CSE 5311 Notes 18: NP-Completeness

Transcription

1 SE 53 Notes 8: NP-ompleteness (Last upate 7//3 8:3 PM) ELEMENTRY ONEPTS Satisfiability: ( p q) ( p q ) ( p q) ( p q ) Is there an assignment? (Deision Problem) Similar to ebugging a logi iruit - Is there an input ase that turns on the output LED? sie: Evaluating one input setting for a iruit is P-omplete har to massively parallelize. NP-omplete means (informally):. The problem may be ompute ( eie ) in noneterministi polynomial time. a. Guess a solution (polynomial time - easy to get) b. hek the solution in polynomial time (eterministi). heking ( verifiation ) is easier than omputing.. ll problems in NP may be transforme ( reue ) to this problem in polynomial time. If an instane of one problem is transforme to an instane of another, the new problem has a solution iff the ol problem has a solution. Instanes of all problems in NP Instanes of new problem

2 Showing that all NP problems reue to new problem is unneessary. Instea, fin another NP-omplete problem: Polynomial Time Instanes of all problems in NP Instanes of ol NP-omplete problem Instanes of new problem Easy proof Diffiult proof tehnique - tehnique require for first NP-omplete problem (satisfiability - ook s theorem) Property without property - problem is just in NP. Example: Is a table sorte? Property without property - problem is sai to be NP-har (at least as iffiult as all other problems in NP). Note: property is usually trivial to establish an is often omitte in proofs. Signifiane of a problem being NP-omplete No polynomial-time algorithm is known for any NP-omplete problem. (Only exponential time) If a polynomial-time algorithm is known for one NP-omplete problem, then there is a polynomial-time algorithm for every NP-omplete problem. Exponential lower boun has never been shown. If iffiult instanes of an NP-omplete problem arise in pratie, then approximation shemes with bouns on the quality of the solution are neee. What about playing hess? Example Problems: Satisfiability Graph (Vertex) oloring Job Sheuling with Penalties - urations, ealines, penalties (single proessor) in Paking - how many fixe-size bins are neee to hol variable-size objets? Knapsak - how many objets with ifferent profits an sizes shoul go into a knapsak? Subset Sums - is there a subset whose sum is a partiular value?

3 Hamiltonian Path - oes a graph (or igraph) have a path inluing eah vertex exatly one? 3 Hamiltonian iruit - is there a yle inluing eah vertex exatly one? Traveling Salesperson - minimize istane for Hamiltonian iruit Steiner subgraph - is there a onnete subgraph (tree) that inlues esignate terminal verties an whose total weight oes not exee a given value? (Euliean version is NP-har) REDUTIONS Important resoure - M.R. Garey an D.S. Johnson, omputers an Intratability: Guie to the Theory of NP-ompleteness, Freeman, 979. Suppose you know irete Hamiltonian iruit is NP-omplete. Show that unirete Hamiltonian iruit is NP-omplete:. Replae eah vertex x by three verties x, x, x 3 onnete as: x x x3. Inlue an ege {u 3, v } for eah ege (u, v) in the irete graph. u u u u3 y v z y y y3 z z z3 v v v3 x w x x x3 w w w3 Leaving out x will not work - allows going in wrong iretion: D D D3 D D D3

4 Show 3-satisfiability is NP-omplete by reution from onjuntive normal form satisfiability. 4 In NF an expression is a onjuntion of several lauses (isjuntions). Eah lause has several literals whih may be asserte or negate. The reution is base on replaing eah lause with k > 3 literals by k - lauses for 3- satisfiability an introuing k - 3 new variables: D E F G,,, D, E, F, G X X X X D X 3 X 3 E X 4 X 4 F G,,, D, E, F, G Note, however, that -satisfiability is in P. onvert to a graph problem by replaing eah P Q by P Q an Q P base on. If there is a path from X to X, then X is true. If there is a path from X to X, then X is false. If X an X are in a yle, then the expression is unsatisfiable. onsier,, : Show that graph 3-olorability is NP-omplete by a reution from 3-sat. This reution is fairly iffiult. Others are muh worse. oneptually, we will all the 3 olors TRUE, FLSE, an RED. Sine oloring is usually viewe as assigning the numbers,, to the verties, for any suessful oloring there are five renamings base on permutations.

