AI Manual (E-Next - In)

T.Y.B.Sc.
IT ARITIFICIAL INTELLIGENCE 2018-19
UNIVERSITY OF MUMBAI
Teacher’s Reference Manual

Subject: Artificial Intelligence
with effect from the academic year

2018 – 2019
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T.Y.B.Sc.IT ARITIFICIAL INTELLIGENCE 2018-19
PRACTICAL NO-1
A. Write a program to implement depth first search algorithm.
B. Write a program to implement breadth first search algorithm
AIM:-
Write a program to implement depth first search algorithm.
GRAPH:-
PYTHON CODE:-
graph1 = {
'A': set(['B', 'C']),
'B': set(['A', 'D', 'E']),
'C': set(['A', 'F']),
'D': set(['B']),
'E': set(['B', 'F']),
'F': set(['C', 'E'])
}
def dfs(graph, node, visited):
if node not in visited:
visited.append(node)
for n in graph[node]:
dfs(graph,n, visited)
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return visited
visited = dfs(graph1,'A', [])
print(visited)
OUTPUT:-
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AIM:-
Write a program to implement breadth first search algorithm.
GRAPH:-
PYTHON CODE:-
# sample graph implemented as a dictionary
graph = {'A': set(['B', 'C']),
'B': set(['A', 'D', 'E']),
'C': set(['A', 'F']),
'D': set(['B']),
'E': set(['B', 'F']),
'F': set(['C', 'E'])
}
#Implement Logic of BFS
def bfs(start):
queue = [start]
levels={} #This Dict Keeps track of levels
levels[start]=0 #Depth of start node is 0
visited = set(start)
while queue:
node = queue.pop(0)
neighbours=graph[node]
for neighbor in neighbours:
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if neighbor not in visited:

queue.append(neighbor)
visited.add(neighbor)
levels[neighbor]= levels[node]+1
print(levels) #print graph level
return visited
print(str(bfs('A'))) #print graph node
#For Finding Breadth First Search Path

def bfs_paths(graph, start, goal):
queue = [(start, [start])]
while queue:
(vertex, path) = queue.pop(0)
for next in graph[vertex] - set(path):
if next == goal:
yield path + [next]
else:
queue.append((next, path + [next]))
result=list(bfs_paths(graph, 'A', 'F'))
print(result)# [['A', 'C', 'F'], ['A', 'B', 'E', 'F']]
#For finding shortest path

def shortest_path(graph, start, goal):
try:
return next(bfs_paths(graph, start, goal))
except StopIteration:
return None
result1=shortest_path(graph, 'A', 'F')
print(result1)# ['A', 'C', 'F']
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OUTPUT:-
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Practical no-2
A. Write a program to simulate 4-Queen / N-Queen problem.
B. Write a program to solve tower of Hanoi problem.
Aim:-
Write a program to simulate 4-Queen / N-Queen problem
DIAGRAM:
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PYTHON CODE:-
class QueenChessBoard:
def __init__(self, size):
# board has dimensions size x size
self.size = size
# columns[r] is a number c if a queen is placed at row r and column c.
# columns[r] is out of range if no queen is place in row r.
# Thus after all queens are placed, they will be at positions
# (columns[0], 0), (columns[1], 1), ... (columns[size - 1], size - 1)
self.columns = []
def place_in_next_row(self, column):
self.columns.append(column)
def remove_in_current_row(self):
return self.columns.pop()
def is_this_column_safe_in_next_row(self, column):
# index of next row
row = len(self.columns)
# check column
for queen_column in self.columns:
if column == queen_column:
return False
# check diagonal
for queen_row, queen_column in enumerate(self.columns):
if queen_column - queen_row == column - row:
return False
# check other diagonal
for queen_row, queen_column in enumerate(self.columns):
if ((self.size - queen_column) - queen_row
== (self.size - column) - row):
return False
return True
def display(self):
for row in range(self.size):
for column in range(self.size):
if column == self.columns[row]:
print('Q', end=' ')
else:
print('.', end=' ')
print()
def solve_queen(size):
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"""Display a chessboard for each possible configuration of placing n

