Completely random selection of possible test questions

All questions used in the past

1. a) (x,i,m) (Secret modifies Unclassified?--definitely wrong!)
      (x,j,b) (Both access must be equal, but compartments don't match)

   b) {(y,h,o),(v,g,m)}  (just the rest of the b-matrix)

2. (member x l) = (if (listp l) 
                       (if (equal x (car l))
                           t
                           (member x (cdr l)))
                       f)

    (dmem x y l) = (and (member x l)
	                (member y l))

3.  purge _u_(w) = { (u',c)* | sid(u') >= sid(u)}

4.  If the two files are in the same company data set, then they are
    automatically in the same COI, but they also may both be seen by
    the same analyst.

5.  Makes each level a virtual machine (no contact between security levels),
    since every action will be ruled by a greater than and less than,
    the levels will always have to be equal.

6.  Assume that plus is defined as follows:

	(plus a b) = (if (zerop a)
                         b
                         (add1 (plus (sub1 x) y)))

i) Prove that (plus x 0) = x

	Base:  Assume x = 0

	(plus 0 0) = 0  subst for x
                0  = 0  expand plus & simplify

	Induction:  Assume x = (add1 k)
                    IH  (plus k 0) = k

	(plus (add1 k) 0) = (add1 k)  subst for x
        (add1 (plus (sub1 (add1 k)) 0)) = (add1 k) expand plus & simplify
        (add1 (plus k 0)) = (add1 k)   use sub1-add1 axiom
        (add1 k) = (add1 k)  use IH

ii) Prove that (plus i (add1 j)) = (add1 (plus i j))

        Base:  assume i = 0
        
	(plus 0 (add1 j)) - (add1 (plus 0 j))  subst for i
	(add1 j) = (add1 j)  expand plus & simplify

	Induction:  Assume i = (add1 k)
		    IH (plus k (add1 j)) = (add1 (plus k j))

	(plus (add1 k) (add1 j)) = (add1 (plus (add1 k) j)) subst for i
        (add1 (plus (sub1 (add1 k)) (add1 j))) = 
                        (add1 (add1 (plus (sub1 (add1 k)) j))) 
					expand plus & simplify
        (add1 (plus k (add1 j))) = (add1 (add1 (plus k j))) sub1-add1 axiom
        (add1 (add1 (plus k j))) = (add1 (add1 (plus k j))) use IH

iii) Use these two facts to prove that (plus x y) = (plus y x)

	Base:  Assume x = 0

	(plus 0 y) = (plus y 0) subst for x
                0  =  (plus y 0) expand plus & simplify
	        0  =  0          use rule from part i)

	Induction:  Assume x = (add1 k)
		    IH  (plus k y) = (plus y k)
	
	(plus (add1 k) y) = (plus y (add1 k))
        (add1 (plus k y)) = (plus y (add1 k)) expand plus, use sub1-add1
        (add1 (plus k y)) = (add1 (plus y k)) use rule from part ii)
        (add1 (plus y k)) = (add1 (plus y k)) use IH

7.  The Kasiski method is used to determine the number of alphabets in
    a polyalphabetic substitution.  Repeating trigraphs are located,
    the distance between the measured, and the least common denominator
    (of the distances) computerd.  This should be the number of alphabets.

8.  Decrypt the following text:
   
	This problem was too easy for a test  (it was just ROT-13)