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ASYMPTOTIC FORMULA FOR DISCONNECTION PROBABILITY OF GRAPH ON TWO DIMENSIONAL MANIFOLD
G. Tsitsiashvili
IAM, FEB RAS, Vladivostok, Russia e-mails: guram@iam.dvo.ru
1. INTRODUCTION
A problem of a calculation of a graph disconnection probability is considered in a lot of papers, see for example [1]-[4]. In [1] upper and low bounds of the graph disconnection probability (a reliability polynomial) are constructed using maximal systems of disjoint cross sections. For a graph with sufficiently small number of arcs in [2] accelerated algorithms are constructed. These algorithms showed good results in a comparison with Maple 11. In [3] this problem is solved using Monte-Carlo method and some specific combinatory indexes and formulas.
But when a number of arcs increases this problem becomes much more complicated. So it is necessary to construct convenient asymptotic formulas for connectivity or disconnection probability of graph with high reliable arcs. In this paper such problem is solved for planar graphs or graphs arranged on two dimensional manifolds. Such graphs appear in honeycombed structures which are widely used in different applications.
2. PRELIMINARIES
Consider unoriented and connected graph G with finite sets of nodes U and of arcs W. Denote l(u, v) the set of all cross sections in G which divide nodes u, v e U, u ^ v
L = U L (u,v ).
u ^v
Put d(L) a number of arcs in cross section L and define
D(u,v)= min(d(L): L e L(u,v)), D = min D(u,v), L* = {L e L: d(L)= D},
' uev ' '
C is a number of cross sections from the set L*
Theorem 1. Suppose that graph arcs w e W fail independently with the probability h then the probability P of the graph G disconnection satisfies the formula
P ~ ChD , h ^ 0. (1)
So to calculate asymptotic of graph disconnection probability it is necessary to find the constants C, D . Suppose that G is two dimensional integer rectangle with the size M x N. If M, N > 1 then the set L* consists of four angle cross sections with two arcs [5] and so
P ~ 4h2, h ^ 0.
In general case for M > 0, N > 0 we obtain the formula
P ~ (4 +1 (M = 1)N +1 (N = 1)M )h2, h ^ 0. But to make asymptotic analysis of disconnection probability in honeycombed structures it is necessary to pass from integer rectangle to more general graphs.
3. MAIN RESULTS
This generalization is based on a concept of a graph G arranged on connected and two dimensional smooth manifold without edge T [6,chapter 1].
Suppose that between two nodes of the graph G there is not more than two arcs and there are not arcs beginning and ending at the same node (loops). Arcs do not intersect and may have only common nodes. Each node and each arc belong to some cycle with more than two arcs and more than two nodes.
Call faces (or cells) areas Sj,i = 0,...,m, of the manifold T limited by its cycles minimal by the set theory inclusions. So faces may have common nodes, common arcs but have not common internal points. Put two faces adjacent if there is their common arc. Each arc belongs to two faces (is adjacent to two faces). Denote by SSi the face Si boundary.
Suppose that faces Si,...,Sm are bounded and call them internal. Then the face
m
S 0 = T \ U Si
i=1
may be called external. The face So may be unbounded if for example the manifold T is a plane. It may be bounded also if for example T is a sphere or a torus.
(A). Suppose that each two internal faces Si, Sj ,1 < i < m , may have no more than single common arc.
Examples of graphs satisfied Condition (A) are connected aggregations of quadrates from rectangular lattice or connected aggregations of hexagons from hexagonal lattice.
Denote Ai j the set of arcs adjacent to faces Si, Sj ,0 < i ^ j < m , and put ni j a number
2
of arcs in the set Ai j . Designate Mt j = Cn , if ni j > 1 and Mi j = 0 if ni j < 1. Define
N = Z1<i<mMi 0 , M = Z0<i< j<mMu j .
Theorem 2. Suppose that Condition (A) and the inequality N > 0 are true then C = N, D = 2 .
An example of a graph satisfied Theorem 2 conditions is integer rectangle. Theorem 3. If M > 0 then the equalities C = M, D = 2 are true.
Remark that Condition (A) is absent in Theorem 3. Denote U3 the set of the graph G
nodes which are connected with three arcs and put K 3 the number of elements in U 3.
Theorem 4. If M = 0, K3 > 0, then C = K3,D = 3.
Examples of graphs which satisfy Theorem 3 conditions are the dodecahedron [3,Chapter 4, Figure 4.2] and integer tube obtained by a gluing of a pair of opposite sides in an integer rectangle with a size M x N M > 1 N > 1.
1 1
Theorem 5. If M = 0, K3 = 0, K4 > 0 then C = K4,D = 4.
An example of a graph satisfies Theorem 5 conditions is a graph arranged on two dimensional torus and obtained by a gluing of two pairs of opposite sides in an integer rectangle with a size M x N M > 1 N > 1.
4. PROOF OF MAIN RESULTS
Theorem 1.
