SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} - \left\langle \mathbf {M} _ {\mathbf {y}} \left(\mathbf {x} _ {k - 1} - \hat {\mathbf {x}} _ {k - 1}\right), \mathbf {y} _ {k - 1} - \mathbf {y} _ {k} \right\rangle \leq \| \mathbf {M} _ {\mathbf {y}} \left(\mathbf {x} _ {k - 1} - \hat {\mathbf {x}} _ {k - 1}\right) \| _ {*} \| \mathbf {y} _ {k - 1} - \mathbf {y} _ {k} \| \\ \leq \left\| \mathbf {M} _ {\mathbf {y}} \right\| _ {*} \left\| \mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2} \right\| \left\| \mathbf {y} _ {k - 1} - \mathbf {y} _ {k} \right\| \\ \leq \frac {\left\| \mathbf {M} _ {\mathbf {y}} \right\| _ {*}}{2} \left(\alpha \left\| \mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2} \right\| ^ {2} + \frac {1}{\alpha} \left\| \mathbf {y} _ {k} - \mathbf {y} _ {k - 1} \right\| ^ {2}\right), \tag {C.6} \\ \end{array}
$$
where the last line is by Young's inequality and holds for any  $\alpha > 0$ .
On the other hand, recalling that  $\mathbf{M}_{\mathbf{x}} = \sum_{t=1}^{s-1} \mathbf{M}_{t,t+1:s}$ , we also have
$$
\begin{array}{l} \eta_ {k} \sum_ {t = 1} ^ {s} \left\langle \mathbf {M} ^ {(t, t + 1: s)} \Big (\mathbf {x} _ {k} - \mathbf {x} _ {k - 1} - \frac {\eta_ {k - 1}}{\eta_ {k}} (\mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2}) \Big) ^ {(t + 1: s)}, \mathbf {v} ^ {(t)} - \mathbf {y} _ {k} ^ {(t)} \right\rangle \\ = \eta_ {k} \sum_ {t = 1} ^ {s} \left\langle \mathbf {M} _ {t, t + 1: s} \left(\mathbf {x} _ {k} - \mathbf {x} _ {k - 1} - \frac {\eta_ {k - 1}}{\eta_ {k}} \left(\mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2}\right)\right), \mathbf {v} - \mathbf {y} _ {k} \right\rangle \\ = \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {k} - \mathbf {x} _ {k - 1} - \frac {\eta_ {k - 1}}{\eta_ {k}} \left(\mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2}\right)\right), \mathbf {v} - \mathbf {y} _ {k} \right\rangle \\ = \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {k} - \mathbf {x} _ {k - 1}\right), \mathbf {v} - \mathbf {y} _ {k} \right\rangle - \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2}\right), \mathbf {v} - \mathbf {y} _ {k} \right\rangle \\ = \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {k} - \mathbf {x} _ {k - 1}\right), \mathbf {v} - \mathbf {y} _ {k} \right\rangle - \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2}\right), \mathbf {v} - \mathbf {y} _ {k - 1} \right\rangle \\ + \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2}\right), \mathbf {y} _ {k} - \mathbf {y} _ {k - 1} \right\rangle . \tag {C.7} \\ \end{array}
$$
Observe that in (C.7) the first two terms telescope and thus we only need to focus on bounding the last term. Applying the definitions of dual and matrix norms from Section 2 and using Young's inequality, we have that for any  $\alpha >0$
$$
