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RANDOMNESS AND COUNTING in .NET Integrate 39 barcode in .NET RANDOMNESS AND COUNTING




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RANDOMNESS AND COUNTING using visual studio .net touse code 3/9 on asp.net web,windows application Customer Bar Code Exercise 6.23 (com Code 39 Extended for .NET puting the permanent of integer matrices): Prove that computing the permanent of matrices with 0/1-entries is computationally equivalent to computing the number of perfect matchings in bipartite graphs.

. Guideline: Given a bipartite graph G = ((X, Y ), E), consider the matrix M representing the edges between X and Y (i.e., the (i, j)-entry in M is 1 if the i th vertex of X is connected to the j th entry of Y ), and note that only perfect matchings in G contribute to the permanent of M.

. Exercise 6.24 (com Code39 for .NET puting the permanent modulo 3): Combining Proposition 6.

21 and Theorem 6.29, prove that for every xed n > 1 that does not divide any power of c, computing the permanent modulo n is NP-hard under randomized reductions. Since Proposition 6.

21 holds for c = 210 , hardness holds for every integer n > 1 that is not a power of 2. (We mention that, on the other hand, for any xed n = 2e , the permanent modulo n can be computed in polynomial time [230, Thm. 3].

). Guideline: Apply t he reduction of Proposition 6.21 to the promise problem of deciding whether a 3CNF formula has a unique satis able assignment or is unsatis able. Note that for any m it holds that cm 0 (mod n).

. Exercise 6.25 (neg Code 39 for .NET ative values in Proposition 6.

21): Assuming P = N P, prove that Proposition 6.21 cannot hold for a set I containing only non-negative integers. Note that the claim holds even if the set I is not nite (and even if I is the set of all non-negative integers).

. Guideline: A reduc tion as in Proposition 6.21 yields a Karp-reduction of 3SAT to deciding whether the permanent of a matrix with entries in I is non-zero. Note that the permanent of a non-negative matrix is non-zero if and only if the corresponding bipartite graph has a perfect matching.

. Exercise 6.26 (hig Code 3 of 9 for .NET h-level analysis of the permanent reduction): Establish the correctness of the high-level reduction presented in the proof of Proposition 6.

21. That is, show that if the clause gadget satis es the three conditions postulated in the said proof, then each satisfying assignment of contributes exactly cm to the SWCC of G whereas unsatisfying assignments have no contribution..

Guideline: Cluster the cycle covers of G according to the set of track edges that they use (i.e., the edges of the cycle cover that belong to the various tracks).

(Note the correspondence between these edges and the external edges used in the de nition of the gadget s properties.) Using the postulated conditions (regarding the clause gadget) prove that, for each such set T of track edges, if the sum of the weights of all cycle covers that use the track edges T is non-zero then the following hold: 1. The intersection of T with the set of track edges incident at each speci c clause gadget is non-empty.

Furthermore, if this set contains an incoming edge (resp., outgoing edge) of some entry vertex (resp., exit vertex) then it also contains an outgoing edge (resp.

, incoming edge) of the corresponding exit vertex (resp., entry vertex). 2.

If T contains an edge that belongs to some track then it contains all edges of this track. It follows that, for each variable x, the set T contains the edges of a single track associated with x..

EXERCISES 3. The tracks pic ked by T correspond to a single truth assignment to the variables of , and this assignment satis es (because, for each clause, T contains an external edge that corresponds to a literal that satis es this clause). Note that different sets of the aforementioned type yield different satisfying assignments, and that each satisfying assignment is obtained from some set of the aforementioned type.

. Exercise 6.27 (ana USS Code 39 for .NET lysis of the implementation of the clause gadget): Establish the correctness of the implementation of the clause gadget presented in the proof of Proposition 6.

21. That is, show that if the box satis es the three conditions postulated in the said proof, then the clause gadget of Figure 6.4 satis es the conditions postulated for it.

. Guideline: Cluster the cycle covers of a gadget according to the set of non-box edges that they use, where non-box edges are the edges shown in Figure 6.4. Using the postulated conditions (regarding the box) prove that, for each set S of non-box edges, if the sum of the weights of all cycle covers that use the non-box edges S is non-zero then the following hold: 1.

The intersection of S with the set of edges incident at each box must contain two (non-self-loop) edges, one incident at each of the box s terminals. Needless to say, one edge is incoming and the other outgoing. Referring to the six edges that connects one of the six designated vertices (of the gadget) with the corresponding box terminals as connectives, note that if S contains a connective incident at the terminal of some box then it must also contain the connective incident at the other terminal.

In such a case, we say that this box is picked by S. 2. Each of the three (literal-designated) boxes that is not picked by S is traversed from left to right (i.

e., the cycle cover contains an incoming edge of the left terminal and an outgoing edge of the right terminal). Thus, the set S must contain a connective, because otherwise no directed cycle may cover the leftmost vertex shown in Figure 6.

4. That is, S must pick some box. 3.

The set S is fully determined by the non-empty set of boxes that it picks. The postulated properties of the clause gadget follow, with c = b5 ..

Exercise 6.28 (ana .net vs 2010 Code39 lysis of the design of a box for the clause gadget): Prove that the 4-by-4 matrix presented in Eq.

(6.4) satis es the properties postulated for the box used in the second part of the proof of Proposition 6.21.

In particular: 1. Show a correspondence between the conditions required of the box and conditions regarding the value of the permanent of certain sub-matrices of the adjacency matrix of the graph. (Hint: For example, show that the rst condition corresponds to requiring that the value of the permanent of the entire matrix equals zero.

The second condition refers to sub-matrices obtained by omitting either the rst row and fourth column or the fourth row and rst column.) 2. Verify that the matrix in Eq.

(6.4) satis es the aforementioned conditions (regarding the value of the permanent of certain sub-matrices). Prove that no 3-by-3 matrix (and thus also no 2-by-2 matrix) can satisfy the aforementioned conditions.

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