我们来产生产生

import ply.lex as lex # pip install ply
import ply.yacc as yacc
from functools import reduce
tokens = ('WORD',); t_WORD = r'\w+'; literals = (':', ';')
t_ignore  = ' \t\r\n'
def t_error(t): raise SyntaxError()
d = {}
def p_1(p): "rules : rule"
def p_2(p): "rules : rule rules"
def p_4(p):
    "rule : WORD ':' words ';'"
    k = p[1]; v = d.get(k, []); v.append([p[3]])
    d[k] = v
def p_5(p): "words : WORD"; p[0] = [p[1]]
def p_6(p): "words : WORD words"; p[0] = [p[1]] + p[2]
lexer = lex.lex()
istr = """
    language : sentence ;
    language : sentence language;
    sentence : subject predicate object ;
    sentence : subject predicate object adverbial ;
    subject : Tom ;
    subject : Jerry ;
    predicate : holds ;
    object : Tom ;
    object : Jerry ;
    adverbial : tightly ;
    adverbial : loosely ;
"""
try: yacc.yacc().parse(istr)
except SyntaxError: quit()
def gen(s):
    if len(s) > 4: return
    expanded = False
    for i in range(len(s)):
        w = s[i]
        for r in d.get(w, []):
            expanded = True
            gen(s[:i] + r[0] + s[i + 1:])
    if not expanded: print(reduce(lambda a,b:a+' '+b, s, '')[1:] + '.')
gen(['language'])
# LISP

#define A(x,y) B(x,y)
#define B(x,y) A(x,y)
#define C(x,y)
int main() { A(1,2); }

gcc说扩展B时A没有定义。#define A(x,y) C(x,y)显然没事。SQL有explain功能，Guide to predefined macros in C++ compilers