\begin{array}{l} \langle \mathbf {M} _ {\mathbf {x}} (\mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2}), \mathbf {y} _ {k} - \mathbf {y} _ {k - 1} \rangle \leq \| \mathbf {M} _ {\mathbf {x}} (\mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2}) \| _ {*} \| \mathbf {y} _ {k} - \mathbf {y} _ {k - 1} \| \\ \leq \left\| \mathbf {M} _ {\mathbf {x}} \right\| _ {*} \left\| \mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2} \right\| \left\| \mathbf {y} _ {k} - \mathbf {y} _ {k - 1} \right. \\ \leq \frac {\left\| \mathbf {M} _ {\mathbf {x}} \right\| _ {*}}{2} \left(\alpha \left\| \mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2} \right\| ^ {2} + \frac {1}{\alpha} \left\| \mathbf {y} _ {k} - \mathbf {y} _ {k - 1} \right\| ^ {2}\right). \tag {C.8} \\ \end{array}
$$
Hence, combining (C.3)-(C.8), we can conclude that
$$
\begin{array}{l} \eta_ {k} \left\langle \mathbf {M} \mathbf {x} _ {k} - \mathbf {g} _ {k}, \mathbf {v} - \mathbf {y} _ {k} \right\rangle \leq \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {y}} \left(\mathbf {x} _ {k} - \hat {\mathbf {x}} _ {k}\right), \mathbf {v} - \mathbf {y} _ {k} \right\rangle - \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {y}} \left(\mathbf {x} _ {k - 1} - \hat {\mathbf {x}} _ {k - 1}\right), \mathbf {v} - \mathbf {y} _ {k - 1} \right\rangle \\ + \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {k} - \mathbf {x} _ {k - 1}\right), \mathbf {v} - \mathbf {y} _ {k} \right\rangle - \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2}\right), \mathbf {v} - \mathbf {y} _ {k - 1} \right\rangle \\ + \eta_ {k - 1} \frac {\| \mathbf {M _ {x}} \| _ {*} + \| \mathbf {M _ {y}} \| _ {*}}{2} \left(\alpha \| \mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2} \| ^ {2} + \frac {1}{\alpha} \| \mathbf {y} _ {k} - \mathbf {y} _ {k - 1} \| ^ {2}\right), \\ \end{array}
$$
completing the proof of the first claim.
Similarly, we observe from Algorithm 1 that
$$
\begin{array}{l} \left(\mathbf {M} ^ {(:, t)}\right) ^ {\top} \mathbf {y} _ {k} - \mathbf {h} _ {k} ^ {(t)} = \left(\mathbf {M} ^ {(1: t - 1, t)}\right) ^ {\top} \left(\mathbf {y} _ {k} - \hat {\mathbf {y}} _ {k} + \frac {\eta_ {k - 1}}{\eta_ {k}} (\mathbf {y} _ {k - 1} - \hat {\mathbf {y}} _ {k - 1})\right) ^ {(1: t - 1)} \\ + \left(\mathbf {M} ^ {(t: s, t)}\right) ^ {\top} \left(\mathbf {y} _ {k} - \mathbf {y} _ {k - 1} + \frac {\eta_ {k - 1}}{\eta_ {k}} (\mathbf {y} _ {k - 1} - \mathbf {y} _ {k - 2})\right) ^ {(t: s)}. \\ \end{array}
$$
Observing that  $\sum_{t=1}^{s} \mathbf{M}_{1:t-1,t} = \mathbf{M}_{\mathbf{x}}$  and  $\sum_{t=1}^{s} \mathbf{M}_{t:s,t} = \mathbf{M}_{\mathbf{y}}$ , using the same sequence of arguments as for bounding (C.3), we can conclude that for any  $\beta > 0$ ,
$$
