COMP202. Complexity of Algorithms. Section 1: Answer all questions. Section 2: Answer 2 questions.

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1 COMP202 Complexity of Algorithms Section 1: Answer all questions. Section 2: Answer 2 questions.

2 2 Section 1 Answer all questions in this section. 1. Define what is meant by the following terms, An n-vertex undirected graph. (3 marks) A strongly connected directed graph. (3 marks) The Time Complexity of a decision problem. (3 marks) (d) The complexity class Polynomial Time. (3 marks) (e) NP-complete decision problem. (3 marks) 2. For the 4-vectors, a = [ a 0, a 1, a 2, a 3 ] = [1,5,3,6] b=[b 0,b 1,b 2,b 3 ]=[2,4,0,7] Give the formal polynomials P a ( x )and P b ( x )represented by a and b. (5marks) Find the convolution - a b - of a and b. (10 marks) c = [ c 0, c 1, c 2, c 3, c 4, c 5, c 6, c 7 ]

3 3 3. In the 10-vertex directed graph, State whether or not each of the following sets of vertices are in the same strongly-connected component and give a brief justification for your answer in each case. Vertices 1 and 7. (5marks). Vertices 3 and 4. (4marks). Vertices 8, 9, and 10 (6 marks). 4. Carefully define each of the following NP-complete decision problems, giving the form that an instance of each problem takes and the question being asked of these instances. The Satisfiability Problem (SAT ). (5 marks) The 3Colouring Problem (3 COL). (5 marks) The Hamiltonian Cycle Problem (HC). (5 marks).

4 4 Section 2 Answer two of Questions 5, 6, and Briefly outline how the Discrete Fourier Transform (DFT )can be used to compute the convolution of 2 n-vectors, a = [ a 0, a 1,...,a n 1 ] b=[b 0,b 1,...,b n 1 ] [Do not describe the details of the Fourier Transform process itself.]. (5 marks) Carefully describe how the recursive Discrete Fourier Transform method computes DFT ( x, n, ω )with x = [4,5,8,6] ; n = 4 ; ω = 4 and arithmetic calcluations being performed modulo M = 17. (10 marks). The run-time of the Schönhage-Strassen multiplication algorithm is measured in terms of the number of bit operations - O B () it performs.. What is the principal difficulty that would arise in attempting to adapt this method to work directly with decimal (rather than binary) representations? (5 marks)

5 5 6. State necessary and sufficient conditions for a connected, undirected graph G( V, E )tohave an Eulerian Tour. (5marks) In the undirected graph below, carefully show how the partial tour <{7,6},{6,2},{2,1},{1,7},{7,8},{8,10},{10, 7 } > that contains all of the edges adjacent to vertex 7,isextended to a complete Eulerian Tour of the graph. (10 marks) The edge-dual of a graph G( V, E )isthe graph Ĝ( W, F )formed by replacing each edge { x, y }ineby a single vertex u {x,y} in W. The edges of Ĝ connect those vertices whose corresponding edges in E have a common end-point, e.g. as in the example below ,2 1,3 3,4 3 2,3 3,5 4,5 4 5 G Ĝ G has an Eulerian Tour if and only if Ĝ has a Hamiltonian cycle. Now consider the following algorithm: given G, construct the graph H for which Ĥ = G and test if H has an Eulerian Tour. Briefly explain why this algorithm not yield a method for deciding if an arbitrary graph has a Hamiltonian cycle. (5 marks)

6 6 7. Given the following instance, F, ofcnf SAT, F( X 8 ) = ( x 1 ) ( x 2 x 8 ) ( x 3 x 4 x 5 x 7 ) ( x 1 x 2 x 4 x 5 x 6 x 7 ) (x 4 x 5 x 7 ) ( x 4 x 7 x 8 ) Identify all of the clauses of F which are not in the form required by the decision problem 3-SAT.(5marks) Carefully show how this instance can be transformed into an instance, G F,of3-SAT that is satisfiable if and only if F is satisfiable. (10 marks) An instance of the decision problem PART 3 SAT consists of an instance F( X n ) = m C j = m ( y j,1 y j,2 y j,3 ) j=1 j=1 of 3 SAT and two positive integers r and s (with r/s <1). The question asked of these being: "is there an assignment to X n that satisfies at least (mr/s) ofthe clauses C j?". Prove that PART 3 SAT is NP-complete, by showing that the restricted case with r = 8and s = 9isNP-complete. [Hint: Use a reduction that transforms instances F( X n ) = m C j of 3 SAT to instances ( G F,8,9) of PART 3 SAT in which G F ( X n, Z t ) = F( X n ) H( Z t ) with H( Z t )having exactly 8m clauses and Z t is a set of 3m new variables.] (5 marks) j=1 END OF PAPER

