1.算法描述
强化学习通常包括两个实体agent和environment。两个实体的交互如下,在environment的statestst下,agent采取actionatat进而得到rewardrtrt 并进入statest+1st+1。Q-learning的核心是Q-table。Q-table的行和列分别表示state和action的值,Q-table的值Q(s,a)Q(s,a)衡量当前states采取actiona到底有多好。
在每一时刻,智能体观测环境的当下状态并选择一个动作,这会导致环境转移到一个新的状态,与此同时环境会返回给智能体一个奖励,该奖励反映了动作所导致的结果。在倒立摆任务中,每一个时间步的奖励均为+1,但是一旦小车偏离中心超过4.8个单位或者杆的倾斜超过15度,任务就会终止。因此,我们的目标是使得该任务能够尽可能地运行得更久,以便获得更多的收益。原始倒立摆任务中,智能体的输入包含4个实数(位置,速度等),但实际上,神经网络可以直接通过观察场景来完成任务,所以我们可以直接使用以小车为中心的屏幕补丁作为输入。严格来说,我们设计的状态是当前屏幕补丁与上一个屏幕补丁的差值,这使得智能体能够从一张图像中推断出杆的速度。
为了训练DQN,我们将使用经验回放池(experience replay memory)来存储智能体所观测到的环境状态转移情况,在之后的训练中我们可以充分利用这些数据。通过对经验回放池中的数据进行随机采样,组成一个批次的转移情况是互不相关(decorrelated)的,这极大地提升了DQN训练的性能和稳定性。
主要步骤如下:
采样得到一个批次的样本,将这些样本对应的张量连接成一个单独的张量;
分别利用策略Q网络与目标Q网络计算 与Q(st,at)与V(st+1)=maxaQ(st+1,a) ,利用它们计算损失函数.。另外,如果 s 为终止状态,则令 V(s)=0
更新Q网络参数。目标Q网络的参数每隔一段时间从主Q网络处固定而来,在本例中,我们在每个episode更新一次目标Q网络。
2.仿真效果预览
matlab2022a仿真结果如下:
3.MATLAB部分代码预览
for trial=1:MaxTr, %外部循环开始 count=0; failure=0; failReason=0; lfts = 1; newSt = inistate; inputs = newSt./NF; lc = Initlc; la = Initla; xhist=newSt; %计算newAction ha = inputs*wa1; g = (1 - exp(-ha))./(1 + exp(-ha)); va = g*wa2; newAction = (1 - exp(-va))./(1 + exp(-va)); %计算J inp=[inputs newAction]; qc=inp*wc1; p = (1 - exp(-qc))./(1 + exp(-qc)); J=p*wc2; Jprev = J; while(lfts<Tit), %内部循环开始 if (rem(lfts,500)==0), disp(['It is ' int2str(lfts) ' time steps now......']); end %生成控制信号 if (newAction >= 0) sgnf = 1; else sgnf = -1; end u = Mag*sgnf; %bang-bang control %Plug in the model [T,Xf]=ode45('cartpole_model',[0 tstep],newSt,[],u); a=size(Xf); newSt=Xf(a(1),:); inputs=newSt./NF; %input normalization %计算newAction ha = inputs*wa1; g = (1 - exp(-ha))./(1 + exp(-ha)); va = g*wa2; newAction = (1 - exp(-va))./(1 + exp(-va)); %calculate new J inp=[inputs newAction]; qc=inp*wc1; p = (1 - exp(-qc))./(1 + exp(-qc)); J=p*wc2; xhist=[xhist;newSt]; %%===========================================================%% %%求取强化信号r(t),即reinf %% %%===========================================================%% if (abs(newSt(1)) > FailTheta) reinf = 1; failure = 1; failReason = 1; elseif (abs(newSt(3)) > Boundary) reinf = 1; failure = 1; failReason = 2; else reinf = 0; end %%================================%% %% learning rate update scheme %% %%================================%% if (rem(lfts,5)==0) lc = lc - 0.05; la = la - 0.05; end if (lc<0.01) lc=0.005; end if (la<0.01) la=0.005; end %%================================================%% %% internal weights updating cycles for critnet %% %%================================================%% cyc = 0; ecrit = alpha*J-(Jprev-reinf); Ec = 0.5 * ecrit^2; while (Ec>Tc & cyc<=Ncrit), gradEcJ=alpha*ecrit; %----for the first layer(input to hidden layer)----------- gradqwc1 = [inputs'; newAction]; for i=1:N_Hidden, gradJp = wc2(i); gradpq = 0.5*(1-p(i)^2); wc1(:,i) = wc1(:,i) - lc*gradEcJ*gradJp*gradpq*gradqwc1; end %----for the second layer(hidden layer to output)----------- gradJwc2=p'; wc2 = wc2- lc*gradEcJ*gradJwc2; %----compute new J---- inp=[inputs newAction]; qc=inp*wc1; p = (1 - exp(-qc))./(1 + exp(-qc)); J=p*wc2; cyc = cyc +1; ecrit = alpha*J-(Jprev-reinf); Ec = 0.5 * ecrit^2; end % end of "while (Ec>0.05 & cyc<=Ncrit)" %normalization weights for critical network if (max(max(abs(wc1)))>1.5) wc1=wc1/max(max(abs(wc1))); end if max(max(abs(wc2)))>1.5 wc2=wc2/max(max(abs(wc2))); end %%=============================================%% %% internal weights updating cycles for actnet %% %%=============================================%% cyc = 0; eact = J - Uc; Ea = 0.5*eact^2; while (Ea>Ta & cyc<=Nact), graduv = 0.5*(1-newAction^2); gradEaJ = eact; gradJu = 0; for i=1:N_Hidden, gradJu = gradJu + wc2(i)*0.5*(1-p(i)^2)*wc1(WC_Inputs,i); end %----for the first layer(input to hidden layer)----------- for (i=1:N_Hidden), gradvg = wa2(i); gradgh = 0.5*(1-g(i)^2); gradhwa1 = inputs'; wa1(:,i)=wa1(:,i)-la*gradEaJ*gradJu*graduv*gradvg*gradgh*gradhwa1; end %----for the second layer(hidden layer to output)----------- gradvwa2 = g'; wa2=wa2-la*gradEaJ*gradJu*graduv*gradvwa2; %----compute new J and newAction------- ha = inputs*wa1; g = (1 - exp(-ha))./(1 + exp(-ha)); va = g*wa2; newAction = (1 - exp(-va))./(1 + exp(-va)); inp=[inputs newAction]; qc=inp*wc1; p = (1 - exp(-qc))./(1 + exp(-qc)); J=p*wc2; cyc = cyc+1; eact = J - Uc; Ea = 0.5*eact^2; end %end of "while (Ea>Ta & cyc<=Nact)" if ~failure Jprev=J; else break; %another trial 即跳出“while(lfts<Tit),” end lfts=lfts+1; end %end of "while(lfts<Tit)" 结束内部循环 msgstr1=['Trial # ' int2str(trial) ' has ' int2str(lfts) ' time steps.']; msgstr21=['Trial # ' int2str(trial) ' has successfully balanced for at least ']; msgstr22=[msgstr21 int2str(lfts) ' time steps ']; A_027
标签:newAction,newSt,%%,Qlearning,qc,matlab,exp,ha,倒立 From: https://www.cnblogs.com/51matlab/p/17033879.html