Topics in Algorithms 2007 Ramesh Hariharan. Tree Embeddings.
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Transcript of Topics in Algorithms 2007 Ramesh Hariharan. Tree Embeddings.
![Page 1: Topics in Algorithms 2007 Ramesh Hariharan. Tree Embeddings.](https://reader036.fdocuments.in/reader036/viewer/2022062409/5697bfa31a28abf838c96ded/html5/thumbnails/1.jpg)
Topics in Algorithms 2007
Ramesh Hariharan
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Tree Embeddings
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Solving Graph Metric Problems
How do we make the problem easier?
Convert into a tree!!
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Projections
Can vertices in a given weighted graph G be mapped to vertices in any edge-weighted tree H so that all distances (i.e., shortest paths) only increase, but not by too much?
Distortion: max stretch over all edges in G
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H vs G
Two cases |G|=|H| |G|<|H|
Two vertices in G cannot map to the same vertex in H
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A Simple Case
G=unweighted cycle of length n and |H|=|G|
How much is the distortion?
>=n-1?
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Proof of the Simple Case
Embedding of G in H must cancel out (why??)
Take each e in G, map to H and then map back to G: two cases Result is e itself G maps to itself contradiction
(why??) Result is some other path in G between the
endpoints of e distortion for e is n-1 (why??)
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Generalization
Embedding of unweighted G into H has distortion at least g-1 if X(H)<X(G) and |H|=|G| (g is girth of G, X() is E-V+1)
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Proof
Take each e in G, map to H and then map back to G: two cases
Result is e itself each cycle in G maps to itself
Result is some other path in G between the endpoints of e distortion for e is g-1 (why??)
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Proof
If each cycle in G maps to itself in G->H ->G
X(G) independent cycles in G map to at most X(H) independent cycles in H
X(H) independent cycles in H cannot map to more than X(H) fundamental cycles in G
Contradiction
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Further Generalization
Embedding of unweighted G into H has distortion at least gk-1 if X(H)<=X(G)-k and |H|=|G| (gk is the length of the kth smallest cycle in G, X() is E-V+1)
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Proof
Take each e in G, map to H and then map back to G: two cases
Result is e itself each cycle in G maps to itself
Result is some other path in G between the endpoints of e the cycle formed by the path + e is NOT amongst the k-1
shortest cycles distortion for e is at least g k -1 (why??) the cycle formed by the path + e IS amongst the k-1 shortest
cycles each cycle in G maps to itself plus a combination of the k-1 shortest cycles
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Proof
If each cycle in G maps (in G->H ->G) either to itself or to itself plus a combination of the smallest k-1 cycles of G
After G->H ->G there are at least X(G)-k+1 independent cycles in G (Why?)
G->H has only X(H)= X(G)-k basis cycles
Contradiction!!
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Steiner Points
Will |H|>|G| help?
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Further Generalization
Embedding of unweighted G into H has distortion at least gk/3-4/3 if X(H)=X(G)-k and |H|>=|G| (gk is the length of the kth smallest cycle in G, X() is E-V+1)
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Problem
Take each e in G, map to H and then map back to G: map back is not defined for vertices in H-G
Add extra degree 2 vertices to G (many choices, pick any one), i.e., artificially define map-back for H-G
Add degree two vertices to H so each edge in H has length at most 1 (technical)
E-V+1 is in G or in H as a result of the above unchanged
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Problem
Distances between new vertices in G may not expand in H
Translate to distances in terms of original vertices in G
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Problem
Take each e=(x,y) in G, map to H and then map back to G:
We get a sequence on vertices x a1 a2 a3….y; two cases
This cancels to just e itself each cycle in G maps to itself contradiction as before
This gives some other path in G between the endpoints of e
the cycle formed by the path + e IS amongst the k-1 shortest cycles each cycle in G maps to itself plus a combination of the k-1 shortest cycles contradiction as before
the cycle formed by the path + e is NOT amongst the k-1 shortest cycles (so length >= g k) cannot claim distortion at least g k -1 as before because the distance between ai and ai+1 in G need not be less than that in H (unless both vertices were in G originally)
Break e into edgelets based on intervening vertices in H Endpoints of an edgelet may not map on to themselves anymore Define G->H->G map for an edgelet xy to be x to the mapback x’ for x, then x’ to y’, then y’ to y Two cases
xy itself each cycle in G maps to itself
some other path in G between x and y
result+e is a cycle of length at least g k distortion is >= g k/3-4/3 (why?) or, result+e is a combination one the k-1 smallest basis cycles (why??)