queens on an n x n chessboard and print the number of such
configurations."""
board = QueenChessBoard(size)
number_of_solutions = 0
row = 0
column = 0
# iterate over rows of board
while True:
# place queen in next row
while column < size:
if board.is_this_column_safe_in_next_row(column):
board.place_in_next_row(column)
row += 1
column = 0
break
else:
column += 1
# if could not find column to place in or if board is full

if (column == size or row == size):
# if board is full, we have a solution
if row == size:
board.display()
print()
number_of_solutions += 1
# small optimization:
# In a board that already has queens placed in all rows except
# the last, we know there can only be at most one position in
# the last row where a queen can be placed. In this case, there
# is a valid position in the last row. Thus we can backtrack two
# times to reach the second last row.
board.remove_in_current_row()
row -= 1
# now backtrack
try:
prev_column = board.remove_in_current_row()
except IndexError:
# all queens removed
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# thus no more possible configurations

break
# try previous row again
row -= 1
# start checking at column = (1 + value of column in previous row)
column = 1 + prev_column
print('Number of solutions:', number_of_solutions)
n = int(input('Enter n: '))
solve_queen(n)
OUTPUT:-
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NOTE:
1. The user is prompted to enter n where n is the number of queens to place and the
size of the board.
2. Solve queens is called on n to display all possible board configurations and the
number of solutions.
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AIM:-
Write a program to solve tower of Hanoi problem.
DIAGRAM:
PYTHON CODE:
def moveTower(height,fromPole, toPole, withPole):
if height >= 1:
moveTower(height-1,fromPole,withPole,toPole)
moveDisk(fromPole,toPole)
moveTower(height-1,withPole,toPole,fromPole)
def moveDisk(fp,tp):
print("moving disk from",fp,"to",tp)
moveTower(3,"A","B","C")
OUTPUT:-
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PRACTICAL NO.-3
A. Write a program to implement alpha beta search.
B. Write a program for Hill climbing problem.
AIM:-
Write a program to implement alpha beta search.
PYTHON CODE
tree = [[[5, 1, 2], [8, -8, -9]], [[9, 4, 5], [-3, 4, 3]]]
root = 0
pruned = 0
def children(branch, depth, alpha, beta):

global tree
global root
global pruned
i=0
for child in branch:
if type(child) is list:
(nalpha, nbeta) = children(child, depth + 1, alpha, beta)
if depth % 2 == 1:
beta = nalpha if nalpha < beta else beta
else:
alpha = nbeta if nbeta > alpha else alpha
branch[i] = alpha if depth % 2 == 0 else beta
i += 1
else:
if depth % 2 == 0 and alpha < child:
alpha = child
if depth % 2 == 1 and beta > child:
beta = child
if alpha >= beta:
pruned += 1
break
if depth == root:
tree = alpha if root == 0 else beta
return (alpha, beta)
def alphabeta(in_tree=tree, start=root, upper=-15, lower=15):

global tree
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global pruned
global root
(alpha, beta) = children(tree, start, upper, lower)
if __name__ == "__main__":
print ("(alpha, beta): ", alpha, beta)
print ("Result: ", tree)
print ("Times pruned: ", pruned)
return (alpha, beta, tree, pruned)
if __name__ == "__main__":
alphabeta(None)
OUTPUT
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AIM:-
Write a program for Hill climbing problem.
DIAGRAM:-
PYTHON CODE:
import math
increment = 0.1
startingPoint = [1, 1]
point1 = [1,5]
point2 = [6,4]
point3 = [5,2]
point4 = [2,1]
def distance(x1, y1, x2, y2):

dist = math.pow(x2-x1, 2) + math.pow(y2-y1, 2)
return dist
def sumOfDistances(x1, y1, px1, py1, px2, py2, px3, py3, px4, py4):
d1 = distance(x1, y1, px1, py1)
return d1 + d2 + d3 + d4
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def newDistance(x1, y1, point1, point2, point3, point4):

d1 = [x1, y1]
d1temp = sumOfDistances(x1, y1, point1[0],point1[1], point2[0],point2[1],
point3[0],point3[1], point4[0],point4[1] )
d1.append(d1temp)
return d1
minDistance = sumOfDistances(startingPoint[0], startingPoint[1],

