BT Research at Martlesham, Suffolk

Transcription

1 Connectivity of random graphs Keith Briggs research.btexact.com/teralab/keithbriggs.html Alan Turing Institute 2004 November typeset 2004 November 17 10:55 in pdflatex on a linux system Connectivity of random graphs 1 of 25

2 BT Research at Martlesham, Suffolk Cambridge-Ipswich high-tech corridor 2000 technologists 15 companies UCL, Univ of Essex Connectivity of random graphs 2 of 25

3 Graphs (simple unlabelled undirected) graph: (simple unlabelled undirected) connected graph: (simple undirected) labelled graph: 3 4 Connectivity of random graphs 3 of 25

4 The Bernoulli random graph model G{n, p} let G be a graph of n nodes let p = 1 q be the probability that each possible edge exists edge events are independent let P (n) be the probability that G{n, p} is connected then P (1) = 1 and P (n) = 1 n 1 ( n 1 ) k=1 k 1 P (k)q k(n k) for n = 2, 3, 4,.... P (1) = 1 P (2) = 1 q P (3) = (2 q+1) (q 1) 2 P (4) = ( 6 q 3 +6 q 2 +3 q+1 ) (1 q) 3 P (5) = ( 24 q q q q q 2 +4 q+1 ) (q 1) 4 as n, we have P (n) 1 nq n 1. Keith Briggs Connectivity of random graphs 4 of 25

5 Connectivity for the Bernoulli model 1.0 2,5,10,20,50,100,200 nodes 0.8 Pr[connected] p x-axis: log(n = number of nodes), n = 2,..., 100 y-axis: p, 0 at top, 1 at bottom blue=0 red=1 Keith Briggs Connectivity of random graphs 5 of 25

6 Probability of connectivity - the G(n, m) model problem: compute the numbers of connected labelled graphs with n nodes and m = n 1, n, n+1, n+2,... edges with this information, we can compute the probability of a randomly chosen labelled graph being connected compute large-n asymptotics for these quantities, where the number of edges is only slightly larger than the number of nodes I did some exact numerical calculations to try to establish the dominant asymptotics I then looked at some earlier papers and found that the required theory to compute exact asymptotics is known I computed the exact asymptotics and got perfect agreement with my exact numerical data Keith Briggs Connectivity of random graphs 6 of 25

7 The inspirational paper [fss04] Philippe Flajolet, Bruno Salvy and Gilles Schaeffer: Airy Phenomena and Analytic Combinatorics of Connected Graphs The claim: the number C(n, n+k) of labelled (étiquetés) connected graphs with n nodes and excess (edges-nodes) = k 2 is A k (1) π ( n e ) n ( n 2)3k 1 2 [ 1 Γ(3k/2) +A k (1)/A k(1) k 2 Γ((3k 1)/2) n +O ( ) ] 1 n k A k (1) 5/24 5/ / / / / / A k (1) 19/24 65/ / / / / /98304 Airy in Playford: ipswich/history/airys Country Retreat.htm Keith Briggs Connectivity of random graphs 7 of 25

8 Some problems with the paper I did some comparisons with exact counts for up to n = 1000 nodes and for excess k = 2, 3,..., 8 The exact data was computed from the generating functions The fit was very bad This formula was found to fit the data much better for k = 2: A k (1) ( [ )3k 1 πn n n Γ(3k/2) A k (1)/A ( ) ] k(1) k 2 1 Γ((3k 1)/2) n +O n Also, on pages 4 and 24, I think S should have the expansion 1 (5/4)α+(15/4)α Keith Briggs Connectivity of random graphs 8 of 25

9 Asymptotic expansion of C(n, n+k)/n n+3k 1 2 ξ 2π green: from [bcm90] red: from [fss04] (with removal of factor e) k type [n 0 ] [n 1/2 ] [n 1 ] [n 3/2 ] 1 tree unicycle ξ bicycle tricycle ξ ξ quadricycle pentacycle ξ π blue: conjectured by KMB from numerical experiments k type [n 0 ] [n 1/2 ] [n 1 ] [n 3/2 ] [n 2 ] [n 5/2 ] 0 unicycle ξ ξ bicycle 5 24 ξ ξ ? ξ ? 2 tricycle ξ ξ quadricycle ξ Keith Briggs Connectivity of random graphs 9 of 25

10 Counting Given a finite set S, how many elements S does it have? The classical method for computing S : S = x S 1 Keith Briggs Connectivity of random graphs 10 of 25

11 Definitions of generating functions generating function (gf): {a 1, a 2, a 3,... } k=1 a k x k exponential generating function (egf): {a 1, a 2, a 3,... } k=1 a k k! xk Keith Briggs Connectivity of random graphs 11 of 25

