# Troubleshooting Guide

#### Gilles Colling

#### 2026-05-18

## Overview

This vignette provides solutions to common issues encountered when using couplr. Each section describes a problem, explains why it occurs, and provides step-by-step solutions.

Quick reference:

Issue | Jump to Section
--- | ---
No valid pairs found | Infeasible Problems
Poor balance after matching | Poor Balance
Matching is too slow | Performance Issues
Memory errors | Memory Errors
Different results with different methods | Non-Determinism
Unexpected NA in results | Missing Values
C++ compilation errors | Installation Issues

## Infeasible Problems

### Symptom

    # This fails: all assignments have Inf cost
    cost <- matrix(c(1, Inf, Inf, Inf, Inf, Inf, Inf, 2, 3), nrow = 3, byrow = TRUE)
    result <- lap_solve(cost)
    #> Error:
    #> ! Infeasible given forbidden edges

Error message: “No feasible assignment found” or total cost is Inf.

### Cause

A feasible assignment requires that every row can be assigned to at least one column with finite cost. When rows have all-Inf or all-NA entries, no valid assignment exists.

### Diagnosis

    # Check for infeasible rows
    check_feasibility <- function(cost_matrix) {
      finite_per_row <- rowSums(is.finite(cost_matrix))
      infeasible_rows <- which(finite_per_row == 0)

      if (length(infeasible_rows) > 0) {
        cat("Infeasible rows (no finite costs):", infeasible_rows, "\n")
        return(FALSE)
      }

      finite_per_col <- colSums(is.finite(cost_matrix))
      infeasible_cols <- which(finite_per_col == 0)

      if (length(infeasible_cols) > 0) {
        cat("Infeasible columns (no finite costs):", infeasible_cols, "\n")
        return(FALSE)
      }

      cat("Problem appears feasible\n")
      return(TRUE)
    }

    # Check the problematic matrix
    cost <- matrix(c(1, Inf, Inf, Inf, Inf, Inf, Inf, 2, 3), nrow = 3, byrow = TRUE)
    check_feasibility(cost)
    #> Infeasible rows (no finite costs): 2
    #> [1] FALSE

### Solutions

1. Remove infeasible rows/columns:

    # Remove rows with no valid assignments
    cost <- matrix(c(1, Inf, Inf, Inf, Inf, Inf, Inf, 2, 3), nrow = 3, byrow = TRUE)
    valid_rows <- rowSums(is.finite(cost)) > 0
    cost_valid <- cost[valid_rows, , drop = FALSE]

    if (nrow(cost_valid) > 0) {
      result <- lap_solve(cost_valid)
      cat("Matched", nrow(result), "of", nrow(cost), "rows\n")
    }
    #> Matched 2 of 3 rows

2. Add fallback costs:

    # Replace Inf with high (but finite) penalty
    cost_with_fallback <- cost
    cost_with_fallback[!is.finite(cost_with_fallback)] <- 1e6  # Large penalty

    result <- lap_solve(cost_with_fallback)
    print(result)
    #> Assignment Result
    #> =================
    #>
    #> # A tibble: 3 × 3
    #>   source target    cost
    #>    <int>  <int>   <dbl>
    #> 1      1      1       1
    #> 2      2      3 1000000
    #> 3      3      2       2
    #>
    #> Total cost: 1000003
    #> Method: bruteforce

3. For matching: check covariate overlap:

    # Simulate poor overlap scenario
    set.seed(123)
    left <- tibble(id = 1:50, age = rnorm(50, mean = 25, sd = 3))
    right <- tibble(id = 1:50, age = rnorm(50, mean = 55, sd = 3))

    # Visualize overlap
    library(ggplot2)
    combined <- bind_rows(
      left |> mutate(group = "Left"),
      right |> mutate(group = "Right")
    )

    ggplot(combined, aes(x = age, fill = group)) +
      geom_density(alpha = 0.5) +
      labs(title = "Poor Covariate Overlap",
           subtitle = "No overlap means no good matches possible") +
      theme_minimal() +
      theme(plot.background = element_rect(fill = "transparent", color = NA),
            panel.background = element_rect(fill = "transparent", color = NA))

## Poor Balance Despite Matching

### Symptom

After matching, `balance_diagnostics()` shows |std_diff| > 0.25 for some variables.