Proof. From the Burtin-Pittel formula [3] the probability P(u, v ) of the nodes u, v disconnection in G satisfies the relation P(u, v) ~ C(u,v)hD(u,v), h ^ 0 where C(u, v) is a number of sections L e L(u, v): d(L)= D(u, v). Assume that VL is random event when all arcs of cross section L fail then
C
P = P
\
U VL V LeL J
P
W
U VL
LeL
U
W
U Vl
V LeL\L* J j
P
V
U Vl
LeL*
, h ^ 0,
as
P(VL) = o(hD), L e L \ L*, P
f \ U VL
V
LeL*
L
ChD.
J
Theorem 2.
Proof. 1. Prove at first that D > 1. If D = 1 then there are nodes u, v eU dividing by single arc w eW Suppose that T is a way in the graph G connecting nodes u, v and the arc w belongs to the face Sj .Without a restriction of a generality assume that the nodes u,v e Sj. Then it is possible to construct a way T in which the arc w is replaced by a way along the face Sj boundary which bypasses the arc w (see fig. 1). Consequently the arc $w$ does not divide the nodes u, v and so minimal number of arcs which create the graph G cross section is larger than one.
V
\ —
w
y —y-é
u
Fig. 1. Item 1 illustration
2. Suppose that w^ w2 is a pair of graph arcs. Prove that if these arcs do not belong to common internal face then the set {w!, w2 } € L . Indeed the arc w! (the arc w2) may be bypassed by its internal face boundary (see fig. 1). And a way around the arc w1 (around the arc w2) does not contain the arcs w1, w2 . So the set of arcs {w1, w2 } € L .
3. Assume that the arcs w1, w2 belong to internal face Sj but do not belong simultaneously to external face So .prove that the set {w1, w2 }€ L.
a) At first consider the case when there are internal faces Sj, Sk, j ^ i, k ^ j so that wl e Sj, w2 e Sk. As any two internal faces have not more than single common arc then j ^ k.
So the arc wj (the arc W2) may be bypassed along ôSj (along ôSk ) without the arcs Wj, W2 . Consequently the set {wj, W2 } € L .
b) Suppose now that the arc Wj = (uj,Vj)e S0 and the arc W2 = («2,v2)€ S0. Then the arc Wj may be bypassed by the way along ôS0. The arc W2 may be bypassed along ôSj from the node V2 to the node Vj. Then this way may be prolonged along ôS0 to the node Uj and then it may return along ôSj from the node Uj to the node «2 (see fig. 2). Consequently the set {w:, W2 }€ L .
Fig. 2. Item 3, subitem b) illustration.
4. Suppose that w: e Sj R S0, w2 e Sj R S0. Prove that {w:, w2 }e L . Take a cycle around the face Sj and suppose that u, v are first nodes of this cycle touching w^ w2 appropriately. Extract from the face Sj the set 8 Sj of arcs which do not belong to S0 Exclude trivial case when the arcs w^ w2 have common node. Contrast each arc w from 8Sj the set of internal faces {S^, k e Jw } defined by the symbols set Jw, i e Jj. This set satisfies recurrent conditions: if k e Jw,t ^ k and Sk R St ^ 0 then t e Jw. Denote Rw = U Sk and obtain
keJw
that if u e Rw> , v E Rw» then Rw> R Rw» = 0 (see fig. 3). In opposite case Wj € Sj R S0 (see
2
fig. 4). This contradiction proves that the set {wj, W2 }e L .
V
R
u
,rr
S, Pi
ir
R
u
,r
Fig. 3. The case R^, C\Rw»=0
Fig. 4. The case Rw> C\Rw»*0
The illustration of the figure 4 may be added by following formal inference. Assume that the arcs w', w1 have common node u and the arcs w", Wj have common node v and w' £ Sj, w" £ Sj. Suppose that the node z e Rw' H Rw".
From the recurrent definition of the set Jw< it is easy to prove that between the nodes u, z there is acyclic way r(u,z) consisting of the faces St,t e Jw< arcs excluding the face Sj arcs. Analogously between the nodes z, v there is acyclic way r(z, v) consisting of the faces St,t e Jw" arcs without the face Sj arcs. Consider a polygon bounded by the way r(u, z ) U r(z, v ) and by the way r(u,v)which connects the nodes u,v and passes along SSj, wj e r(u,v). It is obvious that this polygon consists of internal faces and does not contain the face Sj. So the arc wj is adjacent to some internal face which does not coincide with Sj. This statement contradicts with initial suggestion that wj e Sj. Consequently Rw' H Rw" ^ 0.
Theorem 3.
Proof. This statement may be proved analogously to Theorem 2 statement if to pass from the faces S0, Sj to the faces Sj, Sk, see item 3, subitem a) in Theorem 2 proof.
Theorem 4.
Proof. As M = 0 and each face may have no more than single common arc with the face So then
D > 0. From Theorem 2the graph G which satisfies Theorem 5 conditions has not cross sections with two arcs. Prove that there are graph cross sections with three arcs and all these cross sections {wj, wj, w3}contain arcs connected with a node from U3 . Indeed if arcs wj, wj, w3 are connected with a node from U3 then the set {wj, wj, w3 }e L. Prove that if arcs wj, wj, w3 are not connected with a node from U3 then the set {wj, wj, w3 }£ L. Then Theorem 4 will be proved.