12 Exponential generating functions exponential generating function for all labelled graphs: g(w, z) = n=0 (1+w) (n 2) z n /n! exponential generating function for all connected labelled graphs: c(w, z) = log(g(w, z)) = z+w z2 2 +(3w2 +w 3 ) z3 6 +(16w3 +15w 4 +6w 5 +w 6 ) z4 4! +... Keith Briggs Connectivity of random graphs 12 of 25

13 rooted labelled trees egfs for labelled graphs [jklp93] T (z) = z exp(t (z)) = n n 1zn n! = z+ 2 2! z ! z3 + unrooted labelled trees W 1 (z) = T (z) T (z) 2 /2 = z+ 1 2! z ! z ! z unicyclic labelled graphs W 0 (z) = 1 [ ] 2 log T (z) 2 T (z) 1 4 T (z)2 = 1 3! z ! z ! z ! z 6 +. bicyclic labelled graphs W 1 (z) = T (z)4( 6 T (z) ) 24 ( 1 T (z) ) 3 = 6 4! z ! z ! z Keith Briggs Connectivity of random graphs 13 of 25

14 Theory 1 The previous observations can be proved using theory available in [jklp93] and [fgkp95]. I sketch the computations. Ramanujan's Q-function is defined for n = 1, 2, 3,... : Q(n) k=1 n k n k = 1+n 1 n +(n 1)(n 2) +... n 2 n=1 Q(n)nn 1zn n! = log(1 T (z)) here T is the egf for rooted labelled trees: T (z) = n=1 nn 1 n! z n T (z) = z exp(t (z)) Keith Briggs Connectivity of random graphs 14 of 25

15 Theory 2 to get the large-n asymptotics of Q, we first consider the related function R(n) 1+ n n+1 + n 2 (n+1)(n+2) +..., n = 1, 2, 3,... we have Q(n)+R(n) = n! e n /n n let D(n) = R(n) Q(n) zn (1 T (z))2 n=1 D(n)nn 1 n! = log[ 2(1 ez) ] D(n) k=1 c(k)[zn ](T (z) 1) k, where c(k) [δ k ] log(δ 2 /2/(1 (1+δ)e δ )) this gives D(n) n n n n n 5 + O ( n 6) now using Q(n) = (n! e n /n n D(n))/2, we get Q(n) 1 2 n1/2 2π πn 1/ πn 5/ n πn 7/ πn 11/2 +O ( n 6) n πn 3/ n n n πn 9/2 Keith Briggs Connectivity of random graphs 15 of 25

16 Theory 3 Let W k be the egf for connected labelled (k+1)-cyclic graphs for unrooted trees W 1 (z) = T (z) T 2 (z)/2, [z n ]W 1 (z) = n n 2 for unicycles W 0 (z) = (log(1 T (z))+t (z)+t 2 (2)/2)/2 for bicycles W 1 (z) = 6T 4 (z) T 5 (z) 24(1 T (z)) 3 for k 1, W k (z) = results in [jklp93] A k (T (z)) (1 T (z)) 3k, where A k are polynomials computable from Knuth and Pittel's tree polynomials t n (y) (y 0) are defined by (1 T (z)) y = n=0 t n(y) zn n! we can compute these for y > 0 from t n (1) = 1; t n (2) = n n (1+Q(n)); t n (y+2) = n t n (y)/y+t n (y+1) thanks to this recurrence, the asymptotics for t n the known asymptotics of Q follow from Keith Briggs Connectivity of random graphs 16 of 25

17 Theory 4 Let ξ = 2π. All results agree with numerical estimates on this page. the number of connected unicycles is C(n, n) = n![z n ]W 0 (z) = 1 2 Q(n)nn 1 +3/2+t n ( 1) t n ( 2)/4 C(n,n) n n 1 4 ξn 1/2 7 6 n ξn 3/ n ξn 5/ n ξ ( n 1) 9/ n ξn 11/2 +O ( n 6) 2835 n ξn 7/2 + the number of connected bicycles is C(n, n+1)=n![z n ]W 1 (z)= 5 24 t n(3) t n(2) t n(1) 7 12 t n(0) t n( 1) t n( 2) C(n,n+1) n n 5 24 n 7 24 ξn1/ ξn 1/ n ξn 3/ n ξn 5/ n ξn 7/ n ξn 9/ n ξn 11/2 +O ( n 6) similarly, for the number of connected tricycles we get C(n,n+2) n n ξn5/ n ξn3/ n 1 +O (n 3/2) n ξn1/ ξn 1/2 + Keith Briggs Connectivity of random graphs 17 of 25

