SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Lastly, according to Observation 3.4, there exists a non-zero element in row  $n - (i + 1)$  at position  $n - i$ . Combining this with the Proposition 3.13 and the fact that column  $n - i$  has already been played (according to Item (ii)), we can conclude that the cumulative utility of row  $n - (i + 1)$  must be non-zero. This concludes the proof of Item (iv).
B.2.3 Proof of Theorem 3.15
Proposition B.6. There exists a round  $T_{i}^{4} > T_{i}^{3}$  at which
(i) the strategy profile is  $(n - (i + 1),(i + 1) + 1)$  for the first time,
(ii) for all rounds  $t \in [T_i^3, T_i^4 - 1]$ , the strategy profile is  $(n - (i + 1), n - i)$ ,
(iii) row  $n - i - 1$  admits cumulative utility  $R_{n - (i + 1)}^{(T_i^4)} \geq (4i + 4) \cdot C_{n - i}^{(T_i^3)}$ ,
(iv) all rows  $k \in [(i + 1) + 1, n - (i + 1) - 1]$  admit  $R_k^{(T_i^4)} = 0$  and all columns  $k \in [(i + 1) + 2, n - (i + 1)]$  admit  $C_k^{(T_i^4)} = 0$ .
Proof. We repeat the same reasoning as in the proof of Proposition 3.12. To begin with, we must establish that the new strategy profile chosen by fictitious play will be  $(n - (i + 1), (i + 1) + 1)$ .
We can infer from both Items (i) and (ii) in Proposition 3.14 that the most recent strategy switch was made by the row player. Therefore, based on Corollary 3.11, we are certain that the next strategy switch will be initiated by the column player.
In other words, as long as the row player continues to play their current strategy, strategy  $(n - (i + 1), n - i)$  will be played in every subsequent round, which establishes Item (ii). This implies that the column player's cumulative utility will increase by  $e_{n - (i + 1)}^{\top} A$ . As per Observation 3.3, row  $n - (i + 1)$  only has non-zero elements at positions  $(i + 1) + 1$  and  $n - (i + 1)$ .
$$
e _ {n - (i + 1)} ^ {\top} A = \left[ 0, \dots , \underbrace {4 i + 5} _ {(i + 1) + 1}, 0, \dots , 0, \underbrace {4 i + 4} _ {n - (i + 1) - 1}, 0, \dots \right] \tag {8}
$$
Since strategy  $(i + 1) + 1$  has a higher value, a strategy switch in a later time step is certain. Consequently, there will be a round  $T_{i}^{4}$  in which the strategy  $(n - (i + 1),(i + 1) + 1)$  will be played for the first time, establishing Item (i).
We must determine the point at which the strategy switch  $(n - (i + 1), (i + 1) + 1)$  will take place. According to Equation (8), the value of strategy  $(i + 1) + 1$  is exactly one greater than the value of strategy  $n - i$ . From the Proposition 3.14, we know that column  $(i + 1) + 1$  has a cumulative utility of zero, whereas column  $n - i$  has a cumulative utility of  $C_{n - i}^{(T_i^3)}$ . Therefore, it will take precisely
$C_{n - i}^{(T_i^3)}$  steps for those strategies to have equal cumulative utility values. Consequently, the strategy switch will either occur in that round or the immediate next, as the cumulative utility of column  $(n - (i + 1),(i + 1) + 1)$  will have surpassed that of column  $n - i$ .
In order to proceed with Item (iii), we compute the updated cumulative utility of row  $n - (i + 1)$ . As shown in Equation (8), row  $n - (i + 1)$  has a value of  $(4i + 4)$  at position  $n - i$ . Thus, if column  $n - i$  has been played for a minimum of  $C_{n - i}^{(T_i^3)}$  rounds, then the cumulative utility of row  $n - (i + 1)$  satisfies  $R_{n - (i + 1)}^{(T_i^4)} \geq (4i + 4) \cdot C_{n - i}^{(T_i^3)}$ .
Lastly, according to Observation 3.4, there exists a non-zero element in column  $(i + 1) + 1$  at position  $n - (i + 1)$ . Combining this with the Proposition 3.14 and the fact that row  $n - (i + 1)$  has already been played (according to Item (ii)), we can conclude that the cumulative utility of column  $(i + 1) + 1$  must be non-zero. This concludes the proof of Item (iv).