command-line combinatorial calculator
Command line Combinatorial Calculator for Counting Constrained Collections.
ccc is a calculator that can:
- tell you the probability of possible ways a certain collection of items can meet one or more constraints
- count the number of possible ways a certain collection of items can meet one or more constraints
For example, consider the following problem (asked on stats.stackexchange.com):
There are 232 tickets for an event. 363 people apply for a ticket, 12 of whom are from a particular group (so 351 are not from the group). Each ticket is allocated to one person at random and each person can recieve at most one ticket. What is the probability that at most 2 tickets are given to people in the group?
ccc makes it easy to translate the problem into a single command, handling the tedious computational details from you:
$ ccc probability draw --from 'group=12; rest=351' \ --number 232 \ --constraints 'town <= 2' \ --float 0.00093
That's roughly 1 in 10000.
Or suppose we're designing a Magic: The Gathering deck:
A deck of 60 cards contains 13 mountain cards and 12 swamp cards. What is the probability that we draw 7 cards and get between 1 and 3 mountains and exactly 2 swamps?
Using ccc, we can write these constraints very easily:
$ ccc prob draw --from 'mountain=13; swamp=12; rest=35' \ --number 7 \ --constraints '1 <= mountain <= 3, swamp == 2' \ --float 0.23264
These types of problem are common in everyday life but, depending on how familiar you are with statistics or combinatorics, they can be hard to calculate or even estimate accurately.
See below for numerous other examples and explanations.
Installation requires Python 3.6 or newer.
pip install ccc-calculator
Clone the repo, create a virtual environment, install the development dependencies, install the module in editable mode, install pre-commit hooks:
git clone https://github.com/ajcr/ccc cd ccc python -m venv venv . venv/bin/activate python -m pip requirements-dev.txt python -m pip install -e . pre-commit install
Unit tests can be run with
pytest. Pre-commit hooks will run on each git commit. The hooks can also be invoked with
pre-commit run -a.
The easiest way to introduce ccc is to show various example calculations from the command-line.
ccc is always invoked in the following way:
ccc [computation-type] [collection-type] [--args]
In all of the examples below, notice how easy it is to express constraints that would otherwise necessitate writing complicated code, or performing repetitive arithmetic on paper.
ccc can tell you the probability of selecting particular counts of items from a collection (either with or without replacement).
Here is an example:
A bag contains:
- 3 red marbles (:red_circle::red_circle::red_circle:)
- 5 black marbles (:black_circle::black_circle::black_circle::black_circle::black_circle:)
- 7 blue marbles (:large_blue_circle::large_blue_circle::large_blue_circle::large_blue_circle::large_blue_circle::large_blue_circle::large_blue_circle:)
You must draw at random (and without replacement) 4 of these marbles. You lose if you draw 1 or more of the blue marbles (:blue_circle:).
What is your probability of winning?
To solve this with ccc we can easily specify the collection we draw from, the size of the draw we make, and any constraints on the draw:
ccc probability draw --from "red=3; black=5; blue=5" \ --number 4 \ --constraints "blue == 0"
This puts the probability of winning (not drawing a blue marble) at 2/39, so perhaps we'd win once every 19 or so attempts.
Notice how easy it is to specify the constraints. Just the item's name (blue) and its desired count (0). Any comparison operators can be used (
If we wanted to allow a marble to be replaced after each draw, we would add the
We can specify more complicated constraints on what we want to draw from the bag:
This time, draw 5 marbles without replacement from the same collection, but with 2 white marbles (:white_circle::white_circle:) added.
You win a toy if you draw a collection which comprises
- both white marbles (:white_circle::white_circle:), and
- at least 1 black marble (:black_circle:+)
What is you chance of winning the toy?
ccc probability draw --from "red=3; black=5; blue=7; white=2" \ --number 5 \ --constraints "white == 2, black >= 1"
Our chance of winning is 3/119 according to ccc, so we would expect to win around once every 40 attempts.
You can see that to express multiple constraints on our draw, we simply used commas
, to separate them.
Lastly, it is possible to use
or (any number of times) in constraints:
Draw 3 marbles such that you get:
- 3 red marbles (:red_circle::red_circle::red_circle:) or
- 1 white, 1 black and 1 blue marble (:white_circle::black_circle::large_blue_circle:)
What is the probability of succeeding now?
ccc probability draw --from "red=3; black=5; blue=7; white=2" \ --number 3 \ --constraints "red == 3 or (white == 1, black == 1, blue == 1)"
The probability is 0.1.
Multisets are unordered collections (like sets) in which an item may appear multiple times.
How many ways can we gather up 20 pieces of fruit such that we have:
- fewer than 10 apples (:green_apple:), and
- at least 5 bananas (:banana:), and
- the number of grapes is not 13 (:grapes:), and
- there are are even number of strawberries (:strawberry:)?
ccc count multisets --size 20 \ --constraints 'apples < 10, bananas >= 5, grapes != 13, strawberries % 2 == 0'
The answer is 406.
The modulo (
%) operator used above provides the facility to solve coin change problems, for example:
The UK currency has the following coins:
1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p)
How many different ways can £5 be made using any number of coins?
ccc count multisets --constraints 'a%1==0, b%2==0, c%5==0, d%10==0, e%20==0, f%50==0, g%100==0, h%200==0' \ --size 500
The answer is computed within a couple of seconds as 6,295,434 different ways.
We could also put additional contraints on the coins, such as restricting the number of times a coin may be used.
Permutations of words can be counted as follows:
ccc count permutations MISSISSIPPI
There are 34650 unique permutations of this famous river/state.
What about if we only count permutations where instances of the same letter are not adjacent to each other?
ccc count permutations MISSISSIPPI --constraints no_adjacent
Using the 'no_adjacent' constraint, the answer comes back immediately as 2016. The speed of this calculation is thanks to the use of Generalised Larguerre Polynomials.
We can also treat the letters as distinguishable (the
--same-distinct flag) to solve a related type of problem:
Five couples are to seated in a row. How many ways can they be seated such no couples are seated together?
ccc count permutations AABBCCDDEE --constraints no_adjacent --same-distinct
ccc allows derangements (permutations where no item occupies its original place) to be counted.
Five couples draw names from a hat. What is the probability that nobody draws either their own name, or the name of their partner?
ccc probability permutation AABBCCDDEE --constraints derangement --float
It turns out that there is only a 0.121 probability of this occurring.
Sequences are ordered collections of items.
How many 30 letter sequences can you make using no more than 20 each of A, B and C?
ccc count sequences --size 30 --constraints 'A <= 20, B <= 20, C <= 20'
The answer is a lot: there are 205,863,750,414,990 such sequences.
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