labs/lab2-reductions-and-prefix-sums/README.md

-# Reductions and Prefix Sums
-
-Use the following commands to make a fresh clone of your repository:
-
-```
-git clone -b reductions git@gitlab.epfl.ch:lamp/student-repositories-s21/cs206-GASPAR.git cs206-reductions
-```
-
-## Useful links
-
-  * [A guide to the Scala parallel collections](https://docs.scala-lang.org/overviews/parallel-collections/overview.html)
-  * [The API documentation of the Scala parallel collections](https://www.javadoc.io/doc/org.scala-lang.modules/scala-parallel-collections_2.13/latest/scala/collection/index.html)
-  * [The API documentation of the Scala standard library](https://www.scala-lang.org/files/archive/api/2.13.4)
-  * [The API documentation of the Java standard library](https://docs.oracle.com/en/java/javase/15/docs/api/index.html)
-
-**If you have issues with the IDE, try [reimporting the
-build](https://gitlab.epfl.ch/lamp/cs206/-/blob/master/labs/example-lab.md#ide-features-like-type-on-hover-or-go-to-definition-do-not-work),
-if you still have problems, use `compile` in sbt instead.**
-
-## Introduction
-
-In this assignment, you will implement several variants of reduction and prefix
-sum algorithms. Each of the three parts of the assignment will exercise a
-different aspect of parallel programming:
-
-- choosing the right parallelization threshold
-- identifying the correct reduction operator
-- identifying the correct prefix sum operator
-
-We will use the `parallel` construct, defined in the file `package.scala`, as in the lecture to start parallel
-computations. Every `parallel` construct invocation takes two tasks as input and
-outputs the corresponding results as a tuple of two elements. You are not allowed
-to use the `task` construct in this assignment.
-
-## Parallel Counting Change
-
-If you took the course Functional Programming in Scala, you surely recall the
-assignment in which you had to count the number of ways in which you can make
-the change for a given amount of money.
-The text of that assignment was as follows:
-
-> Write a recursive function that counts how many different ways you can
-> make change for an amount, given a list of coin denominations. For
-> example, there are 3 ways to give change for 4 if you have coins with
-> denomination 1 and 2: 1+1+1+1, 1+1+2, 2+2.
-
-In this assignment, you will repeat the same task, but this time, your
-implementation will be parallel.
-Start with the sequential version of this problem once more -- the `countChange`
-function takes the amount of money and the list of different coin denominations.
-It returns the total number of different ways you can give change:
-
-```scala
-def countChange(money: Int, coins: List[Int]): Int
-```
-
-Note that the solution to this problem is recursive.
-In every recursive call, we either decide to continue subtracting the next coin
-in the `coins` list from the `money` amount, or we decide to drop the coin from the list of coins.
-For example, if we have 4 CHF, and coin denominations of 1 and 2, the call
-graph, in which every node depicts one invocation of the `countChange` method,
-is as follows:
-
-                                     4,[1, 2]
-                       3,[1, 2]          +            4,[2]
-               2,[1, 2]    +     3,[2]          2,[2]   +   4,[]
-          1,[1, 2] + 2,[2]   1,[2] + 3,[]    0,[2] + 2,[]    0
-    0,[1, 2] + 1,[2]   1       0      0       1       0
-        1        0
-
-We can take advantage of this recursive structure by evaluating different
-subtrees in parallel.
-This is the next part of the assignment -- implement the method `parCountChange`
-that counts the amount of change in parallel:
-
-```scala
-def parCountChange(money: Int, coins: List[Int], threshold: Threshold): Int
-```
-
-As we learned in the lectures, the `parCountChange` should not spawn parallel
-computations after reaching the leaf in the call graph -- the synchronization
-costs of doing this are way too high.
-Instead, we need to *agglomerate* parts of the computation.
-We do this by calling the sequential `countChange` method when we decide that
-the amount of work is lower than a certain value, called the *threshold*.
-To separate the concern of deciding on the threshold value from the
-implementation of our parallel algorithm, we implement the threshold
-functionality in a separate function, described by the `Threshold` type alias:
-
-```scala
-type Threshold = (Int, List[Int]) => Boolean
-```
-
-When a `threshold` function returns `true` for a given amount of money and the
-given list of coins, the sequential `countChange` implementation must be called.
-
-Implement `parCountChange`!
-
-Now that we have the `parCountChange` method, we ask ourselves what is the right
-implementation of the `threshold` function?
-Recall the examples from the lectures, such as summing the array values and
-computing the norm, where this was easy -- we exactly knew the amount of work
-required to traverse a subrange of the array, so `threshold` could return `true`
-when the length of the subrange was smaller than a certain value.