\begin{array}{l} - \eta_ {k} \left\langle \mathbf {u} - \mathbf {x} _ {k}, \mathbf {M} ^ {\top} \mathbf {y} _ {k} - \mathbf {h} _ {k} \right\rangle \\ \leq - \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {x}} ^ {\top} (\mathbf {y} _ {k} - \hat {\mathbf {y}} _ {k}), \mathbf {u} - \mathbf {x} _ {k} \right\rangle + \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {x}} ^ {\top} (\mathbf {y} _ {k - 1} - \hat {\mathbf {y}} _ {k - 1}), \mathbf {u} - \mathbf {x} _ {k - 1} \right\rangle \\ - \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {y}} ^ {\top} (\mathbf {y} _ {k} - \mathbf {y} _ {k - 1}), \mathbf {u} - \mathbf {x} _ {k} \right\rangle + \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {y}} ^ {\top} (\mathbf {y} _ {k - 1} - \mathbf {y} _ {k - 2}), \mathbf {u} - \mathbf {x} _ {k - 1} \right\rangle \\ + \eta_ {k - 1} \frac {\left\| \mathbf {M} _ {\mathbf {x}} ^ {\top} + \mathbf {M} _ {\mathbf {y}} ^ {\top} \right\| _ {*}}{2} \Big (\beta \| \mathbf {x} _ {k - 1} - \mathbf {x} _ {k} \| ^ {2} + \frac {1}{\beta} \| \mathbf {y} _ {k - 1} - \mathbf {y} _ {k - 2} \| ^ {2} \Big), \\ \end{array}
$$
Proof Theorem 3.2. Recalling the definition of  $\mathcal{E}_k$ , by Lemma C.1,
$$
\begin{array}{l} \mathcal {E} _ {k} \leq \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {y}} \left(\mathbf {x} _ {k} - \hat {\mathbf {x}} _ {k}\right), \mathbf {v} - \mathbf {y} _ {k} \right\rangle - \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {y}} \left(\mathbf {x} _ {k - 1} - \hat {\mathbf {x}} _ {k -}\right), \mathbf {v} - \mathbf {y} _ {k - 1} \right\rangle \\ + \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {k} - \mathbf {x} _ {k - 1}\right), \mathbf {v} - \mathbf {y} _ {k} \right\rangle - \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2}\right), \mathbf {v} - \mathbf {y} _ {k - 1} \right\rangle \\ + \eta_ {k - 1} \frac {\left\| \mathbf {M} _ {\mathbf {x}} \right\| _ {*} + \left\| \mathbf {M} _ {\mathbf {y}} \right\| _ {*}}{2} \left(\alpha \left\| \mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2} \right\| ^ {2} + \frac {1}{\alpha} \left\| \mathbf {y} _ {k} - \mathbf {y} _ {k - 1} \right\| ^ {2}\right) \\ - \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {x}} ^ {\top} \left(\mathbf {y} _ {k} - \hat {\mathbf {y}} _ {k}\right), \mathbf {u} - \mathbf {x} _ {k} \right\rangle + \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {x}} ^ {\top} \left(\mathbf {y} _ {k - 1} - \hat {\mathbf {y}} _ {k - 1}\right), \mathbf {u} - \mathbf {x} _ {k - 1} \right\rangle \tag {C.9} \\ - \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {y}} ^ {\top} (\mathbf {y} _ {k} - \mathbf {y} _ {k - 1}), \mathbf {u} - \mathbf {x} _ {k} \right\rangle + \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {y}} ^ {\top} (\mathbf {y} _ {k - 1} - \mathbf {y} _ {k - 2}), \mathbf {u} - \mathbf {x} _ {k - 1} \right\rangle \\ + \eta_ {k - 1} \frac {\left\| \mathbf {M} _ {\mathbf {x}} ^ {\top} \right\| _ {*} + \left\| \mathbf {M} _ {\mathbf {y}} ^ {\top} \right\| _ {*}}{2} \left(\beta \| \mathbf {x} _ {k - 1} - \mathbf {x} _ {k} \| ^ {2} + \frac {1}{\beta} \| \mathbf {y} _ {k - 1} - \mathbf {y} _ {k - 2} \| ^ {2}\right) \\ - D _ {\psi} \left(\mathbf {y} _ {k}, \mathbf {y} _ {k - 1}\right) - D _ {\phi} \left(\mathbf {x} _ {k}, \mathbf {x} _ {k - 1}\right). \\ \end{array}
