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· 231-235 · 236-240 · 241-245 · 246-250 · 251-255 · 256-260 · 261-265 · 266-270 · 271-275 · 276-280 · 281-285 ·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
Let T[1..n] be a sorted array of distinct integers, some of which may be negative. Give an algorithm that can find an index i such that 1 <= i <= n and T[i] = i, provided such an index exists. Your algorithm should take a time in Big "O" (log n) in worst case.
Subject:
Math
Topic:
Discrete Structures
Posting ID:
49476
OTA ID:
104455
Solving Recurrence Relations/Difference Equations
See attached file for full problem description.
Subject:
Math
Topic:
Discrete Structures
Posting ID:
49806
OTA ID:
105035
Prove that any binary search algorithm on a sorted array of size n that uses only key comparisons must require at least omega (log n) comparisons in the worst case.
Subject:
Math
Topic:
Discrete Structures
Posting ID:
50555
OTA ID:
101298
Prove that a graph with n nodes and more than n-1 edges must contain at least one cycle.
Subject:
Math
Topic:
Discrete Structures
Posting ID:
50814
OTA ID:
105059
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