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%=Interchanging distance and capacity in probabilistic mappings>>(Uriel Feige
Weizmann Institute"$Harald Racke
Optimal Hierarchical Decompositions for Congestion Minimization in Networks
[STOC 2008]
Presentation based on discussions with Reid Andersen from Microsoft Research.b L
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$#Motivating example Graph Bisection$$(Given a (weighted) graph with an even number of vertices, partition its vertices into two equal sets, minimizing the number of edges cut. #Motivating example Graph Bisection$$(Given a (weighted) graph with an even number of vertices, partition its vertices into two equal sets, minimizing the number of edges cut. Coping with NP-hardnesszBi-criteria approximation: near bisection of nearly optimal width [Leighton and Rao; Linial, London and Rabinovich; Arora, Rao and Vazirani].
Based on metric embeddings into l1 and into (l2)2. O(log1/2 n) pseudo approximation.
Feige and Krauthgamer: O(log3/2 n) approximation. ARV + dynamic programming.
Racke: O(log n) approximation.
Based on probabilistic capacity mappings. ZV Zl Z* ZCH$
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Z8= Intuition
:Unary subset sum can be reduced to deciding whether the bisection width is 0.
Any approximation algorithm for bisection must use dynamic programming. Easy family of graphs}For many NP-hard graph properties, dynamic programming runs in time exponential in treewidth.
Min-bisection on trees is in P.~ ~S "
Plan of attackGiven an input graph, find a low distortion embedding into a tree.
Solve min-bisection optimally on the tree.
Will provide a solution to the original problem, with approximation ratio at most the distortion. A problem
The distortion when embedding a graph into a tree is large (e.g., for the n-cycle).
Solution [Alon, Karp, Peleg and West; Bartal]: a probabilistic embedding into (dominating) trees.
Suffices when objective function is linear, as in bisection., ^"s>^s Modified plan of attack#Given an input graph, find a low distortion probability distribution over embeddings to polynomially many (dominating) trees.
Solve min-bisection optimally on each of these trees.
Best solution will solve the original problem within an approximation ratio of at most the average distortion. $ $X
Another problemJ Standard low distortion embeddings relate to distortion of distances.
Small cut in tree does not imply small cut in graph. E.g., distortion 3 for complete graph.
PD Capacities rather than distances!!Distance distortion 3 for complete graph.
Give every tree edge a capacity equal to the capacity of the associated cut.
n2/2 capacity distortion.
* x Racke s approach
Consider distortion of capacities rather than distances.
Find a low distortion (of capacity) probability distribution of embeddings into (dominating) trees.
Solve min-bisection on every tree (use DP).
Approximation ratio for bisection of best solution is bounded by the distortion. (Racke s main theoremTheorem: there are probability distributions of embeddings into (dominating) trees with O(log n) distortion of capacity.
Proof: based on a reduction to existing results for distortion of distances.
Corollary: an O(log n) approximation for min-bisection. (And other corollaries.)t RI ;,Yx>
In this talk
pAn abstract version of Racke s result.
Applies more generally.
Tighter parameters.
More modular presentation: easier to understand.
(Implicit in Racke s work. We make it explicit.)6' _ 3 ,vThe abstract frameworkE set of items (edges).
P 2E collection of subsets of E ( paths ).
M : E P maps edges to paths. Can be represented as a matrix:
Mij indicates if edge j lies in path M(i) in map M.
M family of admissible mappings (trees).
Probabilistic mapping between E and M: probability distribution over mappings M 2 M.
lM weight of mapping M\ G5E)G,#A mapping M*
Edge 2 is mapped to the path {1,2,4}+ + Distance mapping~Edge i has positive length leni.
distM(i)=Sj Mij lenj length of path M(i).
distM(i)/leni stretch of edge i.
Average stretch of edge i under a probabilistic mapping: weighted average according to lM of stretches of i.
Stretch of probabilistic mapping: maximum over all edges of their average stretches.@ >Wt
NotesFor certain length functions and mappings, the stretch of an edge may be smaller than 1.
The stretch of the shortest edge is at least 1.
Stretch of probabilistic mapping is at least 1. $Capacity mappingPEdge i has positive capacity capi.
loadM(j) = i Mij capi
sum of capacities of edges whose paths under M contain edge j.
Congestion of edge j = loadM(j)/capj.
Average congestion of edge j: average weighted according to lM.
Congestion of probabilistic mapping: maxj[avgM-congestion(j)]. > Z? Z Z2
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mNotesFor certain mappings, the congestion of an edge may be smaller than 1.
The sum of capacities of M(E) is at least as large as sum of capacities of E.
