Answer:
1. mysteryXY(4, 1); = 4
2. mysteryXY(4, 2); = 8, ,4
3. mysteryXY(8, 2); = 16, , 8
4. mysteryXY(4, 3); = 12, , 8
5. mysteryXY(3, 4); = 12, , 9
Step-by-step explanation:
public void mysteryXY(int x, int y) {
if (y == 1) {
System.out.print(x);
}
else
{
System.out.print(x * y + ", ");
mysteryXY(x, y - 1);
System.out.print(", " + x * y); }
}
mysteryXY(4, 1); = 4
On line 2, the value of Y is tested;
Y = 1. So the operation on line 3 will be executed.
The values of X will be printed
X= 1
For question 2 through 5, the value of Y is not 1, so it'll skip line and jump to 6.
The statement on line 6 print x * y appended with a comma
On line 7, the values of y is reduced by 1
On line 8, it prints , and the result of x * y.
So, we have
2. mysteryXY(4, 2); = 8, ,4
4 * 2 = 8
Reduce y by 1
Then, 4 * 1 = 4
Output: 8, , 4
Applying the same logic to 3 to 5
3. mysteryXY(8, 2); =
Output: 16, , 8
4. mysteryXY(4, 3); =
Output: 12, , 8
5. mysteryXY(3, 4); =
Output: 12, , 9
