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    <p>I liked the task, so I wrote an alternative solution.</p> <p>We could build it up, a bit like you would with a pretty printer. Look into the <a href="http://hackage.haskell.org/package/pretty/" rel="noreferrer">pretty</a> package to take these ideas and use them properly, but let's stick to plain old <code>[String]</code> for this.</p> <p>First let's make a blank grid</p> <pre><code>blank :: Int -&gt; [String] blank n = replicate (3^n) $ replicate (3^n) ' ' </code></pre> <p>Then let's define a diamond.</p> <pre><code>diamond :: Int -&gt; [String] diamond 0 = ["*"] diamond n = let o = diamond (n-1) x = blank (n-1) in joinMatrix [[x,o,x] ,[o,x,o] ,[x,o,x]] </code></pre> <p>But how can we join this matrix of <code>[String]</code> together? First get all the <code>String</code>s that should be concatenated together next to each other instead of under each other using <code>transpose</code>, then <code>concat</code> them all:</p> <pre><code>joinLine :: [[String]] -&gt; [String] joinLine = map concat.transpose </code></pre> <p>To do that to a whole matrix we need to join the lines on each row, then concat all the lines together into one list of lines:</p> <pre><code>joinMatrix :: [[[String]]] -&gt; [String] joinMatrix = concat.map joinLine </code></pre> <p>helper functions for printing:</p> <pre><code>put = mapM_ putStrLn d n = put $ diamond n </code></pre> <p>You could argue that the numerical solution is more efficient, and it is, but <code>d 4</code> is the largest that fits on my screen and isn't slow. You could also argue that this solution is clearer.</p> <pre><code>*Main&gt; d 0 * *Main&gt; d 1 * * * * *Main&gt; d 2 * * * * * * * * * * * * * * * * </code></pre> <p>(It works for higher n too, but they would make this post unnecessarily long on the page.)</p>
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    1. CODon't forget to vote for cebewee's [good answer](http://stackoverflow.com/questions/13778632/print-diamond-pattern-using-haskell/13779048#13779048)!
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    2. COThanks :) I very much like how in your approach, you only need to encode the diamond once, whereas in my approach, there are two diamonds: Once in `next 0`, and once in `next n`.
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      1. USLars Noschinski
 

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