List of Python versions Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) The PPCG Site design is on its way - help us make it awesome! Sandbox for Proposed ChallengesTips for golfing in JavaParse a Python string literalGolf Practice: PythonRandomised list, formatted nicelySmoothing out a listPython Code ReviewerThe Stack Exchange Site ListWhich Ubuntu versions are supported?Golfing Python string literalsMost common elements of a list in PythonList all possible titles for the Anno games
What do you call the main part of a joke?
Why wasn't DOSKEY integrated with COMMAND.COM?
Why does it sometimes sound good to play a grace note as a lead in to a note in a melody?
Why should I vote and accept answers?
What is "gratricide"?
A term for a woman complaining about things/begging in a cute/childish way
Is it possible for SQL statements to execute concurrently within a single session in SQL Server?
What is the appropriate index architecture when forced to implement IsDeleted (soft deletes)?
Is there any word for a place full of confusion?
Has negative voting ever been officially implemented in elections, or seriously proposed, or even studied?
What is a fractional matching?
Find 108 by using 3,4,6
As a beginner, should I get a Squier Strat with a SSS config or a HSS?
How much damage would a cupful of neutron star matter do to the Earth?
Time to Settle Down!
How do living politicians protect their readily obtainable signatures from misuse?
Why is it faster to reheat something than it is to cook it?
Why do early math courses focus on the cross sections of a cone and not on other 3D objects?
What would you call this weird metallic apparatus that allows you to lift people?
What is the topology associated with the algebras for the ultrafilter monad?
What initially awakened the Balrog?
Is CEO the "profession" with the most psychopaths?
What is the difference between globalisation and imperialism?
Why do we need to use the builder design pattern when we can do the same thing with setters?
List of Python versions
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
The PPCG Site design is on its way - help us make it awesome!
Sandbox for Proposed ChallengesTips for golfing in JavaParse a Python string literalGolf Practice: PythonRandomised list, formatted nicelySmoothing out a listPython Code ReviewerThe Stack Exchange Site ListWhich Ubuntu versions are supported?Golfing Python string literalsMost common elements of a list in PythonList all possible titles for the Anno games
$begingroup$
Python is the fastest-growing major programming language today. It is the most wanted language for the third year in a row, meaning that developers who do not yet use it say they want to learn it. [1].
The reason for Python's popularity is its many versions.[citation needed]. There are in fact 116 versions of Python, including two development versions.
Your task is to output/print a list of all the Python versions, in whatever order you like, and on whatever format you like. You may not use any built-in functions that have this information stored.
You are free to choose the output format, but each version must be identified on the standard way: 1.1
, 2.3.0
, 2.7.10
and so on.
The complete list2 of Python versions, comma-separated is shown below:
1.1, 1.2, 1.3, 1.4, 1.5, 1.5.1, 1.5.2, 1.6, 2.0, 2.0.1, 2.1, 2.1.1, 2.1.2, 2.1.3, 2.2, 2.2.1, 2.2.2, 2.2.3, 2.3, 2.3.1, 2.3.2, 2.3.3, 2.3.4, 2.3.5, 2.4, 2.4.1, 2.4.2, 2.4.3, 2.4.4, 2.5, 2.5.1, 2.5.2, 2.5.3, 2.5.4, 2.6, 2.6.1, 2.6.2, 2.6.3, 2.6.4, 2.6.5, 2.6.6, 2.6.7, 2.6.8, 2.6.9, 2.7, 2.7.1, 2.7.2, 2.7.3, 2.7.4, 2.7.5, 2.7.6, 2.7.7, 2.7.8, 2.7.9, 2.7.10, 2.7.11, 2.7.12, 2.7.13, 2.7.14, 2.7.15, 2.7.16, 3.0, 3.0.1, 3.1, 3.1.1, 3.1.2, 3.1.3, 3.1.4, 3.1.5, 3.2 , 3.2.1, 3.2.2, 3.2.3, 3.2.4, 3.2.5, 3.2.6, 3.3.0, 3.3.1, 3.3.2, 3.3.3, 3.3.4, 3.3.5, 3.3.6, 3.3.7, 3.4.0, 3.4.1, 3.4.2, 3.4.3, 3.4.4, 3.4.5, 3.4.6, 3.4.7, 3.4.8, 3.4.9, 3.4.10, 3.5.0, 3.5.1, 3.5.2, 3.5.3, 3.5.4, 3.5.5, 3.5.6, 3.5.7, 3.6.0, 3.6.1, 3.6.2, 3.6.3, 3.6.4, 3.6.5, 3.6.6, 3.6.7, 3.6.8, 3.7.0, 3.7.1, 3.7.2, 3.7.3
or by major versions:
1.1
1.2
1.3
1.4
1.5, 1.5.1, 1.5.2
1.6
2.0, 2.0.1
2.1, 2.1.1, 2.1.2, 2.1.3
2.2, 2.2.1, 2.2.2, 2.2.3
2.3, 2.3.1, 2.3.2, 2.3.3, 2.3.4, 2.3.5
2.4, 2.4.1, 2.4.2, 2.4.3, 2.4.4
2.5, 2.5.1, 2.5.2, 2.5.3, 2.5.4
2.6, 2.6.1, 2.6.2, 2.6.3, 2.6.4, 2.6.5, 2.6.6, 2.6.7, 2.6.8, 2.6.9
2.7, 2.7.1, 2.7.2, 2.7.3, 2.7.4, 2.7.5, 2.7.6, 2.7.7, 2.7.8, 2.7.9, 2.7.10, 2.7.11, 2.7.12, 2.7.13, 2.7.14, 2.7.15, 2.7.16
3.0, 3.0.1
3.1, 3.1.1, 3.1.2, 3.1.3, 3.1.4, 3.1.5
3.2, 3.2.1, 3.2.2, 3.2.3, 3.2.4, 3.2.5, 3.2.6
3.3.0, 3.3.1, 3.3.2, 3.3.3, 3.3.4, 3.3.5, 3.3.6, 3.3.7
3.4.0, 3.4.1, 3.4.2, 3.4.3, 3.4.4, 3.4.5, 3.4.6, 3.4.7, 3.4.8, 3.4.9, 3.4.10
3.5.0, 3.5.1, 3.5.2, 3.5.3, 3.5.4, 3.5.5, 3.5.6, 3.5.7
3.6.0, 3.6.1, 3.6.2, 3.6.3, 3.6.4, 3.6.5, 3.6.6, 3.6.7, 3.6.8
3.7.0, 3.7.1, 3.7.2, 3.7.3
The challenge is a fixed output challenge, and very close to a kolmogorov-challenge, except that the output format is optional.
2The list is taken from the official Python website, here and here. There are some versions that aren't included, such as 0.9.0
.. 0.9.9
and 1.5.1p1
. You must use the list above, even if you find versions that aren't included. I've decided to stick with the official lists, since otherwise someone would probably find a 2.1.0.1.2
version or something like that.
code-golf string number kolmogorov-complexity
$endgroup$
add a comment |
$begingroup$
Python is the fastest-growing major programming language today. It is the most wanted language for the third year in a row, meaning that developers who do not yet use it say they want to learn it. [1].
The reason for Python's popularity is its many versions.[citation needed]. There are in fact 116 versions of Python, including two development versions.
Your task is to output/print a list of all the Python versions, in whatever order you like, and on whatever format you like. You may not use any built-in functions that have this information stored.
You are free to choose the output format, but each version must be identified on the standard way: 1.1
, 2.3.0
, 2.7.10
and so on.
