SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Theorem B.13. Under Assumptions 4.1. Given an inexact deterministic gradient oracle in Definition 4 with  $\delta \leq \mathcal{O}(\epsilon)$ . The average iterate  $(\bar{x}_{T + 1 / 2},\bar{y}_{T + 1 / 2})$  of  $EG$  satisfies  $\max_{y\in \mathcal{Y}}F(\bar{x}_{T + 1 / 2},y) - \min_{x\in \mathcal{X}}F(x,\bar{y}_{T + 1 / 2})\leq \mathcal{O}(\epsilon)$  with complexity  $T = \mathcal{O}(1 / (\epsilon \sqrt{\delta}))$  if setting stepsize  $\alpha = \min \{1 / \ell ,(\delta /(2\ell^{2}))^{1 / 2}\}$ . The reproducibility is  $\mathcal{O}(\delta^2 /\epsilon^2)$ .
Proof. The same as Section B.3.3, since  $\alpha \ell \leq 1$  and  $\alpha^2 \leq \delta / (2\ell^2)$ , we have that
$$
\begin{array}{l} \| \bar {x} _ {T + 1 / 2} - \bar {x} _ {T + 1 / 2} ^ {\prime} \| ^ {2} + \| \bar {y} _ {T + 1 / 2} - \bar {y} _ {T + 1 / 2} ^ {\prime} \| ^ {2} \leq 8 \alpha^ {2} \delta^ {2} + \frac {4}{T} \sum_ {t = 0} ^ {T - 1} (2 \alpha^ {3} \ell^ {2} L + 4 \alpha \delta) ^ {2} t ^ {2} \\ \leq 8 \left(\left(2 L \ell^ {2} \alpha^ {2}\right) ^ {2} \alpha^ {2} T ^ {2} + 8 \delta^ {2} \alpha^ {2} T ^ {2} + \delta^ {2} \alpha^ {2}\right) \\ \leq 8 (L ^ {2} + 9) \delta^ {2} (\alpha T) ^ {2}. \\ \end{array}
$$
When the stepsize  $\alpha \leq 1 / \ell$ , we also have
$$
\begin{array}{l} \max  _ {y \in \mathcal {Y}} F (\bar {x} _ {T + 1 / 2}, y) - \min  _ {x \in \mathcal {X}} F (x, \bar {y} _ {T + 1 / 2}) \leq \frac {D ^ {2}}{\alpha T} + 3 \sqrt {2} \delta D \\ \leq \frac {\ell D ^ {2}}{T} + \frac {\sqrt {2} \ell D ^ {2}}{T \sqrt {\delta}} + 3 \sqrt {2} \delta D. \\ \end{array}
$$
To guarantee  $\mathcal{O}(\epsilon)$ -saddle point, we need to ensure  $\delta \leq \mathcal{O}(\epsilon)$  and  $\alpha T = c / \epsilon$  for some constant  $c$ . This means a  $\mathcal{O}(1 / (\epsilon \sqrt{\delta}))$  convergence rate with reproducibility  $\mathcal{O}(\delta^2 /\epsilon^2)$ . Since  $\delta \leq \mathcal{O}(\epsilon)$ , the gradient complexity is  $\mathcal{O}(1 / \epsilon^{3 / 2})$ .
Finally, we want to mention that the analysis can also be extended to reproducibility under stochastic gradient oracle and stability of Extragradient [16] that matches with SGDA. We will not provide all details here. The key is to select stepsize  $\alpha$  to balance the convergence  $\mathcal{O}(1 / (\alpha T))$  in Lemma B.8 and the error term  $\mathcal{O}(\alpha^3 T)$  that appears according to Lemma B.9. Moreover, we also acknowledge that it is unclear whether the analysis of EG is tight since the specific lower-bound is unknown. We leave this problem for future exploration.
C Near-optimal Guarantees in the Minimax Case
This section discusses near-optimal guarantees for algorithmic reproducibility and gradient complexity in smooth convex-concave minimax optimization.
C.1 Useful Lemmas
We first establish the convergence behavior of gradient descent ascent (GDA) and Extragradient (EG) [48] for smooth and strongly-convex-strongly-concave (SC-SC) functions under the inexact gradient oracle in Definition 4. For the sake of simplicity and to enable a general analysis, we slightly abuse notation here to consider the minimax optimization problem
$$
\min  _ {x \in \mathcal {X}} \max  _ {y \in \mathcal {Y}} f (x, y),
$$
where  $f:\mathcal{X}\times \mathcal{Y}\to \mathbb{R}$  satisfies the following assumption.
Assumption C.1. The function  $f(x,y)$  is  $\ell$ -smooth and  $\mu$ -strongly-convex-strongly-concave on the closed convex domain  $\mathcal{X} \times \mathcal{Y}$ .
