Consider the following code.
public void printNumbers(int x, int y) {
if (x < 5) {
System.out.println("x: " + x);
}
if (y > 5) { System.out.println("y: " + y);
}
int a = (int)(Math.random() * 10);
int b= (int)(Math.random() * 10);
if (x != y) printNumbers(a, b);
}
Which of the following conditions will cause recursion to stop with certainty?
x < 5
x < 5 or y > 5
x != y
x == y
In the given code, recursion will stop with certainty if the condition `x == y` is true.
To understand why, let's analyze the code:
The method `printNumbers` takes two integer parameters `x` and `y`. It first checks if `x` is less than 5 using the condition `x < 5`. If this condition is true, it prints the value of `x`.
Next, it checks if `y` is greater than 5 using the condition `y > 5`. If this condition is true, it prints the value of `y`.
After that, it generates two random integers `a` and `b` using `Math.random()`.
Finally, there is a recursive call to `printNumbers` with `a` and `b` as new arguments, but only if `x` is not equal to `y` (`x != y`).
So, if `x` is equal to `y`, the recursive call will not be made, and recursion will stop. Hence, the condition `x == y` causes recursion to stop with certainty.
the only time recursion occurs is if x != y
so ...