couplr (version 1.0.10)

example_costs: Example cost matrices for assignment problems

Description

Small example datasets for demonstrating couplr functionality across different assignment problem types: square, rectangular, sparse, and binary.

Usage

example_costs

Format

A list containing four example cost matrices:

simple_3x3

A 3x3 cost matrix with costs ranging from 2-7. Optimal assignment: row 1 -> col 2 (cost 2), row 2 -> col 1 (cost 3), row 3 -> col 3 (cost 4). Total optimal cost: 9.

rectangular_3x5

A 3x5 rectangular cost matrix demonstrating assignment when rows < columns. Each of 3 rows is assigned to one of 5 columns; 2 columns remain unassigned. Costs range 1-6.

sparse_with_na

A 3x3 matrix with NA values indicating forbidden assignments. Use this to test algorithms' handling of constraints. Position (1,3), (2,2), and (3,1) are forbidden.

binary_costs

A 3x3 matrix with binary (0/1) costs, suitable for testing the HK01 algorithm. Diagonal entries are 0 (preferred), off-diagonal entries are 1 (penalty).

Details

These matrices are designed to test different aspects of LAP solvers:

simple_3x3: Basic functionality test. Any correct solver should find total cost = 9.

rectangular_3x5: Tests handling of non-square problems. The optimal solution assigns all 3 rows with minimum total cost.

sparse_with_na: Tests constraint handling. Algorithms must avoid NA positions while finding an optimal assignment among valid entries.

binary_costs: Tests specialized binary cost algorithms. The optimal assignment uses all diagonal entries (total cost = 0).

See Also

lap_solve, example_df

Examples

# Simple 3x3 assignment
result <- lap_solve(example_costs$simple_3x3)
print(result)
# Optimal: sources 1,2,3 -> targets 2,1,3 with cost 9

# Rectangular problem (3 sources, 5 targets)
result <- lap_solve(example_costs$rectangular_3x5)
print(result)
# All 3 sources assigned; 2 targets unassigned

# Sparse problem with forbidden assignments
result <- lap_solve(example_costs$sparse_with_na)
print(result)
# Avoids NA positions

# Binary costs - test HK01 algorithm
result <- lap_solve(example_costs$binary_costs, method = "hk01")
print(result)
# Finds diagonal assignment (cost = 0)