Assumption C.2. We assume the existence of an inexact gradient oracle that returns a vector  $g(x,y) = (g_x(x,y),g_y(x,y))$  at any querying point  $(x,y)\in \mathcal{X}\times \mathcal{Y}$  such that  $\| \nabla f(x,y) - g(x,y)\| ^2\leq \delta^2$  where  $\nabla f(x,y) = (\nabla_{x}f(x,y),\nabla_{y}f(x,y))$  is the true gradient at  $(x,y)$ .
The lemma below shows the convergence behavior of GDA under the inexact gradient oracle presented above, also referred to as Inexact-GDA.
Lemma C.3. Under Assumption C.1. Let  $z^{*} = (x^{*},y^{*})\in \mathcal{X}\times \mathcal{Y}$  be the unique saddle point of  $f(x,y)$  and  $\kappa \coloneqq \ell /\mu$  be the condition number. Given an inexact gradient oracle in Assumption C.2. Denote  $z_{t} = (x_{t},y_{t})$  and  $\tilde{g} (z_t) = (g_x(x_t,y_t), - g_y(x_t,y_t))$ . Starting from  $z_0\in \mathcal{X}\times \mathcal{Y}$ , GDA that updates for  $t = 0,1,\dots ,T - 1$
$$
z _ {t + 1} = \Pi_ {\mathcal {X} \times \mathcal {Y}} \left(z _ {t} - \alpha \tilde {g} \left(z _ {t}\right)\right), \quad (\text {I n e x a c t - G D A})
$$
with stepsize  $\alpha = \mu /(4\ell^2)$  converges with
$$
\left\| z _ {T} - z ^ {*} \right\| ^ {2} \leq \exp \left(- \frac {T}{8 \kappa^ {2}}\right) \left\| z _ {0} - z ^ {*} \right\| ^ {2} + \left(\frac {1}{\ell^ {2}} + \frac {2}{\mu^ {2}}\right) \delta^ {2}.
$$
Proof. Let  $\tilde{\nabla} f(z_{t}) = (\nabla_{x}f(x_{t},y_{t}), - \nabla_{y}f(x_{t},y_{t}))$ . It holds that  $z^{*} = \Pi_{\mathcal{X}\times \mathcal{Y}}(z^{*} - \alpha \tilde{\nabla} f(z^{*}))$  since the saddle point problem and the projection problem share the same optimality condition when  $f(x,y)$  is convex-concave (see Proposition 1.4.2 in Facchinei and Pang [32]) such that
$$
\tilde {\nabla} f (z ^ {*}) ^ {\top} (z - z ^ {*}) \geq 0, \quad \forall z = (x, y) \in \mathcal {X} \times \mathcal {Y}.
$$
Therefore, similarly to (11), by the GDA updates, we have
$$
\begin{array}{l} \left\| z _ {t + 1} - z ^ {*} \right\| ^ {2} = \left\| \Pi_ {\mathcal {X} \times \mathcal {Y}} \left(z _ {t} - \alpha \tilde {g} \left(z _ {t}\right)\right) - \Pi_ {\mathcal {X} \times \mathcal {Y}} \left(z ^ {*} - \alpha \tilde {\nabla} f \left(z ^ {*}\right)\right) \right\| ^ {2} \\ \leq \left\| \left(z _ {t} - z ^ {*}\right) - \alpha \left(\tilde {g} \left(z _ {t}\right) - \tilde {\nabla} f \left(z ^ {*}\right)\right) \right\| ^ {2} \\ = \left\| z _ {t} - z ^ {*} \right\| ^ {2} - 2 \alpha \left(\tilde {g} \left(z _ {t}\right) - \tilde {\nabla} f \left(z ^ {*}\right)\right) ^ {\top} \left(z _ {t} - z ^ {*}\right) + \alpha^ {2} \left\| \tilde {g} \left(z _ {t}\right) - \tilde {\nabla} f \left(z ^ {*}\right) \right\| ^ {2}. \\ \end{array}
$$
Since  $\tilde{\nabla} f$  is  $\mu$ -strongly-monotone if  $f(x, y)$  is  $\mu$ -strongly-convex-strongly-concave [66, 35], i.e.,  $\forall z_1, z_2 \in \mathcal{X} \times \mathcal{Y}$ ,  $(\tilde{\nabla} f(z_1) - \tilde{\nabla} f(z_2))^{\top}(z_1 - z_2) \geq \mu \| z_1 - z_2\|^2$ , we have that
$$