5 The reution starts with a triangle to establish whih number has whih olor: 5 TRUE FLSE RED For eah variable X, another triangle is neee to onstrain the value: X X RED For eah lause X Y Z the following pattern is use. t least one of X, Y, an Z is fore to be true. X Y Z D E TRUE TRUE Wiget Observe:. X, Y, an Z must have the same olor as TRUE or FLSE.. One of D an E has the same olor as FLSE, the other the same olor as RED. 3. If E has the same olor as FLSE, then Z has the same olor as TRUE. 4. If E has the same olor as RED, then D has the same olor as FLSE.

6 Summary: 6 D E X Y Z TRUE RED??? FLSE RED FLSE FLSE FLSE TRUE RED FLSE RED FLSE FLSE FLSE TRUE TRUE FLSE RED FLSE RED FLSE TRUE FLSE TRUE FLSE RED FLSE RED FLSE TRUE TRUE FLSE TRUE RED FLSE RED TRUE FLSE FLSE FLSE TRUE RED FLSE RED TRUE FLSE TRUE FLSE RED TRUE FLSE RED TRUE TRUE FLSE FLSE RED TRUE FLSE RED TRUE TRUE TRUE Unsatisfiable instane - graph will require 4 olors Wiget Wiget TRUE RED FLSE Wiget 3 Wiget 4 If is remove, 3 oloring is possible.

7 k-lique - omplete subgraph with k verties 7 Show k-lique is NP-omplete by a reution from 3-sat. Eah literal beomes a vertex. onnet eah vertex to the verties for all other lauses, exept for X X Is there a lique with one vertex per lause (i.e. k is the number of lauses)? Vertex over = set of verties suh that every ege has at least one inient vertex in over. Is there a vertex over with no more than p verties? Reue from k-lique:. Take omplement of graph. (Ege is in omplement iff ege is not in graph.). Is there a V - k over? (hoose verties not in the k-lique.) From k-lique example - Is there a 3-over?

8 onsier lique when first lause is remove. 8 3-vertex over Show Steiner subgraph is NP-omplete by a reution from 3-sat. Steiner vertex for eah possible literal on n propositions. Terminal vertex for eah of m lauses, u, an v. Unit-weight eges in subgraph with u, v, an literal verties (for hoosing assignment). Eges with weight n + between eah lause vertex an verties for its literals Is there a subgraph with total weight not exeeing n + m(n + )?

9 9 u v u v PPROXIMTION Goal: Performane guarantees for optimization (NP-har) problems orresponing to NP-omplete problems.. How fast?. pproah: Greey Online Preproessing (e.g. MST, DFS) Ranomization Restrite ases (e.g. spare or ense graphs) (Parallelism) 3. Quality of solution max ratio = Optimal (e.g. knapsak) Solution min ratio = Solution (e.g. TSP) Optimal

10 4. Generality pproximation lgorithm - ahieve max/min ratio in Ο n k time (k fixe) pproximation Sheme - flexible ratio + ε in Ο( f ( n,ε) ) Polynomial-time pproximation Sheme - Ο n f ε ( ) l Fully PTS - Ο n k ε time Examples are presente in asening orer of min/max ratio Ege oloring ( n unusually optimisti situation ( ) Vizing s Theorem ( ) ( ) Δ( G) + (Require number of olors is either egree ( lass ) or ( ) = Χ ( G).) Δ G Χ G egree + ( lass ). For bipartite graphs, Δ G NP-omplete to test if Δ( G) = Χ ( G), but takes only O(VE) to olor with Δ G for bipartite) in an inremental fashion. Thus: ( ) + ( ) min ratio Δ G Χ G ( ) + ( ) Δ G Δ G ( ) + olors ( Δ( G)

11 What if? E E 4 3 x? y 3 5 F G 3 x? y F G D H D H free={,5} free={,4} free={4,5} free={,} ={} = So, steal a olor from another ege an give that ege a new olor. HOW? sie: The simpler problem of oloring a bipartite graph using Δ olors. From p. 347 of H.N. Gabow, Using Euler Partitions to Ege olor ipartite Multigraphs, Int l J. of omputer an Information Sienes 5(4), 976, : Now we esribe a metho for oloring ue to Vizing. Originally eah ege of G is unolore; it must be assigne one of Δ possible olors. n unolore ege (v, w) is olore as follows. t most Δ - eges inient to v are olore, so some olor a is missing at v; similarly, some olor b is missing at w. onstrut an alternating (a, b) path starting at w, as follows. The path begins with the ege inient to w that is olore a (if it exists). onseutive eges in the path are alternately olore a an b. The path ens at the vertex z where the next olor is missing. It is easy to see that z v, w if the graph is bipartite. Interhange olors along the path, swithing a to b an b to a. This makes olor a missing at both v an w, sine z v, w. Ege (v, w) an now be olore a.

