# Computer LEARNS Tic-Tac-Toe

## 1. What is Tic Tac Toe ?

A fairly simple game which most of us have been playing since childhood. In case you are not familiar with the game you can read more about it here or watch this short video.

The simplicity of the game makes it very suitable for pedagogical purposes, that is one of the reason why we find references to Tic-Tac-Toe in Artificial Intelligence literature.

We will discuss about making the computer learn to play Tic-Tac-Toe. We will start with few basic deterministic ways of solving Tic-Tac-Toe and slowly move towards making the computer LEARN playing Tic-Tac-Toe.

At times we might have come across the strategy of putting X or O at the center, considering it a safe move one can simply put X or O at the center and expect to at least draw the game. We will also examine some of our own strategies like this while playing Tic-Tac-Toe and discuss whether a computer can learn these strategies.

## 2. ROTE Learning way (or the NOT so learning way)

One unique feature of this game is that it has less number of possible game states. A game state is a description of the positions of X’s and O’s at a given point in time in the game.

So the initial game state looks like the empty game figure. The next game state look like the figure corresponding to one of the possible game moves made by the first player, say X. Although there are 9 possible moves of which only 3 are distinct if we consider symmetry. All these game states can be depicted in form of a game tree . Game trees allow us to enumerate all possible scenarios in the game. Take look at the figure below for game tree associated with Tic-Tac-Toe.

There are some states which are losing and some other states which are winning. You might have guessed the meaning of winning and losing here, but let’s define it more rigorously. A state is called winning for $X$ if either of the following holds true:

1. Player $X$ has won the game. (the obvious wining state)
2. There exists NO such path in the tree (sub-tree) starting from the given state which will NOT land player $X$ in an obvious winning state as described in 1. (Or we can say every path will land player $X$ into an obvious winning state)

In an analogous way we can define winning states for O’s. Similarly we can also defining a losing state. Now you can try defining losing state precisely by yourself.

### 2.1 The ROTE Learning approach

As the name suggest this is a very straight forward way of solving this problem. Make a tree and figure out all winning states and losing states for a given player. If it is a winning state give a high score to the player, if it is a losing state give a penalizing score and if it is neither (draw) then give a zero.

Algorithm 2.1.1

GetStateScore(state, player):

1. GameTree $\leftarrow$ A representation of game tree.
2. If state is winning for player in GameTree:
• Then return 100
• Else If state is losing for player in GameTree:
• Then return -100
• Else
• return 0

The actual difficulty in the above algorithm is figuring out if a state is winning or not. One possible way is to use the game tree itself and ensure that all paths in all subtrees starting from the given state ends up making the given player win. This is again relatively straight forward if you are familiar with graph theory and understand BFS or DFS.

## 3. The Minimax way

If you have heard of the minimax algorithm, then this is probably one of the most widely used algorithms to solve this type of problem. The core idea behind this class of algorithms is to make sure that you always consider your opponents best possible move before making yours. As the name suggests, Min-Max, we find the minimum of the maximum possible score of your opponent or in other words minimize your damage in worst case.

There is another version of the same algorithm which is maximizing the minimum gain. In case of Tic-Tac-Toe we try to maximize the minimum gain. This allows us to consider the opponent’s optimal moves.

Algorithm 3.1.1

GetMaxScore(currentState, currentPlayer):

1. If currentState is winning for currentPlayer:
• Then return 100
2. If currentState is losing for currentPlayer:
• Then return -100
3. If currentState is a draw:
• Then return 0
4. Else
• nextPlayer $\leftarrow$ Get next player from currentPlayer
• possibleStates $\leftarrow$ Get all possible next states from the currentState
• return $$\max_{\forall \; s \; \in \; possibleStates}(-GetMaxScore(s, nextPlayer))$$

The algorithm 3.1.1 works because it considers the negative scores for the other player, effectively maximizing the minimum gains. Below we have a working example in python for the above algorithm.

