Transcript Slide 1
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).
Notation:
x y ("x dominates y")
x S (" x dominates S ") if x z for all z S
D(S ) {x | x S} (note that D(S ) S )
k paradoxical : D( S ) whenever S is a set of vertices with | 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)
x x 1
x x 2 mod 7
x x 4
Paley tournaments
vertices : Fp where p 3 (mod 4)
G( p) has
2
edges
:
x
x
t
where x Fp and 1 t ( p 1) 2
A Paley tournament on p = 11 vertices
Source: www.ams.jhu.edu/~leslie/paley.eps
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.
Erdős (1963): Non-constructive, probabilistic proof.
Sketch of proof
T random tournament with n vertices, |S | k , x A S Pr x
S 1
1
2k
1
Probability that nobody deafeats all teams from S : Pr (x
S for every x A S ) = 1 k
2
nk
T is not k paradoxical: for some S Pk A , there is no team in A S defeating all teams from S.
S : |S | k and x A S x
S
Probability of a randomly selected tournament T not being k paradoxical
n
1
Pr x A S x
S 1 k
SPk A
k 2
poly
exp
nk
0 (k fixed, n )
Graham and Spencer (1971) used Paley tournaments G(p) and Weil
estimates to provide explicit examples of k – paradoxical tournaments.
In G p , if S Fp with S k , then D(S )
p
O k p for an absolute implied constant.
2k
Proof - preliminaries
Weil estimates:
P x d 1
xFq
: Fq multiplicative character of order r
q , where
P x Fq x of degree d , not a perfect r th power
APPLICATION OF WEIL's ESTIMATES TO DISTRIBUTION OF POWER RESIDUES
p odd prime, r 2, p 1 mod r ; : F multiplicative character of order r ; t 1;
p
2 k i
1 , 2 ,..., t U r exp
, 0 k r 1 ; d1 , d 2 ,..., dt Fp d i d j for i j ;
r
N N p; ; d1 , d 2 ,..., dt ; 1 , 2 ,..., t : # x Fp | x di i for i 1, 2,..., t
THEOREM 1: N
p
O t p , with an absolute implied constant.
rt
Proof
S a1 , a2 ,..., ak
a x
x S x Fp S and i
1 for i 1,..., k
p
x ai
Or x Fp S and
p
1 for i 1,..., k , since p 3 mod 4
With the notation from THEOREM 1,
x ai
D( S ) # x Fp |
p
D( S )
1
for
i
1,
2,...,
k
N
p
;
;
a
,
a
,...,
a
;
1,
1,...,
1
1 2
k
p
p
O k p with an absolute implied constant
2k
D( S )
p
if k is fixed and p is large.
k
2
A MORE EXPLICIT BOUND:
G( p) is k paradoxical if p k 2 4k
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 residues.
Chung and Graham (1992, Quasi - random subsets of
D( S )
N
, Journal Of Combinatorial Theory A)
p
1
behaves as a quasi random subset of Fp of density k
k
2
2
1 2 3 p 1
Historical: Harold Davenport in 1930's considered the 1 sequences of interest: , , ,...,
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
Cumulative sums, p = 17491
Distribution of the dominating set D( S ) in Paley tournaments G( p)
Intuitive understanding: if S Fp , S k and I x Fp | p x p , then we expect
DS I I
I
2k
p
2k
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.
THEOREM 2 Burgess estimates :
N H
P x K
p log p
x N 1
: Fq
multiplicative character of order r, P x Fq x of degree d , not a perfect r th power
holds uniformly for N and H , where K depends on the degree of P( x) only.
EXAMPLE : G(103) with D 0,5, 6
11, 27, 45, 48,53, 75,80,90,94,95,101
0
5
6
103
D S 11 3
2
Uniform distribution Paley Tournaments
S a1, a2 ,..., ak
I N 1,..., N H Fp
p 3 mod 4
Proof
1
DS I k
2
THEOREM 2 Uniform distribution for the dominating sets in Paley tournaments
The number of elements of the dominating set D S in the interval I is given by
DS I
x ai
1
xI S i 1
p
k
x ai
A 1
xI i 1
p
k
H
O
2k
p log p , where the implied constant depends on k only.
1
k A B , where
2
k
x ai
,
and
B
1
xI S i 1
p
The main term A satisfies
H
O
2k
x ai x ai ... x ai
k
t
1
2
k
= I O p log p H O
A 1 1
p
xI
t 1
1i1 i2 ...it k
(the implied constant depends on k only), while the 'small term' B satisfies B 2k 1.
Thus D S I
p log p
p log p , where the implied constant depends on k only.
DS I
H
O
2k
p log p , where the implied constant depends on k only.
Corollary: If k is fixed and H
p log1 p for some 0 then
D S I D S N 1, N 2,..., N H
H H
D S
k
2
p
exactly what one should expect under the hypothesis of a random tournament
k paradoxical oriented graphs with a (relatively) small number of edges
DEFINITION. Let k 1, 0 fixed. Let q 1, q odd, with
1
.
q
From Dirichelet's Theorem: there are infinitely many primes p
of the form p 4mq 2q 1 (note that gcd(4q, 2q 1) 1)
Observation: since q is odd, any such prime satisfies p 3 mod 4 , so G p is defined.
Let : Fp U2q a multiplicative character of order exactly 2q.