-
-Sadly, the total amount of work for a given `parCountChange` invocation is hard
-to evaluate from the remaining amount of money and a list of coins.
-In fact, the amount of work directly corresponds to the count that
-`parCountChange` returns, which is the value that we are trying to compute.
-Counting change is a canonical example of a task-parallel problem in which the
-partitioning the workload across processors is *solution-driven* -- to know how
-to optimally partition the work, we would first need to solve the problem
-itself.
-
-For this reason, many parallel algorithms in practice rely on heuristics to
-assess the amount of work in a subtask.
-We will implement several such heuristics in this exercise, and assess the
-effect on performance.
-First, implement the `moneyThreshold` method, which creates a threshold function
-that returns `true` when the amount of money is less than or equal to `2 / 3` of
-the starting amount:
-
-```scala
-def moneyThreshold(startingMoney: Int): Threshold
-```
-
-Remember that `a / b` will return the integer division of `a` and `b` when both
-operands are `Int`s. To avoid this problem, be sure to always do the multiplication of
-`startingMoney` by `2` before doing the division by `3`.
-
-Now run the `ParallelCountChange` application in sbt and observe the speedup:
-
-```
-> runMain reductions.ParallelCountChangeRunner
-```
-
-The previous heuristic did not take into account how many coins were left on the
-coins list, so try two other heuristics.
-Implement the method `totalCoinsThreshold`, which returns a threshold function
-that returns `true` when the number of coins is less than or equal to the `2 / 3` of the
-initial number of coins:
-
-```scala
-def totalCoinsThreshold(totalCoins: Int): Threshold
-```
-
-Again, be careful about the order of operations.
-
-Then, implement the method `combinedThreshold`, which returns a threshold
-function that returns `true` when the amount of money multiplied with the number
-of remaining coins is less than or equal to the starting money multiplied with
-the initial number of coins divided by `2`:
-
-```scala
-def combinedThreshold(startingMoney: Int, allCoins: List[Int]): Threshold
-```
-
-Which of the three threshold heuristics gives the best speedup?
-Can you think of a heuristic that improves performance even more?
-
-
-## Parallel Parentheses Balancing
-
-In this part of the assignment, we recall the Parenthesis Balancing assignment
-that might be familiar to you from the Functional Programming in Scala course.
-Here, the task is to, given an array of characters, decide if the parentheses in
-the array are balanced.
-
-Let us recall a few examples of strings in which parentheses are correctly
-balanced:
-
-```
-(if (zero? x) max (/ 1 x))
-I told him (that it's not (yet) done). (But he wasn't listening)
-```
-
-Similarly, the parentheses in the following strings are not balanced:
-
-```
-(o_()
-:-)
-())(
-```
-
-Implement a sequential function `balance`, which returns `true` iff the
-parentheses in the array are balanced:
-
-```scala
-def balance(chars: Array[Char]): Boolean
-```
-
-Next, you will implement a parallel version of this method.
-By now, you're already an expert at implementing the structure of a reduction
-algorithm, so you should have no problem there.
-The tricky part in parallel parentheses balancing is choosing the reduction
-operator -- you probably implemented `balance` by keeping an integer
-accumulator, incrementing it for left parentheses and decrementing it for the
-right ones, taking care that this accumulator does not drop below zero.
-Parallel parentheses balancing will require a bit more ingenuity on your part,
-so we will give you a hint -- you will need two integer values for the
-accumulator.
-
-Implement the `parBalance` method, which checks if the parentheses in the input
-array are balanced using two helper methods `reduce` and `traverse`.
-These methods implement the parallel reduction and the sequential traversal
-part, respectively:
-
-```scala
-def parBalance(chars: Array[Char], threshold: Int): Boolean = {
-  def traverse(idx: Int, until: Int, _???_: Int, _???_: Int): ???
-
-  def reduce(from: Int, until: Int): ??? = ???
-
-  reduce(0, chars.length) == ???
-}
-```
-
-In this case, we again use the fixed threshold parameter, as we did in the
-lectures. Sections with size smaller or equal to the threshold should be processed sequentially.
-For maximum performance, use a `while` loop in the `traverse` method, or make
-`traverse` tail-recursive -- do not use a `Range`.
-
-Now, run the `ParallelParenthesesBalancing` application from sbt:
-
-```
-> runMain reductions.ParallelParenthesesBalancingRunner
-```
-
-How large was your speedup?
-
-If you are looking for additional challenges, prove that your reduction operator
-is associative!
-
-
-## Line of Sight
-
-In the last part of the exercise, you will be implementing an entirely new
-parallel algorithm -- you will apply the prefix sum algorithm to computing the
-line-of-sight in two-dimensional terrain.
-
-Imagine that you are standing at the zero of a coordinate system.
