Tower of Hanoi

Animation of an iterative algorithm solving 6-disk problem

A simple solution for the toy puzzle is to alternate moves between the smallest piece and a non-smallest piece. When moving the smallest piece, always move it to the next position in the same direction (to the right if the starting number of pieces is even, to the left if the starting number of pieces is odd). If there is no tower position in the chosen direction, move the piece to the opposite end, but then continue to move in the correct direction. For example, if you started with three pieces, you would move the smallest piece to the opposite end, then continue in the left direction after that. When the turn is to move the non-smallest piece, there is only one legal move. Doing this will complete the puzzle in the fewest moves.[8]

For an even number of disks:

• make the legal move between pegs A and B (in either direction),
• make the legal move between pegs A and C (in either direction),
• make the legal move between pegs B and C (in either direction),
• repeat until complete.

For an odd number of disks:

• make the legal move between pegs A and C (in either direction),
• make the legal move between pegs A and B (in either direction),
• make the legal move between pegs B and C (in either direction),
• repeat until complete.

In each case, a total of 2ⁿ − 1 moves are made.

Another way to generate the unique optimal iterative solution:

Number the disks 1 through n (largest to smallest).

• If n is odd, the first move is from peg A to peg C.
• If n is even, the first move is from peg A to peg B.

Now, add these constraints:

• No odd disk may be placed directly on an odd disk.
• No even disk may be placed directly on an even disk.
• There will sometimes be two possible pegs: one will have disks, and the other will be empty. Place the disk on the non-empty peg.
• Never move a disk twice in succession.

Considering those constraints after the first move, there is only one legal move at every subsequent turn.

The sequence of these unique moves is an optimal solution to the problem equivalent to the iterative solution described above.[9]

Illustration of a recursive solution for the Towers of Hanoi puzzle with 4 disks. In the SVG file, click a grey button to expand or collapse it

The key to solving a problem recursively 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, 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. Thence, for the Towers of Hanoi:

• label the pegs A, B, C,
• let n be the total number of disks,
• 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:

1. 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.
2. 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.
3. 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.
4. The base case is to move 0 disks (in steps 1 and 3), that is, do nothing – which obviously doesn't violate the rules.

The full Tower of Hanoi solution then consists of moving 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.

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 S₃). 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 simply 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:

1. If h > 1, then first use this procedure to move the h − 1 smaller disks from peg A to peg B.
2. Now the largest disk, i.e. disk h can be moved from peg A to peg C.
3. If h > 1, then again use this procedure to move the h − 1 smaller disks from peg B to peg C.

By means of mathematical induction, it is easily proven that the above procedure requires the minimal number of moves possible and that the produced solution is the only one with this minimal number of moves. Using recurrence relations, the exact number of moves that this solution requires can be calculated by: ${\displaystyle 2^{h}-1}$. This result is obtained by noting that steps 1 and 3 take ${\displaystyle T_{h-1}}$ moves, and step 2 takes one move, giving ${\displaystyle T_{h}=2T_{h-1}+1}$.

The following Python code demonstrates the recursive solution.

A = [3, 2, 1]
B = []
C = []

def move(n, source, target, auxiliary):
    if n > 0:
        # Move n - 1 disks from source to auxiliary, so they are out of the way
        move(n - 1, source, auxiliary, target)

        # Move the nth disk from source to target
        target.append(source.pop())

        # Display our progress
        print(A, B, C, '##############', sep='\n')

        # Move the n - 1 disks that we left on auxiliary onto target
        move(n - 1, auxiliary, target, source)

# Initiate call from source A to target C with auxiliary B
move(3, A, C, B)

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.

Disk positions may be determined more directly from the binary (base-2) representation of the move number (the initial state being move #0, with all digits 0, and the final state being with all digits 1), using the following rules:

• There is one binary digit (bit) for each disk.
• The most significant (leftmost) bit represents the largest disk. A value of 0 indicates that the largest disk is on the initial peg, while a 1 indicates that it's on the final peg (right peg if number of disks is odd and middle peg otherwise).
• The bitstring is read from left to right, and each bit can be used to determine the location of the corresponding disk.
• A bit with the same value as the previous one means that the corresponding disk is stacked on top of the previous disk on the same peg.