import math
import copy
import typing
import random

def get_winner(state) -> str:
for i in range(3):
if state[i][0] == state[i][1] == state[i][2] and \
(state[i][0] == 'O' or state[i][0] == 'X'):
return state[i][0]
if state[0][i] == state[1][i] == state[2][i] and \
(state[0][i] == 'O' or state[0][i] == 'X'):
return state[0][i]
if state[0][0] == state[1][1] == state[2][2] and \
(state[0][0] == 'O' or state[0][0] == 'X'):
return state[0][0]
if state[0][2] == state[1][1] == state[2][0] and \
(state[0][2] == 'O' or state[0][2] == 'X'):
return state[0][2]
return '_'

def flip_player(player):
if player == 'X':
return 'O'
else:
return 'X'

class MinMaxPlayer:
cache = [None]*(2**19) # Eases the recursion, creating big enough cache to avoid resizing
def __init__(self, symbol : str):
self.symbol = symbol

def get_score(self, i, j, state): # Gets the score if player with symbol makes a move at i, j
return MinMaxPlayer._get_score(0, i, j, self.symbol, state)

@staticmethod
def _get_score(move, i, j, player, state):
state = copy.deepcopy(state)
state[i][j] = player
hash = MinMaxPlayer._get_hash(player, state)
prev = MinMaxPlayer.cache[hash]
if prev != None:
return prev # Don't recompute score if already computed
winner = get_winner(state)
score = -math.inf
if winner == player:
score = 100
elif winner == '_' and move <= 8:
max_score = -math.inf
for l in range(3):
for m in range(3):
if state[l][m] == '_':
max_score = max(max_score, MinMaxPlayer._get_score(move + 1, l, m, flip_player(player), state))
score = -max_score if max_score != -math.inf else 0
elif winner == '_' and move > 8:
score = 0
else:
score = -100
MinMaxPlayer.cache[hash] = score
return score

@staticmethod
def _get_hash(symb, state):
hash = 0
to_match = 'X'
for q in range(2):
for p in range(9):
(i, j) = (p//3, p%3)
to_match = 'X' if q == 0 else 'O'
hash = hash * 2 + (1 if state[i][j] == to_match else 0)
hash = hash * 2 + (0 if symb == 'X' else 1)
return hash

player_x = MinMaxPlayer('X')
player_o = MinMaxPlayer('O')

state1 = [
['_', 'X', 'X'],
['_', 'O', '_'],
['_', 'O', '_']
]

state2 = [
['O', 'X', 'X'],
['_', 'O', '_'],
['O', '_', 'X']
]

print(player_x.get_score(0, 0, state1)) # 'X' will win if it makes the move at 0,0 so score is 100
print(player_o.get_score(2, 2, state1)) # 2,2  is really a bad move for 'O' so score is -100
print(player_o.get_score(1, 2, state2)) # move 1,2 for 'O' will eventually lead to a draw so score is 0
print(player_x.get_score(1, 0, state2)) # move 1,0 for 'X' will eventually lead to a draw so score is 0

100
-100
0
0


## 4. The Learning Way

Now that we have seen various deterministic ways of solving the problem, we can try something which is more like learning Tic-Tac-Toe. The methods discussed previously will in worst case generate all possible states and assign scores to each state. Even though applying machine learning to a problem which can be easily solved by deterministic methods doesn’t add much value, it can still be considered a good problem for learning ML.

### 4.1 The Data

Before starting to solve any problem the ML way, we need data. We can generate the data using the deterministic algorithms discussed previously. Now at this point you might want to call it unfair, because we are consuming the data which is previously available from deterministic best player. However, towards the end of this section, we will discuss a Q-Learning approach for solving this problem and there we won’t use any previous data.

So lets take a look at the data, we have represented the game state using variables $X_{i} \; \forall \; 0 \le i \le 8$ ; $O_{i} \; \forall \; 0 \le i \le 8$; $IsO$; and $score$. $X_{i}$ tells whether there is symbol ‘$X$’ at position $i$, similarly $O_{i}$ tells whether there is symbol ‘$O$’ at position $i$. In case both $O_i$ and $X_i$ are zero then position $i$ is empty. $IsO$ tells if $O$ made the last move.

