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Exponential Circuit Complexity for NP-Complete Problems • We Shall Prove Exponential Lower Bounds for NP-Complete Problems Using Monotone Circuits

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Exponential Circuit Complexity for NP-Complete Problems • We Shall Prove Exponential Lower Bounds for NP-Complete Problems Using Monotone Circuits Exponential Circuit Complexity for NP-Complete Problems ² We shall prove exponential lower bounds for NP-complete problems using monotone circuits. { Monotone circuits are circuits without : gates. ² Note that this does not settle the P vs. NP problem or any of the conjectures on p. 544. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 692 The Power of Monotone Circuits ² Monotone circuits can only compute monotone boolean functions. ² They are powerful enough to solve a P-complete problem, monotone circuit value (p. 266). ² There are NP-complete problems that are not monotone; they cannot be computed by monotone circuits at all. ² There are NP-complete problems that are monotone; they can be computed by monotone circuits. { hamiltonian path and clique. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 693 cliquen;k ² cliquen;k is the boolean function deciding whether a graph G = (V; E) with n nodes has a clique of size k. ¡n¢ ² The input gates are the 2 entries of the adjacency matrix of G. { Gate gij is set to true if the associated undirected edge f i; j g exists. ² cliquen;k is a monotone function. ² Thus it can be computed by a monotone circuit. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 694 Crude Circuits ² One possible circuit for cliquen;k does the following. 1. For each S ⊆ V with jSj = k, there is a circuit with O(k2) ^-gates testing whether S forms a clique. ¡n¢ 2. We then take an or of the outcomes of all the k subsets S1;S2;:::;S n . (k) ¡ ¢ 2 n ² This is a monotone circuit with O(k k ) gates, which is exponentially large unless k or n ¡ k is a constant. ² A crude circuit CC(X1;X2;:::;Xm) tests if any of Xi ⊆ V forms a clique. { The above-mentioned circuit is CC(S1;S2;:::;S n ). (k) °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 695 The Proof: Positive Examples ² Analysis will be applied to only positive examples and negative examples as inputs. ¡k¢ ² A positive example is a graph that has 2 edges connecting k nodes in all possible ways. ¡n¢ ² There are k such graphs. ² They all should elicit a true output from cliquen;k. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 696 The Proof: Negative Examples ² Color the nodes with k ¡ 1 di®erent colors and join by an edge any two nodes that are colored di®erently. ² There are (k ¡ 1)n such graphs. ² They all should elicit a false output from cliquen;k. { Each set of k nodes must have 2 identically colored nodes; hence there is no edge between them. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 697 Positive and Negative Examples with k = 5 $SRVLWLYHH[DPSOH $QHJDWLYHH[DPSOH °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 698 Sunflowers ² Fix p 2 Z+ and ` 2 Z+. ² A sunflower is a family of p sets fP1;P2;:::;Ppg, called petals, each of cardinality at most `. ² Furthermore, all pairs of sets in the family must have the same intersection (called the core of the sunflower). FRUH °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 699 A Sample Sunflower ff1; 2; 3; 5g; f1; 2; 6; 9g; f0; 1; 2; 11g; f1; 2; 12; 13g; f1; 2; 8; 10g; f1; 2; 4; 7gg: É¿³Ì Æ¿³È ÿ³ÄÄ Ç¿³Ê ÄÅ¿³ÄÆ Ë¿³Äà °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 700 The Erd}os-RadoLemma Lemma 83 Let Z be a family of more than M = (p ¡ 1)``! nonempty sets, each of cardinality ` or less. Then Z must contain a sunflower (with p petals). ² Induction on `. ² For ` = 1, p di®erent singletons form a sunflower (with an empty core). ² Suppose ` > 1. ² Consider a maximal subset D ⊆ Z of disjoint sets. { Every set in Z¡D intersects some set in D. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 701 The Proof of the Erd}os-RadoLemma (continued) For example, Z = ff1; 2; 3; 5g; f1; 3; 6; 9g; f0; 4; 8; 11g; f4; 5; 6; 7g; f5; 8; 9; 10g; f6; 7; 9; 11gg; D = ff1; 2; 3; 5g; f0; 4; 8; 11gg: °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 702 The Proof of the Erd}os-RadoLemma (continued) ² Suppose D contains at least p sets. { D constitutes a sunflower with an empty core. ² Suppose D contains fewer than p sets. { Let C be the union of all sets in D. { j C j · (p ¡ 1)`. { C intersects every set in Z by D's maximality. { There is a d 2 C that intersects more than M `¡1 (p¡1)` = (p ¡ 1) (` ¡ 1)! sets in Z. { Consider Z0 = fZ ¡ fdg : Z 2 Z; d 2 Zg. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 703 The Proof of the Erd}os-RadoLemma (concluded) ² (continued) { Z0 has more than M 0 = (p ¡ 1)`¡1(` ¡ 1)! sets. { M 0 is just M with ` replaced with ` ¡ 1. { Z0 contains a sunflower by induction, say fP1;P2;:::;Ppg: { Now, fP1 [ fdg;P2 [ fdg;:::;Pp [ fdgg is a sunflower in Z. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 704 Comments on the Erd}os-RadoLemma ² A family of more than M sets must contain a sunflower. ² Plucking a sunflower means replacing the sets in the sunflower by its core. ² By repeatedly ¯nding a sunflower and plucking it, we can reduce a family with more than M sets to a family with at most M sets. ² If Z is a family of sets, the above result is denoted by pluck(Z). ² Note: pluck(Z) is not unique. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 705 An Example of Plucking ² Recall the sunflower on p. 700: Z = ff1; 2; 3; 5g; f1; 2; 6; 9g; f0; 1; 2; 11g; f1; 2; 12; 13g; f1; 2; 8; 10g; f1; 2; 4; 7gg ² Then pluck(Z) = ff1; 2gg: °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 706 Razborov's Theorem Theorem 84 (Razborov (1985)) There is a constant c such that for large enough n, all monotone circuits for 1=4 cn1=8 cliquen;k with k = n have size at least n . ² We shall approximate any monotone circuit for cliquen;k by a restricted kind of crude circuit. ² The approximation will proceed in steps: one step for each gate of the monotone circuit. ² Each step introduces few errors (false positives and false negatives). ² But the ¯nal crude circuit has exponentially many errors. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 707 The Proof ² Fix k = n1=4. ² Fix ` = n1=8. a ² Note that µ ¶ ` 2 · k ¡ 1: 2 ² p will be ¯xed later to be n1=8 log n. ² Fix M = (p ¡ 1)``!. { Recall the Erd}os-Radolemma (p. 701). aCorrected by Mr. Moustapha Bande (D98922042) on January 05, 2010. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 708 The Proof (continued) ² Each crude circuit used in the approximation process is of the form CC(X1;X2;:::;Xm), where: { Xi ⊆ V . { jXij · `. { m · M. ² It answers true if any Xi is a clique. ² We shall show how to approximate any circuit for cliquen;k by such a crude circuit, inductively. ² The induction basis is straightforward: { Input gate gij is the crude circuit CC(fi; jg). °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 709 The Proof (continued) ² Any monotone circuit can be considered the or or and of two subcircuits. ² We shall show how to build approximators of the overall circuit from the approximators of the two subcircuits. { We are given two crude circuits CC(X ) and CC(Y). { X and Y are two families of at most M sets of nodes, each set containing at most ` nodes. { We construct the approximate or and the approximate and of these subcircuits. { Then show both approximations introduce few errors. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 710 The Proof: or ² CC(X[Y) is equivalent to the or of CC(X ) and CC(Y). { A set of nodes C 2 X [ Y is a clique if and only if C 2 X is a clique or C 2 Y is a clique. ² Violations in using CC(X[Y) occur when jX [ Yj > M. ² Such violations can be eliminated by using CC(pluck(X[Y)) as the approximate or of CC(X ) and CC(Y). °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 711 The Proof: or ² If CC(Z) is true, then CC(pluck(Z)) must be true. { The quick reason: If Y is a clique, then a subset of Y must also be a clique. { For each Y 2 X [ Y, there must exist at least one X 2 pluck(X[Y) such that X ⊆ Y . { If Y is a clique, then this X is also a clique. ² We now bound the number of errors this approximate or makes on the positive and negative examples. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 712 The Number of False Negatives (concluded) X Y °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 713 The Proof: or (concluded) ² CC(pluck(X[Y)) introduces a false positive if a negative example makes both CC(X ) and CC(Y) return false but makes CC(pluck(X[Y)) return true. ² CC(pluck(X[Y)) introduces a false negative if a positive example makes either CC(X ) or CC(Y) return true but makes CC(pluck(X[Y)) return false.
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