Assignment Chef icon Assignment Chef

[SOLVED] Coms w4701: artificial intelligence homework 2 problem 1: tic-tac-twist (22 points) two players are playing a modified tic-tac-toe game

5.0 1 customer review Digital download

Digital download

$25.00

Availability
In stock
Checkout
One item

Need a hand?

Message us on WhatsApp for payment or download support.

WhatsApp QR code
Two players are playing a modified tic-tac-toe game in which grid spaces have point values, shown
on the left grid below. The players take turns marking a grid space with their own designation (X
or O) until either one player gets three marks in a row or the board has no empty spaces. When
the game ends, a score is computed as the sum of the values of the X spaces minus the sum of the
values of the O spaces. In addition, if X has three in a row, 3 points are added to the score; if O
has three in a row, 3 points are subtracted from the score. X seeks to maximize the total score and
O seeks to minimize it.(a) The right grid shows the current board configuration, whose current value is −3. It is O’s turn
to move. Draw out the entire game tree with the root corresponding to the current board. Use
game tree convention to draw the MAX and MIN nodes. Also sketch out the tic-tac-toe board
configuration for each terminal node (e.g., draw them right below each node).(b) Compute the minimax values of each node. What is the best move for O to make, and what is
the expected score of the game assuming both players play optimally?(c) Suppose we are performing alpha-beta search. In what order would the successors of the root
node have to be processed in order to maximize the number of nodes that can be pruned?
Identify the node(s) in the game tree that can be pruned, and specify the α or β inequality
that allows pruning to take place.(d) Suppose that instead of playing to maximize the score, player X chooses a random valid move
with uniform probability. Explain how the game tree will change (you do not have to redraw
it), and compute the new utility and best move for O starting from the current state.We will consider a simplified version of the game of Yahtzee. We first roll three 4-sided dice (with
results 1, 2, 3, 4 occurring with equal probability), and then we can either reroll one of the dice
or keep the original result. Let S be the sum of the final dice results. We get a score equal to the
max over S and one of the following, if the situation applies:
• 10 points for two-of-a-kind
• 15 points for three-of-a-kind
• 7 points for a series (1-2-3 or 2-3-4)The expectimax tree representing the player’s decision is shown below:
Suppose we initially rolled a 3, 4, and 4. From left to right, the second layer of the tree represents
actions of rerolling the 3, rerolling one 4, rerolling the other 4, and keeping the original dice results.
The third layer represents the outcomes of each of the three die rerolls from 1 to 4 (going left to
right for each).(a) Fill in the values of the nodes of the tree (you can simply write out the values for each node
variable, e.g. M = 1). Start with the utilities of the leaf nodes, followed by the chance nodes
and finally the value at the root. What decision maximizes our expected score?(b) Suppose we change the rules slightly. We can still either reroll a die or keep the initial results,
but choosing the ”reroll” action results in a random die being rerolled (we do not get to choose
which). Describe precisely how the structure of the expectimax tree would change and say what
the new optimal action is. You do not need to draw the new tree or compute new node values.(c) Consider another change of rules. Instead of choosing one die to reroll, the options are to
choose exactly two dice to reroll (we can indicate which two dice we want) or to keep the
initial results. Describe precisely how the structure of the expectimax tree would change. You
do not need to draw the new tree or compute new node values.Sudoku is a logic-based number placement puzzle. The basic rules are as follows: Given a n
2 × n
2
grid and a set of pre-filled cells (clues), fill in the remaining cells such that each row, column,
and n × n (non-overlapping) subgrid contains an instance of each number from 1 to n
2
(inclusive).Classic sudoku has n = 3 and a 9 × 9 grid. Like n-queens, we can define sudoku as a constraint
satisfaction problem, which we can then solve using exhaustive or local search methods.
We are providing a simple Python scaffold for solving sudoku using hill-climbing search. A state is
a filled-in grid (2D NumPy array) with n
2 of each of the numbers from 1 to n
2
.A transition occurs
by swapping two numbers that are not in the clue indices. Finally, we can measure the number of
“conflicts” for a given state by counting the number of errors in each row, column, and subgrid.The code that we provide generates a sudoku puzzle, which you can test out yourself. Given
the order (n) of the desired puzzle and the number of clues (which should be fewer than n
4
,
the total number of cells), generate returns a dictionary (problem) specifying the indices and
values of clue cells.The display function prints out the board with the clues filled in and zeroes
elsewhere. initialize returns a state that satisfies the clues (but is not necessarily a solution), and
successors returns a list of valid successor states given a current state. The latter two functions
will be used by your hill-climbing procedure.3.1: Computing Errors (5 points)
Write the num errors function that will count and return the number of errors in a given state.
The simplest way to do this would be to iterate over each of the n
2
rows, n
2
columns, and n
2
non-overlapping n×n subgrids and count the number of missing values in each. Note that this will
inevitably double- or triple-count individual errors.3.2: Basic Hill-Climbing (10 points)
Implement the basic hill-climbing procedure in hill climb, ignoring the optional arguments for
now. You should only initialize the given problem once, and no sideways moves are permitted.
Either move to a “better” successor state with the lowest number of errors (choose a random one
if there are multiple best states), or return the current state if no successor state has fewer errors
than the current one. Also create a list tracking the number of errors in each iteration and return
that along with the solution.3.3: Preliminary Tests (5 points)
You should find that your basic procedure performs decently on order-2 puzzles, although it is
certainly prone to getting stuck at local minima. Generate a few different puzzles with n = 2
and c = 5 clues (the minimum number of clues for a unique solution, though not sufficient for all
puzzles). For two puzzles, one that hill climb can solve and one that hill climb cannot solve
successfully, show the problem using display, the returned solution state, and a plot showing the
number of errors over each iteration.Run hill climb in batch on 100 random order-2 puzzles with 5 clues each. Report the average
