 251, 359, 22079, 2237, 593, 521, 433, 2663, 6551, 5519, 10949, 2371, 563, 277, 4259, 2543, 1009, 71, 397, 109
Example ( 1, multiple limit cycles)k
( 3, )EXAMPLE k GPF Tribonacci
1 2 3n n n nq gpf q q q
We found four distinct GPF-Tribonacci limit cycles, of lengths 100, 212, 28 and 6
1 126390 1103 , 2n nq gpf q q Maximum cycle element: 18964967822676015504193
Logarithmic plot
Pseudo-random 1 strings from GPF sequences
1, if 3 mod 4
1, if 2 or 1 mod 4i
ii i
qu
q q
A general construction of tournaments from sets of positive integers
1 2 1 2, ,...., define the tournament , ,...., as follows
Vertex set 1, 2,....,
, if gpf 3 mod 4 Edges: for 1 , we have
, otherwise
n n
i j
r r r T r r r
n
j i r ri j n
i j
Using GPF sequences in the above construction
1 2 1 2
1 2
METHOD 2: If , ,..., are GPF sequences, we produce tournaments , ,...,
Extensive computational evidence shows in general low cross correlations between , ,..., .
n ni i i i i i i
ni i i
q q q n T T q q q
q q q
1 1METHOD 1: If is a GPF sequence with a large period, we produce tournaments , ,...,i k k k k nq T T q q q
Preserve the standard order unless
the greatest prime factor of the sum
of the labels is congruent to 3 mod 4
,1
,21
,
,1 ,2 ,
METHOD 3: If = is an dimensional GPF sequence ,
we may use , , , . Again, extensive computational evidence shows in general
low cross correlati
i
ii i i
i d
i i i i n
q
qQ n Q gpf AQ B
q
T T q q q
,1 ,2 ,ons between , , , .i i i ni i iq q q
2
3
5
7
1113
17
19
23
top related