What do the two equations do in the image? Please explain in plain English and with Python or Java example code (or any programming language) if possible and/or provide usage examples with step-by-step explanations for the solution(s). Thanks in advance.

Operations Research : Applications and Algorithms
4th Edition
ISBN:9780534380588
Author:Wayne L. Winston
Publisher:Wayne L. Winston
Chapter17: Markov Chains
Section17.5: Steady-state Probabilities And Mean First Passage Times
Problem 6P
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What do the two equations do in the image? Please explain in plain English and with Python or Java example code (or any programming language) if possible and/or provide usage examples with step-by-step explanations for the solution(s). Thanks in advance.
6
is the total number of colorings of the current guess. For a fixed action a and state st, P(s,, a)
can be constructed by taking guess a conditioned on each potential solution w€ s being the
solution, and recording what the next-state s+1 is. The associated transition probabilities are
computed by counting the number of times a particular next-state st+1 appears while iterating
through w Est, and dividing this count by [s]. For a particular st+1 € P(st, a), we denote this
transition probability as p. (st+1; 8,a).
Given this formulation of Wordle as a Markov Decision Process, we can apply the Bellman equation
to solve for the optimal value of a state via:
Bertsimas and Paskov: An Ezact and Interpretable Solution to Wordle
Vi (st) = mine(st, a) +
which can more explicitly be written as
02+1EP(a)
Σpi(8+18, a) Vi+1(S1+1)
2. Vers(0)}
ot+1EP (st,a)
V (8₁)= min 1+ pe(St+1; S₁, a)V₁+1(8+1)
aEA
if t < 6, se=1,
if t < 6, |st| > 1,
if t = 6.
(1)
Overview of The Algorithm
Firstly, observe that performing traditional back-propagation by enumerating all possible states at
successively earlier times is computationally intractable, as the number of possible states at each
time is exponential in the number of solutions that is, 2231510697. As such, we instead solve the
equation for Vo* (so) directly via recursion, in which so is the starting state with all 2,315 solutions
are present. Specifically, in evaluating V (so), we iteratively search through the different guesses in
A, and recursively solve V+1(8+1) for states that can be transitioned into. Note that while this still
involves a significant amount of computation (because of the large action space and large number of
possible states), a number of important computational optimizations make the problem feasible to
solve, which we outline in the following section.
Finally, we note that a nice property of this algorithm is that it is trivially parallelizable. In
particular, we can enumerate all possible next-states $₁+1 from so for any a € A, and evaluate the
value function of these next-states across many different processes.
Scaling The Algorithm
In this section, we enumerate a number of observations that lead to significant speed-ups in solving
the problem.
Transcribed Image Text:6 is the total number of colorings of the current guess. For a fixed action a and state st, P(s,, a) can be constructed by taking guess a conditioned on each potential solution w€ s being the solution, and recording what the next-state s+1 is. The associated transition probabilities are computed by counting the number of times a particular next-state st+1 appears while iterating through w Est, and dividing this count by [s]. For a particular st+1 € P(st, a), we denote this transition probability as p. (st+1; 8,a). Given this formulation of Wordle as a Markov Decision Process, we can apply the Bellman equation to solve for the optimal value of a state via: Bertsimas and Paskov: An Ezact and Interpretable Solution to Wordle Vi (st) = mine(st, a) + which can more explicitly be written as 02+1EP(a) Σpi(8+18, a) Vi+1(S1+1) 2. Vers(0)} ot+1EP (st,a) V (8₁)= min 1+ pe(St+1; S₁, a)V₁+1(8+1) aEA if t < 6, se=1, if t < 6, |st| > 1, if t = 6. (1) Overview of The Algorithm Firstly, observe that performing traditional back-propagation by enumerating all possible states at successively earlier times is computationally intractable, as the number of possible states at each time is exponential in the number of solutions that is, 2231510697. As such, we instead solve the equation for Vo* (so) directly via recursion, in which so is the starting state with all 2,315 solutions are present. Specifically, in evaluating V (so), we iteratively search through the different guesses in A, and recursively solve V+1(8+1) for states that can be transitioned into. Note that while this still involves a significant amount of computation (because of the large action space and large number of possible states), a number of important computational optimizations make the problem feasible to solve, which we outline in the following section. Finally, we note that a nice property of this algorithm is that it is trivially parallelizable. In particular, we can enumerate all possible next-states $₁+1 from so for any a € A, and evaluate the value function of these next-states across many different processes. Scaling The Algorithm In this section, we enumerate a number of observations that lead to significant speed-ups in solving the problem.
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