Approximations for hard problems

Department of Computer Science | Institute of Theoretical Computer Science

Theory of Combinatorial Algorithms Prof. Emo Welzl

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? £ ?$VX¿¿?ÿ??¦ i ?¨ Approximations for hard problems¡!! ?¢� ��

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? £ ?đX¿¿?ÿ?j??< ?¨Algorithm: List scheduling Basic idea: In a list of jobs, schedule the next one as soon � �� as a machine is free¡* g? ? s ?¿?ÿ??@p°? ? s ?¿?ÿ??@ ?? ? ?¿?ÿ? @°?? ? s ?¿?ÿ? @p ? ? s ?¿?ÿ?? P??¢� ��

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? £ ?4½X¿¿?ÿ?j??< ?¨Algorithm: List scheduling Basic idea: In a list of jobs, schedule the next one as � �� soon as a machine is free¡* g? ? s ?¿?ÿ??@p°? ? s ?¿?ÿ??@ ?? ? ?¿?ÿ? @°?? ? s ?¿?ÿ? @p ? ? s ?¿?ÿ?? � �� P??¢

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? £ ?D?¿¿?ÿ?z?c ?¨ machine 1? ? ³ ?? ?¨xTheorem: k-center is NP-complete (decision version). Proof: � �� Vertex cover reduces to dominating set reduces to k-center.¡yy? ?`

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?d £ ??\¿¿?ÿ? !¤ ?� �� >Example Problem: k-center. Given: G = (V,E); E = V x V; c(i,j) for all edges (i,j) with i ¹?j; positive integer number k of clusters. Compute set S of k cluster centers, S a subset of V, such that the distance of any point to its closest cluster center is minimum.¡: R??¢�

?d £ ??¿¿?ÿ???>? ?¨xTheorem: k-center is NP-complete (decision version). Proof: Vertex cover reduces � � � �� to dominating set reduces to k-center.¡yy?¢

?d £ ? ?¿¿?ÿ? ?� � � �� ?For G=(V,E), find a smallest subset V of V such that every vertex is either itself in V or has a neighbor in V .? ?d ?¿??oÿ?P ???? ?d ?¿??oÿ?p P P ? ?d� ��

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?h £ ??\¿¿?ÿ???>? ?¨D Proof: Vertex cover reduces to dominating set reduces to k-center.¡EEªD?¢� ��

? h £ ?\?¿¿?ÿ? ?� �� ?For G=(V,E), find a smallest subset V of V such that every vertex is either itself in V or has a neighbor in V .? ? h ?¿??oÿ?P ???? ? h ?¿??oÿ?p P P ?B ?� ��

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?l £ ?4X\¿¿?ÿ???>? ?¨D Proof: Vertex cover reduces to dominating set reduces to k-center.¡EEªD?¢� ��

?l £ ?,?¿¿?ÿ? ?� �� ?For G=(V,E), find a smallest subset V of V such that every vertex is either itself in V or has a neighbor in V .? ?l ?¿??oÿ?P ???? ?l ?¿??oÿ?p P P ?B ?l s ?D¿??oÿ?° ° ?2 ?l ?¿?? ÿ?PP ° °?2 ?l � �� Ԕ ?¿?? ÿ?P° °?2 ?l ?¿??oÿ?°0 ?2 ?l ?¿??oÿ?� � �� Ԕ ? ??B ?l?s ?D¿??oÿ??P� � ?¢

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?p £ ??]¿¿?ÿ?jF? ?¨®Non-approximability Theorem: Finding a ratio-M-approximation for fixed M is NP-� � � �� hard for k-center. Proof: Replace 2 in the construction above by more than M.¡ ¯ª? ?p� �

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?t £ ??]¿¿?ÿ?jF?R ?¨QNon-approximability Theorem: Finding a ratio-M-approximation for fixed M is � �� NP-hard for k-center. Proof: Replace 2 in the construction above by more than M. Theorem: Finding a ratio-less-than-2-approximation is NP-hard for k-center with triangle inequality. Proof: Exactly as in the reduction above.¡R² ª>? ?t

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?x £ ?¸+]¿¿?ÿ?¦?� �� ?¨YTheorem: A 2-approximation for k-center with triangle inequality exists.¡ZZ? ?x�

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? � Theorem: A 2-approximation for k-center with triangle inequality exists. Proof: Gonzalez algorithm. Pick v1 arbitrarily as the first cluster center. Pick v2 farthest from v1. Pick v3 farthest from the closer of v1 and v2. & Pick vi farthest from the closest of the v1,& vi-1. & until vk is picked.¡[/ª| ? ?|�

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? £ ?t5X¿¿?ÿ? ? ?� � �� ?Example Problem: Traveling salesperson with triangle inequality. Given: G = (V,E); E = V x V; c(i,j) for all edges (i,j) with i ¹?j. Compute a round trip that visits each vertex exactly once and has minimum total cost. Comment: is NP-hard.¡:þ x|? ?� �

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? £ ??]¿¿?ÿ? ?º ?� � � � �� ¾Example Problem: Traveling salesperson with triangle inequality. Given: G = (V,E); E = V x V; c(i,j) for all edges (i,j) with i ¹?j. Compute a round trip that visits each vertex exactly once and has minimum total cost. Approximation algorithm Find a minimum spanning tree. Run around it and take shortcuts to avoid repeated visits.¡T` xlZª?s? ?� �

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[? ??? ? ?? ? S ??]¿ÿ?°?P??ÿÿÿÿ ] ?¨qApproximation algorithm Find a minimum spanning tree. Run � � � �� around it and take shortcuts to avoid repeated visits.¡"rZªq?¢

? £ ?W]¿¿?ÿ???_^ ?¨?Quality ? Theorem: This is a 2-approximation. Proof: (1) OPT-TSP minus 1 edge is � � � � �� spanning tree. (2) MST is not longer than any spanning tree. (3) APX-TSP <= 2 MST <= 2 ST <= 2 OPT-TSP.¡$? ?? ? �

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? £ ?a]¿¿?ÿ?¦S¶ ?ldº___PPT9F¬> ?� �� Better quality ? Christofides algorithm. Find MST. Find all odd �degree vertices V in MST. Comment: There is an even number of them. Find a minimum cost perfect matching for V in the induced subgraph of G (no even vertices present). Call this M. In MST + M, find an Euler circuit. Take shortcuts in the Euler circuit.¡,20;=HbxH�

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? £ ?e]¿¿?ÿ?v?� �� ? � ?Theorem: Christofides algorithm is a 1.5-approximation of TSP with triangle inequality.¡iiª R? ?�

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? £ ?]¿¿?ÿ?v? ?H@º___PPT9"¬ ?� �� ¢Theorem: Christofides algorithm is a 1.5-approximation of TSP with triangle inequality. Proof: MST <= TSP. M <= TSP / 2 & . as we see soon. 3. Shortcuts make the tour only shorter.¡Br6*j`ª »¦? @` � ? ? �

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?´ £ ?0]¿¿?ÿ?� � �� P?T?XPº___PPT92¬* ?� � � � *Proof of M <= TSP / 2: consider the subgraph induced by odd degree vertices V ; consider TSP restricted to that subgraph, and compare with the sub-TSP(V ) found there: sub-TSP(V ) <= TSP; picking alternate edges in sub-TSP(V ) gives two perfect matchings for V , call them M1 and M2; pick the cheaper of M1, M2, call it M, with M <= TSP / 2.¡H ? ?ª>4 R? ¦? @`� � ? ?´

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? £ ?°]¿¿?ÿ?j¦IR ?xpº___PPT9R¬J ?¨!Notes: Christofides algorithm may be as bad as 1.5 really. It is � � � �� unkown whether this is best possible. For Euclidean TSP, a better bound is known: any quality above 1 can be achieved. For TSP without triangle inequality, no fixed approximation is possible. ¡ a,3< b_V�

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? £ ??¿¿?ÿ? N¤ ?� �� ?Example Problem: Set cover. Given: Universe U of elements e1, e2, & , en; collection of subsets S1, S2, & , Sk of U, nonnegative cost per Si. Find a collection of subsets that cover U with minimum cost.¡? ?ª,x(?? ? �

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? £ ?°?¿¿?ÿ? N¤ ?� �� ?Example Problem: Set cover. Given: Universe U of elements e1, e2, & , en; collection of subsets S1, S2, & , Sk of U, nonnegative cost per Si. Find a collection of subsets that cover U with minimum cost.¡? ?ª,x(??¢

? £ ?¸?¿¿?ÿ?J ? ?¨Idea for an approximation: Repeat greedily choose a best set (cheapest per new covered� �� element) until all of U is covered.¡? ?��

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?¤ £ ?x?¿¿?ÿ??v? ?¨YConsider a step in the iteration. The greedy algorithm has selected some of the Sj, � � � �� with covered C = union of all selected Sj so far. Choose Si in this step. Price per new element in Si is cheapest: price(i) = cost(Si) / (number of new elements of Si) is minimum. Consider the elements in the order they are chosen.ªtQ1(&L? ?¤

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? � ?Consider the elements in the order they are chosen, and rename: e1, e2, e3, & ., ek, & . en Consider ek, call the set in which ek is chosen Si. What is the price of ek? Bound the price from above: Instead of Si, the greedy algorithm could have picked any of the sets in OPT that have not been picked yet (there must be one), but at which price? ? compare with the optimum¡$|aªtP +©? ?¨�

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?¸ £ ?_¿¿?ÿ??^ ?¨Instead of Si, the greedy algorithm could have picked any of the sets in OPT that have � �� not been picked yet (there must be one), but at which price? Take all elements not covered yet, and their total cost will be at most all of OPT. Across all sets in OPT not picked yet, the average cost is therefore at most OPT / size of U-C. Hence, at least one of the sets in OPT not picked yet has at most this average price. This set could have been picked. Hence, the price of ek is at most OPT / size of U-C.ª> ?? ?¸

