Argentine National Olympiad 2015

Marker $1$ needs at least $12$ horizontal operations to reach its final position $13$; by symmetry the same holds for marker $13$. Similarly markers $2$ and $12$ need at least $10$ horizontal operations each. The analogous observation about the pairs of markers $3,11$; $4,10$; $5,9$; $6,8$ implies that at least $2(12+10+8+6+4+2)=84$ horizontal operations are needed to complete the task. As for the vertical operations, we claim that all markers except possibly one must move up vertically at some point, and hence go down by one more vertical operation. Assume on the contrary that markers $i$ and $j$, $i<j$, never leave row $1$. Then their mutual disposition will not change, $i$ will always precede $j$ no matter the remaining operations. However $j$ has to precede $i$ in the final position. The contradiction shows that at least $12$ markers need $2$ vertical operations each. In all, at least $84+2\cdot 12=108$ operations are needed to achieve the goal. Let us show that $108$ operations are enough. Carry out steps (1) – (5) in the order they are described below.

(1) For each $i=1,2,\dots ,6$ let $S_i$ be the following sequence of operations: Move marker $i$ up to the second row, then move it to the right to position $14-i$. Carry out sequences $S_1,S_2,\dots ,S_6$ in this order; this is clearly possible. Positions $8,9,10,11,12,13$ in row $2$ are now occupied by markers $6,5,4,3,2,1$. The number of operations used is $6+(12+10+8+6+4+2)=48$.

(2) Move marker $7$ up to the second row.

(3) For each $i=8,9,10,11,12$ let $S_i$ be the following sequence of operations: move marker $i$ to position $14-i$ in row $1$, then lift it up to row $2$. Carry out the sequences $S_8,S_9,\dots ,S_{12}$ in this order. Positions $2,3,4,5,6,7$ in row $2$ are now occupied by markers $12,11,10,9,8,7$. The number of operations used is $5+(2+4+6+8+10)=35$.

(4) Move marker $13$ to position $1$ in row $1$ without lifting it up; $12$ operations are used.

(5) Move down all $12$ markers in row $2$; $12$ operations are used.

The problem is solved with the minimum possible number of $108$ operations.