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重建个自己的avatar放metaverse里面跑 Human Face Reconstruction ReferencePIFuPIFu分为single-view surface reconstruction及multi-view surface reconstruction，但是实际上multi-view surface reconstruction就是对多个single-view的一个整合(比如简单粗暴的取平均值之类的)，故不多赘述。 PIFu，全称Pixel-Aligned Implicit Function，顾名思义，它与之前的做human digitalization的算法的不同之处在于两点，一个是Pixel-Aligned，一个是implicit function。关于pixel-aligned，之前的算法大多都是voxel-aligned，也就是将一个三维的数据，即voxel，放到model中训练，显然，这费时又费空间，而PIFu只先关注二维数据，再利用implicit function的优势，将深度参数$z(X)$放进隐式函数中求解。 首先声明一下implicit ...

有趣的内容 IntroductionKey takeaways: Network defines the interactions between the components How do we reason about networks? Empirical: Study network data to find organizational principles Mathematical models: Probabilistic, graph theory Algorithms for analyzing graphs 我们study network是为了achieve什么？ Paterns and statistical properites of network data Design principles and models Understand why networks are organized the way they are (Predict behavior of networked systems) 我们要学习network的什么？ Struc ...

NLP Learning Notes#01 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190 ...

Lecture 2: Cryptographic Hash Function (Example: SHA256)$\sum$指所有可能出现的字符. $\sum^* = \sum^0 \cup \sum^1 \cup …$。上面的数字指的是字符序列长度 $f: \sum^* \rightarrow \sum^n$ map from infinite to finite Hash function characteristic: Collision-resistance: It is hard to find $x, y \in \sum^*$, s.t., $x \neq y$ and $f(x) = f(y)$ Hiding: Given a hash value $h$, it is hard to find same $x$ s.t. $f(x) = h$ Puzzle friendliness Problem: Finding Files I have huge files $F_1, F_2, …, F_n$. Given a new file, find i ...

First of all, let me give a basic intuition of $DQN$ (Deep Q-Network). It is composed of $D$ (deep learning) and $QN$ (Q-Network), among which the Q-Learning algorithm is the core and the neural network is just a tool to maintain the matrix in Q algorithm because of the memory limits (a $state \times actioin$ size of matrix is too big for most of the times). Algorithm DescriptionQ-LearningIt is explained in detail in this blog: A Painless Q-Learning Tutorial DQNSimilar to dynamic programming, th ...

Normalized Training ModelStage 1: Load DataDivide the stage into four sub-stage: load data file separate data into groups according to the number of features divide the dataset into training & test dataset normalize data to the range of [0, 1] 1.1.1 load data file 12345678import paddleimport numpy as npimport jsondef load_data (): # load data file datafile = '.../.../.../' data = np.fromfile (datafile, sep = ' ') np.fromfile is used for constructing data from ...

It’s Time To Learn PYTHON! (record of python challenges)Sometimes refer to the solutions to all the passes: HackingNote Python Challenges Solutions The First PassThe gibberish displayed on the screen is a passcode which is coded with Caeser cipeher, one of the simplest encryption techniques. From the Wikipedia, we can know that It is a type of substitution cipher in which each letter in the plaintext is replaced by a letter some fixed number of positions down the alphabet. For example, with a l ...

以一块 $Cu$ 结束了这么多年的 $\text{OI}$ 生涯 $Day ~ ~ 0$ 的笔试十分顺利地拿到了 $100$ 分满分，晚上一直在肝模板，但好像实际上并没有啥用，需要模板的部分分估摸着大多都快到正解了，咱也想不到 $Day ~ ~ 1$ 开始前，按照往年成绩，我给自己定下地分数目标大概是 $Day ~ ~ 1$ 要大于等于 $150$，$Day ~ ~ 2$ 大于等于 $130$ 左右，但我认为这样的目标间接造成了 $Day ~~ 1$ 比较大的失利 $T1$ 看了大概 $20min$ 左右就想到了 $O (kn^3\log n)$ 矩阵乘法的思路，但好像怎么算都最多只能过 $k = 0$ 的点，剩下的怎么算都过不去（实际上我认为这矩乘可以过 $k = 0$ 的时候时间也已经过去了很久），结合暴力这样就有 $65 pts$ 了，但由于给自己定下的目标，我一直在想后续解法，一直犹豫到想到了 $1h45min$ 左右也没啥想法，这时我算是比较慌张了，因为我本来是打算每题整个 $1.5h$，比较从容，但这么优柔寡断一下导致我最简单的暴力都一直打错，最后到 $2.5h$ 才勉勉 ...

CRT用于求解类似$$\left\{\begin{aligned}&x \equiv a_1 \pmod{m_1} \\&x \equiv a_2 \pmod{m_2} \\&… \\&x \equiv a_n \pmod{m_n}\end{aligned}\right.$$其中，$\forall i, j, ~ (m_i, m_j) = 1$ 设 $M = \prod_{i = 1}^n m_i, ~ M_i = \frac{M}{m_i}$，再设 $M_it_i \equiv 1 \pmod{m_i}$，其中 $1 \le i \le n$ 则可构造一解 $x = \sum\limits_{i = 1}^n a_iM_it_i$，任意解 $x_0 = x + k \times M$ ExCRT用于求解类似$$\left\{\begin{aligned}&x \equiv a_1 \pmod{m_1} \\&x \equiv a_2 \pmod{m_2} \\&… \\&x \equiv a_n \pmod{m_n}\e ...

终于搞完了。。 这是第三版的博客迁移 之前由于咕了一段时间OI 搞的原博客服务器过期域名过期数据全没了 然后就重新搞了这么一个玩意儿 至于原因 大概就是咕了半年后去省选摸个鱼一不小心进队了 于是就得准备NOI了 当然，这个“一不小心”不带有任何自傲的意思 我只是觉得自己的水平还没有达到标准。。 还有一个月，出去集训 希望到NOI的时候能达到自己想要的水平吧 保银争金 那么我的博客也得重新回来了 嗯 就这样 $\text{Upd:}$ A.F.O.心路历程&NOI2020游记