    # Example of poor balance
    set.seed(456)
    left <- tibble(
      id = 1:100,
      age = rnorm(100, 30, 5),
      income = rnorm(100, 80000, 20000)  # Very different from right
    )
    right <- tibble(
      id = 1:100,
      age = rnorm(100, 32, 5),
      income = rnorm(100, 40000, 10000)  # Very different income
    )

    result <- match_couples(left, right, vars = c("age", "income"), auto_scale = TRUE,
                            method = "hungarian")
    #> Auto-selected scaling method: standardize
    balance <- balance_diagnostics(result, left, right, vars = c("age", "income"))
    print(balance)
    #>
    #> Balance Diagnostics for Matched Pairs
    #> ======================================
    #>
    #> Matching Summary:
    #>   Method: lap
    #>   Matched pairs: 100
    #>   Unmatched left: 0 (of 100)
    #>   Unmatched right: 0 (of 100)
    #>
    #> Variable-level Balance:
    #> # A tibble: 2 × 7
    #>   Variable `Mean Left` `Mean Right` `Mean Diff` `Std Diff` `Var Ratio` `KS Stat`
    #>   <chr>          <dbl>        <dbl>       <dbl>      <dbl>       <dbl>     <dbl>
    #> 1 age             30.6         32.5       -1.89     -0.386        1.05      0.22
    #> 2 income       77666.       41353.     36313.        2.40         2.15      0.81
    #>
    #> Overall Balance:
    #>   Mean |Std Diff|: 1.391 (Poor)
    #>   Max |Std Diff|: 2.397
    #>   Vars with |Std Diff| > 0.25: 100.0%
    #>
    #> Balance Interpretation:
    #>   |Std Diff| < 0.10: Excellent balance
    #>   |Std Diff| 0.10-0.25: Good balance
    #>   |Std Diff| 0.25-0.50: Acceptable balance
    #>   |Std Diff| > 0.50: Poor balance

### Causes

1. Weak overlap: Groups are too different on key variables

2. Missing confounders: Important variables not included in matching

3. Caliper too loose: Accepting poor matches

4. Wrong scaling: Variables not properly weighted

### Solutions

1. Add more matching variables:

    # Include additional relevant variables
    result <- match_couples(
      left, right,
      vars = c("age", "income", "education", "region"),  # More variables
      auto_scale = TRUE
    )

2. Tighten caliper (fewer but better matches):

    result_strict <- match_couples(
      left, right,
      vars = c("age", "income"),
      max_distance = 0.3,  # Stricter caliper
      auto_scale = TRUE,
      method = "hungarian"
    )
    #> Auto-selected scaling method: standardize

    cat("Original matches:", result$info$n_matched, "\n")
    #> Original matches: 100
    cat("With caliper:", result_strict$info$n_matched, "\n")
    #> With caliper: 23

    balance_strict <- balance_diagnostics(result_strict, left, right, vars = c("age", "income"))
    print(balance_strict)
    #>
    #> Balance Diagnostics for Matched Pairs
    #> ======================================
    #>
    #> Matching Summary:
    #>   Method: lap
    #>   Matched pairs: 23
    #>   Unmatched left: 0 (of 100)
    #>   Unmatched right: 77 (of 100)
    #>
    #> Variable-level Balance:
    #> # A tibble: 2 × 7
    #>   Variable `Mean Left` `Mean Right` `Mean Diff` `Std Diff` `Var Ratio` `KS Stat`
    #>   <chr>          <dbl>        <dbl>       <dbl>      <dbl>       <dbl>     <dbl>
    #> 1 age             31.5         31.5      -0.008     -0.002        1.01     0.13
    #> 2 income       53958.       51791.     2167.         0.296        1.07     0.217
    #>
    #> Overall Balance:
    #>   Mean |Std Diff|: 0.149 (Good)
    #>   Max |Std Diff|: 0.296
    #>   Vars with |Std Diff| > 0.25: 50.0%
    #>
    #> Balance Interpretation:
    #>   |Std Diff| < 0.10: Excellent balance
    #>   |Std Diff| 0.10-0.25: Good balance
    #>   |Std Diff| 0.25-0.50: Acceptable balance
    #>   |Std Diff| > 0.50: Poor balance