success rate (proportion of final states with 0 error) as well as the average error over all final states
(counting both successes and failures).3.4: Sideways Moves (5 points)
The basic hill climbing procedure can be improved in a couple ways. First, we should allow for
sideways moves. If the best neighbors all have the same number of conflicts as the current state,
we can move to a random one. You should also keep a counter and increment it when a sideways
move is taken consecutively. If max sideways moves are made consecutively, quit and return the
current state.Generate a puzzle with n = 2 and c = 5 in which at least one (preferably more) sideways move is
taken and show the corresponding error per iteration plot. The sideways move(s) should clearly be
seen as a horizontal segment on this plot. Then do a batch solve of 100 random puzzles allowing
up to 10 sideways moves each, and report the success rate and average error.3.5: Random Restarts (5 points)
The last modification of hill climbing that you will implement is including random restarts. When
either all current neighbors are worse than the current state or when max sideways moves are
made, we should initialize a new sudoku board and assign it to be the current state. A counter
on the number of restarts should be created and incremented each time this happens. The return
condition would be when we have hit max restarts or when a solution is found.Run the batch of 100 (n = 2, c = 5) puzzles with max sideways and max restarts both equal to 10
and report your findings. Lastly, experiment with (n = 3, c = 40) puzzles; these are standard-sized
sudoku problems, but on the easier side given the number of clues. Run hill climbing on a few of
these and show the result of one puzzle that was successfully solved, including the given problem,
returned solution, and a plot of errors per iteration. This may take a few tries, and you may tweak
the max sideways and max restarts parameters. Write down the parameter values that you used.The game of Connect Four bears some similarities to tic-tac-toe. Two players take turns placing
marks or pieces on a grid, trying to be the first to achieve four consecutive pieces. The difference,
aside from the size of the board (typically six rows, seven columns) and that we want four rather
than three in a row, is that the grid is vertical and affected by gravity. In short, a player’s actions
only consist of the column in which to place their piece, and the piece must then subsequently be
placed in the bottom-most free cell in that column.We are providing a simple Python scaffold that will be developed into a Connect Four-playing agent
using alpha-beta depth-limited search. A game state is represented as a 2D NumPy array. We will
use ’X’ and ’O’ to represent each player just as in tic-tac-toe, and the character ’.’ to represent
an empty cell. To make our implementation more general, we will allow grids of any size (m × n)
and make this a “Connect k” game, rather than just Connect Four. ’X’ will always go first and be
the maximizing player, and ’O’ will be the minimizing player. ’X’ wins have positive utility, ’O’
wins have negative utility, and draws have zero utility.We provide several utility functions; the relevant ones for you are terminal and eval. terminal
takes in a current game state and returns either a utility value if the state is terminal or None
if the game is not terminal (this thus serves as both a utility function and terminal test). eval
takes in the same inputs as terminal and returns an evaluation of a non-terminal state based on
both players’ potential for winning (how this is computed is not important for this assignment,
but we are happy to talk more about it if you are interested). You will use both functions when
implementing alpha-beta search.Finally, given the game parameters, game loop runs a game from start to finish, using your implemented alpha-beta search procedure for both players. While this nominally means that both
players will be playing with identical search capabilities, the optional X params and O params will
allow you to tweak the strength of each player in the form of search depth.4.1: Successor Function (5 points)
Before writing alpha-beta search, we first need to write the successors function. Given both a
state and the player making a move (either ’X’ or ’O’), return a list of possible successor states.
Be sure to follow the rules of Connect k as described above. Note that this function combines the
actions and result functions if following the textbook’s pseudocode.4.2: Depth-Limited Alpha-Beta Search (20 points)
We provide the wrapper function for depth-limited alpha-beta search, which is called repeatedly by
game loop. You will write the two recursive functions, mostly following the textbook pseudocode
but with some additional functionalities. The arguments to each are the state, alpha and beta
values, k value, current node depth, and maximum allowed search depth.Each function should
perform the following tasks:
• Perform the terminal test; if terminal, return the state’s score and None move.
• If current node depth is equal to maximum allowed depth, this state is treated as a leaf. Run
the evaluation function on the state and return the value and None move.• Otherwise we expand the node’s successors and search them using the alternative recursive
function. Before doing so, you must generate and sort the successors such that the “best”
nodes occur first for each player. You can do this by calling eval on each successor and then
sorting them from largest to smallest for MAX and smallest to largest for MIN.• The recursive calls should be similar to the pseudocode. Be sure to increment the current
node depth by 1 when you make a new recursive call.4.3: Testing Connect k (5 points)
Once you are ready to test your implementations, have your agent play itself in Connect 3 and
then Connect 4, both on a 4 × 4 board. Run both without depth limiting; you can simply omit
the two optional arguments to game loop. Show the final state for each.To play larger games, depth limiting will be necessary. Have your agent play the full Connect 4
game on a 6 × 7 board. Show the final game states for maximum search depths of 5 and 6 (for
both players). You should find that one ends in a draw and the other ends in a win for one player.
Then try to change the depth values, making them different if necessary, so that the other player
wins the game. Show the final state for this last run.Submission
You should have one PDF document containing your solutions for problems 1-2, as well as the
information and plots in the relevant parts of 3-4. You should also have the completed code file
implementing problems 3 and 4; make sure that all provided function headers are unchanged.
Submit the document and code files to the respective assignment bins on Gradescope. For full
credit, you must tag your pages for each given problem on the former.