The concept of trigram similarity relies on having any sentence divided into "trigrams" (sequences of three consecutive letters), and treating the result as a SET (i.e.: the order doesn't matter, and you don't have repeated values). Before the sentence is considered, two blank spaces are added at the beginning, and one at the end, and single spaces are replaced by double ones.
Trigrams are a special case of N-grams.
The trigram set corresponding to "Chateau blanc" is found by finding all sequences of three letters that appear on it:
chateau blanc
--- => ' c'
--- => ' ch'
--- => 'cha'
--- => 'hat'
--- => 'ate'
--- => 'tea'
--- => 'eau'
--- => 'au '
--- => 'u '
--- => ' b'
--- => ' bl'
--- => 'bla'
--- => 'lan'
--- => 'anc'
--- => 'nc '
Sorting them, and taking out repetitions gets you:
' b'
' c'
' bl'
' ch'
'anc'
'ate'
'au '
'bla'
'cha'
'eau'
'hat'
'lan'
'nc '
'tea'
This can be computed by PostgreSQL by means of the function show_trgm:
SELECT show_trgm('Chateau blanc') AS A
A = [ b, c, bl, ch,anc,ate,au ,bla,cha,eau,hat,lan,nc ,tea]
... which has 14 trigrams. (Check pg_trgm).
And the trigram set corresponding to "Chateau Cheval Blanc" is:
SELECT show_trgm('Chateau Cheval Blanc') AS B
B = [ b, c, bl, ch,anc,ate,au ,bla,cha,che,eau,evl,hat,hev,la ,lan,nc ,tea,vla]
... which has 19 trigrams
If you count how many trigrams have both sets in common, you find that they have the following ones:
A intersect B =
[ b, c, bl, ch,anc,ate,au ,bla,cha,eau,hat,lan,nc ,tea]
and the ones they have in total are:
A union B =
[ b, c, bl, ch,anc,ate,au ,bla,cha,che,eau,evl,hat,hev,la ,lan,nc ,tea,vla]
That is, both sentences have 14 trigrams in common, and 19 in total.
The similarity is computed as:
similarity = 14 / 19
You can check it with:
SELECT
cast(14.0/19.0 as real) AS computed_result,
similarity('Chateau blanc', 'chateau cheval blanc') AS function_in_pg
and you'll see that you get: 0.736842
... which explains how similarity is computed, and why you get the values you get.
NOTE: You can compute the intersection and union by means of:
SELECT
array_agg(t) AS in_common
FROM
(
SELECT unnest(show_trgm('Chateau blanc')) AS t
INTERSECT
SELECT unnest(show_trgm('chateau chevla blanc')) AS t
ORDER BY t
) AS trigrams_in_common ;
SELECT
array_agg(t) AS in_total
FROM
(
SELECT unnest(show_trgm('Chateau blanc')) AS t
UNION
SELECT unnest(show_trgm('chateau chevla blanc')) AS t
) AS trigrams_in_total ;
And this is a way to explore the similarity of different pair of sentences:
WITH p AS
(
SELECT
'This is just a sentence I''ve invented'::text AS f1,
'This is just a sentence I''ve also invented'::text AS f2
),
t1 AS
(
SELECT unnest(show_trgm(f1)) FROM p
),
t2 AS
(
SELECT unnest(show_trgm(f2)) FROM p
),
x AS
(
SELECT
(SELECT count(*) FROM
(SELECT * FROM t1 INTERSECT SELECT * FROM t2) AS s0)::integer AS same,
(SELECT count(*) FROM
(SELECT * FROM t1 UNION SELECT * FROM t2) AS s0)::integer AS total,
similarity(f1, f2) AS sim_2
FROM
p
)
SELECT
same, total, same::real/total::real AS sim_1, sim_2
FROM
x ;
You can check it at Rextester
