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One of the things I learned by reading AIM 239 is the Game of Life and Cellular Automata. One particular kind of one dimensional cellular automata, Rule 110 popped by my twitter stream the other day, so I thought I could try and code it with the minimal Haskell subset that I can handle.

Rule 110 is special because it is proven to be able to simulate a Turing machine. Head over to its Wikipedia page if you want to learn more about the proof and the interesting story around it.

Rule 110 starts with a string of zeros and ones and a transition table that decides the next state of the automaton. If you put each line of the strings after the other, interesting patterns can emerge. Let’s see the transition state:

Current pattern 111 110 101 100 011 010 001 000
New state for center cell 0 1 1 0 1 1 1 0

If you look closely, you can use a list of eight digits and its index in order to encode the above state transitions:

```rule110 = [
0, -- ((0,0,0), 0)
1, -- ((0,0,1), 1)
1, -- ((0,1,0), 1)
1, -- ((0,1,1), 1)
0, -- ((1,0,0), 0)
1, -- ((1,0,1), 1)
1, -- ((1,1,0), 1)
0  -- ((1,1,1), 0)
] :: [Int]
```

But what about the transitions of the leftmost and rightmost digit you might think. Let’s assume that their missing neighbor is zero. Therefore, given an initial state and a rule that governs the transitions, we may calculate the next state with:

```nextState :: [Int] -> [Int] -> [Int]
nextState state rule =
[ rule !! x |
let t = [0] ++ state ++ [0],
i <- [1..(length(t)-2)],
let x = (t !! (i-1)) * 4 + (t !! i) * 2 + (t !! (i+1))
]

-- construct an infinite sequence of next states
sequenceState :: [Int] -> [Int] -> [[Int]]
sequenceState state rule =
[state] ++ sequenceState (nextState state rule) rule
```

Example:

```*Main> state = [0,1,1,0]
*Main> nextState state rule110
[1,1,1,0]
```

One of the most interesting patterns occurs when we begin with the right most digit being 1 and all the rest being zeros:

```*Main> state = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1] :: [Int]
*Main> x = take 30 \$ sequenceState state rule110
*Main> showState x
*
**
***
** *
*****
**   *
***  **
** * ***
******* *
**     ***
***    ** *
** *   *****
*****  **   *
**   * ***  **
***  **** * ***
** * **  ***** *
******** **   ***
**      ****  ** *
***     **  * *****
** *    *** ****   *
*****   ** ***  *  **
**   *  ***** * ** ***
***  ** **   ******** *
** * ******  **      ***
*******    * ***     ** *
**     *   **** *    *****
***    **  **  ***   **   *
** *   *** *** ** *  ***  **
*****  ** *** ****** ** * ***
**   * ***** ***    ******** *
*Main>
```

The output was somehow pretty printed:

```showState [] = return ()
showState state = do
-- putStrLn \$ show (state !! 0)
putStrLn \$ [ c | d <- (state !! 0), let c = if d == 0 then ' ' else '*' ]
showState \$ tail state
```

I wish I can find time and play more with cellular automata. I kind of find a day every five years or so.

Update: Here is a pattern using Rule 90:

```                             *
*
* *
*
* *
*   *
* * * *
*
* *
*   *
* * * *
*       *
* *     * *
*   *   *   *
* * * * * * * *
*
* *
*   *
* * * *
*       *
* *     * *
*   *   *   *
* * * * * * * *
*               *
* *             * *
*   *           *   *
* * * *         * * * *
*       *       *       *
* *     * *     * *     * *
*   *   *   *   *   *   *   *
```

WordPress.com does not always render Markadown properly, so a copy of this post resides here.

I am coninuing my adventure in Haskell. In order to make it a bit more fun, I decided to code a simple yet very intriguing problem, I first heard of when I read AI Memo 239: The Collatz conjecture.

Construct a sequence of integers where given an arbitrary interger the value of the next is:
* If the number is even, divide it by two.
* If the number is odd, triple it and add one.

This can easily be coded in Haskell as follows:

```collatz :: Int -> Int
collatz 1 = 1
collatz n =
if (even n)
then (n `div` 2)
else (3 * n + 1)
```

But how can one obtain a sequence of numbers from this? A very clever solution is here where the author implements a variation of takeWhile which includes also the first list item that fails the test the first time. So my question became, can it be done in another way? Yes it can:

```collatzSequence :: Int -> [Int]
collatzSequence n =
if n == 1
then [1]
else [n] ++ collatzSequence (collatz n)
```

Let's see some test runs:

```*Main> collatzSequence 5
[5,16,8,4,2,1]
*Main> collatzSequence 50
[50,25,76,38,19,58,29,88,44,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1]
*Main> collatzSequence 500
[500,250,125,376,188,94,47,142,71,214,107,322,161,484,242,121,364,182,91,274,137,412,206,103,310,155,466,233,700,350,175,526,263,790,395,1186,593,1780,890,445,1336,668,334,167,502,251,754,377,1132,566,283,850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,7288,3644,1822,911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106,53,160,80,40,20,10,5,16,8,4,2,1]
*Main> collatzSequence 512
[512,256,128,64,32,16,8,4,2,1]
*Main> collatzSequence 513
[513,1540,770,385,1156,578,289,868,434,217,652,326,163,490,245,736,368,184,92,46,23,70,35,106,53,160,80,40,20,10,5,16,8,4,2,1]
*Main>
```