  (That is to say: a straight sequence of 1s or 0s means that the corresponding disks are all on the same peg.)

• A bit with a different value to the previous one means that the corresponding disk is one position to the left or right of the previous one. Whether it is left or right is determined by this rule:
  • Assume that the initial peg is on the left.
  • Also assume "wrapping" – so the right peg counts as one peg "left" of the left peg, and vice versa.
  • Let n be the number of greater disks that are located on the same peg as their first greater disk and add 1 if the largest disk is on the left peg. If n is even, the disk is located one peg to the right, if n is odd, the disk located one peg to the left (in case of even number of disks and vice versa otherwise).

For example, in an 8-disk Hanoi:

• Move 0 = 00000000.
  • The largest disk is 0, so it is on the left (initial) peg.
  • All other disks are 0 as well, so they are stacked on top of it. Hence all disks are on the initial peg.
• Move 2⁸ − 1 = 11111111.
  • The largest disk is 1, so it is on the middle (final) peg.
  • All other disks are 1 as well, so they are stacked on top of it. Hence all disks are on the final peg and the puzzle is complete.
• Move 216₁₀ = 11011000.
  • The largest disk is 1, so it is on the middle (final) peg.
  • Disk two is also 1, so it is stacked on top of it, on the middle peg.
  • Disk three is 0, so it is on another peg. Since n is odd (n = 1), it is one peg to the left, i.e. on the left peg.
  • Disk four is 1, so it is on another peg. Since n is odd (n = 1), it is one peg to the left, i.e. on the right peg.
  • Disk five is also 1, so it is stacked on top of it, on the right peg.
  • Disk six is 0, so it is on another peg. Since n is even (n = 2), the disk is one peg to the right, i.e. on the left peg.
  • Disks seven and eight are also 0, so they are stacked on top of it, on the left peg.

The source and destination pegs for the mth move can also be found elegantly from the binary representation of m using bitwise operations. To use the syntax of the C programming language, move m is from peg (m & m - 1) % 3 to peg ((m | m - 1) + 1) % 3, where the disks begin on peg 0 and finish on peg 1 or 2 according as whether the number of disks is even or odd. Another formulation is from peg (m - (m & -m)) % 3 to peg (m + (m & -m)) % 3.

Furthermore, the disk to be moved is determined by the number of times the move count (m) can be divided by 2 (i.e. the number of zero bits at the right), counting the first move as 1 and identifying the disks by the numbers 0, 1, 2, etc. in order of increasing size. This permits a very fast non-recursive computer implementation to find the positions of the disks after m moves without reference to any previous move or distribution of disks.

The operation, which counts the number of consecutive zeros at the end of a binary number, gives a simple solution to the problem: the disks are numbered from zero, and at move m, disk number count trailing zeros is moved the minimal possible distance to the right (circling back around to the left as needed).[10]

The binary numeral system of Gray codes gives an alternative way of solving the puzzle. In the Gray system, numbers are expressed in a binary combination of 0s and 1s, but rather than being a standard positional numeral system, Gray code operates on the premise that each value differs from its predecessor by only one (and exactly one) bit changed.

If one counts in Gray code of a bit size equal to the number of disks in a particular Tower of Hanoi, begins at zero and counts up, then the bit changed each move corresponds to the disk to move, where the least-significant bit is the smallest disk, and the most-significant bit is the largest.

Counting moves from 1 and identifying the disks by numbers starting from 0 in order of increasing size, the ordinal of the disk to be moved during move m is the number of times m can be divided by 2.

