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· 216-220 · 221-225 · 226-230 · 231-235 · 236-240 · 241-245 · 246-250 · 251-255 · 256-260 · 261-265 · 266-270 ·Log n!, Summation and Theta Relation
Prove that log n! and sum from i=2 to n log i have a theta relation to n log n.
Subject:
Math
Topic:
Discrete Structures
Posting ID:
48337
OTA ID:
103300
Time Complexity of an Algorithm in Theta Notation : Calculating Time from the Number of Loops
How much time does the following algorithm require as a funciton of n? Express your answer in theta notation in the simplest form. Consider each individual instruction (including loop control) as elementary. l = 0 for i = 1 to n for j = 1 to n^2 for k = 1 to n^3 l = l + 1
Subject:
Math
Topic:
Discrete Structures
Posting ID:
48985
OTA ID:
103300
Time complexity of an algorithm in theta notation
How much time does the following algorithm require as a function of n? Express your answer in "theta notation" in the simplest possible form. Show all work! l = 0 for i = 1 to n for j = 1 to i for k = j to n l = l +1
Subject:
Math
Topic:
Discrete Structures
Posting ID:
48987
OTA ID:
104459
Solve the following recurrence exactly for n of the form 2^2^k. T(2) = 1 T(n) = 2T(n^(1/2)) + log n Express your answer as simply as possible using theta notation. note added ** theta notation is based on big O notation Show all work!
Subject:
Math
Topic:
Discrete Structures
Posting ID:
49037
OTA ID:
101298
I received the following proof, can someone show all steps of how the solution was formed? Proof: Let n=2^2^k, then we have T(n)=T(2^2^k)=2T(n^(1/2))+log n ***How do you get n^(1/2) equals 2^2^(k-1) =2T(2^2^(k-1))+2^k ***How do you get 2T(2^2^(k-1)) equals 2(2T(2^2^(k-2))+2^(k-1)) =2(2T(2^2^(k-2))+2^(k-1))+2^k =2^2*T(2^2^(k-2))+2*2^k =2^2*(2T(2^2^(k-3))+2^(k-2))+2*2^k =2^3*T(2^2^(k-3))+3*2^k =... =2^k*T(2^2^0)+k*2^k =2^k*T(2)+k*2^k =2^k+k*2^k =(k+1)*2^k Since n=2^2^k, then 2^k=lg n and k=lg lg n Thus T(n)=theta((lg n)*(lg lg n)) ***Please show intermediate steps
Subject:
Math
Topic:
Discrete Structures
Posting ID:
49076
OTA ID:
103300
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