The complete list2 of Python versions, comma-separated is shown below:
1.1, 1.2, 1.3, 1.4, 1.5, 1.5.1, 1.5.2, 1.6, 2.0, 2.0.1, 2.1, 2.1.1, 2.1.2, 2.1.3, 2.2, 2.2.1, 2.2.2, 2.2.3, 2.3, 2.3.1, 2.3.2, 2.3.3, 2.3.4, 2.3.5, 2.4, 2.4.1, 2.4.2, 2.4.3, 2.4.4, 2.5, 2.5.1, 2.5.2, 2.5.3, 2.5.4, 2.6, 2.6.1, 2.6.2, 2.6.3, 2.6.4, 2.6.5, 2.6.6, 2.6.7, 2.6.8, 2.6.9, 2.7, 2.7.1, 2.7.2, 2.7.3, 2.7.4, 2.7.5, 2.7.6, 2.7.7, 2.7.8, 2.7.9, 2.7.10, 2.7.11, 2.7.12, 2.7.13, 2.7.14, 2.7.15, 2.7.16, 3.0, 3.0.1, 3.1, 3.1.1, 3.1.2, 3.1.3, 3.1.4, 3.1.5, 3.2 , 3.2.1, 3.2.2, 3.2.3, 3.2.4, 3.2.5, 3.2.6, 3.3.0, 3.3.1, 3.3.2, 3.3.3, 3.3.4, 3.3.5, 3.3.6, 3.3.7, 3.4.0, 3.4.1, 3.4.2, 3.4.3, 3.4.4, 3.4.5, 3.4.6, 3.4.7, 3.4.8, 3.4.9, 3.4.10, 3.5.0, 3.5.1, 3.5.2, 3.5.3, 3.5.4, 3.5.5, 3.5.6, 3.5.7, 3.6.0, 3.6.1, 3.6.2, 3.6.3, 3.6.4, 3.6.5, 3.6.6, 3.6.7, 3.6.8, 3.7.0, 3.7.1, 3.7.2, 3.7.3
or by major versions:
1.1
1.2
1.3
1.4
1.5, 1.5.1, 1.5.2
1.6
2.0, 2.0.1
2.1, 2.1.1, 2.1.2, 2.1.3
2.2, 2.2.1, 2.2.2, 2.2.3
2.3, 2.3.1, 2.3.2, 2.3.3, 2.3.4, 2.3.5
2.4, 2.4.1, 2.4.2, 2.4.3, 2.4.4
2.5, 2.5.1, 2.5.2, 2.5.3, 2.5.4
2.6, 2.6.1, 2.6.2, 2.6.3, 2.6.4, 2.6.5, 2.6.6, 2.6.7, 2.6.8, 2.6.9
2.7, 2.7.1, 2.7.2, 2.7.3, 2.7.4, 2.7.5, 2.7.6, 2.7.7, 2.7.8, 2.7.9, 2.7.10, 2.7.11, 2.7.12, 2.7.13, 2.7.14, 2.7.15, 2.7.16
3.0, 3.0.1
3.1, 3.1.1, 3.1.2, 3.1.3, 3.1.4, 3.1.5
3.2, 3.2.1, 3.2.2, 3.2.3, 3.2.4, 3.2.5, 3.2.6
3.3.0, 3.3.1, 3.3.2, 3.3.3, 3.3.4, 3.3.5, 3.3.6, 3.3.7
3.4.0, 3.4.1, 3.4.2, 3.4.3, 3.4.4, 3.4.5, 3.4.6, 3.4.7, 3.4.8, 3.4.9, 3.4.10
3.5.0, 3.5.1, 3.5.2, 3.5.3, 3.5.4, 3.5.5, 3.5.6, 3.5.7
3.6.0, 3.6.1, 3.6.2, 3.6.3, 3.6.4, 3.6.5, 3.6.6, 3.6.7, 3.6.8
3.7.0, 3.7.1, 3.7.2, 3.7.3
The challenge is a fixed output challenge, and very close to a kolmogorov-challenge, except that the output format is optional.
2The list is taken from the official Python website, here and here. There are some versions that aren't included, such as 0.9.0
.. 0.9.9
and 1.5.1p1
. You must use the list above, even if you find versions that aren't included. I've decided to stick with the official lists, since otherwise someone would probably find a 2.1.0.1.2
version or something like that.
code-golf string number kolmogorov-complexity
$endgroup$
1
$begingroup$
I guess we aren't allowed to output1.1.0
(to make all versions 3 numbers) instead of1.1
?
$endgroup$
– Kevin Cruijssen
2 days ago
2
$begingroup$
You guess correctly @Kevin. I considered allowing it, but went with the official names instead.
$endgroup$
– Stewie Griffin
2 days ago
add a comment |
$begingroup$
Python is the fastest-growing major programming language today. It is the most wanted language for the third year in a row, meaning that developers who do not yet use it say they want to learn it. [1].
The reason for Python's popularity is its many versions.[citation needed]. There are in fact 116 versions of Python, including two development versions.
Your task is to output/print a list of all the Python versions, in whatever order you like, and on whatever format you like. You may not use any built-in functions that have this information stored.
You are free to choose the output format, but each version must be identified on the standard way: 1.1
, 2.3.0
, 2.7.10
and so on.
The complete list2 of Python versions, comma-separated is shown below:
1.1, 1.2, 1.3, 1.4, 1.5, 1.5.1, 1.5.2, 1.6, 2.0, 2.0.1, 2.1, 2.1.1, 2.1.2, 2.1.3, 2.2, 2.2.1, 2.2.2, 2.2.3, 2.3, 2.3.1, 2.3.2, 2.3.3, 2.3.4, 2.3.5, 2.4, 2.4.1, 2.4.2, 2.4.3, 2.4.4, 2.5, 2.5.1, 2.5.2, 2.5.3, 2.5.4, 2.6, 2.6.1, 2.6.2, 2.6.3, 2.6.4, 2.6.5, 2.6.6, 2.6.7, 2.6.8, 2.6.9, 2.7, 2.7.1, 2.7.2, 2.7.3, 2.7.4, 2.7.5, 2.7.6, 2.7.7, 2.7.8, 2.7.9, 2.7.10, 2.7.11, 2.7.12, 2.7.13, 2.7.14, 2.7.15, 2.7.16, 3.0, 3.0.1, 3.1, 3.1.1, 3.1.2, 3.1.3, 3.1.4, 3.1.5, 3.2 , 3.2.1, 3.2.2, 3.2.3, 3.2.4, 3.2.5, 3.2.6, 3.3.0, 3.3.1, 3.3.2, 3.3.3, 3.3.4, 3.3.5, 3.3.6, 3.3.7, 3.4.0, 3.4.1, 3.4.2, 3.4.3, 3.4.4, 3.4.5, 3.4.6, 3.4.7, 3.4.8, 3.4.9, 3.4.10, 3.5.0, 3.5.1, 3.5.2, 3.5.3, 3.5.4, 3.5.5, 3.5.6, 3.5.7, 3.6.0, 3.6.1, 3.6.2, 3.6.3, 3.6.4, 3.6.5, 3.6.6, 3.6.7, 3.6.8, 3.7.0, 3.7.1, 3.7.2, 3.7.3
or by major versions:
1.1
1.2
1.3
1.4
1.5, 1.5.1, 1.5.2
1.6
2.0, 2.0.1
2.1, 2.1.1, 2.1.2, 2.1.3
2.2, 2.2.1, 2.2.2, 2.2.3
2.3, 2.3.1, 2.3.2, 2.3.3, 2.3.4, 2.3.5
2.4, 2.4.1, 2.4.2, 2.4.3, 2.4.4
2.5, 2.5.1, 2.5.2, 2.5.3, 2.5.4
2.6, 2.6.1, 2.6.2, 2.6.3, 2.6.4, 2.6.5, 2.6.6, 2.6.7, 2.6.8, 2.6.9
2.7, 2.7.1, 2.7.2, 2.7.3, 2.7.4, 2.7.5, 2.7.6, 2.7.7, 2.7.8, 2.7.9, 2.7.10, 2.7.11, 2.7.12, 2.7.13, 2.7.14, 2.7.15, 2.7.16
3.0, 3.0.1
3.1, 3.1.1, 3.1.2, 3.1.3, 3.1.4, 3.1.5
3.2, 3.2.1, 3.2.2, 3.2.3, 3.2.4, 3.2.5, 3.2.6
3.3.0, 3.3.1, 3.3.2, 3.3.3, 3.3.4, 3.3.5, 3.3.6, 3.3.7
3.4.0, 3.4.1, 3.4.2, 3.4.3, 3.4.4, 3.4.5, 3.4.6, 3.4.7, 3.4.8, 3.4.9, 3.4.10
3.5.0, 3.5.1, 3.5.2, 3.5.3, 3.5.4, 3.5.5, 3.5.6, 3.5.7
3.6.0, 3.6.1, 3.6.2, 3.6.3, 3.6.4, 3.6.5, 3.6.6, 3.6.7, 3.6.8
3.7.0, 3.7.1, 3.7.2, 3.7.3
The challenge is a fixed output challenge, and very close to a kolmogorov-challenge, except that the output format is optional.
2The list is taken from the official Python website, here and here. There are some versions that aren't included, such as 0.9.0
.. 0.9.9
and 1.5.1p1
. You must use the list above, even if you find versions that aren't included. I've decided to stick with the official lists, since otherwise someone would probably find a 2.1.0.1.2
version or something like that.
code-golf string number kolmogorov-complexity
$endgroup$
Python is the fastest-growing major programming language today. It is the most wanted language for the third year in a row, meaning that developers who do not yet use it say they want to learn it. [1].
The reason for Python's popularity is its many versions.[citation needed]. There are in fact 116 versions of Python, including two development versions.
Your task is to output/print a list of all the Python versions, in whatever order you like, and on whatever format you like. You may not use any built-in functions that have this information stored.
You are free to choose the output format, but each version must be identified on the standard way: 1.1
, 2.3.0
, 2.7.10
and so on.
