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parse/train/PbEHqvFtcS/PbEHqvFtcS.md

@@ -112,13 +112,13 @@ $$
 \| \nabla f ( x _ { t } ) \| \le \varepsilon \quad a n d \quad \nabla ^ { 2 } f ( x _ { t } ) \succeq - \sqrt { \varepsilon } { \bf I } .
 $$
 
-Remark 2.4. If one only wishes to achieve a significantly simpler goal — finding first-order critical points $\| \nabla f ( x _ { t } ) \| \leq \varepsilon$ — the analysis becomes much easier (see Section 3.1). In particular, having one safe guard without perturbation (i.e. $\nu = 0$ ) suffices, and the iteration complexity reduces to T = Oe f(x0)−min f(x)ε4 (α2 + 1m )
. Bulusu et al. (2020) achieves this easier goal but requires an additional assumption: there is one guaranteed good worker known by the master.
+Remark 2.4. If one only wishes to achieve a significantly simpler goal — finding first-order critical points $\| \nabla f ( x _ { t } ) \| \leq \varepsilon$ — the analysis becomes much easier (see Section 3.1). In particular, having one safe guard without perturbation (i.e. $\nu = 0$ ) suffices, and the iteration complexity reduces to T = Oe f(x0)−min f(x)ε4 (α2 + 1m )
. Bulusu et al. (2020) achieves this easier goal but requires an additional assumption: there is one guaranteed good worker known by the master.
 
 Our contribution. We reiterate our theoretical contributions from three perspectives. 1) When $\alpha < 1 / \sqrt { m }$ , our algorithm requires $\begin{array} { r } { m T = \widetilde { O } \big ( \frac { ( f ( x _ { 0 } ) - \operatorname* { m i n } f ( x ) ) d } { \varepsilon ^ { 4 } } \big ) } \end{array}$ stochastic gradient computations. This matches the best known result (Jin et al., 2019) under our minimal assumptions of the nonconvex objective. (There exist other works in the stochastic setting that break the $\varepsilon ^ { - 4 }$ barrier and get rid of the dimension dependence $d$ under stronger assumptions.)5. 2) When $\alpha < 1 / \sqrt { m }$ , our algorithm enjoys linear parallel speed-up: the parallel time complexity reduces by a factor of $\Theta ( m )$ . When $\alpha \in [ 1 / \sqrt { m } , 1 / 2 )$ , our parallel time complexity is $\widetilde { O } ( \alpha ^ { 2 } )$ times that needed when no parallelism is used, still giving noticeable speedup. The $\textstyle { \hat { \alpha } } ^ { 2 }$ factor also appeared in convex Byzantine distributed optimization (and is known to be tight there) (Yin et al., 2018; Alistarh et al., 2018).
 
 Comparison to (Yin et al., 2019). Yin et al. (2019) derived three gradient descent-type algorithms to find points with a weaker (and less standard) guarantee: $\| \nabla f ( x ) \| ~ \leq ~ \varepsilon$ and $\bar { \nabla ^ { 2 } } f ( x ) ~ \succeq ~ - ( \varepsilon ^ { 2 } d ) ^ { \bar { 1 } / 5 } \mathbf { I }$ . Despite practical differences (namely, gradient descent may be less favorable comparing to stochastic gradient descent especially in deep learning applications), the parallel time complexities derived from their result are also generally larger than ours.
 
