HIERATIC Hierarchical Analysis of Complex Dynamical Systems WP6 - - PowerPoint PPT Presentation

HIERATIC – Hierarchical Analysis of Complex Dynamical Systems

WP6 – Formal theoretical results concerning network dynamics John Haslegrave, Chris Cannings The University of Sheffield December 11, 2014

Overview

1 Tournament process 2 Cyclic relationships in Moran process

Rock-paper-scissors Symmetric generalisations Coarse-graining random generalisations

3 Preferential attachment with choice

Tournament process

We consider a random process on a connected graph G with n

• vertices. Each vertex has a strategy taken from {1, . . . , m} at
• random. At each time step an edge is chosen; both vertices take
• n the lower of the two strategies with probability p and the higher

with probability 1 − p. Since G is connected, eventually we will reach a state where only

• ne strategy remains. Write S for the strategy that is left; then

P(S l) can be computed by coarse-graining the strategies into those at most l and those exceeding l.

Probability of survival

Write a0 for the number of vertices initially in {1, . . . , l}, and let ar be the number after the rth time a significant edge is chosen. This is a random walk where ar = ar−1 + 1 with probability p and ar = ar−1 − 1 with probability 1 − p, independent of which edges are chosen, and independent of G. We can find the probability that this random walk hits 0 before n, and take a weighted average over a0 (which is Binomial) to get P(S = l) = (l − 2lp + mp)n − (l − 1 − 2lp + 2p + mp)n (m − mp)n − (mp)n .

Time to reach consensus

The distribution of final results does not depend on the graph, but the time taken to reach consensus does. Conjecture For any n, m and p, the complete graph has the smallest expected time to consensus. Theorem For any n and p, if m = 2 the complete graph has the smallest expected time to consensus of any regular graph.

Time to reach consensus

What about the worst case? Not the path, at least for p close to 0 (or 1). Define a sundew to be a graph formed from Kn−m by adding m pendant edges, distributed as evenly as possible. We can show that the expected time for a sundew is e(G)(log m ± 1). Setting m = cn gives an expected time which is Θ(n2 log n). The worst-case expected time for the path when p = 0 is (n − 1)2, and in fact we can show that the overall expected time for the path is O(n(log n)2).

Time to reach consensus

PRISM simulations for a range of plausible graphs suggest that a sundew is the worst case for p close to 0, and a tadpole (formed by adding a pendant path to Kn−m) is the worst case for p close to 1/2.

Monotonicity

Can we prove that the expected time is always monotonic on [0, 1/2]? Not obvious even in simple cases. For a simple random walk in {0, . . . , n}, moving left or right with probability p and 1 − p, we can show that the time to reach an endpoint is monotonic on [0, 1/2] provided the starting distribution is symmetric. The same applies if there are equal delays at all points – if we move left, move right, or remain with probabilities rp, r(1 − p), 1 − r. What if the value of r depends on where we are, but is symmetric on {0, . . . , n}? This would imply monotonicity for a complete graph.

Rock-paper-scissors

We look at similar process with three strategies in cyclic order: rock always beats scissors, scissors always beats paper, paper always beats rock. For general graphs, the starting position will affect the outcome. We’ll concentrate on the well-mixed situation (complete graph).

Rock-paper-scissors

From simulations, it appears that all strategies are roughly equally likely to eventually dominate unless the starting state is very

• ne-sided.

The eventual winner is determined by the first strategy eliminated; if Rock is initially dominant it is likely Scissors will be eliminated, allowing Paper to win.

Rock-paper-scissors

If almost all the population are of one type, the numbers of the

• ther types change like an urn process which starts with w white

balls and b black. At each step a ball is drawn; if white, it is removed, if black, two black balls are returned. Eventually (at time Tw,b) all white balls are eliminated. E(Tw,b) = ∞, but we can bound the probability that the process has ended by a given time. Lemma For any w, b and t with t ≥ w, wb t + wb ≤ P(Tw,b > t) ≤ wb t + b + 1 − w .

Rock-paper-scissors

Write r for the number of Rock players, etc. It follows that if ps = o(n2/3), Scissors will be eliminated whp before any new Scissors arise, allowing Paper to win. In fact rps is a sub-martingale, and this implies that if ps = o(n) Rock will not be eliminated first, and so either Paper or Rock will win (whp). Would conjecture that if ps = o(n) Paper wins whp but if min(rp, ps, sr) = ω(n) then the probabilities of winning approach (1/3, 1/3, 1/3).

Symmetric generalisations

What if we have more than three strategies? The first elimination no longer determines the result, but once a strategy is eliminated two others can be lumped together. If Rock is eliminated, then we reduce to the three-strategy model on Scissors, Spock, {Paper, Lizard}.

Symmetric generalisations

We can take a similar model for larger numbers of strategies. For any odd n, we take strategies 1, . . . , n where strategy i beats strategy i + j (mod n) for each j = 1, . . . , n−1

2 .

There are no coarse-grainings of this, but if we eliminate one strategy (eg 6), two of the others may be lumped together.

Iterated lumping

Suppose we take this symmetric system for some odd n, and we wish to simulate the Moran process to estimate how often strategy 1 (say) is the last remaining. We could simply run the process until either strategy 1 is eliminated or all others are. A better algorithm is to stop when one strategy is eliminated and lump together the two opposite strategies (provided neither of them is 1), then

• continue. Every time we do this lumping, we get a smaller

symmetric system with the same properties. This speeds the process up considerably.