3. Use blocking on problematic variable:

    # Block on income tertiles to ensure exact balance
    left$income_cat <- cut(left$income, breaks = 3, labels = c("low", "mid", "high"))
    right$income_cat <- cut(right$income, breaks = 3, labels = c("low", "mid", "high"))

    blocks <- matchmaker(left, right, block_type = "group", block_by = "income_cat")
    result_blocked <- match_couples(
      blocks$left, blocks$right,
      vars = c("age"),  # Match on age within income blocks
      block_id = "block_id"
    )

4. Try different scaling:

    # Compare scaling methods
    for (scale_method in c("robust", "standardize", "range")) {
      res <- match_couples(left, right, vars = c("age", "income"),
                           auto_scale = TRUE, scale = scale_method,
                           method = "hungarian")
      bal <- balance_diagnostics(res, left, right, vars = c("age", "income"))
      cat(scale_method, "- max |std_diff|:",
          round(bal$overall$max_abs_std_diff, 3), "\n")
    }
    #> robust - max |std_diff|: 2.397
    #> standardize - max |std_diff|: 2.397
    #> range - max |std_diff|: 2.397

## Performance Issues

### Symptom

`match_couples()` or `lap_solve()` takes too long or doesn’t complete.

### Cause

Optimal matching has \(O(n^3)\) complexity. For n = 5,000, this means ~125 billion operations.

### Diagnosis

    # Estimate runtime
    estimate_runtime <- function(n, seconds_per_billion = 1) {
      ops <- n^3
      time_sec <- ops / 1e9 * seconds_per_billion

      if (time_sec < 60) {
        sprintf("%.1f seconds", time_sec)
      } else if (time_sec < 3600) {
        sprintf("%.1f minutes", time_sec / 60)
      } else {
        sprintf("%.1f hours", time_sec / 3600)
      }
    }

    cat("Estimated runtime for optimal matching:\n")
    #> Estimated runtime for optimal matching:
    for (n in c(100, 500, 1000, 3000, 5000, 10000)) {
      cat(sprintf("  n = %5d: %s\n", n, estimate_runtime(n)))
    }
    #>   n =   100: 0.0 seconds
    #>   n =   500: 0.1 seconds
    #>   n =  1000: 1.0 seconds
    #>   n =  3000: 27.0 seconds
    #>   n =  5000: 2.1 minutes
    #>   n = 10000: 16.7 minutes

### Solutions

1. Use greedy matching for large problems:

    set.seed(789)
    n <- 500
    large_left <- tibble(id = 1:n, x1 = rnorm(n), x2 = rnorm(n))
    large_right <- tibble(id = 1:n, x1 = rnorm(n), x2 = rnorm(n))

    # Greedy is much faster
    time_greedy <- system.time({
      result_greedy <- greedy_couples(
        large_left, large_right,
        vars = c("x1", "x2"),
        strategy = "row_best"
      )
    })

    cat("Greedy matching (n=500):", round(time_greedy["elapsed"], 2), "seconds\n")
    #> Greedy matching (n=500): 0.23 seconds
    cat("Quality (mean distance):", round(mean(result_greedy$pairs$distance), 4), "\n")
    #> Quality (mean distance): 0.2886

2. Use blocking to divide the problem:

    # Create clusters to match within
    blocks <- matchmaker(
      large_left, large_right,
      block_type = "cluster",
      block_vars = c("x1", "x2"),
      n_blocks = 10  # 10 blocks of ~200 each
    )

    # Match within blocks (10 x O(200^3) << O(2000^3))
    result_blocked <- match_couples(
      blocks$left, blocks$right,
      vars = c("x1", "x2"),
      block_id = "block_id"
    )

3. Choose a faster algorithm:

    # For n > 1000, auction algorithm often faster
    result <- match_couples(
      large_left, large_right,
      vars = c("x1", "x2"),
      method = "auction"
    )

    # For sparse problems (many forbidden pairs)
    result <- match_couples(
      left, right,
      vars = vars,
      max_distance = 0.5,  # Creates sparsity
      method = "sap"       # Sparse algorithm
    )

4. Pre-compute and cache distances:

    # Compute once, reuse multiple times
    dist_cache <- compute_distances(
      large_left, large_right,
      vars = c("x1", "x2"),
      scale = "robust"
    )

    # Fast: reuse cached distances
    result1 <- match_couples(dist_cache, max_distance = 0.3, method = "hungarian")
    result2 <- match_couples(dist_cache, max_distance = 0.5, method = "hungarian")
    result3 <- match_couples(dist_cache, max_distance = 1.0, method = "hungarian")