0 1 2
3 4 5
6 7 8

Table 4.1.1: This shows the different positions of $i$ with respect to the locations on actual Tic-Tac-Toe matrix.

import pandas as pd
data


You can find the actual data here ticTacToeScore.csv

X0 X1 X2 X3 X4 X5 X6 X7 X8 O0 O1 O2 O3 O4 O5 O6 O7 O8 IsO score
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0
2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0
3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0
4 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0
5 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0
6 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0
7 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0
8 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0
9 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
10 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0
11 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 -100
12 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 -100
13 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 -100
14 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 1 -100
15 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0
16 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 -100
17 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 1 0
18 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 1 0
19 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0
20 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0
21 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 1 0
22 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 1 0
23 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 1 0
24 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 -100
25 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 -100
26 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 1 -100
27 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 1 100
28 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 -100
29 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
10924 1 1 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 1 0 100
10925 1 1 1 0 1 0 0 0 0 0 0 0 0 0 1 1 0 1 0 100
10926 1 1 1 0 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 100
10927 1 1 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 1 0 100
10928 1 1 1 0 1 0 0 0 0 0 0 0 1 0 0 1 0 1 0 100
10929 1 1 1 0 1 0 0 0 0 0 0 0 1 0 0 1 1 0 0 100
10930 1 1 1 0 1 0 0 0 0 0 0 0 1 0 1 0 0 1 0 100
10931 1 1 1 0 1 0 0 0 0 0 0 0 1 0 1 0 1 0 0 100
10932 1 1 1 0 1 0 0 0 0 0 0 0 1 0 1 0 1 1 0 100
10933 1 1 1 0 1 0 0 0 0 0 0 0 1 0 1 1 0 0 0 100
10934 1 1 1 0 1 0 0 0 0 0 0 0 1 0 1 1 0 1 0 100
10935 1 1 1 0 1 0 0 0 0 0 0 0 1 0 1 1 1 0 0 100
10936 1 1 1 0 1 0 0 0 1 0 0 0 1 0 1 1 1 0 0 100
10937 1 1 1 0 1 0 0 1 0 0 0 0 1 0 1 1 0 1 0 100
10938 1 1 1 0 1 0 1 0 0 0 0 0 1 0 1 0 1 1 0 100
10939 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 100
10940 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 100
10941 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 100
10942 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 100
10943 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 100
10944 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 100
10945 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 100
10946 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 100
10947 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 100
10948 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 100
10949 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 0 1 0 100
10950 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 100
10951 1 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 1 0 0 100
10952 1 1 1 1 0 0 0 1 0 0 0 0 0 1 1 1 0 1 0 100
10953 1 1 1 1 0 0 1 0 0 0 0 0 0 1 1 0 1 1 0 100

10954 rows × 20 columns

Let’s consider the $12^{th}$ data point.

data.iloc[[11]]

X0 X1 X2 X3 X4 X5 X6 X7 X8 O0 O1 O2 O3 O4 O5 O6 O7 O8 IsO score
11 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 -100

The $12^{th}$ data point represents the following Tic-Tac-Toe state:

_ _ _
_ _ _
_ O X

$IsO$ is 1 indicating that last turn was with $O$. Also, $score$ is -100 indicating that $O$ will definitely lose, given $X$ plays optimally. We can actually see that’s true because in the next move ‘X’ can actually mark at position $2$,

_ _ X
_ _ _
_ O X

forcing $O$ to mark at position $5$.

_ _ X
_ _ O
_ O X

After that $X$ makes a move at position $0$, trapping $O$ and making it lose the game.

X _ X
_ _ O
_ O X

Clearly our dataset is good, it represents the actual game states and their corresponding scores accurately.

