CSC300: Reading pure recursive functions [11/13] |
01 |
public static int g (int n) { if (n <= 1) return n; int nMinus1 = g (n-1); int nMinus2 = g (n-2); return nMinus1 + nMinus2; } |
Bottom up (from the base case):
g(0) = 0 g(1) = 1 g(2) = g(0) + g(1) = 0 + 1 = 1 g(3) = g(1) + g(2) = 1 + 1 = 2 g(4) = g(2) + g(3) = 1 + 2 = 3 g(5) = g(3) + g(4) = 2 + 3 = 5 g(6) = g(4) + g(5) = 3 + 5 = 8 g(7) = g(5) + g(6) = 5 + 8 = 13 g(8) = g(6) + g(7) = 8 + 13 = 21 g(9) = g(7) + g(8) = 13 + 21 = 34
Top down (from the initial argument):
g(5) = g(3) + g(4) = (g(2) + g(1)) + g(4) = ((g(0) + g(1)) + g(1)) + g(4) = ((0 + 1 ) + 1 ) + g(4) = 2 + g(4) = 2 + g(3) + g(2) = 2 + (g(2) + g(1)) + g(2) = 2 + ((g(0) + g(1)) + g(1)) + g(2) = 2 + ((0 + 1 ) + 1 ) + g(2) = 2 + 2 + g(2) = 2 + 2 + (g(0) + g(1)) = 2 + 2 + (0 + 1 ) = 2 + 2 + 1 = 5
You can view this as a call tree:
g(5) + g(4) | + g(3) | | + g(2) | | | + g(1) | | | + g(0) | | + g(1) | + g(2) | + g(1) | + g(0) + g(3) + g(2) | + g(1) | + g(0) + g(1)
g(5)
calls g(4)
and g(3)
.
g(4)
calls g(3)
and g(2)
.