Add draft of lazy evaluation post

content/blog/haskell_lazy_evaluation.md

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+---
+title: "Clairvoyance for Good: Using Lazy Evaluation in Haskell"
+date: 2020-05-03T20:05:29-07:00
+tags: ["Haskell"]
+draft: true
+---
+
+While tackling a project for work, I ran across a rather unpleasant problem.
+I don't think it's valuable to go into the specifics here (it's rather
+large and convoluted); however, the outcome of this experience led me to
+discover a very interesting technique for lazy functional languages,
+and I want to share what I learned.
+
+### 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 {linenos=table}
+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
+
+doRepMax :: [Int] -> [Int]
+doRepMax xs = xs'
+  where (largest, xs') = repMax xs largest
+```
+
+In the above snippet, `repMax` expects to receive the maximum value of
+its input list. At the same time, it also computes that maximum value,
+returning it and the newly created list. `doRepMax` is where the magic happens:
+the `where` clauses receives the maximum number from `repMax`, while at the
+same time using that maximum number to call `repMax`!
+
+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 graph. Graphs that
+refer to themselves also violate the properties of a tree.
+{{< /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, unsimplied, in the tree. This is
+what allows us to write the code above: the graph of `repMax xs largest`
+effectively refers to itself. While traversing the list, it places references
+to itself in place of each of the elements, and thanks to laziness, these
+references are not evaluated.
+
+Let's try a more complicated example. How about instead of creating a new list,
+we return a `Map` containing the number of times each number occured, but only
+when those numbers were a factor of the maximum numbers. Our expected output
+will be:
+
+```
+>>> countMaxFactors [1,3,3,9]
+
+fromList [(1, 1), (3, 2), (9, 1)]
+```