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UNIT III DYNAMIC PROGRAMMING ALGORITHMS
Developing a Dynamic Programming Algorithm-Subtle Points- Question for the Little Bird- Sub instances and Sub solutions-Set of Substances-Decreasing Time and Space-Number of Solutions-Code. Reductions and NP-Completeness-Satisfiability-Proving NP-Completeness- 3-Coloring- Bipartite Matching. Randomized Algorithms-Randomness to Hide Worst Cases- Optimization Problems with a Random Structure.
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Dynamic Programming A hard topic.
I try to provide a unified way to think of it and a fixed set of steps to follow. Even if you don’t get the details of the algorithm correct, at least get the right structure. I provide analogies (little bird) to make it hopefully more fun & easier to follow.
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Non-Deterministic Poly-Time Decision Problems (NP)
An optimization problem Each solution is either valid or not (no cost) The output is Yes, it has a valid solution. No, it does not the solution is not returned Eg: Given graph and integer <G,k>, does G have a clique of size k?
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an instance I (= <G,k>) and a solution S (= subset of nodes)
Key: Given an instance I (= <G,k>) and a solution S (= subset of nodes) there is a poly-time alg Valid(I,S) to test whether or not S is a valid solution for I. Poly-time in |I| not in |S|. |S| can't be too big. Not Valid k=4 Valid Formal definition: Prob NP iff poly time Valid such that Prob(I) = S Valid(I,S)
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Valid k=4 Key: If the instance has a valid solution
A non-deterministic (fairy god mother) could prove it to you by giving you such a solution as a witness. You could check that it is valid. You could convince your boss. k=4 Valid
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Key: If the instance does not have a valid solution A non-deterministic (fairy god mother) could prove it to you by giving you ???? You have no way to convince your boss. k=5
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Non-Deterministic Poly-Time (NP)
Example: 3-Col: Instance: A graph G. Solution: Colouring C nodes of G with 3 colours such that every edge has two colours. G is a Yes instance if there is such a colouring. Given an instance G and a solution C, there is a poly-time alg Valid(G,C) to test whether or not C is a valid 3- colouring of G. 3-Col NP.
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Non-Deterministic Poly-Time (NP)
Example: Airplane Wing: Instance: Requirements I of the wing. Solution: A description S of how to make the wing. I is a Yes instance if there is such a wing. Given an instance I and a proof S, there is a poly-time alg Valid(I,S) to test whether or not S is a valid solution for I. Airplane Wing NP. I = [weight, lift, cost, …]
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Reductions Palg ≤poly Poracle
Reduction: Design a fast algorithm for one computational problem, using a supposedly fast algorithm for another problem as a subroutine.
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NP vs Co-NP Preserving Reductions
Definition is asymmetric. There is a witness that a graph has a 3-Col. There is no known witness that a graph has no 3-Col. Cook Reduction: Design any fast algorithm for Palg using a supposed fast algorithm for Poracle as a subroutine. Karp Reduction: The algorithm for Palg calls that for Poracle only once Yes Yes & No No
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Karp Reduction: Yes Yes & No No
We will only consider reductions of this simple form. Because they preserve NP vs Co-NP
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Problem Pnew is NP-Complete Pnew not too hard.
Pnew NP
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Test in poly-time if a given solution is valid
NP-Complete Problems Computable Exp Poly Test in poly-time if a given solution is valid NP Sat complete Pnew Known GCD
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NP-Complete Problems Problem Pnew is NP-Complete Pnew not too hard.
Pnew NP Pnew sufficiently hard. PNP, P ≤poly Pnew Easier: Sat ≤poly Pnew Cook: P ≤poly Sat Computable Exp Poly NP Sat complete Pnew Known GCD
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K-Clique vs K-Independent Set
Clique: Given <G,k>, does G contains a k-clique? Independent Set: Given <G,k>, does G contains a k-Ind Set? A K-independent set is a set of K nodes with no edges between them. A K-clique is a set of K nodes with all edges between them.
