SLIDES AND AUDIO UPDATED! embedded on my site: http://alejandroerickson.com/joomla/research-projects/13-slidecast-negative-correlation-properties-for-graphs A practice run for a presentation in combinatorics. The effective conductance between two points in a network of resistors should not decrease if any of the conductances of the resistors is increased. This is equivalent to the following. If I select a spanning tree with a certain probability, then the chances that my tree contains a desired edge e is not increased if I choose only among spanning trees already containing any other edge, f. We do not know, however, if the same negative correlation property holds for spanning forests of graphs and this talk is about some of the efforts towards that end.

2. Negative correlation properties
in graphs or: How I learned to stop
looking for the edge e among
spanning trees already containing f
and instead look among all
spanning trees.
Alejandro Erickson
Master’s thesis work while at the
Department of Combinatorics and Optimization
University of Waterloo
April 11, 2010

3. Outline
Outline
Kirchhoff’s law and Rayliegh monotonicity
Kirchhoff’s Law
Rayleigh monotonicity
The combinatorics!
Example
Previous work
Rayleigh condition for other stuff
Forest Rayleigh is equivalent to negative correlation
Evidence for forest Rayleigh property
Current work
Series Parallel graphs and 2-sums

4. Kirchhoff’s law
Electrical network of resistors, each with conductance
yg .
a e
e has resistance re ,
conductance ye = r1 e
b
Kirchhoff’s law gives a formula for the conductance
between nodes a and b.

5. Model the network as a graph G (shocking!) and let T
be the generating polynomial for spanning trees of G .
T (G ) = + + +
+ + + +
= ye yf yh + ye yf yg + ye yg yh + yf yg yh
+ ye yh yi + yf yg yi + ye yf yi + yg yh yi
a e
i g
f
h b

6. Let G /ab be G with nodes a and b identified.
Kirchhoff’s Law
T T (G )
ab
T (G /ab)
= =
T( )
a e
i g
f
h b

7. Rayleigh monotonicity
Lord Rayleigh (1842-1919)
observed that increasing
the conductance of any
edge should not decrease
the effective conductance
of the whole network.
T (G )
ab
T (G /ab)
=
ab is non-decreasing
in the direction
of every variable ye .
For any edge e, we have
∂
ab ≥ 0
∂ye

8. Three bits of notation:
T g denotes evaluation at yg = 0. T g = T (G g ).
Tg denotes partial derivative w.r.t yg . Tg = T (G /g ).
Add an edge, G G + f where f = ab.
1
old G

9. Three bits of notation:
T g denotes evaluation at yg = 0. T g = T (G g ).
Tg denotes partial derivative w.r.t yg . Tg = T (G /g ).
Add an edge, G G + f where f = ab.
Now, Kirchhoff’s law is1
T (G ) Tf
ab
T (G /ab) Tf
= =
and the Rayleigh property is that
Tef Tf − T f Tef
≥0
(Tf )2
for each distinct pair of edges e and f and positive yg s
1
old G

10. Forget all that stuff about electrical networks.

11. Selecting trees
Notice that
yg Tg are the spanning tress containing g
T g generates those ones not containing g .

12. Selecting trees
Notice that
yg Tg are the spanning tress containing g
T g generates those ones not containing g .
We select a spanning tree X with probability
proportional to g ∈X yg , with positive yg s.

13. Selecting trees
Notice that
yg Tg are the spanning tress containing g
T g generates those ones not containing g .
We select a spanning tree X with probability
proportional to g ∈X yg , with positive yg s.
The chances our tree contains e are
ye Te
T

14. Selecting trees
Notice that
yg Tg are the spanning tress containing g
T g generates those ones not containing g .
We select a spanning tree X with probability
proportional to g ∈X yg , with positive yg s.
The chances our tree contains e are
ye Te
T
If we restrict ourselves to trees already containing f ,
then the chances our tree contains e are
ye yf Tef
yf Tf

15. Equivalent conditions
Obvious but important:
T consists of those terms not containing g and those
containing g . That is
T = T g + yg Tg

