blog-static/content/blog/haskell_lazy_evaluation/index.md

---
title: "Time Traveling In Haskell: How It Works And How To Use It"
date: 2020-07-30T00:58:10-07:00
tags: ["Haskell"]
---

I recently got to use a very curious Haskell technique
{{< sidenote "right" "production-note" "in production:" >}}
As production as research code gets, anyway!
{{< /sidenote >}} time traveling. I say this with
the utmost seriousness. This technique worked like
magic for the problem I was trying to solve, and so
I thought I'd share what I learned. In addition
to the technique and its workings, I will also explain how 
time traveling can be misused, yielding computations that
never terminate.

### Time Traveling
Some time ago, I read [this post](https://kcsongor.github.io/time-travel-in-haskell-for-dummies/) by Csongor Kiss about time traveling in Haskell. It's
really cool, and makes a lot of sense if you have wrapped your head around
lazy evaluation. I'm going to present my take on it here, but please check out
Csongor's original post if you are interested.

Say that you have a list of integers, like `[3,2,6]`. Next, suppose that
you want to find the maximum value in the list. You can implement such
behavior quite simply with pattern matching:

```Haskell
myMax :: [Int] -> Int
myMax [] = error "Being total sucks"
myMax (x:xs) = max x $ myMax xs
```

You could even get fancy with a `fold`:

```Haskell
myMax :: [Int] -> Int
myMax = foldr1 max
```

All is well, and this is rather elementary Haskell. But now let's look at
something that Csongor calls the `repMax` problem:

> Imagine you had a list, and you wanted to replace all the elements of the
> list with the largest element, by only passing the list once.

How can we possibly do this in one pass? First, we need to find the maximum
element, and only then can we have something to replace the other numbers
with! It turns out, though, that we can just expect to have the future
value, and all will be well. Csongor provides the following example:

```Haskell
repMax :: [Int] -> Int -> (Int, [Int])
repMax [] rep = (rep, [])
repMax [x] rep = (x, [rep])
repMax (l : ls) rep = (m', rep : ls')
  where (m, ls') = repMax ls rep
        m' = max m l
```
In this example, `repMax` takes the list of integers,
each of which it must replace with their maximum element.
It also takes __as an argument__ the maximum element,
as if it had already been computed. It does, however,
still compute the intermediate maximum element,
in the form of `m'`. Otherwise, where would the future
value even come from?

Thus far, nothing too magical has happened. It's a little
strange to expect the result of the computation to be
given to us; it just looks like wishful
thinking. The real magic happens in Csongor's `doRepMax`
function:

```Haskell
doRepMax :: [Int] -> [Int]
doRepMax xs = xs'
  where (largest, xs') = repMax xs largest
```

Look, in particular, on the line with the `where` clause.
We see that `repMax` returns the maximum element of the
list, `largest`, and the resulting list `xs'` consisting
only of `largest` repeated as many times as `xs` had elements.
But what's curious is the call to `repMax` itself. It takes
as input `xs`, the list we're supposed to process... and
`largest`, the value that _it itself returns_.

This works because Haskell's evaluation model is, effectively,
[lazy graph reduction](https://en.wikipedia.org/wiki/Graph_reduction). That is,
you can think of Haskell as manipulating your code as
{{< sidenote "right" "tree-note" "a syntax tree," >}}
Why is it called graph reduction, you may be wondering, if the runtime is
manipulating syntax trees? To save on work, if a program refers to the
same value twice, Haskell has both of those references point to the
exact same graph. This violates the tree's property of having only one path
from the root to any node, and makes our program a DAG (at least). Graph nodes that
refer to themselves (which are also possible in the model) also violate the properties of a
a DAG, and thus, in general, we are working with graphs.
{{< /sidenote >}} performing
substitutions and simplifications as necessary until it reaches a final answer.
What the lazy part means is that parts of the syntax tree that are not yet
needed to compute the final answer can exist, unsimplified, in the tree.
Why don't we draw a few graphs to get familiar with the idea?

### Visualizing Graphs and Their Reduction
__A word of caution__: the steps presented below may significantly differ
from the actual graph reduction algorithms used by modern compilers.
In particular, this section draws a lot of ideas from Simon Peyton Jones' book,
[_Implementing functional languages: a tutorial_](https://www.microsoft.com/en-us/research/publication/implementing-functional-languages-a-tutorial/).
However, modern functional compilers (i.e. GHC) use a much more
complicated abstract machine for evaluating graph-based code,
based on -- from what I know -- the [spineless tagless G-machine](https://www.microsoft.com/en-us/research/wp-content/uploads/1992/04/spineless-tagless-gmachine.pdf).
In short, this section, in order to build intuition, walks through how a functional program
evaluated using graph reduction _may_ behave; the actual details
depend on the compiler. 

