Approximation Algorithms Subset Sum III Instance : X = { x 1 , . . - PowerPoint PPT Presentation

NEW CS 473: Theory II, Fall 2015 Subset Sum Approximation Algorithms Subset Sum III Instance : X = { x 1 , . . . , x n } – n integer positive num- bers, t - target number Question: ∃ subset of X s.t. sum of its elements is t ? Lecture 10 September 24, 2015 Assume x 1 , . . . , x n are all ≤ n . Then this problem can be solved in (A) The problem is still NP-Hard , so probably exponential time. (B) O ( n 3 ) . (C) 2 O (log 2 n ) . (D) O ( n log n ) . (E) None of the above. 1/41 2/41 Subset Sum Subset Sum Efficient algorithm??? Subset Sum 1. Algorithm solving Subset Sum in O ( Mn 2 ) . Instance : X = { x 1 , . . . , x n } – n integer positive num- 2. M might be prohibitly large... bers, t - target number 3. if M = 2 n = ⇒ algorithm is not polynomial time. Question: ∃ subset of X s.t. sum of its elements is t ? 4. Subset Sum is NPC . 5. Still want to solve quickly even if M huge. SolveSubsetSum ( X , t , M ) 6. Optimization version: b [0 . . . Mn ] ⇐ false // b [ x ] is true if x can be Subset Sum Optimization M : Max value // realized by subset of X . input numbers. Instance : ( X , t ) : A set X of n positive integers, b [0] ← true . and a target number t . for i = 1 , . . . , n do � Mn 2 � R.T. O . Question: The largest number γ opt one can repre- for j = Mn down to x i do sent as a subset sum of X which is smaller or equal b [ j ] ← B [ j − x i ] ∨ B [ j ] to t . return B [ t ] 3/41 4/41

Subset Sum Polynomial Time Approximation Schemes 2 -approximation Definition ( PTAS ) Lemma PROB : Max imization problem. 1. ( X , t ) ; Given instance of Subset Sum . γ opt ≤ t : Opt. ε > 0 : approximation parameter. 2. = ⇒ Compute legal subset with sum ≥ γ opt / 2 . A ( I , ε ) is a polynomial time approximation scheme 3. Running time O ( n log n ) . ( PTAS ) for PROB : � � � � � � 1. ∀ I : (1 − ε ) � opt( I ) � ≤ � A ( I , ε ) � ≤ � opt( I ) � , � � � � � � Proof. 2. | opt( I ) | : opt price, 1. Sort numbers in X in decreasing order. 3. | A ( I , ε ) | : price of solution of A . 2. Greedily - add numbers from largest to smallest (if 4. A running time polynomial in n for fixed ε . possible). For minimization problem: 3. s : Generates sum. | opt( I ) | ≤ | A ( I , ε ) | ≤ (1 + ε ) | opt( I ) | . 4. u : First rejected number. s ′ : sum before rejection. 5. s ′ > u > 0 , s ′ < t , and s ′ + u > t = ⇒ t < s ′ + u < s ′ + s ′ = 2 s ′ = ⇒ s ′ ≥ t / 2 . 5/41 6/41 Polynomial Time Approximation Schemes Approximating Subset Sum 1. Example: Approximation algorithm with running time Subset Sum Approx O ( n 1 /ε ) is a PTAS . Instance : ( X , t , ε ) : A set X of n positive integers, a Algorithm with running time O (1 /ε n ) is not. target number t , and parameter ε > 0 . 2. Fully polynomial... Question: A number z that one can represent as a sub- Definition ( FPTAS ) set sum of X , such that (1 − ε ) γ opt ≤ z ≤ γ opt ≤ t . An approximation algorithm is fully polynomial time approximation scheme ( FPTAS ) if it is a PTAS , and its running time is polynomial both in n and 1 /ε . 3. Example: PTAS with running time O ( n 1 /ε ) is not a FPTAS . 4. Example: PTAS with running time O ( n 2 /ε 3 ) is a FPTAS . 7/41 8/41

