bunnyEars challenge

Hi,

I am working on below challenge
http://codingbat.com/prob/p183649


We have a number of bunnies and each bunny has two big floppy ears. We want to compute the total number of ears across all the bunnies recursively (without loops or multiplication).

bunnyEars(0) → 0
bunnyEars(1) → 2
bunnyEars(2) → 4
I am not clear on above desription. what it mean by computing recursively?
 please advise
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gudii9Asked:
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awking00Connect With a Mentor Commented:
Maybe a general description of what this challenge is doing will offer a better understanding of recursion. We know that if there are no bunnies, there will be no ears (the base case). We can also agree that if there is one bunny, it will have two ears. But if you look a little deeper for the case of 1 bunny, you can see that it is equivalent to 2 plus the bunnyEars(0) function call which is equivalent to 2 + bunnyEars(bunnies - 1). If bunnies equal 2, that's equivalent to 2 + bunnyEars(1). If we keep making calls to the bunnyEars function passing in the bunnies value - 1 and adding 2, we eventually will reach the point where bunnies = 0 (i.e. the base case) and the function will "recursively" return the number of ears dictated by the value of bunnies. Hope this helps a little bit.
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CPColinConnect With a Mentor Senior Java ArchitectCommented:
A "recursive method" is one that calls itself in order to compute a value. Explaining much more about the concept would pretty much solve the challenge for you. You have a better start going in one of the other questions you just posted.
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gudii9Author Commented:
The idea of calling one function from another immediately suggests the possibility of a function calling itself. The function-call mechanism in Java supports this possibility, which is known as recursion.
Your first recursive program. The "Hello, World" for recursion is the factorial function, which is defined for positive integers n by the equation
n!=n×(n−1)×(n−2)×…×2×1
n!=n×(n−1)×(n−2)×…×2×1
The quantity n! is easy to compute with a for loop, but an even easier method in Factorial.java is to use the following recursive function:
public static long factorial(int n) {
    if (n == 1) return 1;
    return n * factorial(n-1);
}
We can trace this computation in precisely the same way that we trace any sequence of function calls.
factorial(5)
   factorial(4)
      factorial(3)
         factorial(2)
            factorial(1)
               return 1
            return 2*1 = 2
         return 3*2 = 6
      return 4*6 = 24
   return 5*24 = 120

i am reading from below link
http://introcs.cs.princeton.edu/java/23recursion/
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zzynxConnect With a Mentor Software engineerCommented:
So in fact you're asking what recursion is. I would use Google for that.
Eg.
Explanation of recursion
An introduction to recursion
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gudii9Author Commented:
pseudeo code:
1. if o bunnies return ears 0;
2. if one bunny ears 0 i.e 2+bunnyEars(1-1)
3. if n bunnies 2+bunneyEars(n-1)
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gudii9Author Commented:
public int bunnyEars(int bunnies) {
 if(bunnies==0){
   return 0;
 }
 else{
   return 2+bunnyEars(bunnies-1);
 }
}

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i passed all the tests. any improvements or alternate approach?
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