Unformatted text preview: Theorem 17 ISIT-`2 2 P. i.e., ISIT-`2 can be answered in polynomial time. Proof: Assume that there exists an embedding of f1; 2; : : : ; ng into fv1 = 0; v2; : : : ; vng.
Consider one such embedding. Then, d2i; j = jjvi , vj jj2 = vi , vj vi , vj = vi2 , 2vi vj + vj2:
But vi2 = d21; i and vj2 = d21; j which means that, vi vj = d 1; i + d 1; j , d i; j : 2
2 2 2 We now construct M = mij where, mij = d 1; i + d 1; j , d i; j : 2
2 2 2 Hence if M is not positive semi-de nite then there is no embedding into `2. If M is positive semide nite then we carry out a Cholesky decomposition on M to express M as M = V V T . From the rows of V we can obtain an embedding into Rn; `2. Theorem 18 ISIT-`1 is NP-complete.
This theorem is given without proof. The reduction is from MAX CUT, since as we will see later there is a very close relationship between l1-embeddable metrics and cuts. We also omit the proof of the following theorem. Theorem 19 Let X Rn. X; `2 can be embedded into Rm; `1 for some m.
The converse of this theorem is not true as can be seen from the metric space f0; 0; ,1; 0; 1; 0; 0; 1g; `1 . 8.1 Reducing multicommodity ow cut questions to embedding questions
In this section, we relate Claim 20 Proof: and through the use of metrics. = min `1-embeddable metrics V; ` Approx-41 P u `x; y x;y Pk2E `xy ; t : f s
i=1 i i i Given S , let 'x = 0 if x 62 S . 1 if x 2 S Let ` be the `1 metric on the line, i.e. `a; b = ja , bj. Then, X u S = uxy `x; y f S =
x;y2E k X fi`si ; ti i=1 since `x; y = 1 if and only if x is separated from y by S and 0 otherwise. We can view any `1 embeddable metric ` as a combination of cuts. See gure 11 for the 2-dimensional case.
2 (1,3,4) 4 (1,3) 3 (1) 1 (1) (1,2) (1,2,3) Figure 11: Viewing an `1-metric as a combination of cuts. For any set S de ne a metric 1S by, 1S = 1 if x; y are separated by S 0 otherwise. Then we can write ` as, X `x; y = i 1S x; y ;
i where the i's are nonnegative. Hence, P P u S u S x;y2E uxy `x; y i Pk f `s ; t = P i f S i min f S : S
i=1 i i i i i i Approx-42 Claim 21
= min `1 -embeddable metrics V; d P x;y2E uxy dx; y P f ds ; t : i i i i Note that by theorem 16 we actually minimize over all metrics. Proof: For any metricP let the volume of an edge x; y be uxy dx; y. The total volume d of the graph is x;y2E uxy dx; y. If we send a fraction of the demand then the amount of volume that we use is at least Pi fidsi; ti. Hence Pi fi dsi; ti P
x;y2E uxy dx; y . We use the strong duality of linear programming. can be formulated as a linear program in several di erent ways. Here we use a formulation which works well for the purpose of this proof although it is quite impractical. Enumerate the paths from si to ti, let Pij be the j th such path and let xij be the ow on Pij . The linear program corresponding to multicommodity ow is, Max subject to: X fi , xij 0 X Xj xij ue i 2 f1; : : : ; kg e2E 0 xij 0 i j :e2P ij The dual of this linear program is: X Min ue`e subject to: k X
i=1 e2E e2P hi 0 `e 0
ij fi hi 1 X `e , hi 0 8i; j The second constraint in the dual implies that hi is at most the shortest path length between si and ti with respect to `e . By strong duality if ` is an optimum Approx-43 solution to the dual then, = ue`e e2E P u` Pek2E f eh e P i=1 i ui di; j Pi;j2E dijs ; t ; k f i=1 i i i X where da; b represents the shortest path length with respect to `e . The rst inequality holds because Pk=1 fihi is constrained to be at least 1. i Linial, London, and Rabinovitch and Aumann and Rabani use the following strategy to bound and approximate the minimum multicommodity cut. 1. Using linear programming, nd and the corresponding metric d as given in claim 21. 