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 C.4, Problem 7E
Program Plan Intro
Toshow the given equation.
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Consider a Diffie-Hellman scheme with a common prime q = 17 and a primitive root α = 3.
a) If user A has a private key XA=4, what is A’s public key, YA?
b) A sends YA to B. If B has a private key XB=6, what is the shared secret key, K that B can calculate and share with A?
c) If B computes YB and sends it to A, what is the shared secret Key, K computed by A?
Consider a logistic regression classifier that implements the 2-input OR gate. At iteration t, the parameters are given by w0=0, w1=0, w2=0. Given binary input (x1,x2), output of logistic regression is given by 1/(1+exp(-w0-w1*x1-w2*x2)).
What will be value of the loss function at t?
What will be the values of w0, w1 and w2 at (t+1) with learning rate ɳ=1?
Many forms of cryptography are based on the following discrete log problem. Choose a cyclic group G (produced by one generator g) with finitely many elements. Alice and Bob want to set up a secret communication and need a secret key for that, which becomes a group element from G. They set up the following exchange (the so-called Diffie-Hellman protocol):
1. Alice chooses a secret group element a and sends unsecured g^a to Bob
2.Bob chooses a secret group element b and sends unsecured g^b to Alice
3. Alice calculates (g^b )^a and keeps this group element secret
4.Bob calculates (g^a )^b and keeps this group element secret
Since a cyclic group is abelian, Alice and Bob now share the secret key (g^b )^a=(g^a )^b. The blue group elements are publicly known, the red elements are secret. The discrete log problem presented to an eavesdropper is computing say a from the publicly known data G,g,g^a. The naming of the problem comes from the solution a= glog(ga ) which is generally difficult to…
Chapter C Solutions
Introduction to Algorithms
Ch. C.1 - Prob. 1ECh. C.1 - Prob. 2ECh. C.1 - Prob. 3ECh. C.1 - Prob. 4ECh. C.1 - Prob. 5ECh. C.1 - Prob. 6ECh. C.1 - Prob. 7ECh. C.1 - Prob. 8ECh. C.1 - Prob. 9ECh. C.1 - Prob. 10E
Ch. C.1 - Prob. 11ECh. C.1 - Prob. 12ECh. C.1 - Prob. 13ECh. C.1 - Prob. 14ECh. C.1 - Prob. 15ECh. C.2 - Prob. 1ECh. C.2 - Prob. 2ECh. C.2 - Prob. 3ECh. C.2 - Prob. 4ECh. C.2 - Prob. 5ECh. C.2 - Prob. 6ECh. C.2 - Prob. 7ECh. C.2 - Prob. 8ECh. C.2 - Prob. 9ECh. C.2 - Prob. 10ECh. C.3 - Prob. 1ECh. C.3 - Prob. 2ECh. C.3 - Prob. 3ECh. C.3 - Prob. 4ECh. C.3 - Prob. 5ECh. C.3 - Prob. 6ECh. C.3 - Prob. 7ECh. C.3 - Prob. 8ECh. C.3 - Prob. 9ECh. C.3 - Prob. 10ECh. C.4 - Prob. 1ECh. C.4 - Prob. 2ECh. C.4 - Prob. 3ECh. C.4 - Prob. 4ECh. C.4 - Prob. 5ECh. C.4 - Prob. 6ECh. C.4 - Prob. 7ECh. C.4 - Prob. 8ECh. C.4 - Prob. 9ECh. C.5 - Prob. 1ECh. C.5 - Prob. 2ECh. C.5 - Prob. 3ECh. C.5 - Prob. 4ECh. C.5 - Prob. 5ECh. C.5 - Prob. 6ECh. C.5 - Prob. 7ECh. C - Prob. 1P
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