The complete list2 of Python versions, comma-separated is shown below:
1.1, 1.2, 1.3, 1.4, 1.5, 1.5.1, 1.5.2, 1.6, 2.0, 2.0.1, 2.1, 2.1.1, 2.1.2, 2.1.3, 2.2, 2.2.1, 2.2.2, 2.2.3, 2.3, 2.3.1, 2.3.2, 2.3.3, 2.3.4, 2.3.5, 2.4, 2.4.1, 2.4.2, 2.4.3, 2.4.4, 2.5, 2.5.1, 2.5.2, 2.5.3, 2.5.4, 2.6, 2.6.1, 2.6.2, 2.6.3, 2.6.4, 2.6.5, 2.6.6, 2.6.7, 2.6.8, 2.6.9, 2.7, 2.7.1, 2.7.2, 2.7.3, 2.7.4, 2.7.5, 2.7.6, 2.7.7, 2.7.8, 2.7.9, 2.7.10, 2.7.11, 2.7.12, 2.7.13, 2.7.14, 2.7.15, 2.7.16, 3.0, 3.0.1, 3.1, 3.1.1, 3.1.2, 3.1.3, 3.1.4, 3.1.5, 3.2 , 3.2.1, 3.2.2, 3.2.3, 3.2.4, 3.2.5, 3.2.6, 3.3.0, 3.3.1, 3.3.2, 3.3.3, 3.3.4, 3.3.5, 3.3.6, 3.3.7, 3.4.0, 3.4.1, 3.4.2, 3.4.3, 3.4.4, 3.4.5, 3.4.6, 3.4.7, 3.4.8, 3.4.9, 3.4.10, 3.5.0, 3.5.1, 3.5.2, 3.5.3, 3.5.4, 3.5.5, 3.5.6, 3.5.7, 3.6.0, 3.6.1, 3.6.2, 3.6.3, 3.6.4, 3.6.5, 3.6.6, 3.6.7, 3.6.8, 3.7.0, 3.7.1, 3.7.2, 3.7.3
or by major versions:
1.1
1.2
1.3
1.4
1.5, 1.5.1, 1.5.2
1.6
2.0, 2.0.1
2.1, 2.1.1, 2.1.2, 2.1.3
2.2, 2.2.1, 2.2.2, 2.2.3
2.3, 2.3.1, 2.3.2, 2.3.3, 2.3.4, 2.3.5
2.4, 2.4.1, 2.4.2, 2.4.3, 2.4.4
2.5, 2.5.1, 2.5.2, 2.5.3, 2.5.4
2.6, 2.6.1, 2.6.2, 2.6.3, 2.6.4, 2.6.5, 2.6.6, 2.6.7, 2.6.8, 2.6.9
2.7, 2.7.1, 2.7.2, 2.7.3, 2.7.4, 2.7.5, 2.7.6, 2.7.7, 2.7.8, 2.7.9, 2.7.10, 2.7.11, 2.7.12, 2.7.13, 2.7.14, 2.7.15, 2.7.16
3.0, 3.0.1
3.1, 3.1.1, 3.1.2, 3.1.3, 3.1.4, 3.1.5
3.2, 3.2.1, 3.2.2, 3.2.3, 3.2.4, 3.2.5, 3.2.6
3.3.0, 3.3.1, 3.3.2, 3.3.3, 3.3.4, 3.3.5, 3.3.6, 3.3.7
3.4.0, 3.4.1, 3.4.2, 3.4.3, 3.4.4, 3.4.5, 3.4.6, 3.4.7, 3.4.8, 3.4.9, 3.4.10
3.5.0, 3.5.1, 3.5.2, 3.5.3, 3.5.4, 3.5.5, 3.5.6, 3.5.7
3.6.0, 3.6.1, 3.6.2, 3.6.3, 3.6.4, 3.6.5, 3.6.6, 3.6.7, 3.6.8
3.7.0, 3.7.1, 3.7.2, 3.7.3
The challenge is a fixed output challenge, and very close to a kolmogorov-challenge, except that the output format is optional.
2The list is taken from the official Python website, here and here. There are some versions that aren't included, such as 0.9.0
.. 0.9.9
and 1.5.1p1
. You must use the list above, even if you find versions that aren't included. I've decided to stick with the official lists, since otherwise someone would probably find a 2.1.0.1.2
version or something like that.
code-golf string number kolmogorov-complexity
code-golf string number kolmogorov-complexity
edited 2 days ago
Stewie Griffin
asked 2 days ago
Stewie GriffinStewie Griffin
28.7k11106274
28.7k11106274
1
$begingroup$
I guess we aren't allowed to output1.1.0
(to make all versions 3 numbers) instead of1.1
?
$endgroup$
– Kevin Cruijssen
2 days ago
2
$begingroup$
You guess correctly @Kevin. I considered allowing it, but went with the official names instead.
$endgroup$
– Stewie Griffin
2 days ago
add a comment |
1
$begingroup$
I guess we aren't allowed to output1.1.0
(to make all versions 3 numbers) instead of1.1
?
$endgroup$
– Kevin Cruijssen
2 days ago
2
$begingroup$
You guess correctly @Kevin. I considered allowing it, but went with the official names instead.
$endgroup$
– Stewie Griffin
2 days ago
1
1
$begingroup$
I guess we aren't allowed to output
1.1.0
(to make all versions 3 numbers) instead of 1.1
?$endgroup$
– Kevin Cruijssen
2 days ago
$begingroup$
I guess we aren't allowed to output
1.1.0
(to make all versions 3 numbers) instead of 1.1
?$endgroup$
– Kevin Cruijssen
2 days ago
2
2
$begingroup$
You guess correctly @Kevin. I considered allowing it, but went with the official names instead.
$endgroup$
– Stewie Griffin
2 days ago
$begingroup$
You guess correctly @Kevin. I considered allowing it, but went with the official names instead.
$endgroup$
– Stewie Griffin
2 days ago
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
JavaScript (ES6), 128 125 124 bytes
Saved 1 bytes thanks to @OlivierGrégoire
Outputs each version on a separate line. Ordered from highest to lowest major version, and from lowest to highest revision.
f=(r=v=28)=>v?r<parseInt('0111131000244655ah002678b8940'[v],36)?(1+v/10).toFixed(1)+(r|v>22?'.'+r:'')+`
`+f(r+1):f(+!v--):''
Try it online!
How?
The major and minor versions are held in the variable $v in [0..27]$:
- major = $lfloor v/10+1rfloor$
- minor = $v bmod 10$
The revision is held in the variable $rge0$. The maximum value of $r$ depends on $v$ and is stored in a lookup table encoded in Base-36. Any $0$ in this table means that this version was not released at all.
Besides, we use the test $v>22$ to know whether the revision should be included even when it's $0$ (starting with Python 3.3.0).
$endgroup$
2
$begingroup$
@KevinCruijssen semes like it doesn't work for2.0.1
$endgroup$
– ASCII-only
2 days ago
$begingroup$
Can you gain bytes by startingv
at 27 and going down to zero?
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire This saves 1 byte. Thanks!
$endgroup$
– Arnauld
yesterday
add a comment |
$begingroup$
Java (JDK), 134 bytes
v->for(int a=0,b;;)for(b="0111131000244655:A002678;894".charAt(++a)-48;b-->0;)System.out.printf("%.1f%s ",a*.1+1,b<1&a<23?"":"."+b);
Try it online!
The versions are printed from the highest to the lowest.
Credits
- -1 byte thanks to Kevin Cruijssen
- -3 bytes thanks to Embodiment of Ignorance
$endgroup$
1
$begingroup$
(a>1|b>0)&c<a.valueOf(y,36)
can bea>1|b>0&&c<a.valueOf(y,36)
andc<1&(a<3|b<3)?
can bec<1&&a<3|b<3?
to save 2 bytes. Relevant Java tip - section Combining bit-wise and logical checks instead of using parenthesis
$endgroup$
– Kevin Cruijssen
yesterday
$begingroup$
@KevinCruijssen Thank you, but I made so many changes that your suggestions are now irrelevant... Don't know how to credit you since I don't even use your suggestion anymore :(
$endgroup$
– Olivier Grégoire
yesterday
1
$begingroup$
Np, instead I will suggest a new golf ;)/10d
can be*.1
$endgroup$
– Kevin Cruijssen
yesterday
1
$begingroup$
int a=28
->int a=1
, and remove the condition in the for loop, then add ana++
to save 3 bytes. TIO
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance Since REPL seem accepted in this challenge, that's indeed acceptable. Thank you!
