Question

b) Consider the following code. public static int f(int n) if (n == 1) return 0; else if (n % 2 == 0). return g(n/2); else return g(n+1); public static int g(int n) int r = n % 3; if (r == 0) return f(n/3); else if (r == 1) return f(n+2); else return f(2 * n); // (HERE) public static void main(String[] args) { int x = 3; System.out.println(f(x)); (1) (5 points) Draw the call stack as it would appear when execution reaches the point marked (HERE). In each stack frame, include the name and value of all argumen and local variables (you can omit the command-line args). (II) (5 points) Will f() terminate on all inputs? If so, why? If not, give an example input.

Answer 1

part ii:

function f(3)

f(3) --> g(4)

g(4) --> f(6) (r=1 ,f(n+2))

f(6) --> g(3) (n%2==0 ,g(n/2))

g(3) --> f(1) (r=0 ,f(n/3))

f(1) will return 0

call stack diagram:

f(3) ESSERE 9 (4) 116) 9 (3) +(1) - 0

part 2

no function f wont terminate on all inputs

for e.g

for n=4

function f(4)

f(4) will call g(2)

g(2) will have r=2 so f(2*2) i.e f(4)

f(4) will call g(2)

g(2) will have r=2 so f(2*2) i.e f(4)

so it repeats so its function f(n) will not terminate for all inputs

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