-Their paper focuses on bounding the number of sampled stochastic functions, as opposed to the number of stochastic gradient evaluations like we do. When translated to our language, each of the workers in their setting needs to evaluate T stochastic gradients, where (1) T = Oe α2dε4 $\begin{array} { r } { 1 ) T = \widetilde { O } \big ( \frac { \alpha ^ { 2 } \overline { { d } } } { \varepsilon ^ { 4 } } + \frac { d ^ { 2 } } { \varepsilon ^ { 4 } m } + \frac { \sqrt { d } } { \varepsilon ^ { 3 } } \big ) } \end{array}$ if using coordinate-wise median, (2) T = Oe α2d2ε4 $\begin{array} { r } { T = \widetilde { O } \big ( \frac { \alpha } { \varepsilon ^ { 4 } } + \frac { d } { \varepsilon ^ { 4 } m } \big ) } \end{array}$ if using iterative filt if using trimmed mean, and (3)(1) and (2) are larger than ours (also with a weaker guarantee); the complexity (3) seems incomparable to ours, but when translating to the more standard (ε, √ε) guarantee, becomes T = Oe αd2ε5 $\begin{array} { r } { T = \widetilde { \cal O } \big ( \frac { \alpha d ^ { 2 } } { \varepsilon ^ { 5 } } + \frac { d ^ { 3 } } { \varepsilon ^ { 5 } m } \big ) } \end{array}$ so is also larger than ours. It is worth noting that (3) requires $\alpha < 1 / 4$ so cannot withstand half of the machines being Byzantine.
+Their paper focuses on bounding the number of sampled stochastic functions, as opposed to the number of stochastic gradient evaluations like we do. When translated to our language, each of the workers in their setting needs to evaluate T stochastic gradients, where (1) T = Oe α2dε4 $\begin{array} { r } { 1 ) T = \widetilde { O } \big ( \frac { \alpha ^ { 2 } \overline { { d } } } { \varepsilon ^ { 4 } } + \frac { d ^ { 2 } } { \varepsilon ^ { 4 } m } + \frac { \sqrt { d } } { \varepsilon ^ { 3 } } \big ) } \end{array}$ if using coordinate-wise median, (2) T = Oe α2d2ε4 $\begin{array} { r } { T = \widetilde { O } \big ( \frac { \alpha } { \varepsilon ^ { 4 } } + \frac { d } { \varepsilon ^ { 4 } m } \big ) } \end{array}$ if using iterative filt if using trimmed mean, and (3)(1) and (2) are larger than ours (also with a weaker guarantee); the complexity (3) seems incomparable to ours, but when translating to the more standard (ε, √ε) guarantee, becomes T = Oe αd2ε5 $\begin{array} { r } { T = \widetilde { \cal O } \big ( \frac { \alpha d ^ { 2 } } { \varepsilon ^ { 5 } } + \frac { d ^ { 3 } } { \varepsilon ^ { 5 } m } \big ) } \end{array}$ so is also larger than ours. It is worth noting that (3) requires $\alpha < 1 / 4$ so cannot withstand half of the machines being Byzantine.
 
 Resilience against practical attacks. Our algorithm’s filtering is based upon tracking $B _ { i }$ (resp. $A _ { i , { \bf \Pi } }$ ), the stochastic gradients of each machine $i$ averaged over a window of $T _ { 0 }$ (resp. $T _ { 1 . }$ ) iterations. This is a departure from previous defenses, most of which are history-less, and enables us to be provably Byzantine-resilient against state-of-the-art attacks (Baruch et al., 2019; Xie et al., 2020).
 

parse/train/af_hng9tuNj/af_hng9tuNj.md

@@ -135,13 +135,13 @@ How to set $c _ { i j } \mathbf { \hat { \theta } } ^ { } $ Breaking symmetries.
 
 Final LS assembly for a single bond. We now describe the assembly process depicted in fig. $\boxed { 5 }$ We first predict the LS of node $\mathrm { X }$ as in section $\boxed { 2 . 3 }$ obtaining several 3D coordinates: $\mathbf { p } _ { X } = \mathbf { 0 } , \mathbf { p } _ { Y } , \mathbf { \overline { { \mathbf { p } } } } _ { T _ { i } } \in$ $\mathbb { R } ^ { 3 } , \dot { \forall } i \in [ 1 . . n ]$ , as well as the LS of node $\mathrm { Y }$ : $\mathbf { q } _ { Y } = \bar { \mathbf { 0 } _ { \cdot } } \mathbf { q } _ { X } , \mathbf { q } _ { Z _ { j } } \in \mathbb { R } ^ { 3 } , \forall j \in [ 1 . . m ]$ . By design, we have that $\| \mathbf { q } _ { X } \| = \| \mathbf { p } _ { Y } \|$ . These two sets are currently not aligned. To achieve this, we first rotate the LS of $\mathrm { X }$ such that $\mathbf { p } _ { Y }$ becomes $\left[ \left. \mathbf { p } _ { Y } \right. \quad 0 \quad 0 \right] ^ { \intercal }$ , while $\mathbf { p } _ { X }$ remains 0. Next, we rotate and translate the LS of $\mathrm { Y }$ such that $\mathbf { q } _ { Y }$ becomes $\mathbf { p } _ { Y }$ and $\mathbf { q } _ { X }$ becomes $\mathbf { p } _ { X } = \mathbf { 0 }$ . These two rotations have one degree of freedom each, which we set randomly. Exact formulas are in appendix B. Thus, the bond XY is now matched, but the torsional rotation is still arbitrary/random. The remaining step is to rotate the LS of $\mathrm { X }$ with an angle $\gamma$ such that all dihedrals $\angle ( X Y T _ { i } , X Y Z _ { j } )$ match their true counterparts. This is done by applying to all vectors $\mathbf { p } _ { T _ { i } }$ the same rotation of type: $\mathbf { H } _ { \gamma } : = \left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { 0 } & { \cos ( \gamma ) } & { - \sin ( \gamma ) } \\ { 0 } & { \sin ( \gamma ) } & { \cos ( \gamma ) } \end{array} \right] .$
 
