# 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 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 memoization, dynamic programming, and the fibonacci sequence.

# The Core Questions

• What is the data?
• Is the data mutable?
• If so, how is it shared?

# Code To Use

### Parallel Docs

future(body)
get()

Check out the wiki's reference page to see how Future's can be used!

### BigInteger Docs

fibonacci(47) is greater than Integer.MAX_VALUE which would overflow if we used the primitive int data type. We use BigInteger instead to prevent this problem.

BigInteger.ZERO
BigInteger.ONE
bi.multiply(other)
bi.valueOf(val)
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: $O(\Phi^n)$
 class: RecurrenceRelationSequentialFibonacciCalculator.java methods: fibonacci package: fibonacci.studio source folder: src/main/java

method: public BigInteger fibonacci(int n) (sequential implementation only)

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

$F_n = F_{n-1} + F_{n-2}$

### seed values

$F_0=0$
$F_1=1$

## RecurrenceRelationParallelFibonacciCalculator

 Work: $O(\Phi^n)$ CPL: $O(n)$ Ideal Parallelism: $\dfrac{\Phi^n}{n}$
 class: RecurrenceRelationParallelFibonacciCalculator.java methods: fibonacci package: fibonacci.studio source folder: src/main/java

method: public BigInteger fibonacci(int n) (parallel implementation required)

Now make the solution you just wrote parallel, such that fibonacci(n-1) and fibonacci(n-2) 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.

## MemoizationSequentialFibonacciCalculator

 Work: $O(n)$
 class: MemoizationSequentialFibonacciCalculator.java methods: fibonacciMemo package: fibonacci.studio source folder: src/main/java

method: private BigInteger fibonacciMemo(BigInteger[] memos, int n) (sequential implementation only)

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.

	@Override
public BigInteger fibonacci(int n) {
BigInteger[] memos = new BigInteger[n + 1];
return fibonacciMemo(memos, n);
}

 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: $O(n)$
 class: DynamicIterativeSequentialFibonacciCalculator.java methods: fibonacci package: fibonacci.studio source folder: src/main/java

method: public BigInteger fibonacci(int n) (sequential implementation only)

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. For this class, convert that process of summing the two previous numbers over and over into code. It should be sequential, and not have any recursive calls.

## LinearRecurrenceSequentialFibonacciCalculator

 Work: $O(n)$
note: can be memoized to
 Work: $O(\log{}n)$
 class: LinearRecurrenceSequentialFibonacciCalculator.java methods: fibonacci package: fibonacci.studio source folder: src/main/java

method: public BigInteger fibonacci(int n) (sequential implementation only)

Thanks to how awesome math is, we can actually solve the nth 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
$k = \dfrac{n+1}{2}$
$F_n = (F_{k} \times F_{k}) + (F_{k-1} \times F_{k-1})$
else
$k = \dfrac{n}{2}$
$F_n = ((2 \times F_{k-1}) + F_{k}) \times F_{k}$

### seed values

$F_0=0$
$F_1=1$
 Warning: Do NOT invoke fibonacci(k) or fibonacci(k-1) 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.

# Fun (Optional)

## RecurrenceRelationParallelWithThresholdFibonacciCalculator

Similar to #RecurrenceRelationParallelFibonacciCalculator but with a threshold 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].

## RoundPhiToTheNOverSqrt5SequentialFibonacciCalculator

$\dfrac{\Phi^n}{\sqrt{5}}$