Assignment 3 Solution

Description

1. Okay, let’s get back to the 6 networks we explored in the first assignment. Questions 2 through 4 will focus on them.

Measure the degree-degree assortativity. This is the standard Pearson correlation coefficient and the focus is on links, and then the nodes at the end of each link. For undirected networks, we need to think about how we choose the ordering of an edge’s two degrees when we perform the correlation. Which degree goes first? Or should we include both orderings? How about randomly choosing the ordering? Does it matter?

For directed networks, various correlations are possible (in-in, in-out, etc.). For this question, measure the correlation of the in-degree of the source node and the out-degree of the destination node for each link.

and becomes still broader as

! 1 , in which limit the

min

homo

1 by definition; for ! 0, Jmin=Jhomo ! 0, because t

number of iteration steps diverges as the minima becomes

vanishing

kl allow the remaining

kl ! 1.

less and less steep.

Figure 3(b) shows the behavior of minimal dissipati

1 presents a marginal case. The results of the

rate close to

1. For

smaller than 1, the relaxati

simulation suggest that the minimum is highly degenerate;

method only furnishes a local minimum, the Monte Ca

i.e., there are a large number of conductivity distributions

algorithm searching for the optimal tree topologies gi

2. Produce plots of the adjacency matrices.

lower dissipation values. The different values correspon

yielding the same minimal dissipation.

For

< 1, the output of

the

relaxation algorithm is

ing to different realization indicate

that the employ

3. Using a network visualization tool of your choice, produce plots of the networks (if

qualitatively different [Fig. 2(b)]. A large number of con-

Monte Carlo method does not find the exact global m

possible, depending on size).

ima. For

> 1,

the optimal

tree

obtained

by t

ductivities converge to zero so that no loop remains. The

highly redundant network is

transformed to

spanning

Monte Carlo algorithm is not the optimal solution sin

For the smaller one , please label the

odes numerically.

tree topology and the currents are canalized in a hierarch-

the absolute and only minima has loops. The dissipati

ical manner. This, too, is predicted

y the

results

rate which results from the

relaxation algorithm is then,

4. For river networks,

basin areanalyticalsaredistributed according to

P (a) / a .

[6]. In

contrast to

> 1, the conductivity

distribution

course, lower than the dissipation of any tree. While t

cannot

Determine the exponent in terms of the Horton ratios

R

and

Rs.

1 shows a cl

be interpreted

as a discrete approximation

of a

curve is continuous, the crossover at

conductivity tensor:FollowfortheNsamedia! procedure1,thestructureshownbecomesinlectures forchangeP(ℓ) in ℓthe

.slope of Jmin . One could interpret t

/

fractal.

behavior as a second order phase transition. (The change

5. (3 + 3 + 3) Reproduce Bohn and Magnasco’s Figs. 2a and 2b in [1]:

(a)

(b)

(c)

FIG. 2. Examples of the optimized conductivity distributions obtained by the relaxation method for (a) 2:0 and (b) 0:5.

Steps are given below but please read through the paper to understand how they

< 1, the relaxation leads in general only to a local minimum. The global minimum can be approached by searching in the space tree topologies. Theset thingsresultforup. 0:5 is shown in (c). The arrows indicate the localized inlet.

The full team is encouraged to work together088702on-3 Slack.

1. (3) Construct an adjacency matrix A representing the hexagonal lattice used in [1]. Plot this adjacency matrix.

2. (3 + 3) Run a minimization procedure to construct Figs. 2a and 2b which

correspond to = 2 and = 1/2. Steps:

i. Set each link’s length to unity (the d_kl). The goal then reduces to

minimizing the cost	∑
	C =jIklj
	k;l

where I_kl is the current on link kl and = 2 /( + 1).

2. place a current source of nominal size i₀ at one node.

2. All other nodes are sinks, drawing a current of

ik =	i0	:
	Nnodes 1

4. Suggest setting i₀ = 1000 (arbitrary but useful value given the size of the network).

5. Generate an initial set of random conductances for each link, the f _klg. These must sum to some global constraint as

∑

K = _kl:

k;l

Note: There seems to be no reason not to set K = 1 but the power of

is a bit of a worry. (Also: we now have a lot of k types on deck.)

6. Solve the following to determine the potential U at each node, and hence the current on each link using:

∑

i_k = _kl(U_k U_l);

l

and then

I_kl = _kl(U_l U_k):

Note: the paper erroneously has I_kl = R_kl(U_l U_k) below equation 4; there are a few other instances of similar miswritings of R_kl instead of

_kl.

7. Now, use scaling in equation (10) to compute a new set of f _klg from the I_kl. Everything boils down to

_kl / jI_klj ^{( 2)};

where the constant of proportionality is determined by again making

sure K = ^∑_{k;l kl}:

Bonus: Please see reference 1 in [1] for a random connection to the next assignment’s code name.

Some help—Let’s sort out the key equation:

∑

i_k = _kl(U_k U_l):

l

Each time we loop around through this equation, we know the i_k and the _kl and

must determine the . In matrixology, we love A ^⃗ problems so let’s see if we

U_k ⃗x = b

can fashion one: _∑

i_k = _kl(U_k U_l)

l

∑ ∑

⁼ kl^Uk kl^Ul

l l

∑ ∑

^{= U}k kl Kkl^Ul

l l

• • ⃗

• _kU_k [ U ]_k

where we have set _k = ^∑_{l kl}; the sum of the kth row of the matrix K. We now construct a diagonal matrix with the _k on the diagonal, and obtain:

⃗	⃗
i = (	K) U:
⃗	⃗	using standard features of
The above is in the form A x = b so we can solve for U

R, Matlab, Python, …(hopefully).

References

1. S. Bohn and M. O. Magnasco. Structure, scaling, and phase transition in the optimal transport network. Phys. Rev. Lett., 98:088702, 2007. pdf

4