-How to compute $\gamma$ ? The current dihedrals $\Delta _ { i j } ^ { c u r } \ \stackrel { \mathsf { d e f } } { = } \ \angle ^ { c u r } ( X Y T _ { i } , X Y Z _ { j } )$ depend on the random torsional rotations from the initial assembly step of LS of $\mathbf { X }$ and of $\mathrm { Y }$ . After applying the $\mathbf { H } _ { \gamma }$ rotation, we obtain the new dihedral angles: $[ \Delta _ { i j } ^ { c u r } - \gamma ]$ mod $2 \pi$ that should match the ground truth dihedral angles $\Delta _ { i j } ^ { * } \ { \stackrel { \mathrm { d e f } } { = } } \ \angle ^ { * } ( X Y T _ { i } , X Y Z _ { j } )$ . This is equivalently written as $\mathbf { s } _ { i j } ^ { * } = \mathbf { A } _ { i j } ^ { c u r } \mathbf { s } _ { \gamma }$ , where $\mathbf { s } _ { \gamma } \ { \stackrel { \mathrm { d e f } } { = } } \ \left[ \cos ( \gamma ) \right]$  and Acurij def=  $\begin{array} { r } { \mathbf { A } _ { i j } ^ { c u r } \stackrel { \mathrm { d e f } } { = } \left[ \begin{array} { c c } { \cos ( \Delta _ { i j } ^ { c u r } ) } & { \sin ( \Delta _ { i j } ^ { c u r } ) } \\ { \sin ( \Delta _ { i j } ^ { c u r } ) } & { - \cos ( \Delta _ { i j } ^ { c u r } ) } \end{array} \right] . } \end{array}$ . Let $\begin{array} { r } { \mathbf { s } ^ { * } \overset { \mathrm { d e f } } { = } \sum _ { i , j } c _ { i j } \mathbf { s } _ { i j } ^ { * } } \end{array}$ and $\mathbf { A } ^ { c u r } \ { \stackrel { \mathrm { d e } 1 } { = } }$ $\begin{array} { r } { \sum _ { i , j } c _ { i j } \mathbf { A } _ { i j } ^ { c u r } } \end{array}$ . The necessary condition for becomes $\mathbf { s } _ { \gamma } = \left( \mathbf { A } ^ { c u r } \right) ^ { \top } \mathbf { s } ^ { * }$ , which is also sufficient due i,j  to proposition $\bigstar \bigstar \bigstar$ This implies it is enough to predict only the normalized $\frac { \mathbf { s } ^ { * } } { \| \mathbf { s } ^ { * } \| }$ and, in practice, we do that by predicting $\mathbf { s } _ { \alpha } \triangleq \left[ \cos ( \alpha ) \right]$ using a function commutative in $\mathrm { X }$ and $\mathrm { Y }$ (i.e., swapping $\mathrm { X }$ and $\mathrm { Y }$ does not change $\alpha$ ):
+How to compute $\gamma$ ? The current dihedrals $\Delta _ { i j } ^ { c u r } \ \stackrel { \mathsf { d e f } } { = } \ \angle ^ { c u r } ( X Y T _ { i } , X Y Z _ { j } )$ depend on the random torsional rotations from the initial assembly step of LS of $\mathbf { X }$ and of $\mathrm { Y }$ . After applying the $\mathbf { H } _ { \gamma }$ rotation, we obtain the new dihedral angles: $[ \Delta _ { i j } ^ { c u r } - \gamma ]$ mod $2 \pi$ that should match the ground truth dihedral angles $\Delta _ { i j } ^ { * } \ { \stackrel { \mathrm { d e f } } { = } } \ \angle ^ { * } ( X Y T _ { i } , X Y Z _ { j } )$ . This is equivalently written as $\mathbf { s } _ { i j } ^ { * } = \mathbf { A } _ { i j } ^ { c u r } \mathbf { s } _ { \gamma }$ , where $\mathbf { s } _ { \gamma } \ { \stackrel { \mathrm { d e f } } { = } } \ \left[ \cos ( \gamma ) \right]$  and Acurij def=  $\begin{array} { r } { \mathbf { A } _ { i j } ^ { c u r } \stackrel { \mathrm { d e f } } { = } \left[ \begin{array} { c c } { \cos ( \Delta _ { i j } ^ { c u r } ) } & { \sin ( \Delta _ { i j } ^ { c u r } ) } \\ { \sin ( \Delta _ { i j } ^ { c u r } ) } & { - \cos ( \Delta _ { i j } ^ { c u r } ) } \end{array} \right] . } \end{array}$ . Let $\begin{array} { r } { \mathbf { s } ^ { * } \overset { \mathrm { d e f } } { = } \sum _ { i , j } c _ { i j } \mathbf { s } _ { i j } ^ { * } } \end{array}$ and $\mathbf { A } ^ { c u r } \ { \stackrel { \mathrm { d e } 1 } { = } }$ $\begin{array} { r } { \sum _ { i , j } c _ { i j } \mathbf { A } _ { i j } ^ { c u r } } \end{array}$ . The necessary condition for becomes $\mathbf { s } _ { \gamma } = \left( \mathbf { A } ^ { c u r } \right) ^ { \top } \mathbf { s } ^ { * }$ , which is also sufficient due i,j  to proposition $\bigstar \bigstar \bigstar$ This implies it is enough to predict only the normalized $\frac { \mathbf { s } ^ { * } } { \| \mathbf { s } ^ { * } \| }$ and, in practice, we do that by predicting $\mathbf { s } _ { \alpha } \triangleq \left[ \cos ( \alpha ) \right]$ using a function commutative in $\mathrm { X }$ and $\mathrm { Y }$ (i.e., swapping $\mathrm { X }$ and $\mathrm { Y }$ does not change $\alpha$ ):
 
