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Abstract

Automatic differentiation is a technology to differentiate a computer program automatically. It is known to many people for its use in machine learning in recent decades. Nowadays, researchers are becoming increasingly aware of its importance in scientific computing, especially in the physics simulation. Differentiating physics simulation can help us solve many important issues in chaos theory, electromagnetism, seismic and oceanographic. Meanwhile, it is also challenging because these applications often require a lot of computing time and space. This paper will review several automatic differentiation strategies for physics simulation, and compare their pros and cons. These methods include adjoint state methods, forward mode automatic differentiation, reverse mode automatic differentiation, and reversible programming automatic differentiation.

Keywords:

Authors and contacts

Liu Jin-Guo¹,
Xu Kai-Lai²

Corresponding author:Liu Jin-Guo,jinguoliu@g.harvard.edu

Funds:Project suppored by the Research Program of QuEra Computing Inc.

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References

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Cited By

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输入: 动力学参数$\theta$, 开始时间$t_0$, 结束时间$t_n$, 末态$s_n$, 以及需要回传的导数$\dfrac{\partial {\cal{L} } }{\partial s_n}$

输出: $\dfrac{\partial {\cal{L} } }{\partial s_0}$, $\dfrac{\partial {\cal{L} } }{\partial \theta}$

1 functionaug_dynamics((s,a, –),t, $\theta$)

2 　$q=f(s, t, \theta)$　　 #定义拓展动力学函数

3 　return$\left(q, \;-a^{\rm{T}}\dfrac{\partial q}{\partial s}\right.$, $\left. -a^{\rm{T}}\dfrac{\partial q}{\partial \theta}\right)$

4end

5 ${S_0} = \left( { {s_n}, \dfrac{ {\partial {\cal{L} } } }{ {\partial {s_n} } }, 0} \right)$ #计算拓展动力学函数的初始状态

6 $\left(s_0,\; \dfrac{\partial {\cal{L} } }{\partial s_0}\right.$, $\left.\dfrac{\partial {\cal{L} } }{\partial \theta}\right)$ = ODESolve(aug_dynamics,S₀, $\theta$, $t_n$,t₀)　　　　　　 #对拓展动力学反向积分

DownLoad: CSV

方法	时间	空间	是否严格
伴随状态法	${\cal{O}}(T)$	${\cal{O}}(S)$	否
前向自动微分	${\cal{O}}(NT)$	${\cal{O}}(S)$	是
基于最优检查点的后向自动微分	${\cal{O}}(T\log T)$	${\cal{O}}(S\log T)$	是
基于可逆计算的后向自动微分	${\cal{O}}(T^{1+\epsilon})$	${\cal{O}}(S\log T)$	是

DownLoad: CSV

方法	Julia	ForwardDiff	NiLang	Neural ODE + NiLang
时间/ms	1.90	2.88	6.70	34.30
空间 (估计)	1	6	10⁴	50

DownLoad: CSV

放置规则: 如果第$ i $个格子上有鹅卵石, 则可以从自己堆中取一个鹅卵石放置于第$ i+1 $个格子中,

回收规则: 仅当第$ i $个格子上有鹅卵石, 才可以把第$ i+1 $ 个格子上的鹅卵石取下放入自己的堆中,

结束条件: 第$ n $个格子上有鹅卵石.

游戏目标: 在固定可使用鹅卵石数目为$ S $ (不包括初始鹅卵石) 的前提下, 使用尽可能少的步骤数触发游戏结束.

DownLoad: CSV

输入: 初始状态集合$S=\{0:s_0\}$, 子分块数目$k$, 分块　　起点点$i=0$, 分块长度$L=n$

输出: 末态$S[n]$

1functionbennett (S, $k$, $i$, $L$)

2 　ifL= 1then

3 　$S[i+1]\leftarrow0$

4 　$S[i+1]+=f_iS[i]$

5 　else

6 　　　$l\leftarrow\lceil {L}/{k} \rceil$

7 　　 $k'\leftarrow\lceil {L}/{l} \rceil$

8 　for$j=1,2,\cdots,k'$do

9 　 bennett $(S, k, i+(j-1)l) $, min $(l,L(j-1) l))$

10 　end

11 　for$j=k'-1,k'-2,\cdots 1$do

12 　 bennett $(S, k, i+(j-1)l,l)$

13 　end

14 　end

15 　end

DownLoad: CSV

放置规则: 如果第$ i $个格子上有鹅卵石, 则可以从自己堆中取一个鹅卵石放置于第$ i+1 $个格子中.

回收规则: 可以随意把格子上的鹅卵石取下放入自己的堆中, 收回鹅卵石不计步骤数.

涂鸦规则: 当第$ i $个格子有鹅卵石, 且第$ i+1 $个格子被涂鸦, 可以涂鸦第$ i $个格子, 涂鸦不记入步骤数.

结束条件: 涂鸦完所有的格点.

游戏目标: 在固定可使用鹅卵石数目为$ S $ (不包括初始鹅卵石) 的前提下, 使用尽可能少的步骤数触发游戏结束.

DownLoad: CSV

输入: 状态缓存集合$S=\{0:s_0\}$, 需回传的梯度$\overline{s_n}\equiv $$ \dfrac{\partial {\cal{L} } }{\partial s_n}$, 允许缓存的状态数$\delta$, 扫描次数$\tau$, 分块起点$\beta=0$, 分开终点$\phi=n$内容以及把分块分割为两部分的分割点$\sigma=0$

输出: 回传的梯度$\overline{s_0} \equiv \dfrac{\partial {\cal{L} } }{\partial s_0}$

1functiontreeverse(S, $\overline{s_\phi}$, $\delta$, $\tau$, $\beta$, $\sigma$, $\phi$)

2 　if$\sigma > \beta$then

3 　　$\delta = \delta - 1$

4 　　$s = S[\beta]$　　#加载初始状态 $s_\beta$

5 　　for$j=\beta, \beta+1, ..., \sigma-1$do

6 　　　$s_{j+1} = f_j(s_j)$　　#计算$s_\sigma$

7 　　end

8 　　$S[\sigma] = s_\sigma$

9 　end

10 　#以$\kappa$为最优分割点(二项分布), 递归调用
Treeverse算法

11 　while$\tau>0$and$\kappa={\rm{mid}}(\delta, \tau, \sigma, \phi) < \phi$do

12 　　$\overline{s_{\kappa}}$ = treeverse(S, $\overline{s_\phi}$, $\delta$, $\tau$, $\sigma$, $\kappa$, $\phi$)

13 　　$\tau = \tau-1$

14 　　$\phi = \kappa$

15 　end

16 　$\overline{s_\sigma} \!=\! \overline{f_\sigma}(\overline{s_{\sigma+1} }, s_\sigma)$　　#利用已有的$s_\sigma$和$\overline{s_\phi}$ 回
传导数

17 　if$\sigma>\beta$then

18 　　remove($S[\sigma]$)　# 从缓存的状态集合中移除$s_\sigma$

19 　end

20 　return$\overline{s_\sigma}$

21end

22functionmid($\delta$, $\tau$, $\sigma$, $\phi$) #选取二项分布分割点

23 　$\kappa = \lceil(\delta\sigma + \tau\phi)/(\tau+\delta)\rceil$

24 　if$\kappa \geq \phi$and$\delta > 0$then

25 　　$\kappa = \max(\sigma+1, \phi-1)$

26 　end

27end

DownLoad: CSV

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Vol. 70, Issue 14 - 2021-07-20

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