The field is divided into $25$ squares, with numbers written on each square as shown in the figure.
$$\begin{array}{|c|c|c|c|c|} \hline 5 & 3 & 17 & 18 & 12 \\ \hline 9 & 13 & 1 & 25 & 23 \\ \hline 6 & 19 & 20 & 2 & 21 \\ \hline 16 & 14 & 15 & 8 & 7 \\ \hline 4 & 24 & 10 & 22 & 11 \\ \hline \end{array}$$
Mufasa thinks that it is a good puzzle for Simba to put them all into regular order so that the first line reads $$1 \text{ }2 \text{ }3 \text{ }4 \text{ }5$$ and the second line reads $$6\text{ } 7\text{ } 8 \text{ }9 \text{ }10$$ and so on to the end, by taking up one counter in each hand and making them change places. Thus, Simba might take up $7$ and $1$ and replace them as $1$ and $7$. Then take up $24$ and $2$ and make them also change places, then Simba will have the first two counters properly placed.
The puzzle is for Simba to determine the fewest possible exchanges in which this can be done.
What I did
- The table shown above can be written as the element $$p=(5\text{ }3\text{ }17\text{ }18\text{ }12\text{ }9\text{ }13 \dots 22\text{ }11)$$ of $S_{25}.$
- When we are exchanging two elements, it corresponds to the process of multiplication by a transposition of the form $(a\text{ }b)$ in $S_{25}.$
- So the question reduces to compute the number of transpositions we need to multiply $p$ to obtain $q=(1\text{ }2\text{ }3\text{ }4\text{ }5\text{ }6\text{ }7 \dots 24\text{ }25)$
- So I computed $p^{-1} q$ and it turns out to be product of $3$ cycles of length $16,8$ and $1$. Hence I think the minimum number of exchanges we need to do is $15+7=22$. But this is not the correct answer
Where is the fault in my reasoning? and how can I solve this?