$$
Recalling that  $\psi$  is  $c_{\psi}$ -strongly convex,  $\phi$  is  $c_{\phi}$ -strongly convex, setting  $\alpha = \beta = \sqrt{\frac{c_{\phi}}{c_{\psi}}}$ , and using that  $\eta_{k-1} \leq \frac{\sqrt{c_{\phi} c_{\psi}}}{\|\mathbf{M}_{\mathbf{x}}\|_{*} + \|\mathbf{M}_{\mathbf{y}}\|_{*} + \|\mathbf{M}_{\mathbf{x}}^{\top}\|_{*} + \|\mathbf{M}_{\mathbf{y}}^{\top}\|_{*}} = \frac{\sqrt{c_{\phi} c_{\psi}}}{\mu_x + \mu_y}$ , (C.9) simplifies to
$$
\begin{array}{l} \mathcal {E} _ {k} \leq \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {y}} \left(\mathbf {x} _ {k} - \hat {\mathbf {x}} _ {k}\right), \mathbf {v} - \mathbf {y} _ {k} \right\rangle - \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {y}} \left(\mathbf {x} _ {k - 1} - \hat {\mathbf {x}} _ {k -}\right), \mathbf {v} - \mathbf {y} _ {k - 1} \right\rangle \\ + \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {k} - \mathbf {x} _ {k - 1}\right), \mathbf {v} - \mathbf {y} _ {k} \right\rangle - \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2}\right), \mathbf {v} - \mathbf {y} _ {k - 1} \right\rangle \\ - \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {x}} ^ {\top} \left(\mathbf {y} _ {k} - \hat {\mathbf {y}} _ {k}\right), \mathbf {u} - \mathbf {x} _ {k} \right\rangle + \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {x}} ^ {\top} \left(\mathbf {y} _ {k - 1} - \hat {\mathbf {y}} _ {k - 1}\right), \mathbf {u} - \mathbf {x} _ {k - 1} \right\rangle \\ - \eta_ {k} \left\langle \mathbf {M} _ {\mathbf {y}} ^ {\top} \left(\mathbf {y} _ {k} - \mathbf {y} _ {k - 1}\right), \mathbf {u} - \mathbf {x} _ {k} \right\rangle + \eta_ {k - 1} \left\langle \mathbf {M} _ {\mathbf {y}} ^ {\top} \left(\mathbf {y} _ {k - 1} - \mathbf {y} _ {k - 2}\right), \mathbf {u} - \mathbf {x} _ {k - 1} \right\rangle \tag {C.10} \\ + \frac {c _ {\psi} \mu_ {y}}{2 \left(\mu_ {x} + \mu_ {y}\right)} \left(\| \mathbf {y} _ {k - 1} - \mathbf {y} _ {k - 2} \| ^ {2} - \| \mathbf {y} _ {k} - \mathbf {y} _ {k - 1} \| ^ {2}\right) \\ + \frac {c _ {\phi} \mu_ {x}}{2 (\mu_ {x} + \mu_ {y})} \big (\| \mathbf {x} _ {k - 1} - \mathbf {x} _ {k - 2} \| ^ {2} - \| \mathbf {x} _ {k} - \mathbf {x} _ {k - 1} \| ^ {2} \big). \\ \end{array}
$$
Telescoping (C.10) and recalling that  $\eta_0 = 0$ , we now have
$$
