"... In PAGE 3: ... Theorem 3 improves previous results on tree 3-spanners in interval graphs [20,22,27] andon split graphs [6,20,29]. The complexity status of T REE t-SPANNER on chordal graphs consideredin this paper is summarizedin Table1 andFig. 1.... ..."

"... In PAGE 11: ... Furthermore, both algorithms can execute a cubic number of coalescings, but, in the average, the quantity of copy instructions per program is small when compared to the total number of instructions. Table2 compares the two algorithms when the interference graphs are chordal... ..."

"... In PAGE 4: ... We examine in detail the low-degree cases, lt; 7, and derive best possible upper bounds on the maximum clique and chromatic numbers, as well as inductiveness of squares of outerplanar graphs. These bounds are illustrated in Table1 . We treat the special case of chordal outerplanar graphs separately, and further classify all chordal outerplanar graphs G for which the inductiveness of G2 exceeds or the clique or chromatic number of G2 exceed + 1.... In PAGE 17: ...orollary 4.10 together with Theorems 4.3 and 4.5 complete the proof of Theorem 4.1 as well as the entries in Table1 in the chordal case for 2 f2; 3; 4; 5; 6g. Observation 4.... ..."

"... In PAGE 6: ... Thus there are 2n such cycles. The columns labeled Direct in Table2 show results for enumerating the chord- free cycles for our benchmarks. For each correct microprocessor, we have two graphs: one for which transitivity constraints played no role in the verification, and one (in- dicated with a t at the end of the name) modified to require enforcing transitivity constraints.... In PAGE 8: ... The graph has N(N ? 1) edges and N(N ? 1)(N ? 2)=6 chord-free cycles, yielding a total of N(N ? 1)(N ? 2)=2 = O(N3) transitivity constraints. The columns labeled Dense in Table2 show the complexity of this method for the benchmark circuits. For the smaller graphs 1 DLX-C, 1 DLX-Ct, M4 and M5, this method yields more clauses than direct enumeration of the cycles in the original graph.... In PAGE 8: ... If more than one vertex has minimum degree, we choose one that minimizes the number of new edges added. The columns in Table2 labeled Sparse show the effect of making the benchmark graphs chordal by this method. Observe that this method gives superior results to either of the other two methods.... In PAGE 11: ... Table 4 shows the complexity of the graphs generated by this method for our bench- mark circuits. Comparing these with the full graphs shown in Table2 , we see that we... ..."

"... In PAGE 10: ... In addition, the edges forming the perimeter of each face Fi create a chord-free cycle, giving a total of 2n + n chord-free cycles. The columns labeled Direct in Table2 show results for enumerating the chord-free cycles for our benchmarks. For each correct microprocessor, we have two graphs: one for which transitivity constraints played no role in the verification, and one (indicated with a t at the end of the name) modified to require enforcing transitivity constraints.... In PAGE 13: ... The graph has N(N ? 1) edges and N(N ? 1)(N ? 2)=6 chord-free cycles, yielding a total of N(N ? 1)(N ? 2)=2 = O(N3) transitivity constraints. The columns labeled Dense in Table2 show the complexity of this method for the benchmark circuits. For the smaller graphs 1 DLX-C, 1 DLX-C-t, M4 and M5, this method yields more clauses than direct enumeration of the cycles in the original graph.... In PAGE 14: ... If more than one vertex has minimum degree, we choose one that minimizes the number of new edges added. The columns in Table2 labeled Sparse show the effect of making the benchmark graphs chordal by this method. Observe that this method gives superior results to either of the other two methods.... In PAGE 15: ... One can see that the set of edges forming the border of each face forms a chord-free cycle of Mn. As shown in Table2 , many other cycles are also chord-free, e.g.... In PAGE 21: ... Table 4 shows the complexity of the graphs generated by this method for our benchmark cir- cuits. Comparing these with the full graphs shown in Table2 , we see that we typically reduce the number of relational vertices (i.e.... ..."

"... In PAGE 7: ... Again by the exponential relationship, P n 0;q 0 An;qxnyq=n! = exp(P n 0;q 0 an;qxnyq=n!), which leads to An;q = an;q + q X l=0 1 n n 1 X k=1 k n k ! ak;lAn k;q l !! : (4) Together, (3) and (4) determine the numbers an;q recursively, beginning with a2;1 = 1. Table2 gives the resulting values of an;q for small n. 3.... ..."