Congestion of probabilistic mapping is at least 1. $1-Main theorem
For every r and every admissible family M, the two statements are equivalent:
For every collection of lengths leni there is a probabilistic mapping with stretch at most r.
For every collection of capacities capi there is a probabilistic mapping with congestion at most r.N
E7%9,n]=2.A spanning tree exampleE set of edges in a graph.
P set of paths in the graph.
M Admissible maps correspond to spanning trees of the graph.
M an edge is mapped to the unique path between its endpoints.b <=3/An instructive graph
40Probabilistic mappingsFor any choice of lengths, probabilistic mapping with max-avg(stretch) < 3.
For any choice of capacities, probabilistic mapping with max-avg(congestion) < 3.
Will illustrate both statements when the graph is unweighted. & @ >:D:
51Uniform length, low stretch
62Delete at random one column
731Delete at random one columnMax[avg(stretch)] < 222(
84 Uniform capacity, low congestion!!(
95Keep at random one path
:60Keep at random one pathMax[avg(congestion)] < 311(
;7,Simultaneous low stretch and low congestion--(If top edge is in spanning tree, it is loaded by n1/2.
If top edge is not in spanning tree, it is stretched by n1/2.
Hence for every distribution over spanning trees, either stretch or congestion is n1/2/2. 1:T<8Implication for main theoremFor every r and every admissible family M, the two statements are equivalent:
For every collection of lengths leni there is a probabilistic mapping with stretch at most r.
For every collection of capacities capi there is a probabilistic mapping with congestion at most r.
The probabilistic mappings achieving the two guarantees are sometimes very different from each other.N f
H7(:f,n]/A useful equivalence [Alon, Karp, Peleg, West]&0 ,Lemma: for every r 1, admissible family M and length function len, there is a probabilistic mapping of stretch at most r iff for every nonnegative coefficients ai (with S ai = 1) for the edges, there is a mapping M 2 M s.t.
S ai distM(i)/leni r 6 ( % C
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>Proof use the minimax theorem ( DView as a 0-sum two player game. Map player chooses an admissible mapping M. Edge player chooses an edge i.
Value for edge player: stretch of i under M.
Probabilistic mapping
= a mixed strategy for the map player.
Nonnegative coefficients ai
= a mixed strategy of the edge player.d# !
"-If direction
Inequality holds.
Edge player cannot ensure stretch larger than r.
Map player can limit edge player to stretch of r.
Probabilistic mapping is a mixed strategy for map player.
Wins against every edge.D @1UOnly if directionNo probabilistic mapping of stretch at most r.
Map player cannot limit value to r.
Edge player can get expected value larger than r.
Mixed strategy for edge player gives ai.
Map player cannot choose a map M satisfying the inequality.~ ,#1'>Likewise& ((Lemma 2: for every r 1, admissible family M and capacity function cap, there is a probabilistic mapping of congestion at most r iff for every nonnegative coefficients bi (with S bi = 1) for the edges, there is a mapping M 2 M s.t.
S bi loadM(i)/capi r
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Back to main theoremFor every r and every admissible family M, the two statements are equivalent:
For every collection of lengths leni there is a probabilistic mapping with stretch at most r.
For every collection of capacities capi there is a probabilistic mapping with congestion at most r.N
H7(:,n]=!-Proof that low congestion implies low stretch..(\Assume probabilistic mapping with congestion r.
By Lemma 2, for every b, there is M with S bj loadM(j)/capj r.
Rewriting the load as a sum:
Si,j bj Mij capi/capj r. -
P_
& "Need to prove that there is a probabilistic mapping with stretch at most r.
By Lemma 1, suffices to show that for every a, there is a mapping M with S ai distM/ leni r.
Rewriting distance as sum:
Si,j ai Mij leni/lenj r. I, 0
b$ #@Know that for all b and cap there is M
Si,j bj Mij capi/capj r.
Find for every a and len an M satisfying
Si,j ai Mij leni/lenj r.
Choose bj = aj and capi = ai/leni (and then of course capj = aj/lenj), find M that satisfies the first inequality, and it will satisfy the second. !
P) P$ -Proof that low stretch implies low congestion..(Z
Same as before, this time choosing ai = bi and leni = bi/capi, finding M that satisfies the second inequality, which will necessarily also satisfy the first inequality. ($$$$$$$
$$$$$
$l$,4l%!The proof does not assume that: (The matrix M is a 0/1 matrix.
Lengths and capacities of edges are nonnegative (though they cannot be 0).
Paths are actual paths in graphs.
Edges are edges of graphs.
Hence the theorem applies in more general situations.& 7 D;The algorithmic aspectKThe proof of the main theorem as presented is existential.
