What will be the output of the following C program?
void count (int n) {
static int d=1;
printf ("%d",n);
printf ("%d",d);
d++;
if (n>1) count (n1);
printf ("%d",d);
}
void main(){
count (3);
}
A language with string manipulation facilities uses the following operations
head(s): first character of a string
tail(s): all but exclude the first character of a string
concat(s1, s2): s1s2
For the string "acbc" what will be the output of
concat(head(s), head(tail(tail(s))))
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Consider the following C function:
int f(int n)
{
static int i = 1;
if(n >= 5) return n;
n = n+i;
i++;
return f(n);
}
The value returned by f(1) is
double foo(int n)
{
int i;
double sum;
if(n == 0)
{
return 1.0;
}
else
{
sum = 0.0;
for(i = 0; i < n; i++)
{
sum += foo(i);
}
return sum;
}
}
The space complexity of the above code is?
double foo(int n)
{
int i;
double sum;
if(n == 0)
{return 1.0;
}
else
{
sum = 0.0;
for(i = 0; i < n; i++)
{
sum += foo(i);
}
return sum;
}
}
Suppose we modify the above function foo() and stores the value of foo(i) 0 <= i < n, as and when they are computed. With this modification the time complexity for function foo() is significantly reduced. The space complexity of the modified function would be:
int f(int n)
{
static int r = 0;
if (n <= 0) return1;
if (n > 3)
{ r = n;
return f(n2) + 2;
}
return f(n1) + r;
}
What is the value of f(5)?
The function f is defined as follows:
int f (int n) {
if (n <= 1) return 1;
else if (n % 2 == 0) return f(n/2);
else return f(3n  1);
}
Assuming that arbitrarily large integers can be passed as a parameter to the function, consider the following statements.
i. The function f terminates for finitely many different values of n ≥ 1.
ii. The function f terminates for infinitely many different values of n ≥ 1.
iii. The function f does not terminate for finitely many different values of n ≥ 1.
iv. The function f does not terminate for infinitely many different values of n ≥ 1.
Which one of the following options is true of the above?
A. i and iii
B. i and iv
C. ii and iii
D. ii and iv
Consider the following two functions.
void fun1(int n) {
if(n == 0) return;
printf("%d", n);
fun2(n  2);
printf("%d", n);
}
void fun2(int n) {
if(n == 0) return;
printf("%d", n);
fun1(++n);
printf("%d", n);
}
The output printed when fun1 (5) is called is
Consider the following computation rules. Paralleloutermost rule: Replace all the outermost occurrences of F (i.e., all occurrences of F which do not occur as arguments of other F's) simultaneously. Parallel  innermost rule: Replace all the innermost occurrences of F (i.e.,all occurrences of F with all arguments free of F's) simultaneously. Now consider the evaluations of the recursive program over the integers.
F(x, y) <== if x = 0 then 0 else
[ F(x + 1, F(x, y)) * F(x  1, F(x, y))]
where the multiplication functions * is extended as follows:
0 * w & w * 0 are 0
a * w & w * a are w (for any nonzero integer a)
w * w is w
We say that F(x, y) = w when the evaluation of F(x, y) does not terminate. Computing F(1, 0) using the parallel  innermost and parallel  outermost rule yields
Consider the class of recursive and iterative programs. Which of the following is false?
55 docs215 tests

55 docs215 tests