Let's start with something that doesn't have anything fancy. We can
take a look at the graph of the expression:

```Haskell
length [1]
```

Stripping away Haskell's syntax sugar for lists, we can write this expression as:

```Haskell
length (1:[])
```

Then, recalling that `(:)`, or 'cons', is just a binary function, we rewrite:

```Haskell
length ((:) 1 [])
```

We're now ready to draw the graph; in this case, it's pretty much identical
to the syntax tree of the last form of our expression:

{{< figure src="length_1.png" caption="The initial graph of `length [1]`." class="small" >}}

In this image, the `@` nodes represent function application. The
root node is an application of the function `length` to the graph that
represents the list `[1]`. The list itself is represented using two
application nodes: `(:)` takes two arguments, the head and tail of the
list, and function applications in Haskell are
[curried](https://en.wikipedia.org/wiki/Currying). Eventually,
in the process of evaluation, the body of `length` will be reached,
and leave us with the following graph:

{{< figure src="length_2.png" caption="The graph of `length [1]` after the body of `length` is expanded." class="small" >}}

Conceptually, we only took one reduction step, and thus, we haven't yet gotten
to evaluating the recursive call to `length`. Since `(+)`
is also a binary function, `1+length xs` is represented in this
new graph as two applications of `(+)`, first to `1`, and then
to `length []`.

But what is that box at the root? This box _used to be_ the root of the
first graph, which was an application node. However, it is now a
an _indirection_. Conceptually, reducing
this indirection amounts to reducing the graph
it points to. But why have we {{< sidenote "right" "altered-note" "altered the graph" >}}
This is a key aspect of implementing functional languages.
The language itself may be pure, while the runtime
can be, and usually is, impure and stateful. After all,
computers are impure and stateful, too!
{{< /sidenote >}} in this manner? Because Haskell is a pure language,
of course! If we know that a particular graph reduces to some value,
there's no reason to reduce it again. However, as we will
soon see, it may be _used_ again, so we want to preserve its value.
Thus, when we're done reducing a graph, we replace its root node with
an indirection that points to its result. 

When can a graph be used more than once? Well, how about this:

```Haskell
let x = square 5 in x + x
```

Here, the initial graph looks as follows:

{{< figure src="square_1.png" caption="The initial graph of `let x = square 5 in x + x`." class="small" >}}

As you can see, this _is_ a graph, but not a tree! Since both
variables `x` refer to the same expression, `square 5`, they
are represented by the same subgraph. Then, when we evaluate `square 5`
for the first time, and replace its root node with an indirection,
we end up with the following:

{{< figure src="square_2.png" caption="The graph of `let x = square 5 in x + x` after `square 5` is reduced." class="small" >}}

There are two `25`s in the graph, and no more `square`s! We only
had to evaluate `square 5` exactly once, even though `(+)`
will use it twice (once for the left argument, and once for the right).

Our graphs can also include cycles.
A simple, perhaps _the most_ simple example of this in practice is Haskell's
`fix` function. It computes a function's fixed point,
{{< sidenote "right" "fixpoint-note" "and can be used to write recursive functions." >}}
In fact, in the lambda calculus, <code>fix</code> is pretty much <em>the only</em>
way to write recursive functions. In the untyped lambda calculus, it can
be written as: $$\lambda f . (\lambda x . f (x \ x)) \  (\lambda x . f (x \ x))$$
In the simply typed lambda calculus, it cannot be written in any way, and
needs to be added as an extension, typically written as \(\textbf{fix}\).
{{< /sidenote >}}
It's implemented as follows:

```Haskell
fix f = let x = f x in x
```

See how the definition of `x` refers to itself? This is what
it looks like in graph form:

{{< figure src="fixpoint_1.png" caption="The initial graph of `let x = f x in x`." class="tiny" >}}

I think it's useful to take a look at how this graph is processed. Let's
pick `f = (1:)`. That is, `f` is a function that takes a list,
and prepends `1` to it. Then, after constructing the graph of `f x`,
we end up with the following:

{{< figure src="fixpoint_2.png" caption="The graph of `fix (1:)` after it's been reduced." class="small" >}}

We see the body of `f`, which is the application of `(:)` first to the
constant `1`, and then to `f`'s argument (`x`, in this case). As
before, once we evaluated `f x`, we replaced the application with
an indirection; in the image, this indirection is the top box. But the
argument, `x`, is itself an indirection which points to the root of `f x`,
thereby creating a cycle in our graph. Traversing this graph looks like
traversing an infinite list of `1`s.