\begin{array}{l} \sum_ {k = 1} ^ {K} \mathcal {E} _ {k} \leq \eta_ {K} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {K} - \mathbf {x} _ {K - 1}\right), \mathbf {v} - \mathbf {y} _ {K} \right\rangle - \eta_ {K} \left\langle \mathbf {M} _ {\mathbf {y}} ^ {\top} \left(\mathbf {y} _ {K} - \mathbf {y} _ {K - 1}\right), \mathbf {u} - \mathbf {x} _ {K} \right\rangle \\ + \eta_ {K} \left\langle \mathbf {M} _ {\mathbf {y}} \left(\mathbf {x} _ {K} - \hat {\mathbf {x}} _ {K}\right), \mathbf {v} - \mathbf {y} _ {K} \right\rangle - \eta_ {K} \left\langle \mathbf {M} _ {\mathbf {x}} ^ {\top} \left(\mathbf {y} _ {K} - \hat {\mathbf {y}} _ {K}\right), \mathbf {u} - \mathbf {x} _ {K} \right\rangle \tag {C.11} \\ - \frac {c _ {\psi} \mu_ {y}}{2 \left(\mu_ {x} + \mu_ {y}\right)} \left\| \mathbf {y} _ {K} - \mathbf {y} _ {K - 1} \right\| ^ {2} \\ - \frac {c _ {\phi} \mu_ {x}}{2 (\mu_ {x} + \mu_ {y})} \| \mathbf {x} _ {K} - \mathbf {x} _ {K - 1} \| ^ {2}. \\ \end{array}
$$
Observe that  $\mathrm{Gap}^{\mathbf{u},\mathbf{v}}(\cdot ,\cdot)$  is linear in both its arguments. Hence,  $\mathrm{Gap}^{\mathbf{u},\mathbf{v}}(\bar{\mathbf{x}}_K,\bar{\mathbf{y}}_K) = \frac{1}{H_K}\sum_{k = 1}^{K}\eta_k\mathrm{Gap}^{\mathbf{u},\mathbf{v}}(\mathbf{x}_k,\mathbf{y}_k)$ . Applying Proposition 3.1 and combining with (C.11), we now have
$$
\begin{array}{l} H _ {K} \operatorname {G a p} ^ {\mathbf {u}, \mathbf {v}} \left(\bar {\mathbf {x}} _ {k}, \bar {\mathbf {y}} _ {k}\right) \leq D _ {\phi} \left(\mathbf {u}, \mathbf {x} _ {0}\right) + D _ {\psi} \left(\mathbf {v}, \mathbf {y} _ {0}\right) \\ + \eta_ {K} \left\langle \mathbf {M} _ {\mathbf {x}} \left(\mathbf {x} _ {K} - \mathbf {x} _ {K - 1}\right), \mathbf {v} - \mathbf {y} _ {K} \right\rangle - \eta_ {K} \left\langle \mathbf {M} _ {\mathbf {y}} ^ {\top} \left(\mathbf {y} _ {K} - \mathbf {y} _ {K - 1}\right), \mathbf {u} - \mathbf {x} _ {K} \right\rangle \\ + \eta_ {K} \left\langle \mathbf {M _ {y}} \left(\mathbf {x} _ {K} - \hat {\mathbf {x}} _ {K}\right), \mathbf {v} - \mathbf {y} _ {K} \right\rangle - \eta_ {K} \left\langle \mathbf {M _ {x} ^ {\top}} \left(\mathbf {y} _ {K} - \hat {\mathbf {y}} _ {K}\right), \mathbf {u} - \mathbf {x} _ {K} \right\rangle \\ - \frac {c _ {\psi} \mu_ {y}}{2 \left(\mu_ {x} + \mu_ {y}\right)} \left\| \mathbf {y} _ {K} - \mathbf {y} _ {K - 1} \right\| ^ {2} - \frac {c _ {\phi} \mu_ {x}}{2 \left(\mu_ {x} + \mu_ {y}\right)} \left\| \mathbf {x} _ {K} - \mathbf {x} _ {K - 1} \right\| ^ {2} \\ - D _ {\phi} \left(\mathbf {u}, \mathbf {x} _ {K}\right) - D _ {\psi} \left(\mathbf {v}, \mathbf {y} _ {K}\right). \tag {C.12} \\ \end{array}
$$
To complete bounding the gap, it remains to argue that the right-hand side of (C.12) is bounded by  $D_{\phi}(\mathbf{u}, \mathbf{x}_0) + D_{\psi}(\mathbf{v}, \mathbf{y}_0) - \frac{\mu_x}{\mu_x + \mu_y} D_{\phi}(\mathbf{u}, \mathbf{x}_K) - \frac{\mu_y}{\mu_x + \mu_y} D_{\psi}(\mathbf{v}, \mathbf{y}_K)$ . This is done using the same sequence of arguments as in bounding  $\mathcal{E}_k$  and is omitted.
