Melden

ruckfulesnishuFolgen

21. Mar 2023•0 gefällt mir•2 views

21. Mar 2023•0 gefällt mir•2 views

Melden

Bildung

5.8: 1. (a) Does the recurrence (5.8) hold for the graph of Figure 5.7? Why? Pigure 5.7 Graph for Exercise 1 (b) Why does Equation 5.8 not hold for graphs with cycles of negative length?Recurrence (5.8) can be solved for An by first computing A1, then A2, then A3, and so on. Since there is no vertex in G with index greater than n, A(i,j)=An(i,j). Function AllPaths computes A n(i,j). The computation is done inplace so the superscript on A is not needed. The reason this computation can be carried out in-place is that Ak(i,k)=Ak1(i,k) and Ak(k,j)=Ak1(k,j). Hence, when Ak is formed, the k th column and row do not change. Consequently, when Ak(i,j) is computed in line 11 of Algorithm 5.3, A(i,k)=Ak1(i,k)=Ak(i,k) and A(k,j)=Ak1(k,j)=Ak(k,j). So, the old values on which the new values are based do not change on this iteration..

6 Add records to the database a This query will add a new.pdfruckfulesnishu

6 Assessed at both Level 6 and 7 Assume that in an experim.pdfruckfulesnishu

6 Aviden makes at most I mintines in The fictisaicy elat a .pdfruckfulesnishu

6 A quantity q is related to other measuring variables .pdfruckfulesnishu

5 Which of the following is the correct order of the main s.pdfruckfulesnishu

5 Two traits are observed in an organism A and B When a f.pdfruckfulesnishu

- 1. 5.8: 1. (a) Does the recurrence (5.8) hold for the graph of Figure 5.7? Why? Pigure 5.7 Graph for Exercise 1 (b) Why does Equation 5.8 not hold for graphs with cycles of negative length?Recurrence (5.8) can be solved for An by first computing A1, then A2, then A3, and so on. Since there is no vertex in G with index greater than n, A(i,j)=An(i,j). Function AllPaths computes A n(i,j). The computation is done inplace so the superscript on A is not needed. The reason this computation can be carried out in-place is that Ak(i,k)=Ak1(i,k) and Ak(k,j)=Ak1(k,j). Hence, when Ak is formed, the k th column and row do not change. Consequently, when Ak(i,j) is computed in line 11 of Algorithm 5.3, A(i,k)=Ak1(i,k)=Ak(i,k) and A(k,j)=Ak1(k,j)=Ak(k,j). So, the old values on which the new values are based do not change on this iteration.