"Prosas march" stands for probability solver of absorbing states in Markov chain. This is my solution for an online programming game.
As the title clearly indicates, the simple code in this repository provides a method to compute probabilities for the absorbing states in a given Markov chain. This solution will process matrices of fractions in order to avoid float approximations, for this reason it doesn't use external libraries such as numpy.
The method prosasmarch.solve(n) gets a list of list m as input, representing the transition matrix of a Markov Chain, and returns an array containing the probability of each absorbing state or None if there is no abs. state. If there are K absorbing states, the returned array will contain K+1 integer numbers, where the first K items are the numerators of the abs. states' probabilities, and the last one is the denominator
Things to keep in mind
- An absorbing state with transition prob. of 1.0 towards itself is equivalent to a terminal state.
- The first state (first row of the input) is considered as the starting one.
Some simple examples. Also, try the demo.py script.
Let's consider the following Markov chain:
The state s0 is considered the starting state, while the absorbing states are: s2, s3 and s4.
To solve this chain let's call:
prosasmarch.solve([[0, 2, 1, 0, 0], [0, 0, 0, 3, 4], [0, 0, 0, 0, 0], [0, 0, 0, 0,0], [0, 0, 0, 0, 0]])The output will be:
[7, 6, 8, 21]
which means that the probabilities for each absorbing state to be reached from s0 are: 7/21 for s2, 6/21 for s3 and 8/21 for s4.
Let's consider the following Markov chain:
The state s0 is the starting state, while the absorbing states are: s2, s3, s4 and s5. Please, note that s2 is an unreachable state.
To solve this chain let's call:
prosasmarch.solve([[0, 1, 0, 0, 0, 1], [4, 0, 0, 3, 2, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]])The output will be:
[0, 3, 2, 9, 14]
which means that the probabilities for each absorbing state to be reached from s0 are: 0 for s2 (unreachable), 3/14 for s3, 2/14 for s4 and 9/14 for s5.
Copyright(c) 2020 Sebastiano Campisi - ianovir.com. Read the LICENSE file for more details.