Will give a polynomial time algorithm that finds a probabilistic mapping whenever the associated 0-sum two player game can be solved efficiently.
Standard low regret online algorithms (e.g., multiplicative weight update) solve 0-sum games under fairly general conditions.L LH=Approximation algorithm( Let r be a desired target distortion for probabilistic capacity embedding. To efficiently find such a mapping, it suffices that for every nonnegative coefficients bi (with S bi = 1) for the edges, can efficiently find a mapping M 2 M s.t.
S bi loadM(i)/capi r
6 C
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J> Alternative sufficient condition&! ((By main theorem, it also suffices that for every length function len and every nonnegative coefficients ai (with S ai = 1) for the edges, can efficiently find a mapping M 2 M satisfying
S ai distM(i)/leni rf A $ 1 C
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F:)Back to min-bisection simplified version**(E is a set of edges in a graph G.
P is the collection of all paths.
M is collection of spanning trees.
Every edge of G is mapped to its unique path in the tree.
Load on tree edge equals the capacity of the cut.Jh l ""K?4Load of tree bisection is lower than graph bisection55(_The load on a cut in the tree is at least as high as the capacity of the same cut in the graph.` `L@Approximating min-bisectionConsider minimum bisection in graph. Its average load in the dominating trees (average weighted by lM) is at most r times larger than its capacity.
Hence the minimum of all minimum load bisections of trees is a r approximation for minimum bisection.h c
a&MAThe value of r(
For spanning trees, max-avg(stretch) is at most O*(log n) [Elkin, Emek, Spielman and Teng; Abraham, Bartal and Neiman].
By our main theorem, the same holds for congestion.
Gives O*(log n) approximation for bisection.@ ;dboNB^Racke s approximation for min-bisection00()SUse probabilistic mappings into trees (that may use non-graph edges). There, max-avg(stretch) is at most O(log n) [Fakcharoenphol, Rao, and Talwar].
Every edge of the tree is a shortest path in G. Every edge of G is mapped to its unique path in the tree, and hence to a sequence of paths of G.
Hence M has nonnegative integer entries..T ZwbU
OCSummarynProbabilistic mappings of low congestion are useful algorithmic tools for cut problems (and for oblivious routing & ).
Given the same family of admissible mappings, the maximum stretch (over all length functions) equals the maximum congestion (over all capacity functions).
Proof is algorithmic (in most cases).8 8
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;Arialcmsy10SymbolDefault DesignMicrosoft Equation 3.0>Interchanging distance and capacity in probabilistic mappingsSlide 2$Motivating example Graph Bisection$Motivating example Graph BisectionCoping with NP-hardness
IntuitionEasy family of graphsPlan of attack
A problemModified plan of attackAnother problem!Capacities rather than distancesRacke’s approachRacke’s main theorem
In this talkThe abstract frameworkA mapping MDistance mappingNotesCapacity mappingNotes
Main theoremA spanning tree exampleAn instructive graphProbabilistic mappingsUniform length, low stretchDelete at random one column2Delete at random one column Max[avg(stretch)] < 2!Uniform capacity, low congestionKeep at random one path1Keep at random one path Max[avg(congestion)] < 3-Simultaneous low stretch and low congestionImplication for main theorem0A useful equivalence [Alon, Karp, Peleg, West]Proof – use 0-sum gamesThe minimax theorem
If directionOnly if direction LikewiseBack to main theorem.Proof that low congestion implies low stretch Slide 42 Slide 43.Proof that low stretch implies low congestion The proof does not assume that:The algorithmic aspectApproximation algorithm!Alternative sufficient condition*Back to min-bisection simplified version5Load of tree bisection is lower than graph bisectionApproximating min-bisectionThe value of 2Racke’s approximation for min-bisectionSummaryFonts UsedDesign TemplateEmbedded OLE Servers
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%ԃ=Interchanging distance and capacity in probabilistic mappings>>(Uriel Feige
Weizmann Institute"$Harald Racke
Optimal Hierarchical Decompositions for Congestion Minimization in Networks
[STOC 2008]
Presentation based on discussions with Reid Andersen from Microsoft Research.b L
'
$#Motivating example Graph Bisection$$(Given a (weighted) graph with an even number of vertices, partition its vertices into two equal sets, minimizing the number of edges cut. #Motivating example Graph Bisection$$(Given a (weighted) graph with an even number of vertices, partition its vertices into two equal sets, minimizing the number of edges cut. Coping with NP-hardnesszBi-criteria approximation: near bisection of nearly optimal width [Leighton and Rao; Linial, London and Rabinovich; Arora, Rao and Vazirani].
