CMPSCI 190DM: A Mathematical Foundation for Informatics

Solutions for Practice Final Exam

David Mix Barrington

13 December 2015

Directions:

Answer the problems on the exam pages.
There are six problems for 125 total points. Scale will be determined after the exam but a good guess is A = 105, C = 70.
If you need extra space use the back of a page.
No books, notes, calculators, or collaboration.

  Q1: 15 points
  Q2: 15 points
  Q3: 25 points
  Q4: 30 points
  Q5: 15 points
  Q6: 25 points
Total: 100 points

Question text is in black, solutions in blue.

Question 1 (15): Briefly explain the difference between the terms or concepts in each of the following pairs (3 points each):

(a) the inclusive or and exclusive or operations
Both are binary operators on propositions. Both are true if one argument is true and the other false, but the inclusive or is also true if both are true.
(b) a counterexample and a contradiction
A counterexample is a value of x making some statement P(x) false -- its existence proves that the statement ∀x:P(x) is false. A contradiction is a statement that is false for all values of its variables -- we often prove a statement to be false by showing that it implies a contradiction.
(c) a one-to-one function and an onto function
A one-to-one-function maps each input to a separate output. An onto function is one where every possible output value is mapped to by some input.
(d) the Product Rule and the Sum Rule for counting
Suppose A and B are finite sets with a and b elements respectively. The Product Rule says that the size of A × B is ab, that is, that there are exactly ab pairs (x, y) such that x is in A and y is in B. The Sum Rule says that if A and B are disjoint, the size of their union is a + b. (The Sum Rule With Overlap says that the size of the union is a + b - c, where c is the size of the intersection.
(e) a directed graph and an undirected graph
Both have nodes and edges. An edge in an undirected graph is between two nodes, with no direction specified. An edge in a directed graph goes from some node to some node (possibly the same node).

Question 2 (15): Determine whether each of these five statements is true or false. No explanation is needed or wanted, and there is no penalty for guessing (3 points each).

(a) If p and q are any propositions, and q and (p → ¬q) are both true, then p must be false.
TRUE. The contrapositive of "p → ¬q" is "q → ¬p", and this together with q implies ¬p by Modus Ponens.
(b) If I prove P(1), and I prove that P(n) implies P(n+2) for any n, then P(n) must be true for all n.
FALSE. These premises prove P(n) only for odd n. If P(n) were "n is odd", we could prove the two premises but P(n) would still be false for all even n.
(c) There are exactly 2³ = 8 ordered lists of length 2 from the set {a, b, c}.
FALSE. The correct number is 3² = 9.
(d) If I roll two six-sided dice, the probability that the sum of the numbers is in the set {7, 11} is exactly 2/11.
FALSE. Six of the possible 36 throws of 2D6 yield a sum of 7 and two yield a sum of 11, and thus the probability is 8/36 = 2/9. There are 11 possible sums from 2D6, but they are not all equally likely and thus that 11 cannot be used as the denominator in the probability formula.
(e) If G is a directed graph with ten nodes and five edges, there must exist nodes x and y in G such that there is no path in G from x to y.
TRUE. A connected graph with ten nodes must have at least nine edges.

Question 3 (25): Vandals have sabotaged the traffic light in my neighborhood so that it now behaves randomly according to a Markov process. During every ten-second interval it is either red, yellow, or green. After an interval in which it is red, it is red in the following interval 90% of the time and green 10% of the time. After it is yellow, it is red 50% of the time and green 50% of the time. After it is green, it is green 90% of the time and yellow 10% of the time.

(a, 5) Write down the 3 by 3 matrix T of transition probabilities for this Markov process. Use the row order red-yellow-green.
```
0.9   0.0   0.1
0.5   0.0   0.5
0.0   0.1   0.9
```

(b, 5) Compute the matrix T².

0.81+0.00+0.00   0.00+0.00+0.01  0.09+0.00+0.09
0.45+0.00+0.00   0.00+0.00+0.05  0.05+0.00+0.45
0.00+0.05+0.00   0.00+0.00+0.09  0.05+0.05+0.81

which evaluates to

0.81  0.01  0.18
0.45  0.05  0.50
0.05  0.09  0.86

(c, 5) If the light is now yellow, what is the probability that it is green 20 seconds (two time steps) later? Remember that you do not have to evaluate an arithmetic expression.
This is the Y-G entry of the matrix T², which is 0.50. There is an 0.05 probability of its going Y-R-G and an 0.45 probability of its going Y-G-G.
(d, 10) If the light is now red, what is the expected number of time steps that it will remain red before changing?

There is a 0.9 chance that it will remain red each turn. If X is the desired expected value, we can compute X in terms of itself by looking at the two possible events. If it remains red on the first turn, the expected number of red turns is 1 + X. If it changes on the first term, the number of red turns is 0. So we have that X = (0.9)(1 + X) + (0.1)(0). This solves to X = 0.9 + 0.9X, 0.1X = 0.9, and thus X = 9.