point1[0],point1[1], point2[0],point2[1],
point3[0],point3[1], point4[0],point4[1] )
flag = True
def newPoints(minimum, d1, d2, d3, d4):

if d1[2] == minimum:
return [d1[0], d1[1]]
elif d2[2] == minimum:
return [d2[0], d2[1]]
return [d3[0], d3[1]]
return [d4[0], d4[1]]
i=1
while flag:
d1 = newDistance(startingPoint[0]+increment, startingPoint[1], point1, point2,
point3, point4)
d2 = newDistance(startingPoint[0]-increment, startingPoint[1], point1, point2,
point3, point4)
d3 = newDistance(startingPoint[0], startingPoint[1]+increment, point1, point2,
point3, point4)
d4 = newDistance(startingPoint[0], startingPoint[1]-increment, point1, point2,
point3, point4)
print (i,' ', round(startingPoint[0], 2), round(startingPoint[1], 2))
minimum = min(d1[2], d2[2], d3[2], d4[2])
if minimum < minDistance:
startingPoint = newPoints(minimum, d1, d2, d3, d4)
minDistance = minimum
#print i,' ', round(startingPoint[0], 2), round(startingPoint[1], 2)
i+=1
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else:
flag = False
OUTPUT
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Practical no-4
A. Write a program to implement A* algorithm.
B. Write a program to implement AO* algorithm.
Aim:-
Write a program to implement A* algorithm.
Note:
Install 2 package in python scripts directory using pip command.
1. pip install simpleai
2. pip install pydot flask
PYTHON CODE:-
from simpleai.search import SearchProblem, astar
GOAL = 'HELLO WORLD'

class HelloProblem(SearchProblem):
def actions(self, state):
if len(state) < len(GOAL):
return list(' ABCDEFGHIJKLMNOPQRSTUVWXYZ')
else:
return []
def result(self, state, action):

return state + action
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def is_goal(self, state):

return state == GOAL
def heuristic(self, state):

# how far are we from the goal?
wrong = sum([1 if state[i] != GOAL[i] else 0
for i in range(len(state))])
missing = len(GOAL) - len(state)
return wrong + missing
problem = HelloProblem(initial_state='')
result = astar(problem)
print(result.state)
print(result.path())
OUTPUT:-
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Practical no-5
A. Write a program to solve water jug problem.
B. Design the simulation of tic – tac – toe game using min-max algorithm.
Aim:-
Write a program to solve water jug problem.
Diagram:-
Python Code:-
# 3 water jugs capacity -> (x,y,z) where x>y>z
# initial state (12,0,0)
# final state (6,6,0)
capacity = (12,8,5)
# Maximum capacities of 3 jugs -> x,y,z
x = capacity[0]
y = capacity[1]
z = capacity[2]
# to mark visited states

memory = {}
# store solution path

ans = []
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def get_all_states(state):
# Let the 3 jugs be called a,b,c
a = state[0]
b = state[1]
c = state[2]
if(a==6 and b==6):

