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    <p>Edit: Nevermind all of this:</p> <p>There's something wrong here, and I can't quite get my head around what exactly you are thinking, so I'll keep reading. Basically, my qualm is that in the second equation for Mk(n) your recursive solution only considers Mk-1(some other n).</p> <p>This could possibly because you do not suggest a concrete ordering, only a runtime ( Mk(n) ). Ideally, you would specify some ordering and then argue on the grounds of a function mapping some your ordering to a runtime. Reading between the lines, your 'ordering' is always of the form move i discs to A, move n - i discs to B, move i discs from A to B. This is exactly the solution to the original problem (M2(n)). Yet you claim faster runtimes? </p> <p>Consider the obvious solution to the case M3(3). One simply spreads the discs out, solving in 5 moves. Your "expanded sum" gives optimally a 7 move solution. I don't think this is quite the error in your logic, but it is the most obvious error in your exposition. Please elaborate what you mean by Mk(n) = Mk-1(i) + 2Mk(n-i). As I understand it, this implies you solve Mk(n) by first moving i discs, then n - i dsics, then i discs again (recursively - this is of course legitimate). The problem is expanding in this way certainly does not represent ever legal solution. It is plain to see this is <strong>not</strong> exactly the order you have in your head. The order you have in your head involves a series of decrasing k values in the recursion.</p> <p>Look at M4(4) to see the difference. The expansion you have shown with Mk-1 gives a similar solution to M2(n). Now consider the solution you want to have written, spreading them out. after the first two discs have been moved, the next discs have fewer places to go. Specifically their movements is an instance of the M2(2) problem. Read that carefully. <strong>2</strong>. Not k-1 = 3.</p>
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