7 7 Model Solutions 1. An n-vertex undirected graph, G( V, E ) is defined by a set V = { v 1,...,v n }ofnvertices, and a set E of unordered pairs of distinct vertices called the graph edges. A directed graph H(W, F) isstrongly connected if for each pair of distinct vertices x and y in W there is a directed path from x to y and a directed path from y to x. The time complexity of a decision problem, f, is a function T : N N, such that there is a deterministic Turing machine, M, that correctly decides all instances of length n in at most T (n) moves. Also accept run-time of fastest algorithm for f orsimilar. (d) A decision problem f is in the complexity class Polynomial Time (P), if there is a deterministic Turing machine program, M deciding f and aconstant k such that for any input x of length n, M makes at most n k moves onx. (e) f is NP-complete if f NP and g NP, g p f. 2. P a ( x ) = 1 + 5x + 3x 2 + 6x 3 P b ( x ) = 2 + 4x + 0x 2 + 7x 3 or P b (x) = 2 + 4x + 7x 3 The component c k (0 k 7) of a b is: c k = Σ k a j * b k j with a i = b i = 0ifi<0or i >3. Hence, c 0 = a 0 b 0 = 2 c 1 = a 0 b 1 + a 1 b 0 = = 14 c 2 = a 0 b 2 + a 1 b 1 + a 2 b 0 = = 26 c 3 = a 0 b 3 + a 1 b 2 + a 2 b 1 + b 3 a 0 = = 31 c 4 = a 1 b 3 + a 2 b 2 + a 3 b 1 = = 59 c 5 = a 2 b 3 + a 3 b 2 = = 21 c 6 = a 3 b 3 = 42 c 7 = 0 j=0

8 Vertices 1 and 7 are in the same component: there is a directed cycle Vertices 3 and 4 are in the same component: there is a directed cycle Vertices 8, 9, 10 re each in distinct components: the cycle in can be extended to include 8 (using instead of 7 5). This cannot be extended to include { 3, 4, 10 } since there are no edges directed into this set from any any of the other vertices: thus A = {3,4,10} and B = {1,2,5,6,7,8} are in separate components. It now follows that 9 must also be in a component distinct from A and B: 9has no edges directed into A, and no edges directed from B. Satisfiability Instance: Boolean expression, F( X n )over aset of Boolean variables X n = < x 1, x 2,...,x n >. Question: Is there an assignment, α, of Boolean values to the variables X n under which F( α )istrue? 3 Colouring Instance: n-vertex undirected graph G( V, E ). Question: Can each vertex of G be assigned one of 3 available colours, in such a way that no two vertices joined by an edge in E receive the same colour? Hamiltonian Cycle Instance: n-vertex undirected graph G( V, E ). Question: Is there an ordering, π :{1,2,...,n} {1,2,...,n} of the vertices such that i (1 i<n) {π(i), π (i + 1) } E { π (1), π (n)} E i.e. does G contain a simple cycle that includes each vertex?

9 5. Let F k (x) denote the DFT of a k-vector x and Fk 1 (x) the inverse DFT. To compute the convolution of a and b. i. Form the 2n-vectors a 0n = [ a 0,...,a n 1,0,0,...,0] b 0n =[b 0,...,b n 1,0,0,...,0] ii. Compute x = F 2n ( a 0n )and y = F 2n ( b 0n ). iii. Form the 2n-vector z = [ z 0,...,z 2n 1 in which z i = x i * y i. iv. Return the 2n-vector F2n( 1 z )asthe convolution of a and b. Given x = [x 0, x 1, x 2, x 3 ] = [4, 5, 8, 6 ]. Step 1: Form the 2-vectors x02 = [x 0, x 2 ] = [4, 8] and x 13 = [x 1, x 3 ] = [5, 6]. Step 2: Recursively compute a 01 = DFT (x 02,2,4 2 )Since arithmetic is performed modulo 17, 4 2 = 16 = 1. a 01 = [ a 0, a 1 ] = [x 0 + x 2, x 0 x 2 ] Hence, a 01 = [4+8, 4 8]=[12, 4]. Step 3: Recursively compute b 01 = DFT (x 13,2,4 2 )Since arithmetic is performed modulo 17, 4 2 = 16 = 1. b 01 = [ b 0, b 1 ] = [x 1 + x 3, x 1 x 3 ] Hence, b 01 = [5+6, 5 6]=[11, 1]. Step 4: Combine a 01 and b 01 to give the 4-vector y = [y 0, y 1, y 2, y 3 ] = DFT (x,4,4). [ y 0, y 2 ] = [ a 0 + b 0, a 0 b 0 ] Giving [ y 0, y 2 ] = [12+11, ] = [6,1] [y 1,y 3 ]=[a 1 +ωb 1,a 1 ωb 1 ] Giving [ y 1, y 3 ] = [ 4 4, 4 + 4]=[ 8, 0 ] So that DFT ([4, 5, 8, 6], 4, 4) = [6, 8, 1, 0]. The number of bit operations relies critically on the value ω underpinning the computation. Working modulo some value M, this must satisfy ω n/2 = 1for n-bit values (technically ω is a primitive n th root of unity in the ring of integers modulo M). In binary M can be chosen as 2 k + 1(kapower of 2) so that ω is a power of 2. This reduces the cost of operations (at bit level) since just shifting is involved. To parallel this process in decimal operations would require very large values of M, and ω. M = 10 k + 1with k apower of 10. For similar operations on decimal representations 9

