Question

3. Consider the mystery method given. public static int mystery ( int n) [ if (n == 0 ) { return 1; How do we get the values for recurse? else if (n%2 == 0 ) { int recurse = mystery ( n - 1); int result = recurse + n; return result; since n =5, we go to the else statement and do int recurse = mystery(5-1) which equals 4? why is 3 written? else { int recurse = mystery (n-1); int result = recurse - n; return result; Fill in the recursive stack frame for a call to mystery(5) from main. Use as many of the provided stack frames as appropriate. Cross out any unused frames or variables. main mystery recurse = 3 result = -2 mystery css II ! recurse = 1 mystery recurse = recurse = 2 0 result = result = result = result = 3 -1 2 0 ! mystery ! mystery recurse = mystery D. recurse a result = n= n = mystery recurse = result = mystery n = recurse = result =

Answer 1

EXECUTION SEQUENCE rec=recurse! Mes = result; Met = retorn n=5; rec = mystery (4) ; res rec - 5; return resa - res= -2; return - 21 n=4; rec a mystory (3) Mes = rec+4 : retorn res; res= 3 1 return 3 n=3, rec = mystery (2); res - rec-3; return ries' f res = -1; return 1 I = 27 rec a mystery (1); res a rect2; Mes = 2 ; return res : return 2; n=1; ree = mystery (0); res a sec-1; retulin res. I res=o; return o; no; setrom 1