CSE 531 Homework 1 Solution. Jiayi Xian

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1 CSE 53 Homework Solution Jiayi Xian Fall 08

2 . (a) True. Since f(n) O(f(n)), clearly we have O(f(n)) O(O(f(n))). For the opposite, suppose g(n) O(O(f(n))), that is c > 0, h(n) O(f(n)), s.t g(n) c h(n) as n. Since h(n) O(f(n)), we have some c > 0, s.t h(n) c f(n) as n, thus g(n) c c f(n) as n, i.e g(n) O(f(n)). (b) True. Since f(n) Θ(f(n)), clearly we have O(f(n)) O(Θ(f(n))). For the opposite, suppose g(n) O(Θ(f(n))), according to definition, h(n) Θ(f(n)), c, c, c 3 > 0, s.t g(n) c h(n), c f(n) h(n) c 3 f(n) as n. It gives g(n) c c 3 f(n) as n, i.e. g(n) O(f(n)). (c) False. The relation should be Θ(O(f(n)) Θ(f(n)). Let f(n) = n, g(n) = O(f(n)). clearly Θ() Θ(O(f(n))) but / Θ(n) = Θ(f(n)) (since there doesn t exist c > 0 s.t > c n as n ). (d) True. g(n), let h(n) = max{g(n), f(n)}, l(n) = min{g(n), f(n)}. We know g(n) max{g(n), f(n)} Ω(f(n)), so g(n) O(Ω(f(n))). g(n) min{g(n), f(n)} Ω(O(f(n))), so g(n) Ω(O(f(n)). So Ω(O(f(n))) = O(Ω(f(n))) = {all the functions : N + R + } (e) True. By definition, c > 0 s.t. c(g(n) + h(n)) f(n), since both g(n) and h(n) are positive, we have c g(n) f(n), c h(n) f(n) respectively. Therefore f(n) = Ω(g(n)),f(n) = Ω(h(n)). (f) False. Take f(n) = + n, g(n) =. f(n) = Θ() but log f(n) = Θ( n ) (g) True. Since both f(n) and g(n) are positive, clearly we have max (f(n), g(n)) f(n) + g(n). For the other direction, WLOG we assume f(n) g(n), then f(n) + g(n) f(n) = max (f(n), g(n)). (a) When n =, it s trivially true. Now suppose when the statement holds for n = k: then for n = k +, we have:! + 3! + + k (k + )! = (k + )!! + 3! + + k (k + )! + k + (k + )! = (k + )! + k + (k + )! = k + (k + )! + k + (k + )! = (k + )! It completes our proof.

3 (b) Proof. First we use mathematical induction to prove the maximum number of regions generated by the intersections of n unit circles on a plane is O(n ): Given arbitrary n circles, let r(k) be the number of regions generated by the first k circles. Then it s sufficient to show r(k) r(k )+Θ(k) for all k n. Consider the intersection of the k-th circle (denoted as R k ) and the first k circles. Since any two circle intersects at at most points, there are at most (k ) intersection points on R k, which divides R k into at most (k ) many arcs. Any newly generated region must include at least one arc of R k as its boundary, otherwise it s an old region. But for any arc of R k, there are at most two regions include it as their boundary. Therefore, the newly generated region is upper-bounded by 4(k ). Namely, r(k) r(k ) + 4(k ), implies r(n) = O(n ). Second, to show the maximum number of regions generated by the intersections of n unit circles on a plane is Ω(n ), we just need to provide a certain arrangement. For convenience let s consider n unit circles under Cartesian coordinate system. The coordinates of the center of a circle solely determines its position. Place the first n circles so that the i-th circle centered at ( + i/n, 0). Place the rest of n circles so that the n + i-th circle centered at (0, + i/n ). For sufficient large n, there are Θ(n ) many regions inside the rectangle bounded by (0, 0) and (/n, /n). 3. (a) Stirling formula: n! πn( n e )n = O(n n ) (b) n x log xdx = n x log x n xdx = n log n + o(n log n) = Θ(n log n) (c) = log k + log k log + log 0 i=0 = log( k k... ) = log( k+(k ) ) = log( k(k+) ) = k + k = (log n + log n) = Θ(log n)

