Table of Contents

After the madness of the language for homework 1, the solution to the second homework offers a moment of respite. Let’s get right into the problems, shall we?

### Homework 2

Besides some free-response questions, the homework contains two problems. The first:

From hw2.txt, lines 29 through 34
 29 30 31 32 33 34  1. Implement mergesort. >>> mergesort([4, 2, 5, 1, 6, 3]) [1, 2, 3, 4, 5, 6] Filename: msort.py

And the second:

From hw2.txt, lines 36 through 44
 36 37 38 39 40 41 42 43 44  2. Calculate the number of inversions in a list. >>> num_inversions([4, 1, 3, 2]) 4 >>> num_inversions([2, 4, 1, 3]) 3 Filename: inversions.py Must run in O(nlogn) time.

At first glance, it’s not obvious why these problems are good for us. However, there’s one key observation: num_inversions can be implemented using a slightly-modified mergesort. The trick is to maintain a counter of inversions in every recursive call to mergesort, updating it every time we take an element from the [note: If this nomenclature is not clear to you, recall that mergesort divides a list into two smaller lists. The "right list" refers to the second of the two, because if you visualize the original list as a rectangle, and cut it in half (vertically, down the middle), then the second list (from the left) is on the right. ] while there are still elements in the [note: Why this is the case is left as an exercise to the reader. ] . When we return from the call, we add up the number of inversions from running num_inversions on the smaller lists, and the number of inversions that we counted as I described. We then return both the total number of inversions and the sorted list.

So, we either perform the standard mergesort, or we perform mergesort with additional steps added on. The additional steps can be divided into three general categories:

1. Initialization: We create / set some initial state. This state doesn’t depend on the lists or anything else.
2. Effect: Each time that an element is moved from one of the two smaller lists into the output list, we may change the state in some way (create an effect).
3. Combination: The final state, and the results of the two sub-problem states, are combined into the output of the function.

This is all very abstract. In the concrete case of inversions, these steps are as follows:

1. Initializtion: The initial state, which is just the counter, is set to 0.
2. Effect: Each time an element is moved, if it comes from the right list, the number of inversions is updated.
3. Combination: We update the state, simply adding the left and right inversion counts.

We can make a language out of this!

### A Language

Again, let’s start by visualizing what the solution will look like. How about this:

From hw2.lang, entire file
state 0;

effect {
if(SOURCE == R) {
STATE = STATE + |LEFT|;
}
}

combine {
STATE = STATE + LSTATE + RSTATE;
}


We divide the code into the same three steps that we described above. The first section is the initial state. Since it doesn’t depend on anything, we expect it to be some kind of literal, like an integer. Next, we have the effect section, which has access to the variables below:

• STATE, to manipulate or check the current state.
• LEFT and RIGHT, to access the two lists being merged.
• L and R, constants that are used to compare against the SOURCE variable.
• SOURCE, to denote which list a number came from.
• LSTATE and RSTATE, to denote the final states from the two subproblems.

We use an if-statement to check if the element that was popped came from the right list (by checking SOURCE == R). If it did, we increment the counter (state) by the proper amount. In the combine step, which has access to the same variables, we simply increment the state by the counters from the left and right solutions, stored in LSTATE and RSTATE. That’s it!

#### Implementation

The implementation is not tricky at all. We don’t need to use monads like we did last time, and nor do we have to perform any fancy Python nested function declarations.

To keep with the Python convention of lowercase variables, we’ll translate the uppercase “global” variables to lowercase. We’ll do it like so:

From LanguageTwo.hs, lines 167 through 176
 167 168 169 170 171 172 173 174 175 176  translateExpr (Var s) | s == "SOURCE" = Py.Var "source" | s == "LEFT" = Py.Var "left" | s == "RIGHT" = Py.Var "right" | s == "STATE" = Py.Var "state" | s == "LSTATE" = Py.Var "ls" | s == "RSTATE" = Py.Var "rs" | s == "L" = Py.IntLiteral 1 | s == "R" = Py.IntLiteral 2 | otherwise = Py.Var s

Note that we translated L and R to integer literals. We’ll indicate the source of each element with an integer, since there’s no real point to representing it with a string or a variable. We’ll need to be aware of this when we implement the actual, generic mergesort code. Let’s do that now:

From LanguageTwo.hs, lines 101 through 161
 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161  baseFunction :: Py.PyExpr -> [Py.PyStmt] -> [Py.PyStmt] -> Py.PyStmt baseFunction s e c = Py.FunctionDef "prog" ["xs"] $[Py.IfElse (Py.BinOp Py.LessThan (Py.FunctionCall (Py.Var "len") [Py.Var "xs"]) (Py.IntLiteral 2)) [Py.Return$ Py.Tuple [s, Py.Var "xs"]] [] Nothing , Py.Assign (Py.VarPat "leng") (Py.BinOp Py.FloorDiv (Py.FunctionCall (Py.Var "len") [Py.Var "xs"]) (Py.IntLiteral 2)) , Py.Assign (Py.VarPat "left") (Py.Access (Py.Var "xs") [Py.Slice Nothing $Just (Py.Var "leng")]) , Py.Assign (Py.VarPat "right") (Py.Access (Py.Var "xs") [Py.Slice (Just (Py.Var "leng")) Nothing]) , Py.Assign (Py.TuplePat [Py.VarPat "ls", Py.VarPat "left"]) (Py.FunctionCall (Py.Var "prog") [Py.Var "left"]) , Py.Assign (Py.TuplePat [Py.VarPat "rs", Py.VarPat "right"]) (Py.FunctionCall (Py.Var "prog") [Py.Var "right"]) , Py.Standalone$ Py.FunctionCall (Py.Member (Py.Var "left") "reverse") [] , Py.Standalone $Py.FunctionCall (Py.Member (Py.Var "right") "reverse") [] , Py.Assign (Py.VarPat "state") s , Py.Assign (Py.VarPat "source") (Py.IntLiteral 0) , Py.Assign (Py.VarPat "total") (Py.ListLiteral []) , Py.While (Py.BinOp Py.And (Py.BinOp Py.NotEqual (Py.Var "left") (Py.ListLiteral [])) (Py.BinOp Py.NotEqual (Py.Var "right") (Py.ListLiteral [])))$ [ Py.IfElse (Py.BinOp Py.LessThanEq (Py.Access (Py.Var "left") [Py.IntLiteral $-1]) (Py.Access (Py.Var "right") [Py.IntLiteral$ -1])) [ Py.Standalone $Py.FunctionCall (Py.Member (Py.Var "total") "append") [Py.FunctionCall (Py.Member (Py.Var "left") "pop") []] , Py.Assign (Py.VarPat "source") (Py.IntLiteral 1) ] []$ Just [ Py.Standalone $Py.FunctionCall (Py.Member (Py.Var "total") "append") [Py.FunctionCall (Py.Member (Py.Var "right") "pop") []] , Py.Assign (Py.VarPat "source") (Py.IntLiteral 2) ] ] ++ e ] ++ c ++ [ Py.Standalone$ Py.FunctionCall (Py.Member (Py.Var "left") "reverse") [] , Py.Standalone $Py.FunctionCall (Py.Member (Py.Var "right") "reverse") [] , Py.Return$ Py.Tuple [ Py.Var "state" , foldl (Py.BinOp Py.Add) (Py.Var "total") [Py.Var "left", Py.Var "right"] ] ]

This is probably the ugliest part of this assignment: we handwrote a Python AST in Haskell that implements mergesort with our augmentations. Note that this is a function, which takes a Py.PyExpr (the initial state expression), and two lists of Py.PyStmt, which are the “effect” and “combination” code, respectively. We simply splice them into our regular mergesort function. The translation is otherwise pretty trivial, so there’s no real reason to show it here.

### The Output

What’s the output of our solution to num_inversions? Take a look for yourself:

def prog(xs):
if len(xs)<2:
return (0, xs)
leng = len(xs)//2
left = xs[:(leng)]
right = xs[(leng):]
(ls,left) = prog(left)
(rs,right) = prog(right)
left.reverse()
right.reverse()
state = 0
source = 0
total = []
while (left!=[])and(right!=[]):
if left[-1]<=right[-1]:
total.append(left.pop())
source = 1
else:
total.append(right.pop())
source = 2
if source==2:
state = state+len(left)
state = state+ls+rs
left.reverse()
right.reverse()
return (state, total+left+right)


Honestly, that’s pretty clean. As clean as left.reverse() to allow for $O(1)$ pop is. What’s really clean, however, is the implementation of mergesort in our language. It goes as follows:

state 0;
effect {}
combine {}


To implement mergesort in our language, which describes mergesort variants, all we have to do is not specify any additional behavior. Cool, huh?

That’s the end of this post. If you liked this one (and the previous one!), keep an eye out for more!

### Appendix (Missing Homework Question)

I should not view homework assignments on a small-screen device. There was a third problem on homework 2:

From hw2.txt, lines 46 through 65
 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65  3. [WILL BE GRADED] Length of the longest path in a binary tree (number of edges). We will use the "buggy qsort" representation of binary trees from HW1: [left_subtree, root, right_subtree] >>> longest([[], 1, []]) 0 >>> longest([[[], 1, []], 2, [[], 3, []]]) 2 >>> longest([[[[], 1, []], 2, [[], 3, []]], 4, [[[], 5, []], 6, [[], 7, [[], 9, []]]]]) 5 Note the answer is 5 because the longest path is 1-2-4-6-7-9. Filename: longest.py Must run in O(n) time.

This is not a mergesort variant, and adding support for it into our second language will prevent us from making it the neat specialized [note: DSL is a shortened form of "domain specific language", which was briefly described in another sidenote while solving homework 1. ] that was just saw. We’ll do something else, instead: we’ll use the language we defined in homework 1 to solve this problem:

empty() = [0, 0];
longest(xs) =
if |xs| != 0
then _longest(longest(xs[0]), longest(xs[2]))
else empty();
_longest(l, r) = [max(l[0], r[0]) + 1, max(l[0]+r[0], max(l[1], r[1]))];


[note: This is probably true with any program written in our first language. ] In these 6 lines of code, there are two hacks to work around the peculiarities of the language.

At each recursive call, we want to keep track of both the depth of the tree and the existing longest path. This is because the longest path could be found either somewhere down a subtree, or from combining the largest depths of two subtrees. To return two values from a function in Python, we’d use a tuple. Here, we use a list.

Alarm bells should be going off here. There’s no reason why we should ever return an empty list from the recursive call: at the very least, we want to return [0,0]. But placing such a list literal in a function will trigger the special case insertion. So, we have to hide this literal from the compiler. Fortunately, that’s not too hard to do - the compiler is pretty halfhearted in its inference of types. Simply putting the literal behind a constant function (empty) does the trick.

The program uses the subproblem depths multiple times in the final computation. We thus probably want to assign these values to names so we don’t have to perform any repeated work. Since the only two mechanisms for [note: To bind a variable means to assign a value to it. ] in this language are function calls and list selectors, we use a helper function _longest, which takes two subproblem solutions an combines them into a new solution. It’s pretty obvious that _longest returns a list, so the compiler will try insert a base case. Fortunately, subproblem solutions are always lists of two numbers, so this doesn’t affect us too much.