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K-Clique vs K-Independent Set
Clique: Given <G,k>, does G contains a k-clique? Independent Set: Given <G,k>, does G contains a k-Ind Set? Brute Force: Try out all n choose k possible subsets. If k = (n) then 2(n) subsets to check If k=3 then only O(n3)
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K-Clique vs K-Independent Set
Clique ≤poly Indep Set <G,k> G has a k clique or not GIVEN: Indep. Set Oracle BUILD: Clique Oracle <G*, k> G* has a k Indep. set or not
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K-Clique vs K-Independent Set
Clique ≤poly Indep Set <G,k> G has a k clique or not GIVEN: Indep. Set Oracle BUILD: Clique Oracle <G*, k> Proof of correctness: Our oracle says yes to <G,k> iff Old oracle says yes to <G*,k> iff G* has a k indep. set iff G has a k clique G* has a k Indep. set or not
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K-Clique vs K-Independent Set
G* G G* has edge if and only if G does not
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K-Clique vs K-Independent Set
This graph contains a clique of size 4. G if and only if G* This graph contains an independent set of size 4.
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12 Steps Steps for proving that Pnew is NP-Complete
1) I am addicted to solving NP-Complete Problems 2) I trust in my higher power to help A witness
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12 Steps Steps for proving that Pnew is NP-Complete 0) Pnew NP
1) What to reduce it to 2) What is what 3) Direction of reduction & Code 4) Look for similarities 5) Instance Map 6) Solution Map 7) Valid to Valid 8) Reverse Solution Map 9) Valid to Valid 10) Working Algorithm 11) Running Time
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Design poly-time algorithm Valid(I,S)
12 Steps Step 0: Pnew NP Formal definition: Pnew NP iff poly time Valid such that Pnew(I) = S Valid(I,S) Design poly-time algorithm Valid(I,S) which determines whether a given solution S is valid for a given instance I.
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Step 1: What to reduce it to
12 Steps Step 1: What to reduce it to Palg ≤poly Poracle Choose a problem Pis NP-comp that is as similar to Pnew as possible and that is already known to be NP-Complete. ..
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= Independent Set = Clique k=3 k=4 12 Steps Step 2: What is what Pnew
Pis NP-comp = Clique Inew Snew Iis NP-comp Sis NP-comp k=3 k=4 ..
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Step 3: Direction of reduction & Code
12 Steps Step 3: Direction of reduction & Code Reduce Pnew to Pis NP-comp or Pis NP-comp to Pnew? Pis NP-comp ≤poly Pnew Palg ≤poly Poracle ..
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Step 3: Direction of reduction & Code
12 Steps Step 3: Direction of reduction & Code ..
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Step 4: Look for similarities
12 Steps Step 4: Look for similarities Clique Independent Set .. Both instances are graphs: nodes & edges A clique solution is a subset of the nodes with edges. An Ind Set solution is a subset of the nodes without edges.
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12 Steps Step 5: Instance Map ..
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12 Steps Step 5: Instance Map Ialg Ioracle = InstanceMap(Ialg)
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no yes Yes mapped to Yes No to No no yes 12 Steps Step 5: Instance Map
Ialg Yes mapped to Yes No to No Ioracle no yes
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12 Steps Step 6: Solution Map ..
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Potential Solution = SolutionMap(S). Potential Solution S. 12 Steps
Step 6: Solution Map Potential Solution = SolutionMap(S). Ialg Ioracle Potential Solution S.
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Salg = SolutionMap(Soracle) Valid
12 Steps Step 7: Valid to Valid Salg = SolutionMap(Soracle) Ialg Valid Ioracle = InstanceMap(Ialg) If Soracle is valid solution for Ioracle, then Salg is valid solution for Ialg Soracle Valid
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Step 8: Reverse Solution Map
12 Steps Step 8: Reverse Solution Map Potential Solution S. Ialg Potential Solution = ReverseSolutionMap(S). Ioracle Not part of code, but of proof.
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Soracle = ReverseSolutionMap(Salg)
12 Steps Step 9: Valid to Valid Salg Ialg Valid Valid Ioracle = InstanceMap(Ialg) If Salg is valid solution for Ialg, then Soracle is valid solution for Ioracle Soracle = ReverseSolutionMap(Salg)
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Step 10: Working Algorithm
12 Steps Step 10: Working Algorithm Our Algalg says yes to Ialg iff old Algoracle says yes to Ioracle iff Ioracle has a valid solution Soracle iff Ialg has a valid solution Salg iff Ialg is a yes instance. .. Steps 8&9 Steps 6&7
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Salg = SolutionMap(Soracle)
12 Steps Step 11: Running Time Algalg(Ialg) is poly time if the following are poly time: Algoracle(Ioracle) Assumed Ioracle = InstanceMap(Ialg) Your job Salg = SolutionMap(Soracle)
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Circuit Satisfiability
Palg ≤poly Poracle or Palg Poracle Graph Colouring Scheduling Circuit Satisfiability Any NP-Problem Clique Independent Set This will prove that Cir-SAT is NP-Complete.
