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For n=1 the problem is trivial, namely S(1,h,t) = "move a disk from rod h to rod t" (there is only one disk left).
The number of moves required by this solution is 2<sup>''n''</sup> − 1. If the objective is to '''maximize''' the number of moves (without cycling) then the dynamic programming [[Bellman equation|functional equation]] is slightly more complicated and 3<sup>''n''</sup> − 1 moves are required.<ref>{{Citation |author=Moshe Sniedovich |title= OR/MS Games: 2. The Towers of Hanoi Problem |journal=INFORMS Transactions on Education |volume=3 |issue=1 |year=2002 |pages=34–51 |url=
=== Egg dropping puzzle ===
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: ''W''(''n'',''k'') = minimum number of trials required to identify the value of the critical floor under the worst-case scenario given that the process is in state ''s'' = (''n'',''k'').
Then it can be shown that<ref name="sniedovich_03">Sniedovich, M. (2003). [
: ''W''(''n'',''k'') = 1 + min{max(''W''(''n'' − 1, ''x'' − 1), ''W''(''n'',''k'' − ''x'')): ''x'' = 1, 2, ..., ''k'' }
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