10 10 6. A connected, undirected graph G(V, E) contains an Eulerian Tour if and only every vertex ofghas even degree (> 0), i.e. is the end-point of an even number of edges. The partial tour, T 7 = <{7,6},{6,2},{2,1},{1,7},{7,8},{8,10},{10, 7 } > does not contain every edge of G. Find the first vertex ont 7,for which there is some adjacent edge not included in T 7.(Vertex 6). Find a partial tour (T 6 )that contains all of those edges adjacent to 6 that do not occur in T 7.For example T 6 = <{6, 9}, {9, 10}, {10, 4}, {4, 3}, {3, 1}, {1, 4}, {4, 5}, {5, 6} > T 6 can be inserted into T 7 between the edges { 7, 6 } and { 6, 2 }. giving a partial tour T 6,7. The unused edges remaining in G form the graph, Vertex 9isthe first vertex encountered in T 6,7 which has unused edges adjacent to it. The partial tour T 9 = <{9, 4}, {4, 2}, {2, 5}, {5, 9} > can be inserted between the edges {6, 9} and {9, 10} of T 6,7 to give the completed Eulerian Tour. The suggested algorithm fails in general because many graphs are not the image of some graph using the edge-dual transformation, i.e. the graph H for which Ĥ = G does not exist. This is both for Hamiltonian and non-hamiltonian G, e.g.

11 SAT instances require all clauses to contain exactly 3 literals. Thus the clauses: (x 1 ), (x 2 x 8 ), ( x 3 x 4 x 5 x 7 ) and ( x 1 x 2 x 4 x 5 x 6 x 7 )ornot of the correct form. The first stage is to replace the unit clause ( x 1 )by4clauses (x 1 z 1 z 2 ) (x 1 z 1 z 2 ) (x 1 z 1 z 2 ) (x 1 z 1 z 2 ) with { z 1, z 2 }being new variables. Then replace the 2 literal clause ( x 2 x 8 )bythe 2 clauses ( x 2 x 8 z 3 ) ( x 2 x 8 z 3 ) The clause ( x 3 x 4 x 5 x 7 )isreplaced by ( x 3 x 4 z 4 ) ( x 5 x 7 z 4 ) The clause ( x 1 x 2 x 4 x 5 x 6 x 7 )by ( x 1 x 2 z 5 ) (x 4 z 5 z 6 ) (x 6 x 7 z 6 ) The final clauses - ( x 4 x 5 x 7 ) and ( x 4 x 7 x 8 ) - require no changes since these already contain 3 literals. Given F( X n ) = m C j an instance of 3 SAT,aninstance ( G F,8,9) j=1 of PART 3 SAT is formed as follows: G F uses n + 3m variables and has 9m clauses, G F ( X n, Z 3m ) = F( X n ) m 1 H k ( Z k ) the sub-formula H k consists of all eight possible clauses over the variables Z k = { z 3k+1, z 3k+2, z 3k+3 }. An assignment to Z 3m can satisfy at most 7m of the clauses in m 1 H k k=0 thus if the instance ( G F,8,9) of PART 3 SAT is true then there must be an assignment to X n that satisfies all m clauses in F, i.e. F is satisfiable. On the other hand, if F( X n )issatisfiable, then setting z i = false for each 1 i 3m will satisfy all but the clauses ( z 3k+1 z 3k+2 z 3k+3 )and so exactly 8m of the 9m clauses in G F are satisfied, i.e. the instance ( G F,8,9) of PART 3 SAT is true. k=0