2. Embed d into Rm; `1 with distortion c. Let ` be the resulting metric. By claim 20 this shows that c since, P P x;y x;y P 2E uxy `x; y c P 2E uxy dx; y = c : k f `s ; t k f ds ; t i=1 i i i i=1 i i i In order to approximate the minimum multicommodity cut, we can use the proof of claim 20 to decompose ` into cuts. If S is the best cut among them then, P u S x;y2E uxy `x; y : Pk f `s ; t f S i=1 i i i Our remaining two questions are: How do we get an embedding of d into `? Equivalently, how can we embed `1 into `1. What is c? 8.2 Embedding metrics into `1 The following theorem is due to Bourgain. Theorem 22 For all metrics d on n points, there exists an embedding of d into `1 which satis es: dx; y jjx , yjj1 Olog ndx; y: Approx-44 Let k range over f1; 2; 4; 8; : : : ; 2j ; : : : ; 2pg where p = blog nc. Hence we have p + 1 = Olog n di erent for k. Now choose nk sets of size k. At rst values take all sets of size k, i.e., nk = n . Introduce a coordinate for every such set. This k implies that points are mapped into a space of dimension Pp=0 n2 2n . For a set A j of size k the corresponding coordinate of a point x is, dx; A nk where dx; A = minz2A dx; z and is a constant which we shall determine later. Suppose that dx; A = dx; s and dy; A = dy; t, where s and t are in A. Then, dx; A , dy; A = dx; s , dy; t dx; t , dy; t dx; y: Exchanging the roles of x and y, we deduce that jdx; A , dy; Aj dx; y. Hence, X jjx , yjj1 = n jdx; A , dy; Aj A jAj X 1 n dx; y
j Proof: = p + 1dx; y = Olog ndx; y: We now want to prove that jjx , yjj1 dx; y. Fix two points x and y and de ne, B x; r = fz : dx; z rg; B x; r = fz : dx; z rg; 0 = 0; t t t = inf fr : jB x; rj 2 ; jB y; rj 2 g: A jAj Let ` be the least index such that ` dx;y . Rede ne ` so that it is equal to dx;y . 4 4 Observe that for all t either jBx; tj 2t or jBy; tj 2t. Since B x; `,1 B y; `,1 = ; we have 2`,1 +2`,1 n ` p. Now x k = 2j where p,1 j p,` and let t = p , j thus, 1 t l. By our observation we can assume without loss of generality that jBx; tj 2t . Let A be a set of size k and consider the following two conditions. 1. A Bx; t = ;. 2. A B y; t,1 6= ;. If 1. and 2. hold then dx; A t and dy; A t,1 and so jdx; A , dy; Aj t , t,1 . Let Rk = fA : jAj = k and A satis es conditions 1. and 2.g Approx-45 Lemma 23 For some constant 1 independent of k, there are at least of size k which satisfy conditions 1. and 2, i.e. jRk j n . From this lemma we derive, p,1 X X jjx , yjj1 jdx; A , dy; Aj R nk j =p,`;k=2 p,1 X nk nk p,j , p,j,1 j =p,`;k=2 p,1 X p,j , p,j,1 =
k j k j n k sets = j =p,` ` = 4 dx; y: Hence if we choose = 4 then we have jjx , yjj1 dx; y. We now have to prove lemma 23. Proof of lemma 23: Since jBx; tj 2t, jB y; t,1j 2t,1 and we are considering all sets of size k the following is a restatement of the lemma: Given disjoint sets P and Q with a = jP j 2t and b = jQj 2t,1, if E is the event that a uniformly selected A misses P and intersects Q then Pr E 1 . We calculate this probability as follows: n,a n,a,b , Pr E = k n k !n k !n k = nn, aa , k, n! , nn, aa, bb , k, n! , ! ! , , ! ! a a k = n , a n ,,, 1 n ,,, ++ 1 , n n 1 n k 1 n , a, bn , a, b , 1 n , a ,b , k + 1 k n a n , 1 n , a + 1 a 1 , = 1, n 1, n,1 n , k + 1 ,! ! ! a+ b 1 , a + b 1 , a + b : 1, n n,1 n,k +1 a As an approximation this can be made formal, we replace 1 , n,j by e,a=n, and a 1 , n+b by e,a+b=n. Thus, ,j Pr E e, , e, e, 1 , e, :
ak n k a+bk
n ak n bk n Approx-46 This for example shows that if a; b and k are all n then this probability is a constant, which may seem a bit paradoxical. Using our bounds on a and b, we get , 1 , e, Pr E e, 1 , e, e, , , , e,1 1 , e, : = e 1,e
ak n bk n p 2t 2j
n 2t 1 2j n 2p
n 2p 1 n 1 4 ,1 and the proof is complete. We now choose e,1 1 , e, Bourgain's proof is not quite algorithmic since the dimension is exponential. Linial, London and Rabinovitch just sample uniformly with nk = Olog n and show that with high probability the embedding has the required properties. This follows from a Cherno bound. We have thus shown that the distortion c can be chosen to be Olog n. We can do even better by proving the following variant to Bourguain's theorem.
1 4 Theorem 24 Let d be a metric on a set V of n points. Suppose that T V and jT j = k. Then there exists an embedding of d into `1 which satis es: jjx , yjj1 Olog kdx; y 8x; y 2 V: jjx , yjj1 dx; y 8x; y 2 T:
In order to prove this theorem we restrict the metric to T and then embed the restricted metric. If we look at the entire vertex set V then the rst part of the original proof still works. This new theorem is enough to show that Olog k and to approximate the multicommodity cut to within Olog k. This result is best possible in the sense that we can have = log k. References
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