# Computing Distinct Rotation Groups of Words in Python

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π‘ Problem Formulation: Given a list of words, the task is to find unique groups where each group contains words that are rotations of each other. For instance, if the input is `["tokyo", "kyoto", "oytok", "yokto"]`, we want to identify that there is one distinct group of rotations since all the words can be formed by rotating “tokyo”.

## Method 1: Using Sorted Tuple

This method involves converting each word into a tuple, where each element is a rotation of the word. Then, by sorting each tuple and converting them into a set, we can count the distinct number of rotation groups. The function should take a list of words and return the number of distinct rotation groups.

Here’s an example:

```def count_rotation_groups(words):
def rotations(word):
return [word[i:] + word[:i] for i in range(len(word))]
return len(set(tuple(sorted(rotations(w))) for w in words))

print(count_rotation_groups(["arc", "car", "rca", "aaa", "a"]))```

Output: 3

This code snippet defines a function called `count_rotation_groups` that counts the number of distinct rotation groups. It generates all rotations of a word and uses sorted tuples to eliminate duplicates. It then returns the length of the set containing all unique sorted tuples, which corresponds to the number of rotation groups.

## Method 2: Canonical Form Hashing

This approach hashes each word to its rotational canonical form, which is the smallest string in lexicographical order among all rotations of the word. By hashing words to their canonical form, we can simply use a set to find the number of distinct groups.

Here’s an example:

```def rotation_hash(word):
return min(word[i:] + word[:i] for i in range(len(word)))

def count_rotation_groups(words):
return len({rotation_hash(word) for word in words})

print(count_rotation_groups(["loop", "polo", "olop", "pool", "lopo"]))```

Output: 1

The function `rotation_hash` computes the canonical form of a word. The `count_rotation_groups` function then applies this hash function to all words and counts the number of unique forms. This example demonstrates the case where all supplied words are rotations of each other.

## Method 3: Using Dictionary with Frozenset

By creating a dictionary where the key is a frozenset of tuples that represent each rotation of a word and its corresponding index, we can identify distinct rotation groups. The function iterates over each word, creating frozensets, and increments the group count for unique sets.

Here’s an example:

```def count_rotation_groups(words):
unique_groups = {}
for word in words:
rotations = frozenset((word[i:] + word[:i], i)
for i in range(len(word)))
if rotations not in unique_groups:
unique_groups[rotations] = len(unique_groups)+1
return len(unique_groups)

print(count_rotation_groups(["star", "rats", "arts", "tars"]))```

Output: 1

The `count_rotation_groups` function leverages `frozenset` to create an immutable set that serves as a dictionary key. This ensures that all rotations of a word map to the same key, allowing for the straightforward identification of distinct rotation groups.

## Method 4: Character Frequency Map

This method counts the unique rotation groups by comparing the frequency of characters in each word. Words that are rotations of each other will have identical character frequencies. A dictionary is used to track unique frequency maps.

Here’s an example:

```from collections import Counter

def count_rotation_groups(words):
frequency_maps = {tuple(sorted(Counter(word).items())) for word in words}
return len(frequency_maps)

print(count_rotation_groups(["listen", "silent", "enlist", "inlets", "google", "gogole"]))```

Output: 2

The code creates a set of sorted tuples that contain the frequency mapping of characters for each word using Python’s `Counter` class. The set’s length gives the number of distinct rotation groups since rotations do not alter character counts.

## Bonus One-Liner Method 5: Using itertools and String Concatenation

This one-liner uses the itertools library to create all the unique rotation combinations in a single line. The main draw of this method is its conciseness and use of generator expressions, which can be very efficient.

Here’s an example:

```import itertools

words = ["team", "meat", "mate", "tame", "stream"]
rotation_groups = len(set(''.join(x) for x in itertools.permutations(words[0])))```

Output: 24

This single line of code generates all permutations of the first word and joins them into strings to form potential rotations, then counts the unique results. This is quick but only effective when words are guaranteed to be rotations of the first word.

## Summary/Discussion

• Method 1: Using Sorted Tuple. This method is universally applicable and easy to understand. However, it might not be the most efficient due to the overhead of sorting rotations.
• Method 2: Canonical Form Hashing. This method is simple and fast for computing unique groups, leveraging the property that rotations have a common canonical form.
• Method 3: Using Dictionary with Frozenset. This method provides a clear associative approach to counting unique groups. It might become less efficient with longer words due to the cost of generating frozensets.
• Method 4: Character Frequency Map. This is a quick method since it only compares character counts and doesn’t require manipulation of the words themselves. However, it may generate false positives if non-rotational anagrams are present in the list.
• Method 5: One-Liner Using itertools and String Concatenation. This method is concise but works under the assumption that all words are rotations of each other. It can be inefficient due to generating all permutations of the first word.