Let  $(\mathbf{x}^{*},\mathbf{y}^{*})\in \mathcal{X}\times \mathcal{Y}$  be any primal-dual solution to (PD). Then  $\mathrm{Gap}^{(\mathbf{x}^{*},\mathbf{y}^{*})}(\bar{\mathbf{x}}_K,\bar{\mathbf{y}}_K)\geq 0$  and we can conclude that
$$
\frac {\mu_ {x}}{\mu_ {x} + \mu_ {y}} D _ {\phi} (\mathbf {x} ^ {*}, \mathbf {x} _ {K}) + \frac {\mu_ {y}}{\mu_ {x} + \mu_ {y}} D _ {\psi} (\mathbf {y} ^ {*}, \mathbf {y} _ {K}) \leq D _ {\phi} (\mathbf {x} ^ {*}, \mathbf {x} _ {0}) + D _ {\psi} (\mathbf {y} ^ {*}, \mathbf {y} _ {0}).
$$
Further, using that  $D_{\phi}(\cdot ,\cdot)\geq 0$ $D_{\psi}(\cdot ,\cdot)\geq 0$  , we can also conclude that
$$
\begin{array}{l} \sup  _ {\mathbf {u} \in \mathcal {X}, \mathbf {v} \in \mathcal {Y}} \operatorname {G a p} ^ {\mathbf {u}, \mathbf {v}} (\bar {\mathbf {x}} _ {K}, \bar {\mathbf {y}} _ {K}) = \sup  _ {\mathbf {u} \in \mathcal {X}, \mathbf {v} \in \mathcal {Y}} \left\{\langle \mathbf {M} \bar {\mathbf {x}} _ {K}, \mathbf {v} \rangle - \langle \mathbf {M} \mathbf {u}, \bar {\mathbf {y}} _ {K} \rangle \right\} \\ \leq \frac {\operatorname* {s u p} _ {\mathbf {u} \in \mathcal {X} , \mathbf {v} \in \mathcal {Y}} \left\{D _ {\phi} (\mathbf {u} , \mathbf {x} _ {0}) + D _ {\psi} (\mathbf {v} , \mathbf {y} _ {0}) \right\}}{H _ {K}}. \\ \end{array}
$$
Finally, setting  $\eta_{k} = \frac{\sqrt{c_{\phi}c_{\psi}}}{\mu_{x} + \mu_{y}}$  for all  $k\geq 1$  immediately leads to the conclusion that  $H_{K} = K\frac{\sqrt{c_{\phi}c_{\psi}}}{\mu_{x} + \mu_{y}}$  as  $H_K = \sum_{k = 1}^{K}\eta_k$  , by definition. The last bound is by setting  $\frac{\sup_{\mathbf{u}\in\mathcal{X},\mathbf{v}\in\mathcal{Y}}\{D_{\phi}(\mathbf{u},\mathbf{x}_0) + D_{\psi}(\mathbf{v},\mathbf{y}_0)\}}{H_K}\leq \epsilon ,$  and solving for  $K$
D Algorithm Implementation Details
In Algorithm 2, we present an implementation-specific version of ECyclicPDA, in order to make it clear that our algorithm can be implemented without any extra computation compared to the
computation needed for gradient and prox updates in MP. Note that MP performs two gradient computations and two prox computations per player, due to how it achieves "extrapolation"; we want to argue that we perform an equivalent number operations as needed for a single gradient computation and prox computation per player. Note that the overall complexity of first-order methods when applied to EFGs is dominated by the gradient and prox update computations; this is why we compare our algorithm to MP on this basis. The key differences from Algorithm 1 are that we explicitly use  $\hat{\mathbf{x}}_k$  and  $\hat{\mathbf{y}}_k$  to represent the behavioral strategy that is computed via the partial prox updates (which are then scaled at the end of a full iteration of our method to  $\mathbf{x}_k$  and  $\mathbf{y}_k$ ), and that we use  $\hat{\mathbf{h}}_k^j$  and  $\hat{\mathbf{g}}_k^j$  to accumulate gradient contributions from decision points that occur underneath  $j$ , to make the partial prox update explicit.
In Lines 8 and 13, we are only dealing with the columns and rows, respectively, of the payoff matrix that correspond to the current block number  $t$ , which means that as  $t$  ranges from 1 to  $s$ , for the computation of the gradient, we will only consider each column and row, respectively, once, as would have to be done in a full gradient computation for MP.
The more difficult aspect of the implementation is ensuring that we do the same number of operations for the prox computation in ECyclicPDA as an analogous single prox computation in MP. We achieve this by applying the updates in Proposition 2.2 only for the decision points in the current block, in Lines 9 to 12 for  $\mathbf{x}$  and 14 to 17 for  $\mathbf{y}$ .