12 b free={a} v? a b w free={b} a b a b... b a b z? z? ak to non-bipartite Most general ase - maximal fan with free(x) free(f) =. If no ege inient to X is olore with last free olor, immeiately rotate fan: f {,} f {}? {,} {,} {,,} X k... Y {,} {3,} {k,} {,} X 3... k Y {,} {,} Z {,} Z {k,}

13 While maximizing the fan, suppose the next olor () is alreay on an earlier fan ege: 3 f {,} X? j {,} {,} {=j+,} {,} =j+ {j+,} k Y {k,} Z {=j+,}. Fin (alternating) -path starting with X-. -path (above) reahes no more than one of or Z. (Why?). Invert olors ( ) along entire -path. a. Neither or Z reahe - fan stops at. f {,} X? j {,} {,} {=j+,} {,} {j+,} k Y {k,} Z {=j+,}

14 b. Z reahe - fan stops at. 4 f {,} X? j {,} {,} {=j+,} {,} k Y {j+,} {k,} Z {,}. reahe - fan keeps all verties. f {,} {,} X? j {,} {,} {} {j+,} k Y {k,} Z {=j+,} 3. Rotate fan.

15 Vertex over - pproximation lgorithm 5 V := for eah ege {u, v} // arbitrary orer if u V an v V V := V {u, v}. t termination, V is a vertex over.. Polynomial time - obvious. 3. a. V OPT must over the set of eges proesse base on the if. b. V OPT must inlue at least one of {u, v} for eah of these eges, so: V V OPT min ratio V V OPT ggressive strategy of hoosing one vertex from an unovere ege is vulnerable to stars. Minimum ipartite Vertex over - Exat Solution (not NP-omplete unless P = NP) Instane of max-flow is isomorphi to bipartite mathing max-flow, exept apaities from V to V are. Set of eges from V to V with (unit) flow after max-flow is foun is a maximum mathing. The size of a maximum mathing gives an (obvious) lower boun on the size of a minimum bipartite vertex over. (Showing that the size of a maximum mathing is also an upper boun is more involve an omitte). Theorem: If a minimum S-T ut is known, then V T vertex over. Proof: Suppose the bipartite graph has an ege v,v Sine the apaity of { v,v } is, v,v prevent { v,v } from being unovere. *** ( ) ( V S) is a minimum bipartite { } with v V an v V an v S. { } is an ege in the resiual network an v S to

16 s / / / / / / / / / / t 8 8 / / 9 S = {s} T = {t,,,, 3, 4, 5, 6, 7, 8, 9} V T = {,,,3 } V S = F G / F G / / H / / H / D I J K S / / / D / I / / J / K T E L M / E / / L / M S = {s,,, H} T = {t,, D, E, F, G, I, J, K, L, M} V T = {,D,E} V S = { H} in Paking (one imensional) Minimize number of unit size bins to hol objets with sizes < si. Next Fit Online Use one bin, then seal when next item oesn t fit. Worst ase sequene,, N,, N,, N,, N Repeate to get 4N elements OPT (offline) Next Fit NF OPT

17 First-fit Dereasing 7 Sort n sizes esening For eah objet, go through bins left-to-right to fin first bin that objet fits in. hieves FF.5 (not har to improve ratio to 4/3, iffiult to get /9) OPT laim: Objets plae in extra bins have size / OPT OPT+ OPT+ Proof: Suppose otherwise. laim: Number of objets in extra bins OPT(S) - Proof: Suppose OPT(S) extra objets are use.. Waste in every OPT bin < size of every objet in extra bins.. OPT onsier ( OPT i sum + Extra Objet i ) i= ut sine eah OPTi sum + Extra Objeti >, this sum exees OPT, a ontraition. ase on the two laims, the number of extra bins OPT/ an the ratio is.5. Set over n Input: Set S an subsets suh that S = S i i= Output: Small set of subsets overing S. Greey Tehnique: hoose subset with largest number of unovere elements. (Implementation: Doubly-linke list for eah element in S. Doubly-linke list for eah subset. Orere table for priority queue.) hieves: Greey OPT ln(largest subset ) + See LRS, p. 9- for etaile proof.

