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Tower of Hanoi – Wikipedia

The Tower of Hanoi (also called The problem of Benares Temple or Tower of Brahma or Lucas’ Tower and sometimes pluralized as Towers, or simply pyram …

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• Date Published: 2013. 3. 19.
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» Pyramid Path Rising vs Path Origin Bowling Ball Reaction Video Ball Review

Pyramid technology is the culmination of years of research, observation, core and coverstock analysis. With ever changing variables within the sport of bowling, (e.g., lane conditions, ball speed, rev rate) this technology allows any bowler, from stroker to two-handed, to achieve peak performance on medium to heavy oil conditions.

Published on December 23, 2013 by videoballreviews

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Tower of Hanoi

Mathematical game or puzzle

A model set of the Tower of Hanoi (with 8 disks)

Tower of Hanoi puzzle for T(4, 3) An animated solution of thepuzzle for(4, 3)

Tower of Hanoi interactive display at Mexico City’s Universum Museum

The Tower of Hanoi (also called The problem of Benares Temple[1] or Tower of Brahma or Lucas’ Tower[2] and sometimes pluralized as Towers, or simply pyramid puzzle[3]) is a mathematical game or puzzle consisting of three rods and a number of disks of various diameters, which can slide onto any rod. The puzzle begins with the disks stacked on one rod in order of decreasing size, the smallest at the top, thus approximating a conical shape. The objective of the puzzle is to move the entire stack to the last rod, obeying the following rules:

Only one disk may be moved at a time. Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack or on an empty rod. No disk may be placed on top of a disk that is smaller than it.

With 3 disks, the puzzle can be solved in 7 moves. The minimal number of moves required to solve a Tower of Hanoi puzzle is 2n − 1, where n is the number of disks.

Origins [ edit ]

The puzzle was introduced to the West by the French mathematician Édouard Lucas in 1883. Numerous myths regarding the ancient and mystical nature of the puzzle popped up almost immediately,[4] including one about an Indian temple in Kashi Vishwanath containing a large room with three time-worn posts in it, surrounded by 64 golden disks. Acting out the command of an ancient prophecy, Brahmin priests have been moving these disks in accordance with the immutable rules of Brahma since that time. The puzzle is therefore also known as the Tower of Brahma. According to the legend, when the last move of the puzzle is completed, the world will end.[5]

If the legend were true, and if the priests were able to move disks at a rate of one per second, using the smallest number of moves, it would take them 264 − 1 seconds or roughly 585 billion years to finish,[6] which is about 42 times the current age of the universe.

There are many variations on this legend. For instance, in some tellings, the temple is a monastery, and the priests are monks. The temple or monastery may be in various locales including Hanoi, and may be associated with any religion. In some versions, other elements are introduced, such as the fact that the tower was created at the beginning of the world, or that the priests or monks may make only one move per day.

Solution [ edit ]

The puzzle can be played with any number of disks, although many toy versions have around 7 to 9 of them. The minimal number of moves required to solve a Tower of Hanoi puzzle is 2n − 1, where n is the number of disks.[7] This is precisely the nth Mersenne number without primality requirements.[1]

Iterative solution [ edit ]

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]

Simpler statement of iterative solution [ edit ]

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.

Equivalent iterative solution [ edit ]

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.

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]

Recursive solution [ edit ]

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,

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:

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 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.

Logical analysis of the recursive solution [ edit ]

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 S 3 ). 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:

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 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: 2 h − 1 {\displaystyle 2^{h}-1} . This result is obtained by noting that steps 1 and 3 take T h − 1 {\displaystyle T_{h-1}} moves, and step 2 takes one move, giving T h = 2 T h − 1 + 1 {\displaystyle T_{h}=2T_{h-1}+1} .

Recursive implementation [ edit ]

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 = ‘

‘ ) # 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 )

Non-recursive solution [ edit ]

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.

is even, the remaining third peg during successive moves is , , , , , , 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.

Binary solution [ edit ]

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 2 8 − 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.

− 1 = 11111111. Move 216 10 = 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.

= 11011000.

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]

Gray-code solution [ edit ]

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.