16. Equivalent conditions
Obvious but important:
T consists of those terms not containing g and those
containing g . That is
T = T g + yg Tg
Back to the Rayleigh condition
(Te )Tf − ( T ) Tef
=(Tef + yf Tef )Tf − ( T f + yf Tf ) Tef
(Rayleigh) =(Tef )Tf − ( T f ) Tef ≥ 0

17. So what!?!
We showed that
Te Tf − TTef = Tef Tf − T f Tef ≥ 0
The missing piece
Te Tf − TTef ≥ 0 if and only if
ye yf (Te Tf − TTef ) ≥ 0 if and only if
ye Te ye yf Tef
≥
T yf Tf
So the chances of selecting a spanning tree with e are
not increased by choosing among those already
containing f !

18. Time for an example
a e
i g
f
h b
T = + + + + + + +
Te = + + + +
Tf = + + + +
Tef = + +
and ye yf (Te Tf − TTef ) = ye yg yh × yf yg yh = ×

19. Where is the proof?
The classical proof using electrical networks is
printed in Grimmett’s book.
The most often cited proof is due to Brooks Smith
Stone and Tutte (1940).
A stronger property was shown by Choe and
Wagner (2006).
A combinatorical (bijective) proof is given by
Cibulka, Hladky, LaCroix and Wagner (2008).
So if this stuff has been done over at least four times,
what’s all the fuss?

20. Mathematicians love variations!
Let’s replace T by the spanning forests, F .
This was proposed in print in the early 90s.
Considerable evidence has been published but, as of
yet, no proof that
Fe Ff − FFef ≥ 0
for positive yg s and pair of distint edges e and f .

21. A “weaker” version
Special case of Rayeligh, Fe Ff − FFef
Set each yg to 1.
ie, choose spanning forest uniformly at random.
Special case ≡ Rayleigh
(independently: Cocks and E., 2008)
All graphs are forest Rayleigh iff
all graphs satisfy the special case.
proof idea: Suppose a graph is not Rayleigh, then
Fe Ff − FFef < 0 for certain yg s. Replace edges by
certain disjoint paths to create a graph that is not
negatively correlated.

22. Evidence for the conjecture
– Small graphs
are negatively correlated
(Grimmett, Winkler, 2004).
– Two-sums
of Rayleigh graphs
are Rayleigh (Wagner,
Semple, Welsh 2008)
Smaller graphs are
Rayleigh (E., Wagner, 2008)
Series parallel graphs are
Rayleigh (E., Wagner, 2008)

23. SOS conjecture (Wagner)
The spanning forest Rayleigh difference,
∆F {e f } = Fe Ff − FFef
is a sum of monomials times squares of polynomials,
∆F {e f } = yS A(S)2
S

24. SOS conjecture (Wagner)
The spanning forest Rayleigh difference,
∆F {e f } = Fe Ff − FFef
is a sum of monomials times squares of polynomials,
∆F {e f } = yS A(S)2
S
The Rayleigh property, Fe Ff − FFef ≥ 0 for positive yg s,
follows immediately.
One major hangup: the signs of the terms in A(S) are
unknown.

25. S-sets and A-sets
∆F {e f } = yS A(S)2
S
An S-set is a set of edges S so that S ∪ {e f } is
contained in a cycle.
The A-sets of S are those spanning forests A so that
A ∪ {e f } contains a unique cycle which contains S.

26. S-sets and A-sets
∆F {e f } = yS A(S)2
S
An S-set is a set of edges S so that S ∪ {e f } is
contained in a cycle.
The A-sets of S are those spanning forests A so that
A ∪ {e f } contains a unique cycle which contains S.
e f

27. S-sets and A-sets
∆F {e f } = yS A(S)2
S
An S-set is a set of edges S so that S ∪ {e f } is
contained in a cycle.
The A-sets of S are those spanning forests A so that
A ∪ {e f } contains a unique cycle which contains S.
e f

28. S-sets and A-sets
∆F {e f } = yS A(S)2
S
An S-set is a set of edges S so that S ∪ {e f } is
contained in a cycle.
The A-sets of S are those spanning forests A so that
A ∪ {e f } contains a unique cycle which contains S.
e f

29. S-sets and A-sets
∆F {e f } = yS A(S)2
S
An S-set is a set of edges S so that S ∪ {e f } is
contained in a cycle.
The A-sets of S are those spanning forests A so that
A ∪ {e f } contains a unique cycle which contains S.
e f

30. Given an S-set, S with S ∪ {e f } contained in a cycle C ,
A(S) = c(S e f C )yA−S
A
There they are! The signs c(S e f C ). And there are
MANY of them.

