对着规则核一句

（每道题开头都有同一段：T() 是上一站 lexer 的替身——源码里 token 之间用空格分开，它切成 (kind, text, line, col)；show() 把树写成一行：二元运算全加括号，语句之间用 |。）

还没写完整 parser，先用最笨的办法核「是不是一条 let 语句」：五个 token 逐个对照产生式。看三句各得什么：

KEYWORDS = ("let", "if", "else", "while")
CMP = (">=", "<=", "==", "!=", "<", ">")

def T(src):
    toks = []
    for ln, line in enumerate(src.split("\n"), 1):
        pos = 0
        for w in line.split():
            pos = line.index(w, pos)
            if w in KEYWORDS:
                k = "KEYWORD"
            elif w[0] == '"':
                k = "STRING"
            elif w.isdigit():
                k = "NUMBER"
            elif w[0].isalpha() or w[0] == "_":
                k = "IDENT"
            else:
                k = "OP"
            toks.append((k, w, ln, pos + 1))
            pos += len(w)
    toks.append(("EOF", "", ln, pos + 1))
    return toks

def is_let(toks):
    return (len(toks) == 6 and toks[0][1] == "let" and toks[1][0] == "IDENT"
            and toks[2][1] == "=" and toks[3][0] in ("NUMBER", "STRING", "IDENT") and toks[4][1] == ";")

for s in ["let rate = 12 ;", "let = rate 12 ;", "let msg = \"hi\" ;"]:
    print(is_let(T(s)), end=",")