Question

1. (20 pts) To understand the value of recursion in a programming language: implement the binary search algorithm first as a recursive function and again using a conditional loop. Your program should create an array of characters holding the letters ‘A’ – ‘Z’ and find the index in the array where the letter ‘K’ is stored. You may use any programming language that supports recursion.

2. (5pts) Define syntax and semantics and give an example.

3. (5pts) Why is it important for a grammar to be unambiguous?

4. (5pts) Write an EBNF description for a C switch statement.

5. (5pts) Rewrite the BNF grammar below to give + precedence over * and force + to be right associative

<assign> à<id> = <expr>

<id> àA | B | C

<expr> à<expr> + <expr>

             | <expr> * <expr>

             | ( <expr> )

             | <id>

6. (6pts) Consider the following BNF, with start symbol <S>.

         <S> à[ <B>, <S> ]  | <B>

        <B> à<C>  |  ( <B> )

        <C>  à a  |  b  |  c

Which of the following are valid sentences in this language?   Give parse trees if possible.  

a)  (((b) ))                  

b) [ a, (b) ]   

c)  [ [c,c], [b,b] ]  

7. (5pts) What are attribute grammars used for?

8. (5pts) Compute the weakest precondition for b = (c + 10)/3 {b > 6}

9. (5pts) Computer the weakest precondition for the statement sequence:

a = 2b + 1;

b = a – 3 {b < 0}

Answer 1

Answer to the first question is provided below. Please ask other questions in a separate post.

Binary search algorithm:

# Iterative Binary Search
def binarySearch(arr, l, r, x):
while l <= r:
mid = int(l + (r - l)/2)
print(ord(arr[mid]),ord(x) )
if arr[mid] == x:
return mid
elif arr[mid] > x:
l = mid + 1
else:
r = mid - 1
return -1

arr = [ 'A', 'C', 'K', 'L', 'M' ]
x = 'K'
result = binarySearch(arr, 0, len(arr)-1, x)
if result != -1:
print("Element is present at index ",result)
else:
print("Element is not present in array")

# Recursive binary search.
def binarySearch (arr, l, r, x):
if r >= l:
mid = int(l + (r - l)/2)
if arr[mid] == x:
return mid
elif arr[mid] > x:
return binarySearch(arr, l, mid-1, x)
else:
return binarySearch(arr, mid+1, r, x)
else:
return -1

# Test array
arr = [ 'A', 'C', 'K', 'L', 'M' ]
x = 'K'

# Function call
result = binarySearch(arr, 0, len(arr)-1, x)
if result != -1:
print("Element is present at index ",result)
else:
print("Element is not present in array")

OUTPUT:

In the recursive search algorithm, the number of comparisons is one while in the iterative algorithm of binary search, the number of comparisons is 3. Thus, the recursive is faster than iterative one in worst cases. But in bast cases, like if the element is present at the first index then the iterative algorithm will be faster.