CMPSCI 190DM: A Mathematical Foundation for Informatics

Solutions to First Midterm Exam

David Mix Barrington

28 September 2015

Directions:

Answer the problems on the exam pages.
There are four problems for 100 total points. Actual scale was A = 90, C = 60.
If you need extra space use the back of a page.
No books, notes, calculators, or collaboration.

  Q1: 20 points
  Q2: 20 points
  Q3: 25 points
  Q4: 35 points
Total: 100 points

Question text is in black, solutions in blue.

Question 1 (20): Briefly identify the following terms or concepts (4 points each):

(a) the summation of a number sequence
The summation of a number sequence a₁, a₂, a₃,... is the sequence s₁, s₂, s₃,... where each s_n is the sum of all terms a_i for i from 1 through n.
(b) the inverse of an implication
The inverse of an implication "p → q" is the implication "¬p → ¬q". It is not equivalent to the original implication.
(c) the Double Negative rule in propositional logic
This rule says that for any proposition p, "¬¬p" is equivalent to p.
(d) an existential quantifier
Written "∃x" for some variable x, it means that at least one value of x exists making the following statement true.
(e) for a compound proposition to be a tautology
A compound proposition is a tautology if every possible setting of its atomic variables makes the compound proposition true.

Question 2 (20): For each of these two number sequences, give the next three numbers in the sequence, find a recursive formula, and find a closed formula. Don't forget that the recursive formula must specify the first element. (10 points each)

a₁ = -7, a₂ = -3, a₃ = 1, a₄ = 5, a₅ = 9
The recursive form is a₁ = -7, a_k = a_k-1 + 4.
The next three terms are a₆ = 13, a₇ = 17, and a₈ = 21.
The closed form is a_n = 4n - 11. We can find this because we can see that the function is a linear one with slope 4, so that we only need to find the y-intercept.
b₁ = -1, b₂ = 0, b₃ = 3, b₄ = 12, b₅ = 39
There are at least two possible recursive forms, both with first term b₁ = -1. The one I intended was b_k = 3b_k-1 + 3. But an equally good one is b_k = b_k-1 + 3^k-1.
The next three terms are b₆ = 120, b₇ = 363, and b₈ = 1092. (These may be calculated from either of the given recursive rules.)
To find the closed form, we use the fact that the tripling on every step is going to create a term of the form 3ⁿ. If we look for part of the sequence resembling (1, 3, 9, 27, 81, 243, 729,...) we can see that some of the numbers in our sequence are around half of these numbers. From this we can get b_n = (1/2)(3^n-1 - 3), so that b₁ = (1/2)(3⁰ - 3) = -1 and b₂ = (1/2)(3¹ - 3) = 0.

Question 3 (25): Translate the following statements as indicated. The dogs Cardie and Duncan are denoted symbolically by c and d respectively. The proposition p means "Cardie eats paper", and q means "Duncan barks too much". If x and y are dogs, the predicate S(x, y) means "dog x is sillier than dog y". All variables in quantifiers are of type "dog". (5 points each)

(a) (to symbols) It is not the case that if Cardie eats paper, then Duncan barks too much.
¬(p → q)
(b) (to English) ∀x: S(d, x) ∨ ¬S(x, c)
Every dog is either sillier than Duncan or less silly than Cardie. (Many people said instead "Either every dog is siller than Duncan, or every dog is less silly than Cardie", which is not the same thing -- that would be the translation of (∀x: S(d, x)) ∨ (∀y: S(y, c)).)
(c) (to symbols) If Cardie eats paper and Duncan barks too much, then both are sillier than all other dogs.
(p ∧ q) → ∀x: ((x ≠ c) ∧ (x ≠ d)) → (S(c, x) ∧ S(d, x)) (We need the reference to x not being equal to c or d in order to capture the phrase "all other dogs".)
(d) (to English) S(d, c) ∨ (q ∧ S(c, d) ∧ ¬p)
Either Duncan is sillier than Cardie, or Duncan barks too much, Cardie is sillier than Duncan, and Cardie does not eat paper.
(e) (to symbols) There is a dog that is sillier than every dog that is not sillier than either Cardie or Duncan.
∃x:∀y:(¬S(y, c) ∧ ¬S(y, d)) → S(x, y)

Question 4 (35): Establish the following facts with truth tables.

(a, 15) The compound proposition "((p → ¬q) ∧ q) → ¬p)" is a tautology.

   ((p --> not q) and q) --> not p
-----------------------------------
     0  1   1  0   0  0   1   1  0
     0  1   0  1   1  1   1   1  0
     1  1   1  0   0  0   1   0  1
     1  0   0  1   0  1   1   0  1

(b, 20) The compound propositions "(p ∧ ¬q ∧ ¬r) ∨ (q ∧ r)" and "(q ↔ r) ∧ (p ∨ q)" are logically equivalent. Remember that "↔" is the symbol for equivalence or "if and only if".

((p and not q and not r) or (q and r)) <--> ((q <--> r) and (p or q))
----------------------------------------------------------------------
  0  0   1  0  0   1  0  0   0  0  0     1    0   1  0   0   0  0 0
  0  0   1  0  0   0  1  0   0  0  1     1    0   0  1   0   0  0 0
  0  0   0  1  0   1  0  0   1  0  0     1    1   0  0   0   0  1 1
  0  0   0  1  0   0  1  1   1  1  1     1    1   1  1   1   0  1 1
  1  1   1  0  1   1  0  1   0  0  0     1    0   1  0   1   1  1 0
  1  1   1  0  0   0  1  0   0  0  1     1    0   0  1   0   1  1 0
  1  0   0  1  0   1  0  0   1  0  0     1    1   0  0   0   1  1 1
  1  0   0  1  0   0  1  1   1  1  1     1    1   1  1   1   1  1 1

Last modified 13 October 2015