Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter 25.2, Problem 4E
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To explain the updating in FLOYD-WARSHALL
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Fermat's "Little" Theorem states that whenever n is prime and a is an integer, a^n−1≡1modn Then 281825≡ 263 mod827. c) If a=146 and n=331, then efficiently compute 146332≡ mod331Use The Extended Euclidean Algorithm to compute281^1≡362 mod827.
Then 281^825≡???mod827.
c) If a=146 and n=331, then efficiently compute146^332≡ ???mod331
let us use x as an integer, construct an algorithm that determines how many with repetitions the integer × be written as the sum of 1,2,4 and prove its time complexity phsudocode
Let n be a positive integer, and consider the following algorithm segment.
for i := 1 to n
for j := 1 to i
[Statements in body of inner loop.
None contain branching statements
that lead outside the loop.]
next j
next i
How many times will the inner loop be iterated when the algorithm is implemented and run?
Chapter 25 Solutions
Introduction to Algorithms
Ch. 25.1 - Prob. 1ECh. 25.1 - Prob. 2ECh. 25.1 - Prob. 3ECh. 25.1 - Prob. 4ECh. 25.1 - Prob. 5ECh. 25.1 - Prob. 6ECh. 25.1 - Prob. 7ECh. 25.1 - Prob. 8ECh. 25.1 - Prob. 9ECh. 25.1 - Prob. 10E
Ch. 25.2 - Prob. 1ECh. 25.2 - Prob. 2ECh. 25.2 - Prob. 3ECh. 25.2 - Prob. 4ECh. 25.2 - Prob. 5ECh. 25.2 - Prob. 6ECh. 25.2 - Prob. 7ECh. 25.2 - Prob. 8ECh. 25.2 - Prob. 9ECh. 25.3 - Prob. 1ECh. 25.3 - Prob. 2ECh. 25.3 - Prob. 3ECh. 25.3 - Prob. 4ECh. 25.3 - Prob. 5ECh. 25.3 - Prob. 6ECh. 25 - Prob. 1PCh. 25 - Prob. 2P
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- Recall that Fibonacci numbers form a sequenceFnwhereF0= 0,F1= 1, andFn=Fn−1+Fn−2. The standard algorithm for finding the n-th Fibonacci number takes O(n) time. The goal of this question is to design a significantly faster algorithm for this problem. (a) Prove by induction that for alln≥1:[1 11 0]n=[Fn+1FnFnFn−1]. (b) Use the first part to design an algorithm that findsFninO(logn) time.arrow_forwardPlease solve sections, Find the asymptotic (large-Θ) limits for the running times of the algorithms whose running time is given iteratively. 1. T (n) = 4T (n/4) + 5n2. T (n) = 4T (n/5) + 5n3. T (n) = 5T (n/4) + 4n4. T (n) = T (n/2) + 2T (n/5) + T (n/10) + 4narrow_forwardFor each of the following algorithm, find running Time complexity T(n) and Space complexity S(n). Also show in big-O notation.Code Segment 1Algorithm Sumint a=23;int b=5;int sum=a+b;cout<<sum;arrow_forward
- Use previous solution to Derive Big Oh for the algorithms from Problem 1. This is the pervious solution starting with Step 1 and Step 2: Step 1 - We need to provide the asymptotic equation that will describe the complexity of the provided code snippets. Step 2 :: 1. First code :: -> In this we have 3 operations and all the operation take constant time. We are considering (input = n). Lets take time taken for :: x = 2 , y = 2 -> a (Constant time) x = ((y + z) * 80) /4 -> b (Constant time) print x ,y, z -> c (Constant time) Here all are constant operations. So, T(n) = a + b + c , which can also be written as :: T(n) = k (Constant) 2. Second code :: -> In this code we have a loop and so the operation will take linear time. We are considering (input = n). Lets take time taken for :: x = 1 , y = 5 -> a (Constant time) loop, (y = y + 5) , (print x, y) -> 4*n (Linear time) T(n) = a + 4*n or, T(n) = a + b*n ( a, b =…arrow_forwardTwo algorithms A, B sort the same problem. When you go through each algorithm and break them down into their primitive operations, each can be represented as below A = n4 + 100n2 + 10n + 50 B = 10n3 + 2n2 + nlogn + 200 For very large values of n which of these algorithms explain why B will run in the shortest time to solve the problemarrow_forwardTake into account the MAXMIN methodology below. It uses how many comparisons? Is it more likely to operate quicker or slower in practise than the divide-and-conquer algorithm? procedure The remark maxmin2(S) calculates the maximum and minimum of S[1..n] in max and min, respectively. 1. if n is odd then max:=S[n]; min:=S[n] 2. else max:=−∞; min:=∞ 3. for i := 1 to n/2 do 4. if S[2i − 1] ≤ S[2i] 5. then small:=S[2i − 1]; large:=S[2i] 6. else small:=S[2i]; large:=S[2i − 1] 7. if small < min then min:=small 8. if large > max then min:=smallarrow_forward
- Correct answer will be upvoted else downvoted. Computer science. Allow us to signify by d(n) the amount of all divisors of the number n, for example d(n)=∑k|nk. For instance, d(1)=1, d(4)=1+2+4=7, d(6)=1+2+3+6=12. For a given number c, track down the base n to such an extent that d(n)=c. Input The principal line contains one integer t (1≤t≤104). Then, at that point, t experiments follow. Each experiment is characterized by one integer c (1≤c≤107). Output For each experiment, output: "- 1" in case there is no such n that d(n)=c; n, in any case.arrow_forwardConsidering the following algorithm, analyze its best case, worst case and average case time complexity in terms of a polynomial of n and the asymptotic notation of ɵ. You need to show the steps of your analysis.arrow_forwardAnalyze the complexity of the algorithm below by determining a function f(n) such that T(n), the worst-case complexity of the algorithm, is Θ(f(n)). input(n) x <- 0 for i <- 1 to n do for j <- i + 1 to n do for k <- 1 to j - i do x <- x +1arrow_forward
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