\begin{array}{l} \left. \left(\tilde {g} \left(z _ {t}\right) - \tilde {\nabla} f \left(z ^ {*}\right)\right) ^ {\top} \left(z _ {t} - z ^ {*}\right) = \left(\tilde {\nabla} f \left(z _ {t}\right) - \tilde {\nabla} f \left(z ^ {*}\right)\right) ^ {\top} \left(z _ {t} - z ^ {*}\right) + \left(\tilde {g} \left(z _ {t}\right) - \tilde {\nabla} f \left(z _ {t}\right)\right) ^ {\top} \left(z _ {t} - z ^ {*}\right) \right. \\ \geq \mu \| z _ {t} - z ^ {*} \| ^ {2} - \delta \| z _ {t} - z ^ {*} \| \\ \geq \frac {\mu}{2} \| z _ {t} - z ^ {*} \| ^ {2} - \frac {\delta^ {2}}{2 \mu}, \\ \end{array}
$$
where we use Assumption C.2 such that  $\| \tilde{g}(z_t) - \tilde{\nabla} f(z_t) \| \leq \delta$  and fact (iii) in Lemma A.1. Then by  $\ell$ -smoothness of  $f(x, y)$ , we can obtain that
$$
\begin{array}{l} \| \tilde {g} (z _ {t}) - \tilde {\nabla} f (z ^ {*}) \| ^ {2} \leq 2 \| \tilde {g} (z _ {t}) - \tilde {\nabla} f (z _ {t}) \| ^ {2} + 2 \| \tilde {\nabla} f (z _ {t}) - \tilde {\nabla} f (z ^ {*}) \| ^ {2} \\ \leq 2 \delta^ {2} + 2 \ell^ {2} \| z _ {t} - z ^ {*} \| ^ {2}. \\ \end{array}
$$
Combining all three results together, when choosing the stepsize  $\alpha = \mu /(4\ell^2)$ , we get that
$$
\begin{array}{l} \left\| z _ {t + 1} - z ^ {*} \right\| ^ {2} \leq \left(1 - \alpha \mu + 2 \alpha^ {2} \ell^ {2}\right) \left\| z _ {t} - z ^ {*} \right\| ^ {2} + \left(\frac {\alpha}{\mu} + 2 \alpha^ {2}\right) \delta^ {2} \tag {20} \\ = \left(1 - \frac {1}{8 \kappa^ {2}}\right) \| z _ {t} - z ^ {*} \| ^ {2} + \frac {\delta^ {2}}{4 \ell^ {2}} \left(1 + \frac {1}{2 \kappa^ {2}}\right). \\ \end{array}
$$
Unrolling the recursion, we thus obtain
$$
\begin{array}{l} \left\| z _ {T} - z ^ {*} \right\| ^ {2} \\ \leq \left(1 - \frac {1}{8 \kappa^ {2}}\right) ^ {T} \| z _ {0} - z ^ {*} \| ^ {2} + \frac {\delta^ {2}}{4 \ell^ {2}} \left(1 + \frac {1}{2 \kappa^ {2}}\right) \left(1 + \left(1 - \frac {1}{8 \kappa^ {2}}\right) + \dots + \left(1 - \frac {1}{8 \kappa^ {2}}\right) ^ {T - 1}\right) \\ \leq \exp \left(- \frac {T}{8 \kappa^ {2}}\right) \| z _ {0} - z ^ {*} \| ^ {2} + \left(\frac {1}{\ell^ {2}} + \frac {2}{\mu^ {2}}\right) \delta^ {2}. \\ \end{array}
$$
This means a  $\mathcal{O}(\kappa^2)$  convergence rate to a  $\mathcal{O}(\delta^2)$  neighborhood, where  $\kappa = \ell/\mu$  is the condition number.
The lemma below establishes the convergence performance of EG under Assumption C.1 and C.2.
Lemma C.4. Under Assumption C.1. Let  $z^{*} = (x^{*},y^{*})\in \mathcal{X}\times \mathcal{Y}$  be the unique saddle point of  $f(x,y)$  and  $\kappa \coloneqq \ell /\mu$  be the condition number. Given an inexact gradient oracle in Assumption C.2. Denote  $z_{t} = (x_{t},y_{t})$  and  $\tilde{g} (z_t) = (g_x(x_t,y_t), - g_y(x_t,y_t))$ . Starting from  $z_0\in \mathcal{X}\times \mathcal{Y}$ , Extragradient that updates for  $t = 0,1,\dots ,T - 1$
$$
z _ {t + 1 / 2} = \Pi_ {\mathcal {X} \times \mathcal {Y}} (z _ {t} - \alpha \tilde {g} (z _ {t})),
$$
$$
z _ {t + 1} = \Pi_ {\mathcal {X} \times \mathcal {Y}} \left(z _ {t} - \alpha \tilde {g} \left(z _ {t + 1 / 2}\right)\right), \tag {Inexact-EG}
$$
with stepsize  $\alpha = 1 / (2\ell)$  converges with
$$
\left\| z _ {T} - z ^ {*} \right\| ^ {2} \leq \exp \left(- \frac {T}{8 \kappa}\right) \left\| z _ {0} - z ^ {*} \right\| ^ {2} + \frac {8 \delta^ {2}}{\mu} \left(\frac {2}{\ell} + \frac {1}{\mu}\right).
