CS556 HW 3 Solution- Fall 2012
4.3. The adversary queries for the message m1 |m2 and recieves the tag
< Fk (m1 ), Fk (Fk (m2 ) >. He then queries for the message Fk (m1 )|m2 and
receives the tag Fk (F
CS556 HW 4 Solution- Fall 2012
7.5) Observe that 3 Z as gcd(3, 100) = 1.
100
Further, (100) = 2.1.4.5 = 40. Thus
31000 = 340.25 = (340 )25 = 125 = 1
mod 100
As 31000 = 1 mod 100, the nal two digits of
CS556 HW 5 Solution- Fall 2012
9.3 Alice outputs k . Bob outputs
w t = (u r) t = (s t) r) t = (k r) t) r) t = k
An adversary who listens to the protocol has access to s, u and w. He can
thus compute
s
CS556 Midterm Exam
October 17, 2012
Professor Ming-Deh Huang
Duration: 80 minutes. Closed books and notes. Calculators allowed.
1. Consider the following variant of one-time pad with the message space
CS556 Midterm Exam
Spring 2010
Professor Ming-Deh Huang
Duration: 80 minutes
Closed books and notes. Calculators allowed. Each question carries 20 points.
In the following questions, whenever an encry
CSE 581 Introduction to Database Management Systems
Syllabus
Spring 2017 Draf
Instructor: Dusan Palider, MSCE ([email protected])
Course Scope:
The course will discuss the following topics: database de
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Lab4
Computer Engineering
JIAHAO PEI
Task1 Matrix Traverse
1. The original D1 miss rate:
2. The optimized program:
3. The D1 miss rate of the optimized program:
4. Explanation of my optimization:
For
CS556 Homework 2 solution
3.16 (b) (YES) Observe that G(s) = G (s|r), where r cfw_0, 1n .
Here | denotes concatenation. Assume that G is not pseuodorandom.
Thus, there exists a polynomial time disting
cs556 Homework 1 solution
1.3. Breaking the improved Vigenere cipher: As with the original Vigenere
cipher, once the length t of the key is known, the cipher can be broken
the same way a normal mono-a
CS556 Midterm Solution- Spring 2010
1.) The decryption function is Deck (c) := c k mod 2n .
Deck (c) = (m + k mod 2n ) k ) mod 2n = m + k k mod 2n =
m mod 2n .
Pr[C = c|M = m] = Pr[c = M + K
mod 2n |M