View source for Fibonacci

=Motivation=
It is important to always reduce the work, then parallelize if possible.  In this studio you will build a beautiful, elegant solution to Fibonacci which can be easily parallelized yielding off-the-charts ideal parallelism.  Sadly, it performs terribly since the work is exponential.  It is far better to build a linear or log time algorithm whether or not it can be parallelized.

We will gain experience with the future construct.

We will convert recurrence relations into code.

Additionally, (and of minor importance) we gain experience working with the [https://docs.oracle.com/javase/8/docs/api/java/math/BigInteger.html BigInteger] class.

=Background=
The fibonacci sequence is a mathematical concept often used in computer science as a means to demonstrate iteration and recursion. Although you should be familiar with it from CSE 131, we will use the fibonacci sequence to demonstrate memoization and dynamic programming. Follow these links for a quick recap on [http://classes.engineering.wustl.edu/cse231/core/index.php/FAQ#Unit_2:_Futures.2FAccumulators.2FData_Races.2FMemoization memoization, dynamic programming], and the [https://en.wikipedia.org/wiki/Fibonacci_number fibonacci sequence].

==Ted Talk==
<youtube>SjSHVDfXHQ4</youtube>

==Scientific American==
[https://www.scientificamerican.com/video/the-mind-blowing-mathematics-of-sunflowers/ Mathematics of Sunflowers]

[[File:Fibonacci spiral.jpg|400px|link=https://www.scientificamerican.com/video/the-mind-blowing-mathematics-of-sunflowers/]]

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=Lecture=
[https://docs.google.com/presentation/d/1Q4gzoclVBFmkKvX3W80rpGBBBPkXDe3q6IPW0iTKpSw/pub slides]

<youtube>l1hZiAXAcqI</youtube>

<youtube>mRZpq4KPMkY</youtube>
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=Code To Use=
==BigInteger Docs==
fibonacci(47) is greater than [https://docs.oracle.com/javase/8/docs/api/java/lang/Integer.html#MAX_VALUE Integer.MAX_VALUE] which would [https://en.wikipedia.org/wiki/Integer_overflow overflow] if we used the [https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html primitive int data type].  We use [https://docs.oracle.com/javase/8/docs/api/java/math/BigInteger.html BigInteger] instead to prevent this problem.

[https://docs.oracle.com/javase/8/docs/api/java/math/BigInteger.html BigInteger]
: [https://docs.oracle.com/javase/8/docs/api/java/math/BigInteger.html#ZERO BigInteger.ZERO]
: [https://docs.oracle.com/javase/8/docs/api/java/math/BigInteger.html#ONE BigInteger.ONE]
: [https://docs.oracle.com/javase/8/docs/api/java/math/BigInteger.html#add-java.math.BigInteger- bi.add(other)]
: [https://docs.oracle.com/javase/8/docs/api/java/math/BigInteger.html#multiply-java.math.BigInteger- bi.multiply(other)]
: [https://docs.oracle.com/javase/8/docs/api/java/math/BigInteger.html#valueOf-long- bi.valueOf(val)]
: [https://docs.oracle.com/javase/8/docs/api/java/math/BigInteger.html#shiftLeft-int- bi.shiftLeft(n)]

=Studio (Required)=

For the required part of the studio, you will be implementing a Fibonacci calculator five different ways. Some will be in parallel, some won't. The goal for each is to simply have it return the nth Fibonacci number that is initially passed through 

==RecurrenceRelationSequentialFibonacciCalculator==
{{Work|<math>O(\Phi^n)</math>}}
{{CodeToImplement|RecurrenceRelationSequentialFibonacciCalculator|fibonacci|fibonacci.studio}}

{{Sequential|public BigInteger fibonacci(int n)}}

The first file to complete is the basic recursive solution to Fibonacci. This should be done sequentially, and won't look too different from when you did it back in 131. Remember that instead of adding integers, we are using BigIntegers, so make sure to properly handle those.

===recurrence relation===
: <math>F_n = F_{n-1} + F_{n-2}</math>

===seed values===
: <math>F_0=0</math>

: <math>F_1=1</math>

==RecurrenceRelationParallelFibonacciCalculator==
{{WorkCPLIdealParallelism|<math>O(\Phi^n)</math>|<math>O(n)</math>|<math>\dfrac{\Phi^n}{n}</math>}}
{{CodeToImplement|RecurrenceRelationParallelFibonacciCalculator|fibonacci|fibonacci.studio}}
{{Parallel|public BigInteger fibonacci(int n)}}

Now make the solution you just wrote parallel, such that <code>fibonacci(n-1)</code> and <code>fibonacci(n-2)</code> are being solved at the same time. Do this by using futures. Make sure to call future.get() at the right time so there is sufficient parallelism.

