代写作业:数学代写COMP 9020 – Assignment 3北美math代写刷题
Due: 14th of October 2018 at 11:59pm
COMP 9020 – Assignment 3
Note: In your assignment, how you arrived at your solution is as important (if not more so) than the solution itself and will be assessed accordingly. There may be more than one way to find a solution, and your approach should contain enough detail to justify its correctness. Lecture content can be assumed to be common knowledge.
1. Let (T, ∧, ∨,j , 0, 1) be a Boolean Algebra.
Define ∗ : T × T → T and ◦ : T × T → T as follows:
x ∗ y := (x ∨ y)j x ◦ y := (x ∧ y)j
(a) Show, using the laws of Boolean Algebra, how to define x ∗ y using only x, y, ◦ and parentheses.
(b) Show, using the laws of Boolean Algebra, how to define x ◦ y using only x, y, ∗ and parentheses.
Define R ⊆ T × T as follows:
(x, y) ∈ R if, and only if, (x ∧ y) ∨ (xj ∧ yj) = 1
(c) Show, using the laws of Boolean Algebra, that R is an equivalence relation. Hint: You may want to use the observation that if A = B = 1 then A ∧ B ∧ C = A ∧ B implies C = 1 (why?)
2. Let P F denote the set of well-formed propositional formulas made up of propositional variables, , , and the connectives , , and . Recall from Quiz 7 the definitions of dual and flip as functions from P F to P F :
• dual(p) = p • flip(p) = ¬p
• dual(T) = ⊥; dual(⊥) = T • flip(T) = T; flip(⊥) = ⊥
• dual(¬ϕ) = ¬dual(ϕ) • flip(¬ϕ) = ¬flip(ϕ)
• dual(ϕ ∧ ψ) = dual(ϕ) ∨ dual(ψ) • flip(ϕ ∧ ψ) = flip(ϕ) ∧ flip(ψ)
• dual(ϕ ∨ ψ) = dual(ϕ) ∧ dual(ψ) • flip(ϕ ∨ ψ) = flip(ϕ) ∨ flip(ψ)
(a) For the formula ϕ = p ∨ (q ∧ ¬r):
(i) What is dual(ϕ)?
(ii) What is flip(ϕ)?
(b) Prove that for all ϕ ∈ P F : flip(ϕ) is logically equivalent to ¬dual(ϕ).
3. Let P (n) be the proposition that: for all k, with 1 ≤ k ≤ n,
.nΣ = .n − 1Σ + .n − 1Σ.
k k − 1 k
(a) Prove that P (n) holds for all n 1. (Note: it is possible to do this without using induction)
We can compute n from the formula given in lectures, however this can often require computing unnecessarily large numbers. For example,
100 = 253338471349988640 which can be expressed as a 64-bit integer,
but 100! is larger than a 512-bit integer. We can, however, make use of the equation above to compute n more efficiently. Here are two algorithms
for doing this:
chooseRec(n, k) : chooseIter(n, k) :
if k = 0 or k = n : return 1 Let C be a n n array else : for m = 1 to n :
x := chooseRec(n 1, k 1) C[m]=C[m][m]=1
y := chooseRec(n 1, k) for j = 1 to m 1 :
return x + y C[m][j]=C[m 1][j 1]
+ C[m 1][j]
Let trec(n, k) be the running time for chooseRec(n, k), and let titer(n) be the running time for chooseIter(n, k). Let Trec(n) = max0≤k≤n trec(n, k) and Titer(n) = max0≤k≤n titer(n, k) (so Trec(n) ≥ trec(n, k) for all k, and likewise for Titer(n)).
(b) Give an asymptotic upper bound for Trec(n). Justify your answer.
(c) Give an asymptotic upper bound for Titer(n). Justify your answer.
Advice on how to do the assignment
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