Based on metric embeddings into l1 and into (l2)2. O(log1/2 n) pseudo approximation.
Feige and Krauthgamer: O(log3/2 n) approximation. ARV + dynamic programming.
Racke: O(log n) approximation.
Based on probabilistic capacity mappings. ZV Zl Z* ZCH$
:P
Z8= Intuition
:Unary subset sum can be reduced to deciding whether the bisection width is 0.
Any approximation algorithm for bisection must use dynamic programming. Easy family of graphs}For many NP-hard graph properties, dynamic programming runs in time exponential in treewidth.
Min-bisection on trees is in P.~ ~S "
Plan of attackGiven an input graph, find a low distortion embedding into a tree.
Solve min-bisection optimally on the tree.
Will provide a solution to the original problem, with approximation ratio at most the distortion. A problem
The distortion when embedding a graph into a tree is large (e.g., for the n-cycle).
Solution [Alon, Karp, Peleg and West; Bartal]: a probabilistic embedding into (dominating) trees.
Suffices when objective function is linear, as in bisection., ^"s>^s Modified plan of attack#Given an input graph, find a low distortion probability distribution over embeddings to polynomially many (dominating) trees.
Solve min-bisection optimally on each of thes
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%ԃ=Interchanging distance and capacity in probabilistic mappings>>(Uriel Feige
Weizmann Institute"$Harald Racke
Optimal Hierarchical Decompositions for Congestion Minimization in Networks
[STOC 2008]
Presentation based on discussions with Reid Andersen from Microsoft Research.b L
'
$#Motivating example Graph Bisection$$(Given a (weighted) graph with an even number of vertices, partition its vertices into two equal sets, minimizing the number of edges cut. #Motivating example Graph Bisection$$(Given a (weighted) graph with an even number of vertices, partition its vertices into two equal sets, minimizing the number of edges cut. Coping with NP-hardnesszBi-criteria approximation: near bisection of nearly optimal width [Leighton and Rao; Linial, London and Rabinovich; Arora, Rao and Vazirani].
Based on metric embeddings into l1 and into (l2)2. O(log1/2 n) pseudo approximation.
Feige and Krauthgamer: O(log3/2 n) approximation. ARV + dynamic programming.
Racke: O(log n) approximation.
Based on probabilistic capacity mappings. ZV Zl Z* ZCH$
:P
Z8= Intuition
:Unary subset sum can be reduced to deciding whether the bisection width is 0.
Any approximation algorithm for bisection must use dynamic programming. Easy family of graphs}For many NP-hard graph properties, dynamic programming runs in time exponential in treewidth.
Min-bisection on trees is in P.~ ~S "
Plan of attackGiven an input graph, find a low distortion embedding into a tree.
Solve min-bisection optimally on the tree.
Will provide a solution to the original problem, with approximation ratio at most the distortion. A problem
The distortion when embedding a graph into a tree is large (e.g., for the n-cycle).
Solution [Alon, Karp, Peleg and West; Bartal]: a probabilistic embedding into (dominating) trees.
Suffices when objective function is linear, as in bisection., ^"s>^s Modified plan of attack#Given an input graph, find a low distortion probability distribution over embeddings to polynomially many (dominating) trees.
Solve min-bisection optimally on each of these trees.
Best solution will solve the original problem within an approximation ratio of at most the average distortion. $ $X
Another problemJ Standard low distortion embeddings relate to distortion of distances.
Small cut in tree does not imply small cut in graph. E.g., distortion 3 for complete graph.
PD Capacities rather than distances!!Distance distortion 3 for complete graph.
Give every tree edge a capacity equal to the capacity of the associated cut.
n2/2 capacity distortion.
* x Racke s approach
Consider distortion of capacities rather than distances.
Find a low distortion (of capacity) probability distribution of embeddings into (dominating) trees.
Solve min-bisection on every tree (use DP).
Approximation ratio for bisection of best solution is bounded by the distortion. (Racke s main theoremTheorem: there are probability distributions of embeddings into (dominating) trees with O(log n) distortion of capacity.
Proof: based on a reduction to existing results for distortion of distances.
Corollary: an O(log n) approximation for min-bisection. (And other corollaries.)t RI ;,Yx>
In this talk
pAn abstract version of Racke s result.
Applies more generally.
Tighter parameters.
More modular presentation: easier to understand.
(Implicit in Racke s work. We make it explicit.)6' _ 3 ,vThe abstract frameworkE set of items (edges).
P 2E collection of subsets of E ( paths ).
M : E P maps edges to paths. Can be represented as a matrix:
Mij indicates if edge j lies in path M(i) in map M.