ans.append(state)
return True
# if current state is already visited earlier

if((a,b,c) in memory):
return False
memory[(a,b,c)] = 1
#empty jug a
if(a>0):
#empty a into b
if(a+b<=y):
if( get_all_states((0,a+b,c)) ):
ans.append(state)
return True
else:
if( get_all_states((a-(y-b), y, c)) ):
ans.append(state)
return True
#empty a into c
if(a+c<=z):
if( get_all_states((0,b,a+c)) ):
ans.append(state)
return True
else:
if( get_all_states((a-(z-c), b, z)) ):
ans.append(state)
return True
#empty jug b
if(b>0):
#empty b into a
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if(a+b<=x):
if( get_all_states((a+b, 0, c)) ):
ans.append(state)
return True
else:
if( get_all_states((x, b-(x-a), c)) ):
ans.append(state)
return True
#empty b into c
if(b+c<=z):
if( get_all_states((a, 0, b+c)) ):
ans.append(state)
return True
else:
if( get_all_states((a, b-(z-c), z)) ):
ans.append(state)
return True
#empty jug c
if(c>0):
#empty c into a
if(a+c<=x):
if( get_all_states((a+c, b, 0)) ):
ans.append(state)
return True
else:
if( get_all_states((x, b, c-(x-a))) ):
ans.append(state)
return True
#empty c into b
if(b+c<=y):
if( get_all_states((a, b+c, 0)) ):
ans.append(state)
return True
else:
if( get_all_states((a, y, c-(y-b))) ):
ans.append(state)
return True
return False
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initial_state = (12,0,0)
print("Starting work...\n")
get_all_states(initial_state)
ans.reverse()
for i in ans:
print(i)
Output:-
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Aim:-
Design the simulation of TIC – TAC –TOE game using min-max algorithm
Diagram:-
Python Code:
import os
import time
board = [' ',' ',' ',' ',' ',' ',' ',' ',' ',' ']
player = 1
########win Flags##########
Win = 1
Draw = -1
Running = 0
Stop = 1
###########################
Game = Running
Mark = 'X'
#This Function Draws Game Board
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def DrawBoard():
print(" %c | %c | %c " % (board[1],board[2],board[3]))
print("___|___|___")
print("___|___|___")
print(" | | ")
#This Function Checks position is empty or not

def CheckPosition(x):
if(board[x] == ' '):
return True
else:
return False
#This Function Checks player has won or not

def CheckWin():
global Game
#Horizontal winning condition
if(board[1] == board[2] and board[2] == board[3] and board[1] != ' '):
Game = Win
elif(board[4] == board[5] and board[5] == board[6] and board[4] != ' '):
Game = Win
Game = Win
#Vertical Winning Condition
Game = Win
Game = Win
Game=Win
#Diagonal Winning Condition
Game = Win
Game=Win
#Match Tie or Draw Condition
elif(board[1]!=' ' and board[2]!=' ' and board[3]!=' ' and board[4]!=' ' and
board[5]!=' ' and board[6]!=' ' and board[7]!=' ' and board[8]!=' ' and board[9]!=' '):
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Game=Draw
else:
Game=Running
print("Tic-Tac-Toe Game")
print("Player 1 [X] --- Player 2 [O]\n")
print()
print()
print("Please Wait...")
time.sleep(1)
while(Game == Running):
os.system('cls')
DrawBoard()
if(player % 2 != 0):
print("Player 1's chance")
Mark = 'X'
else:
print("Player 2's chance")
Mark = 'O'
choice = int(input("Enter the position between [1-9] where you want to mark :
"))
if(CheckPosition(choice)):
board[choice] = Mark
player+=1
CheckWin()
os.system('cls')
DrawBoard()
if(Game==Draw):
print("Game Draw")
elif(Game==Win):
player-=1
if(player%2!=0):
print("Player 1 Won")
else:
print("Player 2 Won")
NOTE:-
Game Rules
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1. Traditionally the first player plays with "X". So you can decide who wants
to go with "X" and who wants go with "O".
2. Only one player can play at a time.
3. If any of the players have filled a square then the other player and the same
player cannot override that square.
4. There are only two conditions that may match will be draw or may win.
5. The player that succeeds in placing three respective marks (X or O) in a
horizontal, vertical or diagonal row wins the game.
OUTPUT:-
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PRACTICAL No.-6
A. Write a program to solve Missionaries and Cannibals problem.
B. Design an application to simulate number puzzle problem.
Aim:-
Write a program to solve Missionaries and Cannibals problem.
Diagram:-
Python Code:-
import math
# Missionaries and Cannibals Problem
class State():
def __init__(self, cannibalLeft, missionaryLeft, boat, cannibalRight,
missionaryRight):
self.cannibalLeft = cannibalLeft
self.missionaryLeft = missionaryLeft
self.boat = boat
self.cannibalRight = cannibalRight
self.missionaryRight = missionaryRight
self.parent = None
def is_goal(self):
if self.cannibalLeft == 0 and self.missionaryLeft == 0:
return True
else:
return False
def is_valid(self):
if self.missionaryLeft >= 0 and self.missionaryRight >= 0 \
and self.cannibalLeft >= 0 and self.cannibalRight >= 0 \
and (self.missionaryLeft == 0 or self.missionaryLeft >=
self.cannibalLeft) \
and (self.missionaryRight == 0 or self.missionaryRight >=
self.cannibalRight):
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return True
else:
return False
def __eq__(self, other):

