This is the simplest explanation of cartesian product.
You write a simple web crawler. Where to start? You pick most popular TLDs (com, net, org) and a list of English words. Combine them to get list of domains. This is cartesian product of three sets.
#!/usr/bin/env python3
import itertools
TLDs=["com", "net", "org"]
words=["apple", "banana", "orange"]
http=["http", "https"]
for _http, _word, _TLD in itertools.product(http, words, TLDs):
print (_http+"://"+_word+"."+_TLD)
http://apple.com http://apple.net http://apple.org http://banana.com http://banana.net http://banana.org http://orange.com http://orange.net http://orange.org https://apple.com https://apple.net https://apple.org https://banana.com https://banana.net https://banana.org https://orange.com https://orange.net https://orange.org
Size of resulting set (domains) = TLDs * words * 2. This is why 'product'.
All this you can achieve with three nested for loops. And it may be more efficient! But calling itertools.product() is more readable and concise.
Compile your code for various OS, arch, using two compiler flags (optimizations).
#!/usr/bin/env python3
import itertools
OS=["Windows", "macOS", "Linux"]
arch=["x86_64", "arm"]
compiler_flags=["O1", "O3"]
for _OS, _arch, _compiler_flags in itertools.product(OS, arch, compiler_flags):
print (_OS, _arch, _compiler_flags)
Windows x86_64 O1 Windows x86_64 O3 Windows arm O1 Windows arm O3 macOS x86_64 O1 macOS x86_64 O3 macOS arm O1 macOS arm O3 Linux x86_64 O1 Linux x86_64 O3 Linux arm O1 Linux arm O3
If you know that password can have only a/b/c/d characters. Password length is 3 characters.
>>> import itertools
>>> a=['a','b','c','d']
>>> list(itertools.product(a,a,a))
[('a', 'a', 'a'), ('a', 'a', 'b'), ('a', 'a', 'c'), ('a', 'a', 'd'), ('a', 'b', 'a'), ('a', 'b', 'b'), ('a', 'b', 'c'), ('a', 'b', 'd'), ('a', 'c', 'a'), ('a', 'c', 'b'), ('a', 'c', 'c'), ('a', 'c', 'd'), ('a', 'd', 'a'), ('a', 'd', 'b'), ('a', 'd', 'c'), ('a', 'd', 'd'), ('b', 'a', 'a'), ('b', 'a', 'b'), ('b', 'a', 'c'), ('b', 'a', 'd'), ('b', 'b', 'a'), ('b', 'b', 'b'), ('b', 'b', 'c'), ('b', 'b', 'd'), ('b', 'c', 'a'), ('b', 'c', 'b'), ('b', 'c', 'c'), ('b', 'c', 'd'), ('b', 'd', 'a'), ('b', 'd', 'b'), ('b', 'd', 'c'), ('b', 'd', 'd'), ('c', 'a', 'a'), ('c', 'a', 'b'), ('c', 'a', 'c'), ('c', 'a', 'd'), ('c', 'b', 'a'), ('c', 'b', 'b'), ('c', 'b', 'c'), ('c', 'b', 'd'), ('c', 'c', 'a'), ('c', 'c', 'b'), ('c', 'c', 'c'), ('c', 'c', 'd'), ('c', 'd', 'a'), ('c', 'd', 'b'), ('c', 'd', 'c'), ('c', 'd', 'd'), ('d', 'a', 'a'), ('d', 'a', 'b'), ('d', 'a', 'c'), ('d', 'a', 'd'), ('d', 'b', 'a'), ('d', 'b', 'b'), ('d', 'b', 'c'), ('d', 'b', 'd'), ('d', 'c', 'a'), ('d', 'c', 'b'), ('d', 'c', 'c'), ('d', 'c', 'd'), ('d', 'd', 'a'), ('d', 'd', 'b'), ('d', 'd', 'c'), ('d', 'd', 'd')]
... or ...
>>> list(itertools.product(['a','b','c','d'], repeat=3))
[('a', 'a', 'a'), ('a', 'a', 'b'), ...
This code can be scaled.
What if a 3-characters password, where first character can be uppercase/lowercase, but rest is always lowercase?