$$
Proof. Let  $\tilde{\nabla} f(z_t) = (\nabla_x f(x_t, y_t), -\nabla_y f(x_t, y_t))$  and  $\Delta(z_t) = \tilde{g}(z_t) - \tilde{\nabla} f(z_t)$ . By (15) in the proof of Lemma B.8, setting  $z = z^*$ , we have that,
$$
\begin{array}{l} \left\| z _ {t + 1} - z ^ {*} \right\| ^ {2} \leq \left\| z _ {t} - z ^ {*} \right\| ^ {2} - \left\| z _ {t + 1 / 2} - z _ {t} \right\| ^ {2} - \left\| z _ {t + 1 / 2} - z _ {t + 1} \right\| ^ {2} + 2 \alpha \bar {\nabla} f (z _ {t + 1 / 2}) ^ {\top} (z ^ {*} - z _ {t + 1 / 2}) \\ + 2 \alpha (\tilde {\nabla} f (z _ {t}) - \tilde {\nabla} f (z _ {t + 1 / 2})) ^ {\top} (z _ {t + 1} - z _ {t + 1 / 2}) + 2 \alpha \Delta (z _ {t + 1 / 2}) ^ {\top} (z ^ {*} - z _ {t + 1 / 2}) \\ + 2 \alpha \left(\Delta \left(z _ {t}\right) - \Delta \left(z _ {t + 1 / 2}\right)\right) ^ {\top} \left(z _ {t + 1} - z _ {t + 1 / 2}\right). \tag {21} \\ \end{array}
$$
By strong-convexity-strong-concavity of the function  $f(x,y)$ , we know that
$$
f (x ^ {*}, y _ {t + 1 / 2}) \geq f (x _ {t + 1 / 2}, y _ {t + 1 / 2}) + \nabla_ {x} f (x _ {t + 1 / 2}, y _ {t + 1 / 2}) ^ {\top} (x ^ {*} - x _ {t + 1 / 2}) + \frac {\mu}{2} \| x _ {t + 1 / 2} - x ^ {*} \| ^ {2},
$$
$$
- f (x _ {t + 1 / 2}, y ^ {*}) \geq - f (x _ {t + 1 / 2}, y _ {t + 1 / 2}) - \nabla_ {y} f (x _ {t + 1 / 2}, y _ {t + 1 / 2}) ^ {\top} (y ^ {*} - y _ {t + 1 / 2}) + \frac {\mu}{2} \| y _ {t + 1 / 2} - y ^ {*} \| ^ {2}.
$$
Summing up the above two inequalities, using the definition of saddle points, we have
$$
\begin{array}{l} \tilde {\nabla} f \left(z _ {t + 1 / 2}\right) ^ {\top} \left(z ^ {*} - z _ {t + 1 / 2}\right) + \Delta \left(z _ {t + 1 / 2}\right) ^ {\top} \left(z ^ {*} - z _ {t + 1 / 2}\right) \\ \leq f \left(x ^ {*}, y _ {t + 1 / 2}\right) - f \left(x _ {t + 1 / 2}, y ^ {*}\right) - \frac {\mu}{2} \left\| z _ {t + 1 / 2} - z ^ {*} \right\| ^ {2} + \Delta \left(z _ {t + 1 / 2}\right) ^ {\top} \left(z ^ {*} - z _ {t + 1 / 2}\right) \\ \leq - \frac {\mu}{2} \| z _ {t + 1 / 2} - z ^ {*} \| ^ {2} + \| \Delta \left(z _ {t + 1 / 2}\right) \| \| z ^ {*} - z _ {t + 1 / 2} \| \tag {22} \\ \leq - \frac {\mu}{4} \| z _ {t + 1 / 2} - z ^ {*} \| ^ {2} + \frac {\delta^ {2}}{\mu} \\ \leq - \frac {\mu}{8} \| z _ {t} - z ^ {*} \| ^ {2} + \frac {\mu}{4} \| z _ {t} - z _ {t + 1 / 2} \| ^ {2} + \frac {\delta^ {2}}{\mu}, \\ \end{array}
$$
where we use fact (iii) in Lemma A.1 and  $\| z_{t} - z^{*}\|^{2}\leq 2\| z_{t} - z_{t + 1 / 2}\|^{2} + 2\| z_{t + 1 / 2} - z^{*}\|^{2}$ . By smoothness of  $f(x,y)$  and fact (iii) in Lemma A.1, we also have that
