How do I create a list of all possible anagrams of a word/string in PostgreSQL - sql

Question

How do I create a list of all possible anagrams of a word/string in PostgreSQL.
For example if String is 'act'
then the desired output should be:
act,
atc,
cta,
cat,
tac,
tca
I have one Table 'tbl_words' which contains million of words.
Then I want to check/search for only valid words in my database table from this anagrams list.
Like from above list of anagrams valid words are : act, cat.
Is there any way to do this?
Update 1:
I need output like this:
(all permutation for given word )
any idea ??

Answer 1

The query generates all permutations of 3 elements set:
with recursive numbers as (
select generate_series(1, 3) as i
),
rec as (
select i, array[i] as p
from numbers
union all
select n.i, p || n.i
from numbers n
join rec on cardinality(p) < 3 and not n.i = any(p)
)
select p as permutation
from rec
where cardinality(p) = 3
order by 1
permutation
-------------
{1,2,3}
{1,3,2}
{2,1,3}
{2,3,1}
{3,1,2}
{3,2,1}
(6 rows)
Modify the final query to generate permutations of the letters of a given word:
with recursive numbers as (
select generate_series(1, 3) as i
),
rec as (
select i, array[i] as p
from numbers
union all
select n.i, p || n.i
from numbers n
join rec on cardinality(p) < 3 and not n.i = any(p)
)
select a[p[1]] || a[p[2]] || a[p[3]] as result
from rec
cross join regexp_split_to_array('act', '') as a
where cardinality(p) = 3
order by 1
result
--------
act
atc
cat
cta
tac
tca
(6 rows)

Answer 2

Here is a solution:
with recursive params as (
select *
from (values ('cata')) v(str)
),
nums as (
select str, 1 as n
from params
union all
select str, 1 + n
from nums
where n < length(str)
),
pos as (
select str, array[n] as poses, array_remove(array_agg(n) over (partition by str), n) as rests, 1 as lev
from nums
union all
select pos.str, array_append(pos.poses, nums.n), array_remove(rests, nums.n), lev + 1
from pos join
nums
on pos.str = nums.str and array_position(pos.rests, nums.n) > 0
where cardinality(rests) > 0
)
select distinct pos.str , string_agg(substr(pos.str, thepos, 1), '')
from pos cross join lateral
unnest(pos.poses) thepos
where cardinality(rests) = 0
group by pos.str, pos.poses;
This is quite tricky, particularly when there are repeated letters in the string. The approach taken here generates all permutations of the numbers from 1 to n, where n is the length of the string. It then uses these as indexes to extract characters from the original string.
Those who are keen will notice that this uses select distinct with group by. That seems like the easiest way to avoid duplication in the resultant strings.