31. Testing on small graphs
Wagner had some guesses for the signs and we tested
yS A(S)2 = Fe Ff − FFef
S
in Maple, for graphs up to 7 vertices.
He also found signs that worked for the
cube and Möbius ladder on 8 vertices.
Necessary conditions
Next, we “show” the SOS-conjecture
should hold for two sums and that
it does hold for series parallel graphs.

32. Series parallel graphs and 2-sums
My presentation The details

33. Suppose G = H ⊕g K and let C be a cycle of G . Then
either C is contained in H − g or K − g or
C = CH ∪ CK − g for cycles through g in H and K .

34. Suppose G = H ⊕g K and let C be a cycle of G . Then
either C is contained in H − g or K − g or
C = CH ∪ CK − g for cycles through g in H and K .
Facts about 2 sums and ∆F {e f } = Fe Ff − FFef
If e ∈ H and f ∈ K , then
∆F (G ){e f } := ∆F (H){e g }∆F (K ){f g }
If e f ∈ H, then ∆F (G ){e f } = F (K )2 ∆F (H){e f }
The Rayleigh difference factors over the factors of the
2-sum.

35. How does the SOS-form factor?
yS A(S)2
S
is all about the cycles of G , through e and f
If e ∈ H and f ∈ K , these cycles come from
C = CH ∪ CK − g .
∆F (G ){e f } := ∆F (H){e g }∆F (K ){f g }
In the same way, A-sets of G come from A-sets of H
and K , so the SOS-form factors.
H e K
g f

36. How does the SOS-form factor?
Show yS A(S)2 = F (K )2 ∆F (H){e f } = F (K )2 ySH AH (SH )2
S SH
If e f ∈ H, we sum over
S-sets in H not containing g .
careful that A-sets of H and forests of K do not
form extra cycles in G .
f
e g

37. How does the SOS-form factor?
Show yS A(S)2 = F (K )2 ∆F (H){e f } = F (K )2 ySH AH (SH )2
S SH
If e f ∈ H, we sum over
S-sets in H not containing g .
careful that A-sets of H and forests of K do not
form extra cycles in G .
f
e g

38. How does the SOS-form factor?
Show yS A(S)2 = F (K )2 ∆F (H){e f } = F (K )2 ySH AH (SH )2
S SH
If e f ∈ H, we sum over
S-sets in H not containing g .
careful that A-sets of H and forests of K do not
form extra cycles in G .
S-sets containing g in H and edges of K .
snag! The forests we use from K need to make a
unique cycle with g and satisfy another SOS form.
f
e g
(K g − Kg )Kg = Q yQ B(Q)2 ?

39. Series parallel graphs.
∆-SOS
∆F {e f } = Fe Ff − FFef = yS A(S)2
S
Φ-SOS
ΦF {g } = (F g − Fg )Fg = yQ B(Q)2
Q
If K is series parallel then it is Φ-SOS.
Hope for 2-sum
K is ∆-SOS by inductive hypothesis. Can we show that
if K is ∆-SOS, then it is Φ-SOS?
This reduces to yet a third “SOS” form (see paper).

40. Summary
Goal: Prove
∆F {e f } = Fe Ff − FFef ≥ 0
Method: Prove
Fe Ff − FFef = yS A(S)2
S
Next step: Prove
ΦF {g } = (F g − Fg )Fg = yQ B(Q)2
Q
Many thanks to David Wagner and my classmates
from Waterloo for their ideas and encouragements.