No, Lyosha always wins. After every Lyosha's turn, their chips lie in different parts. This implies that Pasha cannot stop Lyosha's chip from moving to a new field. On the other hand, since Pasha starts so Pasha's chip will finish visiting all the fields a step earlier than Lyosha. Thus, Pasha cannot win.

shinichiman wrote: No, Lyosha always wins. After every Lyosha's turn, their chips lie in different parts. This implies that Pasha cannot stop Lyosha's chip from moving to a new field. On the other hand, since Pasha starts so Pasha's chip will finish visiting all the fields a step earlier than Lyosha. Thus, Pasha cannot win. Oh, it's not so easy. Indeed, the first player cannot block a move of the second one. But, a position cannot be repeated, which means the pair (violet chip position, purple chip position). For example, $A$ goes to a field $a_1$, then $B$ to $b_1$ ; further if $A$ returns to its initial position, $B$ cannot do the same.

For the sake of sanity, I'll call the players $A$ and $B$, with $A$ going first, and their names interchangeable with their respective chips. For further disambiguation, $A$ is female, and $B$ is male. I'll let the board be a graph $G$, with two parts $X$ and $Y$, such that $A$ begins in $X$ and $B$ begins in $Y$. Call the vertex $A$ starts on $A_0$. First, some basic facts. Note that after the $t$th move by $A$, both $A$ and $B$ are in $X$ iff $t$ is even, and otherwise $A$ and $B$ are both in $Y$. Similarly, after the $t$th move by $B$, $A$ is in $X$ and $B$ is in $Y$ iff $t$ is even, otherwise, $A$ is in $Y$ and $B$ is in $X$. This implies that, before $B$'s move, both chips are in the same part. Hence, the only constraint on $B$'s move is that he cannot create a position that has previously occurred. Now we claim that $B$ always wins with the following strategy: $B$ considers a path from his starting location to $A_0$. $B$ will move back to the position he was in in before his last move, unless he's not allowed to, in which case he advances along the path. Note that he always advances or retreats one step along the path. First - $B$ cannot retreat twice consecutively, if he's following the strategy. After retreating once, both options given in the strategy instruct him to advance. Hence, once $B$ reaches any point along the path, he cannot ever return to any point at least $2$ steps back from it. Next - $B$ can always advance along the path. Suppose it's been true up to some point. This point will be the present. If the destination has yet to be occupied by $B$ at any previous point, no position with $B$ occupying that point could have existed, so $B$ is free to move there. Otherwise, that means that at some previous point, $B$ has been at the destination - the point ahead of his current location along the path, and $A$ was in the position that she's currently at, right after her move. If we go back to that previous instance of the position and consider $B$'s move immediately preceding it (it must have been $B$'s move due to the 'basic fact' above), it must have been an advance. Were it a retreat, $B$ would have been retreating from a position two steps ahead of the position he's currently at, a contradiction to the first fact. But the position immediately preceding the advance is equivalent to the position at present time, right after $A$'s move, so $A$'s move just before the present time was illegal. Now we've shown that the strategy is well-defined, and the move suggested is legal, up till the point where $B$ has reached the end of the path, and is forced to advance. We'll eventually show that this cannot occur. Note that the moves $B$ makes can be partitioned into chunks (which I'll call 'cycles') consisting several (possibly $0$) pairs of advances and retreats, followed by 2 advances. Within a cycle, $A$ may not return to any vertex she's previously been in, except for the vertex she was in when the cycle began. Suppose otherwise, and consider the position just after she does so. $B$ will have been alternating between two vertices, one in $X$, and one in $Y$ (let's call these $x$ and $y$). Consider the previous time $A$ was in the vertex she's currently in ($n$), specifically the position just before $B$'s move. $A$ would have been in $n$, and $B$ would be in whichever of $x$ and $y$ are in the same part as $n$, preparing to move to the other. This is the same position as in present time, hence $A$'s move was illegal. The reason why it's possible if $A$ moved back to the starting vertex is because in that case, $B$ would have been in a vertex from the previous cycle instead of $y$, as would be the case with other vertices in the part. Thus, $A$ can only return to a vertex she's previously been in within a cycle, if that vertex is the vertex she began the cycle in. Since the two vertices $B$ alternates between during a cycle are unique to that cycle, the only way a position is repeated, and $B$ is forced to advance is if $A$ returns to the starting vertex. Hence, $A$ always starts a cycle on the same vertex, $A_0$. Now consider the final cycle, the cycle where $B$ alternates between $A_0$, and the vertex just before that on the path. Every time $A$ moves into a vertex in $Y$, $B$ moves to $A_0$. Hence, $A$ can never return to $A_0$, as it'll always be occupied by $B$ when she moves into $Y$. Thus, she has to keep visiting new vertices indefinitely, which is impossible.

The official solution. $B$ always wins. There is a path from the initial field of $A$ to the initial field of $B$. Enumerate the fields of that path $0,1,2,\dots, n$ , so at the beggining $A$ is upon $0$ and $B$ - upon $n$. At every next move $B$ sticks to the following strategy. If $B$ is at an even field, he/she moves at the next field. If $B$ is upon an odd field he/she retreats back with one exeption. Namely, when $A$ occupies $n$. In that case $B$ cannot move back, so he/she moves again forward. It can be checked $B$ cannot lose. The above solution apparently follows that strategy. @joyce_tan: Nice solution. Btw, both names of the players are Russian male names.

Quote: If $B$ is upon an odd field he/she retreats back with one exeption. Namely, when $A$ occupies $n$. I don't understand.When $B$ retreats will he repeat a position twice? And it seems impossible for $A$ to occupy $n$ because it has been occupied by $B$.

The word "position" here means the combination (purple counter field, violet counter field). So, to repeat a position means to repeat twice that disposition. It is exactly the same as it is in the chess game.