Course Wintersemester 2015/16 Cop and Robber Game Cont./Randomizations
Elmar Langetepe
University of Bonn
November 24th, 2015
Proof:
Two cops protect some paths, the third cop can proceed!
c2 v1
P2 P1
c1
v2
Ri
Lemma 39:Consider a graphG and a shortest path
P =s,v1,v2, . . . ,vn,t between two verticess andt inG, assume that we have two cops. After a finite number of moves the path is protected by the cops so that after a visit of the robberR of a vertex ofP the robber will be catched.
Move copc onto some vertexc =vi ofP
Assuming, r closer to somex in s,v1, . . . ,vi−1 andsomey in vi+1, . . . ,vn,t. Contradiction shortest path fromx and y d(x,c) +d(y,c)≤d(x,r) +d(r,y)
Move towardx, finally: d(r,v)≥d(c,v) for all v∈P Now robot moves, but we can repair all the time r goes to some vertexr0 and we have
d(r0,v)≥d(r,v)−1≥d(c,v)−1 for allv ∈P.
Somev0∈P with d(c,v0)−1 =d(r0,v0) exists, move tov0
Theorem 40:For any planar graphG we havec(G)≤3.
Proof:
Case 1: All three cops occupy a single vertexc and the robber is located in one componentRi ofG \ {c} Case 2: There are two different pathsP1 andP2 fromv1 to
v2 that are protected in the sense of Lemma 39 by copsc1 andc2. In this caseP1∪P2 subdividedG into an interior,I, and an exterior regionE. That is G\(P1∪P2) has at least two components. W.l.o.g.
we assume thatR is located in the exteriorE =Ri.
Theorem 40:For any planar graphG we havec(G)≤3.
Case 1 and Case 2
c Ri
c2 v1
P2 P1
c1
v2
Ri
Theorem 40:For any planar graphG we havec(G)≤3.
Case 1: Number of neighbors!
c one neighbor inRi: Move all cops to this neighbor c0 and ConsiderRi+1=Ri \ {c0}. Case 1 again.
c more than one neighbor inRi: a andb be two neighbors, P(a,b) a shortest path inRi between aand b. One cop remains inc, another cop protects the path P(a,b) by Lemma 39. Thus P1 =a,c,b and P2 =P(a,b). Case 2 withRi+1⊂Ri.
Theorem 40:For any planar graphG we havec(G)≤3.
Case 2:
c2 v1
P2 c1
v2 R1i+1
P1 x1
x2 r1
r3
r1 r4 a x1
xi
R2i+1
x2
r2 r3
r2 R1i+1
P3
P3 x
Theorem 40:For any planar graphG we havec(G)≤3.
Case 2:
1 There is a another shortest pathP0(v1,v2) inP1∪P2∪Ri but different fromP1 and P2. Leaves P1∪P2 at x1, hitsP1∪P2
again at x2.
2 There is no such path! There is a single vertexx of P1∪P2 so that R is in the component behind x. Move all three cops to x. Case 1 again!
Shortest pathP0(v1,v2) in P1∪P2∪Ri but different from P1 and P2. Leaves P1∪P2 at x1, hitsP1∪P2 again at x2.
c2 v1
P2 c1
v2 R1i+1
P1 x1
x2 r1
r3
r1 r4 a x1
xi
R2i+1
x2
r2 r3
r2 R1i+1
P3
P3 x
Letc3 protectP3 =v1, . . . ,x1,r1, . . . ,rk,x2, . . . ,v2 whilec1 andc2
protectP1∪P2.
Case 2 again:c3 protectsP3,c1 or c2 the remaining one!
Examples for the use of randomizations Context of decontaminations
Randomization for a strategy Beat the greedy algorithm for trees Randomization as part of the variant Probability distribution for the root Expected number of vertices saved
Integer LP formlation for trees (Exercise):
Minimize X
v∈V
xvwv
so that xr = 0 = 0 X
v≤u
xv ≤ 1 : for every leafu X
v∈Li
xv ≤ 1 : for every level Li,i ≥1 xv ∈ {0,1} : ∀v ∈V
optILP ≤optRLP
optRLP in polynomial time!
Subtree Tv with xv =a≤1 is a-saved, a portion a·wv of the subtree is saved
v1 is ancestor ofv2 andxv1 =a1 andxv2 =a2
Vertices ofTv2 are (a1+a2)-saved. The remaining vertices of Tv1 are onlya1-saved.
Randomized rounding scheme for every level
Sum of the xv =a-values for level i: Probability distribution for choosing v. Shuffle and setxv to 1.
Sum up to less than 1: Probability of not choosing a vertex at
double-protections: Choose vertices on the same path to a leaf! We only use the predecessor! Skip the higher level!
No suchdouble-protections: The expected approximation value would be indeed 1.
Intuitive idea: TreeTvi at leveli is fullysaved by the fractional strategy!
Worst-case: Fractional strategy has assigned a 1/i fraction to all vertices on the path fromr to vi. This gives 1 for Tvi. Probability of saving vi is: 1−(1−1/i)i ≥1− 1e. Formal general proof!
w.r.t. the probability distribution given byoptRLP. The expected approximation ratio of the above strategy for the number of vertices protected is 1−e1
.
SF fractional solution for optRLP
Probabilistic rounding scheme:SI outcome of this assignment Show: Expected protection ofSI is larger than 1−1e
times the value ofSF
xvF value ofxv for the fractional strategy xvI value {0,1} of integer strategy yv =P
u≤vxu∈ {0,1}indicate whetherv is finally saved yF =P
xF ≤1 fraction of v saved by fractional strategy
Theorem 41:Consider an algorithm that protects the vertices w.r.t. the probability distribution given byoptRLP. The expected approximation ratio of the above strategy for the number of vertices protected is 1−e1
.
Foryv = 1 it suffices that one of the predecessor ofv was chosen.
Letr =v0,v1,v2, . . . ,vk =v be the path fromr to v Pr[yv = 1] = 1−
k
Y
i=1
(1−xvFi). Explanation: The probability thatv2 is safe is x1+ (1−x1)x2 = 1−(1−x1)(1−x2) The probability thatv3 is safe is
1−(1−x1)(1−x2) + (1−x1)(1−x2)x3 = 1−(1−x1)(1−x2)(1−x3) and so on.
RLP
approximation ratio of the above strategy for the number of vertices protected is 1−e1
. Pr[yv = 1] = 1−
k
Y
i=1
(1−xvFi)
≥ 1− Pk
i=1(1−xvFi) k
!k
= 1− k−Pk i=1xvFi k
!k
= 1−
k−yvF k
k
= 1−
1− yvFk
≥1−e−yvF ≥
1− 1 yvF.
Theorem 41:Consider an algorithm that protects the vertices w.r.t. the probability distribution given byoptRLP. The expected approximation ratio of the above strategy for the number of vertices protected is 1−e1
.
E(|SI|= X
v∈V
Pr[yv = 1]≥
1−1 e
X
v∈V
yvF =
1−1 e
|SF|.
G = (V,E) fixed numberk of agents
k-surviving rate,sk(G), is the expectation of the proportion of vertices saved
Any vertex is root vertex with the same probability Classes, C, of graphs G: For constant,sk(G)≥ Given G,k,v ∈V let:
snk(G,v):number of vertices that can be protected by k agents, if the fire starts at v
1
|V|
P
v∈V snk(G,v)≥|V|
ClassC: let the minimum number k that guarantees sk(G)> for anyG ∈C be denoted as the
firefighter-number, ffn(C), ofC.