Approximating Subset Sum Trim the lists... Looking again at the exact algorithm Definition L ′ : Inc. sorted list of numbers For two positive real ExactSubsetSum ( S , t ) Trim ( L ′ , δ ) numbers z ≤ y , the n ← | S | L = � y 1 . . . y m � number y is a P 0 ← { 0 } curr ← y 1 δ -approximation to z if for i = 1 . . . n do y L out ← { y 1 } 1 + δ ≤ z ≤ y . P i ← P i − 1 ∪ ( P i − 1 + x i ) for i = 2 . . . m do Remove from P i all elements > t if y i > curr · (1 + δ ) Observation Append y i to L out If x ∈ L ′ then there exists return largest element in P n curr ← y i a number y ∈ L out such return L out that y ≤ x ≤ y (1 + δ ) , 1. S = { a 1 , . . . , a n } where L out ← Trim( L ′ , δ ) . x + S = { a 1 + x , a 2 + x , . . . a n + x } 2. Lists might explode in size. 9/41 10/41 Trim the lists... Understanding trimming ApproxSubsetSum ( S , t ) Trim ( L ′ , δ ) // S = { x 1 , . . . , x n } , L = � y 1 . . . y m � // x 1 ≤ x 2 ≤ . . . ≤ x n curr ← y 1 n ← | S | , L 0 ← { 0 } , L out ← { y 1 } δ = ε/ 2 n for i = 2 . . . m do for i = 1 . . . n do if y i > curr · (1 + δ ) E i ← L i − 1 ∪ ( L i − 1 + x i ) Append y i to L out L i ← Trim( E i , δ ) curr ← y i Remove from L i elems > t . return L out return largest element in L n E i : Computed by merging two sorted lists in linear time. 11/41 12/41

Remark Analysis 1. E i list generated by algorithm in i th iteration. 1. Can assume that trimmed lists L i are sorted... 2. P i : list of numbers (no trimming). 2. Algorithm: E i ← L i − 1 ∪ ( L i − 1 + x i ) Claim 3. So, this is just copy, shift, and merge of two sorted lists. For any x ∈ P i there exists y ∈ L i such that 4. ... resulting in a sorted lest. y ≤ x ≤ (1 + δ ) i y . 5. takes linear time in size of lists. Proof 1. If x ∈ P 1 then follows by observation above. ⇒ (induction) ∃ y ′ ∈ L i − 1 s.t. 2. If x ∈ P i − 1 = y ′ ≤ x ≤ (1 + δ ) i − 1 y ′ . 3. By observation ∃ y ∈ L i s.t. y ≤ y ′ ≤ (1 + δ ) y , As such, y ≤ y ′ ≤ x ≤ (1 + δ ) i − 1 y ′ ≤ (1 + δ ) i y . 13/41 14/41 Proof continued Running time of ApproxSubsetSum Proof continued Lemma For x ∈ [0 , 1] , it holds exp( x / 2) ≤ (1 + x ) . 1. If x ∈ P i \ P i − 1 = ⇒ x = α + x i , for some α ∈ P i − 1 . 2. By induction, ∃ α ′ ∈ L i − 1 s.t. α ′ ≤ α ≤ (1 + δ ) i − 1 α ′ . Lemma 3. Thus, α ′ + x i ∈ E i . For 0 < δ < 1 , and x ≥ 1 , we have 4. ∃ x ′ ∈ L i s.t. x ′ ≤ α ′ + x i ≤ (1 + δ ) x ′ . log 1+ δ x ≤ 2 ln x � ln x � = O . 5. Thus, δ δ x ′ ≤ α ′ + x i ≤ α + x i = x ≤ (1 + δ ) i − 1 α ′ + x i ≤ (1 + δ ) i − 1 ( α ′ + x i ) ≤ (1 + δ ) i x ′ . See notes for a proof of lemmas. 15/41 16/41