4 (d) Wrong. By limit test, n n lim n n = lim n log n n n = lim n n(log n ) = 4. n lg n, lg (lg(n)), (lg(n)), (lg n) lg(lg n), lg n, n 5, [lg(lg n)] lg n, n0.00, n!, n n lg n = ( lg n ) lg n = Denote k = lg lg n, f(k) = lg n, f(k ) = lg lg(n), f(k ) = lg lg lg(n).... Then we have f(i ) = lg(f(i)), f(i) = f(i ), for i (,..., k). First, it s easy to see by induction k f(k): (a) k =, f() = x. (b) Suppose k = n >, n f(n). (c) For k = n +, f(n + ) = f(n) > n > n +. Second, according to the definition of lg lg n, we have 0 < f(0) < and lg n = f(k ). So k < (k )/ < f(k )/ = lg n for n. The above proves n lg n lg (lg(n)) (lg(n)). Let x = lg(n) (lg(n)) = x / / lg x = (lg n) lg(lg n) = (lg(lg n)) lg(lg n) = (lg(x)) lg n = x n 5 = 5 lg n = 5x [lg(lg n)] lg n = [lg(lg(lg n))] lg n [lg(lg x)] x = [lg(lg n)] lg n = [lg(lg(lg n))] lg n n0.00 Remember lg x x ε for ε > 0, we can get (lg(n)) (lg n) lg(lg n) lg n n 5 [lg(lg n)] lg n n0.00 By Stirling s formula, n! πn( n e )n. lg n0.00 = n

5 lg n! = lg(πn) + n lg(n e ) lg n = n Since n 0.00 n lg( n e ), lg(πn) + n lg( n e ) n when n, we have n0.00 n! n. 5. (a) T (n) = T (n ) + 3 n T (n) = T (n ) + 3 n = T (n ) + 3 n + 3 n = T () + = + 3n+ 3 = 3n+ (b) Assume n = 3 k, from T (n) = 4T (n/3) + n.5 we have: T (n) = T (3 k ) = 4T (3 k ) + 3 k = 4(4T (3 k ) + 3.5(k ) ) + 3.5k = 4 T (3 k ) (k ) + 3.5k = 4 k T () + = 4 k ( + ( k i 3.5i ( ) k+ = 4 k = Θ(3.5k ) = Θ(n.5 ) or using master theorem. ) i ) n i= 3 i 4

6 (c) Assume n = 8 k, from T (n) = 5T (n/8) + n we have: T (n) = T (8 k ) = 5T (8 k ) + 8 k = 5(5T (8 k ) + 8 k ) + 8 k = 5 T (8 k ) k + 8 k = 5 k T () + = 5 k ( + 5 k i 8 i (8) i) 5 ( 8 ) k+ = 5 k = Θ(8 k ) = Θ(n) or using master theorem. (d) Assume n = a k, then log n = k log a T (n) = T ( n) + log n = T (n 4 ) + log n + log n = T (n k ) + log n k + log n k + + log n + log n = T (a) + log n ( k i=0 ( )k )i = T (a) + log n = T (a) + log n log a = T (a) + ( ( )k ) log n (e) T (n) T (n ) = T (n ), so T (n) = 3T (n ) = 3 T (n ) = = 3 n T () = 3 n. 5

7 (f) Assume n = k, from T (n) = 3T (n/) + n log n we have: T ( k ) = 3T ( k ) + k k = 3(3T ( k ) + k (k ) + k k = 3 T ( k ) + 3 k (k ) + k k = 3 k T () + = 3 k + 3 k 3 k i i i = 3 k ( + o(3 k )) = Θ(3 k ) = Θ(n log 3 ) () i i 3 or using master theorem. (g) Assume n = k, from T (n) = T (n/) + n log log n we have: T ( k ) = T ( k ) + k log k = (T ( k k ) + log (k ) ) + k log k = T ( k ) + = k T () + = k + k k log (k ) + k log k k log i log i Using integral method, k ln xdx which is called the Logarithmic integral function, or li(x). It s asymptotic behavior is li(x) = Θ(x/ ln x). T (n) = Θ(n log n/ log log n). log i is bounded by ln k Therefore k log i = Θ(k/ ln k). 6

8 (h) Assume n = k, from T (n) = nt ( n) + n we have: T ( k ) = k T ( k ) + k + k = k ( k T ( k ) + k ) + k + k = k + k T ( k ) + k + k + k + k = k k T ( k ) + k + k + k + k = k k ( k 3 T ( k 3 ) + k ) + k + k + k + k = k k 3 T ( k 3 ) + k + k + k + k + k + k = k T () + k + i = k / T () + k = n/ T () + n Θ(n) = Θ(n ) i 6. The characteristic equation is x = 8x 5. The roots are x = 3 and x = 5. So a n = A 3 n + B 5 n. Plug in a 0 = 3 and a = 5, we have A + B = 3 3A + 5B = 5 So A = 5, B =. So a n = 5 3 n 5 n. 7