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Reduction I I don’t even know what problem I am trying to solve!!! ?
Any NP-problem ≤poly Cir-SAT I I don’t even know what problem I am trying to solve!!! BUILD: Oracle for arbitrary NP Problem GIVEN: Oracle for Cir-Sat ?
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We need to solve some unknown NP-Problem.
What do we know about it? Formal definition: Parbitrary NP iff poly time Valid such that Parbitrary(I) = S Valid(I,S) We have a poly-time alg Valid(I,S) to test whether or not S is a valid solution for I. Valid k=4
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Reduction I Please, give me a solution S such that Valid(I,S) is true.
Any NP-problem ≤poly Cir-SAT I Please, give me a solution S such that Valid(I,S) is true. BUILD: Oracle for arbitrary NP Problem GIVEN: Oracle for Cir-Sat That looks like a Turing Machine. I only know about circuits
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The Circuit Satisfiability Problem
An instance is a circuit C. A solution is an assignment X = [F,T,F…]. C(X) evaluates to T or F. x3 x2 x1 OR AND NOT F T F T One bit output No feedback
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The Circuit Satisfiability Problem
An instance is a circuit C. A solution is an assignment X = [F,T,F…]. A valid solution has C(X) = True. x3 x2 x1 OR AND NOT F F T Given a circuit, does it have a satisfying assignment?
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The Circuit Satisfiability Problem
Turing (and friends) prove that any algorithm computed by a JAVA program in poly-time can be computed by a Turing Machine in poly-time. Very Powerful Cook proves that any algorithm computed by a Turing Machine in time T(n) can be computed by a family of circuits of size [T(n)]2. But clearly, circuits compute.
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Reduction I I build a circuit VI(S) equivalent to Valid(I,S)
Any NP-problem ≤poly Cir-SAT I I build a circuit VI(S) equivalent to Valid(I,S) Please, give me a solution S such that VI(S) is true. BUILD: Oracle for arbitrary NP Problem GIVEN: Oracle for Cir-Sat Thanks for the circuit. But what is S? I can only find assignments.
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Reduction I I build a circuit VI(S) equivalent to Valid(I,S)
Any NP-problem ≤poly Cir-SAT I I build a circuit VI(S) equivalent to Valid(I,S) I decode X into S and am done. Please, give me an assignment X such that VI(X) is true. BUILD: Oracle for arbitrary NP Problem GIVEN: Oracle for Cir-Sat My pleasure. Here: X
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12 Step Program Let's be more formal. And flow the 12 steps.
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Step 0: Cir-SAT NP Design poly-time algorithm ValidCir-SAT(C,X)
which determines whether a given assignment X is valid for a given circuit C. Easy: Evaluate C(X).
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Step 1: What to reduce it to
Palg ≤poly Cir-Sat Choose a problem Pis NP-comp that is as similar to Cir-Sat as possible and that is already known to be NP-Complete. Have none. .. Any NP-problem ≤poly Cir-SAT We have a poly-time alg Valid(I,S) to test whether or not S is a valid solution for I.
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Step 2: What is what x3 x2 x1 = Circuit-Sat = I = S F AND NOT OR Pnew
Parbitrary = some NP problem Inew Snew Iarbitrary Sarbitrary = I = S x3 x2 x1 OR AND NOT F ..
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Step 3: Direction of reduction & Code
Given oracle for Cir-Sat, we need to be able to solve any NP-Complete problem. Parbitrary .. Cir-Sat
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Step 4: Look for similarities
x3 x2 x1 OR AND NOT F .. ?
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Step 5: Instance Map ..
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Step 5: Instance Map I circuit
We have a poly-time alg Valid(I,S) to test whether or not S is a valid solution for I. Let Validn(I,S) be a circuit: I is a bit string representing an instance I. S is a bit string representing a solution S. Outputs T if S is an encoding of a valid solution of I. G,k Eg: Clique S I {u.v} E OR uS OR vS |S| k And over pairs of nodes
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Outputs T if S is an encoding of a valid solution S of I
Step 5: Instance Map I circuit Given an instance I Circuit VI(S) = Valid(I,S) I S hard wired Outputs T if S is an encoding of a valid solution S of I Validn(I,S) I VI(S)
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Step 6: Solution Map ..