$endgroup$
– Olivier Grégoire
yesterday
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 109 bytes
for(int j,k=1;;k++)for(j=@" [SOH][SOH][SOH][SOH][ETX][SOH][NUL][NUL][NUL][STX][EOT][EOT][ACK][ENQ][ENQ]
[DC1][NUL][NUL][STX][ACK][BEL][BS][VT][BS][TAB][EOT]"[k];j-->0;)Print($"k*.1+1:N1"+(j<1&k<17?"":"."+j));
Contains many unprintables, whose codes are shown in the brackets. This is a full program. The null bytes are replaced by s in the TIO link, since my device can't copy and paste them.
Saved one byte thanks to @OlivierGregoire.
Try it online! (Thanks to @OlivierGregoire for implanting the null bytes)
Explanation
Each character in the string represents how many minor versions in the major position. For example, the character at index 5(ETX
) has an ASCII value of three, and corresponds to the major version 1.5.x
which has three minor versions. The program takes the ascii value of the current character and loops that many times, printing the minor versions before moving on to the next major version.
For some versions, there are gaps to the next versions. To fix that, the string contains null bytes, so that the program loops zero times when it encounters those.
The unprintable string contains these character values:
1,1,1,1,3,1,0,0,0,2,4,4,6,5,5,10,17,0,0,2,6,7,8,11,8,9,4
$endgroup$
$begingroup$
Can be shortened withj="..."[k];j-->0;
, especially since the order has no importance. Also, can you explain the difference in size between the TIO (115 bytes) and the entry (110 bytes)?
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire Probably the five null bytes which tio represent as
$endgroup$
– Sefa
yesterday
$begingroup$
@Sefa yes, probably... But I'm asking for certainty.
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire Exactly what Sefa said, I can't really copy-paste the null bytes. If thes were replaced by null bytes, it would be 110 bytes
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
Then, here you are, with a nul-byte TIO
$endgroup$
– Olivier Grégoire
yesterday
add a comment |
$begingroup$
Retina, 105 bytes
11* 111131 244655TS 2678E894
L$`.
$&_$.`
T
10
E
11
S
17
.+_
*
Lv$`_+(.)(.)
$1.$2.$.%`
,16`(...).0
$1
Try it online! Loosely based on @Arnauld's solution. Explanation:
11* 111131 244655TS 2678E894
Insert the string consisting of 11 spaces followed by the given characters.
L$`.
$&_$.`
For each character, list it suffixed with a _
and its column number.
T
10
E
11
S
17
Convert the three letters to numeric values.
.+_
*
Convert the numeric values to unary.
Lv$`_+(.)(.)
$1.$2.$.%`
For each value up to the given value, use that as the suffix for the version number, extracting the major and minor from the column number.
,16`(...).0
$1
Delete the zero suffix for the first 16 versions that have one.
$endgroup$
add a comment |
$begingroup$
Pyth, 52 bytes
.emj.+W|d>k18,h/k8%k8dbxLG"abbbbdbaceegffkrcghilije
Try it online here.
Output is a nested list, with elements grouped by major and minor version. There is an empty list at the start of the output, and another one after 1.6
. Full output is as follows:
[[], ['1.1'], ['1.2'], ['1.3'], ['1.4'], ['1.5', '1.5.1', '1.5.2'], ['1.6'], [], ['2.0', '2.0.1'], ['2.1', '2.1.1', '2.1.2', '2.1.3'], ['2.2', '2.2.1', '2.2.2', '2.2.3'], ['2.3', '2.3.1', '2.3.2', '2.3.3', '2.3.4', '2.3.5'], ['2.4', '2.4.1', '2.4.2', '2.4.3', '2.4.4'], ['2.5', '2.5.1', '2.5.2', '2.5.3', '2.5.4'], ['2.6', '2.6.1', '2.6.2', '2.6.3', '2.6.4', '2.6.5', '2.6.6', '2.6.7', '2.6.8', '2.6.9'], ['2.7', '2.7.1', '2.7.2', '2.7.3', '2.7.4', '2.7.5', '2.7.6', '2.7.7', '2.7.8', '2.7.9', '2.7.10', '2.7.11', '2.7.12', '2.7.13', '2.7.14', '2.7.15', '2.7.16'], ['3.0', '3.0.1'], ['3.1', '3.1.1', '3.1.2', '3.1.3', '3.1.4', '3.1.5'], ['3.2', '3.2.1', '3.2.2', '3.2.3', '3.2.4', '3.2.5', '3.2.6'], ['3.3.0', '3.3.1', '3.3.2', '3.3.3', '3.3.4', '3.3.5', '3.3.6', '3.3.7'], ['3.4.0', '3.4.1', '3.4.2', '3.4.3', '3.4.4', '3.4.5', '3.4.6', '3.4.7', '3.4.8', '3.4.9', '3.4.10'], ['3.5.0', '3.5.1', '3.5.2', '3.5.3', '3.5.4', '3.5.5', '3.5.6', '3.5.7'], ['3.6.0', '3.6.1', '3.6.2', '3.6.3', '3.6.4', '3.6.5', '3.6.6', '3.6.7', '3.6.8'], ['3.7.0', '3.7.1', '3.7.2', '3.7.3']]
If this isn't acceptable, prepend .n
to the code to have output as a flattened list, at a cost of 2 bytes.
$endgroup$
add a comment |
$begingroup$
Jelly, 51 bytes
+⁵D;ⱮḶ}j€”.
“øṄƇịɱ⁽Ɱj>⁶7,Ẉ¢’b18Ė0ị$Ƈç/€ḣ3$€1¦€17R¤¦
Try it online!
A niladic link that outputs a list of lists of .
-separated integers, grouped by major version. On TIO, there’s some footer code to print these prettily.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "200"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f183284%2flist-of-python-versions%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
JavaScript (ES6), 128 125 124 bytes
Saved 1 bytes thanks to @OlivierGrégoire
Outputs each version on a separate line. Ordered from highest to lowest major version, and from lowest to highest revision.
f=(r=v=28)=>v?r<parseInt('0111131000244655ah002678b8940'[v],36)?(1+v/10).toFixed(1)+(r|v>22?'.'+r:'')+`
`+f(r+1):f(+!v--):''
Try it online!
How?
The major and minor versions are held in the variable $v in [0..27]$:
- major = $lfloor v/10+1rfloor$
- minor = $v bmod 10$
The revision is held in the variable $rge0$. The maximum value of $r$ depends on $v$ and is stored in a lookup table encoded in Base-36. Any $0$ in this table means that this version was not released at all.
Besides, we use the test $v>22$ to know whether the revision should be included even when it's $0$ (starting with Python 3.3.0).
$endgroup$
2
$begingroup$
@KevinCruijssen semes like it doesn't work for2.0.1
$endgroup$
– ASCII-only
2 days ago
$begingroup$
Can you gain bytes by startingv
at 27 and going down to zero?
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire This saves 1 byte. Thanks!
$endgroup$
– Arnauld
yesterday
add a comment |
$begingroup$
JavaScript (ES6), 128 125 124 bytes
Saved 1 bytes thanks to @OlivierGrégoire
Outputs each version on a separate line. Ordered from highest to lowest major version, and from lowest to highest revision.
f=(r=v=28)=>v?r<parseInt('0111131000244655ah002678b8940'[v],36)?(1+v/10).toFixed(1)+(r|v>22?'.'+r:'')+`
`+f(r+1):f(+!v--):''
Try it online!
How?
The major and minor versions are held in the variable $v in [0..27]$:
- major = $lfloor v/10+1rfloor$
- minor = $v bmod 10$
The revision is held in the variable $rge0$. The maximum value of $r$ depends on $v$ and is stored in a lookup table encoded in Base-36. Any $0$ in this table means that this version was not released at all.
Besides, we use the test $v>22$ to know whether the revision should be included even when it's $0$ (starting with Python 3.3.0).
$endgroup$
2
$begingroup$
@KevinCruijssen semes like it doesn't work for2.0.1
$endgroup$
– ASCII-only
2 days ago
$begingroup$
Can you gain bytes by startingv
at 27 and going down to zero?
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire This saves 1 byte. Thanks!
$endgroup$
– Arnauld
yesterday
add a comment |
$begingroup$
JavaScript (ES6), 128 125 124 bytes
Saved 1 bytes thanks to @OlivierGrégoire
Outputs each version on a separate line. Ordered from highest to lowest major version, and from lowest to highest revision.
f=(r=v=28)=>v?r<parseInt('0111131000244655ah002678b8940'[v],36)?(1+v/10).toFixed(1)+(r|v>22?'.'+r:'')+`
`+f(r+1):f(+!v--):''
Try it online!
How?
The major and minor versions are held in the variable $v in [0..27]$:
- major = $lfloor v/10+1rfloor$
- minor = $v bmod 10$
The revision is held in the variable $rge0$. The maximum value of $r$ depends on $v$ and is stored in a lookup table encoded in Base-36. Any $0$ in this table means that this version was not released at all.
