/// <summary>
/// Constructs the graph from a set of words.
/// </summary>
/// <remarks>
/// The algorithm first buckets the words by length. Since only words whose
/// lengths differ by at most one can be Levenshtein-adjacent, we only need
/// to consider edges between words in buckets i and i+1.
/// </remarks>
/// <param name="allWords">the set of words that will become nodes in the graph</param>
public WordGraph(List <string> allWords)
{
Validate.IsNotNull(allWords, "allWords");
Graph = new Dictionary <string, List <string> >();
var buckets = BucketByLength(allWords);
foreach (var pair in buckets)
{
int bucketSize = pair.Key;
var bucket = pair.Value;
var arrangements = new Arrangement <string>(bucket);
AddValidEdges(arrangements.GetPairs());
if (buckets.TryGetValue(bucketSize + 1, out List <string> nextBucket))
{
AddValidEdges(Combinatorics.CartesianProduct(bucket, nextBucket));
}
}
}
