Your browser does not support the canvas element.

Jincheng Yang

The University of Texas at Austin

%--Paired Delimiters-- $ \newcommand{\abs}[1]{\left\lvert #1 \right\rvert} \newcommand{\ang}[1]{\left\langle #1 \right\rangle} \newcommand{\bkt}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\nor}[1]{\left\lVert #1 \right\rVert} \newcommand{\pth}[1]{\left( #1 \right)} \newcommand{\set}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\abset}[1]{\abs{\set{#1}}} \newcommand{\ptset}[1]{\pth{\set{#1}}} $ %--Operators-- $ \newcommand{\grad}{\nabla} \newcommand{\La}{\Delta} \renewcommand{\div}{\operatorname{div}} \newcommand{\curl}{\operatorname{curl}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\Hess}{\operatorname{Hess}} \newcommand{\mm}{\mathcal{M}} \newcommand{\inv}{^{-1}} \newcommand{\tensor}{\otimes} \newcommand{\cross}{\times} \newcommand{\weak}[1]{\xrightharpoonup{#1}} $ %--Notations-- $ \newcommand{\Vol}{\mathrm{Vol}} \newcommand{\loc}{\mathrm{loc}} \renewcommand{\dim}{\mathrm{dim}} \newcommand{\Span}{\mathrm{Span}} \newcommand{\diam}{\mathrm{diam}} \newcommand{\Id}{\mathrm{Id}} \newcommand{\Lip}{\mathrm{Lip}} \newcommand{\dist}{\mathrm{dist}} \newcommand{\inv}{^{-1}} $ %--Algebra-- $ \newcommand{\hfsq}[1]{\frac{\abs{#1} ^2}{2}} $ %--Sets-- $ \newcommand{\R}{\mathbb{R}} \newcommand{\RR}[1]{\R ^{#1}} \newcommand{\Rd}{\RR d} \newcommand{\Rn}{\RR n} \renewcommand{\S}{\mathbb{S}} \renewcommand{\Z}{\mathbb{Z}} \newcommand{\T}{\mathbb{T}} \newcommand{\ssubset}{\subset \subset} \newcommand{\spt}{\operatorname{spt}} \newcommand{\supp}{\operatorname{supp}} \newcommand{\diam}{\mathrm{diam}} \newcommand{\diag}{\operatorname{diag}} \renewcommand{\dim}{\mathrm{dim}} \newcommand{\dimH}{\mathrm{dim} _\mathcal{H}} \newcommand{\Sing}{\mathrm{Sing}} \newcommand{\Vol}{\mathrm{Vol}} \newcommand{\inds}[1]{\mathbf1_{\set{#1}}} \newcommand{\ind}[1]{\mathbf1_{#1}} $ %--Summations-- $ \newcommand{\SUM}[3]{\sum \cnt{#1}{#2}{#3}} \newcommand{\PROD}[3]{\prod \cnt{#1}{#2}{#3}} \newcommand{\cnt}[3]{ _{#1 = #2} ^{#3} } \newcommand{\seq}[2]{\set{#1 _{#2}} _{#2}} \newcommand{\seqi}[2]{\set{#1 _{#2}} \cnt{#2}1i} $ %--Superscript and Subscript $ \newcommand{\pp}[1]{^{(#1)}} \newcommand{\bp}[1]{^{(#1)}} \newcommand{\pps}[2][1]{^{#2 + #1}} \newcommand{\bps}[2][1]{_{#2 + #1}} \newcommand{\pms}[2][1]{^{#2 - #1}} \newcommand{\bms}[2][1]{_{#2 - #1}} $ %--Notations-- $ \newcommand{\loc}{\mathrm{loc}} \newcommand{\Span}{\operatorname{Span}} \newcommand{\argmin}{\operatorname{argmin}} \newcommand{\argmax}{\operatorname{argmax}} \newcommand{\osc}{\operatorname{osc}} \newcommand{\Id}{\mathrm{Id}} \newcommand{\Lip}{\operatorname{Lip}} \newcommand{\Leb}{\operatorname{Leb}} \newcommand{\PV}{\mathrm{P.V.}} \newcommand{\dist}{\operatorname{dist}} \newcommand{\at}[1]{\bigr\rvert _{#1}} \newcommand{\At}[1]{\biggr\rvert _{#1}} \newcommand{\half}{\frac12} $ %--Text-- $ \newcommand{\inn}{\text{ in }} \newcommand{\onn}{\text{ on }} \renewcommand{\ae}{\text{ a.e. }} \newcommand{\st}{\text{ s.t. }} \newcommand{\forr}{\text{ for }} \newcommand{\as}{\text{ as }} $ %--Differential-- $ \newcommand{\d}{\mathop{\kern0pt\mathrm{d}}\!{}} \newcommand{\dt}{\d t} \newcommand{\dx}{\d x} \newcommand{\dy}{\d y} \newcommand{\ptil}{\partial} \newcommand{\pt}{\ptil _t} \newcommand{\pfr}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\dfr}[2]{\frac{\mathrm{d} #1}{\mathrm{d} #2}} \newcommand{\ddt}{\dfr{}t} \newcommand{\pthf}[2]{\pth{\frac{#1}{#2}}} $ %--Integral-- $ \newcommand{\intR}{\int _0 ^\infty} \newcommand{\intRn}{\int _{\Rn}} \newcommand{\intRd}{\int _{\Rd}} \newcommand{\fint}{-\!\!\!\!\!\!\int} \newcommand{\intset}[1]{\int _{\set{#1}}} $ %--Greek-- $ \newcommand{\e}{\varepsilon} \newcommand{\vp}{\varphi} $ %--Norm-- $ \newcommand{\nmL}[2]{\nor{#2} _{L ^#1}} $

Research Blog

Guessing Letter

2017, September 01

This is a very simple game called Guessing Letter, similar to the more popular four digit number guessing game. I wrote this as a practice of javaScript.

The secret password is a four-letter string without spaces. Your job is to find out this word. Input your guess and click “Compare”, then you will get a comparason in the form of $m$A$n$B, where $m$ is the number of slots in the password that you input correctly, and $n$ is the number of slots in the password that you used the letter in that slot, possibly elsewhere.

I wrote a Matlab program to solve this question in as few steps as possible, based on the principle that one should maximize the negentropy of each guess.

Recent Posts

22 Jan 2021

De Giorgi

14 Apr 2020

Lorentz Space