Besides, we use the test $v>22$ to know whether the revision should be included even when it's $0$ (starting with Python 3.3.0).
$endgroup$
JavaScript (ES6), 128 125 124 bytes
Saved 1 bytes thanks to @OlivierGrégoire
Outputs each version on a separate line. Ordered from highest to lowest major version, and from lowest to highest revision.
f=(r=v=28)=>v?r<parseInt('0111131000244655ah002678b8940'[v],36)?(1+v/10).toFixed(1)+(r|v>22?'.'+r:'')+`
`+f(r+1):f(+!v--):''
Try it online!
How?
The major and minor versions are held in the variable $v in [0..27]$:
- major = $lfloor v/10+1rfloor$
- minor = $v bmod 10$
The revision is held in the variable $rge0$. The maximum value of $r$ depends on $v$ and is stored in a lookup table encoded in Base-36. Any $0$ in this table means that this version was not released at all.
Besides, we use the test $v>22$ to know whether the revision should be included even when it's $0$ (starting with Python 3.3.0).
edited yesterday
answered 2 days ago
ArnauldArnauld
81.6k797336
81.6k797336
2
$begingroup$
@KevinCruijssen semes like it doesn't work for2.0.1
$endgroup$
– ASCII-only
2 days ago
$begingroup$
Can you gain bytes by startingv
at 27 and going down to zero?
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire This saves 1 byte. Thanks!
$endgroup$
– Arnauld
yesterday
add a comment |
2
$begingroup$
@KevinCruijssen semes like it doesn't work for2.0.1
$endgroup$
– ASCII-only
2 days ago
$begingroup$
Can you gain bytes by startingv
at 27 and going down to zero?
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire This saves 1 byte. Thanks!
$endgroup$
– Arnauld
yesterday
2
2
$begingroup$
@KevinCruijssen semes like it doesn't work for
2.0.1
$endgroup$
– ASCII-only
2 days ago
$begingroup$
@KevinCruijssen semes like it doesn't work for
2.0.1
$endgroup$
– ASCII-only
2 days ago
$begingroup$
Can you gain bytes by starting
v
at 27 and going down to zero?$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
Can you gain bytes by starting
v
at 27 and going down to zero?$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire This saves 1 byte. Thanks!
$endgroup$
– Arnauld
yesterday
$begingroup$
@OlivierGrégoire This saves 1 byte. Thanks!
$endgroup$
– Arnauld
yesterday
add a comment |
$begingroup$
Java (JDK), 134 bytes
v->for(int a=0,b;;)for(b="0111131000244655:A002678;894".charAt(++a)-48;b-->0;)System.out.printf("%.1f%s ",a*.1+1,b<1&a<23?"":"."+b);
Try it online!
The versions are printed from the highest to the lowest.
Credits
- -1 byte thanks to Kevin Cruijssen
- -3 bytes thanks to Embodiment of Ignorance
$endgroup$
1
$begingroup$
(a>1|b>0)&c<a.valueOf(y,36)
can bea>1|b>0&&c<a.valueOf(y,36)
andc<1&(a<3|b<3)?
can bec<1&&a<3|b<3?
to save 2 bytes. Relevant Java tip - section Combining bit-wise and logical checks instead of using parenthesis
$endgroup$
– Kevin Cruijssen
yesterday
$begingroup$
@KevinCruijssen Thank you, but I made so many changes that your suggestions are now irrelevant... Don't know how to credit you since I don't even use your suggestion anymore :(
$endgroup$
– Olivier Grégoire
yesterday
1
$begingroup$
Np, instead I will suggest a new golf ;)/10d
can be*.1
$endgroup$
– Kevin Cruijssen
yesterday
1
$begingroup$
int a=28
->int a=1
, and remove the condition in the for loop, then add ana++
to save 3 bytes. TIO
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance Since REPL seem accepted in this challenge, that's indeed acceptable. Thank you!
$endgroup$
– Olivier Grégoire
yesterday
add a comment |
$begingroup$
Java (JDK), 134 bytes
v->for(int a=0,b;;)for(b="0111131000244655:A002678;894".charAt(++a)-48;b-->0;)System.out.printf("%.1f%s ",a*.1+1,b<1&a<23?"":"."+b);
Try it online!
The versions are printed from the highest to the lowest.
Credits
- -1 byte thanks to Kevin Cruijssen
- -3 bytes thanks to Embodiment of Ignorance
$endgroup$
1
$begingroup$
(a>1|b>0)&c<a.valueOf(y,36)
can bea>1|b>0&&c<a.valueOf(y,36)
andc<1&(a<3|b<3)?
can bec<1&&a<3|b<3?
to save 2 bytes. Relevant Java tip - section Combining bit-wise and logical checks instead of using parenthesis
$endgroup$
– Kevin Cruijssen
yesterday
$begingroup$
@KevinCruijssen Thank you, but I made so many changes that your suggestions are now irrelevant... Don't know how to credit you since I don't even use your suggestion anymore :(
$endgroup$
– Olivier Grégoire
yesterday
1
$begingroup$
Np, instead I will suggest a new golf ;)/10d
can be*.1
$endgroup$
– Kevin Cruijssen
yesterday
1
$begingroup$
int a=28
->int a=1
, and remove the condition in the for loop, then add ana++
to save 3 bytes. TIO
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance Since REPL seem accepted in this challenge, that's indeed acceptable. Thank you!
$endgroup$
– Olivier Grégoire
yesterday
add a comment |
$begingroup$
Java (JDK), 134 bytes
v->for(int a=0,b;;)for(b="0111131000244655:A002678;894".charAt(++a)-48;b-->0;)System.out.printf("%.1f%s ",a*.1+1,b<1&a<23?"":"."+b);
Try it online!
The versions are printed from the highest to the lowest.
Credits
- -1 byte thanks to Kevin Cruijssen
- -3 bytes thanks to Embodiment of Ignorance
$endgroup$
Java (JDK), 134 bytes
v->for(int a=0,b;;)for(b="0111131000244655:A002678;894".charAt(++a)-48;b-->0;)System.out.printf("%.1f%s ",a*.1+1,b<1&a<23?"":"."+b);
Try it online!
The versions are printed from the highest to the lowest.
Credits
- -1 byte thanks to Kevin Cruijssen
- -3 bytes thanks to Embodiment of Ignorance
edited yesterday
answered 2 days ago
Olivier GrégoireOlivier Grégoire
9,46511944
9,46511944
1
$begingroup$
(a>1|b>0)&c<a.valueOf(y,36)
can bea>1|b>0&&c<a.valueOf(y,36)
andc<1&(a<3|b<3)?
can bec<1&&a<3|b<3?
to save 2 bytes. Relevant Java tip - section Combining bit-wise and logical checks instead of using parenthesis
$endgroup$
– Kevin Cruijssen
yesterday
$begingroup$
@KevinCruijssen Thank you, but I made so many changes that your suggestions are now irrelevant... Don't know how to credit you since I don't even use your suggestion anymore :(
$endgroup$
– Olivier Grégoire
yesterday
1
$begingroup$
Np, instead I will suggest a new golf ;)/10d
can be*.1
$endgroup$
– Kevin Cruijssen
yesterday
1
$begingroup$
int a=28
->int a=1
, and remove the condition in the for loop, then add ana++
to save 3 bytes. TIO
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance Since REPL seem accepted in this challenge, that's indeed acceptable. Thank you!
$endgroup$
– Olivier Grégoire
yesterday
add a comment |
1
$begingroup$
(a>1|b>0)&c<a.valueOf(y,36)
can bea>1|b>0&&c<a.valueOf(y,36)
andc<1&(a<3|b<3)?
can bec<1&&a<3|b<3?
to save 2 bytes. Relevant Java tip - section Combining bit-wise and logical checks instead of using parenthesis
$endgroup$
– Kevin Cruijssen
yesterday
$begingroup$
@KevinCruijssen Thank you, but I made so many changes that your suggestions are now irrelevant... Don't know how to credit you since I don't even use your suggestion anymore :(
$endgroup$
– Olivier Grégoire
yesterday
1
$begingroup$
Np, instead I will suggest a new golf ;)/10d
can be*.1
$endgroup$
– Kevin Cruijssen
yesterday
1
$begingroup$
int a=28
->int a=1
, and remove the condition in the for loop, then add ana++
to save 3 bytes. TIO
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance Since REPL seem accepted in this challenge, that's indeed acceptable. Thank you!
