De Bruijn Graphs and their Applications to Fault Tolerant Networks

More documents

Recommendations

Info

$Math 664 Homework #2: Solutions 1. Let Î© = R, F = {A â R : either A ...$

22 JOEL BAKERWe see that if we let δ = c j − c i then,c i − c i+1 = c j − c j+1 → c i+1 = c j+1 − (c j − c i ) = c j+1 − δc i+1 − c i+2 = c j+1 − c j+2 → c i+2 = c j+2 − (c j+1 − c i+1 ) = c j+2 − δ.c i+n−1 − c i+n = c j+n−1 − c j+n → c i+n = c j+n − δThis gives us that the cycles C and C δ have the edge (c i c i+1 · · · c i+n−1 )(c i+1 · · · c i+n ) incommon and thus it must be the case that δ = 0. This implies thatc i = c jc i+1 = c j+1. = .c i+n = c j+nand therefore we see that ψ(C) can have no repeated nodes.Example. Consider the cycle C = [2, 1, 1, 0, 1] in B(3, 2). We generateψ(C) = [2 − 1, 1 − 1, 1 − 0, 0 − 1, 1 − 2] = [1, 0, 1, 2, 2]We see that ψ(C) repeats no nodes and thus we know that the set of cycles {C α |0 ≤ α ≤d − 1} are pairwise edge disjoint. We can list these cycles,C 0 = [2, 1, 1, 0, 1]C 1 = [0, 2, 2, 1, 2]C 2 = [1, 0, 0, 2, 0]and verify that no edge (3-tuple) is used more than once.primitive.□Thus we see that C is edgeExample. We can instead consider the cycle C = [1, 2, 0, 0, 1] in B(3, 2). First, we generateψ(C) = [1 − 2, 2 − 0, 0 − 0, 0 − 1, 1 − 1] = [2, 2, 0, 2, 0]Notice that the node (20) shows up two times and therefore by Lemma 4 the set of cycles{C α | 0 ≤ α ≤ d − 1} are not pairwise edge disjoint. Here we will list these cycles,C 0 = [1, 2, 0, 0, 1]C 1 = [2, 0, 1, 1, 2]C 2 = [0, 1, 2, 2, 0]We see that the edge (11)(12) shows up in both C 0 and in C 1 and thus verifies that C isnot edge primitive.
DE BRUIJN GRAPHS AND THEIR APPLICATIONS TO FAULT TOLERANT NETWORKS 23The next lemma comes from examining the relationship between cycles in B(d, n) andlinear recurrence relations over F d .If we have a sequence S defined by the recurrence relation,(4) s n+i = a n−1 s n−1+i + · · · + a o s i for i ≥ 0with a i ∈ F d and a 0 ≠ 0, then the period of S is the smallest k > 0 such that s i = s i+kfor i ≥ 0. We see then that a sequence with period k corresponds to a cycle of length k inthe De Bruijn graph B(d, n), specifically the cycle [s 0 , s 1 , . . . , s k−1 ].We will call the polynomial(5) p(x) = x n − a n−1 x n−1 − · · · − a 1 x − a 0the characteristic polynomial of S.Definition 19. If p(x) is a polynomial over F d then the order of p(x) is the least positiveinteger k such that p(x) divides 1 − x k . An irreducible polynomial of degree n withcoefficients in F d is primitive if its order is d n − 1.Remark. If p(x) is irreducible then from [RB91] the period of S is exactly the order of p(x).From [HP03] we know that there is a primitive polynomial for every choice of n and d (aprime power).Definition 20. We will say that C = [c 0 , c 1 , . . . , c k−1 ] is a maximal cycle if it is a cyclewith length k = d n −1 and it satisfies a recurrence relation whose characteristic polynomialis primitive.Lemma 5. If C is a maximal cycle, then C is edge primitive.Proof. Let C be a maximal cycle which is not edge primitive. Since it is not edge primitive,there exist two cycles, C α and C β which share an edge. This means that the node x =(x 0 , x 1 , . . . , x n−1 ) is followed by the node (x 1 , . . . , x n−1 , x n ) in both C α and C β . Then sinceC is generated by the recurrence relation (4) we see that in C α this implies,which impliesx n =a n−1 (x n−1 − α) + · · · + a 0 (x 0 − α) + α= a n−1 x n−1 + · · · + a 0 x 0 + (−a n−1 α − · · · − a 0 α + α)dx n − a n−1 x n−1 − · · · − a 0 x 0 =−a n−1 α − · · · − a 0 α + α= α(1 − (a n−1 + · · · + a 0 )).Likewise, in C βx n = a n−1 (x n−1 − β) + · · · + a 0 (x 0 − β) + β= a n−1 x n−1 + · · · + a 0 x 0 + (−a n−1 β − · · · − a 0 β + β).and sox n − a n−1 x n−1 − · · · − a 0 x 0 = −a n−1 β − · · · − a 0 β + β= β(1 − (a n−1 + · · · + a 0 )).
Page 9 and 10: DE BRUIJN GRAPHS AND THEIR APPLICAT
Page 21: DE BRUIJN GRAPHS AND THEIR APPLICAT
Page 29: DE BRUIJN GRAPHS AND THEIR APPLICAT

De Bruijn Graphs and their Applications to Fault Tolerant Networks

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?