Leaf-to-Leaf path distances in Complete Tree Graphs
Andrew M. Goldsborough , S. Alex Rautu , Rudolf A. Römer
Department of Physics and Centre for Scientific Computing
The University of Warwick
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Motivation - Correlation Functions
In tensor networks, correlation functions are related to path length through
the network.
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What about in a tree?
A regular tree tensor network (TTN) has the geometry of a complete
binary tree
Can we find an analogue to the correlation functions that can be solved
analytically?
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Definitions
Level n
Root node
Vertex
1
2
3
4
Separation r
Leaf
Binary, as each vertex has two daughters.
Complete, as all of the leaves are the same depth (number of vertices to
the root).
Let the path length d be the number of vertices that connects two leaves.
Let the system Length L be the number of leaves. L = 2n .
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The problem
We want to find the average path length An (r) for a given separation r in
a tree with n levels.
Let Pn (r) be the sum of the path lengths connecting leaves of separation
r in a tree with n levels.
For separation r there are 2n − r possible paths. Thus:
An (r) =
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Pn (r)
2n − r
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n = 4, r = 1
r=1
1
3
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n = 4, higher r
r=1
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Pn(r)
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n = 4, higher r
r=1
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Look at just the maximal paths
n-1
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n-1
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Look at just the maximal paths
n-1
Pn (r) =
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n-1
2Pn−1 (r) + (2n − 1)r
(2n − 1)(2n − r)
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r < 2n−1
r ≥ 2n−1
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Recurrence relation
n-1
Pn (r) =
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n-1
2Pn−1 (r) + (2n − 1)r
(2n − 1)(2n − r)
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r < 2n−1
r ≥ 2n−1
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Solving the recursion relation
Pn (r) = 2Pn−1 (r) + (2n − 1)r
2Pn−1 (r) = 22 Pn−2 (r) + 2(2n − 3)r
22 Pn−2 (r) = 23 Pn−3 (r) + 22 (2n − 5)r
..
.
2ν Pn−ν (r) = 2ν+1 Pn−ν−1 (r) + 2ν (2(n − ν) − 1)r
..
.
2n−nc Pnc (r) = 2n−nc +1 Pnc −1 (r) + 2n−nc (2nc − 1)r
summed gives
Pn (r) = 2n−nc +1 Pnc −1 (r) +
n−n
Xc
2i (2(n − i) − 1)r
i=0
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Solving the recursion relation
The first part is from the non-recursive part of the equation for Pn (r):
Pnc −1 (r) = (2nc − 3)(2nc −1 − r)
When the final bracket is expanded, it is simply the sum of three
geometric series:
n−n
Xc
2i (2(n − i) − 1)r = rn
i=0
n−n
c +1
X
2i − r
i=1
n−n
Xc
i=1
i2i+1 − r
n−n
Xc
2i
i=0
= r(2n−nc +1 (2nc + 1) − (3 + 2n))
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Finding nc
From the limits r < 2n−1 :
n > log2 r + 1
The critical value for n is the smallest integer that satisfies this condition
or:
{nc ∈ Z : log2 r + 1 < nc ≤ log2 r + 2}
hence:
nc = blog2 rc + 2 ≡ q + 2
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The result
Putting it all together:
2r
Pn (r) = 2n 2q + 1 + q − (3 + 2n)r
2
1
2r
n
2 2q + 1 + q − (3 + 2n)r
An (r) = n
2 −r
2
In the limit of n → ∞:
lim An (r) = 2q + 1 +
n→∞
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2r
2q
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Results
50
n = 20
n→∞
40
An(r)
30
20
10
0
1
10
100
1000
10000
1e+05
1e+06
|r|
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What about ternary trees?
Ternary trees have three daughters at each vertex
P3,n (r) =
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3P3,n−1 (r) + 2(2n − 1)r
(2n − 1)(3n − r)
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r < 3n−1
r ≥ 3n−1
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What about m-ary trees?
Pm,n (r) =
r < mn−1
r ≥ mn−1
mPm,n−1 (r) + (m − 1)(2n − 1)r
(2n − 1)(mn − r)
Pm,n (r) = m 2qm + 1 +
n
2r
q
m
m (m − 1)
1+m
− 2n −
1−m
r
1
2r
1+m
n
Am,n (r) = n
m 2qm + 1 + qm
− 2n −
r
m −r
m (m − 1)
1−m
lim Am,n (r) = 2qm + 1 +
n→∞
2r
mqm (m
− 1)
where qm = blogm rc.
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Results for n → ∞
50
m=2
m=3
m=5
m = 10
m = 50
m = 100
40
An(r)
30
20
10
0
1
10
100
1000
10000
1e+05
1e+06
|r|
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Variance
Want a measure of uncertainty for the averages. Variance can be defined
as
Qm,n (r)
Pm,n (r) 2
Vm,n (r) = n
−
m −r
mn − r
Where Qm,n (r) is the sum of the squares of the path lengths:
mQm,n−1 (r) + (m − 1)(2n − 1)2 r
r < mn−1
Qm,n (r) =
2
n
(2n − 1) (m − r)
r ≥ mn−1
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Variance
In the end:
Vm,n (r) =
h
4r
2n
qm
m
m
(m
+
1)
−
r
m2qm (mn − r)2 (m − 1)2
− mn m2qm +2 (n − qm )2 − m2qm +1 (2n2 − n(4qm + 2)
+ 2qm (qm + 1) − 1) + m2qm (qm − n + 1)2
i
− mqm (2n − 2qm − 1)(m − 1)r + m2qm +1 r
and
lim Vm,n (r) =
n→∞
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4r (mqm (m + 1) − r)
m2qm (m − 1)2
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Variance for binary trees
8
Vn(r)
6
4
2
n = 20
n→∞
0
1
100
10000
1e+06
|r|
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Various m with n → ∞
10
9
8
m=2
m=3
m=5
m = 10
m = 50
m = 100
7
Vn(r)
6
5
4
3
2
1
0
1
100
10000
1e+06
|r|
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General raw moments
The k-th raw statistical moment is:
E[X k ] = Am,k,n (r) =
0
Pm,k,n (r)
mn − r
0
Pm,k,n (r) = mn−nc (2n0c − 1)k (mnc − r)
1
1
0
+ r(m − 1)(−2)k Φ m, −k, − n − mn−nc Φ m, −k, − n0c
2
2
h
0
0
lim Am,k,n (r) = m−nc (2n0c − 1)k (mnc − r)
n→∞
1
0
k
r(m − 1)(−2) Φ m, −k, − nc
2
where n0c = nc − 1 = q + 1 and Φ is the Hurwitz-Lerch transcendent or
Hurwitz-Lerch zeta-function.
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Periodic Systems
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General raw moments for periodic systems
P BC
OBC
OBC
Pm,k,n
(r) = Pm,k,n
(r) + Pm,k,n
(mn − r)
0
0
0
0
= mn−nc (2n0c − 1)k (mnc − r) + mn−ñc (2ñ0c − 1)k (mñc − (mn − r))
1
1
n−n0c
0
n
Φ m, −k, − nc
+ (m − 1)(−2) m Φ m, −k, − n − rm
2
2
k
1
0
− (mn − r)mn−ñc Φ m, −k, − ñ0c
2
where n0c = blogm rc + 1 as before and ñ0c = blogm (2n − r)c + 1.
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Conclusions
We can find an analytic expression for average path length for a given
separation for binary trees.
This can be generalised to m-ary trees, periodic boundaries and
general raw statistical moments.
The general equation can be concisely written in terms of the
Hurwitz-Lerch Transcendent.
Maths is fun!
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Thank you!
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