$endgroup$
– Olivier Grégoire
yesterday
1
1
$begingroup$
(a>1|b>0)&c<a.valueOf(y,36)
can be a>1|b>0&&c<a.valueOf(y,36)
and c<1&(a<3|b<3)?
can be c<1&&a<3|b<3?
to save 2 bytes. Relevant Java tip - section Combining bit-wise and logical checks instead of using parenthesis$endgroup$
– Kevin Cruijssen
yesterday
$begingroup$
(a>1|b>0)&c<a.valueOf(y,36)
can be a>1|b>0&&c<a.valueOf(y,36)
and c<1&(a<3|b<3)?
can be c<1&&a<3|b<3?
to save 2 bytes. Relevant Java tip - section Combining bit-wise and logical checks instead of using parenthesis$endgroup$
– Kevin Cruijssen
yesterday
$begingroup$
@KevinCruijssen Thank you, but I made so many changes that your suggestions are now irrelevant... Don't know how to credit you since I don't even use your suggestion anymore :(
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@KevinCruijssen Thank you, but I made so many changes that your suggestions are now irrelevant... Don't know how to credit you since I don't even use your suggestion anymore :(
$endgroup$
– Olivier Grégoire
yesterday
1
1
$begingroup$
Np, instead I will suggest a new golf ;)
/10d
can be *.1
$endgroup$
– Kevin Cruijssen
yesterday
$begingroup$
Np, instead I will suggest a new golf ;)
/10d
can be *.1
$endgroup$
– Kevin Cruijssen
yesterday
1
1
$begingroup$
int a=28
->int a=1
, and remove the condition in the for loop, then add an a++
to save 3 bytes. TIO$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
int a=28
->int a=1
, and remove the condition in the for loop, then add an a++
to save 3 bytes. TIO$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance Since REPL seem accepted in this challenge, that's indeed acceptable. Thank you!
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@EmbodimentofIgnorance Since REPL seem accepted in this challenge, that's indeed acceptable. Thank you!
$endgroup$
– Olivier Grégoire
yesterday
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 109 bytes
for(int j,k=1;;k++)for(j=@" [SOH][SOH][SOH][SOH][ETX][SOH][NUL][NUL][NUL][STX][EOT][EOT][ACK][ENQ][ENQ]
[DC1][NUL][NUL][STX][ACK][BEL][BS][VT][BS][TAB][EOT]"[k];j-->0;)Print($"k*.1+1:N1"+(j<1&k<17?"":"."+j));
Contains many unprintables, whose codes are shown in the brackets. This is a full program. The null bytes are replaced by s in the TIO link, since my device can't copy and paste them.
Saved one byte thanks to @OlivierGregoire.
Try it online! (Thanks to @OlivierGregoire for implanting the null bytes)
Explanation
Each character in the string represents how many minor versions in the major position. For example, the character at index 5(ETX
) has an ASCII value of three, and corresponds to the major version 1.5.x
which has three minor versions. The program takes the ascii value of the current character and loops that many times, printing the minor versions before moving on to the next major version.
For some versions, there are gaps to the next versions. To fix that, the string contains null bytes, so that the program loops zero times when it encounters those.
The unprintable string contains these character values:
1,1,1,1,3,1,0,0,0,2,4,4,6,5,5,10,17,0,0,2,6,7,8,11,8,9,4
$endgroup$
$begingroup$
Can be shortened withj="..."[k];j-->0;
, especially since the order has no importance. Also, can you explain the difference in size between the TIO (115 bytes) and the entry (110 bytes)?
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire Probably the five null bytes which tio represent as
$endgroup$
– Sefa
yesterday
$begingroup$
@Sefa yes, probably... But I'm asking for certainty.
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire Exactly what Sefa said, I can't really copy-paste the null bytes. If thes were replaced by null bytes, it would be 110 bytes
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
Then, here you are, with a nul-byte TIO
$endgroup$
– Olivier Grégoire
yesterday
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 109 bytes
for(int j,k=1;;k++)for(j=@" [SOH][SOH][SOH][SOH][ETX][SOH][NUL][NUL][NUL][STX][EOT][EOT][ACK][ENQ][ENQ]
[DC1][NUL][NUL][STX][ACK][BEL][BS][VT][BS][TAB][EOT]"[k];j-->0;)Print($"k*.1+1:N1"+(j<1&k<17?"":"."+j));
Contains many unprintables, whose codes are shown in the brackets. This is a full program. The null bytes are replaced by s in the TIO link, since my device can't copy and paste them.
Saved one byte thanks to @OlivierGregoire.
Try it online! (Thanks to @OlivierGregoire for implanting the null bytes)
Explanation
Each character in the string represents how many minor versions in the major position. For example, the character at index 5(ETX
) has an ASCII value of three, and corresponds to the major version 1.5.x
which has three minor versions. The program takes the ascii value of the current character and loops that many times, printing the minor versions before moving on to the next major version.
For some versions, there are gaps to the next versions. To fix that, the string contains null bytes, so that the program loops zero times when it encounters those.
The unprintable string contains these character values:
1,1,1,1,3,1,0,0,0,2,4,4,6,5,5,10,17,0,0,2,6,7,8,11,8,9,4
$endgroup$
$begingroup$
Can be shortened withj="..."[k];j-->0;
, especially since the order has no importance. Also, can you explain the difference in size between the TIO (115 bytes) and the entry (110 bytes)?
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire Probably the five null bytes which tio represent as
$endgroup$
– Sefa
yesterday
$begingroup$
@Sefa yes, probably... But I'm asking for certainty.
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire Exactly what Sefa said, I can't really copy-paste the null bytes. If thes were replaced by null bytes, it would be 110 bytes
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
Then, here you are, with a nul-byte TIO
$endgroup$
– Olivier Grégoire
yesterday
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 109 bytes
for(int j,k=1;;k++)for(j=@" [SOH][SOH][SOH][SOH][ETX][SOH][NUL][NUL][NUL][STX][EOT][EOT][ACK][ENQ][ENQ]
[DC1][NUL][NUL][STX][ACK][BEL][BS][VT][BS][TAB][EOT]"[k];j-->0;)Print($"k*.1+1:N1"+(j<1&k<17?"":"."+j));
Contains many unprintables, whose codes are shown in the brackets. This is a full program. The null bytes are replaced by s in the TIO link, since my device can't copy and paste them.
Saved one byte thanks to @OlivierGregoire.
Try it online! (Thanks to @OlivierGregoire for implanting the null bytes)
Explanation
Each character in the string represents how many minor versions in the major position. For example, the character at index 5(ETX
) has an ASCII value of three, and corresponds to the major version 1.5.x
which has three minor versions. The program takes the ascii value of the current character and loops that many times, printing the minor versions before moving on to the next major version.
For some versions, there are gaps to the next versions. To fix that, the string contains null bytes, so that the program loops zero times when it encounters those.
The unprintable string contains these character values:
1,1,1,1,3,1,0,0,0,2,4,4,6,5,5,10,17,0,0,2,6,7,8,11,8,9,4
$endgroup$
C# (Visual C# Interactive Compiler), 109 bytes
for(int j,k=1;;k++)for(j=@" [SOH][SOH][SOH][SOH][ETX][SOH][NUL][NUL][NUL][STX][EOT][EOT][ACK][ENQ][ENQ]
[DC1][NUL][NUL][STX][ACK][BEL][BS][VT][BS][TAB][EOT]"[k];j-->0;)Print($"k*.1+1:N1"+(j<1&k<17?"":"."+j));
Contains many unprintables, whose codes are shown in the brackets. This is a full program. The null bytes are replaced by s in the TIO link, since my device can't copy and paste them.
Saved one byte thanks to @OlivierGregoire.
Try it online! (Thanks to @OlivierGregoire for implanting the null bytes)
Explanation
Each character in the string represents how many minor versions in the major position. For example, the character at index 5(ETX
) has an ASCII value of three, and corresponds to the major version 1.5.x
which has three minor versions. The program takes the ascii value of the current character and loops that many times, printing the minor versions before moving on to the next major version.
For some versions, there are gaps to the next versions. To fix that, the string contains null bytes, so that the program loops zero times when it encounters those.
The unprintable string contains these character values:
1,1,1,1,3,1,0,0,0,2,4,4,6,5,5,10,17,0,0,2,6,7,8,11,8,9,4
edited yesterday
answered 2 days ago
Embodiment of IgnoranceEmbodiment of Ignorance
3,024127
3,024127
$begingroup$
Can be shortened withj="..."[k];j-->0;
, especially since the order has no importance. Also, can you explain the difference in size between the TIO (115 bytes) and the entry (110 bytes)?
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire Probably the five null bytes which tio represent as
$endgroup$
– Sefa
yesterday
$begingroup$
@Sefa yes, probably... But I'm asking for certainty.
