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Mathematics-Online problems:

Interactive Problem 231: Sliding Block Puzzle

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Given a sliding block puzzle with 24 pieces in a box of $ 5 \times 5$ squares. One of the squares keeps empty. Adjoining pieces can be moved vertical or horizontal onto the empty square.

\includegraphics[width=.45\linewidth]{g71_bild_sorted.eps}

\includegraphics[width=.45\linewidth]{g71_bild_permutation.eps}

The pieces are numerated in the way shown in the left picture, which is the starting position.The numeration starts in the upper left corner following each line from the left to the right. The right picture shows a slid puzzle. Identifying the number of each piece with its new position, one gets the according permutation $ \pi$.

Complete the permutation

$\displaystyle \pi=\left(\begin{array}{rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr... ... \ & & \ & \ & \ & \ & 14 & 10 & & & & & \ldots& & 4 & \end{array}\right) \; . $

Determine the number of elements in the longest cycle and the sum over all elements in the second longest cycle. Is the puzzle on the right side solvable, i.e. is there a finite sequence of slides which transform the puzzle into the starting position shown on the left?


Solution:

Number of elements in the longest cycle:

Sum over the elements in the second longest cycle:

Sliding block puzzle solvable: no answer ,     yes ,     no
   

(Authors: Wipper/Höfert)
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  automatically generated: 3/12/2018