Define the oriented graph Gq p as follows
Vertex set Fp
Edges:
x
x
t
where
x
F
and
t
F
a nonzero power of order 2q.
p
p
x y iff y x =1.
Gq p subgraph of the Paley tournament G p
p 3 mod 4 , and any nonzero power of order 2q is also a nonzero perfect square
p 1
2m 1 emerging edges per vertex
2q
(and the same # of incoming edges)
REGULARITY: there are
Small (relatively) number of edges:
E Gq p =
p p 1 E G p
E G p
2q
q
Dominating sets for G q p : Dq S x Fp | y x 1 for all y S
Example: the oriented regular graph G3: (43).
q m 3, p 4mq 2q 1 43.
It has 43 vertices, out degree 2m 1 7 each.
Example: G3 (127) fragment
q 3, m 10, 4mq 2q 1 127.
127 vertices, out degree 2m 1 21 each.
Shown: vertices emerging from 4, 37, 77, 98
Main uniform distribution result for the oriented graphs Gq p
THEOREM 3: If k , q are fixed, and q is odd, then for all large enough primes p in the arithmetic progression
p 4mq 2q 1, the oriented graphs Gq p are k paradoxical. Moreover, for every integer interval
I N 1, N 2,..., N H modulo p, we have Dq S I
p
2q
k
O
p log p where the implied
constant depends on k and q only.
Therefore the oriented graphs Gq p are k paradoxical for all large enough p 2q 1 mod 4q
Indeed
p
2q
k
O
p log p 0 if p is large enough and H p log p for some 0.
PROOF: similar to that of THEOREM 2 only that Burgess' character sum estimates
are used with a character of order 2q instead of a quadratic character.
KEY: if x Fp S Fp a1 , a2 ,..., ak , the quantity
2q
2 q 1
1,if x Dq S
j
1
a
x
is
i
j 1
i 1
0, else
k
1
k
COROLLARY: For every positive integer k and every 0 there exists a k paradoxical subgraph of a Paley tournament,
with a number of edges less than a fraction of out of the number of edges of the underlying tournament.
NOTE: Dq S are quasirandom subsets of
since the incomplete exponential sums
p
in the sense described by F.R.K. Chung and R. Graham
x with nontrivial additive characters evaluate as o p .
xDq S
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 k paradoxical tournaments.
The quasi random behavior of power residues well known (e.g. Chung and Graham, Quasi - random subsets of
Journal of Combinatorial Theory 61(1992)). Not surprisingly, the Graham and Spencer construction works.
OUR ALTERNATIVE : uses one of the animating ideas of the undergraduate research program
at ONU: the greatest prime factor sequences ('GPF sequences')
GPF sequence of order k: A prime sequence qn n0 with q0 , q1 ,..., qk 1 given
and qn gpf a0 a1 xn 1 a2 xn 2 ... ak xn k for n k (a j , not all zero)
THE GPF CONJECTURE: Every GPF sequence is ultimately periodic.
Proved for k 1 in the special case qn gpf (aqn1 b) with a | b.
Proved for k 2 in the special case qn gpf (qn 1 qn 2 )
('GPF Fibonacci ', G. Back and M. Caragiu, Fibonacci Quarterly, 2010).
Unique limit cycle 7,3,5,2
Higher dimensional analogues Qn gpf AQn 1 B investigated ( ' gpf ' applies componentwise).
Qn are prime vectors. A a nonnegative square matrix. UP proved in some special cases.
Computational evidence: UP appears to hold true in general.
N
,
Example (k 1, multiple limit cycles)
qn P 123qn1 1
I. q0 658545674551
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
II. q0 6599
LC: 587, 2777, 1109, 59, 191, 691, 467, 373,
37, 569, 17497, 31649, 1259, 7039, 2243
EXAMPLE (k 3, GPF Tribonacci)
qn gpf qn1 qn2 qn3
We found four distinct GPF-Tribonacci limit cycles, of lengths 100, 212, 28 and 6
qn gpf 26390qn1 1103 , q1 2
Maximum cycle element: 18964967822676015504193
Logarithmic plot
Pseudo-random 1 strings from GPF sequences
1, if qi 3 mod 4
ui
1, if qi 2 or qi 1 mod 4
A general construction of tournaments from sets of positive integers
r1 , r2 ,...., rn
define the tournament T r1 , r2 ,...., rn as follows
Vertex set 1, 2,...., n
j i, if gpf ri rj 3 mod 4
Edges: for 1 i j n, we have
i j , otherwise
Preserve the standard order unless
the greatest prime factor of the sum
of the labels is congruent to 3 mod 4
Using GPF sequences in the above construction
METHOD 1: If qi is a GPF sequence with a large period, we produce tournaments Tk T qk , qk 1,..., qk n1
METHOD 2: If qi1 , qi 2 ,..., qi n are n GPF sequences, we produce tournaments Ti T qi1 , qi 2 ,..., qi n
Extensive computational evidence shows in general low cross correlations between qi , qi ,..., qi .
1
2
n
qi ,1
q
i ,2
METHOD 3: If Qi = is an n dimensional GPF sequence Qi gpf AQi 1 B ,
qi ,d
we may use Ti T qi ,1 , qi ,2 ,
, qi ,n . Again, extensive computational evidence shows in general
low cross correlations between
q ,q , ,q .
i ,1 i
i ,2 i
i ,n i
2
23
3
19
5
7
17
13
11