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Discrete Mathematics with Graph Theory (3rd Edition) 230

# Discrete Mathematics with Graph Theory (3rd Edition) 230 -...

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228 Solutions to Exercises (c) There exists a positive integer no such that for all n ~ no, I ~i~? - LI < 1 and hence I ~i~? I - ILl < 1 (using the fact that lal-Ibl ~ la - bl for any real numbers a and b). Thus, for n ~ no, I ~f~~ I < ILl + 1 so, with c = ILl + 1 we see that f = 0(9). We now must show that 9 is also 0(1). But from the theory of limits, lim n --+ oo J~~~ = ± (since L #- 0). So the argument which showed f = 0(9) also shows that 9 = 0(1). Exercises 8.3 1. (a) [BB] Since n = 9 #- 1, we set m = l~J = 4. Since 2 = x ~ a4 = 4, we set n = m = 4 and change the list to 1,2,3,4. Since 4 = n #- 1, we set m = l ~ J = 2. Since 2 = x ~ a2 = 2 we set n = m = 2 and change the list to 1,2. Since 2 = n #- 1, we set m = r~l = 1. Since 2 = x> al = 1, we replace n by n - m = 1 and change the list to 2. Since n = 1 and x = aI, we output "true" and stop. This search required four comparisons of x with an element in the list; a linear search would have used two comparisons. (b) Since n = 9 #- 1, we set m = l~J = 4. Since 7 = x > a4 = 4, we replace n by n - m = 5 and change the list to 5,6,7,8,9. Since 5 = n #- 1, we set m = l~J = 2. Since 7 = x > a2 = 6, we replace n by n - m = 3 and
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