return self.cannibalLeft == other.cannibalLeft and self.missionaryLeft
== other.missionaryLeft \
and self.boat == other.boat and self.cannibalRight ==
other.cannibalRight \
and self.missionaryRight == other.missionaryRight
def __hash__(self):
return hash((self.cannibalLeft, self.missionaryLeft, self.boat,
self.cannibalRight, self.missionaryRight))
def successors(cur_state):
children = [];
if cur_state.boat == 'left':
new_state = State(cur_state.cannibalLeft, cur_state.missionaryLeft -
2, 'right',
cur_state.cannibalRight, cur_state.missionaryRight + 2)
## Two missionaries cross left to right.
if new_state.is_valid():
new_state.parent = cur_state
children.append(new_state)
new_state = State(cur_state.cannibalLeft - 2,
cur_state.missionaryLeft, 'right',
cur_state.cannibalRight + 2, cur_state.missionaryRight)
## Two cannibals cross left to right.
new_state = State(cur_state.cannibalLeft - 1, cur_state.missionaryLeft
- 1, 'right',
cur_state.cannibalRight + 1, cur_state.missionaryRight + 1)
## One missionary and one cannibal cross left to right.
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new_state = State(cur_state.cannibalLeft, cur_state.missionaryLeft -

1, 'right',
cur_state.cannibalRight, cur_state.missionaryRight + 1)
## One missionary crosses left to right.
new_state = State(cur_state.cannibalLeft - 1,
cur_state.missionaryLeft, 'right',
cur_state.cannibalRight + 1, cur_state.missionaryRight)
## One cannibal crosses left to right.
else:
new_state = State(cur_state.cannibalLeft, cur_state.missionaryLeft +
2, 'left',
cur_state.cannibalRight, cur_state.missionaryRight - 2)
## Two missionaries cross right to left.
new_state = State(cur_state.cannibalLeft + 2,
cur_state.missionaryLeft, 'left',
cur_state.cannibalRight - 2, cur_state.missionaryRight)
## Two cannibals cross right to left.
new_state = State(cur_state.cannibalLeft + 1, cur_state.missionaryLeft
+ 1, 'left',
cur_state.cannibalRight - 1, cur_state.missionaryRight - 1)
## One missionary and one cannibal cross right to left.
new_state = State(cur_state.cannibalLeft, cur_state.missionaryLeft +
1, 'left',
cur_state.cannibalRight, cur_state.missionaryRight - 1)
## One missionary crosses right to left.
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new_state = State(cur_state.cannibalLeft + 1,
cur_state.missionaryLeft, 'left',
cur_state.cannibalRight - 1, cur_state.missionaryRight)
## One cannibal crosses right to left.
return children
def breadth_first_search():
initial_state = State(3,3,'left',0,0)
if initial_state.is_goal():
return initial_state
frontier = list()
explored = set()
frontier.append(initial_state)
while frontier:
state = frontier.pop(0)
if state.is_goal():
return state
explored.add(state)
children = successors(state)
for child in children:
if (child not in explored) or (child not in frontier):
frontier.append(child)
return None
def print_solution(solution):
path = []
path.append(solution)
parent = solution.parent
while parent:
path.append(parent)
parent = parent.parent
for t in range(len(path)):
state = path[len(path) - t - 1]
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print ("(" + str(state.cannibalLeft) + "," +

str(state.missionaryLeft) \
+ "," + state.boat + "," + str(state.cannibalRight) + "," + \
str(state.missionaryRight) + ")")
def main():
solution = breadth_first_search()
print ("Missionaries and Cannibals solution:")
print ("(cannibalLeft,missionaryLeft,boat,cannibalRight,missionaryRight)")
print_solution(solution)
# if called from the command line, call main()