This technique identifies which disk to move, but not where to move it to. For the smallest disk, there are always two possibilities. For the other disks there is always one possibility, except when all disks are on the same peg, but in that case either it is the smallest disk that must be moved or the objective has already been achieved. Luckily, there is a rule that does say where to move the smallest disk to. Let f be the starting peg, t the destination peg, and r the remaining third peg. If the number of disks is odd, the smallest disk cycles along the pegs in the order f → t → r → f → t → r, etc. If the number of disks is even, this must be reversed: f → r → t → f → r → t, etc.[11]

The position of the bit change in the Gray code solution gives the size of the disk moved at each step: 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, ... (sequence A001511 in the OEIS),[12] a sequence also known as the ruler function, or one more than the power of 2 within the move number. In the Wolfram Language, IntegerExponent[Range[2^8 - 1], 2] + 1 gives moves for the 8-disk puzzle.

If all moves must be between adjacent pegs (i.e. given pegs A, B, C, one cannot move directly between pegs A and C), then moving a stack of n disks from peg A to peg C takes 3ⁿ − 1 moves. The solution uses all 3ⁿ valid positions, always taking the unique move that does not undo the previous move. The position with all disks at peg B is reached halfway, i.e. after (3ⁿ − 1) / 2 moves.

In Cyclic Hanoi, we are given three pegs (A, B, C), which are arranged as a circle with the clockwise and the counterclockwise directions being defined as A – B – C – A and A – C – B – A respectively. The moving direction of the disk must be clockwise.[13] It suffices to represent the sequence of disks to be moved. The solution can be found using two mutually recursive procedures:

To move n disks counterclockwise to the neighbouring target peg:

1. move n − 1 disks counterclockwise to the target peg
2. move disk #n one step clockwise
3. move n − 1 disks clockwise to the start peg
4. move disk #n one step clockwise
5. move n − 1 disks counterclockwise to the target peg

To move n disks clockwise to the neighbouring target peg:

1. move n − 1 disks counterclockwise to a spare peg
2. move disk #n one step clockwise
3. move n − 1 disks counterclockwise to the target peg

Let C(n) and A(n) represent moving n disks clockwise and counterclockwise, then we can write down both formulas:

      C(n) = A(n-1) n A(n-1)    and      A(n) = A(n-1) n C(n-1) n A(n-1).

Thus  C(1) = 1                  and      A(1) = 1 1,
      C(2) = 1 1 2 1 1          and      A(2) = 1 1 2 1 2 1 1.

The solution for the Cyclic Hanoi has some interesting properties:

1)The move-patterns of transferring a tower of disks from a peg to another peg are symmetric with respect to the center points.

2)The smallest disk is the first and last disk to move.

3)Groups of the smallest disk moves alternate with single moves of other disks.

4)The number of disks moves specified by C(n) and A(n) are minimal.

Although the three-peg version has a simple recursive solution long been known, the optimal solution for the Tower of Hanoi problem with four pegs (called Reve's puzzle) was not verified until 2014, by Bousch.[14]

However, in case of four or more pegs, the Frame–Stewart algorithm is known without proof of optimality since 1941.[15]

For the formal derivation of the exact number of minimal moves required to solve the problem by applying the Frame–Stewart algorithm (and other equivalent methods), see the following paper.[16]

For other variants of the four-peg Tower of Hanoi problem, see Paul Stockmeyer's survey paper.[17]

The so-called Towers of Bucharest and Towers of Klagenfurt game configurations yield ternary and pentary Gray codes.[18]

The Frame–Stewart algorithm is described below:

• Let ${\displaystyle n}$ be the number of disks.
• Let ${\displaystyle r}$ be the number of pegs.
• Define ${\displaystyle T(n,r)}$ to be the minimum number of moves required to transfer n disks using r pegs.

The algorithm can be described recursively:

1. For some ${\displaystyle k}$, ${\displaystyle 1\leq k<n}$, transfer the top ${\displaystyle k}$ disks to a single peg other than the start or destination pegs, taking ${\displaystyle T(k,r)}$ moves.
2. Without disturbing the peg that now contains the top ${\displaystyle k}$ disks, transfer the remaining ${\displaystyle n-k}$ disks to the destination peg, using only the remaining ${\displaystyle r-1}$ pegs, taking ${\displaystyle T(n-k,r-1)}$ moves.
3. Finally, transfer the top ${\displaystyle k}$ disks to the destination peg, taking ${\displaystyle T(k,r)}$ moves.

