Add files using upload-large-folder tool

parse/train/ByJDAIe0b/ByJDAIe0b.md

@@ -431,7 +431,7 @@ $$$$
 \begin{array} { r l } & { \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad } \\ & { \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad } \\ & { \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad } \\ & & { \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad } \\ & & { \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad } \\ & { \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad } \\ &  \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  \end{array}
 $$
 
-It is relatively straightforward to show that the lemma holds in this last form. To do so note that except for terms for which $\tilde { T }$ and ${ \tilde { T } } ^ { \prime }$ are identical on the left (which are not possible on the right), each term on the left of the inequality is also present on the right, however the number of repetitions of each term varies between the left and right. In the left sum if a term includes m values shared between $\tilde { T }$ and ${ \tilde { T } } ^ { \prime }$ , this term will appear $\binom { 2 \bar { ( } n - i - 1 - m ) } { n - i - 1 - m }$ times. This is because we can choose $n -$ $i - m$ non-duplicate values to be in $\tilde { T }$ and place the rest in ${ \tilde { T } } ^ { \prime }$ , each of these permutations will correspond to a term in the sum. On the other hand in the left sum if a term includes m values shared between $\tilde { T }$ and ${ \tilde { T } } ^ { \prime }$ , this term will appear 2(n−i−1−m)n−i−2−m 
 times. Similarly this is because in this case we can choose $n - i - 1 - m$ non-duplicate values to be in $\tilde { T }$ and place the rest in ${ \tilde { T } } ^ { \prime }$ , each of these permutations will correspond to a different term in the sum.
+It is relatively straightforward to show that the lemma holds in this last form. To do so note that except for terms for which $\tilde { T }$ and ${ \tilde { T } } ^ { \prime }$ are identical on the left (which are not possible on the right), each term on the left of the inequality is also present on the right, however the number of repetitions of each term varies between the left and right. In the left sum if a term includes m values shared between $\tilde { T }$ and ${ \tilde { T } } ^ { \prime }$ , this term will appear $\binom { 2 \bar { ( } n - i - 1 - m ) } { n - i - 1 - m }$ times. This is because we can choose $n -$ $i - m$ non-duplicate values to be in $\tilde { T }$ and place the rest in ${ \tilde { T } } ^ { \prime }$ , each of these permutations will correspond to a term in the sum. On the other hand in the left sum if a term includes m values shared between $\tilde { T }$ and ${ \tilde { T } } ^ { \prime }$ , this term will appear 2(n−i−1−m)n−i−2−m 
 times. Similarly this is because in this case we can choose $n - i - 1 - m$ non-duplicate values to be in $\tilde { T }$ and place the rest in ${ \tilde { T } } ^ { \prime }$ , each of these permutations will correspond to a different term in the sum.
 
 Since $\binom { 2 N } { N } > \binom { 2 N } { N - 1 }$ , $\forall N$ every term which is present on the right side is present on the left with more repetitions and thus the left side must be greater than the right and the lemma holds. □
 

