A Note on On-Line Ramsey Numbers for Some Paths

: We consider the important generalisation of Ramsey numbers, namely on-line Ramsey numbers. It is easiest to understand them by considering a game between two players, a Builder and Painter, on an inﬁnite set of vertices. In each round, the Builder joins two non-adjacent vertices with an edge, and the Painter colors the edge red or blue. An on-line Ramsey number ˜ r ( G , H ) is the minimum number of rounds it takes the Builder to force the Painter to create a red copy of graph G or a blue copy of graph H , assuming that both the Builder and Painter play perfectly. The Painter’s goal is to resist to do so for as long as possible. In this paper, we consider the case where G is a path P 4 and H is a path P 10 or P 11 .


Introduction
The terminology, definitions and some descriptions are taken from two previous works by the first author, namely [1,2].
Ramsey numbers and their generalizations have been a fundamentally important area of study in combinatorics for many years. Particularly well-studied are Ramsey numbers for graphs. Here, the Ramsey number of two graphs G and H, denoted by r(G, H), is the least t such that any red-blue edge-coloring of K t contains a red copy of G or a blue copy of H.
In this paper, we consider the following generalization of Ramsey numbers defined independently by Beck [3] and Kurek and Ruciński [4]. Let G and H be two graphs. Consider a game played on the edge set of the infinite clique K N with two players, a Builder and Painter. In each round of the game, the Builder chooses an edge and the Painter colors it red or blue. The Builder wins by creating either a red copy of G or a blue copy of H, and wishes to do so in as few rounds as possible. The Painter wishes to delay the Builder for as many rounds as possible. (Note that the Painter may not delay the Builder indefinitely-for example, the Builder may simply choose every edge of K r(G,H) ).
The on-line Ramsey numberr(G, H) is the minimum number of rounds it takes the Builder to win, assuming that both the Builder and Painter play optimally. We call this game ther(G, H)-game. Note thatr(G, H) ≥ e(G) + e(H) − 1 for all graphs G and H, as the Painter may simply colour the first e(G) − 1 edges red and all subsequent edges blue.
Intuitively, it is not surprising that determining on-line Ramsey numbers exactly has proved even more difficult than determining classical Ramsey numbers exactly (the former are a generalization of the latter). The consequence of this is that there are very few known exact values of on-line Ramsey numbers. A significant amount of effort has been focused on the special case where G is a path P 3 . Cyman, Dzido, Lapinskas and Lo [1] have determined r(P 3 , P +1 ) andr(P 3 , C ) exactly (where P s is a path on s vertices). Theorem 1. In [1]. For all ≥ 2, we haver(P 3 , P +1 ) = 5 /4 . As well, The best known bounds onr(P 4 , P +1 ) were also proved in [1].
This provides more evidence for the conjecture that the latter holds for all l ≥ 3.
In the following discussion we take on the role of the Builder, and we will assume for clarity that the Painter will not voluntarily lose the game by creating a red P 4 . We first observe that the Builder can obtain a long blue path by using the strategy for shorter paths twice. Lemma 1. We haver(P 4 , P n ) ≤r(P 4 , P n 2 ) +r(P 4 , P n 2 ) + 3 Proof. First, the Builder will use at mostr(P 4 , P n 2 ) andr(P 4 , P n 2 ) moves to construct two vertex-disjoint blue paths P n 2 and P n 2 , respectively. Then, the Builder will join their endpoints together to form a blue P n in at most 3 rounds. Lemma 1 implies thatr(P 4 , P 10 ) ≤ 2r(P 4 , P 5 ) + 3 = 15. However, the Builder may join the shorter paths more carefully than in the proof of Lemma 1, resulting in the following. Theorem 3. We haver(P 4 , P 10 ) = 13.
The Builder starts with 2 disjointr(P 4 , P 5 )-games. Recall that both the Builder and Painter play optimally, so the Painter wants to avoid a red P 4 and the Builder will force the Painter to create two separate blue P 5 . At the beginning, let's observe that if the Builder was able to construct a blue P 5 in at most 5 moves and a second, separate blue P 5 in at most 5 moves, then using similar reasoning as in the proof of Lemma 1 we have the result. Now we will be very carefully considering the strategy for ther(P 4 , P 5 )-game described by Prałat in [5]. We will use this strategy for the two above-mentionedr(P 4 , P 5 )-games.
In this strategy, the Builder first shows a path P 4 . Therefore, one of the four possible color patterns appears: bbb, bbr, brb, and rrb. The Builder has to avoid the pattern rbr, otherwise, the Painter has a strategy to 'survive' to the end of the sixth round. In order to do that, the Builder can use the same strategy as for the R(P 3 , P 5 ) case described by Prałat in [5]. Finally, the Builder obtains a blue P 5 in the next three moves (the details as shown in Figure 3 in [5]).
The Builder's strategy forr(P 4 , P 5 )-game will be to build up one of the five nonisomorphic structures independent of the Painter's choices, as shown in Figure 1.  Recall that the Builder's start of the strategy forr(P 4 , P 10 )-game is to play two separatẽ r(P 4 , P 5 )-games with the strategy described in [5].
Lemma 2. Suppose that in ther(P 4 , P 10 )-game, the Builder has already obtained a structure S 1 or S 5 in the firstr(P 4 , P 5 )-game. Then, regardless of the strategy used by the Painter in the second r(P 4 , P 5 )-game, after the end of this game and one more move there is either a red copy of P 4 or a blue copy of P 10 .
Proof. The Builder can join an endpoint of a red P 3 in S 1 or S 5 , which is at the same time the endpoint of a blue path P 5 , with an endpoint of a blue P 5 in the structure obtained after the end of the secondr(P 4 , P 5 )-game. Lemma 3. Suppose that in ther(P 4 , P 10 )-game, the Builder has obtained a structure S 3 or S 4 in bothr(P 4 , P 5 )-games. Then, after one move there is either a red copy of P 4 or a blue copy of P 10 .
Proof. The Builder can join an endpoint of a blue P 5 in the first structure, which is at the same time the middle of a red path P 3 , with the vertex of the same type in the structure obtained in the secondr(P 4 , P 5 )-game.
Note that the structure S 2 could have occurred when the Painter started ther(P 4 , P 5 )game from the configuration rrb or bbr.

