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01-why-maximum-entropy.html
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uniform :{zh:'均勻',en:'Uniform',color:P.lime,q:u=>1},
unimodal:{zh:'單峰',en:'Unimodal',color:P.cyan,q:u=>6*u*(1-u)},
bimodal :{zh:'雙峰',en:'Bimodal',color:P.violet,q:u=>gs(u,.25,.11)+gs(u,.75,.11)},
sharp :{zh:'高尖峰',en:'Sharp peak',color:P.coral,q:u=>gs(u,.5,.08)}};
const SHAPE_KEYS=['uniform','unimodal','bimodal','sharp'];
(function(){const N=4000;for(const k of SHAPE_KEYS){const s=SHAPES[k];let Z=0;
for(let i=0;i<N;i++)Z+=s.q((i+.5)/N)/N;
s.Z=Z;s.qn=u=>(u<0||u>1)?0:s.q(u)/Z;
let h=0;for(let i=0;i<N;i++){const p=s.qn((i+.5)/N);if(p>1e-300)h-=p*Math.log(p)/N}
s.HQ=h}})(); /* entropy on the unit interval; H(W)=HQ+ln W */
const pdf=(k,x,W)=>SHAPES[k].qn(x/W)/W;
const entropy=(k,W)=>SHAPES[k].HQ+Math.log(W);
function areaOf(k,W){const N=2000;let a=0;for(let i=0;i<N;i++)a+=pdf(k,(i+.5)*W/N,W)*W/N;return a}
function peakOf(k,W){let m=0;for(let i=0;i<=400;i++)m=Math.max(m,pdf(k,i*W/400,W));return m}
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/* ===== Page script ===== */
/* 01 — Motivation: a friend somewhere on a 4-km trail, and only a range is known */
let sel='uniform';
const DRAW=[['uniform','畫法一 平平淡淡','Uniform'],['unimodal','畫法二 中間隆起','Unimodal'],['bimodal','畫法三 兩端雙峰','Bimodal']];
const STORY={
uniform :['沒有偷偷新增任何故事','No extra story: every kilometre is treated equally. Nothing is claimed beyond "somewhere in [0, 4]".',P.lime],
unimodal:['偷偷假裝朋友最愛待在中間','Pretends the friend prefers the middle of the trail, yet no message, GPS or history says so.',P.coral],
bimodal :['偷偷假裝朋友在起點與終點左右為難','Pretends the friend hovers near both ends, an invented dramatic twist with zero evidence.',P.coral]};
const M=mapper(880,266,620,222,0,4,0,1.0);
function render(){
reset();title('只知道範圍，該怎麼畫機率？','Knowing only a range, how should we draw the probability?',1);
/* left: the trail scenario */
rr(60,170,700,430,24,P.panel,'#ffffff14',2);
txt('在 4 公里步道旁等朋友',90,205,26,P.ink,'left',850);
txt('Waiting for a friend on a 4-km trail',90,236,18,P.cyan,'left',650);
rr(90,262,640,104,22,P.panel2,P.amber,3);
txt('簡訊 · SMS',118,290,18,P.amber,'left',800);
txt('「我人在 0 到 4 公里之間。」',118,324,28,P.ink,'left',850);
txt('"I am somewhere between 0 and 4 km."',118,352,18,P.muted,'left',600);
const tx=k=>130+k*140,ty=466;
line(tx(0),ty,tx(4),ty,'#ffffff45',12);
for(let k=0;k<=4;k++){line(tx(k),ty-14,tx(k),ty+14,'#ffffff80',3);txt(k+' km',tx(k),ty+38,18,P.muted,'center',650)}
rr(tx(2.6)-28,ty-70,56,56,28,P.amber+'28',P.amber,3);txt('?',tx(2.6),ty-42,32,P.amber,'center',900);
arrow(tx(2.6),ty-12,tx(2.6),ty-4,P.amber,3);
wrap('沒有 GPS 小圓點、沒有打卡紀錄、沒有任何其他線索。',90,528,640,28,20,P.muted,600);
txt('No GPS, no history, no other clue.',90,584,17,'#7f9bb0','left',600);
/* right: three ways to draw the density */
rr(790,170,750,430,24,P.panel,'#ffffff14',2);
txt('同一條步道，三種畫法',820,205,26,P.ink,'left',850);
txt('Three ways to draw p(x) on [0, 4] km',820,236,18,P.cyan,'left',650);
axes(M,[0,1,2,3,4],[0,.25,.5,.75,1]);
const ref=x=>pdf('uniform',x,4);
if(sel!=='uniform')plotCurve(ref,M,0,4,P.lime,3,null,[10,8]);
plotCurve(x=>pdf(sel,x,4),M,0,4,SHAPES[sel].color,5,'33');
txt('均勻基準 0.25',M.px+M.pw-6,M.Y(.25)-18,16,P.lime,'right',700);
DRAW.forEach((d,i)=>btn(d[1],815+i*245,535,225,50,()=>{sel=d[0];render()},sel===d[0],i==0?P.lime:i==1?P.cyan:P.violet,19));
/* verdict */
const s=STORY[sel];
rr(60,620,1480,100,22,P.panel2,s[2],3);
txt(s[0],90,654,26,s[2],'left',900);
wrap(s[1],90,676,1420,26,19,P.muted,600);
/* principle */
rr(60,740,1480,112,22,'#4fd1c514',P.cyan,3);
txt('最大熵原理 · Principle of Maximum Entropy',90,774,26,P.cyan,'left',900);
wrap('在所有滿足已知限制的分佈中,選出最不偏袒任何位置、最不武斷的那一個。',90,796,1420,26,22,P.ink,700);
txt('Among all distributions that satisfy what we know, choose the one that makes the fewest arbitrary commitments.',90,832,18,P.muted,'left',600);
}
render();
</script></body></html>