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{zh:'kNN 密度估計在做什麼？',en:'What does kNN density estimation do?',col:P.cyan,
items:[['固定鄰居數 k，讓體積自己伸縮','\\hat p(x)=\\dfrac{k}{N\\,V(x)}']],
text:'反轉 Parzen 窗：密的地方圈子小、疏的地方圈子大，再用倒數換算成密度。'},
{zh:'兩類分類怎麼判斷？',en:'How is a two-class decision made?',col:C1,
items:[['比拼包住 k 個鄰居所需的體積','\\dfrac{V_2}{V_1}\\ \\underset{\\omega_2}{\\overset{\\omega_1}{\\gtrless}}\\ \\dfrac{N_1}{N_2}\\cdot\\dfrac{P(\\omega_2)}{P(\\omega_1)}\\cdot\\dfrac{\\lambda_{21}-\\lambda_{22}}{\\lambda_{12}-\\lambda_{11}}']],
text:'體積較小代表局部密度較高；再用 N、先驗與損失做校正。第 4 頁的例子裡，V₂/V₁ = 0.25，判給 ω₂。'},
{zh:'維度詛咒在說什麼？',en:'What is the curse of dimensionality?',col:C2,
items:[['平均最近鄰距離','d(l,N)=2\\left(\\dfrac{l\\,\\Gamma(l/2)}{2\\pi^{l/2}N}\\right)^{1/l}'],['N = 10¹⁰ 時的數值',`d(2)\\approx${texSci(dForm(2,1e10))},\\ d(10)\\approx${dForm(10,1e10).toFixed(3)},\\ d(40)\\approx${dForm(40,1e10).toFixed(2)}`]],
text:'相同數量的樣本攤到高維超立方體，點與點的平均距離暴增，甚至超過邊長 1。'},
{zh:'多灑資料能解決嗎？',en:'Can we just collect more data?',col:P.amber,
items:[['l = 2','1000^{1/2}\\approx'+Math.pow(1000,1/2).toFixed(1)+'\\ \\text{倍}'],['l = 10','1000^{1/10}\\approx'+Math.pow(1000,1/10).toFixed(2)+'\\ \\text{倍}'],['l = 40','1000^{1/40}\\approx'+Math.pow(1000,1/40).toFixed(2)+'\\ \\text{倍}']],
text:'低維立竿見影，高維幾乎無效；所需樣本數隨維度指數成長。唯一生路是降維（PCA、LDA、特徵選取）。'}];
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