DDIM公式草稿2

q(xt1xt,x0)=αt1x0+1αt1 εεN(0,I)=αt11αt(xt1αt  εθ(xt,t))+1αt1 ε=αt1x0t^+1αt1 ε=αt1x0t^+1αt1 εθ(xt,t)=αt1x0t^+1αt1σt2εθ+σtε使q(xtxt1),q(xt1xt,x0)q(xsxk,x0)q(xsxk,x0)=αsx0k^+1αsσk2εθ+σkεT=1000999998...3210(DDPM)T=1000888666...1005010(DDIM)q(xtxt1)=N(xt,αtxt1,(1αt)I)trainphaseq(xtx0)=N(xt;αtxt,1αtI)samplingphasex0t^=1αt(xt1αtεθ(xt,t))xt1=μ(xt,x0t^)+σtε\begin{aligned} q(x_{t-1}|x_t,x_0)&=\sqrt{\overline{\alpha_{t-1}}}x_0 + \sqrt{1-\overline{\alpha_{t-1}}}\ \varepsilon \qquad \varepsilon\sim N(0,I)\\ &=\sqrt{\overline{\alpha_{t-1}}} \frac{1}{\sqrt{\overline{\alpha_{t}}}}(x_t-\sqrt{1-\overline{\alpha_{t}}}\ \ \varepsilon_{\theta}(x_t,t) ) +\sqrt{1-\overline{\alpha_{t-1}}} \ \varepsilon \\ &=\sqrt{\overline{\alpha_{t-1}}} \hat{x_{0|t}} +\sqrt{1-\overline{\alpha_{t-1}}} \ \varepsilon \\ &=\sqrt{\overline{\alpha_{t-1}}} \hat{x_{0|t}} +\sqrt{1-\overline{\alpha_{t-1}}} \ \varepsilon_{\theta}(x_t,t)\\ &= \sqrt{\overline{\alpha_{t-1}}} \hat{x_{0|t}} + \sqrt{1-\overline{\alpha_{t-1}}-\sigma_t^2} \varepsilon_{\theta}+\sigma_t\varepsilon \\ 上述推导并未使用 &q(x_t|x_{t-1})条件,即q(x_{t-1}|x_t,x_0)可以写做q(x_s|x_k,x_0) \\ q(x_s|x_k,x_0)&=\sqrt{\overline{\alpha_{s}}} \hat{x_{0|k}} + \sqrt{1-\overline{\alpha_{s}}-\sigma_k^2} \varepsilon_{\theta}+\sigma_k\varepsilon \\ T=1000 &\qquad 999 \qquad998 \qquad ... \qquad 3 \qquad 2 \qquad 1 \qquad 0 \quad(DDPM)\\ T=1000 &\qquad 888 \qquad666 \qquad ... \qquad 100 \qquad 50 \qquad 1 \qquad 0 \quad(DDIM)\\ \end{aligned} \quad \\ q(x_t|x_{t-1})=N(x_t,\sqrt{\alpha_t}x_{t-1},(1-\alpha_t)I) \\ ①train\quad phase \qquad \qquad q(x_t|x_0)=N(x_t;\sqrt{\overline{\alpha_t}}x_t,\sqrt{1-\overline{\alpha_t}}I) \\ \quad \\ \begin{aligned} ② sampling \quad phase \qquad \qquad \hat{x_{0|t}}&=\frac{1}{\sqrt{\overline{\alpha_t}}}(x_t-\sqrt{1-\overline{\alpha_t}}\varepsilon_{\theta}(x_t,t) )\\ x_{t-1}&=\mu(x_t,\hat{x_{0|t}})+\sigma_t\varepsilon \end{aligned}
全部评论

相关推荐

神哥了不得:放平心态,再找找看吧,主要现在计算机也变卷了,然后就比较看学历了,之前高中毕业你技术强,都能找到工作的
点赞 评论 收藏
分享
评论
点赞
收藏
分享

创作者周榜

更多
牛客网
牛客企业服务