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    <p>Here's my attempt. It sounds pretty easy but as usual the devil is in the detail.</p> <p>First, <code>fold</code>:</p> <pre><code> type Matrix = IndexedSeq[IndexedSeq[IndexedSeq[Int]]] def initial(folds: Int): Matrix = { val sideLen = math.pow(2, folds / 2).toInt (1 to sideLen * sideLen) map (Vector(_)) grouped sideLen toIndexedSeq } def leftFold (m: Matrix): Matrix = m map { r =&gt; val (a, b) = r splitAt (r.size / 2) (a.reverse, b).zipped map (_.reverse ++ _) } def rightFold(m: Matrix): Matrix = m map { r =&gt; val (a, b) = r splitAt (r.size / 2) (b.reverse, a).zipped map (_.reverse ++ _) } def topFold (m: Matrix): Matrix = leftFold(m.transpose).transpose def bottomFold(m: Matrix): Matrix = rightFold(m.transpose).transpose def fold(input: String): Seq[Int] = { def go(in: String, m: Matrix): Seq[Int] = in match { case "" =&gt; m(0)(0) case s =&gt; go(s.tail, s.head match { case 'L' =&gt; leftFold(m) case 'R' =&gt; rightFold(m) case 'T' =&gt; topFold(m) case 'B' =&gt; bottomFold(m) }) } go(input, initial(input.length)) } </code></pre> <p>Second, <code>unfold</code>:</p> <pre><code> def unfold(input: Seq[Int]): String = { val m0: Matrix = Vector(Vector(Vector(input: _*))) val sideLen = math.sqrt(input.length).toInt def go(m: Matrix, out: String): String = { val tl = m.head.head val bl = m.last.head val tr = m.head.last if (m.length == sideLen &amp;&amp; m.head.length == sideLen) out.reverse else if (tl.head == tl.last - 1) go(leftUnfold(m), out + 'L') else if (tr.head == tr.last + 1) go(rightUnfold(m), out + 'R') else if (tl.head == tl.last - sideLen) go(topUnfold(m), out + 'T') else if (bl.head == bl.last + sideLen) go(bottomUnfold(m), out + 'B') else sys.error("no fold found " + m) } go(m0, "") } def leftUnfold (m: Matrix): Matrix = m map { r =&gt; val (a, b) = r.map(e =&gt; e.splitAt(e.size / 2)).unzip a.map(_.reverse).reverse ++ b } def rightUnfold(m: Matrix): Matrix = m map { r =&gt; val (a, b) = r.map(e =&gt; e.splitAt(e.size / 2)).unzip b ++ a.map(_.reverse).reverse } def topUnfold (m: Matrix): Matrix = leftUnfold (m.transpose).transpose def bottomUnfold(m: Matrix): Matrix = rightUnfold(m.transpose).transpose </code></pre> <p>I ran a few tests and it passed...</p>
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    1. PO(Un)folding a sheet of paper according to a pattern and giving the order of the layers
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    1. USLuigi Plinge
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    1. COVery elegant solution, I also like the idea of transposing the matrix and applying perpendicular folds. For your unfolding solution, it appears to me that you do not need "b1" and "tr", "tl" is enough.
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      1. USSerkan Eryilmaz
    2. CO@Serkan (tl = top-left, bl = bottom-left, tr = top-right). I did consider that it's probably possible to just look at the top-left Vector[Int] element, but then you need to take into account the current horizontal and vertical lengths in determining what the last Int should be, which I think is more complicated than just using different corners as above. But, if you can improve this I would be interested to see!
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    3. COhttp://ideone.com/n0bsY looks fine to me ;)
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