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# Fibonacci challenge

Posted on 2016-09-09
Medium Priority
165 Views
Hi,

I am working on below challenge
http://codingbat.com/prob/p120015

Psedo code:
1. if first two numbers return same
2. else call same fibonacci method by passing n-1 and n-2 and sum it up

``````public int fibonacci(int n) {

if ((n == 0) || (n == 1)) // base cases
return n;
else
// recursion step
return fibonacci(n - 1) + fibonacci(n - 2);

}
``````

I am passing all tests
Expected      Run
fibonacci(0) → 0      0      OK
fibonacci(1) → 1      1      OK
fibonacci(2) → 1      1      OK
fibonacci(3) → 2      2      OK
fibonacci(4) → 3      3      OK
fibonacci(5) → 5      5      OK
fibonacci(6) → 8      8      OK
fibonacci(7) → 13      13      OK
other tests
OK

How to improve/modify my design, code and any other alternate approaches. please advise
0
Question by:gudii9
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LVL 36

Assisted Solution

mccarl earned 200 total points
ID: 41792555
No improvement necessary, that's about as simple as it gets!
0

LVL 16

Accepted Solution

gurpsbassi earned 1000 total points
ID: 41792556
You could get rid of the extra brackets
1

LVL 35

Assisted Solution

ste5an earned 200 total points
ID: 41792630
This again: These tests and tasks are pretty poor. Fibonacci is definied here over N0. Thus a test covering the input -1 should exist. And you should test the input first.
1

LVL 28

Assisted Solution

dpearson earned 200 total points
ID: 41792714
No changes needed - you've nailed this one.

Doug
0

LVL 29

Assisted Solution

pepr earned 200 total points
ID: 41792752
I agree with ste5an. On the other hand, this is for learning the basics of programming, and the totally correct solution (with checking of everything, with unit test, and so on) is not necessary to understand the principle (of Fibonacci, and of the recursive solution).

What is important in this case is that you should not remember this recursive solution as the right approach. The reason is that it is extremely ineficient. The non-recursive solution is not more difficult here, and it is much more efficient.
0

LVL 32

Assisted Solution

phoffric earned 200 total points
ID: 41792832
More advanced ways to compute fibonacci(n) really fast - no recursion, and no loops to get to the nth case.
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibformproof.html

The way I learned the closed form solution was to take the z-transform of the equation. Here is a link that describes how to do this.
http://mathforum.org/library/drmath/view/51540.html
``````The difference equation for the Fibonacci series is:
u(n+2) = u(n+1) + u(n)   with   u(0) = 0,  u(1) = 1
So:

u(n+2) - u(n+1) - u(n) = 0

Taking the z-transforms:

z^2[u(z) - u(0) - u(1)/z] - z[u(z) - u(0)] - u(z) = 0

u(z)[z^2 - z - 1] - z^2*u(0) - z*u(1) + z*u(0) = 0

Putting u(0) = 0 and u(1) = 1 this becomes:

u(z)[z^2 - z - 1] =  z

z
u(z) =  -------------
z^2 - z - 1

The denominator factorizes to:

[z-1/2 -sqrt(5)/2)][z-1/2 +sqrt(5)/2]

The trick here is to express u(z)/z in partial fractions:

1                     1
u(z)/z =   1/sqrt(5)[ ----------------   - ---------------- ]
z-1/2 - sqrt(5)/2     z-1/2 +sqrt(5)/2

and so:

z                     z
u(z) = 1/sqrt(5)[----------------  - ------------------ ]
z-1/2-sqrt(5)/2)      z-1/2+sqrt(5)/2

Then from table of inverse transforms:

z
-------  =  a^n
z - a

where a is constant. Then:

z                       z
u(z) = 1/sqrt(5) [-------------------- - --------------------]
z - (1/2+sqrt(5)/2)    z - (1/2-sqrt(5)/2)

Now from the table of inverse transforms:
u(n) = 1/sqrt(5)[(1/2+sqrt(5)/2)^n - (1/2-sqrt(5)/2)^n]
``````
2

LVL 32

Expert Comment

ID: 41792846
Here's a graph of the continuous function (i.e., let n be a continuous real positive number) that I asked wolframalpha to plot.
`````` u(n) = [(1/2+sqrt(5)/2)^n - (1/2-sqrt(5)/2)^n]/sqrt(5)
``````
http://www.wolframalpha.com/input/?i=plot+%5B(1%2F2%2Bsqrt(5)%2F2)%5En+-+(1%2F2-sqrt(5)%2F2)%5En%5D%2Fsqrt(5),+where+n+%3D+1+to+8

When n is a positive integer, then u(n) is also a positive integer. Pretty amazing!
In practice, when computing, you probably won't get a positive integer, but a real number that is very close. To get the positive integer, you need to use a round function.
http://www.tutorialspoint.com/java/number_round.htm
1

LVL 32

Expert Comment

ID: 41792858
As n gets larger, the speed at which you compute fibonacci(n) really rocks over the other approaches, whether recursive or just looping. However, if you want a table of the first N values of fibonacci(n), then just adding the previous two integers and storing it into an array of length N will be faster than using fibonacci(n) for all values of n = 1..N, where n is an integer.
1

LVL 7

Author Comment

ID: 41793631
``````public int fibonacci(int n) {

if ( n == 0 || n == 1) // base cases
return n;
else
// recursion step
return fibonacci(n - 1) + fibonacci(n - 2);

}
``````

got rid of extra brackets. when i have to put and when i can get rid of brackets?
0

LVL 36

Expert Comment

ID: 41793635
They make absolutely NO difference (in this case) to the execution of the code, if it is easy for you to read with them there, leave them there. That's all that it comes down to, how easy it is to read for YOU.
0

LVL 28

Expert Comment

ID: 41793704
Yes +1 for what mccarl said - I actually think it's a little better with the extra brackets.  You should always prefer clarity and to me the extra with the extra braces is clearer.

Doug
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