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01-evidence-and-query.html
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/* ===== Page script ===== */
/* ===== Shared helpers for KP18: DAG drawing, parent sets, CPT parameter counting ===== */
const C1='#63b3ed',C2='#ff7b72';
const NM=['x₁','x₂','x₃','x₄','x₅','x₆'];
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const SUPM={'0':'⁰','1':'¹','2':'²','3':'³','4':'⁴','5':'⁵','6':'⁶','7':'⁷','8':'⁸','9':'⁹','-':'⁻'};
const sup=n=>String(n).split('').map(c=>SUPM[c]||c).join('');
/* ===== KP19: the four-variable chain x -> y -> z -> w with the lecture's CPTs ===== */
const CPC={px:.6,py:[.3,.4],pz:[.6,.25],pw:[.3,.45]}; /* P(child=1 | parent=0/1) */
const NC=['x','y','z','w'],POS4=[[.07,.5],[.36,.5],[.64,.5],[.93,.5]];
const bern=(p,t)=>t?p:1-p;
function jointC(v,th){th=th||CPC;return bern(th.px,v[0])*bern(th.py[v[0]],v[1])*bern(th.pz[v[1]],v[2])*bern(th.pw[v[2]],v[3])}
/* exact posterior P(q=1 | evidence) by full enumeration; ev maps node index -> 0/1 */
function postC(q,ev,th){let n1=0,n0=0;for(let m=0;m<16;m++){const v=[(m>>3)&1,(m>>2)&1,(m>>1)&1,m&1];
if(Object.keys(ev).some(k=>v[k]!==ev[k]))continue;const j=jointC(v,th);if(v[q])n1+=j;else n0+=j}return n1/(n1+n0)}
/* 01 — Evidence and query on the chain x -> y -> z -> w (exact answers by enumeration) */
let evI=0,evV=1,qI=3;
function render(){
reset();title('核心叩問：推論，難道每次都要重算聯合機率？','The core question: must every inference recompute the full joint?',1);
rr(60,170,700,680,24,P.panel,'#ffffff14',2);
[['傳話遊戲','The telephone game','每一站只需記住「上一位說了什麼」，不必跑回起點看紙條。',P.amber],['證據 Evidence E','Observed variables','推論時已親眼觀察到的變數值，例如確診抽菸 x = 1。',P.cyan],['查詢 Query P(Q | E)','The target posterior','在證據之下，我們想知道的其他變數的後驗機率。',P.lime]].forEach((c,i)=>{
const y=196+i*214;rr(86,y,648,196,20,P.panel2,c[3]+'88',2);txt(c[0],112,y+34,24,c[3],'left',900);txt(c[1],112,y+64,15,P.muted,'left',650);wrap(c[2],112,y+92,596,30,20,P.ink,650)});
rr(790,170,750,680,24,P.panel,'#ffffff14',2);
txt('在鏈 x → y → z → w 上自己問一個問題',818,205,22,P.ink,'left',850);txt('Pick the evidence and the query; the answer is computed exactly',818,232,15,P.cyan,'left',650);
const box={x:840,y:262,w:660,h:100},ps=POS4.map(p=>[box.x+p[0]*box.w,box.y+p[1]*box.h]);
for(let i=0;i<3;i++)arrow(ps[i][0]+30,ps[i][1],ps[i+1][0]-34,ps[i+1][1],'#9fb7c9',4);
ps.forEach((p,i)=>{const isE=i===evI,isQ=i===qI;ctx.beginPath();ctx.arc(p[0],p[1],28,0,Math.PI*2);ctx.fillStyle=isE?P.cyan+'44':(isQ?P.lime+'33':P.panel2);ctx.fill();
ctx.strokeStyle=isE?P.cyan:(isQ?P.lime:'#8fb2c9');ctx.lineWidth=isE||isQ?4:2.5;ctx.stroke();txt(NC[i],p[0],p[1]+1,22,P.ink,'center',900);
if(isE)txt('證據 '+NC[i]+'='+evV,p[0],p[1]-46,15,P.cyan,'center',800);if(isQ)txt('查詢',p[0],p[1]+50,15,P.lime,'center',800)});
txt('證據在哪個節點？',818,418,18,P.cyan,'left',800);
['無','x','y','z','w'].forEach((l,i)=>btn(l,818+i*98,436,90,44,()=>{evI=i-1;render()},evI===i-1,P.cyan,17));
if(evI>=0){btn(NC[evI]+' = 1',1318,436,100,44,()=>{evV=1;render()},evV===1,P.amber,16);btn(NC[evI]+' = 0',1428,436,90,44,()=>{evV=0;render()},evV===0,P.amber,16)}
txt('想查詢哪個節點？',818,520,18,P.lime,'left',800);
NC.forEach((l,i)=>btn(l,818+i*98,538,90,44,()=>{qI=i;render()},qI===i,P.lime,17));
const ev=evI>=0?{[evI]:evV}:{},p1=postC(qI,ev),pr=postC(qI,{});
rr(818,610,700,126,18,P.panel2,P.amber+'88',2);
txt(`P(${NC[qI]}=1${evI>=0?` | ${NC[evI]}=${evV}`:''}) = ${p1.toFixed(4)}`,844,646,26,P.amber,'left',900);
txt(`無證據時的邊際機率 P(${NC[qI]}=1) = ${pr.toFixed(4)}`,844,684,17,P.muted,'left',650);
const dir=evI<0?'沒有證據：只需順著鏈做邊際化':(qI===evI?'查詢就是證據本身':(qI>evI?'順流下傳（預測）：證據在上游':'逆流上傳（歸因）：證據在下游'));
txt(dir,844,716,18,qI>evI&&evI>=0?C2:(qI<evI?C1:P.ink),'left',800);
wrap('暴力法要對中間所有變數窮舉求和；傳話式的訊息傳遞，每一站只收一個數、查自己的 CPT。',818,760,700,26,17,P.ink,650);
txt('Brute force sums over every intermediate state; message passing does one small step per node.',818,832,14,P.muted,'left',600);
showF([]);
}
render();
</script></body></html>