18 Probability of connectivity 1 we now have all the results needed to calculate the asymptotic probability P (n, n+k) that a randomly chosen graph with n nodes and n+k edges is connected (for n and small fixed k) the total number of graphs is g(n, n+k) ( ( n ) 2) n+k. This can be asymptotically expanded: g(n,n 1) 2π e n 2 ( n 2 ) n n 3/ n n n n n n 6 +O ( n 7) g(n,n+0) 2π e n 2 ( n 2 ) n 1 n 1/ n n n n n n n 7 +O ( n 8) g(n,n+1) 2π e n 2 ( n 2 ) n 1 n 3/ n n n n n n n n 8 +O ( n 9)... g(n, n+k) 2 π en 2( ) n nn k 1/2 ( 2 2 k 1 +O(n 1 ) ) Keith Briggs Connectivity of random graphs 18 of 25

19 P (n,n 1) 2 n e 2 n n 1/2 ξ n n Probability of connectivity n n n n n 7 +O ( n 8) check: n = 10, exact= , asymptotic= P (n,n+0) 2 n e 2 n ξ 1 4 ξ 7 6 n 1/ ξn n 3/ ξn 2 +O ( n 3) check: n = 10, exact=0.276, asymptotic=0.319 P (n,n+1) 2 n e 2 n n 1/2 ξ ξn 1/ n ξn 3/ n ξn 5/2 + O ( n 3) check: n = 10, exact=0.437, asymptotic=0.407 check: n = 20, exact= , asymptotic= check: n = 100, exact= , asymptotic= Keith Briggs Connectivity of random graphs 19 of 25

20 Example of unlabelled case - unicycles for n = 7 Unicyclic graphs - 7 nodes Keith Briggs 2004 Sep 05 09: Keith Briggs Connectivity of random graphs 20 of 25

21 The unlabelled case - unicycles Christian Bower's idea: A connected unicyclic graph is an undirected cycle of 3 or more rooted trees. Start with a single undirected cycle (or polygon) graph. It must have at least 3 nodes. Hanging from each node in the cycle is a tree (a tree is of course a connected acyclic graph). The node where the tree intersects the cycle is the root, thus it is (combinatorially) a rooted tree. A is undirected cycles of exactly 1 rooted tree A is undirected cycles of 3 or more rooted trees A is undirected cycles of exactly 2 rooted trees A is undirected cycles of 1 or more rooted trees this gives formulae which should be amenable to analysis, but let's first do some brute-force numerics to get a feel for the behaviour I calculated the counts up to n = nodes, and tried to guess the form of the asymptotic expansion and then the values of the coefficients Keith Briggs Connectivity of random graphs 21 of 25

22 Transforms A is undirected cycles of exactly 2 rooted trees A (x) = 1 ( A81 (x) 2 +A 81 (x 2 ) ) 2 A is undirected cycles of 1 or more rooted trees A (x) = 1 2 k=1 φ(k) k ( log(1 A81 (x k )) ) + 2A 81 (x)+a 81 (x) 2 +A 81 (x 2 ) 4(1 A 81 (x 2 )) A is undirected cycles of 3 or more rooted trees A 1429 (x) = A (x) A (x) A 81 (x) Keith Briggs Connectivity of random graphs 22 of 25

23 Asymptotics of A81 (unlabelled rooted trees) [fin03] d = A81(n) d n n 3/ n n n n n Keith Briggs Connectivity of random graphs 23 of 25

24 Asymptotics for unlabelled unicycles I get that the number c(n, n) of connected unlabelled unicyclic graphs behaves like ( c(n, n) d n 1 ) 4 n n 3/ n n 2 +O(n 3 ) the coefficients are poorly determined (unlike in a similar analysis of the labelled case) Keith Briggs Connectivity of random graphs 24 of 25

25 References [bcm90] E A Bender, E R Canfield & B D McKay: The asymptotic number of labeled connected graphs with a given number of vertices and edges, Random structures and algorithms 1, (1990) [slo03] N J A Sloane, ed. (2003), The on-line encyclopedia of integer sequences njas/sequences/ [jklp93] S Janson, D E Knuth, T Luczak & B G Pittel: The birth of the giant component Random Structures and Algorithms, 4, (1993) www-cs-faculty.stanford.edu/ knuth/papers/bgc.tex.gz [fss04] Philippe Flajolet, Bruno Salvy and Gilles Schaeffer: Airy Phenomena and Analytic Combinatorics of Connected Graphs 11/Abstracts/v11i1r34.html [gil59] E N Gilbert: Random graphs Ann. Math. Statist., 30, (1959) Keith Briggs Connectivity of random graphs 25 of 25