We focus on the updates for  $\mathbf{x}$ ; the argument is analogous for  $\mathbf{y}$ . When applying this local prox update for decision point  $j \in \mathcal{J}_{\mathcal{X}}^{(t)}$ , we have already correctly computed  $\hat{\mathbf{h}}^j$ , the contributions to the gradient for the local prox update that originate from the children of  $j$ , again because the blocks represent the treeplex ordering; in particular, whenever we have encountered a child decision point of  $j$  in the past, we accumulate its contribution to the gradient for its parent at  $\hat{\mathbf{h}}^{p_j}$ . Since the prox update decomposition from Proposition 2.2 has to be applied for every single decision point in  $\mathcal{J}_{\mathcal{X}}$  in a full prox update (as done in MP), we again do not incur any dependence on the number of blocks.
E Description of EFG Benchmarks
We provide game descriptions of the games we run our experiments on below. Our game descriptions are adapted from Farina et al. [15]. In Table 2, we provide the number of sequences for player  $\mathbf{x}(n)$ , the number of sequences for player  $\mathbf{y}(m)$ , and the number of leaves in the game (NNZ(M)).
Table 2: Number of sequences for both players and number of leaves for each game. These correspond to the dimensions  $n$  and  $m$  of  $\mathbf{M}$ , and the number of nonzero entries of  $\mathbf{M}$ , respectively.
Game	Num. of x sequences	Num. of y sequences	Num. of leaves
Goofspiel (4 ranks)	21329	21329	13824
Liar's Dice	24571	24571	147420
Leduc (13 ranks)	6007	6007	98956
Battleship	73130	253940	552132
E.1 Goofspiel (4 ranks)
Goofspiel is a card-based game that is a standard benchmark in the EFG-solving community [36]. In the version that we test on, there are 4 unique cards (ranks), and there are 3 copies of each rank, divided into 3 separate decks. Each player gets a deck, and the third deck is known as the prize deck. Cards are randomly drawn from the prize deck, and each player submits a bid for the drawn card by submitting a card from one of their respective decks, the value of which represents their bid. Whoever submits the higher bid wins the card from the prize deck. Once all the cards from the prize deck have been drawn, bid on, and won by one of the players, the game terminates, and the payoffs for players are given by the sum of the prize cards they won.
Algorithm 2 Extrapolated Cyclic Primal-Dual EFG Solver (Implementation Version)
1: Input: M, m, n  
2: Initialization:  $\mathbf{x}_0 \in \mathcal{X}, \mathbf{y}_0 \in \mathcal{Y}, \eta_0 = H_0 = 0, \eta = \frac{\sqrt{c_\phi c_\psi}}{\mu_x + \mu_y}$ ,  $\bar{\mathbf{x}}_0 = \mathbf{x}_0$ ,  $\bar{\mathbf{y}}_0 = \mathbf{y}_0$ ,  $\mathbf{g}_0 = \mathbf{0}$ ,  $\mathbf{h}_0 = \mathbf{0}$   