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire Exactly what Sefa said, I can't really copy-paste the null bytes. If thes were replaced by null bytes, it would be 110 bytes
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
Then, here you are, with a nul-byte TIO
$endgroup$
– Olivier Grégoire
yesterday
add a comment |
$begingroup$
Can be shortened withj="..."[k];j-->0;
, especially since the order has no importance. Also, can you explain the difference in size between the TIO (115 bytes) and the entry (110 bytes)?
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire Probably the five null bytes which tio represent as
$endgroup$
– Sefa
yesterday
$begingroup$
@Sefa yes, probably... But I'm asking for certainty.
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire Exactly what Sefa said, I can't really copy-paste the null bytes. If thes were replaced by null bytes, it would be 110 bytes
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
Then, here you are, with a nul-byte TIO
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
Can be shortened with
j="..."[k];j-->0;
, especially since the order has no importance. Also, can you explain the difference in size between the TIO (115 bytes) and the entry (110 bytes)?$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
Can be shortened with
j="..."[k];j-->0;
, especially since the order has no importance. Also, can you explain the difference in size between the TIO (115 bytes) and the entry (110 bytes)?$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire Probably the five null bytes which tio represent as
$endgroup$
– Sefa
yesterday
$begingroup$
@OlivierGrégoire Probably the five null bytes which tio represent as
$endgroup$
– Sefa
yesterday
$begingroup$
@Sefa yes, probably... But I'm asking for certainty.
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@Sefa yes, probably... But I'm asking for certainty.
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
@OlivierGrégoire Exactly what Sefa said, I can't really copy-paste the null bytes. If the
s were replaced by null bytes, it would be 110 bytes$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@OlivierGrégoire Exactly what Sefa said, I can't really copy-paste the null bytes. If the
s were replaced by null bytes, it would be 110 bytes$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
Then, here you are, with a nul-byte TIO
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
Then, here you are, with a nul-byte TIO
$endgroup$
– Olivier Grégoire
yesterday
add a comment |
$begingroup$
Retina, 105 bytes
11* 111131 244655TS 2678E894
L$`.
$&_$.`
T
10
E
11
S
17
.+_
*
Lv$`_+(.)(.)
$1.$2.$.%`
,16`(...).0
$1
Try it online! Loosely based on @Arnauld's solution. Explanation:
11* 111131 244655TS 2678E894
Insert the string consisting of 11 spaces followed by the given characters.
L$`.
$&_$.`
For each character, list it suffixed with a _
and its column number.
T
10
E
11
S
17
Convert the three letters to numeric values.
.+_
*
Convert the numeric values to unary.
Lv$`_+(.)(.)
$1.$2.$.%`
For each value up to the given value, use that as the suffix for the version number, extracting the major and minor from the column number.
,16`(...).0
$1
Delete the zero suffix for the first 16 versions that have one.
$endgroup$
add a comment |
$begingroup$
Retina, 105 bytes
11* 111131 244655TS 2678E894
L$`.
$&_$.`
T
10
E
11
S
17
.+_
*
Lv$`_+(.)(.)
$1.$2.$.%`
,16`(...).0
$1
Try it online! Loosely based on @Arnauld's solution. Explanation:
11* 111131 244655TS 2678E894
Insert the string consisting of 11 spaces followed by the given characters.
L$`.
$&_$.`
For each character, list it suffixed with a _
and its column number.
T
10
E
11
S
17
Convert the three letters to numeric values.
.+_
*
Convert the numeric values to unary.
Lv$`_+(.)(.)
$1.$2.$.%`
For each value up to the given value, use that as the suffix for the version number, extracting the major and minor from the column number.
,16`(...).0
$1
Delete the zero suffix for the first 16 versions that have one.
$endgroup$
add a comment |
$begingroup$
Retina, 105 bytes
11* 111131 244655TS 2678E894
L$`.
$&_$.`
T
10
E
11
S
17
.+_
*
Lv$`_+(.)(.)
$1.$2.$.%`
,16`(...).0
$1
Try it online! Loosely based on @Arnauld's solution. Explanation:
11* 111131 244655TS 2678E894
Insert the string consisting of 11 spaces followed by the given characters.
L$`.
$&_$.`
For each character, list it suffixed with a _
and its column number.
T
10
E
11
S
17
Convert the three letters to numeric values.
.+_
*
Convert the numeric values to unary.
Lv$`_+(.)(.)
$1.$2.$.%`
For each value up to the given value, use that as the suffix for the version number, extracting the major and minor from the column number.
,16`(...).0
$1
Delete the zero suffix for the first 16 versions that have one.
$endgroup$
Retina, 105 bytes
11* 111131 244655TS 2678E894
L$`.
$&_$.`
T
10
E
11
S
17
.+_
*
Lv$`_+(.)(.)
$1.$2.$.%`
,16`(...).0
$1
Try it online! Loosely based on @Arnauld's solution. Explanation:
11* 111131 244655TS 2678E894
Insert the string consisting of 11 spaces followed by the given characters.
L$`.
$&_$.`
For each character, list it suffixed with a _
and its column number.
T
10
E
11
S
17
Convert the three letters to numeric values.
.+_
*
Convert the numeric values to unary.
Lv$`_+(.)(.)
$1.$2.$.%`
For each value up to the given value, use that as the suffix for the version number, extracting the major and minor from the column number.
,16`(...).0
$1
Delete the zero suffix for the first 16 versions that have one.
answered 2 days ago
NeilNeil
83k745179
83k745179
add a comment |
add a comment |
$begingroup$
Pyth, 52 bytes
.emj.+W|d>k18,h/k8%k8dbxLG"abbbbdbaceegffkrcghilije
Try it online here.
Output is a nested list, with elements grouped by major and minor version. There is an empty list at the start of the output, and another one after 1.6
. Full output is as follows:
[[], ['1.1'], ['1.2'], ['1.3'], ['1.4'], ['1.5', '1.5.1', '1.5.2'], ['1.6'], [], ['2.0', '2.0.1'], ['2.1', '2.1.1', '2.1.2', '2.1.3'], ['2.2', '2.2.1', '2.2.2', '2.2.3'], ['2.3', '2.3.1', '2.3.2', '2.3.3', '2.3.4', '2.3.5'], ['2.4', '2.4.1', '2.4.2', '2.4.3', '2.4.4'], ['2.5', '2.5.1', '2.5.2', '2.5.3', '2.5.4'], ['2.6', '2.6.1', '2.6.2', '2.6.3', '2.6.4', '2.6.5', '2.6.6', '2.6.7', '2.6.8', '2.6.9'], ['2.7', '2.7.1', '2.7.2', '2.7.3', '2.7.4', '2.7.5', '2.7.6', '2.7.7', '2.7.8', '2.7.9', '2.7.10', '2.7.11', '2.7.12', '2.7.13', '2.7.14', '2.7.15', '2.7.16'], ['3.0', '3.0.1'], ['3.1', '3.1.1', '3.1.2', '3.1.3', '3.1.4', '3.1.5'], ['3.2', '3.2.1', '3.2.2', '3.2.3', '3.2.4', '3.2.5', '3.2.6'], ['3.3.0', '3.3.1', '3.3.2', '3.3.3', '3.3.4', '3.3.5', '3.3.6', '3.3.7'], ['3.4.0', '3.4.1', '3.4.2', '3.4.3', '3.4.4', '3.4.5', '3.4.6', '3.4.7', '3.4.8', '3.4.9', '3.4.10'], ['3.5.0', '3.5.1', '3.5.2', '3.5.3', '3.5.4', '3.5.5', '3.5.6', '3.5.7'], ['3.6.0', '3.6.1', '3.6.2', '3.6.3', '3.6.4', '3.6.5', '3.6.6', '3.6.7', '3.6.8'], ['3.7.0', '3.7.1', '3.7.2', '3.7.3']]
If this isn't acceptable, prepend .n
to the code to have output as a flattened list, at a cost of 2 bytes.
$endgroup$
add a comment |
$begingroup$
Pyth, 52 bytes
.emj.+W|d>k18,h/k8%k8dbxLG"abbbbdbaceegffkrcghilije
Try it online here.