if __name__ == "__main__":
main()
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AIM:-
Design an application to simulate number puzzle problem.
PYHTON CODE:-
'''
8 puzzle problem, a smaller version of the fifteen puzzle:
States are defined as string representations of the pieces on the puzzle.
Actions denote what piece will be moved to the empty space.
States must allways be inmutable. We will use strings, but internally most of
the time we will convert those strings to lists, which are easier to handle.
For example, the state (string):
'1-2-3
4-5-6
7-8-e'
will become (in lists):
[['1', '2', '3'],
['4', '5', '6'],
['7', '8', 'e']]
'''
from __future__ import print_function

from simpleai.search import astar, SearchProblem
from simpleai.search.viewers import WebViewer
GOAL = '''1-2-3
4-5-6
7-8-e'''
INITIAL = '''4-1-2
7-e-3
8-5-6'''
def list_to_string(list_):
return '\n'.join(['-'.join(row) for row in list_])
def string_to_list(string_):
return [row.split('-') for row in string_.split('\n')]
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def find_location(rows, element_to_find):

'''Find the location of a piece in the puzzle.
Returns a tuple: row, column'''
for ir, row in enumerate(rows):
for ic, element in enumerate(row):
if element == element_to_find:
return ir, ic
# we create a cache for the goal position of each piece, so we don't have to
# recalculate them every time
goal_positions = {}
rows_goal = string_to_list(GOAL)
for number in '12345678e':
goal_positions[number] = find_location(rows_goal, number)
class EigthPuzzleProblem(SearchProblem):
def actions(self, state):
'''Returns a list of the pieces we can move to the empty space.'''
rows = string_to_list(state)
row_e, col_e = find_location(rows, 'e')
actions = []
if row_e > 0:
actions.append(rows[row_e - 1][col_e])
if row_e < 2:
actions.append(rows[row_e + 1][col_e])
if col_e > 0:
actions.append(rows[row_e][col_e - 1])
if col_e < 2:
actions.append(rows[row_e][col_e + 1])
return actions
def result(self, state, action):

'''Return the resulting state after moving a piece to the empty space.
(the "action" parameter contains the piece to move)
'''
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row_e, col_e = find_location(rows, 'e')
row_n, col_n = find_location(rows, action)
rows[row_e][col_e], rows[row_n][col_n] = rows[row_n][col_n],

rows[row_e][col_e]
return list_to_string(rows)
def is_goal(self, state):

'''Returns true if a state is the goal state.'''
return state == GOAL
def cost(self, state1, action, state2):

'''Returns the cost of performing an action. No useful on this problem, i
but needed.
'''
return 1
def heuristic(self, state):

'''Returns an *estimation* of the distance from a state to the goal.
We are using the manhattan distance.
'''
distance = 0
for number in '12345678e':
row_n, col_n = find_location(rows, number)
row_n_goal, col_n_goal = goal_positions[number]
distance += abs(row_n - row_n_goal) + abs(col_n - col_n_goal)
return distance
result = astar(EigthPuzzleProblem(INITIAL))
for action, state in result.path():
print('Move number', action)
print(state)
OUTPUT:
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PRACTICAL No.-7
A. Write a program to shuffle Deck of cards.
B. Solve traveling salesman problem using artificial intelligence technique.
Aim:-
Write a program to shuffle Deck of cards.
Diagram:-
Python Code:-
#first let's import random procedures since we will be shuffling
import random
#next, let's start building list holders so we can place our cards in there:
cardfaces = []
suits = ["Hearts", "Diamonds", "Clubs", "Spades"]
royals = ["J", "Q", "K", "A"]
deck = []
#now, let's start using loops to add our content:

for i in range(2,11):
cardfaces.append(str(i)) #this adds numbers 2-10 and converts them to string
data
for j in range(4):
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cardfaces.append(royals[j]) #this will add the royal faces to the cardbase
for k in range(4):
for l in range(13):
card = (cardfaces[l] + " of " + suits[k])
#this makes each card, cycling through suits, but first through faces
deck.append(card)
#this adds the information to the "full deck" we want to make
#now let's shuffle our deck!
random.shuffle(deck)
#now let's see the cards!

