De Bruijn Graphs and their Applications to Fault Tolerant Networks

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

24 JOEL BAKERNow from C α and C β we get thatand since α ≠ β then we seeand then from (5) we see thatα(1 − (a n−1 + · · · + a 0 )) = β(1 − (a n−1 + · · · + a 0 ))1 − (a n−1 + · · · + a 0 ) = 0p(1) = 1 − (a n−1 + · · · + a 0 ) = 0and thus (1 − x) divides p(x) which contradicts the fact that p(x) is irreducible. Thus C αand C β have no common edges and therefore C is edge primitive.□This gives us some cool hardware, so let’s look at an example of this. We would like toconstruct a maximal cycle in B(3, 3). We know that p(x) = x 3 + 2x 2 + 1 is primitive overF d and so the linear recurrence(6)x i+3 = −2x i+2 − x i= x i+2 + 2x iwith seeds x 0 = x 1 = 0 and x 2 = 1 should generate our cycle. We continuex 3 =x 2 + 2x 0=1 + 2 · 0=1x 4 =x 3 + 2x 1=1 + 2 · 0=1x 5 =x 4 + 2x 2=1 + 2 · 1=0. . . = . . .and so on, continuing to use our recurrence relation to get the next values. This generatesthe cycleC 0 = [00111021121010022201221202].Then we can construct C 1 and C 2 by adding 1 and 2 to each value, respectively, givingC 1 =[11222102202121100012002010]C 2 =[22000210010202211120110121].We see by inspection that C 0 , C 1 and C 2 are edge disjoint.
DE BRUIJN GRAPHS AND THEIR APPLICATIONS TO FAULT TOLERANT NETWORKS 25Theorem 5. The De Bruijn graph B(d, n) with at most d − 1 faulty edges admits a cycleof length d n − 1 when d is a prime power.Proof. We know that every maximal cycle is a cycle of length d n −1 and also that there ared edge disjoint cycles that we can generate in this manner. This means that these cyclescoverd(d n − 1) = d n+1 − dedges. We also notice that none of them use the loop edges since then they would visit anode twice.Additionally, we know from Procedure 1 in Section 2 that there are d n+1 total edges andd loop edges, thus these maximal cycles partition all the d n+1 − d edges which are not loopedges. This means that since we have d edge disjoint cycles, that removing d−1 edges fromthe whole graph cannot break all of these cycles and therefore when d is a prime power wesee that B(d, n) still admits at least one of these cycles.□Rowley and Bose expand this result to provide a lower bound for De Bruijn graphs whichdo not have alphabets with prime power sizes in [RB93a] . They prove by similar meansin the following Lemmas and Theorem.Lemma 6. The graph B(d, n) admits d − 1 disjoint Hamiltonian cycles when d is a powerof 2.Lemma 7. The graph B(d, n) admits d−12disjoint Hamiltonian cycles when d is an oddprime power.We notice that these differ from our previous result in that they are full Hamiltoniancycles and not just cycles of length d n −1 as before. It is this small difference which reducesthe disjoint cycle count.Lemma 8. If B(d, n) admits k disjoint Hamiltonian cycles, B(d ′ , n) admits k ′ Hamiltoniancycles and d and d ′ are relatively prime, then B(dd ′ , n) admits at least kk ′ Hamiltoniancycles.Theorem 6. The number of disjoint Hamiltonian cycles in B(d, n) is at leastk∏2 −k e(p i i − 1)i=1where p e 11 · · · pe kkis the prime factorization of d.Proof. The proof of this theorem follows directly from Lemmas 6, 7 and 8.□
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De Bruijn Graphs and their Applications to Fault Tolerant Networks

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