Output is a nested list, with elements grouped by major and minor version. There is an empty list at the start of the output, and another one after 1.6
. Full output is as follows:
[[], ['1.1'], ['1.2'], ['1.3'], ['1.4'], ['1.5', '1.5.1', '1.5.2'], ['1.6'], [], ['2.0', '2.0.1'], ['2.1', '2.1.1', '2.1.2', '2.1.3'], ['2.2', '2.2.1', '2.2.2', '2.2.3'], ['2.3', '2.3.1', '2.3.2', '2.3.3', '2.3.4', '2.3.5'], ['2.4', '2.4.1', '2.4.2', '2.4.3', '2.4.4'], ['2.5', '2.5.1', '2.5.2', '2.5.3', '2.5.4'], ['2.6', '2.6.1', '2.6.2', '2.6.3', '2.6.4', '2.6.5', '2.6.6', '2.6.7', '2.6.8', '2.6.9'], ['2.7', '2.7.1', '2.7.2', '2.7.3', '2.7.4', '2.7.5', '2.7.6', '2.7.7', '2.7.8', '2.7.9', '2.7.10', '2.7.11', '2.7.12', '2.7.13', '2.7.14', '2.7.15', '2.7.16'], ['3.0', '3.0.1'], ['3.1', '3.1.1', '3.1.2', '3.1.3', '3.1.4', '3.1.5'], ['3.2', '3.2.1', '3.2.2', '3.2.3', '3.2.4', '3.2.5', '3.2.6'], ['3.3.0', '3.3.1', '3.3.2', '3.3.3', '3.3.4', '3.3.5', '3.3.6', '3.3.7'], ['3.4.0', '3.4.1', '3.4.2', '3.4.3', '3.4.4', '3.4.5', '3.4.6', '3.4.7', '3.4.8', '3.4.9', '3.4.10'], ['3.5.0', '3.5.1', '3.5.2', '3.5.3', '3.5.4', '3.5.5', '3.5.6', '3.5.7'], ['3.6.0', '3.6.1', '3.6.2', '3.6.3', '3.6.4', '3.6.5', '3.6.6', '3.6.7', '3.6.8'], ['3.7.0', '3.7.1', '3.7.2', '3.7.3']]
If this isn't acceptable, prepend .n
to the code to have output as a flattened list, at a cost of 2 bytes.
$endgroup$
add a comment |
$begingroup$
Pyth, 52 bytes
.emj.+W|d>k18,h/k8%k8dbxLG"abbbbdbaceegffkrcghilije
Try it online here.
Output is a nested list, with elements grouped by major and minor version. There is an empty list at the start of the output, and another one after 1.6
. Full output is as follows:
[[], ['1.1'], ['1.2'], ['1.3'], ['1.4'], ['1.5', '1.5.1', '1.5.2'], ['1.6'], [], ['2.0', '2.0.1'], ['2.1', '2.1.1', '2.1.2', '2.1.3'], ['2.2', '2.2.1', '2.2.2', '2.2.3'], ['2.3', '2.3.1', '2.3.2', '2.3.3', '2.3.4', '2.3.5'], ['2.4', '2.4.1', '2.4.2', '2.4.3', '2.4.4'], ['2.5', '2.5.1', '2.5.2', '2.5.3', '2.5.4'], ['2.6', '2.6.1', '2.6.2', '2.6.3', '2.6.4', '2.6.5', '2.6.6', '2.6.7', '2.6.8', '2.6.9'], ['2.7', '2.7.1', '2.7.2', '2.7.3', '2.7.4', '2.7.5', '2.7.6', '2.7.7', '2.7.8', '2.7.9', '2.7.10', '2.7.11', '2.7.12', '2.7.13', '2.7.14', '2.7.15', '2.7.16'], ['3.0', '3.0.1'], ['3.1', '3.1.1', '3.1.2', '3.1.3', '3.1.4', '3.1.5'], ['3.2', '3.2.1', '3.2.2', '3.2.3', '3.2.4', '3.2.5', '3.2.6'], ['3.3.0', '3.3.1', '3.3.2', '3.3.3', '3.3.4', '3.3.5', '3.3.6', '3.3.7'], ['3.4.0', '3.4.1', '3.4.2', '3.4.3', '3.4.4', '3.4.5', '3.4.6', '3.4.7', '3.4.8', '3.4.9', '3.4.10'], ['3.5.0', '3.5.1', '3.5.2', '3.5.3', '3.5.4', '3.5.5', '3.5.6', '3.5.7'], ['3.6.0', '3.6.1', '3.6.2', '3.6.3', '3.6.4', '3.6.5', '3.6.6', '3.6.7', '3.6.8'], ['3.7.0', '3.7.1', '3.7.2', '3.7.3']]
If this isn't acceptable, prepend .n
to the code to have output as a flattened list, at a cost of 2 bytes.
$endgroup$
Pyth, 52 bytes
.emj.+W|d>k18,h/k8%k8dbxLG"abbbbdbaceegffkrcghilije
Try it online here.
Output is a nested list, with elements grouped by major and minor version. There is an empty list at the start of the output, and another one after 1.6
. Full output is as follows:
[[], ['1.1'], ['1.2'], ['1.3'], ['1.4'], ['1.5', '1.5.1', '1.5.2'], ['1.6'], [], ['2.0', '2.0.1'], ['2.1', '2.1.1', '2.1.2', '2.1.3'], ['2.2', '2.2.1', '2.2.2', '2.2.3'], ['2.3', '2.3.1', '2.3.2', '2.3.3', '2.3.4', '2.3.5'], ['2.4', '2.4.1', '2.4.2', '2.4.3', '2.4.4'], ['2.5', '2.5.1', '2.5.2', '2.5.3', '2.5.4'], ['2.6', '2.6.1', '2.6.2', '2.6.3', '2.6.4', '2.6.5', '2.6.6', '2.6.7', '2.6.8', '2.6.9'], ['2.7', '2.7.1', '2.7.2', '2.7.3', '2.7.4', '2.7.5', '2.7.6', '2.7.7', '2.7.8', '2.7.9', '2.7.10', '2.7.11', '2.7.12', '2.7.13', '2.7.14', '2.7.15', '2.7.16'], ['3.0', '3.0.1'], ['3.1', '3.1.1', '3.1.2', '3.1.3', '3.1.4', '3.1.5'], ['3.2', '3.2.1', '3.2.2', '3.2.3', '3.2.4', '3.2.5', '3.2.6'], ['3.3.0', '3.3.1', '3.3.2', '3.3.3', '3.3.4', '3.3.5', '3.3.6', '3.3.7'], ['3.4.0', '3.4.1', '3.4.2', '3.4.3', '3.4.4', '3.4.5', '3.4.6', '3.4.7', '3.4.8', '3.4.9', '3.4.10'], ['3.5.0', '3.5.1', '3.5.2', '3.5.3', '3.5.4', '3.5.5', '3.5.6', '3.5.7'], ['3.6.0', '3.6.1', '3.6.2', '3.6.3', '3.6.4', '3.6.5', '3.6.6', '3.6.7', '3.6.8'], ['3.7.0', '3.7.1', '3.7.2', '3.7.3']]
If this isn't acceptable, prepend .n
to the code to have output as a flattened list, at a cost of 2 bytes.
edited 2 days ago
answered 2 days ago
SokSok
4,237925
4,237925
add a comment |
add a comment |
$begingroup$
Jelly, 51 bytes
+⁵D;ⱮḶ}j€”.
“øṄƇịɱ⁽Ɱj>⁶7,Ẉ¢’b18Ė0ị$Ƈç/€ḣ3$€1¦€17R¤¦
Try it online!
A niladic link that outputs a list of lists of .
-separated integers, grouped by major version. On TIO, there’s some footer code to print these prettily.
$endgroup$
add a comment |
$begingroup$
Jelly, 51 bytes
+⁵D;ⱮḶ}j€”.
“øṄƇịɱ⁽Ɱj>⁶7,Ẉ¢’b18Ė0ị$Ƈç/€ḣ3$€1¦€17R¤¦
Try it online!
A niladic link that outputs a list of lists of .
-separated integers, grouped by major version. On TIO, there’s some footer code to print these prettily.
$endgroup$
add a comment |
$begingroup$
Jelly, 51 bytes
+⁵D;ⱮḶ}j€”.
“øṄƇịɱ⁽Ɱj>⁶7,Ẉ¢’b18Ė0ị$Ƈç/€ḣ3$€1¦€17R¤¦
Try it online!
A niladic link that outputs a list of lists of .
-separated integers, grouped by major version. On TIO, there’s some footer code to print these prettily.
$endgroup$
Jelly, 51 bytes
+⁵D;ⱮḶ}j€”.
“øṄƇịɱ⁽Ɱj>⁶7,Ẉ¢’b18Ė0ị$Ƈç/€ḣ3$€1¦€17R¤¦
Try it online!
A niladic link that outputs a list of lists of .
-separated integers, grouped by major version. On TIO, there’s some footer code to print these prettily.
edited yesterday
answered yesterday
Nick KennedyNick Kennedy
1,82149
1,82149
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f183284%2flist-of-python-versions%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
I guess we aren't allowed to output
1.1.0
(to make all versions 3 numbers) instead of1.1
?$endgroup$
– Kevin Cruijssen
2 days ago
2
$begingroup$
You guess correctly @Kevin. I considered allowing it, but went with the official names instead.
$endgroup$
– Stewie Griffin
2 days ago