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@@ -10,7 +10,7 @@
 
 \begin{document}
 \maketitle
-A recent paper shows that \expt{} algorithm for adversarial bandits can be implemented in $O(1)$ expected time per round. \url{https://arxiv.org/pdf/2512.11201v1}
+A recent paper\footnote{\url{https://arxiv.org/pdf/2512.11201v1}} shows that \expt{} algorithm for adversarial bandits can be implemented in $O(1)$ expected time per round (under RAM model).
 
 \section{Problem Settings}
 
@@ -26,6 +26,56 @@ The player's goal is to minimize the expected pseudo-regret $\bar R_T$ defined a
 
 \expt{} Algorithm works like multiplicative weight update method for linear programs.
 
+\begin{algo}
+\underline{\expt{}}:\\
+learning rate $\eta=\sqrt{\frac{2 \ln K}{KT}}$\\
+initial weights $\{w_{1,i}=1\}_{i\in[K]}$\\
+in each round $t$:\\
+\quad select arm $i$ with probability $p_{t,i}=\frac{w_{t,i}}{\sum_{j\in[K]} w_{t,j}}$\\
+\quad update weights:\\ 
+\quad \quad $w_{t+1,a_t}=w_{t,a_t}\exp(-\eta \frac{\ell_{t,a_t}}{p_{t,a_t}})$,
+                        $w_{t+1,i}=w_{t,i}$ for $i\neq a_t$.
+\end{algo}
 
+\expt{} has regret $\bar R_T\leq \sqrt{2}\cdot \sqrt{TK\ln K}$.
+Note that unlike MWU, for \expt{} the weight update happens for only 1 entry in each round. So the hard part is sampling.
+
+\section{$O(\log K)$ Implementation}
+
+We can build a segment tree on the array $\set{w_{t,i}}_{i\in[K]}$. Single update in the segment tree takes $O(\log K)$ time.
+Hence, we forcus on the sampling part.
+
+Let $W_t$ be $\sum_{i\in [K]} w_{t,i}$ and let $p$ be a random number sampled uniformly from $[0,W_t]$. 
+Note that the prefix sums of array $\set{w_{t,i}}_{i\in[K]}$ divides the interval $[0,W_t]$ into $K$ parts and there is a bijection from these parts to bandit arms.
+The probability of $r$ falling in part $i$ is exactly $\frac{w_{t,i}}{W_t}$.
+One can see that the sampling part also takes $O(\log K)$ time on the segment tree.
+
+\section{Alias Method}
+
+Alias method is a static data structure which after a $O(K)$ preprocessing samples element with probability $\frac{w_{t,i}}{W_t}$ in worst case $O(1)$ time.
+
+We drop the subscripts $t$ since alias method is static. Let $\bar W$ be the average weight $W/K$.
+The preprocess works as follows.
+\begin{enumerate}
+\item Divide elements in $\set{w_i}_{i\in [K]}$ into 2 groups. Large group contains elements with $w_i\geq \bar W$. The rest is Small group.
+\item Repeat the following $K$ times. Pick an element $s$ from Small group and an element $l$ from the Large group. Pack $s$ with weight $w_s$ and $l$ with weight $\bar W-w_s$ into a new ``bin''. Remove $s$ from Small group. Then update the weight of $l$ to $w_l-(\bar W-w_s)$. If $w_l$ becomes smaller than $\bar W$, move $l$ to the Small group.
+\end{enumerate}
+
+\begin{remark}
+The preprocess steps are well-defined. One can see in each step we remove one element from the Small group and the total weight decrease by $\bar W$. So the average weight of remaining elements is always $\bar W$ and the Large group is non-empty throughout the preprocessing. After $K$ steps both the Large group and the Small group are empty.
+\end{remark}
+
+When sampling elements, we first uniformly select one ``bin'' (which takes constant time) and then choose element $s$ with $w_s/\bar W$ or $l$ with $1-w_s/\bar W$.
+
+One can see that for small element $s$,
+\[\Pr[\text{$s$ is selected}]=\frac{1}{K}\cdot\frac{w_s}{\bar W}=\frac{w_s}{W}\]
+and for large element $l$,
+\[\Pr[\text{$l$ is selected}]=
+\sum_{\text{bin containing $l$}}\frac{1}{K} \frac{\bar W-w_{s_i}}{\bar W}=\frac{w_l}{W}\]
+which are the desired propabilities.
+
+\section{$O(1)$ Implementation}
+
+The fun part ends here. The idea is standard. They rebuild the alias method data structure every $K$ rounds and always use the newly built one for sampling. Since the alias method structure is always outdated, they use reject sampling with accept chance $\frac{W_t}{W_\tau}$, where $\tau$ is when the last alias structure was built. This gives a $O(1)$ amortized expected time sampling per round.
 
 \end{document}