## Memory Errors

### Symptom

R crashes or shows: “Error: cannot allocate vector of size X GB”

### Cause

A full distance matrix for nxn requires 8n² bytes:

    # Memory requirements
    memory_needed <- function(n) {
      bytes <- 8 * n^2
      if (bytes < 1e6) {
        sprintf("%.1f KB", bytes / 1e3)
      } else if (bytes < 1e9) {
        sprintf("%.1f MB", bytes / 1e6)
      } else {
        sprintf("%.1f GB", bytes / 1e9)
      }
    }

    cat("Memory for full distance matrix:\n")
    #> Memory for full distance matrix:
    for (n in c(1000, 5000, 10000, 20000, 50000)) {
      cat(sprintf("  n = %5d: %s\n", n, memory_needed(n)))
    }
    #>   n =  1000: 8.0 MB
    #>   n =  5000: 200.0 MB
    #>   n = 10000: 800.0 MB
    #>   n = 20000: 3.2 GB
    #>   n = 50000: 20.0 GB

### Solutions

1. Use greedy matching (avoids full matrix):

    # Greedy computes distances on-the-fly
    result <- greedy_couples(
      left, right,
      vars = covariates,
      strategy = "row_best"  # Most memory-efficient
    )

2. Use blocking to create smaller subproblems:

    # Each block is much smaller
    blocks <- matchmaker(left, right, block_type = "cluster", n_blocks = 20)
    result <- match_couples(blocks$left, blocks$right, vars = vars, block_id = "block_id")

3. Use caliper to create sparse matrix:

    # Caliper excludes distant pairs (sparse representation)
    result <- match_couples(
      left, right,
      vars = covariates,
      max_distance = 0.5,
      method = "sap"  # Sparse-optimized algorithm
    )

4. Increase R’s memory limit (Windows):

    # Increase to 16 GB (if available)
    memory.limit(size = 16000)

## Different Results with Different Methods

### Symptom

Different algorithms return different assignments:

    cost <- matrix(c(1, 2, 2, 2, 1, 2, 2, 2, 1), nrow = 3, byrow = TRUE)

    result_jv <- lap_solve(cost, method = "jv")
    result_hungarian <- lap_solve(cost, method = "hungarian")

    cat("JV assignment:       ", result_jv$target, "\n")
    #> JV assignment:        1 2 3
    cat("Hungarian assignment:", result_hungarian$target, "\n")
    #> Hungarian assignment: 1 2 3
    cat("JV total cost:       ", get_total_cost(result_jv), "\n")
    #> JV total cost:        3
    cat("Hungarian total cost:", get_total_cost(result_hungarian), "\n")
    #> Hungarian total cost: 3

### Cause

When multiple optimal solutions exist (tied costs), different algorithms may find different ones. The total cost should be the same; if not, report a bug.

### Diagnosis

    # Check for ties
    check_ties <- function(cost_matrix) {
      n <- nrow(cost_matrix)
      # Check if diagonal dominates (trivial ties)
      diag_costs <- diag(cost_matrix)
      if (length(unique(diag_costs)) < n) {
        cat("Tied costs on diagonal - multiple optima likely\n")
      }

      # Check cost uniqueness
      unique_costs <- length(unique(as.vector(cost_matrix)))
      total_entries <- length(cost_matrix)
      if (unique_costs < total_entries * 0.5) {
        cat("Many repeated costs - ties possible\n")
      }
    }

    check_ties(cost)
    #> Tied costs on diagonal - multiple optima likely
    #> Many repeated costs - ties possible

### Solutions

1. Verify total costs match:

    # Total cost should be identical
    stopifnot(get_total_cost(result_jv) == get_total_cost(result_hungarian))
    cat("Both methods found optimal solutions (same total cost)\n")
    #> Both methods found optimal solutions (same total cost)

2. Use Hungarian for deterministic tie-breaking:

    # Hungarian has consistent tie-breaking
    result <- lap_solve(cost, method = "hungarian")

3. Add small noise to break ties:

    set.seed(42)
    cost_perturbed <- cost + matrix(rnorm(9, 0, 1e-10), 3, 3)
    result <- lap_solve(cost_perturbed)

## Missing Values in Data

### Symptom

Matching fails or produces unexpected results due to NA values.

    left <- tibble(id = 1:5, age = c(25, 30, NA, 35, 40))
    right <- tibble(id = 1:5, age = c(28, 32, 33, 36, 42))

    # This may fail or give unexpected results
    result <- match_couples(left, right, vars = "age")
    #> Error:
    #> ! Missing values (NA) not allowed in matching variables

### Cause

couplr requires complete cases for distance computation. NA values in matching variables cause issues.