{{Warning|The parallelism tests assume that you limit your base cases to 0 and 1, so please do NOT add a base case for 2.}}

== join vs future.get() ==

{{Warning|RecurrenceRelationParallelFibonacciCalculator.java does not contain the static import for join.}}

You can either include the static import:

 <code>import static fj.FJ.join;</code>

and use <code>join</code> or

invoke <code>future.get()</code>:

[https://docs.oracle.com/javase/8/docs/api/java/util/concurrent/Future.html interface Future<V>].  
: [https://docs.oracle.com/javase/8/docs/api/java/util/concurrent/Future.html#get-- get()]

==MemoizationSequentialFibonacciCalculator==
{{Work|<math>O(n)</math>}}
{{CodeToImplement|MemoizationSequentialFibonacciCalculator|fibonacciMemo|fibonacci.studio}}
{{Sequential|private BigInteger fibonacciMemo(BigInteger[] memos, int n)}}

The recurrence relation algorithm for Fibonacci has a lot of repeated calculations. For example, if we want fibonacci(10), we need to calculate fibonacci(9) and fibonacci(8), but during the calculations of fibonacci(9), we also calculate fibonacci(8), so that part is getting repeated. The bigger the number, the more repetition there is and the larger impact it has on our performance. The smart way to get around this is memoization. By storing our calculation results (in this case, with an array), we can simply refer to the work we've already done rather than doing it again.

Implement the recurrence relation algorithm again (sequentially), but employ memoization this time for the win. Make sure to store your work, and only do more work if it hasn't yet been done.

'''provided fibonacci method: '''
 <nowiki>	@Override
	public BigInteger fibonacci(int n) {
		BigInteger[] memos = new BigInteger[n + 1];
		return fibonacciMemo(memos, n);
	}</nowiki>

{{Tip | you need not edit the <code>fibonacci</code> method.  You should only need to edit the <code>fibonacciMemo</code> method.}}

{{Warning | Do NOT invoke fibonacci from fibonacciMemo.  Invoke fibonacciMemo from fibonacciMemo. }}
{{Warning | The entire point of memoization is to avoid recalculating the same computation.  Be sure to store the value for each calculation as it is made even if that means splitting up a line of code.}}

==DynamicIterativeSequentialFibonacciCalculator==
{{Work|<math>O(n)</math>}}
{{CodeToImplement|DynamicIterativeSequentialFibonacciCalculator|fibonacci|fibonacci.studio}}
{{Sequential|public BigInteger fibonacci(int n)}}

In the previous algortihms, we've been tackling this problem from the top-down. But that isn't always the easiest way to think about the problem. Imagine you are simply writing out Fibonacci numbers:

: 0, 1, 1, 2, 3, 5, 8, 13...

You (assuming you aren't a human super-computer) start from the very bottom, and keep working the sequence up to your answer. 

Recall from the lecture where we described calculating each subsequent value based solely on the previous two.

[[File:Dynamic fibonacci slide.png|400px]] [https://docs.google.com/presentation/d/1Q4gzoclVBFmkKvX3W80rpGBBBPkXDe3q6IPW0iTKpSw/pub?slide=id.g71e76e37ea_0_271 slide]

How would you go about converting this process into code?  Note: it should be sequential, and ''not have any recursive calls''.

=Challenge (Optional)=
==LinearRecurrenceSequentialFibonacciCalculator==
{{Work|<math>O(n)</math>}} 

note: can be memoized to {{Work|<math>O(\log{}n)</math>}}
{{CodeToImplement|LinearRecurrenceSequentialFibonacciCalculator|fibonacci|fibonacci.studio}}
{{Sequential|public BigInteger fibonacci(int n)}}

Thanks to how awesome math is, we can actually solve the ''n''th Fibonacci number without knowing those for n-1 and n-2. We have provided the algorithm below. It is your job to translate this into code. Keep it sequential. 

Once you finish with this, make sure to go back and compare the performance speed of all five of these algorithms. The results are very thought-provoking!

===recurrence relation===

: if n is odd

:: <math>k = \dfrac{n+1}{2}</math>

:: <math>F_n = (F_{k} \times F_{k}) + (F_{k-1} \times F_{k-1})</math>

:else

:: <math>k = \dfrac{n}{2}</math>

:: <math>F_n = ((2 \times F_{k-1}) + F_{k}) \times F_{k}</math>


===seed values===

: <math>F_0=0</math>
: <math>F_1=1</math>

{{Warning | Do NOT invoke <code>fibonacci(k)</code> or  <code>fibonacci(k-1)</code> multiple times if you can catch the result in a variable and reuse it. }}
{{Tip | There is no need to store k in a BigInteger.  Since it is smaller than n, it will fit just fine in an int. }}
{{Tip | You can shift left by 1 if you would prefer to not have to multiply by the big integer value of 2.}}

=More Fun (Optional)=
==RecurrenceRelationParallelWithThresholdFibonacciCalculator==
Similar to [[#RecurrenceRelationParallelFibonacciCalculator]] but with a predicate in place to prevent creating too many tasks for the number of processors.

==MemoizationParallelFibonacciCalculator==
Create all of your futures [0..N] up front and then call get on memo[n].
==DynamicRecursiveSequentialFibonacciCalculator==
==RoundPhiToTheNOverSqrt5SequentialFibonacciCalculator==
<math>\dfrac{\Phi^n}{\sqrt{5}}</math>

=Testing Your Solution=
==Correctness==
===Studio (Required)===
{{TestSuite|_FibonacciTestSuite|fibonacci.exercise}}
===Challenge (Optional)===
{{TestSuite|ChallengeFibonacciTestSuite|fibonacci.challenge}}
===More Fun (Optional)===
{{TestSuite|OptionalFunFibonacciTestSuite|fibonacci.fun}}

==Performance==
{{Performance|FibonacciTiming|fibonacci.performance}}

{{Performance|FibonacciIterations|fibonacci.performance}}