Editing Tower of Hanoi (section)

=== Recursive solution ===
{{More citations needed section|date=January 2024}}
[[File:Tower of Hanoi recursion SMIL.svg|thumb|link={{filepath:Tower_of_Hanoi_recursion_SMIL.svg}}|Illustration of a recursive solution for the Towers of Hanoi puzzle with 4 disks. In [{{filepath:Tower_of_Hanoi_recursion_SMIL.svg}} the SVG file,] click a grey button to expand or collapse it]]
The key to solving a problem [[recursive]]ly is to recognize that it can be broken down into a collection of smaller sub-problems, to each of which ''that same general solving procedure that we are seeking'' applies{{Citation needed|date=January 2024}}, and the total solution is then found in some ''simple'' way from those sub-problems' solutions. Each of these created sub-problems being "smaller" guarantees that the base case(s) will eventually be reached. For the Towers of Hanoi:
* label the pegs A, B, C,
* let ''n'' be the total number of disks, and
* number the disks from 1 (smallest, topmost) to ''n'' (largest, bottom-most).

Assuming all ''n'' disks are distributed in valid arrangements among the pegs; assuming there are ''m'' top disks on a ''source'' peg, and all the rest of the disks are larger than ''m'', so they can be safely ignored; to move ''m'' disks from a source peg to a ''target'' peg using a ''spare'' peg, without violating the rules:
# Move ''m'' − 1 disks from the '''source''' to the '''spare''' peg, by ''the same general solving procedure''. Rules are not violated, by assumption. This leaves the disk ''m'' as a top disk on the source peg.
# Move the disk ''m'' from the '''source''' to the '''target''' peg, which is guaranteed to be a valid move, by the assumptions — ''a simple step''.
# Move the ''m'' − 1 disks that we have just placed on the spare, from the '''spare''' to the '''target''' peg by ''the same general solving procedure'', so they are placed on top of the disk ''m'' without violating the rules.
# The base case is to move ''0'' disks (in steps 1 and 3), that is, do nothing—which does not violate the rules.

The full Tower of Hanoi solution then moves ''n'' disks from the source peg A to the target peg C, using B as the spare peg.

This approach can be given a rigorous mathematical proof with [[mathematical induction]] and is often used as an example of recursion when teaching programming.

==== Logical analysis of the recursive solution ====
{{More citations needed section|date=January 2024}}
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As in many mathematical puzzles, finding a solution is made easier by solving a slightly more general problem: how to move a tower of ''h'' (height) disks from a starting peg ''f'' = '''A''' (from) onto a destination peg ''t'' = '''C''' (to), '''B''' being the remaining third peg and assuming ''t'' ≠ ''f''. First, observe that the problem is symmetric for permutations of the names of the pegs ([[Symmetric group|symmetric group ''S''<sub>3</sub>]]). If a solution is known moving from peg '''A''' to peg '''C''', then, by renaming the pegs, the same solution can be used for every other choice of starting and destination peg. If there is only one disk (or even none at all), the problem is trivial. If ''h'' = 1, then move the disk from peg '''A''' to peg '''C'''. If ''h'' > 1, then somewhere along the sequence of moves, the largest disk must be moved from peg '''A''' to another peg, preferably to peg '''C'''. The only situation that allows this move is when all smaller ''h'' − 1 disks are on peg '''B'''. Hence, first all ''h'' − 1 smaller disks must go from '''A''' to '''B'''. Then move the largest disk and finally move the ''h'' − 1 smaller disks from peg '''B''' to peg '''C'''. The presence of the largest disk does not impede any move of the ''h'' − 1 smaller disks and can be temporarily ignored. Now the problem is reduced to moving ''h'' − 1 disks from one peg to another one, first from '''A''' to '''B''' and subsequently from '''B''' to '''C''', but the same method can be used both times by renaming the pegs. The same strategy can be used to reduce the ''h'' − 1 problem to ''h'' − 2, ''h'' − 3, and so on until only one disk is left. This is called recursion. This algorithm can be schematized as follows.

Identify the disks in order of increasing size by the natural numbers from 0 up to but not including ''h''. Hence disk 0 is the smallest one, and disk ''h'' − 1 the largest one.

The following is a procedure for moving a tower of ''h'' disks from a peg '''A''' onto a peg '''C''', with '''B''' being the remaining third peg:
# If ''h'' > 1, then first use this procedure to move the ''h'' − 1 smaller disks from peg '''A''' to peg '''B'''.
# Now the largest disk, i.e. disk ''h'' can be moved from peg '''A''' to peg '''C'''.
# If ''h'' > 1, then again use this procedure to move the ''h'' − 1 smaller disks from peg '''B''' to peg '''C'''.

By [[mathematical induction]], it is easily proven that the above procedure requires the minimum number of moves possible and that the produced solution is the only one with this minimal number of moves. Using [[recurrence relation]]s, the exact number of moves that this solution requires can be calculated by: <math>2^h - 1</math>. This result is obtained by noting that steps 1 and 3 take <math>T_{h-1}</math> moves, and step 2 takes one move, giving <math>T_h = 2T_{h-1} + 1</math>.

==== Non-recursive solution ====
{{More citations needed section|date=January 2024}}
The list of moves for a tower being carried from one peg onto another one, as produced by the recursive algorithm, has many regularities. When counting the moves starting from 1, the ordinal of the disk to be moved during move ''m'' is the number of times ''m'' can be divided by 2. Hence every odd move involves the smallest disk. It can also be observed that the smallest disk traverses the pegs ''f'', ''t'', ''r'', ''f'', ''t'', ''r'', etc. for odd height of the tower and traverses the pegs ''f'', ''r'', ''t'', ''f'', ''r'', ''t'', etc. for even height of the tower. This provides the following algorithm, which is easier, carried out by hand, than the recursive algorithm.

In alternate moves:
* Move the smallest disk to the peg it has not recently come from.
* Move another disk legally (there will be only one possibility).
For the very first move, the smallest disk goes to peg ''t'' if ''h'' is odd and to peg ''r'' if ''h'' is even.

Also observe that:
* Disks whose ordinals have even parity move in the same sense as the smallest disk.
* Disks whose ordinals have odd parity move in opposite sense.
* If ''h'' is even, the remaining third peg during successive moves is ''t'', ''r'', ''f'', ''t'', ''r'', ''f'', etc.
* If ''h'' is odd, the remaining third peg during successive moves is ''r'', ''t'', ''f'', ''r'', ''t'', ''f'', etc.

With this knowledge, a set of disks in the middle of an optimal solution can be recovered with no more state information than the positions of each disk:
* Call the moves detailed above a disk's "natural" move.
* Examine the smallest top disk that is not disk 0, and note what its only (legal) move would be: if there is no such disk, then we are either at the first or last move.
* If that move is the disk's "natural" move, then the disk has not been moved since the last disk 0 move, and that move should be taken.
* If that move is not the disk's "natural" move, then move disk 0.