In all there are $5478$ distinct possible valid states in a Tic-Tac-Toe. A trivial analysis can tell us that maximum number of cases will be capped by $3^9 = 19683$ because every position possible can take three different values ($X$, $O$ or $_$ ), but this clearly is not the best estimate as out of 19683 only 5478 are possible. This also tells us why a representation consisting of more columns, having $blank_{i}$ similar to $X_{i}$s and $O_{i}$s to represent $_$, is not a good idea. It will increase the space in which the learning algorithm has to search for solution. Our current representation has 19 binary columns and hence the search space is constrained by $2^{19} = 524288$ size. But if we have more columns, considering $blank_{i}$ as binary columns, then the total search space size increases to $2^{28} = 268435456$. There are some other possible representations like having 9 columns, one for each position, each of which can contain three values ($0$ for $X$, $1$ for $O$ and $2$ for $_$ ). It is not hard to see that this representation is bad because it is attaching a weighted preference to each of possible outcomes ($X$, $O$ or $_$ ). It will make it hard for some of the simple algorithms like logistic regression to extract the meaning from this representation, even for complex models such as ANNs it may take more time to converge. So a good representation is one which allows us to represent data in a simple form and at the same time not increasing the search space drastically. If you would have noticed the number of rows in our training data is $10954 = 2 * (5478 - 1)$. This is because there are two possible values of $IsO$ for every state and we are not considering empty state at all so $2 * (5478 - 1)$.

### 4.2 Preparing the Data

First we extract out the training data for ‘X’ and ‘O’ separately. This is simply a filter based on the column $IsO$.

#x_training_data_full has training examples in which 'X' made the last move
x_training_data_full = (data[(data.IsO == 0)]).drop(columns=["IsO", "score"])
#o_training_data_full has training examples in which 'O' made the last move
o_training_data_full = (data[(data.IsO == 1)]).drop(columns=["IsO", "score"])
#Below lines make sure that older index is dropped from the new data frame formed
x_training_data_full.reset_index(drop=True, inplace=True)
o_training_data_full.reset_index(drop=True, inplace=True)
#This simply stores score separately
x_score_full = [score for score in data[(data.IsO == 0)]["score"]]
o_score_full = [score for score in data[(data.IsO == 1)]["score"]]


### 4.3 Linear Regression

Clearly we have some features and a score which we want to predict so how about trying Linear Regression first. In this case, Linear Regression tries to learn function which is of form:

$$\sum_{i = 0}^{8}(w_{x_i} X_i) + \sum_{i = 0}^{8}(w_{o_i} O_i) + C$$ … (Eq 4.3.1)

The core idea of linear regression is to search for functions like Eq 4.3.1 which closely approximates the original score function as used by deterministic best player. You can read more about Linear Regression.

We are using sklearn, it is a popular ML library in python.

from sklearn.linear_model import LinearRegression
x_full_linear_reg_model = LinearRegression().fit(x_training_data_full, x_score_full)
x_full_linear_reg_acc = x_full_linear_reg_model.score(x_training_data_full, x_score_full)
o_full_linear_reg_model = LinearRegression().fit(o_training_data_full, o_score_full)
o_full_linear_reg_acc = o_full_linear_reg_model.score(o_training_data_full, o_score_full)

print("Function for X: ",x_full_linear_reg_model.coef_, " * state + ", x_full_linear_reg_model.intercept_)
print("Function for O: ", o_full_linear_reg_model.coef_, " * state + ", o_full_linear_reg_model.intercept_)
print("Accuracy for X = {}, Accuracy for O = {}"
.format(x_full_linear_reg_acc, o_full_linear_reg_acc))

