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copied!<p>I saw your previous question on the same topic, and I start to work on it.<br> As you doesn't want the direct solution, I can provide you my reflexion about your problem, I guess it could help you. </p> <p>Some basic property :<br> 1. The number of movement is alway egal to the length of the list <strong>m = length A</strong><br> 2. The number of starting point is egal to the length of the head of the list <strong>n = length (head A)</strong><br> 3. The current position could never be negative, then :<br> - if the current position is egal to 0 you can either go down or right<br> - else you can go to left, down or right </p> <p>Which lead us to this pseudo code </p> <pre><code>generate_path :: [[Int]] -> [[Int]] generate_path [] = [[]] generate_path A = ... -- You have to put something here where m = length A n = length (head A) </code></pre> <p>This things should look like something as this </p> <pre><code>move pos0 count0 | count0 == 0 = | pos0 == 0 = move (down count) ++ move (right count) | otherwise = move (left count) ++ move (down count) ++ move (right count) where count = count0 - 1 down = position0 left = position0 - 1 right = position0 + 1 </code></pre> <p>In fact keeping all of this in mind and adding the <strong>(!!)</strong> operator, we shouldn't be so far of the solution. To convince you play with <strong>A + list comprehension + !!</strong>, as </p> <pre><code>[A !! x !! y | x <- [1..2], y <- [0..2]] -- I take random range </code></pre> <p>Or play with another version : </p> <pre><code>[[A !! x !! y | x <- [1..2]] | y <- [0..2]]] -- I take random range </code></pre> <p>In fact you have two recursion the main one working on the parameter n = length (head A), you repeat the same action from 0 to (n-1) at (n-1) retrieve the result, this recursion embedded another one which work on m, repeat the same action from 0 to (m-1). </p> <p>Hope it help. Good luck.</p>

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