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.

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-hardness

• Bi-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.

Intuition

Unary subset sum can be reduced to deciding whether the bisection width is 0.

Any approximation algorithm for bisection “must” use dynamic programming.

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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.

Plan of attack

Given 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.

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.

Another problem

“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.

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.

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 theorem

Theorem: 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.)

In this talk

An 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.)

The abstract framework

E set of items (edges).

P ½ 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.

M weight of mapping M

A mapping M

Edge 2 is mapped to the path {1,2,4}

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Distance mapping

• Edge i has positive length leni.

• distM(i)=j 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 M of stretches of i.

• Stretch of probabilistic mapping: maximum over all edges of their average stretches.

Notes

For 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 mapping

• Edge 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 M.• Congestion of probabilistic mapping:

maxj[avgM-congestion(j)].

Notes

For 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.

Main theorem

For every 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 .

• For every collection of capacities capi there is a probabilistic mapping with congestion at most .

A spanning tree example

E 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.

An instructive graph

Probabilistic mappings

• For 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.

Uniform length, low stretch

Delete at random one column

Delete at random one columnMax[avg(stretch)] < 2

Uniform capacity, low congestion

Keep at random one path

Keep at random one pathMax[avg(congestion)] < 3

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.

Implication for main theorem

For every 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 .

• For every collection of capacities capi there is a probabilistic mapping with congestion at most .

The probabilistic mappings achieving the two guarantees are sometimes very different from each other.

A useful equivalence [Alon, Karp, Peleg, West]

Lemma: for every 1, admissible family M and length function len, there is a probabilistic mapping of stretch at most iff for every nonnegative coefficients i (with i = 1) for the edges, there is a mapping M 2 M s.t.

i distM(i)/leni

Proof – use 0-sum games

View 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 i

= a mixed strategy of the edge player.

The minimax theorem

The lemma is equivalent to the minimax theorem:

minMmaxE(stretch)=maxminM(stretch)

If direction

Inequality holds.

Edge player cannot ensure stretch larger than .

Map player can limit edge player to stretch of .

Probabilistic mapping is a mixed strategy for map player.

Wins against every edge.

Only if direction

No probabilistic mapping of stretch at most .

Map player cannot limit value to .

Edge player can get expected value larger than .

Mixed strategy for edge player gives i.

Map player cannot choose a map M satisfying the inequality.

Likewise

Lemma 2: for every 1, admissible family M and capacity function cap, there is a probabilistic mapping of congestion at most iff for every nonnegative coefficients i (with i = 1) for the edges, there is a mapping M 2 M s.t.

i loadM(i)/capi

Back to main theorem

For every 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 .

• For every collection of capacities capi there is a probabilistic mapping with congestion at most .

Proof that low congestion implies low stretch

Assume probabilistic mapping with congestion .

By Lemma 2, for every , there is M with j loadM(j)/capj .

Rewriting the load as a sum:

i,j j Mij capi/capj .

Need to prove that there is a probabilistic mapping with stretch at most .

By Lemma 1, suffices to show that for every , there is a mapping M with i distM/ leni .

Rewriting distance as sum:

i,j i Mij leni/lenj .

Know that for all and cap there is M

i,j j Mij capi/capj .

Find for every and len an M satisfying

i,j i Mij leni/lenj .

Choose j = j and capi = i/leni (and then of course capj = j/lenj), find M that satisfies the first inequality, and it will satisfy the second.

Proof that low stretch implies low congestion

Same as before, this time choosing i = i and leni = i/capi, finding M that satisfies the second inequality, which will necessarily also satisfy the first inequality.

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.

The algorithmic aspect

• The 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.

Approximation algorithm

Let be a desired target distortion for probabilistic capacity embedding. To efficiently find such a mapping, it suffices that for every nonnegative coefficients i (with i = 1) for the edges, can efficiently find a mapping M 2 M s.t.

i loadM(i)/capi

Alternative sufficient condition

By main theorem, it also suffices that for every length function len and every nonnegative coefficients i (with i = 1) for the edges, can efficiently find a mapping M 2 M satisfying

i distM(i)/leni

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.

Load of tree bisection is lower than graph bisection

The load on a cut in the tree is at least as high as the capacity of the same cut in the graph.

Approximating min-bisection

Consider minimum bisection in graph. Its average load in the dominating trees (average weighted by M) is at most times larger than its capacity.

Hence the minimum of all minimum load bisections of trees is a approximation for minimum bisection.

The value of

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.

Racke’s approximation for min-bisection

Use 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.

Summary

Probabilistic 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).