-The curve to your right describes the terrain that you are facing.
-This is shown in the following figure:
-
-![terrain](terrain.png)
-
-The task of the line-of-sight algorithm is to compute the visibility of each
-point of the terrain, as shown in the following figure, where the visible area
-is above of the full line, and the obscured terrain is shown with a dotted
-line.
-
-![visibility](visibility.png)
-
-What is the necessary and sufficient condition for a point on the terrain to be
-visibile from the zero of the coordinate system, where you are standing?
-Imagine that the terrain heights are represented with an array of numbers.
-We can compute (the tangent of) the viewing angle of each point on the terrain by dividing the
-height of the terrain `xs(i)` with the distance from the viewing point `i`,
-as shown in the following figure:
-
-![angle](angle.png)
-
-It turns out that if the viewing angle of some point B is **lower** than the
-viewing angle of an earlier point A, then the point B is not visible, as shown
-in the following figure:
-
-![angle](angle2.png)
-
-This simple realization allows us to easily compute the line-of-sight on the
-terrain -- if you were a sequential programmer, you would traverse the array of
-height values from the beginning to the end, and write the maximum angle seen so
-far into the output array.
-
-Implement the sequential `lineOfSight` method, which, for each height entry in
-the `input` array (except for input(0) which is the location of the observer and
-is always zero), writes the maximum angle until that point into the `output`
-array (output(0) should be 0):
-
-```scala
-def lineOfSight(input: Array[Float], output: Array[Float]): Unit
-```
-
-We keep things simple -- instead of outputting an array of booleans denoting the
-visibilities, we only output the angles.
-
-Note that what we call an angle in this assignment is actually the tangent
-of the angle. Indeed, `xs(i)` is the opposing side of the angle and `i` the adjacent side.
-The ratio `xs(i) / i` that you compute is in fact the tangent of the angle!
-Since the tangent of an angle is strictly increasing between
-0° and 90°, it is a perfectly good replacement for the actual angle in our use case.
-Keep this in mind and make sure that you do not apply any trigonometic functions on the tangent!
-
-When we see a sequential algorithm that produces a sequence of values by
-traversing the input from left to right, this is an indication that the
-algorithm might have a parallel prefix sum variant.
-So let's try to implement one!
-
-Recall what you learned in the lectures -- the first phase of the parallel
-prefix sum algorithm is the *upsweep* phase.
-Here, the algorithm constructs the reduction tree by traversing parts of the
-input array in parallel.
-Implement the method `upsweepSequential`, which returns the maximum angle in a
-given part of the array, and the method `upsweep`, which returns the reduction
-tree over parts of the input array. If the length of the given part of the input
-array is less than or equal to `threshold`, then `upsweep` calls `upsweepSequential`.
-Note that the part of the input array that needs to traversed is represented
-using indices 'from' (inclusive) and 'until' (or 'end') (exclusive).
-
-```scala
-def upsweepSequential(input: Array[Float], from: Int, until: Int): Float
-```
-
-```scala
-def upsweep(input: Array[Float], from: Int, end: Int, threshold: Int): Tree
-```
-
-The `Tree` data type is either a `Leaf` or an inner `Node`, and it contains the
-maximum angle in the corresponding part of the array.
-Note that when the number of elements in a part of the input array,
-which is `(end - from)`, is smaller or equal to the threshold, the sequential `upsweepSequential`
-has to be invoked, and you should return a `Leaf`.
-Otherwise, you should process the part of the input array in parallel, and return
-a `Node`. Make sure that the work is evenly distributed between the parallel computations.
-
-The second phase is called *downsweep* -- here, the algorithm uses the tree to
-push the maximum angle in the corresponding *prefix* of the array to the leaves
-of the tree, and outputs the values.
-Implement the methods `downsweep` which processes parts of the tree in parallel,
-and the method `downsweepSequential`, which traverses the parts of the array
-corresponding to leaves of the tree and writes the final angles into the
-`output` array:
-
-```scala
-def downsweep(input: Array[Float], output: Array[Float],
-  startingAngle: Float, tree: Tree): Unit
-
-def downsweepSequential(input: Array[Float], output: Array[Float],
-  startingAngle: Float, from: Int, until: Int): Unit
-```
-
-Finally, implement `parLineOfSight` using the `upsweep` and `downsweep` methods:
-
-```scala
-def parLineOfSight(input: Array[Float], output: Array[Float],
-  threshold: Int): Unit
-```
-
-Now, run the `LineOfSight` application in sbt and observe the relative speedups:
-
-```
-> runMain reductions.LineOfSightRunner
-```
-
-How large is the speedup compared to the number of cores in your processor?
-Can you explain your results?

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