Lemma 4.
Suppose that in ther(P 4 , P 10 )-game, the Builder has obtained a structure rrb or bbr in bothr(P 4 , P 5 )-games after 3 moves. Then, after 7 moves there is either a red copy of P 4 or a blue copy of P 10 .
Proof. There are only three possible patterns that can appear. Let us consider these three cases depending on the Painter's choice. Lemma 5. Suppose that after 3 rounds forr(P 4 , P 10 )-game, the Builder has obtained a structure rrb or bbr in the firstr(P 4 , P 5 )-game and after 3 rounds he has obtained a structure bbb or brb in the secondr(P 4 , P 5 )-game. Then, after next 7 moves there is either a red copy of P 4 or a blue copy of P 10 .
Proof. First, the Builder continues the second game and he forces the Painter to construct one of the structures S 1 , S 4 or S 5 . If he obtains structure S 1 or S 5 , then by applying Lemma 2, we have the result. So we may assume that the Builder has structure S 4 after secondr(P 4 , P 5 )-game. The Builder now is able to finish the game in the next 4 moves, as shown in Figure 2. The final edge is drawn with a dotted line and a circled number means that the Painter had a choice in that move, which led to branching into subcases.
Case 1 for rrb Case 2a for bbr Case 2b for bbr Finally, notice that sincer(P 4 , P 5 ) = 6 and Lemmas 2-5 exhaust all possible situations of playingr(P 4 , P 5 )-games, thenr(P 4 , P 10 ) ≤ 13. Taking into account the lower bound, the proof is complete. Now we prove that the Builder can obtain either a longer blue path or a red P 4 by simply extending an existing blue path. Lemma 6. Suppose Q is a non-trivial blue path with endpoints a and b. Then the Builder can force the Painter to construct either a red P 4 or a blue path of length e(Q) + 1 in at most 3 moves.
Proof. Let c and d be the new vertices. The Builder chooses the edges ac, bc and bd. If the Painter colors any of the edges blue, then we have a blue path of length e(Q) + 1. Then the Painter colors them all red and we have a red P 4 . Theorem 4. We haver(P 4 , P 11 ) = 15.
The Builder starts withr(P 4 , P 10 )-game and he uses it to force a blue copy of P 10 in at most 13 moves. If the Builder has achieved this goal in 12 or fewer moves, then by using Lemma 6 we have the result. The case that remains to be considered is when ther(P 4 , P 10 )game ends by forcing the Painter to create a blue P 10 in the 13th round. We will apply the strategy described in the proof of Theorem 3 and prove that in each of the cases considered in Lemmas 2-5, two moves are enough to force the Painter to create a red P 4 or a blue P 11 . The result is achieved by case-by-case analysis of the last two moves as shown in Figure 3. The final edge is drawn with a dotted line and a circled number means that the Painter had a choice in that move, which led to branching into subcases. Table 1. New upper bounds for the numbersr(P 4 , P n ) where 12 ≤ n ≤ 22