22C:044 Homework #4 Solutions

DISPLAYS BETTER WITH INTERNET EXPLORER

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1. Order of functions (slowest growth first):

log(logn), 1000logn, (logn)², 3n^.5, n + cosn, nlog(n²), 100n² + logn, n³,

400n – n³ + n⁵, 2ⁿ⁺¹, 3ⁿ, 2²ⁿ, n!

where

lim 3ⁿ/2²ⁿ = lim 3ⁿ/4ⁿ = lim (0.75)ⁿ = 0

^{n®infinity n®infinity n®infinity}

and by L’Hopital’s Rule

lim (logn)²/(n^.5) = 4 lim logn/(n^.5) = 8 lim 1/n^.5 = 0

^{n®infinity n®infinity n®infinity}

2. Show (n+4)² = O(n²) ie find c,n₀ > 0 such that 0 <= (n+4)² <= cn² for all n >= n₀

n² + 8n +16 <= cn² Try c = 2 (Other choices are also acceptable)

Then calculate n₀

n₀² + 8n₀ +16 <= 2n₀² , any n₀ >= 4(2^.5) + 4 will work for c = 2

3.

a) Yes. A runs no slower than c_an², so it is possible that A always runs at say 1n or faster. B runs at best c_bn so B could always run at 2n or slower.

Then A would run faster than B for all inputs.

b) Yes. It could be that A never runs faster than 5nlogn and that B runs at worst

as say 3nlogn. Then B always faster than A.

c) No. B has a TIGHT upper bound of nlogn in the worst case, so B will run

as cnlogn for many inputs. C always runs as 2n so there will be inputs for which B is not faster than C.

d) Yes. B has a lower bound of n, so B will never run better than c_bn. Then it’s

possible B never runs faster than say 3n. But C always runs as 2n, so it’s possible C is always faster than B.

e) Yes. Worst case of D is W(1). This is a LOOSE lower bound on the worst case

running time, D could be much worse than this in the worst case and could in fact be much worse than this for all inputs. So it could be the case that D is slower

than cn² for all inputs which would make it slower than A,B,C for all inputs.