### Solutions

1. Remove rows with NA (before matching):

    left_clean <- left |> filter(!is.na(age))
    right_clean <- right |> filter(!is.na(age))

    result <- match_couples(left_clean, right_clean, vars = "age")
    cat("Matched", result$info$n_matched, "pairs (excluded 1 left unit with NA)\n")
    #> Matched 4 pairs (excluded 1 left unit with NA)

2. Impute missing values:

    # Simple mean imputation
    left_imputed <- left |>
      mutate(age = if_else(is.na(age), mean(age, na.rm = TRUE), age))

    result <- match_couples(left_imputed, right, vars = "age")
    cat("Matched", result$info$n_matched, "pairs (imputed 1 NA with mean)\n")
    #> Matched 5 pairs (imputed 1 NA with mean)

3. Use preprocessing to diagnose:

    health <- preprocess_matching_vars(
      left, right,
      vars = "age"
    )
    print(health)
    #> Preprocessing Result
    #> ====================
    #>
    #> Variables: 1
    #> Scaling method: none

## Installation Issues

### C++ Compilation Errors

#### Symptom

    Error in .Call("_couplr_assignment_impl"): object not found

or

    Error: package 'couplr' was built under R version X.X

#### Cause

C++ code not compiled properly, or object files are stale.

#### Solutions

1. Clean and reinstall:

    # In R:
    remove.packages("couplr")
    install.packages("couplr")

    # Or from GitHub:
    devtools::install_github("gcol33/couplr", force = TRUE)

2. Clean compiled files (development):

    # PowerShell
    Remove-Item src\*.o, src\*.dll -Force

    # Bash
    rm src/*.o src/*.so

Then rebuild:

    devtools::clean_dll()
    devtools::load_all()

3. Check Rtools/compiler (Windows):

    # Check if Rtools is properly configured
    Sys.which("make")
    #>                               make
    #> "c:\\rtools45\\usr\\bin\\make.exe"
    Sys.which("g++")
    #>                                          g++
    #> "c:\\rtools45\\X86_64~1.POS\\bin\\G__~1.EXE"

Make sure Rtools is installed and PATH is configured correctly.

## Common Error Messages

### “method_used attribute is NULL”

Cause: Result object wasn’t created properly.

Solution: Use `lap_solve()` instead of `assignment()` for the tidy interface.

### “Cost matrix contains non-finite values”

Cause: NA or NaN values in cost matrix (not Inf).

Solution: Replace NA with Inf for forbidden assignments, or remove rows/cols.

    cost <- matrix(c(1, NA, 3, 4), 2, 2)
    cost[is.na(cost)] <- Inf  # Mark as forbidden
    result <- lap_solve(cost)

### “All rows assigned to Inf columns”

Cause: Caliper is too strict; all potential matches exceed threshold.

Solution: Increase `max_distance` or check for overlap issues.

### “subscript out of bounds”

Cause: Empty left or right data frame, or ID mismatch.

Solution: Check data dimensions and ID column consistency.

    # Always verify data before matching
    stopifnot(nrow(left) > 0, nrow(right) > 0)

## Getting Help

If you encounter an issue not covered here:

1. Check function documentation: `?match_couples`, `?lap_solve`

2. Search GitHub issues: github.com/gcol33/couplr/issues

3. Create a minimal reproducible example:

    # Minimal example template
    library(couplr)

    # Minimal data that reproduces the issue
    set.seed(123)
    left <- tibble(id = 1:10, x = rnorm(10))
    right <- tibble(id = 1:10, x = rnorm(10))

    # Code that causes the error
    result <- match_couples(left, right, vars = "x")

    # Expected vs actual behavior
    # Expected: ...
    # Actual: [error message]

    # Session info
    sessionInfo()

4. Open an issue with your example at github.com/gcol33/couplr/issues/new

## See Also

- `vignette("getting-started")` - Basic usage

- `vignette("matching-workflows")` - Production matching pipelines

- `vignette("algorithms")` - Algorithm selection guide

- `vignette("comparison")` - Comparison with other packages