A Problem Squared Episode 103 - Caesar Cypher Cycles
Context.
On the episode, a listener poses the problem:
How many English words are also words after they’ve been put through a caesar cypher?
Is there a pair of words which are caesar cyphers of each other, where there’s a semantic link between the words? Or common theme to them, or some other interesting relationship?
I thought the episode was great (especially where there’s terrible Python code, so relatable ❤️) - though to my ear, the bit about words cyphering into each other sounded more like finding cycles (so something where Word A -> Word B -> Word C -> Word A), so I had a go at answering that.
Approach
While I was listening along, I was thinking about how I’d code it up, and came up with a way to calculate this fairly simply. I.e. words can be grouped together by how each letter of the word changes from the preceding one.
It’s probably easier to give an example, from the episode ROT4(bec) = fig.
The code below handles this a little differently, but explaining it this way round is a little simpler.
b to e is +3e to c is +24 (looping around z to a)
So the “signature” vector for bec is [3, 24].
Similarly:
So fig is also [3, 24].
However, there’s no way to turn cypher cat into bec:
c to a is 24a to t is 19
So cat is [24, 19].
I haven’t proved it rigorously, but it makes intuitive sense if you think about building the word letter by letter, the two inputs to making a word are the starting letter, and then how to move from letter to letter, until the end of the word.
This approach means that we can group all the words up into groups that can be cyphered into each other with a single pass through the dictionary, by working out this vector for every word.
Then, once we’ve grouped up, we can examine the contents of each group for cycles.
Answers
As a caveat, the words list I’m using is whatever the default on Ubuntu is (/usr/share/dict/words), I’m not quite sure where it’s actually from originally - it seems pretty good, but your answers may vary, and I’m certainly missing some of the words in Matt’s dictionary.
I’ve also excluded any words shorter than 3 letters, as they don’t seem to be that interesting.
cat /usr/share/dict/words | python3 caesar.py
- How many English words are also words after they’ve been put through a caesar cypher?
- Is there a pair of words which are caesar cyphers of each other
- I’ve found 56 pairs of words that cycle from one to the other, and back using
ROT13
ROT13 is significant here as doing it twice takes you back to the original input
- My favourites here are green terra, and Typhoid Mary’s lesser known rival, Ebola Robyn
- I don’t think there are any cycles longer than 2 words, as the prime factors of the number of letters in the alphabet (
26) are 2, and 13
- The only cycle lengths possible are divisors of
26, if you want to get back to where you started (so 1, 2, 13, 26).
- This means the next shortest cycle will be 13 items long, and the largest actual group of words with the same signature is 6 words (
aol,bpm,esp,max,rfc,the being the closest to not gibberish here)
The Code
import csv
from pathlib import Path
import sys
from typing import Generator, List
# Vector generation
def distance(c1: chr, c2: chr) -> int:
"Rotation distance from here to there."
return (ord(c1) - ord(c2)) % 26
def diffs(word: str) -> str:
"Work out a key to group words on for cypher-equality."
char_diffs = [distance(c, word[i + 1]) for i, c in enumerate(word[:-1].lower())]
return ";".join(str(c) for c in char_diffs)
# Text processing
def clean_word(word: str) -> str:
"Tidy up a word so we don't get messed up on capital letters etc."
return word.strip().lower()
def words_from_stdin(min_word_length=3) -> Generator[str, None, None]:
"Read data piped in, normalise it, and filter down to actual words."
for word in map(clean_word, sys.stdin):
if not word.isalpha():
continue
if len(word) < min_word_length:
continue
yield word
# Answering problems
def find_cycles(word_list: List[str]) -> Generator[List[str], None, None]:
"Spit out any cycles in a word list."
# While this isn't generally the case, realistically, we're only looking for ROT-13.
# Otherwise, the next shortest cycle will be 13 items long, and it gets worse from there.
# Once we ignore words 2 characters or less, the biggest group is 6 items.
for i, word in enumerate(word_list):
for other_word in word_list[i + 1 :]:
if word == other_word:
continue
if distance(word[0], other_word[0]) == 13:
yield [word, other_word]
def answer_question(word_list: List[str], report_file: Path):
"Actually answer the question."
unique_words = set()
for words in word_list:
for word in words:
unique_words.add(word)
with open(report_file, mode="w") as f:
w = csv.writer(f)
w.writerow(["How many words?"])
w.writerow([len(unique_words)])
def output_cycles(word_list: List[str], report_file: Path):
"Write out lists of words where cyphering loops."
# Sort by word length
grouped = sorted(word_list, key=lambda x: -1 * len(x[0]))
with open(report_file, mode="w") as f:
w = csv.writer(f)
for g in grouped:
for w1, w2 in find_cycles(g):
w.writerow([w1, w2])
def write_all(word_list: List[str], report_file: Path):
"Write out all of the groups of cypherable words"
# Sort by group length
grouped = sorted(word_list, key=lambda x: -1000 * len(x) + -1 * len(x[0]))
with open(report_file, mode="w") as f:
w = csv.writer(f)
for g in grouped:
w.writerow(g)
if __name__ == "__main__":
print("Starting")
grouped = {}
for i, word in enumerate(words_from_stdin(min_word_length=3)):
cypher_id = diffs(word)
if cypher_id not in grouped:
grouped[cypher_id] = set()
grouped[cypher_id].add(word)
grouped = [sorted(v) for v in grouped.values() if len(v) > 1]
print(f"Checked {i} words, and found {len(grouped)} clusters of words")
report_dir = Path()
print(f"Writing results out to {report_dir.absolute()}")
answer_question(grouped, Path(report_dir, "answers.csv"))
write_all(grouped, Path(report_dir, "cypher-groups.csv"))
output_cycles(grouped, Path(report_dir, "cycles.csv"))
print("Done")
