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    <pre><code>change [] = 1 : repeat 0 change (d : ds) = c where (a, b) = splitAt d (change ds) c = a ++ zipWith (+) b c </code></pre> <p>Then,</p> <pre><code>result = (!! 100) $ xs where xs = change [1, 5, 10, 15, 20, 25, 50] = let -- g = (\(a,b)-&gt; fix ((a++) . zipWith (+) b)) g (a,b) = let c = a ++ zipWith (+) b c in c in g . splitAt 1 . change $ [5, 10, 15, 20, 25, 50] = g . splitAt 1 . g . splitAt 5 . change $ [10, 15, 20, 25, 50] = .... = let h n = g . splitAt n in h 1 . h 5 . h 10 . h 15 . h 20 . h 25 . h 50 . (1:) . repeat $ 0 </code></pre> <p>or, simpler,</p> <pre><code>Prelude&gt; (!! 100) $ foldr h (1:cycle [0]) [1, 5, 10, 15, 20, 25, 50] 1239 </code></pre> <p>(which is a correct answer, BTW). This is arguably easier to comprehend. Your question is thus localized to the <code>g</code> definition,</p> <pre><code> g (a,b) = let c = a ++ zipWith (+) b c in c </code></pre> <p>The thing about Haskell's definitions is that they are <em>recursive</em> (they are equivalent to Scheme's <code>letrec</code>, not <code>let</code>). </p> <p>Here it works, because when <code>c</code> is <em>lazily</em> consumed, it's definition says it's built from two parts, <code>a ++ ...</code> and so first <code>a</code> is consumed. And this works because <code>a</code> does not depend on <code>c</code>. Calculating <code>a</code> does not demand any knowledge of <code>c</code>.</p> <p>In <code>zipWith (+) b c</code>, <code>c</code> is essentially a <strong><em>pointer</em></strong> into the sequence being defined, <code>length a</code> notches <strong><em>back</em></strong> from the production point, <code>rest</code>, in this re-write:</p> <pre><code> g (a,b) = let c = a ++ rest rest = zipWith (+) b c in c </code></pre> <p>We have <code>h n xs = g (splitAt n xs)</code> and this is describing then the sum of the input list with the result, moved <code>n</code> notches forward:</p> <pre><code> x1 x2 x3 x4 x5 x6 x7 x8 ................ xs A y1 y2 y3 y4 y5 .......... ys B -------- y1 y2 y3 y4 y5 y6 y7.................... ys == A + B </code></pre> <p>This suggests <code>h</code> can be re-written with improved locality of access,</p> <pre><code>change ds n = foldr h (1:cycle [0]) ds !! n -- [1, 5, 10, 15, 20, 25, 50] 100 where h n xs = ys where ys = zipWith (+) xs (replicate n 0 ++ ys) -- = fix (zipWith (+) xs . (replicate n 0 ++)) </code></pre>
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