$$
\begin{array}{l} \left(\tilde {\nabla} f (z _ {t}) - \tilde {\nabla} f \left(z _ {t + 1 / 2}\right)\right) ^ {\top} \left(z _ {t + 1} - z _ {t + 1 / 2}\right) + \left(\Delta (z _ {t}) - \Delta \left(z _ {t + 1 / 2}\right)\right) ^ {\top} \left(z _ {t + 1} - z _ {t + 1 / 2}\right) \\ \leq \ell \| z _ {t} - z _ {t + 1 / 2} \| \| z _ {t + 1} - z _ {t + 1 / 2} \| + 2 \delta \cdot \| z _ {t + 1} - z _ {t + 1 / 2} \| \tag {23} \\ \leq \frac {\ell}{2} \| z _ {t} - z _ {t + 1 / 2} \| ^ {2} + \ell \| z _ {t + 1} - z _ {t + 1 / 2} \| ^ {2} + \frac {2 \delta^ {2}}{\ell}. \\ \end{array}
$$
Plugging (22) and (23) back into (21), choosing  $\alpha = 1 / (2\ell)$ , we obtain that
$$
\begin{array}{l} \left\| z _ {t + 1} - z ^ {*} \right\| ^ {2} \leq \left(1 - \frac {\mu \alpha}{4}\right) \left\| z _ {t} - z ^ {*} \right\| ^ {2} - \left(1 - \frac {\mu \alpha}{2} - \alpha \ell\right) \left\| z _ {t + 1 / 2} - z _ {t} \right\| ^ {2} \\ - \left(1 - 2 \alpha \ell\right) \| z _ {t + 1 / 2} - z _ {t + 1} \| ^ {2} + 2 \alpha \delta^ {2} \left(\frac {2}{\ell} + \frac {1}{\mu}\right) \tag {24} \\ \leq \left(1 - \frac {\mu}{8 \ell}\right) \| z _ {t} - z ^ {*} \| ^ {2} + \frac {\delta^ {2}}{\ell} \left(\frac {2}{\ell} + \frac {1}{\mu}\right). \\ \end{array}
$$
Unrolling the recursion, since  $1 + \eta \leq e^{\eta}$ ,  $\forall \eta \in \mathbb{R}$ , we get that
$$
\begin{array}{l} \left\| z _ {T} - z ^ {*} \right\| ^ {2} \leq \left(1 - \frac {\mu}{8 \ell}\right) ^ {T} \left\| z _ {0} - z ^ {*} \right\| ^ {2} + \frac {\delta^ {2}}{\ell} \left(\frac {2}{\ell} + \frac {1}{\mu}\right) \left(1 + \left(1 - \frac {\mu}{8 \ell}\right) + \dots + \left(1 - \frac {\mu}{8 \ell}\right) ^ {T - 1}\right) \\ \leq \left(1 - \frac {\mu}{8 \ell}\right) ^ {T} \| z _ {0} - z ^ {*} \| ^ {2} + \frac {8 \delta^ {2}}{\mu} \left(\frac {2}{\ell} + \frac {1}{\mu}\right). \\ \end{array}
$$
This means a  $\mathcal{O}(\kappa)$  convergence rate to a  $\mathcal{O}(\delta^2)$  neighborhood, where  $\kappa = \ell/\mu$  is the condition number.
Lemma 4.5 directly follows from Lemma C.4 observing that  $G(x,y) + r(x - x_0,y_0 - y)$  is a  $\delta$ -inexact gradient of  $F_{r}(x,y)$ . Next, we provide a useful lemma showing how to satisfy the stopping criteria for the auxiliary smooth SC-SC sub-problem in Algorithm 2 when presented with inexact gradients. The results are motivated from Yang et al. [74].