3: for  $k = 1:K$  do  
4: Choose  $\eta_k \leq \eta$ ,  $H_k = H_{k-1} + \eta_k$   
5:  $\mathbf{g}_k = \mathbf{0}$ ,  $\mathbf{h}_k = \mathbf{0}$ ,  $\hat{\mathbf{g}}_k = \mathbf{0}$ ,  $\hat{\mathbf{h}}_k = \mathbf{0}$   
6:  $\tilde{\mathbf{x}}_k = \mathbf{x}_{k-1} + \frac{\eta_{k-1}}{\eta_k} (\mathbf{x}_{k-1} - \mathbf{x}_{k-2})$ ,  $\tilde{\mathbf{y}}_k = \mathbf{y}_{k-1} + \frac{\eta_{k-1}}{\eta_k} (\mathbf{y}_{k-1} - \mathbf{y}_{k-2})$   
7: for  $t = 1:s$  do  
8:  $\mathbf{h}_k^{(t)} = (\mathbf{M}^{(:,t)})^\top \tilde{\mathbf{y}}_k$   
9: for  $j \in \mathcal{J}_{\chi}^{(t)}$  do  
10:  $\hat{\mathbf{x}}_k^j = \mathrm{argmin}_{\mathbf{b}^j \in \Delta^{n_j}}\left\{\left\langle \hat{\mathbf{h}}_k^j + \mathbf{h}_k^j, \mathbf{b}^j \right\rangle + D_{\phi^j}\left(\mathbf{b}^j, \hat{\mathbf{x}}_{k-1}^j\right)\right\}$   
11:  $(j', a) = p_j$   
12:  $\hat{h}_k^{p_j} += [\phi^{\downarrow j^*}\left(-\mathbf{h}_k^{\downarrow j} + \nabla \phi^{\downarrow j}\left(x_{k-1}^{\downarrow j}\right)\right) - \phi^j\left(\hat{\mathbf{x}}_{k-1}^j\right) + \left\langle\nabla \phi^j\left(\hat{\mathbf{x}}_{k-1}^j\right), \hat{\mathbf{x}}_{k-1}^j\right\rangle]$   
13:  $\mathbf{g}_k^{(t)} = \mathbf{M}^{(t,:)}\tilde{\mathbf{x}}_k$   
14: for  $j \in \mathcal{J}_{\chi}^{(t)}$  do  
15:  $\hat{\mathbf{y}}_k^j = \mathrm{argmin}_{\mathbf{b}^j \in \Delta^{n_j}}\left\{\left\langle\hat{\mathbf{g}}_k^j + \mathbf{g}_k^j,\mathbf{b}^j\right\rangle + D_{\psi^j}\left(\mathbf{b}^j,\hat{\mathbf{y}}_{k-1}^j\right)\right\}$   
16:  $(j', a) = p_j$   
17:  $\hat{h}_k^{p_j} += [\psi^{\downarrow j^*}\left(-\mathbf{g}_k^{\downarrow j} + \nabla \psi^{\downarrow j}\left(y_{k-1}^{\downarrow j}\right)\right) - \psi^j\left(\hat{\mathbf{y}}_{k-1}^j\right) + \left\langle\nabla \psi^j\left(\hat{\mathbf{y}}_{k-1}^j\right), \hat{\mathbf{y}}_{k-1}^j\right\rangle]$   
18: for  $j \in \mathcal{J}_{\chi}^{(t)}$  do  
19:  $\tilde{\mathbf{x}}_k^j = [\hat{\mathbf{x}}_k^j x_{k-1}^{p_j} + \frac{\eta_{k-1}}{\eta_k}\left(\mathbf{x}_{k-1}^j - \hat{\mathbf{x}}_{k-1}^j x_{k-2}^{p_j}\right)]$   
20: for  $j \in \mathcal{J}_{\chi}^{(t)}$  do  
21:  $\tilde{\mathbf{y}}_k^j = [\hat{\mathbf{y}}_k^j y_{k-1}^{p_j} + \frac{\eta_{k-1}}{\eta_k}\left(\mathbf{y}_{k-1}^j - \hat{\mathbf{y}}_{k-1}^j y_{k-2}^{p_j}\right)]$   
22: for  $j \in \mathcal{J}_{\chi}$  do  
23:  $\mathbf{x}_k^j = x_k^{p_j} \cdot \hat{\mathbf{x}}_k^j$   
24: for  $j \in \mathcal{J}_{\chi}$  do  
25:  $\hat{\mathbf{y}}_k^j = y_k^{p_j} \cdot \hat{\mathbf{y}}_k^j$   
26:  $\bar{\mathbf{x}}_k = \frac{H_k - \eta_k}{H_k} \bar{\mathbf{x}}_{k-1} + \frac{\eta_k}{H_k} x_k, \bar{\mathbf{y}}_k = \frac{H_k - \eta_k}{H_k} \bar{\mathbf{y}}_{k-1} + \frac{\eta_k}{H_k} y_k$   
27: Return:  $\bar{\mathbf{x}}_K, \bar{\mathbf{y}}_K$
E.2 Liar's Dice