By H. H. Schaefer

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N}, i 6= j aij = 0 or aji = 0 ∀i, j ∈ {1, 2, . . n}, i 6= j [aij = 1 and aji = 1] implies i = j ∀i, j ∈ {1, . . n} iRj and jRi imply i = j ∀i, j ∈ {1, 2, . . n} R is antisymmetric. Shortest Paths in Weighted Graphs 6. (b) agbhf e is a shortest (a, e)-path, and thus bhf e must be a shortest (b, e)-path (for if there were a shorter (b, e)-path, appending it to agb would give a shorter (a, e)-path, contrary to agbhf e being a shortest (a, e)-path). 8. (a) The shortest (d, v)-paths found by the implementation of Dijkstra’s algorithm below are dcba, dcb, dc, d, dgf e, dgf , dg.

If an = f (n) = i=0 i2 , then the sequence of first diﬀerences is (02 , 12 , 22 , 32 , . ) and thus the sequence of third diﬀerences is constantly 2! = 2. a = 2, we 2 have a = 3! = 13 . Since f (0) = 0 = d, we now have f (n) = 13 n3 + bn2 + cn. From 2 f (1) = 1 = 13 + b + c and f (2) = 12 + 22 = 5 = 83 + 4b + 2c, we find that b = 12 and c = 16 , so f (n) = 12 + · · · + n2 = n3 n2 n n n(n + 1)(2n + 1) + + = (2n2 + 3n + 1) = . 3 2 6 6 6 th 7. If the kth diﬀerences of (ai )1 degree polynomial, then the i=1 are generated by an n th th 1 n diﬀerences of the k diﬀerences of of (ai )i=1 are a nonzero constant.

Dn , . ). If D = {d1 , d2 , d3 , . ) is countably infinite, then any sequence in D is a subsequence of (d1 , d1 , d2 , d1 , d2 , d3 , d1 , d2 , d3 , d4 , d1 , d2 , d3 , d4 , d5 , d1 , d2 , . ). 45 46 CHAPTER 8. 2 Finite Diﬀerences 1. (a) The sequence of second diﬀerences is constantly 4. This tells us that the sequence is generated by a second degree polynomial p(n) = an2 + bn + c. a = 4, we find that a = 2. Since the first term 4 is p(0) = c, we have p(n) = 2n2 + bn + 4. Now p(1) = 3 = 2(12 ) + b(1) + 4 implies b = −3, so p(n) = 2n2 − 3n + 4.