As M = 0 so each two faces of the graph G may have no more than single common arc and each node connects with more than two arcs. Assume that arcs wj, wj, w3 belong to the faces Sj, Sj, S3 appropriately and some of these faces may coincide.
1. Suppose that there is not a pair of arcs from wj, wj, w3 in the same face. Then each arc
wj may be bypassed by a way along SSj (see fig. !). So any way r(u, v) from u to v which contains arcs from wj,wj,w3 may be replaced by a way r " (u, v) which does not contain arcs from wj,wj,w3 . For this aim it is enough to replace arc from this set by a way which bypasses this arc. Consequently the set {wj, wj, w3 }£ L .
2. Suppose that all three arcs from wj,wj,w3 belong to common face S. Then each arc
wj may be bypassed by a way along SSSj without arcs from wj,wj,w3 (see fig. 1). Consequently the set {wj, w2, w3 }£ L .
3. Suppose that faces Sj,Sj,S3,S are different and wj e Sj, wj e S3, w3 e S. Then the arc wj may be bypassed by a way along SSj, the arc w3 may be bypassed by a way along SS (see fig. 1) and the arc wj - by a way from SS3 to the arc w3, then around the arc w3 along SS
and then along 8S3 to the arc w2 second node (see fig. 2). And the way around wj does not contain arcs from w^ w2, w3 . Consequently the set {wi, w2, w3 } € L .
4. Suppose that the faces S^S2,S3,S are different and w1 e S, w2 e S, w3 € S. Then
each arc wj may be bypassed by a way along 8Sj without arcs from the set w^ w2, w3 (see fig. 1). Consequently the set {w:, w2, w3 } € L .
Theorem 5.
Proof. From Theorem 4 we have that the graph G satisfying Theorem 5 has not cross sections with three arcs. Prove that in G there are cross sections with four arcs w^ w2, w3, w4 and all these cross sections consist of arcs connected with some node from U4. Indeed if the arcs w^ w2, w3, w4 are connected with some node from U4 t {w1, w2, w3 } € L hen the set w^ w2, w3, w4 e L .Prove that all other sets are not cross sections in G.
From M = 0 we have that each two faces in G may have no more than single common arc. Denote faces S^ S2, S3, S4 which contain arcs w^ w2, w3, w4 appropriately.
1. Suppose that there is not a pair of arcs from the set ^, w2, w3, w4 } which belong to the
same face. Then each arc wj may be bypassed by a way along 8Sj without arcs from the set {w1, w2, w3, w4 }. Consequently each way r(u, v ) from u to v which contains some arcs from the set {w^ w2, w3, w4 }may be replaced by a way r'(u, v) without arcs from the set {w:, w2, w3, w4 } (see fig. 1). Consequently the set {w:, w2, w3, w4 } € L.
2. Suppose that all arcs from {w^ w2, w3, w4 } belong to the same face S. Then each arc
wj may be bypassed by a way along 8Sj adjacent with wj (see fig. 1) without arcs from the set {wj, w2, w3, w4 }. Consequently the set {w:, w2, w3, w4 } € L.
3. Suppose that the faces S^S2,S3,S4,S are different and w1 e S2, w3 e S4, w4 € S. Then the arc w1 may be bypassed by a way along S1 the arc w2 - by parts of 8S2 and along 8S1 around w1 (see fig. 2), the arc w4 - along S and the arc w3- by a way along parts of S4 and along S around the arc w4 (see fig. 2). Consequently the set {w1, w2, w3, w4 } € L.
4. Suppose that faces S^S2,S3,S4,S are different and
w: e S, w2 e S, w3 e S, w: € S4, w2 € S4, w3 € S4.
Then each arc from wj, i = 1,2,3, may be bypassed by a way along 8Sj without arcs from {w^ w2, w3, w4 }. Analogously the arc w4 may be bypassed along S4 (see fig. 1). Consequently the set {w:, w2, w3, w4 }€ L.
5. Suppose that the faces S^ S2, S3, S4, S are different and
w: e S, w2 e S, w3 e S, w4 e S:. Then the arc w2 may be bypassed by a way along 8S2 the arc w3 - along 8S3 the arc w4 - along 8S4 (see fig. 1). The arc w1 may be bypassed along parts of 8S1 and around the arc w4 along
SS4 (see fig. 2). Each way bypassing the arc wj does not contain arcs from the set {wj, w2, w3, w4 }. Consequently {wj, w2, w3, w4 }£ L. 5. CONCLUSION
So we obtain asymptotic formulas for disconnection probability of a wide variety of graphs with high reliable arcs. This problem is sufficiently complicated especially for the coefficient C because it is necessary to solve some N-P problem. But a consideration of planar graphs or graphs arranged on two dimensional smooth manifold without edge simplifies this calculations significantly. It takes place because though a problem to arrange a graph on a manifold is sufficiently complicated but in different applications this problem is solved by a designer without any calculations.
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