Function for X:  [ 69.71383658  48.3715394   69.71383658  48.3715394   95.21678634
48.3715394   69.71383658  48.3715394   69.71383658 -48.13167897
-35.40737807 -48.13167897 -35.40737807 -53.08796566 -35.40737807
-48.13167897 -35.40737807 -48.13167897]  * state +  -107.77964267296042
Function for O:  [-48.13167897 -35.40737807 -48.13167897 -35.40737807 -53.08796566
-35.40737807 -48.13167897 -35.40737807 -48.13167897  69.71383658
48.3715394   69.71383658  48.3715394   95.21678634  48.3715394
69.71383658  48.3715394   69.71383658]  * state +  -107.77964267296349
Accuracy for X = 0.2359100795453818, Accuracy for O = 0.23591007954538165


We get a very bad training accuracy of just $23.5\%$ . But if you notice the function for $X$ and $O$ have coefficients which are opposite in sign. It means that the move which makes $X$ win is bad for $O$. But is this data good enough for a linear model to work ? Can we do better with the linear model ?

Let’s try and reduce the dimension using PCA. One may also want to do scaling, but it is not of much use as we only have binary columns.

from sklearn.decomposition import PCA
import matplotlib.pyplot as plt

x_pca2_training_data = PCA(n_components=2).fit_transform(x_training_data_full)

import numpy as np
x_pca2_training_data_win = np.array([x_pca2_training_data[idx] for idx, score in enumerate(x_score_full) if str(x_score_full[idx]) == "100"])
x_pca2_training_data_lose = np.array([x_pca2_training_data[idx] for idx, score in enumerate(x_score_full) if str(x_score_full[idx]) == "-100"])
x_pca2_training_data_draw = np.array([x_pca2_training_data[idx] for idx, score in enumerate(x_score_full) if str(x_score_full[idx]) == "0"])
plt.scatter(x_pca2_training_data_win[:,0], x_pca2_training_data_win[:,1], color="green", label="win", alpha=0.8)
plt.scatter(x_pca2_training_data_lose[:,0], x_pca2_training_data_lose[:,1], color="red", label="lose", alpha=0.8)
plt.scatter(x_pca2_training_data_draw[:,0], x_pca2_training_data_draw[:,1], color="yellow", label="draw", alpha=0.8)
plt.title('\'X\' training data')
plt.legend()
plt.show()


The plot pretty much explains why Linear Regression performs so bad, clearly the data is not linear. Is there a way to improve accuracy ? Even on such a complex decision boundary $23.5\%$ seems too bad. We will address this question later.

### 4.4 Decision Trees Regression

As we already saw the linear regression performed badly, so we will consider more complex models. So let us consider decision tree regression first. You read more about decision tree on one my blogs.

import os
from sklearn.tree import DecisionTreeRegressor
from sklearn import tree
from graphviz import Source
os.environ["PATH"] += os.pathsep + 'C:/Program Files (x86)/Graphviz2.38/bin/'
#This will need installation of Graphviz to work

x_full_dt_reg_model = DecisionTreeRegressor(random_state=0, max_leaf_nodes=15).fit(x_training_data_full, x_score_full)
o_full_dt_reg_model = DecisionTreeRegressor(random_state=0, max_leaf_nodes=15).fit(o_training_data_full, o_score_full)
graph = Source(tree.export_graphviz(x_full_dt_reg_model, out_file=None, feature_names=x_training_data_full.columns))
graph


As you can see even with 15 nodes it is still checking the $X4$ value which is the center position. So one of our childhood assumptions about making a move at center position takes the game towards draw is not without a reason. However, this particular decision tree with just 15 node doesn’t have a good accuracy. In fact, the accuracy improves with the number of nodes in the decision tree and after a certain number of nodes it saturates to $1.0$, when the decision tree over-fits the training data.

x_full_dt_reg_acc = []
o_full_dt_reg_acc = []
num_nodes = [i for i in range(10, 6000, 250)]

for i in num_nodes:
x_full_dt_reg_model = DecisionTreeRegressor(random_state=0, max_leaf_nodes=i).fit(x_training_data_full, x_score_full)
x_full_dt_reg_acc.append(x_full_dt_reg_model.score(x_training_data_full, x_score_full))
for i in num_nodes:
o_full_dt_reg_model = DecisionTreeRegressor(random_state=0, max_leaf_nodes=i).fit(o_training_data_full, o_score_full)
o_full_dt_reg_acc.append(o_full_dt_reg_model.score(o_training_data_full, o_score_full))

plt.plot(num_nodes, x_full_dt_reg_acc, label='o')
plt.plot(num_nodes, o_full_dt_reg_acc, label='x')
plt.legend()
plt.show()


