Asymptotic Analysis and Fibonacci Number
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Q.1 For a number x, with 1< x < p, the number x^n mod p can be computed with at most 2log2 n modulo p multiplications.
Asymptotic notation questions
Q.2 2^(2n) = O(2^n)
Q.3 log*n = O(log*(log n))
Q.4 The sqrt n th Fibonacci number can be computed and written in O(log n) time
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Asymptotic Analysis and Fibonacci Number are investigated.
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Q.1
For a number x, with 1 x p, the number can be computed with at most
modulo p multiplications.
It is true. By Modular Arithmetic, we know that
mod p) (b mod p)] mod p
mod ] mod p
So, when we compute , it suffices to write n in the form of
Since any positive integer can have a binary code ...
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
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- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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