M family of admissible mappings (trees).
Probabilistic mapping between E and M: probability distribution over mappings M 2 M.
lM weight of mapping M\ G5E)G,#A mapping M*
Edge 2 is mapped to the path {1,2,4}+ + Distance mapping~Edge i has positive length leni.
distM(i)=Sj Mij lenj length of path M(i).
distM(i)/leni stretch of edge i.
Average stretch of edge i under a probabilistic mapping: weighted average according to lM of stretches of i.
Stretch of probabilistic mapping: maximum over all edges of their average stretches.@ >Wt
NotesFor certain length functions and mappings, the stretch of an edge may be smaller than 1.
The stretch of the shortest edge is at least 1.
Stretch of probabilistic mapping is at least 1. $Capacity mappingPEdge i has positive capacity capi.
loadM(j) = i Mij capi
sum of capacities of edges whose paths under M contain edge j.
Congestion of edge j = loadM(j)/capj.
Average congestion of edge j: average weighted according to lM.
Congestion of probabilistic mapping: maxj[avgM-congestion(j)]. > Z? Z Z2
-t[
mNotesFor certain mappings, the congestion of an edge may be smaller than 1.
The sum of capacities of M(E) is at least as large as sum of capacities of E.
Congestion of probabilistic mapping is at least 1. $1-Main theorem
For every r and every admissible family M, the two statements are equivalent:
For every collection of lengths leni there is a probabilistic mapping with stretch at most r.
For every collection of capacities capi there is a probabilistic mapping with congestion at most r.N
E7%9,n]=2.A spanning tree exampleE set of edges in a graph.
P set of paths in the graph.
M Admissible maps correspond to spanning trees of the graph.
M an edge is mapped to the unique path between its endpoints.b <=3/An instructive graph
40Probabilistic mappingsFor any choice of lengths, probabilistic mapping with max-avg(stretch) < 3.
For any choice of capacities, probabilistic mapping with max-avg(congestion) < 3.
Will illustrate both statements when the graph is unweighted. & @ >:D:
51Uniform length, low stretch
62Delete at random one column
731Delete at random one columnMax[avg(stretch)] < 222(
84 Uniform capacity, low congestion!!(
95Keep at random one path
:60Keep at random one pathMax[avg(congestion)] < 311(
;7,Simultaneous low stretch and low congestion--(If top edge is in spanning tree, it is loaded by n1/2.
If top edge is not in spanning tree, it is stretched by n1/2.
Hence for every distribution over spanning trees, either stretch or congestion is n1/2/2. 1:T<8Implication for main theoremFor every r and every admissible family M, the two statements are equivalent:
For every collection of lengths leni there is a probabilistic mapping with stretch at most r.
For every collection of capacities capi there is a probabilistic mapping with congestion at most r.
The probabilistic mappings achieving the two guarantees are sometimes very different from each other.N f
H7(:f,n]/A useful equivalence [Alon, Karp, Peleg, West]&0 ,Lemma: for every r 1, admissible family M and length function len, there is a probabilistic mapping of stretch at most r iff for every nonnegative coefficients ai (with S ai = 1) for the edges, there is a mapping M 2 M s.t.
S ai distM(i)/leni r 6 ( % C
PA8_
.Proof use 0-sum gamesDView as a 0-sum two player game. Map player chooses an admissible mapping M. Edge player chooses an edge i.
Value for edge player: stretch of i under M.
Probabilistic mapping
= a mixed strategy for the map player.
Nonnegative coefficients ai
= a mixed strategy of the edge player.d# !
"-QEThe minimax theorem The lemma is equivalent to the minimax theorem:
minMmaxE(stretch)=maxaminM(stretch)V 2$
$$
$
$$$
$
$>If direction
Inequality holds.
Edge player cannot ensure stretch larger than r.
Map player can limit edge player to stretch of r.
Probabilistic mapping is a mixed strategy for map player.
Wins against every edge.D @1UOnly if directionNo probabilistic mapping of stretch at most r.
Map player cannot limit value to r.
Edge player can get expected value larger than r.
Mixed strategy for edge player gives ai.
Map player cannot choose a map M satisfying the inequality.~ ,#1'>Likewise& ((Lemma 2: for every r 1, admissible family M and capacity function cap, there is a probabilistic mapping of congestion at most r iff for every nonnegative coefficients bi (with S bi = 1) for the edges, there is a mapping M 2 M s.t.
S bi loadM(i)/capi r
9 ( * C
>_
Back to main theoremFor every r and every admissible family M, the two statements are equivalent:
For every collection of lengths leni there is a probabilistic mapping with stretch at most r.
