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:
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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