You may have observed we only run it on positive integers. When we run it with negative integers, there are a few more cycles that we need to take into account. Here is the updated sequence function, written with guards:

```collatzSequence :: Int -> [Int]
collatzSequence n
| n == 1 = [1]
| n == (-2) = [(-2)]
| n == (-5) = [(-5)]
| n == (-17) = [(-17)]
| otherwise = [n] ++ collatzSequence (collatz n)
```

Test:

```*Main> collatzSequence (-50)
[-50,-25,-74,-37,-110,-55,-164,-82,-41,-122,-61,-182,-91,-272,-136,-68,-34,-17]
```

Now if there was also a way to prove the conjecture…

Please note that in Haskell the unary minus is a function and not an operator, hence, you need to parenthesize. This also works:

```*Main> collatzSequence \$ -50
[-50,-25,-74,-37,-110,-55,-164,-82,-41,-122,-61,-182,-91,-272,-136,-68,-34,-17]
```

Update: A friend posted me his own elegant version of the Collatz sequence:

```collatz :: Int -> [Int]
collatz 1 = [1]
collatz n
| even n =  n : collatz (n `div` 2)
| odd n  =  n : collatz (n * 3 + 1)

main = do
putStrLn \$ show \$ collatz 1
putStrLn \$ show \$ collatz 6
putStrLn \$ show \$ collatz 23
```

a newbie does list comprehensions

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The year is 1998 and @mtheofy then at Glasgow tells me about a relatively new (then) language called Haskell. I’m intrigued but do not do much. A few years later I buy The Haskell School of Expression since The Craft of Functional Programming did not seem enough to motivate me. Time passes and around 2007 I try yet another start. Nothing. I promised my self yet another restart for a 2017 new year’s resolution. Still nothing. So when the current employer offered Haskell classes I could not say no. Armed with the weekly classes and a Safari Learning Path I am trying to correct this. And I am having some fun with list comprehensions. Because as a friend says, if it makes you feel good, go.

So how do you write an infinite list? Let’s say you want list x to include all numbers from 0 to infinity. stack ghci is my friend. Others might try repl.it:

```x = [ n | n <- [0..]]
```

Now you can have the first 20 items of x:

```Prelude> x = [ n | n <- [0..]]
Prelude> take 20 x
[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
Prelude>
```

So next I wanted to make an infinite list of the same character. Enter the underscore variable:

```Prelude> x = [ 'a' | _ <- [0..]]
Prelude> take 20 x
"aaaaaaaaaaaaaaaaaaaa"
Prelude>
```

OK, so now let’s try to cycle infinitely characters from a string. I end up with:

```Prelude> x = [ c | i  take 20 x
"abcdabcdabcdabcdabcd"
Prelude>
```

I am kind of unsure why the let statements are needed since I am ~10 days into typing stuff and posted my creation to twitter. What my expression says is that x is comprised of characters from string “abcd”, where given a sequence of numbers, each time a character is chosen based on the sequence number modulo 4. Strings are lists of characters in Haskell and list indexing starts from zero.  Helpful comments come my way. Like the obvious cycle (there is a cycle function? Yes ):

```Prelude> take 20 (cycle "abcd")
"abcdabcdabcdabcdabcd"
Prelude> take 20 \$ cycle "abcd"
"abcdabcdabcdabcdabcd"
Prelude>
```

Is not the dollar operator nice to get rid of parentheses? Here is another suggestion about cycling a string:

```Prelude> x = [ "abcd" !! (i `mod` 4) | i  take 20 x
"abcdabcdabcdabcdabcd"
Prelude>
```

This one is more concise and does the same thing, always picking a character from "abcd". If the infix notation for mod confuses you, you can:

```Prelude> x = [ "abcd" !! (mod i 4) | i  take 20 x
"abcdabcdabcdabcdabcd"
Prelude>
```

But the Internet does not stop there. It comes back with more helpful suggestions:

Welcome! A little feedback then if I may: the `!!` operator should be used VERY cautiously it is not typesafe and lists are not random access anyway. Opt for a function returning `Maybe x` and for a random access datastructure (strings are by default lists).

Which made me think: How about an infinite string randomly chosen from “abcd”?

```\$ stack install random
\$ stack ghci
:
Prelude> import System.Random
Prelude System.Random> g <- newStdGen
Prelude System.Random> x = [ "abcd" !! i | i <- randomRs (0,3) g ]
Prelude System.Random> take 10 x
"bcbbddcdab"
Prelude System.Random>
```

If you want a sequence with a different order, you need to reinitialise both g and x. I will figure out a better way some other time when …I have time.

Adventures with Maybe maybe in another post.

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