Question 4 (30): Let G be a directed graph with node set {A, B, C} and edge set {(A, B), (A, C), (B, A), (C, A)}.

(a, 5) Write down the adjacency matrix M for G, using the row order A-B-C.
```
0  1  1
1  0  0
1  0  0
```

(b, 5) Compute the matrices M² and M⁴.

	  0+1+1 0+0+0 0+0+0           2  0  0     
    M^2	= 0+0+0 1+0+0 1+0+0    =      0  1  1
          0+0+0 1+0+0 1+0+0           0  1  1

          4+0+0 0+0+0 0+0+0           4  0  0
    M^4 = 0+0+0 0+1+1 0+1+1    =      0  2  2
          0+0+0 0+1+1 0+1+1           0  2  2

(c, 5) How many four-step paths are there in G from node B to node C?
There are two, as we can read off of the B-C entry of the matrix M⁴. The paths are B-A-B-A-C and B-A-C-A-C.
(d, 15) Prove, by induction on all positive numbers n, that there are exactly 2ⁿ paths of length 2n in G from node A to itself.
Our base case is n = 1. From the A-A entry of the matrix M² above, or by inspection, we can see that there are two two-step paths from A to A, namely A-B-A and A-C-A. This equals 2¹ and so the n = 1 case is proved.
Now for the inductive hypothesis assume that there are exactly 2ⁿ paths of length 2n from A to A. Our inductive goal is to prove that there are 2ⁿ⁺¹ paths of length 2(n+1) from A to A. Let's look at any such path, starting from A. Its first 2n edges form a path of length 2n. In fact this path must go to A, but even if we don't know that we don't have to consider paths that end at B or C after 2n steps, since they could not reach A in two more steps. We know that there are 2ⁿ paths of length 2n from A to A. Each can be extended to a path of length 2n+2 in exactly two ways, by going A-B-A or A-C-A with the last two steps. So the number of paths of length 2n+2 is 2 times 2ⁿ which is 2ⁿ⁺¹.

Question 5 (15): You are dealt three cards from a standard 52-card deck.

(a, 5) How many different possible three-card hands are there?
C(52, 3) = 52*51*50/1*2*3 = 26*17*50 = 17*1300 = 22100.
(b, 5) How many of these hands have three cards of the same suit?
There are four choices of suit and C(13, 3) = 13*12*11/1*2*3 = 13*2*11 = 286 choices of which three cards of that suit. The total number of such hands is thus 4*286 = 1144.
(c, 5) How many of these hands have three cards of three different ranks?
We choose the three ranks in C(13, 3) = 286 ways. Then we pick the suit of each of the three cards independently. The total number of choices is 286*4³ = 18304.

Question 6 (25): Let c be the proposition "Cardie is hungry", d be the proposition "Duncan is asleep", and q be the proposition "both dogs are quiet".

(a, 5) Translate the statement "If both dogs are quiet, then Duncan must be asleep and Cardie is not hungry" into logical symbols.
q → (d ∧ ¬c)
(b, 5) Translate the statement "¬(¬q ∧ ¬c ∧ d)" into English.
It is not the case that one of the dogs is not quiet, Cardie is not hungry, and Duncan is asleep.
(c, 15) With a truth table or otherwise, show that the statements of parts (a) and (b) together are equivalent to the statement "q ↔ ¬(d → c)". This said "q ↔ ¬(c → d)", which is wrong, until 13 December 2015.
Symbolically, we could use the DeMorgan rule to change (b)'s ¬(¬q ∧ ¬c ∧d) to (q ∨ c ∨ ¬d), then use the Definition of Implication rule to change this phrase to (¬c ∧ d) → q. Since this is the converse of the statement of (a), the desired "if and only if" statement follows.
Let's look at the truth table for (a) and (b):
```
  [[(q -> (d and not c)] and [not(not q and not c and d)]]

     0  1  0  0   1  0    1    1   1  0  1   1  0  0  0
     0  1  0  0   0  1    1    1   1  0  0   0  1  0  0
     0  1  1  1   1  0    0    0   1  0  1   1  0  1  1
     0  1  1  0   0  1    1    1   1  0  0   0  1  0  1
     1  0  0  0   1  0    0    1   0  1  0   1  0  0  0
     1  0  0  0   0  1    0    1   0  1  0   0  1  0  0
     1  1  1  1   1  0    1    1   0  1  0   1  0  0  1
     1  0  1  0   0  1    0    1   0  1  0   0  1  0  1
	      
```
Then we look at the truth table for the new statement, making sure that we use the same order q-d-c for the variables.
```
     q <--> not (d --> c)
     0   1   0   0  1  0
     0   1   0   0  1  1
     0   0   1   1  0  0
     0   1   0   1  1  1
     1   0   0   0  1  0
     1   0   0   0  1  1
     1   1   1   1  0  0
     1   0   0   1  1  1
```
Since the two principal columns are the same (the central "and" in the first truth table and the "<-->" in the second), the two statements are logically equivalent.

Last modified 13 December 2015