>>> u=['A','a','b','B']
>>> l=['a','b']
>>> list(itertools.product(u+l,l,l))
[('A', 'a', 'a'), ('A', 'a', 'b'), ('A', 'b', 'a'), ('A', 'b', 'b'), ('a', 'a', 'a'), ('a', 'a', 'b'), ('a', 'b', 'a'), ('a', 'b', 'b'), ('b', 'a', 'a'), ('b', 'a', 'b'), ('b', 'b', 'a'), ('b', 'b', 'b'), ('B', 'a', 'a'), ('B', 'a', 'b'), ('B', 'b', 'a'), ('B', 'b', 'b'), ('a', 'a', 'a'), ('a', 'a', 'b'), ('a', 'b', 'a'), ('a', 'b', 'b'), ('b', 'a', 'a'), ('b', 'a', 'b'), ('b', 'b', 'a'), ('b', 'b', 'b')]
Other ideas about password cracking with itertools, you can find here.
Get list of year/months in shell:
% echo {2020..2022}-{01..12}
2020-01 2020-02 2020-03 2020-04 2020-05 2020-06 2020-07 2020-08 2020-09 2020-10
2020-11 2020-12 2021-01 2021-02 2021-03 2021-04 2021-05 2021-06 2021-07 2021-08
2021-09 2021-10 2021-11 2021-12 2022-01 2022-02 2022-03 2022-04 2022-05 2022-06
2022-07 2022-08 2022-09 2022-10 2022-11 2022-12
Add day. (Some months are incorrect!)
% echo {2020..2022}-{01..12}-{01..31}
2020-01-01 2020-01-02 2020-01-03 2020-01-04 2020-01-05 2020-01-06 2020-01-07 2020-01-08
2020-01-09 2020-01-10 2020-01-11 2020-01-12 2020-01-13 2020-01-14 2020-01-15 2020-01-16
2020-01-17 2020-01-18 2020-01-19 2020-01-20 2020-01-21 2020-01-22 2020-01-23 2020-01-24
2020-01-25 2020-01-26 2020-01-27 2020-01-28 2020-01-29 2020-01-30 2020-01-31 2020-02-01
2020-02-02 2020-02-03 2020-02-04 2020-02-05 2020-02-06 2020-02-07 2020-02-08 2020-02-09
...
Cartesian product (AKA cross product) is what you have in SQL as inner join.
Let's try MySQL.
create table `YEAR` ( `YEAR` INT NOT NULL UNIQUE ); create table `MONTH` ( `MONTH` INT NOT NULL UNIQUE ); insert into YEAR (YEAR) values (2020); insert into YEAR (YEAR) values (2021); insert into YEAR (YEAR) values (2022); insert into MONTH (MONTH) values (1); insert into MONTH (MONTH) values (2); insert into MONTH (MONTH) values (3); insert into MONTH (MONTH) values (4); insert into MONTH (MONTH) values (5); insert into MONTH (MONTH) values (6); insert into MONTH (MONTH) values (7); insert into MONTH (MONTH) values (8); insert into MONTH (MONTH) values (9); insert into MONTH (MONTH) values (10); insert into MONTH (MONTH) values (11); insert into MONTH (MONTH) values (12);
mysql> select * from YEAR; +------+ | YEAR | +------+ | 2020 | | 2021 | | 2022 | +------+ 3 rows in set (0.00 sec) mysql> select * from MONTH; +-------+ | MONTH | +-------+ | 1 | | 2 | | 3 | ... | 10 | | 11 | | 12 | +-------+ 12 rows in set (0.00 sec) mysql> SELECT * FROM YEAR INNER JOIN MONTH; +------+-------+ | YEAR | MONTH | +------+-------+ | 2022 | 1 | | 2021 | 1 | | 2020 | 1 | | 2022 | 2 | | 2021 | 2 | ... | 2021 | 11 | | 2020 | 11 | | 2022 | 12 | | 2021 | 12 | | 2020 | 12 | +------+-------+ 36 rows in set (0.00 sec)
Yes, I know about these lousy Disqus ads. Please use adblocker. I would consider to subscribe to 'pro' version of Disqus if the signal/noise ratio in comments would be good enough.