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Proof
Consider G
Take images of ai and ai+1 on e=xy (distances between images are proportional to those in H)
Distance bi ai is <= gk/3-4/3 (Why?)
Distance bi ai is <= gk/3-4/3 (Why?)
Distance ai ai+1 is <= gk/3-1/3+2, likewise for bi bi+1 (Why?)
Total is <= gk-1, i.e., the cycle bi ai ai+1
bi+1 bi IS amongst the smallest k-1 cycles
Each cycle maps to itself plus a linear combination of small cycles contradiction as before
x y
ai+1ai
bi bi+1
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Exercise
Complete the proof
Is it tight for a cycle G? Is there a graph H with |H|>|G| so that the embedding has distortion as low as |G|/3?
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Way Forward
How about embedding not on to a single less complex graph but to a probability distribution of less complex graphs
What is the maximum expected stretch?
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Cycle to Paths
Take all n paths, each with prob 1/n
How large is the expected stretch for any edge? <=2!!
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Application
K-medians
Find k centers is a graph so sum of all vertices of distance to nearest center is minimized
Suppose you have an A approximate algorithm for trees
And also an embedding on to a prob dist of trees so that the maximum expected stretch is B
Then we can claim an expected approx factor AB on graphs (Why?)
How do we convert expectation to a high probability bound?
How about running time?
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Probabilistic Lower Bounds
Why do the previous arguments fail for probability distributions over graphs in H?
In each graph in H, some edge has a large distortion, but the prob weight on this graph is low.
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Two Player Games
Fix G and some algorithm to embed G into a tree
For all prob dist over trees. there exists an edge for which expected distortion >= c
2 player game You choose prob dist P over trees I choose an edge e Value of this game is the expected distortion for e wrt P I win if the value >=c otherwise you win
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Two Player Games
Flip the game
There exists an edge e, such that for all prob distributions over trees, the expected distortion for e >= c
2 player game I choose an edge e You choose prob dist P over trees Value of this game is the expected distortion for e wrt P I win if the value >=c otherwise you win
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Two Games
Game A You choose prob dist P over trees I choose an edge e Value of this game is the expected distortion for e wrt P I win if the value >=c otherwise you win
Game B I choose an edge e You choose prob dist P over trees Value of this game is the expected distortion for e wrt P I win if the value >=c otherwise you win
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Which game has a higher value?
Game A >= Game B
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Two Games
Game A You choose prob dist P over trees I choose an edge e Value of this game is the expected distortion for e wrt P I win if the value >=c otherwise you win
Game B I choose a probability distribution Q over edges You choose a tree Value of this game is the expected distortion for e wrt Q I win if the value >=c otherwise you win
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Which game has a higher value?
Game A >= Game B !!
Von Neuman’s Principle, Yao’s Lemma
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Probabilistic Lower Bounds
For all prob dist P over trees, there exists edge e with large expected distortion
There exists prob dist Q over edges, such that for all trees, the expected distortion is large
Pick Q, show that for all trees, the expected distortion is large
Eg, Q is uniform, simply show that the average distortion when embedding into any tree is large
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Another Example: RandomizedAverage Case
Show that no randomized algorithm can have good performance for all inputs
Show that for any deterministic algorithm, the average performance over all inputs is not good
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Probabilistic Lower Bounds
Find a graph G such that the average distortion when embedding into any tree is large, say log n/3
Show this for any graph which has >=2n edges Smallest cycle > K At least half the edges must have distortion >
K/3 Average distortion must be greater than K/3