Fancy UDF based on the Levenshtein distance:
CREATE FUNCTION edit_distance(@s1 nvarchar(3999), @s2 nvarchar(3999))
RETURNS int
AS
BEGIN
DECLARE @s1_len int, @s2_len int, @i int, @j int, @s1_char nchar, @c int, @c_temp int,
@cv0 varbinary(8000), @cv1 varbinary(8000)
SELECT @s1_len = LEN(@s1), @s2_len = LEN(@s2), @cv1 = 0x0000, @j = 1, @i = 1, @c = 0
WHILE @j <= @s2_len
SELECT @cv1 = @cv1 + CAST(@j AS binary(2)), @j = @j + 1
WHILE @i <= @s1_len
BEGIN
SELECT @s1_char = SUBSTRING(@s1, @i, 1), @c = @i, @cv0 = CAST(@i AS binary(2)), @j = 1
WHILE @j <= @s2_len
BEGIN
SET @c = @c + 1
SET @c_temp = CAST(SUBSTRING(@cv1, @j+@j-1, 2) AS int) +
CASE WHEN @s1_char = SUBSTRING(@s2, @j, 1) THEN 0 ELSE 1 END
IF @c > @c_temp SET @c = @c_temp
SET @c_temp = CAST(SUBSTRING(@cv1, @j+@j+1, 2) AS int)+1
IF @c > @c_temp SET @c = @c_temp
SELECT @cv0 = @cv0 + CAST(@c AS binary(2)), @j = @j + 1
END
SELECT @cv1 = @cv0, @i = @i + 1
END
RETURN @c
END
(Cribbed from SQLTeam.com forums.)
Unfortunately, this only works when provided two strings, and in O(nm) time, where n and m are the length of the two strings. However, you could improve upon this by modifying the function to cut out as a "NO MATCH" when @c is greater than, say, 3.
But then you'd still need a cursor to walk through each user name in the table and compare it to every other user. Not very efficient, that's for sure.