Almost there! A node can refer to itself, and, when evaluated, it
is replaced with its own value. Thus, a node can effectively reference
its own value! The last piece of the puzzle is how a node can access
_parts_ of its own value: recall that `doRepMax` calls `repMax`
with only `largest`, while `repMax` returns `(largest, xs')`.
I have to admit, I don't know the internals of GHC, but I suspect
this is done by translating the code into something like:

```Haskell
doRepMax :: [Int] -> [Int]
doRepMax xs = snd t
  where t = repMax xs (fst t)
```

#### Detailed Example: Reducing `doRepMax`

If the above examples haven't elucidated how `doRepMax` works,
stick around in this section and we will go through it step-by-step.
This is a rather long and detailed example, so feel free to skip
this section to read more about actually using time traveling.

If you're sticking around, why don't we watch how the graph of `doRepMax [1, 2]` unfolds.
This example will be more complex than the ones we've seen
so far; to avoid overwhelming ourselves with notation,
let's adopt a different convention of writing functions. Instead
of using application nodes `@`, let's draw an application of a
function `f` to arguments `x1` through `xn` as a subgraph with root `f`
and children `x`s. The below figure demonstrates what I mean:

{{< figure src="notation.png" caption="The new visual notation used in this section." class="small" >}}

Now, let's write the initial graph for `doRepMax [1,2]`:

{{< figure src="repmax_1.png" caption="The initial graph of `doRepMax [1,2]`." class="small" >}}

Other than our new notation, there's nothing too surprising here.
The first step of our hypothetical reduction would replace the application of `doRepMax` with its
body, and create our graph's first cycle. At a high level, all we want is the second element of the tuple
returned by `repMax`, which contains the output list. To get
the tuple, we apply `repMax` to the list `[1,2]` and the first element
of its result. The list `[1,2]` itself
consists of two uses of the `(:)` function.

{{< figure src="repmax_2.png" caption="The first step of reducing `doRepMax [1,2]`." class="small" >}}

Next, we would also expand the body of `repMax`. In
the following diagram, to avoid drawing a noisy amount of
crossing lines, I marked the application of `fst` with
a star, and replaced the two edges to `fst` with
edges to similar looking stars. This is merely
a visual trick; an edge leading to a little star is
actually an edge leading to `fst`. Take a look:

{{< figure src="repmax_3.png" caption="The second step of reducing `doRepMax [1,2]`." class="medium" >}}

Since `(,)` is a constructor, let's say that it doesn't
need to be evaluated, and that its
{{< sidenote "right" "normal-note" "graph cannot be reduced further" >}}
A graph that can't be reduced further is said to be in <em>normal form</em>,
by the way.
{{< /sidenote >}} (in practice, other things like
packing may occur here, but they are irrelevant to us).
If `(,)` can't be reduced, we can move on to evaluating `snd`. Given a pair, `snd`
simply returns the second element, which in our
case is the subgraph starting at `(:)`. We
thus replace the application of `snd` with an
indirection to this subgraph. This leaves us
with the following:

{{< figure src="repmax_4.png" caption="The third step of reducing `doRepMax [1,2]`." class="medium" >}}

Since it's becoming hard to keep track of what part of the graph
is actually being evaluated, I marked the former root of `doRepMax [1,2]` with
a blue star. If our original expression occured at the top level,
the graph reduction would probably stop here.  After all,
we're evaluating our graphs using call-by-need, and there
doesn't seem to be a need for knowing the what the arguments of `(:)` are.
However, stopping at `(:)` wouldn't be very interesting,
and we wouldn't learn much from doing so. So instead, let's assume
that _something_ is trying to read the elements of our list;
perhaps we are trying to print this list to the screen in GHCi.

In this case, our mysterious external force starts unpacking and
inspecting the arguments to `(:)`. The first argument to `(:)` is
the list's head, which is the subgraph starting with the starred application
of `fst`. We evaluate it in a similar manner to `snd`. That is,
we replace this `fst` with an indirection to the first element
of the argument tuple, which happens to be the subgraph starting with `max`:

{{< figure src="repmax_5.png" caption="The fourth step of reducing `doRepMax [1,2]`." class="medium" >}}

Phew! Next, we need to evaluate the body of `max`. Let's make one more
simplification here: rather than substitututing `max` for its body
here, let's just reason about what evaluating `max` would entail.
We would need to evaluate its two arguments, compare them,
and return the larger one. The argument `1` can't be reduced
any more (it's just a number!), but the second argument,
a call to `fst`, needs to be processed. To do so, we need to
evaluate the call to `repMax`. We thus replace `repMax`
with its body:

