#! /usr/bin/python3
# Last edited on 2026-09-07 08:31:42 by stolfi

# Solves the puzzle posed by #ololololo on Voynich.Ninja on 2026-09-05
# Namely looks for a parag that has the second word repeated two more times,
# 6 positions apart.

# wp res/starps-gd-wp-par.ivt
# wc res/starps-gd-wc-par.ivt
# ch res/bencao-fu-ch-par.ivt

import sys, re, os, string, glob
from sys import stderr as err
from math import sqrt, hypot, exp, log, floor, ceil, isfinite, isnan, inf, nan

def main():
  utype = sys.argv[1]
  ivt_file = sys.argv[2]
  rd = open(ivt_file, "r")
  lines = rd.readlines()
  
  for line in lines:
    trysolve(line, utype) 
  return
  # ----------------------------------------------------------------------
  
def trysolve(line, utype):
  line = line.strip()
  line = re.sub(r" *[#].*$", "", line)
  if line == "": return
  
  line = re.sub(r"[;][A-Z]", "", line)
  m = re.fullmatch(r"<([a-z0-9.]+)> *(.*)", line)
  assert m is not None, f"bad format [[{line = !r}]]"
  loc = m.group(1)
  text = m .group(2)
  text = re.sub(r"[«=»]", "", text)
  text = re.sub(r"<[%$]>", "", text)
  
  if utype == "wc":
    # Consider commas:
    text = re.sub(r"[-,]", " ", text)
  elif utype == "wp":
    # Ignore commas:
    text = re.sub(r"[-.]", " ", text)
    text = re.sub(r"[,]", "", text)
  elif utype == "ch":
    # Remove ideographic punctuation:
    text = re.sub(r"[］［；：、，。（）　]", "", text)
  else:
    assert False, f"bad {utype = !r}"
    
  if utype == "ch":
    toks = [ c for c in text ]
  else:
    toks = text.split();
  nt = len(toks)
  if nt < 20: return
  key = toks[1]
  occs = []
  for i in range(nt):
    if toks[i] == key and i != 1:
      occs.append(i)
  if len(occs) == 2:
    err.write(f"{loc:<10s} {utype = } {nt = } {key = !r} {occs = }")
    if occs[1] == nt - 3 and occs[0] == occs[1] - 6:
      err.write("***")
    err.write("\n")
  return
  # ----------------------------------------------------------------------
  
main()