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:
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And the second:
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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:
- Initialization: We create / set some initial state. This state doesn’t depend on the lists or anything else.
- 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).
- 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:
- Initializtion: The initial state, which is just the counter, is set to 0.
- Effect: Each time an element is moved, if it comes from the right list, the number of inversions is updated.
- 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:
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. -
LEFTandRIGHT, to access the two lists being merged. -
LandR, constants that are used to compare against theSOURCEvariable. -
SOURCE, to denote which list a number came from. -
LSTATEandRSTATE, 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:
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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:
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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 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:
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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.