18 Example to motivate logarithmi approximation ratio: 8 Traveling Salesperson (omplete graph) No ρ-approximation in P time (unless P = NP). Suppose a graph is to be teste for a Hamiltonian yle: Weight eah real ege with. Imaginary eges are weighte with ρ V +. If ρ-approximation gives TSP with length V, then performane is better than guarantee an have a Hamiltonian yle. If ρ-approximation gives TSP with length > ρ V, then performane guarantee has not been met. If ege weights obey triangle inequality, the sale-up problem is avoie. a b a+b

19 -approximate 9. Fin minimum spanning tree.. Depth-first searh - orer verties by isovery time. 3. Return to start vertex. 4. Remove ege rossings - optional a a b b approximate proof: 9. MST TOPT est ase - removing largest ege in OPT (a yle) gives MST.. TΔ MST Sine MST short uts are no longer than subpath skippe. T Δ MST T OPT, so T Δ T OPT (sie: see for a 3/- approximate tehnique.)

20 Subset Sums - exat solution by ynami programming (ssnew.) Given n numbers S, S n. fin a subset (e.g. an inex set hosen from n) suh that sum is an exat value.. Use table with entries [i] = j, where j is the smallest inex suh that i = S j + values w/inex < j. Initialize table an then go forwar = : Now suppose eah S i is multiplie by. D.P. table grows by fator of.

21 Orinary Subset Sum (LRS, p. 8-9) Saves spae by maintaining lists of reahable sums _ Target=36 L={/} L={/, /} 3 5 L={/, /, 3/, 5/} L3={/, /, 3/, 5/, 6/3, 8/3, 9/3, /3} L4={/, /, 3/, 5/, 6/3, 8/3, 9/3, /3, 3/4, /4, 6/4, 7/4, 9/4, /4, /4} ? L5={/, /, 3/, 5/, 6/3, 8/3, 9/3, /3, 3/4, ? 4/4, 5/5, 6/4, 7/4, 8/5, 9/4, /4, /5, /4, 3/5, 4/5, 6/5, 8/5, 9/5, 3/5, 3/5, 34/5, 35/5} L6={/, /, 3/, 5/, 6/3, 8/3, 9/3, /3, 3/4, 4/4, 5/5, 6/4, 7/4, 8/5, 9/4, /4, /5, /4, 3/5, 4/5, 5/6, 6/5, 7/6, 8/5, 9/5, 3/6, 3/5, 3/5, 33/6, 34/5, 35/5, 36/6}

22 Subset Sum with Range of Target Values Uses intervals to ahieve approximation to within ε of esire value. LRS, p. 3-33, gives fully PTS base on ε _ Target= L={..3/}..5 L={..3/, 3..5/} L={..3/, 3..5/, 5..8/} L3={..3/, 3..5/, 5..8/, 8..4/3} L4={..3/, 3..5/, 5..8/, 8..4/3, 4..5/4} ? L5={..3/, 3..5/, 5..8/, 8..4/3, 4..5/4, 5..36/5} ? L6={..3/, 3..5/, 5..8/, 8..4/3, 4..5/4, 5..36/5}

23 Graph (Vertex) oloring - no effiient heuristi (unless P = NP) 3 Suppose a P time algorithm exists to olor every graph G with Χ G 4 Χ G 3 ( ) olors, then 3-olorability woul be in P. Proof:. k is the omplete graph with k verties.. G is an instane of 3-olorability. 3. Graph H = k [ G] (omposition of graphs). ( ) k using fewer than G k Eah vertex in a opy of G is onnete to all verties in all other opies of G. 4. If Χ( G) = 3, eah opy of G requires 3 olors, so Χ( H) = 3k. lgorithm must use fewer than 4 ( 3 3k ) = 4k olors for G to be 3-olorable. Otherwise, Χ( G) > 3 so eah opy of G nees at least 4 olors. Χ( H) 4k lgorithm must then use at least 4k olors to olor H. Thus, suh an algorithm is unlikely.