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?¬ £ ??_¿¿?ÿ?:?V ?¨zFor the k-th picked element, the size of U-C is at least n-k+1. Therefore, price(ek) � �� <= OPT / (n-k+1) for each ek. The sum of all prices of ek gives the total cost of the greedy solution: SUM(k=1,..,n) price(ek) <= SUM(k=1,..,n) OPT / (n-k+1) <= OPT SUM(k=1,..,n) 1 / k <= OPT (1 + ln n) ªt EC? ?¬�

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?¨£Theorem: Greedy set cover gives a (1+ln n)- approximation. Notes: It can really be that bad. That is �also a best possible approximation for set cover.¡2¤<gª%|? ?¼� �

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?° £ ??_¿¿?ÿ? ?� � � �� tExample Problem: Knapsack. Given: n objects with weights and values, and weight bound: positive integers w1,w2, & , wn, W (weights, total weight); positive integers v1, v2, & , vn (values). Find a subset of the set of objects with total weight at most W and maximum total value.¡; 2ª,Gp?¢�

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?? £ ?1_¿¿?ÿ?)D ?� � � �� >An exact algorithm for knapsack A 1 2 3 v n vmax 1 2 3 j nª,h( ?B ?? c ?D¿?ÿ??º??�B ?? c ?D¿?ÿ?vªª&?B ?? c ?D¿?ÿ?vjj&?B ?? c ?D¿?ÿ?v* * &?B ?? c ?D¿?ÿ?vJJ&?B ? ? c ?D¿?ÿ?6 º6 ?� � � � �B ? ? c ?D¿?ÿ? º ?B ? ? c ?D¿?ÿ?&?J&?¢� � � �

? £ ?¨9_¿¿?ÿ?0?P ?� �� A(j,v ) = smallest weight subset of objects 1,& ,j with total value =v .? ?? � ?��¿??� ÿ?Ԕ0`?P? ?? ?¿?? ÿ?P ? p ? ??� �� Ԕ

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?? £ ?¤D_¿¿?ÿ?ª¶[x ?� �� binductively: A(1,v) = If v = v1 then w1 else infinity A(i+1,v) = min ( A(i,v) , A(i, v v(i+1)) + w(i+1) ) if >= 0 & the result is: max v such that A(n,v) <= W & the runtime is: O(n2 vmax) & . pseudopolynomial¡d2| = ÿþFÿþª>I?? ?? ?¿??oÿ?@`� ��

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?? £ ?P^_¿¿?ÿ?:F?¤ ?¨? pseudopolynomial ? polynomial if numbers are small = value is polynomial in � �� input length Idea: scale numbers down, i.e., ignore less significant digits.¡*?<`@ª?? ??

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?? £ ?j_¿¿?ÿ?Z¦z ?H@º___PPT9"¬ ?¨*Approximation algorithm for knapsack (FPTAS) Given error � � �� bound eps > 0, define K := eps vmax / n . For each object i, define vi := rounded down (vi / K). Use dynamic programming to find optimal solution S for the rounded problem version. 4. Return max (original value for S , vmax).¡n.£Z.RMO ªP@?¦? @` � ? ??

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܌? £ ? �c¿¿?ÿ?4 ?� � �� [B] Lemma: Given items with size >= e, there is an algorithm with runtime polynomial in n and approximation factor 1+e. Proof: let input instance be I = {a1, a2, & . , an}. - Sort items by increasing size. - Partition items into K= ceiling(1/e^2) groups:¡Jw8 &?2 ? s ?¿?ÿ?@?p?2 ? s ?¿?ÿ?@Pp?�� 2 ? s ?¿?ÿ?@�� ?p?2 ? s ?¿?ÿ?@0`p?2 ? s ?¿?ÿ?@? p?2 ? s ?¿?ÿ?@ P p?2 ? s ?¿?ÿ?@0p?2 ?��

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?8 £ ??U¿¿?ÿ? ??TLº___PPT9.¬& ?� �� 0Algorithm (Sekanina) Build MST T. Build T^3. Find round trip in T^3 such that each of T appears exactly twice (induction, and a single extra edge at the very end). Quality for TSP with slight violation of triangle inequality? cost(i,j) <= (1+r) (cost(i,k) + cost(k,j)) for all i,j,k Shortcut increases length by factor <= (1+r)^2 Approximation ratio 2 (1+r)^2 & stability of the approximation¡¦Y¹O?"?¨ÿþÿþ./Bª, ?¦? @`� � � � � ? ?8

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?H £ ?¤?¿¿?ÿ?``? ?� �� ?Summary Problems in NPO have very different approximability properties: & some are impossible to approximate (k-center, TSP) & some are hard, with a bound depending on the input size (set cover) & some can be approximated with some constant ratio (vertex cover, k-center with triangle inequality, TSP with triangle inequality) & and some can be approximated as closely as you like (knapsack) Approximability has its own hierarchy of complexity classes¡*?´<ª,-,? ?H�

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Approximations for hard problems

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