I have just started practicing with recursion I have this very simple practice program problem.
There are bunnies standing in a line, numbered 1,2,3… The odd bunnies (1,2,3..) have normal 2 ears.
The even bunnies (2,4,6…) have 3 ears, because they each have a raised foot. Recursively return the number of ears in the bunny line.
I have the solution. However, I am a bit confused on certain things. For one, it is my understanding that each even numbered rabbit has 3 feet. So bunnyEars2(2), should produce 6 instead of 5?
Also, I notice if I remove certain intricacies like ‘(bunnyEars2(bunnies)’ instead of adding in the ‘-1’ at the end. I get this duplicitous message “at bunnyEarsTwo.bunnyEars2(bunnyEarsTwo.java:13).
Any explanation and breakdown of this problem and recursion in general is very much appreciated. I am determined to learn, I just want to be pointed in the right direction!
public static int bunnyEars2(int bunnies){ if(bunnies==0) return 0; return (bunnies % 2 == 0) ? (3+bunnyEars2(bunnies-1)) : (2+bunnyEars2(bunnies-1)); }
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Answer
I hope you know factorial by recursive. because this is very similar.
int factorial(int n){ if (n == 0) return 1; else return(n * factorial(n-1)); }
Here we are returning `n * n-1 * n-1-1 * n-1-1-1 and so on until it n-1…. is 0.
Likewise,
public static int bunnyEars2(int bunnies){ if(bunnies==0) return 0; if (bunnies % 2 == 0) return 3+bunnyEars2(bunnies-1); else return 2+bunnyEars2(bunnies-1); }
Here, follows same logic, but the differece is ,
when its even, it return 3 + bunnyEars2(bunnies-1)
when its odd, it return 2 + bunnyEars2(bunnies-1)
for example: bunnyEars2(4)
is 10
here our bunnies value will be 4,3,2,1,0
as 4 is even it returns 3+
, 3 is odd it returns 2+
, 2 returns 3+
, 1 returns 2+
and 0 returns 0
3 + 2 + 3 + 2 + 0 = 10.
bunnyEars2(2)
will be 5
, here bunnies value will be 2,1,0
which returns 3 + 2 + 0 = 5
Also removing -1
from bunnyEars2(bunnies-1)
will result in infinite recursion(stack overflow error). It’s like removing -1
from n * factorial(n)
, it won’t end.