{{< figure src="repmax_6.png" caption="The fifth step of reducing `doRepMax [1,2]`." class="medium" >}}

We've reached one of the base cases here, and there
are no more calls to `max` or `repMax`. The whole reason
we're here is to evaluate the call to `fst` that's one
of the arguments to `max`. Given this graph, doing so is easy.
We can clearly see that `2` is the first element of the tuple
returned by `repMax [2]`. We thus replace `fst` with
an indirection to this node:

{{< figure src="repmax_7.png" caption="The sixth step of reducing `doRepMax [1,2]`." class="medium" >}}

This concludes our task of evaluating the arguments to `max`.
Comparing them, we see that `2` is greater than `1`, and thus,
we replace `max` with an indirection to `2`:

{{< figure src="repmax_8.png" caption="The seventh step of reducing `doRepMax [1,2]`." class="medium" >}}

The node that we starred in our graph is now an indirection (the
one that used to be the call to `fst`) which points to
another indirection (formerly the call to `max`), which
points to `2`. Thus, any edge pointing to a star now
points to the value 2. 

By finding the value of the starred node, we have found the first
argument of `(:)`, and returned it to our mysterious external force.
If we were printing to GHCi, the number `2` would appear on the screen
right about now. The force then moves on to the second argument of `(:)`,
which is the call to `snd`. This `snd` is applied to an instance of `(,)`, which
can't be reduced any further. Thus, all we have to do is take the second
element of the tuple, and replace `snd` with an indirection to it:

{{< figure src="repmax_9.png" caption="The eighth step of reducing `doRepMax [1,2]`." class="medium" >}}

The second element of the tuple was a call to `(:)`, and that's what the mysterious
force is processing now. Just like it did before, it starts by looking at the
first argument of this list, which is the list's head. This argument is a reference to
the starred node, which, as we've established, eventually points to `2`.
Another `2` pops up on the console.

Finally, the mysterious force reaches the second argument of the `(:)`,
which is the empty list. The empty list also cannot be evaluated any
further, so that's what the mysterious force receives. Just like that,
there's nothing left to print to the console. The mysterious force ceases.
After removing the unused nodes, we are left with the following graph:

{{< figure src="repmax_10.png" caption="The result of reducing `doRepMax [1,2]`." class="small" >}}

As we would have expected, two `2`s were printed to the console, and our
final graph represents the list `[2,2]`.

### Using Time Traveling
Is time tarveling even useful? I would argue yes, especially
in cases where Haskell's purity can make certain things
difficult.

As a first example, Csongor provides an assembler that works
in a single pass. The challenge in this case is to resolve
jumps to code segments occuring _after_ the jump itself;
in essence, the address of the target code segment needs to be
known before the segment itself is processed. Csongor's
code uses the [Tardis monad](https://hackage.haskell.org/package/tardis-0.4.1.0/docs/Control-Monad-Tardis.html),
which combines regular state, to which you can write and then
later read from, and future state, from which you can
read values before your write them. Check out
[his complete example](https://kcsongor.github.io/time-travel-in-haskell-for-dummies/#a-single-pass-assembler-an-example) here.

Alternatively, here's an example from my research, which my
coworker and coauthor Kai helped me formulate. I'll be fairly
vague, since all of this is still in progress. The gist is that
we have some kind of data structure (say, a list or a tree),
and we want to associate with each element in this data
structure a 'score' of how useful it is. There are many possible
heuristics of picking 'scores'; a very simple one is 
to make it inversely propertional to the number of times
an element occurs. To be more concrete, suppose
we have some element type `Element`:

{{< codelines "Haskell" "time-traveling/ValueScore.hs" 5 6 >}}

Suppose also that our data structure is a binary tree:

{{< codelines "Haskell" "time-traveling/ValueScore.hs" 14 16 >}}

We then want to transform an input `ElementTree`, such as:

```Haskell
Node A (Node A Empty Empty) Empty
```

Into a scored tree, like:

```Haskell
Node (A,0.5) (Node (A,0.5) Empty Empty) Empty
```

Since `A` occured twice, its score is `1/2 = 0.5`. 

Let's define some utility functions before we get to the
meat of the implementation:

{{< codelines "Haskell" "time-traveling/ValueScore.hs" 8 12 >}}

The `addElement` function simply increments the counter for a particular
element in the map, adding the number `1` if it doesn't exist. The `getScore`
function computes the score of a particular element, defaulting to `1.0` if
it's not found in the map.