Lemma C.5. Under Assumption C.1 and C.2. Suppose the domain  $\mathcal{X}$  and  $\mathcal{Y}$  have a diameter of  $D$ . Denote  $z^{*} = (x^{*},y^{*})$  be the unique saddle point of  $f(x,y)$ . For any  $\hat{z} = (\hat{x},\hat{y})\in \mathcal{X}\times \mathcal{Y}$ , we let  $\tilde{g} (\hat{z}) = (g_{x}(\hat{x},\hat{y}), - g_{y}(\hat{x},\hat{y}))$  and define  $[\hat{z}]_{\beta} = ([\hat{x}]_{\beta},[\hat{y}]_{\beta})$  for  $\beta \geq 2\ell$  to be
$$
[ \hat {z} ] _ {\beta} = \Pi_ {\mathcal {X} \times \mathcal {Y}} \left(\hat {z} - \frac {1}{\beta} \tilde {g} (\hat {z})\right),
$$
which is obtained through one step of GDA starting from  $\hat{z}$  with inexact gradients. Denote the true gradient as  $\tilde{\nabla} f([\hat{z}]_{\beta}) = (\nabla_x f([\hat{x}]_{\beta}, [\hat{y}]_{\beta}), -\nabla_y f([\hat{x}]_{\beta}, [\hat{y}]_{\beta}))$ . Then we have that  $\forall z = (x, y) \in \mathcal{X} \times \mathcal{Y}$ ,
$$
\tilde {\nabla} f ([ \hat {z} ] _ {\beta}) ^ {\top} ([ \hat {z} ] _ {\beta} - z) \leq 2 \sqrt {2} \beta D \| \hat {z} - z ^ {*} \| + \sqrt {2} \delta D \left((2 + \sqrt {2}) \sqrt {\frac {\beta}{\mu}} + 3\right).
$$
Moreover, it also holds that  $\| [\hat{z}]_{\beta} - z^{*}\| \leq (1 + \sqrt{2}\ell /\beta)\| \hat{z} -z^{*}\| +\delta (1 / \sqrt{\beta\mu} +\sqrt{2} /\beta).$
Proof. We construct a "ghost" point  $\hat{z}_1 = (\hat{x}_1, \hat{y}_1) \in \mathcal{X} \times \mathcal{Y}$  to be
$$
\hat {z} _ {1} = \Pi_ {\mathcal {X} \times \mathcal {Y}} \left(\hat {z} - \frac {1}{\beta} \tilde {g} ([ \hat {z} ] _ {\beta})\right).
$$
$\hat{z}_1$  can be regarded as performing one update of inexact-EG with stepsize  $1 / \beta$  starting from  $\hat{z}$ . Therefore, by (16) and (19) in the convergence analysis of EG, since  $1 / \beta \leq 1 / \ell$ , we obtain that  $\forall z = (x,y) \in \mathcal{X} \times \mathcal{Y}$ ,
$$
\begin{array}{l} \tilde {\nabla} f ([ \hat {z} ] _ {\beta}) ^ {\top} ([ \hat {z} ] _ {\beta} - z) \leq \frac {\beta}{2} \| \hat {z} - z \| ^ {2} - \frac {\beta}{2} \| \hat {z} _ {1} - z \| ^ {2} + 3 \sqrt {2} \delta D \\ = \frac {\beta}{2} (\hat {z} - \hat {z} _ {1}) ^ {\top} (\hat {z} - z + \hat {z} _ {1} - z) + 3 \sqrt {2} \delta D \\ \leq \frac {\beta}{2} (\| \hat {z} - z ^ {*} \| + \| \hat {z} _ {1} - z ^ {*} \|) \cdot \| \hat {z} - z + \hat {z} _ {1} - z \| + 3 \sqrt {2} \delta D. \\ \end{array}
$$
By (24) in the proof of Lemma C.4, since  $\beta \geq 2\ell$  and  $\mu \leq \ell$ , we have that
$$
\| \hat {z} _ {1} - z ^ {*} \| ^ {2} \leq \| \hat {z} - z ^ {*} \| ^ {2} + \frac {4 \delta^ {2}}{\beta \ell} + \frac {2 \delta^ {2}}{\beta \mu}.
$$
Therefore, we can obtain that  $\| \hat{z}_1 - z^*\| \leq \| \hat{z} -z^*\| +2\delta /\sqrt{\beta\ell} +\sqrt{2}\delta /\sqrt{\beta\mu}$ , and thus,
$$
\begin{array}{l} \tilde {\nabla} f ([ \hat {z} ] _ {\beta}) ^ {\top} ([ \hat {z} ] _ {\beta} - z) \leq \sqrt {2} \beta D (\| \hat {z} - z ^ {*} \| + \| \hat {z} _ {1} - z ^ {*} \|) + 3 \sqrt {2} \delta D \\ \leq 2 \sqrt {2} \beta D \| \hat {z} - z ^ {*} \| + \sqrt {2} \delta D \left(2 \sqrt {\frac {\beta}{\ell}} + \sqrt {\frac {2 \beta}{\mu}} + 3\right). \\ \end{array}
$$
For the last statement, since  $[\hat{z}]_{\beta}$  is obtained through 1-step of GDA with inexact gradients, by (20) in the proof of GDA for SC-SC problems before, we have that
$$
\| [ \hat {z} ] _ {\beta} - z ^ {*} \| ^ {2} \leq \left(1 + \frac {2 \ell^ {2}}{\beta^ {2}}\right) \| \hat {z} - z ^ {*} \| ^ {2} + \delta^ {2} \left(\frac {1}{\beta \mu} + \frac {2}{\beta^ {2}}\right).