Graphical representation [ edit ]

The game can be represented by an undirected graph, the nodes representing distributions of disks and the edges representing moves. For one disk, the graph is a triangle:

The graph for two disks is three triangles connected to form the corners of a larger triangle.

A second letter is added to represent the larger disk. Clearly, it cannot initially be moved.

The topmost small triangle now represents the one-move possibilities with two disks:

The nodes at the vertices of the outermost triangle represent distributions with all disks on the same peg.

For h + 1 disks, take the graph of h disks and replace each small triangle with the graph for two disks.

For three disks the graph is:

The game graph of level 7 shows the relatedness to the Sierpiński triangle

call the pegs a, b, and c

list disk positions from left to right in order of increasing size

The sides of the outermost triangle represent the shortest ways of moving a tower from one peg to another one. The edge in the middle of the sides of the largest triangle represents a move of the largest disk. The edge in the middle of the sides of each next smaller triangle represents a move of each next smaller disk. The sides of the smallest triangles represent moves of the smallest disk.

In general, for a puzzle with n disks, there are 3n nodes in the graph; every node has three edges to other nodes, except the three corner nodes, which have two: it is always possible to move the smallest disk to one of the two other pegs, and it is possible to move one disk between those two pegs except in the situation where all disks are stacked on one peg. The corner nodes represent the three cases where all the disks are stacked on one peg. The diagram for n + 1 disks is obtained by taking three copies of the n-disk diagram—each one representing all the states and moves of the smaller disks for one particular position of the new largest disk—and joining them at the corners with three new edges, representing the only three opportunities to move the largest disk. The resulting figure thus has 3n+1 nodes and still has three corners remaining with only two edges.

As more disks are added, the graph representation of the game will resemble a fractal figure, the Sierpiński triangle. It is clear that the great majority of positions in the puzzle will never be reached when using the shortest possible solution; indeed, if the priests of the legend are using the longest possible solution (without re-visiting any position), it will take them 364 − 1 moves, or more than 1023 years.

The longest non-repetitive way for three disks can be visualized by erasing the unused edges:

Incidentally, this longest non-repetitive path can be obtained by forbidding all moves from a to c.

The Hamiltonian cycle for three disks is:

The graphs clearly show that:

From every arbitrary distribution of disks, there is exactly one shortest way to move all disks onto one of the three pegs.

Between every pair of arbitrary distributions of disks there are one or two different shortest paths.

From every arbitrary distribution of disks, there are one or two different longest non selfcrossing paths to move all disks to one of the three pegs.

Between every pair of arbitrary distributions of disks there are one or two different longest non self-crossing paths.

Let N h be the number of non-self-crossing paths for moving a tower of h disks from one peg to another one. Then: N 1 = 2 N h +1 = ( N h ) 2 + ( N h ) 3

be the number of non-self-crossing paths for moving a tower of disks from one peg to another one. Then:

This gives N h to be 2, 12, 1872, 6563711232, … (sequence A125295 in the OEIS)

Variations [ edit ]

Adjacent pegs [ edit ]

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.[citation needed]

Cyclic Hanoi [ edit ]

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:

move n − 1 disks counterclockwise to the target peg move disk #n one step clockwise move n − 1 disks clockwise to the start peg move disk #n one step clockwise move n − 1 disks counterclockwise to the target peg

To move n disks clockwise to the neighbouring target peg:

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

With four pegs and beyond [ edit ]

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]

Frame–Stewart algorithm [ edit ]

The Frame–Stewart algorithm is described below:

Let n {\displaystyle n}

Let r {\displaystyle r}

Define T ( n , r ) {\displaystyle T(n,r)}

The algorithm can be described recursively:

For some k {\displaystyle k} 1 ≤ k < n {\displaystyle 1\leq k

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YouTube에서 pyramid path origin 주제의 다른 동영상 보기

주제에 대한 기사를 시청해 주셔서 감사합니다 bowlingball.com Pyramid Path Origin Bowling Ball Reaction Video Review | pyramid path origin, 이 기사가 유용하다고 생각되면 공유하십시오, 매우 감사합니다.

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