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    <p>Here is what I understand so far: The algorithm is to process a sequence of values {R,B}. It may choose to put the value in the current container or the next, if there is a next.</p> <p>I first would ask a couple of questions to clarify the things I don't know yet:</p> <ul> <li><p>Is k and n known to the algorithm in advance? I assume so. </p></li> <li><p>Do we know the full sequence of buttons in advance?</p></li> <li><p>If we don't know the sequence in advance, should the average value minimized? Or the maximum (the worst case)?</p></li> </ul> <p>Idea for a proof for the algortihm by Mark Peters</p> <p>Edit: Idea for a proof (sorry, couldn't fit it in a comment)</p> <p>Let L(i) be the length of the ith group. Let d(i) be the diff you get by collapsing container i and i+1 => d(i) = L(i)*L(i+1).</p> <p>We can define a distribution by the sequence of containers collapsed. As index we use the maximum index of the original containers contained in the collapsed container containing the containers with the smaller indexes.</p> <p>A given sequence of collapses I = [i(1), .. i(m)] results in a value which has a lower bound equal to the sum of d(i(m)) for all m from 1 to n-k.</p> <p>We need to proof that there can't be a sequence other then the one created by the algorithm with a smaller diff. So let the sequence above be the one resulting from the algorithm. Let J = [j(1), .. j(m)]. </p> <p>Here it gets skimpy: I think it should be possible to proof that the lower limit of J is larger then the actual value of I because in each step we choose by construction the collapse operation from I so it must be smaller then the matching collapse from the alternate sequence</p> <p>I think we might assume that the sequences are disjunct, but I'm not completely sure about it.</p>
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