$$
Therefore, we obtain that  $\| [\hat{z}]_{\beta} - z^{*}\| \leq (1 + \sqrt{2}\ell /\beta)\| \hat{z} -z^{*}\| +\delta (1 / \sqrt{\beta\mu} +\sqrt{2} /\beta).$
The above lemma also applies to the case when exact gradients are available setting  $\delta = 0$  and  $[\hat{z}]_{\beta} = \Pi_{\mathcal{X}\times \mathcal{Y}}\left(\hat{z} -\frac{1}{\beta}\tilde{\nabla} f(\hat{z})\right)$  for the true gradients  $\tilde{\nabla} f(\hat{z})$ . This implies the stopping criteria  $\tilde{\nabla} f(\hat{z})^{\top}(\hat{z} -z)\leq \hat{\epsilon},\forall z\in \mathcal{X}\times \mathcal{Y}$  in Algorithm 2 and 3 can be translated to  $\| \hat{z} -z^{*}\|^{2}\leq \mathcal{O}(\hat{\epsilon}^{2})$  which can be satisfied within  $\mathcal{O}(\log (1 / \hat{\epsilon}))$  complexity using Lemma C.3 and C.4 with  $\delta = 0$  (or existing results in Tseng [72] or Facchinei and Pang [32]).
C.2 Regularization Helps!
Proof of Theorem 4.4 and 4.6 for the near-optimal guarantees of Algorithm 2 is provided here.
C.2.1 Inexact Initialization Oracle
We also use  $(x_0, y_0)$  as the initialization point when solving the auxiliary strongly-convex problem. Note that the gradient steps starting from  $(x_0, y_0)$  remain the same on  $F(x, y)$  and  $F_r(x, y)$ .
Proof of Theorem 4.4. We first show the convergence guarantee. Let  $z_{r} = (x_{r},y_{r})$ . By fact (i) in Lemma A.1, we have that  $\forall z = (x,y)\in \mathcal{X}\times \mathcal{Y}$ ,
$$
\begin{array}{l} \tilde {\nabla} F (z _ {r}) ^ {\top} (z _ {r} - z) = \left(\tilde {\nabla} F _ {r} (z _ {r}) - r (z _ {r} - z _ {0})\right) ^ {\top} (z _ {r} - z) \\ = \tilde {\nabla} F _ {r} \left(z _ {r}\right) ^ {\top} \left(z _ {r} - z\right) + \frac {r}{2} \| z _ {0} - z \| ^ {2} - \frac {r}{2} \| z _ {r} - z _ {0} \| ^ {2} - \frac {r}{2} \| z _ {r} - z \| ^ {2} \tag {25} \\ \leq \epsilon_ {r} + r D ^ {2}. \\ \end{array}
$$
According to Lemma B.2, this means  $\max_{y\in \mathcal{Y}}F(x_r,y) - \min_{x\in \mathcal{X}}F(x,y_r)\leq \epsilon_r + rD^2$
We then show the reproducibility guarantee. Denote the saddle point of  $F_{r}(x,y)$  given  $(x_0, y_0)$  as  $(x_{r}^{*}, y_{r}^{*})$ , and the saddle point of  $F_{r}'(x,y) = F(x,y) + (r/2)\| x - x_{0}'\|^{2} - (r/2)\| y - y_{0}'\|^{2}$  given  $(x_{0}', y_{0}')$  as  $((x_{r}^{*})', (y_{r}^{*})')$ . By Lemma B.4 in Appendix B.3 of Zhang et al. [78], we have that
$$
\left\| x _ {r} ^ {*} - (x _ {r} ^ {*}) ^ {\prime} \right\| ^ {2} + \left\| y _ {r} ^ {*} - (y _ {r} ^ {*}) ^ {\prime} \right\| ^ {2} \leq \left\| x _ {0} - x _ {0} ^ {\prime} \right\| ^ {2} + \left\| y _ {0} - y _ {0} ^ {\prime} \right\| ^ {2}.
