3. (10) Convert the following EBNF into BNF: A := CB [x] B ::= C (xly|z) D C:= y {y D} z D:= x | y| null
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- Not handwritten convert the following EBNF to BNF S --> A[bA] A --> a{bc} B --> a{bB}|aConvert the following EBNF to BNF: <S> -> <Y> { <Z> } <X> <X> -> x ( + | * | / ) x <Y> -> y { y } <Z> -> <Y> [ z ] w where {, }, (, ), |, [, ] are EBNF metasymbolsterminals are: w, x, y, z, +, *, /nonterminals are: <S>, <X>, <Y>, <Z>start symbol is: <S>Let L₁ = {aa, bb}, L₂ = {A, a, aa}, L3 = {x | x = {a,b}* ^ |x| ≤ 3} = {A, a, b, aa, ab, ba, bb, aaa, aab, aba, abb, baa, bab, bba, bbb}. Evaluate: (L₁n L3) · L2 = •
- Data Abstraction and Problem Solving with C++ 6th ed page 303 Consider the following C++ function f, which calls the function swap. Assume that swap exists and simply swaps the contents of its two arguments. Do not be concerned with f’s purpose. void f(int theArray[], int n) { for (int j = 0; j < n; ++j) { int i = 0; while (i <= j) { if (theArray[i] < theArray[j]) swap(theArray[i], theArray[j]); i++; } // end while } // end for }// end What is the big-O for the number of comparisons performed in f?4. A={3,4}, B={4,5}. An B= AUB= A-B= A OB= AXB=Homework Explain the ANN implementation of Logical XOR.
- Homework -4 1- Implement the following function by : Type 1 MUX ( D partitioned variable) Type 2 MUX (C,D) are partitioned variables • Type 3 MUX (B,C,D) are partitioned variables F(A,B,C,D)={m(4,5,6,7,10,14) 2- Implement the following function by: Type 1 MUX (D) partitioned variable • Type 1 MUX (C) partitioned variable (8X1) MUX's • (4X1) MUX's F(A,B,C,D)=_m(0,1,5,6,8,9,12,13)+d(3,11)a) Draw the RE of the following DFA defined over Σ={a, b}* In the Fortran language as follows A (the matrix is defined (3,4 .) ? Integer: a(3,4) Integer, dimension (3,4): a Real, Dimension (3,4)::a real::a(3,4)