parse/train/zoQJBVrhnn3/zoQJBVrhnn3.md

@@ -251,7 +251,7 @@ $$
 $$
 
 199 where $\begin{array} { r } { | k | _ { m - 1 } ^ { * } = \sum _ { i = 1 } ^ { m - 1 } \mathcal { T } ( z _ { i } ^ { * } = k ) } \end{array}$ . Here, the hard assignment $z _ { m } ^ { * }$ for $\tau _ { m }$ is based on previous   
-200 assignments $z _ { 1 : m - 1 } ^ { * }$ and policies $\phi _ { k } ^ { m - 1 }$ , which is equivalent to applying assumed density filtering   
+200 assignments $z _ { 1 : m - 1 } ^ { * }$ and policies $\phi _ { k } ^ { m - 1 }$ , which is equivalent to applying assumed density filtering   
 201 (ADF) [41] to approximate the true posterior in Eq. (3) with a Delta distribution $\delta \big ( z _ { 1 : m } ^ { * } \big )$ . The hard   
 202 assignment prevents creating a new policy at each step if $\tau _ { m }$ is assigned to an existing policy, which   
 203 significantly reduces the memory usage. Furthermore, the MAP estimations for all existing policies,   
@@ -318,7 +318,7 @@ output :Policy $\bar { \pi } _ { 1 , \ldots , E } ^ { i }$ and meta-policy $\sig
 Initialize learning agent $i$ ’s policy $\pi _ { 0 } ^ { i }$   
 Initialize a memory buffer $\mathbf { B }$   
 Initialize opponent meta-policy $\sigma ^ { - i } ( \cdot ) = 1$   
-for epoch e in $\{ 1 , 2 , \ldots , E \}$ do for episode $h \in \{ 1 , 2 , \dots , H \}$ do Play an episode against the opponent with strategy $\sigma _ { R N R } ^ { 1 }$ Collect the trajectory $\tau _ { e , h }$ and save them into $\mathbf { B }$ end $\begin{array} { r l } & { \tilde { \sigma } ^ { 2 } , \tilde { \Pi } ^ { 2 } = \mathrm { o p p o n e n t } \underline { { \mathrm { { m o d e l i n g } } } } ( \mathbf { B } ) } \\ & { \bar { p } = \frac { 1 } { | \tilde { \Pi } ^ { 2 } | } \sum _ { j } p ^ { j } \tilde { \sigma } ^ { 2 } ( j ) } \end{array}$ Compute missing entries in $U ^ { \tilde { \Pi } }$ from $\tilde { \Pi } = \Pi ^ { 1 } \times \tilde { \Pi } ^ { 2 }$ by simulations ${ \bf \underline { { \ } } } , \sigma _ { R N R } ^ { 2 } = \mathrm { R N R } _ { - } \mathrm { s o l v e r } ( U ^ { \tilde { \Pi } } , \bar { p } , \tilde { \sigma } ^ { 2 } )$ $h \in \{ 1 , 2 , \dots , H \}$ Sample ⇡˜2 from 2RNR Train oracle $\pi ^ { 1 }$ over $\rho \sim \left( \pi ^ { 1 } , \tilde { \pi } ^ { 2 } \right)$ end $\Pi ^ { 1 } = \Pi ^ { 1 } \cup \left\{ \pi ^ { 1 } \right\}$ Compute missing entries in $U ^ { \tilde { \Pi } }$ from $\tilde { \Pi } = \Pi ^ { 1 } \times \tilde { \Pi } ^ { 2 }$ by simulations $\sigma _ { R N R } ^ { 1 } , \ l _ { - } = \mathrm { R N R \_ s o l v e r } ( U ^ { \tilde { \Pi } } , \bar { p } , \tilde { \sigma } ^ { 2 } )$   
+for epoch e in $\{ 1 , 2 , \ldots , E \}$ do for episode $h \in \{ 1 , 2 , \dots , H \}$ do Play an episode against the opponent with strategy $\sigma _ { R N R } ^ { 1 }$ Collect the trajectory $\tau _ { e , h }$ and save them into $\mathbf { B }$ end $\begin{array} { r l } & { \tilde { \sigma } ^ { 2 } , \tilde { \Pi } ^ { 2 } = \mathrm { o p p o n e n t } \underline { { \mathrm { { m o d e l i n g } } } } ( \mathbf { B } ) } \\ & { \bar { p } = \frac { 1 } { | \tilde { \Pi } ^ { 2 } | } \sum _ { j } p ^ { j } \tilde { \sigma } ^ { 2 } ( j ) } \end{array}$ Compute missing entries in $U ^ { \tilde { \Pi } }$ from $\tilde { \Pi } = \Pi ^ { 1 } \times \tilde { \Pi } ^ { 2 }$ by simulations ${ \bf \underline { { \ } } } , \sigma _ { R N R } ^ { 2 } = \mathrm { R N R } _ { - } \mathrm { s o l v e r } ( U ^ { \tilde { \Pi } } , \bar { p } , \tilde { \sigma } ^ { 2 } )$ $h \in \{ 1 , 2 , \dots , H \}$ Sample ⇡˜2 from 2RNR Train oracle $\pi ^ { 1 }$ over $\rho \sim \left( \pi ^ { 1 } , \tilde { \pi } ^ { 2 } \right)$ end $\Pi ^ { 1 } = \Pi ^ { 1 } \cup \left\{ \pi ^ { 1 } \right\}$ Compute missing entries in $U ^ { \tilde { \Pi } }$ from $\tilde { \Pi } = \Pi ^ { 1 } \times \tilde { \Pi } ^ { 2 }$ by simulations $\sigma _ { R N R } ^ { 1 } , \ l _ { - } = \mathrm { R N R \_ s o l v e r } ( U ^ { \tilde { \Pi } } , \bar { p } , \tilde { \sigma } ^ { 2 } )$   
 end
 
 255 To address the above issues, we combine DO with RNR to solve a meta-game built from EGTA where