Every function from int to int can be implemented as a Java method or C function.
The poset (N, |) of non-negative integers N under the dividability relation | is a lattice but is not bounded since it has no greatest element.
Which of the following propositional formulas is not logical equivalent to the implication p → q?
(~p) ∨ q
(~q) → (~p)
~(p ∧ ~q)
p ↔ (~q)
Let G = (N, T, S, P) be a context free grammar, where N = {S, A, B} is the set of nonterminal symbols, T = {0, 1} is the set of terminal symbols, S is the start symbol and the set of productions P is given as follows: {S → 0A, S → 1A, A → 0B, B → 1, B → 1A}. Then which of the followings is the language generated by G?
{01}{0, 1}*{01}
{0, 1}{01}*{01}
{0,1}{01}*{0,1}
{01}{0, 1}*
Let R = {(1, 1), (2, 1), (3, 2), (2, 3), (4, 3)}. Then which of the following statements is not correct?
(2, 2) ∈ R2
(4, 3) ∈ R3
(2, 4) ∈ R*
(4, 1) ∈ R+
In the expansion of (w + x + y + z)20, the coefficient of the term w2x4y6z8 is _______ and there are totally ________ different terms.
The sum of the degrees of all vertices of a tree with n vertices is ______.
What is the least positive integer x satisfying the following system of congruence equations: x ≡ 5 (mod 7), x ≡ 4 (mod 9), and x ≡ 3 (mod 13)? ______
Let Kn denote a complete simple graph with n > 0 vertices. Then what is the length of a shortest circuit containing all edges of the graph K2n? _____
If f is an increasing function satisfying the recurrence relation: f(n) = 9f(n/3) + 2n2, then the asymptotic order of f(n) = Θ( ______ ).
A deterministic finite automata (DFA) M is a tuple (Q, Σ, δ, s, F), where Q is the set of states, Σ is the set of input symbols, s ∈ Q is the initial state, F ⊆ Q is the set of final states and δ: Q × Σ → Q is a state transition function. Suppose now |Q| = n and |Σ| = m. Then there are ______ different DFAs in which Q is the set of states and Σ the set of input symbols.
Let Σ = {0, 1} and A ⊆ Σ* be a set of bit strings defined inductively by the following rules:
Basis case: ε ∈ A, where ε is the empty string.
Inductive case: If x ∈ A and y ∈ A, then so are 0x1, 1x0 and xy.
Moreover, for each bit string z, let f0(z) and f1(z) denote the number of 0s and 1s
occurring in z respectively. So, for example, if z = '1001', then z ∈ A and f0(z) = f1(z) = 2.
According to the above definition explain briefly why the string '1001" belongs to A.
Let Σ = {0, 1} and A ⊆ Σ* be a set of bit strings defined inductively by the following rules:
Basis case: ε ∈ A, where ε is the empty string.
Inductive case: If x ∈ A and y ∈ A, then so are 0x1, 1x0 and xy.
Moreover, for each bit string z, let f0(z) and f1(z) denote the number of 0s and 1s
occurring in z respectively. So, for example, if z = '1001', then z ∈ A and f0(z) = f1(z) = 2.
Prove by structural induction that for all bit strings z, if z ∈ A then f0(z) = f1(z).
Let Σ = {0, 1} and A ⊆ Σ* be a set of bit strings defined inductively by the following rules:
Basis case: ε ∈ A, where ε is the empty string.
Inductive case: If x ∈ A and y ∈ A, then so are 0x1, 1x0 and xy.
Moreover, for each bit string z, let f0(z) and f1(z) denote the number of 0s and 1s
occurring in z respectively. So, for example, if z = '1001', then z ∈ A and f0(z) = f1(z) = 2.
On the other hand, show that if z is a bit string such that f0(z) = f1(z), then z ∈ A.
Let G = (V, E) be a directed multigraph such that E ≠ ∅ and for all vertices x ∈ V, in-degree(x) = out-degree(x). Show that there exists a simple circuit of length > 0 in G.
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