Running time of ApproxSubsetSum Running time of ApproxSubsetSum Observation Proof. In a list generated by Trim , for any number x , there are no 1. L i − 1 + x i ⊆ [ x i , ix i ] . two numbers in the trimmed list between x and (1 + δ ) x . 2. Trimming L i − 1 + x i results in list of size Lemma � � ix i � ln i � � ln n � | L i | = O ( n /ε ) log n , for i = 1 , . . . , n . log 1+ δ = O = O , x i δ δ 3. Now, δ = ε/ 2 n , and � ln n � n ln n � � | L i | ≤ | L i − 1 | + O ≤ | L i − 1 | + O δ ε � n 2 log n � = O . ε 17/41 18/41 Running time of ApproxSubsetSum ApproxSubsetSum Lemma Theorem � � n 3 The running time of ApproxSubsetSum is O ε log n . ApproxSubsetSum returns u ≤ t , s.t. γ opt 1+ ε ≤ u ≤ γ opt ≤ t , Proof. γ opt : opt solution. � ( n 3 /ε ) log n � Running time is O . 1. Running time of ApproxSubsetSum dominated by total length of L 1 , . . . , L n . Proof. 2. Above lemma implies 1. Running time from above. n × n 2 � n 3 � � � � | L i | = O ε log n = O ε log n . 2. γ opt ∈ P n : optimal solution. i 3. ∃ z ∈ L n , such that z ≤ opt ≤ (1 + δ ) n z 3. Trim runs in time proportional to size of lists. � ε 4. (1 + δ ) n = (1 + ε/ 2 n ) n ≤ exp � ≤ 1 + ε, since � � n 3 2 4. Overall, R.T. O ε log n . 1 + x ≤ e x for x ≥ 0 . 5. γ opt / (1 + ε ) ≤ z ≤ opt ≤ t . 19/41 20/41

Maximal matching An example of the greedy algorithm... 1. G = (V , E) 2. Compute maximal matching... 1 1 6 6 3. X ⊆ E which is maxim al and independent. 10 10 4. Maximal = can not improved by adding an edge. 20 20 20 3 3 3 5. Maximum = largest possible set among all possible sets. 11 11 6. Computing the maximum is hard then computing 2 2 9 9 maximal solution. 13 13 13 7. Q: Find maximal matching quickly and of large size... 15 15 15 4 4 17 17 14 14 12 8 12 8 5 5 19 19 19 16 7 18 16 7 18 16 21/41 22/41 Maximal matching: Algorithm Maximal matching: Result Theorem 1. Algorithm: Repeatedly pick an arbitrary edge and remove it. Given a graph G one can compute in O ( n + m ) time, a maximal matching with at least | X | / 2 edges, where X is the 2. M : Generated matching. X : Maximal matching. size of the maximum (optimal) matching. 3. Clearly a maximal matching... 4. This is a 2 -approximation to the maximum matching. 5. Because... 6. Every edge in M “kills” two edges of X in the worst case. 23/41 24/41

Bin packing Bin packing: First fit Problem definition Analysis Bin Packing Instance : v : Bin size. S = { α 1 , . . . , α n } : n items α i : size of i th item. Target: Find min # B , and a decomposition S 1 , . . . , S B Lemma of S , such that ∀ j � x ∈ S j ≤ v . First fit is a 2 -approximation. Proof. 1. ∪ i S i = S and ∀ i � = j S i ∩ S j = ∅ . Observe that only one bin can have less than v / 2 content in 2. NP-Hard from Partition . it... 3. NP-Hard to approximate within 3 / 2 . 4. Natural problem... 5. How to approximate? 6. First fit: Have a row of bins, insert items greedily into the first bin that fits them. 7. First fit decreasing: Sort the elements first... 25/41 26/41 An example Independent set of rectangles Algorithm: Divide & Conquer Input Independent set of rectangles. Assume: Open rectangles. 27/41 28/41