For every collection of capacities capi there is a probabilistic mapping with congestion at most r.N
H7(:,n]=!-Proof that low congestion implies low stretch..(\Assume probabilistic mapping with congestion r.
By Lemma 2, for every b, there is M with S bj loadM(j)/capj r.
Rewriting the load as a sum:
Si,j bj Mij capi/capj r. -
P_
& "Need to prove that there is a probabilistic mapping with stretch at most r.
By Lemma 1, suffices to show that for every a, there is a mapping M with S ai distM/ leni r.
Rewriting distance as sum:
Si,j ai Mij leni/lenj r. I, 0
&!
b$ #@Know that for all b and cap there is M
Si,j bj Mij capi/capj r.
Find for every a and len an M satisfying
Si,j ai Mij leni/lenj r.
Choose bj = aj and capi = ai/leni (and then of course capj = aj/lenj), find M that satisfies the first inequality, and it will satisfy the second. !
P) P$ -Proof that low stretch implies low congestion..(Z
Same as before, this time choosing ai = bi and leni = bi/capi, finding M that satisfies the second inequality, which will necessarily also satisfy the first inequality. ($$$$$$$
$$$$$
$l$,4l%!The proof does not assume that: (The matrix M is a 0/1 matrix.
Lengths and capacities of edges are nonnegative (though they cannot be 0).
Paths are actual paths in graphs.
Edges are edges of graphs.
Hence the theorem applies in more general situations.& 7 D;The algorithmic aspectKThe proof of the main theorem as presented is existential.
Will give a polynomial time algorithm that finds a probabilistic mapping whenever the associated 0-sum two player game can be solved efficiently.
Standard low regret online algorithms (e.g., multiplicative weight update) solve 0-sum games under fairly general conditions.L LH=Approximation algorithm( Let r be a desired target distortion for probabilistic capacity embedding. To efficiently find such a mapping, it suffices that for every nonnegative coefficients bi (with S bi = 1) for the edges, can efficiently find a mapping M 2 M s.t.
S bi loadM(i)/capi r
6 C
,
J> Alternative sufficient condition&! ((By main theorem, it also suffices that for every length function len and every nonnegative coefficients ai (with S ai = 1) for the edges, can efficiently find a mapping M 2 M satisfying
S ai distM(i)/leni rd A $ 1 C
,A
F:)Back to min-bisection simplified version**(E is a set of edges in a graph G.
P is the collection of all paths.
M is collection of spanning trees.
Every edge of G is mapped to its unique path in the tree.
Load on tree edge equals the capacity of the cut.Jh l ""K?4Load of tree bisection is lower than graph bisection55(_The load on a cut in the tree is at least as high as the capacity of the same cut in the graph.` `L@Approximating min-bisectionConsider minimum bisection in graph. Its average load in the dominating trees (average weighted by lM) is at most r times larger than its capacity.
Hence the minimum of all minimum load bisections of trees is a r approximation for minimum bisection.h c
a&MAThe value of r(
For spanning trees, max-avg(stretch) is at most O*(log n) [Elkin, Emek, Spielman and Teng; Abraham, Bartal and Neiman].
By our main theorem, the same holds for congestion.
Gives O*(log n) approximation for bisection.@ ;dboNB^Racke s approximation for min-bisection00()SUse probabilistic mappings into trees (that may use non-graph edges). There, max-avg(stretch) is at most O(log n) [Fakcharoenphol, Rao, and Talwar].
Every edge of the tree is a shortest path in G. Every edge of G is mapped to its unique path in the tree, and hence to a sequence of paths of G.
Hence M has nonnegative integer entries..T ZwbU
OCSummarynProbabilistic mappings of low congestion are useful algorithmic tools for cut problems (and for oblivious routing & ).
Given the same family of admissible mappings, the maximum stretch (over all length functions) equals the maximum congestion (over all capacity functions).
Proof is algorithmic (in most cases).8 8
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.-e trees.
Best solution will solve the original problem within an approximation ratio of at most the average distortion. $ $X
Another problemJ Standard low distortion embeddings relate to distortion of distances.
Small cut in tree does not imply small cut in graph. E.g., distortion 3 for complete graph.
PD Capacities rather than distances!!Distance distortion 3 for complete graph.
Give every tree edge a capacity equal to the capacity of the associated cut.
n2/2 capacity distortion.
* x Racke s approach
Consider distortion of capacities rather than distances.
Find a low distortion (of capacity) probability distribution of embeddings into (dominating) trees.
Solve min-bisection on every tree (use DP).
