浙江大学离散数学 期末考试题 英文班

(Bj) for the set of arrangements with row i (resp., column j) monochromatic. Then the
desired number is |A \ (A1 ∪ A2 ∪ A3 ∪ B1 ∪ B2 ∪ B3)|.
2
We use the PIE for the computation. The number of unrestricted arrangements is
Question 3 (15 marks)
Suppose a1, a2, . . . , a10 are integers satisfying 1 ≤ a1 ≤ a2 ≤ · · · ≤ a10 ≤ 40. Show that there exist distinct 3-subsets {i, j, k} and {r, s, t} of {1, 2, . . . , 10} such that
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Final Examination (A) 11193011 Discrete Mathematics (Prof. Honold)
Nov 20, 2014 8.00–10.00
a) Write (using a programming language or some form of pseudocode) a boolean function has zero subsum(a) with parameter a as above, which returns true if a has a zero subsum and false otherwise.
Final Examination (A) 11193011 Discrete Mathematics (Prof. Honold)
Nov 20, 2014 8.00–10.00
Question 1 (15 marks)
a) Is (p1 ↔ p2) ∨ p1 a tautology? b) Are (p1 → p2) → p3 and p1 → (p2 → p3) logically equivalent? c) Fermat’s Last Theorem (FLT) states that the equation xn + yn = zn has
assigments of truth values (in fact for exactly three).
4
c) An appropriate formula is
¬∃x ∃y ∃z ∃n ¬(xyz = 0) ∧ n ≥ 3 ∧ (xn + yn = zn) .
5
= 15
1
2 Write A for the set of all arrangements of the 9 balls (without any restriction) and Ai
All other intersections between the sets Ai, Bj involve at least one Ai and at least
one Bj. Since monochromatic rows and columns exclude each other, these intersections
Question 2 (15 marks)
Three black balls, three red balls and three yellow balls are arranged in a 3 × 3 matrix in such a way that there is no monochromatic row or column (“monochromatic” refers to balls of the same color). Balls of the same color are considered as indistinguishable. In how many ways can this be done?
Department of Information Science and Electronic Engineering Zhejiang University
Final Examination
11193011 Discrete Mathematics
Assoc. Prof. Dr. Thomas Honold M.Sc. Jingmei Ai Fall Semester, 2014
=
= 120.
3 3·2·1
= 15
2
5
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Final Examination (A) 11193011 Discrete Mathematics (Prof. Honold)
Nov 20, 2014 8.00–10.00
The 120 sums ai + aj + ak are integers in the range [3, 120]. There are 118 such integers.
insufficient, please continue on the blank sheets provided. • This is a CLOSED BOOK examination,
except that you may bring 1 sheet of A4 paper (hand-written only) and a ChineseEnglish dictionary (paper copy only) to the examination.
c) Does there exist a more efficient algorithm for this task? Briefly justify your answer.
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Final Examination (A) 11193011 Discrete Mathematics (Prof. Honold)
b) Determine the worst-case time complexity of your algorithm, using additions and comparisons of list elements as basic operations. Both the exact number of basic operations and an asymptotic growth estimate should be given.
ai + aj + ak = ar + as + at.
Question 4 (20 marks)
Let G be the simple graph whose vertices are the subsets S ⊆ {1, 2, 3, 4, 5, 6, 7} with |S| = 3, two vertices S and T being adjacent if and only if S ∩ T = ∅.
a) How many vertices and edges does G have? b) Is G regular? If applicable, determine its degree. c) Is G connected? d) Is G bipartite?
Question 5 (15 marks)
Nov 20, 2014 8.00–10.00
Solutions
1 a) No. 1 This can be proved either by a truth table or, e.g., as follows: If p1 is false
and p2 is true then p1 ↔ p2 is false. Hence the disjunction (p1 ↔ p2) ∨ p1 is false as
Question 6 (20 marks)
A list a = (a0, a1, . . . , an−1) of integers is said to have a (contiguous) zero subsum if there exist 0 ≤ i ≤ j ≤ n − 1 such that ai + ai+1 + · · · + aj = 0.
well.
4
b) No. 1 For the proof one can use again a truth table or, e.g., the following shorter
argument: p1 → (p2 → p3) is false for exactly one assigment of truth values to p1, p2, p3, viz. p1 = p2 = T, p3 = F. On the other hand, (p1 → p2) → p3 is false whenever p1 = p3 = F (since p1 → p2 = T in this case), and hence for at least two
|A| =
9 3,3,3
=
9 3
6 3
= 84 · 20 = 1680.
3
Further we have
6
|Ai| = |Bj| = 3 3 = 60,
2
|Ai,j| = |Bi,j| = 3! = 6,
2
|A1,2,3| = |B1,2,3| = 6,
3
since 2 monochromatic rows force the 3rd row to be monochromatic as well.
are all empty.
1
It follows that the desired number is
1680 − 6 · 60 + 6 · 6 − 2 · 6 = 1344.
2
3 The number of distinct 3-subsets of {1, 2, . . . , 10} is
10 10 · 9 · 8
Name: Matric. No.:
Question
123456
பைடு நூலகம்
Marks Obtained Maximal Marks 15 15 15 20 15 20 100
Time allowed: 120 minutes
Instructions to candidates: • This examination paper contains six (6) questions. • Please answer every question and subquestion, and justify your answers. • For your answers please use the space provided after each question. If this space is
no integer solution with xyz = 0 and n ≥ 3. Express FLT as a formula of predicate calculus with domain Z. Note: You may use the basic operations of ordinary arithmetic, including powers ‘xn’ and the comparison operators ’=’, ’≤’. No further abbreviations are allowed. The domain (universe of discourse) must be the set of all integers.
a) Find a particular solution of the recurrence relation
an = 3an−1 − 4an−3 + 4n.
Hint: The recurrence relation has a solution of the form an = cn + d, where c, d are constants. b) Determine the general solution of the recurrence relation in a).