The entire process takes ${\displaystyle 2T(k,r)+T(n-k,r-1)}$ moves. Therefore, the count ${\displaystyle k}$ should be picked for which this quantity is minimum. In the 4-peg case, the optimal ${\displaystyle k}$ equals ${\displaystyle n-\left\lfloor {\sqrt {2n+1}}\right\rceil +1}$, where ${\displaystyle \left\lfloor \cdot \right\rceil }$ is the nearest integer function.[19] For example, in the UPenn CIS 194 course on Haskell, the first assignment page[20] lists the optimal solution for the 15-disk and 4-peg case as 129 steps, which is obtained for the above value of k.

This algorithm is presumed to be optimal for any number of pegs; its number of moves is 2^{Θ(n1/(r−2))} (for fixed r).

A curious generalization of the original goal of the puzzle is to start from a given configuration of the disks where all disks are not necessarily on the same peg and to arrive in a minimal number of moves at another given configuration. In general, it can be quite difficult to compute a shortest sequence of moves to solve this problem. A solution was proposed by Andreas Hinz and is based on the observation that in a shortest sequence of moves, the largest disk that needs to be moved (obviously one may ignore all of the largest disks that will occupy the same peg in both the initial and final configurations) will move either exactly once or exactly twice.

The mathematics related to this generalized problem becomes even more interesting when one considers the average number of moves in a shortest sequence of moves between two initial and final disk configurations that are chosen at random. Hinz and Chan Tat-Hung independently discovered[21][22] (see also [23]^{: Chapter 1, p. 14}) that the average number of moves in an n-disk Tower is given by the following exact formula:

${\displaystyle {\frac {466}{885}}\cdot 2^{n}-{\frac {1}{3}}-{\frac {3}{5}}\cdot \left({\frac {1}{3}}\right)^{n}+\left({\frac {12}{59}}+{\frac {18}{1003}}{\sqrt {17}}\right)\left({\frac {5+{\sqrt {17}}}{18}}\right)^{n}+\left({\frac {12}{59}}-{\frac {18}{1003}}{\sqrt {17}}\right)\left({\frac {5-{\sqrt {17}}}{18}}\right)^{n}.}$

For large enough n, only the first and second terms do not converge to zero, so we get an asymptotic expression: ${\displaystyle 466/885\cdot 2^{n}-1/3+o(1)}$, as ${\displaystyle n\to \infty }$. Thus intuitively, we could interpret the fraction of ${\displaystyle 466/885\approx 52.6\%}$ as representing the ratio of the labor one has to perform when going from a randomly chosen configuration to another randomly chosen configuration, relative to the difficulty of having to cross the "most difficult" path of length ${\displaystyle 2^{n}-1}$ which involves moving all the disks from one peg to another. An alternative explanation for the appearance of the constant 466/885, as well as a new and somewhat improved algorithm for computing the shortest path, was given by Romik.[24]

In Magnetic Tower of Hanoi, each disk has two distinct sides North and South (typically colored "red" and "blue"). Disks must not be placed with the similar poles together—magnets in each disk prevent this illegal move. Also, each disk must be flipped as it is moved.

Initial configuration of bicolor Towers of Hanoi (n=4)

This variation of the famous Tower of Hanoi puzzle was offered to grade 3–6 students at 2ème Championnat de France des Jeux Mathématiques et Logiques held in July 1988.[25]

Final configuration of bicolor Towers of Hanoi (n=4)

The rules of the puzzle are essentially the same: disks are transferred between pegs one at a time. At no time may a bigger disk be placed on top of a smaller one. The difference is that now for every size there are two disks: one black and one white. Also, there are now two towers of disks of alternating colors. The goal of the puzzle is to make the towers monochrome (same color). The biggest disks at the bottom of the towers are assumed to swap positions.

A variation of the puzzle has been adapted as a solitaire game with nine playing cards under the name Tower of Hanoy.[26][27] It is not known whether the altered spelling of the original name is deliberate or accidental.[28]