# metaperl's blog

## Why is Clean faster than Haskell?

If you compare Clean and Haskell in the Debian shootout, Clean is consistently faster than Haskell. These languages are spitting images of each other, save that Clean has some way of avoiding Monads. In addition, Clean I believe is based on graphs instead of lambda calculus - but both of these compute the same things so why would that matter?

The summary table (at bottom, horribly rendered in Firefox), shows Clean winning 10-0.

Also, which language came first? Clean or Haskell?

## that's enough of SJT

i'm tired of beating around the bush. i have played with not being able to think long enough.

learning haskell is not an obligation for me. it is a hobby. i can't ever expect it to compete with C++ in terms of popularity. I chose haskell because it forces you to think clearly. I need to see this as a hobby which might become a profitable job one day... Perl did. Haskell might as well. Haskell is gaining corporate attention. Microsoft has always had its eye on it.

better to master something I like than to keep rushing through this book. Funny "The Haskell Road..." does not cover monads, but I feel confident I will be able to handle them if/when I make it through Doets/van Eijck's book.

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## Someone help me. The Haskell Road is Calling...

For some reason, I keep picking up "The Haskell Road to Logic, Maths and Programming"

All my life I have avoided proofs and mathematics, but perhaps now is the moment to meet my Maker. I keep going back to this book.

For some reason, it is infectious. A beginner introduction to how thought is structured. A beginner introduction to what math is all about.

I had been wondering what I was going to do with Haskell once I finished learning it. I am a Perl professional by trade and even though some interplay has been going on, I never really knew why I was learning Haskell. I just liked the conciseness and elegance of the language and the mathematical purity.

The most important thing is to have passion behind what you do. If I want to go deeper into math for math's sake, then so be it... no need to artificially create projects for myself. Just enjoy haskell and math. It is a very very good fit.

## The most important thing you can know about Haskell and Functional Programming

The Haskell equations `f x y = ...`

used in the definition
of a function `f`

are genuine mathematical equations. They
state that the left hand side of the right hand side of the equation
have the same value. This is *very* different from the use of
`=`

in imperative languages like C or Java. In a C or Java
program the statement `x = x*y`

does *not mean* that
`x`

and `x*y`

have the same value, but rather it
is a command to throw away the old value of `x`

and put the
value of `x*y`

in its place. It is a so-called
*destructive assignment statement:* the old value of a variable
is destroyed and replaced by a new one.

Reasoning about Haskell definitions is a lot easier than reasoning
about programs that use destructive assignment. In Haskell, standard
reasoning about mathematical equations applies. E.g. after the Haskell
declarations `x= 1`

and `y = 2`

, the Haskell
declaration `x = x + y`

will raise an error ```
"x"
multiply defined
```

. **... = in Haskell has
the meaning "is by definition equal to"...**

This was a huge landslide victory for me. Because I quit trying to write programs to get data here, data there. Values here, values there. Instead, I simply began to rewrite the original function as a new definition.

I became so confident that I was able to write a program to return all the leaves of a tree. and here it is:

Looking back at "Algorithms in Haskell" by Rabhi and "Craft of FP" by Simon Thompson, they do both make this same statement, but somehow it never really hit me right.

## How to create an infinite lazy list of function applications?

I would like to create an infinite lazy list whose contents consist of:

[ x, f x, f (f x), f (f (f x)) ... ] -- ad infinitum.

How might I do this?

## overlapping pattern matching instances: help needed

I am being told that I have overlapping pattern matches. I'm pretty sure it has to do with the `otherwise`

clause and the following `subString (x:xs) []`

clause, but I don't know how to fix it