Just as before -- noticing that passing around the future values is getting awfully
bothersome -- we write our scoring function as though we have
a 'future value'.

{{< codelines "Haskell" "time-traveling/ValueScore.hs" 18 24 >}}

The actual `doAssignScores` function is pretty much identical to
`doRepMax`:

{{< codelines "Haskell" "time-traveling/ValueScore.hs" 26 28 >}}

There's quite a bit of repetition here, especially in the handling
of future values - all of our functions now accept an extra
future argument, and return a work-in-progress future value.
This is what the `Tardis` monad, and its corresponding
`TardisT` monad transformer, aim to address. Just like the
`State` monad helps us avoid writing plumbing code for
forward-traveling values, `Tardis` helps us do the same
for backward-traveling ones.

#### Cycles in Monadic Bind

We've seen that we're able to write code like the following:

```Haskell
(a, b) = f a c
```

That is, we were able to write function calls that referenced
their own return values. What if we try doing this inside
a `do` block? Say, for example, we want to sprinkle some time
traveling into our program, but don't want to add a whole new
transformer into our monad stack. We could write code as follows:

```Haskell
do
    (a, b) <- f a c
    return b
```

Unfortunately, this doesn't work. However, it's entirely
possible to enable this using the `RecursiveDo` language
extension:

```Haskell
{-# LANGUAGE RecursiveDo #-}
```

Then, we can write the above as follows:

```Haskell
do
    rec (a, b) <- f a c
    return b
``` 

This power, however, comes at a price. It's not as straightforward
to build graphs from recursive monadic computations; in fact,
it's not possible in general. The translation of the above
code uses `MonadFix`. A monad that satisfies `MonadFix` has
an operation `mfix`, which is the monadic version of the `fix`
function we saw earlier:

```Haskell
mfix :: Monad m => (a -> m a) -> m a
-- Regular fix, for comparison
fix :: (a -> a) -> a
```

To really understand how the translation works, check out the
[paper on recursive do notation](http://leventerkok.github.io/papers/recdo.pdf).

### Beware The Strictness
Though Csongor points out other problems with the
time traveling approach, I think he doesn't mention
an important idea: you have to be _very_ careful about introducing
strictness into your programs when running time-traveling code.
For example, suppose we wanted to write a function,
`takeUntilMax`, which would return the input list,
cut off after the first occurence of the maximum number.
Following the same strategy, we come up with:

{{< codelines "Haskell" "time-traveling/TakeMax.hs" 1 12 >}}

In short, if we encounter our maximum number, we just return
a list of that maximum number, since we do not want to recurse
further. On the other hand, if we encounter a number that's
_not_ the maximum, we continue our recursion.

Unfortunately, this doesn't work; our program never terminates.
You may be thinking:

> Well, obviously this doesn't work! We didn't actually
compute the maximum number properly, since we stopped
recursing too early. We need to traverse the whole list,
and not just the part before the maximum number.

To address this, we can reformulate our `takeUntilMax`
function as follows:

{{< codelines "Haskell" "time-traveling/TakeMax.hs" 14 21 >}}

Now we definitely compute the maximum correctly! Alas,
this doesn't work either. The issue lies on lines 5 and 18,
more specifically in the comparison `x == m`. Here, we 
are trying to base the decision of what branch to take
on a future value. This is simply impossible; to compute
the value, we need to know the value!

This is no 'silly mistake', either! In complicated programs
that use time traveling, strictness lurks behind every corner.
In my research work, I was at one point inserting a data structure into
a set; however, deep in the structure was a data type containing
a 'future' value, and using the default `Eq` instance!
Adding the data structure to a set ended up invoking `(==)` (or perhaps
some function from the `Ord` typeclass),
which, in turn, tried to compare the lazily evaluated values.
My code therefore didn't terminate, much like `takeUntilMax`.

Debugging time traveling code is, in general,
a pain. This is especially true since future values don't look any different
from regular values. You can see it in the type signatures
of `repMax` and `takeUntilMax`: the maximum number is just an `Int`!
And yet, trying to see what its value is will kill the entire program.
As always, remember Brian W. Kernighan's wise words:

> Debugging is twice as hard as writing the code in the first place.
Therefore, if you write the code as cleverly as possible, you are,
by definition, not smart enough to debug it.

### Conclusion
This is about it! In a way, time traveling can make code performing
certain operations more expressive. Furthermore, even if it's not groundbreaking,
thinking about time traveling is a good exercise to get familiar
with lazy evaluation in general. I hope you found this useful!