LAST NAME: 

FIRST NAME: 


Note: * is times, ^ is exponentiation, + is plus, - is minus, / is division, 
      all log's are base 2
Note: problems are either 1 point or 2 points each, as labeled.


(1) Let x = 12.  List the positive integers less than x which have 
    multiplicative inverses mod x.  1 point.

Answer: all numbers x between 2 and 12 with gcd(x,12)=1.  So 5,7,11


(2) Let b = 11, e = 10, n = 30.  Compute (b^e) mod n.  1 point.

Answer: use fast-mod-exp by repeated squaring.  check answer on wolfram alpha


(3) Let a = 5, b = 8.  Compute a^(-1) mod b.  1 point.

Answer: use extended euclidean algorithm, or guess and check.  Make sure you
	get an answer y such that y*a = 1 mod b.  Answer is y=5


(4) Let n = p*q with p and q prime.  What are the requirements for e and d, 
    the encryption exponent and decryption exponent?  1 point.

Answer: Let phi(n)=(p-1)*(q-1).  Then e is such that 1<e<phi(n) and 
	gcd(e,phi(n))=1.  Note that e being prime is not the same as 
	gcd(e,phi(n))=1.  Then d = e^(-1) mod phi(n).


(5) Let n, e, d be 1024 bit numbers in RSA.  What is the approximate 
    running time of encrypting a message, in terms of arithmetic operations?  1/2 point.

Answer: It is the running time of mod-exp on 1024 bit integers.  This is 
	O(1024) many arithmetic operations.

Note: remember to pay attention to bit length!  


    What is the running time of decrypting a message if you already know n and d?  1/2 point.

Answer: same, it's the time of mod-exp.


(6) Given two numbers x and y that are 1024 bit integers.  Give a formula for 
    how these numbers are multiplied using Karatsuba multiplication.  Just give
    the formula for the first set of recursive calls.  You can let x1 be the 
    first 512 bits of x and x2 be the second 512 bits of x, and similarly y1 
    and y2 for y.  1 point.

Answer: see wikipedia...


(7) Either prove the following or give a counter example.  1 point.
    For positive integers a, b with a > b, (a*a) mod b = ((a mod b) * (a mod b)) mod b

Note: this is true in general.  giving an example is not sufficient to show
      it is true for every a and b.  That is not a proof that it always
      works!!!!!!!!!!!!!!!!!!  See how many ! I used.  This is important!!!


(8) Either prove the following or give a counter example.  1 point.
    For positive integers a, b with a > b, (a*a) mod b = (a*b) mod a

Answer: this does not work in general.  To show it doesn't always work, you 
	just need to give an example where it doesn't work.  Pick 
	a=2, b=3.  Then a*a mod b = 1.  And a*b mod a = 0.

Note: if you gave an example where they're the same, you got half credit.


(9) Put the following in order from fastest to slowest, in terms of big-O running time, 
    where x and y are positive integers having bit-length n.  2 points total
    (a) x*y using Karatsuba multiplication
    (b) x+y, addition
    (c) x*y using grade-school multiplication
    (d) List all binary strings of length x
    (e) Fermat prime test of x, using grade-school multiplication
    (f) gcd(x,y), using grade-school multiplication for multiplication
    (g) Prime testing x using trial division

Correct order is b,a,c,e,f,g,d.  e and f are about the same, so you could have
put those in either order.  I counted the largest subset that you had in 
the right order.  You got full credit if you only had one misplaced, 
3/4 credit if 2 or 3 misplaced, 1/2 credit if 4 misplaced, 1/4 credit 
if all but 1 or 2 misplaced.

Note: need to pay attention to running time as a function of bit length!!


(10) Give C code for the following functions.  2 points each

// return the gcd of a and b, using Euclid's algorithm
int gcd(int a, b) {

Answer: see wikipedia, etc.  

Notes: mod doesn't work in C, need %.  if doing recursive version, need to 
       return gcd(b,a%b), not just call gcd(b,a%b).  also need to make sure
       to return a if b==0.
Notes: -1/2 if base case was wrong, -1 if logic is wrong.

}


// return b^e mod n, using fast modexp by repeated squaring
int modExp(int b, int e, int n) {

Answer: see wikipedia, etc.  Similar grading as in last question.

Notes: you don't need to compute the binary representation first.  Just
       look at the bits of e as you shift it to the right.  Computing 
       the binary representation first is wasteful, essentially doubling
       the running time; and the way most of you did it limits it to work
       only for 64 bit integers.

}


(11) Evaluate the following.  1/2 point each.

     1 + 3 + 9 + 27 + ... 3^10
     
     Answer: (3^11-1)/(3-1)


     5 + 10 + 15 + ... + 5*100

     Answer: 5*(101*100/2)


     Number of bits in one gigabyte

     Answer: 8*2^30


     (2^30 / 2^10)^4
     
     Answer: 2^80


     log(16^3)

     Answer: 12*log(2) = 12, since we're assuming base 2 log.