Approximation ratio for bisection of best solution is bounded by the distortion. (Racke s main theoremTheorem: there are probability distributions of embeddings into (dominating) trees with O(log n) distortion of capacity.
Proof: based on a reduction to existing results for distortion of distances.
Corollary: an O(log n) approximation for min-bisection. (And other corollaries.)t RI ;,Yx>
In this talk
pAn abstract version of Racke s result.
Applies more generally.
Tighter parameters.
More modular presentation: easier to understand.
(Implicit in Racke s work. We make it explicit.)6' _ 3 ,vThe abstract frameworkE set of items (edges).
P 2E collection of subsets of E ( paths ).
M : E P maps edges to paths. Can be represented as a matrix:
Mij indicates if edge j lies in path M(i) in map M.
M family of admissible mappings (trees).
Probabilistic mapping between E and M: probability distribution over mappings M 2 M.
lM weight of mapping M\ G5E)G,#A mapping M*
Edge 2 is mapped to the path {1,2,4}+ + Distance mapping~Edge i has positive length leni.
distM(i)=Sj Mij lenj length of path M(i).
distM(i)/leni stretch of edge i.
Average stretch of edge i under a probabilistic mapping: weighted average according to lM of stretches of i.
Stretch of probabilistic mapping: maximum over all edges of their average stretches.@ >Wt
NotesFor certain length functions and mappings, the stretch of an edge may be smaller than 1.
The stretch of the shortest edge is at least 1.
Stretch of probabilistic mapping is at least 1. $Capacity mappingPEdge i has positive capacity capi.
loadM(j) = i Mij capi
sum of capacities of edges whose paths under M contain edge j.
Congestion of edge j = loadM(j)/capj.
Average congestion of edge j: average weighted according to lM.
Congestion of probabilistic mapping: maxj[avgM-congestion(j)]. > Z? Z Z2
-t[
mNotesFor certain mappings, the congestion of an edge may be smaller than 1.
The sum of capacities of M(E) is at least as large as sum of capacities of E.
Congestion of probabilistic mapping is at least 1. $1-Main theorem
For every r and every admissible family M, the two statements are equivalent:
For every collection of lengths leni there is a probabilistic mapping with stretch at most r.
For every collection of capacities capi there is a probabilistic mapping with congestion at most r.N
E7%9,n]=2.A spanning tree exampleE set of edges in a graph.
P set of paths in the graph.
M Admissible maps correspond to spanning trees of the graph.
M an edge is mapped to the unique path between its endpoints.b <=3/An instructive graph
40Probabilistic mappingsFor any choice of lengths, probabilistic mapping with max-avg(stretch) < 3.
For any choice of capacities, probabilistic mapping with max-avg(congestion) < 3.
Will illustrate both statements when the graph is unweighted. & @ >:D:
51Uniform length, low stretch
62Delete at random one column
731Delete at random one columnMax[avg(stretch)] < 222(
84 Uniform capacity, low congestion!!(
95Keep at random one path
:60Keep at random one pathMax[avg(congestion)] < 311(
;7,Simultaneous low stretch and low congestion--(If top edge is in spanning tree, it is loaded by n1/2.
If top edge is not in spanning tree, it is stretched by n1/2.
Hence for every distribution over spanning trees, either stretch or congestion is n1/2/2. 1:T<8Implication for main theoremFor every r and every admissible family M, the two statements are equivalent:
For every collection of lengths leni there is a probabilistic mapping with stretch at most r.
For every collection of capacities capi there is a probabilistic mapping with congestion at most r.
The probabilistic mappings achieving the two guarantees are sometimes very different from each other.N f
H7(:f,n]/A useful equivalence [Alon, Karp, Peleg, West]&0 ,Lemma: for every r 1, admissible family M and length function len, there is a probabilistic mapping of stretch at most r iff for every nonnegative coefficients ai (with S ai = 1) for the edges, there is a mapping M 2 M s.t.
S ai distM(i)/leni r 6 ( % C
PA8_
.Proof use 0-sum gamesDView as a 0-sum two player game. Map player chooses an admissible mapping M. Edge player chooses an edge i.
Value for edge player: stretch of i under M.
Probabilistic mapping
= a mixed strategy for the map player.
Nonnegative coefficients ai
= a mixed strategy of the edge player.d# !
"-QEThe minimax theorem The lemma is equivalent to the minimax theorem:
minMmaxE(stretch)=maxaminM(stretch)V 2$
$$
$
$$$
$
$>If direction
Inequality holds.
Edge player cannot ensure stretch larger than r.
Map player can limit edge player to stretch of r.
Probabilistic mapping is a mixed strategy for map player.