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Step 6: Solution Map S assignment X=[T,F,F,T,F,T]
X is viewed as a bit string S representing a solution S. If X is not a bit string representing a solution then “what ever” Outputs T if S is an encoding of a valid solution S of I I S hard wired VI(S)
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Outputs T if S is an encoding of a valid solution S of I
Step 6: Solution Map S assignment X=[T,F,F,T,F,T] X is viewed as a bit string S representing a solution S. solution S Outputs T if S is an encoding of a valid solution S of I I S hard wired VI(S)
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Outputs T if S is an encoding of a valid solution S of I
Step 7: Valid Valid VI(X) = T Valid(I,S) = Validn(I,S) = VI(X) = T S is a valid solution of I. Outputs T if S is an encoding of a valid solution S of I I S hard wired VI(S)
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Outputs T if S is an encoding of a valid solution S of I
Step 8: Rev. Sol. Map solution assignment solution S S bit string representing a solution S X=[T,F,F,T,F,T] Outputs T if S is an encoding of a valid solution S of I I S hard wired VI(S)
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Outputs T if S is an encoding of a valid solution S of I
Step 9: Valid Valid S is a valid solution of I. VI(X) = Validn(I,S) = Valid(I,S) = T VI(X) = T Outputs T if S is an encoding of a valid solution S of I I S hard wired VI(S)
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GIVEN: Alg for circuit problem
Reduction Any NP-problem ≤poly Cir-SAT I Yes/No i.e. S, S is a valid solution for I BUILD: Opt. problem GIVEN: Alg for circuit problem VI satisfiable or not i.e. X, VI(X)
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Circuit Satisfiability
Palg ≤poly Poracle or Palg Poracle 3-Col ? Graph Colouring Circuit Satisfiability Scheduling Any NP-Problem Clique Independent Set
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Graph Col: Given graph G & k can G be coloured with k colours?
can G be coloured with 3 colours? 3-Col ≤poly Graph Col If you can decide whether you can colour with k colours, then you can decide whether you can colour with 3 colours. G,k C G C
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Circuit Satisfiability
Palg ≤poly Poracle or Palg Poracle 3-Col 3-SAT ? Graph Colouring Circuit Satisfiability Scheduling Any NP-Problem Clique Independent Set
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Cir Sat: Given circuit C does it have a satisfying assignment X?
3-SAT: Given an expression: A circuit consisting of a big AND of clauses Each clause is the OR of at most 3 literals Each literal is a variable or its negation. F T F T xoryorz AND xorwora AND …
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Cir Sat: Given circuit C does it have a satisfying assignment X?
3-SAT: Given an expression: 3-Sat ≤poly Cir Sat If you can decide whether any circuit is satifiable, then you can decide whether a circuit that happens to be an expression is satisfiable. F T F T xoryorz AND xorwora AND …
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Circuit Satisfiability
Palg ≤poly Poracle or Palg Poracle 3-Col 3-SAT ? Graph Colouring Circuit Satisfiability Scheduling Any NP-Problem Clique Independent Set 3-Col ≤poly 3-SAT Graph Col ≤poly Cir SAT
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Outputs T if C is an encoding of a valid 3-colouring C of G
Step 5: Instance Map Graph G Expression V Given an instance G Circuit VG(C) = Valid(G,C) Outputs T if C is an encoding of a valid 3-colouring C of G hard wired C G Encoding C of C: u is a node r is a colour x<u,r> = T if node u is colour r.
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Step 5: Instance Map Given an instance G Graph G Expression V G C
Circuit VG(C) = Valid(G,C) hard wired C G Encoding C of C: u is a node r is a colour x<u,r> = T if node u is colour r. x<u, r> = F OR x<v, r> = F And over edges <u,v> and colours r clauses with 2 literals
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x<u, r> =T OR x<u, g>=T OR x<u, b>=T
Step 5: Instance Map Graph G Expression V Given an instance G Circuit VG(C) = Valid(G,C) hard wired C G Encoding C of C: u is a node r is a colour x<u,r> = T if node u is colour r. x<u, r> =T OR x<u, g>=T OR x<u, b>=T And over node u clauses with 3 literals
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