So as you can see the accuracy touches $1.0$, in both overlapping plots, around $1600$ nodes. We may also note that tic-tac-toe has only $765$ essentially different positions (with rotations and symmetry considered), which also hints that one will at least need $765$ nodes in the tree.

### 4.5 ANN Regression

How about applying neural networks to this problem ? Since, neural networks can solve complex problems like face recognition, handwriting recognition etc. with ease, so it appears to be reasonably simple thing for ANN Regression.

from sklearn.neural_network import MLPRegressor
x_full_ann_reg_model = MLPRegressor(random_state=1, max_iter=3500, solver='lbfgs', early_stopping=True, activation='logistic').fit(x_training_data_full, x_score_full)
x_full_ann_reg_model.score(x_training_data_full, x_score_full)

0.773482169075109


Never expected ANN to be this bad, just $77.3\%$ ?

Usually ANN outperforms any other model. This might make us wonder if it is a consequence of No free lunch theorem that the decision tree outperforms ANN.

### 4.6 Fill the missing gap

What if model this as a classification problem rather than regression problem. The possible scores are anyways limited to three values $0$, $100$ and $-100$. All of them simply corresponds to three different scenarios or classes. Sounds like an interesting idea. So let’s convert our data to make it a classification problem.

#This simply coverts -100 to 0; 0 to 1 and 100 to 2
#Basically making 0,1 and 2 classes from scores
x_class_full = [int(score/100 + 1) for score in x_score_full]
o_class_full = [int(score/100 + 1) for score in o_score_full]


#### 4.6.1 Logistic Regression

Let’s replace the linear regression model with logistic regression and see what do we gain.

from sklearn.linear_model import LogisticRegression
x_full_logistic_reg_model = LogisticRegression(multi_class='auto', solver='lbfgs').fit(x_training_data_full, x_class_full)
x_full_logistic_reg_acc = x_full_logistic_reg_model.score(x_training_data_full, x_class_full)
o_full_logistic_reg_model = LogisticRegression(multi_class='auto', solver='lbfgs').fit(o_training_data_full, o_class_full)
o_full_logistic_reg_acc = o_full_logistic_reg_model.score(o_training_data_full, o_class_full)

print("Function for X: ",x_full_logistic_reg_model.coef_, " * state + ", x_full_logistic_reg_model.intercept_)
print("Function for O: ", x_full_logistic_reg_model.coef_, " * state + ", x_full_logistic_reg_model.intercept_)
print("Accuracy for X = {}, Accuracy for O = {}"
.format(x_full_logistic_reg_acc, o_full_logistic_reg_acc))