Iterated lumping

We did this starting from 2k individuals playing strategy 1 and k playing each other strategy. The table below gives time taken (mins) for 1000 trials of the naive (red) and the coarse-graining (blue) algorithms. k = 5 k = 10 6.9 24.5 n = 25 2.1 8.1 70% 67% 63.6 229.6 n = 49 14.0 57.3 78% 75% 252 785–912 n = 73 45 192 82% 75–9%

Random orientations

In more generality, we could use any orientation of the complete graph to define relationships between the strategies (x beats y if the edge is oriented from x to y). If we orient the edges completely at random, how likely is a coarse-graining of the strategies to exist? How do we find one if it does? A coarse-graining will exist if and only if there is some set S with 1 < |S| < n, such that for any x, y ∈ S and z ∈ S, xz and yz are

• riented the same way.

Random orientations

An algorithm for finding a coarse-graining is as follows. If all edges from 1 are oriented the same way, then all other strategies can be lumped together; if not split the other strategies into two sets according to whether they beat 1. This gives us a list of two sets; any lump which doesn’t include 1 must be contained in some set in the list. Now iteratively take a set A from the list, give each strategy in A a signature based on the orientations to strategies outside A, then split A by signature. Repeat this operation until either no further changes can be made (in which case we have a coarse-graining) or every set in the list is a singleton (in which case no coarse-graining without 1 exists). We can then repeat this process starting by removing 2, 3, etc., and will eventually find all coarse-grainings.

Random orientations

Simulation results for how likely a coarse-graining is to exist (of 10000 trials). 3 4 5 6 7 8 9 10 7489 10000 7460 6897 5345 4059 2753 1807 11 12 13 14 15 16 17 18 1176 682 397 250 154 90 38 22 Theorem 1 Write pn for the probability that a random orientation of Kn has a coarse-graining. Then pn = n2 + n + o(1) 2n−1 .

Random orientations

How likely is it that there is a coarse-graining either initially, or when some strategy is eliminated? 3 4 5 6 7 8 9 10 7489 10000 10000 9911 9901 9667 9120 8155 11 12 13 14 15 16 17 18 6787 5417 3832 2655 1846 1142 700 394 Theorem 2 Write qn for the probability that a coarse graining exists either initially or after eliminating some strategy. Then qn = n3 + n + o(1) 2n−1 .

Balanced orientations

If we restrict our orientation to be balanced, that is to say every vertex has exactly half its edges going out, the chance of a coarse-graining existing initially is very low, but there is still a significant chance that a coarse-graining exists after one strategy is removed. initially

• ne removed

neither 7 vertices 2 1 9 vertices 1 11 3 11 vertices 3 729 491 13 vertices 428 532000 962869

Preferential attachment with choice

(joint with Jonathan Jordan, Sheffield) Preferential attachment is a class of evolving graph processes which aims to model behaviour found in real-world growing networks, eg internet, facebook. The simplest model was introduced by Barab´ asi and Albert (1999). Grow a tree, starting from two connected vertices. At each step, a new vertex is added, and it forms a single edge to an existing

• vertex. The vertex to connect to is chosen at random, with

probability proportional to its existing degree. There is a natural coarse-graining of this process, by reducing the graph to its degree sequence. The proportion of vertices of degree k tends to a specific limit πk with high probability. Bollob´ as, Riordan, Spencer and Tusn´ ady proved that πk = Θ(k−3).

Preferential attachment with choice

Malyshkin and Paquette recently introduced a new class of preferential attachment tree models. New nodes first make two independent choices (again proportional to degree). In the max choice model, the new vertex links to the higher-degree of its two choices; in the min-choice model, to the lower-degree. Max choice Pref attach Min choice

Preferential attachment with choice

To generalise, we might make r ≥ 2 choices and then attach to that of rank s. Rather than the proportion of vertices of degree k, it is more convenient to consider the probability of a single preferential choice giving a vertex of degree at most k. Again, this approaches a fixed limit pk (which depends on r, s). Malyshkin and Paquette showed that if r > 2 and s = 1, there is a vertex with degree growing linearly, so pk is bounded away from 1. Conversely, they showed that if r = s = 2, 1 − pk decays doubly-exponentially. Conjecture (Krapivsky and Redner) Doubly-exponential decay happens whenever s ≥ 2, and is of the form Csk.

Preferential attachment with choice

Theorem 1 If r ≥ s ≥ 2, pk → p∗, where p∗ is the smallest positive root of P(Bin(r, p) > r − s) − 2p + 1 = 0 . For s = 2, p∗ = 1 only if r < 7. In general, for fixed s there exists r(s) such that p∗ < 1 iff r < r(s).

Preferential attachment with choice

Theorem 2 For every s ≥ 2, r(s) < ∞ and r(s) ≥ 2s. Also, for any c1, c2 > 0, if s is sufficiently large then 2s + c1 √s < r(s) < 2s + c2s . We also show that doubly-exponential decay occurs whenever possible. Theorem 3 If p∗ = 1 then − log(1 − pk) = Θ(sk).

Preferential attachment with choice

So in all cases when s ≥ 2, either 1 − pk is bounded away from 0

• r it decays doubly-exponentially. If s = 1 and r > 2 we also have

the former behaviour. There is one exceptional case. Theorem 4 If r = 2 and s = 1, pk = 1 − 2 + o(1) log(k + 1) .