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hw8 - p.239#1 The function h(x = is not a good hash...

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(4.1) (5.1) p.239 #1. The function h(x) = is not a good hash function since it is not strongly collision free: h( ) = h( ) = h( ) = h( ) for all x and all integers k. p. 239 #2. a) In general, it is hard to find a square root modulo n of a given integer. We could use the CRT if we knew how to factor n into primes p and q, but in the RSA situation which n = pq implies, we do not in general know p and q. Nor, does the factorization yield easy square roots except in the case of p and q both congruent to 3 mod 4. Hence, h is preimage resistant with y = x 2 . However, for all x, we have h(x) = = = h(x+n) mod n, so h is not strongly collision free. p. 239 #3. The function h is easy to compute, but neither preimage resistant nor strongly collision free. If z is the string of 160 zeros and s is an arbitrary string of 160 bits, then we have that h(s4 z4 ... 4 z) = s, so h is not preimage resistant with y = s. Moreover, if we represent by 1 the strong of 160 consecutive ones, then , and ( both have the same hash, so the function is not strongly collision resistant. p. 239 #4. restart: product(1-i/12,i=1..3); 55 96 p. 239 #5. a) If x is nonnegative and less than 1, then in the given interval the numerator of the derivative is nonnegative, while the denominator is negative. Hence f ' is nonpositive. f:=ln(1-x)+x; diff(f,x); simplify(%); Similarly, if x is nonnegative and less than or equal to 1/2, then the numerator is nonpositive, while
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(5.2) (5.3) (5.4) the the denominator is negative, so the derivative g' is nonnegative. g:=ln(1-x)+x+x^2; diff(g,x); simplify(%); b) if Since , and f is non increasing, then in the given interval we have . It follows that . Similarly, since and g(x) is non decreasing, it follows that on the interval. Therefore . subs(x=0,f); simplify(%); subs(x=0,g); simplify(%); 0 0 c) Setting , and factoring the N out of the sum we get on the right -1/N*Sum(j,j=1..r-1); value(%); simplify(%); Similarly, setting on the left, and factoring the N out of the sum on the left, we get -1/N*Sum(j,j=1..r-1)-1/N^2*Sum(j^2,j=1..r-1); value(%); simplify(op(1,%))+simplify(op(2,%));
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(5.6) (5.5) (5.7) Since , if follows that , so ignoring those terms makes the rightmost of the two term on the left hand side even smaller, and if , so that , then the result follows from part b. 3*r^2-r; factor(%); d) Using the definition of and , c1:=sqrt(lambda/2)-1/3*(2*lambda)^(3/2); c2:=sqrt(lambda/2); we get the desired left hand side. (In a sum, op(1,%) will give us the first term, op(2,%) will give the second sum, etc. The Maple command collect simply collects terms.
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