Wins against every edge.D @1UOnly if directionNo probabilistic mapping of stretch at most r.
Map player cannot limit value to r.
Edge player can get expected value larger than r.
Mixed strategy for edge player gives ai.
Map player cannot choose a map M satisfying the inequality.~ ,#1'>Likewise& ((Lemma 2: for every r 1, admissible family M and capacity function cap, there is a probabilistic mapping of congestion at most r iff for every nonnegative coefficients bi (with S bi = 1) for the edges, there is a mapping M 2 M s.t.
S bi loadM(i)/capi r
9 ( * C
>_
Back to main theoremFor every r and every admissible family M, the two statements are equivalent:
For every collection of lengths leni there is a probabilistic mapping with stretch at most r.
For every collection of capacities capi there is a probabilistic mapping with congestion at most r.N
H7(:,n]=!-Proof that low congestion implies low stretch..(\Assume probabilistic mapping with congestion r.
By Lemma 2, for every b, there is M with S bj loadM(j)/capj r.
Rewriting the load as a sum:
Si,j bj Mij capi/capj r. -
P_
& "Need to prove that there is a probabilistic mapping with stretch at most r.
By Lemma 1, suffices to show that for every a, there is a mapping M with S ai distM/ leni r.
Rewriting distance as sum:
Si,j ai Mij leni/lenj r. I, 0
b$ #@Know that for all b and cap there is M
Si,j bj Mij capi/capj r.
Find for every a and len an M satisfying
Si,j ai Mij leni/lenj r.
Choose bj = aj and capi = ai/leni (and then of course capj = aj/lenj), find M that satisfies the first inequality, and it will satisfy the second. !
P) P$ -Proof that low stretch implies low congestion..(Z
Same as before, this time choosing ai = bi and leni = bi/capi, finding M that satisfies the second inequality, which will necessarily also satisfy the first inequality. ($$$$$$$
$$$$$
$l$,4l%!The proof does not assume that: (The matrix M is a 0/1 matrix.
Lengths and capacities of edges are nonnegative (though they cannot be 0).
Paths are actual paths in graphs.
Edges are edges of graphs.
Hence the theorem applies in more general situations.& 7 D;The algorithmic aspectKThe proof of the main theorem as presented is existential.
Will give a polynomial time algorithm that finds a probabilistic mapping whenever the associated 0-sum two player game can be solved efficiently.
Standard low regret online algorithms (e.g., multiplicative weight update) solve 0-sum games under fairly general conditions.L LH=Approximation algorithm( Let r be a desired target distortion for probabilistic capacity embedding. To efficiently find such a mapping, it suffices that for every nonnegative coefficients bi (with S bi = 1) for the edges, can efficiently find a mapping M 2 M s.t.
S bi loadM(i)/capi r
6 C
,
J> Alternative sufficient condition&! ((By main theorem, it also suffices that for every length function len and every nonnegative coefficients ai (with S ai = 1) for the edges, can efficiently find a mapping M 2 M satisfying
S ai distM(i)/leni rd A $ 1 C
,A
F:)Back to min-bisection simplified version**(E is a set of edges in a graph G.
P is the collection of all paths.
M is collection of spanning trees.
Every edge of G is mapped to its unique path in the tree.
Load on tree edge equals the capacity of the cut.Jh l ""K?4Load of tree bisection is lower than graph bisection55(_The load on a cut in the tree is at least as high as the capacity of the same cut in the graph.` `L@Approximating min-bisectionConsider minimum bisection in graph. Its average load in the dominating trees (average weighted by lM) is at most r times larger than its capacity.
Hence the minimum of all minimum load bisections of trees is a r approximation for minimum bisection.h c
a&MAThe value of r(
For spanning trees, max-avg(stretch) is at most O*(log n) [Elkin, Emek, Spielman and Teng; Abraham, Bartal and Neiman].
By our main theorem, the same holds for congestion.
Gives O*(log n) approximation for bisection.@ ;dboNB^Racke s approximation for min-bisection00()SUse probabilistic mappings into trees (that may use non-graph edges). There, max-avg(stretch) is at most O(log n) [Fakcharoenphol, Rao, and Talwar].
Every edge of the tree is a shortest path in G. Every edge of G is mapped to its unique path in the tree, and hence to a sequence of paths of G.
Hence M has nonnegative integer entries..T ZwbU
OCSummarynProbabilistic mappings of low congestion are useful algorithmic tools for cut problems (and for oblivious routing & ).
Given the same family of admissible mappings, the maximum stretch (over all length functions) equals the maximum congestion (over all capacity functions).
Proof is algorithmic (in most cases).8 8r
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