latin-rectangles 0.1.0

Count the number of one-row extensions of Latin rectangles.

Latin Rectangles Extension Counter

A high-performance Python library for counting the number of ways to extend a 2×n Latin rectangle to a 3×n Latin rectangle using rook polynomial methods and cycle decomposition theory.

Overview

A Latin rectangle is an r×n array filled with n different symbols such that each symbol occurs exactly once in each row and at most once in each column.

Extension Problem

Given a 2×n Latin rectangle:

1  2  3  4  5  6  7  8
p[1]  p[2]  p[3]  p[4]  p[5]  p[6]  p[7]  p[8]

where p is a derangement, the problem is to count how many valid third rows can be added such that the resulting 3×n rectangle remains a Latin rectangle. This library provides an efficient algorithm for computing Latin rectangle extensions.

Key Features

• High Performance: Approximate O(n^2) time complexity, tested up to n=800.
• Memory Efficient: Approximate O(n^1.36) memory complexity
• Mathematically Rigorous: Based on rook polynomial theory and cycle decomposition
• Easy to Use: Simple command-line interface and Python API
• Well Tested: Comprehensive test suite with complexity analysis

Installation

git clone https://github.com/ionmich/latin-rectangles.git
cd latin-rectangles

Quick Start

Command Line (CLI) Usage

Generate random derangement:

> uv run python -m latin_rectangles --n 42
🎲 Generated Random Derangement for n=42
📊 Cycle structure: [2, 2, 4, 8, 26]
🔢 Number of extensions: 185,566,788,772,996,286,199,647,931,971,186,844,003,087,641,029,824

Use specific cycle structure:

> uv run python -m latin_rectangles --c "2,2,4"
⚙️  Specific Cycle Structure for n=8
📊 Cycle structure: [2, 2, 4]
🔢 Number of extensions: 4,744

Enumerate all possible cycle structures:

> uv run python -m latin_rectangles --n 8 --all
🔍 All Cycle Structures for n=8
📊 Found 7 possible structures with non-zero extensions:

 1. [2, 2, 2, 2] → 4,752 extensions
 2. [2, 2, 4] → 4,744 extensions
 3. [2, 3, 3] → 4,740 extensions
 4. [2, 6] → 4,740 extensions
 5. [4, 4] → 4,740 extensions
 6. [3, 5] → 4,738 extensions
 7. [8] → 4,738 extensions

Get help

uv run latin-rectangles --help

Python Library Usage

from latin_rectangles import count_extensions, count_random_extensions
from latin_rectangles import generate_random_derangement

# Method 1: One-liner for random derangement
extensions = count_random_extensions(n=12)
print(f"Extensions: {extensions:,}")

# Method 2: Step-by-step with custom derangement
derangement = generate_random_derangement(n=10)
extensions = count_extensions(derangement)
print(f"Derangement {derangement[1:]} has {extensions:,} extensions")

# Method 3: With predefined derangement (1-indexed with dummy 0)
p = [0, 2, 3, 4, 5, 6, 7, 8, 1]  # 8-cycle for n=8
extensions = count_extensions(p)
print(f"8-cycle has {extensions:,} extensions")  # Output: 4,738

Algorithm Details

Mathematical Foundation

The algorithm leverages rook polynomial theory to solve the Latin rectangle extension problem:

1. Input: A derangement (permutation with no fixed points) representing the second row
2. Cycle Decomposition: Decompose the derangement into disjoint cycles
3. Rook Polynomials: Compute rook polynomial for each cycle structure
4. Polynomial Multiplication: Combine rook polynomials to get the final count

API Reference

Core Functions

count_extensions(permutation: list[int]) -> int

Counts the number of extensions for a given derangement.

Parameters:

• permutation: 1-indexed list representing a derangement (p[0] is dummy value)

Returns: Integer number of possible third rows

Raises: ValueError if input is not a derangement

count_random_extensions(n: int) -> int

Convenience function that generates a random derangement and counts its extensions.

Parameters:

• n: Size of the derangement (must be > 1)

Returns: Number of extensions for the randomly generated derangement

generate_random_derangement(n: int) -> list[int]

Generates a random derangement of size n.

Parameters:

• n: Size of the derangement

Returns: 1-indexed list representing the derangement

find_cycle_decomposition(permutation: list[int]) -> list[list[int]]

Finds the cycle decomposition of a permutation.

Parameters:

• permutation: 1-indexed permutation

Returns: List of cycles (each cycle is a list of indices)

Examples

Basic Usage Examples

from latin_rectangles import count_extensions

# Example 1: Single 8-cycle
p_8_cycle = [0, 2, 3, 4, 5, 6, 7, 8, 1]
print(f"8-cycle: {count_extensions(p_8_cycle):,} extensions")
# Output: 8-cycle: 4,738 extensions

# Example 2: Two 4-cycles  
p_4_4 = [0, 2, 3, 4, 1, 6, 7, 8, 5]
print(f"4,4-cycles: {count_extensions(p_4_4):,} extensions")
# Output: 4,4-cycles: 4,740 extensions

# Example 3: Four 2-cycles
p_2_2_2_2 = [0, 2, 1, 4, 3, 6, 5, 8, 7]
print(f"2,2,2,2-cycles: {count_extensions(p_2_2_2_2):,} extensions")
# Output: 2,2,2,2-cycles: 4,752 extensions

Advanced Usage

from latin_rectangles import generate_random_derangement, find_cycle_decomposition, count_extensions

# Generate and analyze a random derangement
n = 15
derangement = generate_random_derangement(n)
cycles = find_cycle_decomposition(derangement)
cycle_lengths = sorted([len(c) for c in cycles])
extensions = count_extensions(derangement)

print(f"n={n}")
print(f"Derangement: {derangement[1:]}")
print(f"Cycle structure: {cycle_lengths}")
print(f"Extensions: {extensions:,}")

Batch Processing

from latin_rectangles import count_random_extensions

# Process multiple sizes
results = []
for n in range(5, 21):
    extensions = count_random_extensions(n)
    results.append((n, extensions))
    print(f"n={n:2d}: {extensions:,} extensions")

# Find the size with the most extensions in this batch
max_n, max_extensions = max(results, key=lambda x: x[1])
print(f"Maximum: n={max_n} with {max_extensions:,} extensions")

Development

Running Tests

# Run the test suite
uv run pytest

# Run with coverage
uv run pytest --cov=latin_rectangles

# Run specific test
uv run pytest tests/test_main.py -v

Code Quality

# Type checking
uv run mypy src/

# Linting
uv run ruff check src/

# Formatting
uv run ruff format src/

Benchmarking

# Run performance benchmarks
uv run python benchmark.py

# Analyze complexity
uv run python complexity_analysis.py

Contributing

Contributions are welcome! Please see DEVELOPMENT.md for development guidelines.

1. Fork the repository
2. Create a feature branch
3. Add tests for new functionality
4. Ensure all tests pass
5. Submit a pull request

License

This project is licensed under the MIT License - see the LICENSE file for details.

Citation

If you use this library in your research, please cite:

@software{latin_rectangles,
  title={Latin Rectangles Extension Counter},
  author={Ioannis Michaloliakos},
  year={2025},
  url={https://github.com/ionmich/latin-rectangles}
}