```
```prefix [] ys = True

prefix xs [] = False

prefix [x] [y] = (x == y)

prefix (x:xs) (y:ys) = (x == y) && prefix xs ys

subString xs ys

| prefix xs ys = True

| otherwise = subString xs ys'

where ys' = tail ys

subString (x:xs) [] = False

## Haskell Bookmarks

- Web Apps Framework

- cgi and database options for Haskell.

Drupal said I needed 10 words, so here are some more.

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## I am confused on the implementation of (\\)

From the Haskell Online Report we have:

(\\) :: Eq a => [a] -> [a] -> [a]

(\\) = foldl (flip delete)

The `(flip delete)`

changes the type signature to the following:

*Main> :t (flip delete)

forall a. (Eq a) => [a] -> a -> [a]

and `foldl`

**usually** is obvious: it folds the list from the left, or the beginning. Meaning, it takes the seed element and the first element of the list and produces a result. This result is the new seed to be used with the next element of the list and so on until a result is produced.

But what is the flipped version of foldl doing in this case? It receives an entire list in the position it normally just receives an element.

## Recursion and strong static typing can help you build functions!

There is a synergy between recursion and strong static typing that I never

realized and here it is:

If you express problem solution recursively, every choice made on every

branch of the recursion at any time in the problem space has to match

the type signature of the promised outcome.

Take this function to delete a node from a tree for example:

delete :: Ord a => a -> Tree a -> Tree a

So one thing we know for dead certain, is that delete will be returning a

Tree. So, if we code delete with a number of cases for different scenarios

each of those cases **must** return a tree.

Given this, one way to try to find the solution for this is take a look at

all the possible trees that you know about and ask when they would apply as

a solution. It kind of gets you going towards thinking about the mechanics

of the problem. First let's rewrite the type signature as the function head:

delete val (Node v t1 t2)

and now let's ask the pivotal question once again: "what sorts of trees can

I think of?" Well, we are looking at two of them in the second argument

to delete() - t1 and t2. When is t1 the solution to this problem? Well,

that's the same as asking when would we only return the left subtree after hitt\ing a node. And when would we do that? Well, the answer to that is:

"if val == v and there is no left subtree, then we

simply need to return t1." Nice! Codewise, it looks like this:

` (val == v) && (isNil t2) = t1`

Now of course we have the "mirror" situation

of when do we only return t2? And the answer is very similar to the other one:

` (val == v) && (isNil t1) = t2`

Ok, so we've nailed down two cases just by taking a look at the head

of the rule. So, let's see both of those cases had to do with being at a

node where val == v. The only difference was that the left or right subtree

was not there.

So obviously, we want to start thinking about when val == v and both

subtrees are there. Well in that case, we need to remove the root node and

join the 2 subtrees, maintaining the required binary tree ordering. So

that clause looks like this, assuming we use the fallback properties of

guards:

` otherwise = join t1 t2`

So we have thorougly cased out all of the scenarios where we have reached

the node of interest. What happens in the current node has a value greater

than the one we want to delete? Well, the node we are looking for must

exist down the left subtree:

` val < v = Node v (delete val t1) t2`

and conversely, if the current node's value is less than the one we are

looking for then we need to try to delete down the right subtree:

` val > v = Node v t1 (delete val t2)`

So now that we have cased all deletion scenarios out, we have our

final delete function:

```
```delete val (Node v t1 t2)

| val < v = Node v (delete val t1) t2

| val > v = Node v t1 (delete val t2)

| isNil t2 = t1

| isNil t1 = t2

| otherwise = join t1 t2

**conclusion**

SJT says that one should start with an informal description of a problem, then come up with your types. I add to that: once you get your types, then create type signatures for your functions and then once you get your type signatures, build the head of your function with pattern matching and then use the required types to express various cases.

## SJT's book has a limited definition of relation... why?

In 16.9 of SJT's book "Craft of Functional Programming" he says "A binary relation relates together certain elements of a set"

My problem with this is that the elements do not have to be in the same set. A relation is a subset of the cross product of **sets** A and B, where A and B are not of the same type.

Next he gives a type definition

`type Relation a = Set (a,a)`

And I think it should be:

`type Relation a,b = Set (a,b)`