$$
Let  $z_{r} = (x_{r},y_{r})$ ,  $z_{r}^{*} = (x_{r}^{*},y_{r}^{*})$  and  $z_{0} = (x_{0},y_{0})$  for simplicity of the notation.  $z_r^{\prime}$ ,  $(z_r^*)^\prime$  and  $z_0^\prime$  can be defined in the same way. Similarly to the minimization case, we have
$$
\begin{array}{l} \left\| z _ {r} - z _ {r} ^ {\prime} \right\| \leq \left\| z _ {r} - z _ {r} ^ {*} \right\| + \left\| z _ {r} ^ {*} - \left(z _ {r} ^ {*}\right) ^ {\prime} \right\| + \left\| \left(z _ {r} ^ {*}\right) ^ {\prime} - z _ {r} ^ {\prime} \right\| \\ \leq \delta + 2 \sqrt {\frac {2 \epsilon_ {r}}{r}}, \\ \end{array}
$$
where we use  $\| z_r^* - (z_r^*)' \| \leq \| z_0 - z_0' \| \leq \delta$  and optimality of  $z_r^*$  by  $r$  strong-convexity-strong-concavity (SC-SC) of  $F_r(x, y)$  (the same holds true for  $z_r'$  and  $(z_r^*)'$  as well):
$$
\begin{array}{l} \frac {r}{2} \| x _ {r} - x _ {r} ^ {*} \| ^ {2} + \frac {r}{2} \| y _ {r} - y _ {r} ^ {*} \| ^ {2} \leq F _ {r} (x _ {r}, y _ {r} ^ {*}) - F _ {r} (x _ {r} ^ {*}, y _ {r} ^ {*}) + F _ {r} (x _ {r} ^ {*}, y _ {r} ^ {*}) - F _ {r} (x _ {r} ^ {*}, y _ {r}) \\ \leq \max  _ {y \in \mathcal {Y}} F _ {r} (x _ {r}, y) - \min  _ {x \in \mathcal {X}} F _ {r} (x, y _ {r}) \\ \leq \epsilon_ {r}. \\ \end{array}
$$
Thus setting  $r = \epsilon /D^2$  and  $\epsilon_r = \epsilon \cdot \min \{1,\delta^2 /(8D^2)\}$ , we guarantee that  $\max_{y\in \mathcal{Y}}F(x_r,y) - \min_{x\in \mathcal{X}}F(x,y_r)\leq 2\epsilon$  and  $\| x_r - x_r'\| ^2 +\| y_r - y_r'\| ^2\leq 4\delta^2$ . Applying Lemma C.5 with  $\delta = 0$ , the complexity using Extragradient (EG) [72, 57] to achieve  $\epsilon_r$ -error on  $r$ -SC-SC  $(\ell +r)$ -smooth minimax optimization is  $\mathcal{O}((\ell /r + 1)\log (1 / \epsilon_r)) = \tilde{\mathcal{O}} (\ell D^2 /\epsilon)$ , where  $\tilde{\mathcal{O}}$  hides logarithmic terms.
C.2.2 Inexact Deterministic Gradient Oracle
This section contains proof of Theorem 4.6 for the near-optimal guarantees in the inexact deterministic gradient case. The proof is based on Lemma 4.5 (restated and proved as Lemma C.4 in Section C.1) and Lemma C.5.
Proof of Theorem 4.6. For the convergence guarantee, the same as (25), we have that
$$
\max  _ {y \in \mathcal {Y}} F (x _ {r}, y) - \min  _ {x \in \mathcal {X}} F (x, y _ {r}) \leq \epsilon_ {r} + r D ^ {2}.
$$
For the reproducibility guarantee, we can obtain that
$$
\left\| z _ {r} - z _ {r} ^ {\prime} \right\| \leq \left\| z _ {r} - z _ {r} ^ {*} \right\| + \left\| z _ {r} ^ {*} - z _ {r} ^ {\prime} \right\|.
$$
Let  $z_{T}$  be the output of  $T$ -step Extragradient with initialization  $z_{0}$ . By Lemma 4.5, we have that
$$
\begin{array}{l} \left\| z _ {T} - z _ {r} ^ {*} \right\| ^ {2} \leq \exp \left(- \frac {T}{8} \frac {r}{\ell + r}\right) \left\| z _ {0} - z _ {r} ^ {*} \right\| ^ {2} + \frac {8 \delta^ {2}}{r} \left(\frac {2}{\ell + r} + \frac {1}{r}\right) \\ \leq \exp \left(- \frac {T}{1 6} \frac {r}{\ell}\right) \| z _ {0} - z _ {r} ^ {*} \| ^ {2} + \frac {1 6 \delta^ {2}}{r ^ {2}}. \\ \end{array}
$$
Setting  $T \geq (32\ell / r) \log(rD / \delta)$  and  $r = \epsilon / D^2$ , this means the algorithm converges to  $\| z_T - z_r^* \| \leq 3\sqrt{2} D^2(\delta / \epsilon)$ . Therefore, according to Lemma C.5, if we choose  $z_r = [z_T]_{2\ell}$ , since  $1 \leq \ell D^2 / \epsilon$ , we can guarantee that  $\| z_r - z_r^* \| \leq 3(2\sqrt{2} + 1)D^2(\delta / \epsilon)$  and that