 $$
 \alpha = [ \phi ( \mathbf { h } _ { X } , \mathbf { h } _ { Y } , \mathbf { h } _ { m o l } ) + \phi ( \mathbf { h } _ { Y } , \mathbf { h } _ { X } , \mathbf { h } _ { m o l } ) ] { \bmod { 2 \pi } }
 $$
 
-where  is a neural network (e.g., MLP). Finally, s = $\begin{array} { r } { { \bf s } _ { \gamma } = \left[ \begin{array} { c c } { \cos ( \gamma ) } \\ { \sin ( \gamma ) } \end{array} \right] = \frac { 1 } { \Vert ( { \bf A } ^ { c u r } ) ^ { \top } { \bf s } _ { \alpha } \Vert } \left( { \bf A } ^ { c u r } \right) ^ { \top } { \bf s } _ { \alpha } . } \end{array}$
+where  is a neural network (e.g., MLP). Finally, s = $\begin{array} { r } { { \bf s } _ { \gamma } = \left[ \begin{array} { c c } { \cos ( \gamma ) } \\ { \sin ( \gamma ) } \end{array} \right] = \frac { 1 } { \Vert ( { \bf A } ^ { c u r } ) ^ { \top } { \bf s } _ { \alpha } \Vert } \left( { \bf A } ^ { c u r } \right) ^ { \top } { \bf s } _ { \alpha } . } \end{array}$
 
 # 2.5 An optimal transport (OT) loss function for diverse conformer generation