Function for X:  [[-1.08117725 -0.77468022 -1.08117725 -0.77468022 -1.56284699 -0.77468022
-1.08117725 -0.77468022 -1.08117725  0.91620433  0.64007097  0.91620433
0.64007097  0.91947109  0.64007097  0.91620433  0.64007097  0.91620433]
[-0.12796582  0.0266686  -0.12796582  0.0266686   0.0954271   0.0266686
-0.12796582  0.0266686  -0.12796582 -0.42224774 -0.23918628 -0.42224774
-0.23918628 -0.19057815 -0.23918628 -0.42224774 -0.23918628 -0.42224774]
[ 1.20914307  0.74801161  1.20914307  0.74801161  1.46741989  0.74801161
1.20914307  0.74801161  1.20914307 -0.49395659 -0.40088469 -0.49395659
-0.40088469 -0.72889294 -0.40088469 -0.49395659 -0.40088469 -0.49395659]]  * state +  [ 1.61844104  0.56889277 -2.18733381]
Function for O:  [[-1.08117725 -0.77468022 -1.08117725 -0.77468022 -1.56284699 -0.77468022
-1.08117725 -0.77468022 -1.08117725  0.91620433  0.64007097  0.91620433
0.64007097  0.91947109  0.64007097  0.91620433  0.64007097  0.91620433]
[-0.12796582  0.0266686  -0.12796582  0.0266686   0.0954271   0.0266686
-0.12796582  0.0266686  -0.12796582 -0.42224774 -0.23918628 -0.42224774
-0.23918628 -0.19057815 -0.23918628 -0.42224774 -0.23918628 -0.42224774]
[ 1.20914307  0.74801161  1.20914307  0.74801161  1.46741989  0.74801161
1.20914307  0.74801161  1.20914307 -0.49395659 -0.40088469 -0.49395659
-0.40088469 -0.72889294 -0.40088469 -0.49395659 -0.40088469 -0.49395659]]  * state +  [ 1.61844104  0.56889277 -2.18733381]
Accuracy for X = 0.6016067190067556, Accuracy for O = 0.6016067190067556


Wow!! the accuracy increased from $23.5\%$ to $60.1\%$ , however, this is not surprising. Converting a regression problem to classification problem allows more room for the algorithm to make mistakes. Logistic Regression creates constraints on the problem making it less complex than Linear Regression where the possible outcome could be anything in $(-\infty, +\infty)$ .

#### 4.6.2 Decision Tree Classifier

Let’s see if something improves in case of decision tree classification. You read more about decision tree on one my blogs.

from sklearn.tree import DecisionTreeClassifier
x_full_dt_class_acc = []
o_full_dt_class_acc = []
num_nodes = [i for i in range(10, 6000, 250)]

for i in num_nodes:
x_full_dt_class_model = DecisionTreeClassifier(random_state=0, max_leaf_nodes=i).fit(x_training_data_full, x_class_full)
x_full_dt_class_acc.append(x_full_dt_class_model.score(x_training_data_full, x_class_full))
for i in num_nodes:
o_full_dt_class_model = DecisionTreeClassifier(random_state=0, max_leaf_nodes=i).fit(o_training_data_full, o_class_full)
o_full_dt_class_acc.append(o_full_dt_class_model.score(o_training_data_full, o_class_full))

plt.plot(num_nodes, x_full_dt_class_acc, label='o')
plt.plot(num_nodes, o_full_dt_class_acc, label='x')
plt.legend()
plt.show()


It seems like this doesn’t affect the results much. It kind of makes sense, decision trees usually tend to over-fit the training data itself and hence we cannot see any significant improvements here.

#### 4.6.3 Classification using Neural Networks

Let’s see if converting it into neural networks classification problem improves accuracy.

from sklearn.neural_network import MLPClassifier
x_full_ann_class_model = MLPClassifier(random_state=1, max_iter=1000, activation='relu').fit(x_training_data_full, x_class_full)
o_full_ann_class_model = MLPClassifier(random_state=1, max_iter=1000, activation='relu').fit(o_training_data_full, o_class_full)

x_full_ann_class_acc = x_full_ann_class_model.score(x_training_data_full, x_class_full)
o_full_ann_class_acc = o_full_ann_class_model.score(o_training_data_full, o_class_full)
print("Accuracy for X = {}, Accuracy for O = {}"
.format(x_full_ann_class_acc, o_full_ann_class_acc))

Accuracy for X = 1.0, Accuracy for O = 1.0


Finally we have something, the accuracy is 1 now!! Neural Networks learns to play Tic-Tac-Toe. We can finally brag about it.

## 4.7 The fair way

Till this point we have considered that we have scores corresponding to all states. In true sense this is not really ML, in many cases we don’t even have such exhaustive training data. Also, we ideally divide the available data between test and train. If you note we didn’t do that here. In real world the problems are complex enough and we might not even have any way to compute these scores before hand.

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