for m in range(52):
print(deck[m])
OR
# Python program to shuffle a deck of card using the module random and
draw 5 cards
# import modules
import itertools, random
# make a deck of cards
deck = list(itertools.product(range(1,14),['Spade','Heart','Diamond','Club']))
# shuffle the cards
random.shuffle(deck)
# draw five cards
print("You got:")
for i in range(5):
print(deck[i][0], "of", deck[i][1])
Output:-
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OR
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PRACTICAL No.-8
A. Solve the block of World problem.
B. Solve constraint satisfaction problem
Aim:-
Implementation Of Constraints Satisfactions Problem
PYTHON CODE:
from __future__ import print_function
from simpleai.search import CspProblem, backtrack, min_conflicts,

MOST_CONSTRAINED_VARIABLE, HIGHEST_DEGREE_VARIABLE,
LEAST_CONSTRAINING_VALUE
variables = ('WA', 'NT', 'SA', 'Q', 'NSW', 'V', 'T')
domains = dict((v, ['red', 'green', 'blue']) for v in variables)
def const_different(variables, values):

return values[0] != values[1] # expect the value of the neighbors to be different
constraints = [
(('WA', 'NT'), const_different),
(('WA', 'SA'), const_different),
(('SA', 'NT'), const_different),
(('SA', 'Q'), const_different),
(('NT', 'Q'), const_different),
(('SA', 'NSW'), const_different),
(('Q', 'NSW'), const_different),
(('SA', 'V'), const_different),
(('NSW', 'V'), const_different),
]
my_problem = CspProblem(variables, domains, constraints)
print(backtrack(my_problem))
print(backtrack(my_problem,
variable_heuristic=MOST_CONSTRAINED_VARIABLE))
variable_heuristic=HIGHEST_DEGREE_VARIABLE))
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value_heuristic=LEAST_CONSTRAINING_VALUE))
variable_heuristic=MOST_CONSTRAINED_VARIABLE,
variable_heuristic=HIGHEST_DEGREE_VARIABLE,
print(min_conflicts(my_problem))
OUTPUT:-
Note:
Following practical will be update soon:
• Practical No-4-B
• Practical No.-8-A
• Practical No-9
• Practical No.10.
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57 | P a g e
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T.Y.B.Sc.

IT ARITIFICIAL INTELLIGENCE 2018-19

Teacher’s Reference Manual

with effect from the academic year

if neighbor not in visited:

#For Finding Breadth First Search Path

#For finding shortest path

"""Display a chessboard for each possible configuration of placing n

# if could not find column to place in or if board is full

# thus no more possible configurations

def children(branch, depth, alpha, beta):

def alphabeta(in_tree=tree, start=root, upper=-15, lower=15):

(alpha, beta) = children(tree, start, upper, lower)

return (alpha, beta, tree, pruned)

def distance(x1, y1, x2, y2):

def newDistance(x1, y1, point1, point2, point3, point4):

minDistance = sumOfDistances(startingPoint[0], startingPoint[1],

def newPoints(minimum, d1, d2, d3, d4):

GOAL = 'HELLO WORLD'

def result(self, state, action):

def is_goal(self, state):

def heuristic(self, state):

# to mark visited states

# store solution path

if(a==6 and b==6):

# if current state is already visited earlier

#This Function Draws Game Board

#This Function Checks position is empty or not

#This Function Checks player has won or not

def __eq__(self, other):

new_state = State(cur_state.cannibalLeft, cur_state.missionaryLeft -

print ("(" + str(state.cannibalLeft) + "," +

# if called from the command line, call main()

from __future__ import print_function

def find_location(rows, element_to_find):

def result(self, state, action):

rows[row_e][col_e], rows[row_n][col_n] = rows[row_n][col_n],

def is_goal(self, state):

def cost(self, state1, action, state2):

def heuristic(self, state):

#now, let's start using loops to add our content:

cardfaces.append(royals[j]) #this will add the royal faces to the cardbase

#now let's see the cards!

from simpleai.search import CspProblem, backtrack, min_conflicts,

variables = ('WA', 'NT', 'SA', 'Q', 'NSW', 'V', 'T')

domains = dict((v, ['red', 'green', 'blue']) for v in variables)

def const_different(variables, values):

my_problem = CspProblem(variables, domains, constraints)

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def eq(self, other):

from future import print_function