From f4101c76e7e185011fc889c85e0e8a72ecdcff87 Mon Sep 17 00:00:00 2001
From: Leonora Tindall <nora@nora.codes>
Date: Thu, 6 Mar 2025 17:01:57 -0600
Subject: [PATCH] Comments

---
 gen_rhymes.py | 15 ++++++++++++++-
 1 file changed, 14 insertions(+), 1 deletion(-)

diff --git a/gen_rhymes.py b/gen_rhymes.py
index f3b5aef..ce9890d 100644
--- a/gen_rhymes.py
+++ b/gen_rhymes.py
@@ -4,7 +4,7 @@ import sys as trans
 from pprint import pp
 
 # We're taking advantage of the fact that toki pona pronounciation is very
-# orderly. We can determine rhymes entirely lexicographically.
+# orderly. We can determine rhymes entirely lexically.
 VOWELS = ["a", "e", "i", "o", "u"]
 
 words = []
@@ -12,6 +12,8 @@ rhyme_classes = {}
 rhymes = {}
 
 def matching_substring_r(str1, str2):
+    """Return the longest matching substring of two strings which touches the
+    end of both."""
     shortest_len = min(len(str1), len(str2))
     if shortest_len == 0:
         return ""
@@ -26,9 +28,12 @@ def matching_substring_r(str1, str2):
         current_index -= 1
     return current_match
 
+# Need a dict file to work from
 if not len(trans.argv) == 2:
     print("Please provide a single dictionary file to work from.")
+    trans.exit(1)
 
+# Read words, one per line
 with open(trans.argv[1]) as dictionary:
     for line in dictionary.readlines():
         word = line.rstrip("\n")
@@ -40,21 +45,29 @@ for word in words:
     rhymes[word] = []
 
     for possible_rhyme in words:
+        # lili doesn't rhyme with lili
         if word == possible_rhyme:
             continue
+
+        # find the rhyme class represented by this relationship,
+        # which is a nonempty substring touching the end of the words
+        # which starts with a vowel
         rhyme_class = matching_substring_r(word, possible_rhyme)
         while len(rhyme_class) > 1 and\
                 rhyme_class[0] not in VOWELS:
                     rhyme_class = rhyme_class[1:]
 
+        # i don't care that lili rhymes with monsi, that's too much
         if not len(rhyme_class) > 1:
             continue
 
+        # we know that these two rhyme
         rhymes[word].append(possible_rhyme)
 
         if rhyme_class not in rhyme_classes:
             rhyme_classes[rhyme_class] = []
 
+        # we add only this word; the other will be added in its own outer loop
         if word not in rhyme_classes[rhyme_class]:
             rhyme_classes[rhyme_class].append(word)