26.2 Practice Let's Connect (Pyhton Data Structures)

Question

Problem:

Graphs

You are making a social network called X.
Connections between the users are stored as a graph.
The given code declares an X class with its add_friend() method and creates some connections for 5 users.

You need to take a number as input and output the number of connections of the corresponding user.

Each user's connections are stored in the adj matrix.

class X():
def __init__(self, size):
self.adj = [ [0] * size for i in range(size)]
self.size = size

def add_friend(self, x, y):
if x > self.size or y > self.size or x < 0 or y < 0:
print("Error")
else:
self.adj[x-1][y-1] = 1
self.adj[y-1][x-1] = 1

def remove_friend(self, x, y):
if x > self.size or y > self.size or x < 0 or y < 0:
print("Error")
else:
self.adj[x-1][y-1] = 0
self.adj[y-1][x-1] = 0

x = X(5)

x.add_friend(1, 3)
x.add_friend(1, 5)
x.add_friend(2, 5)
x.add_friend(2, 4)
x.add_friend(4, 5)

n = int(input())
#your code goes here

I hate doing this, but I'm just not sure what to do for this

Answer 1

Another one you could try is this one:

class X():
def __init__(self, size):
self.adj = [[0] * size for i in range(size)]
self.size = size

def add_friend(self, x, y):
if x > self.size or y > self.size or x < 0 or y < 0:
print("Error")
else:
self.adj[x - 1][y - 1] = 1
self.adj[y - 1][x - 1] = 1

def remove_friend(self, x, y):
if x > self.size or y > self.size or x < 0 or y < 0:
print("Error")
else:
self.adj[x - 1][y - 1] = 0
self.adj[y - 1][x - 1] = 0

def get_connections(self, user):
if user > self.size or user < 1:
print("Error")
else:
return sum(self.adj[user - 1])

x = X(5)

x.add_friend(1, 3)
x.add_friend(1, 5)
x.add_friend(2, 5)
x.add_friend(2, 4)
x.add_friend(4, 5)

n = int(input("Enter the user number: "))
connections = x.get_connections(n)
print("Number of connections:", connections)

Answer 2

#your code goes here
a =0
for i in range (len(x.adj)):
if i+1==n:
for j in x.adj[i]:
if j==1:
a += 1
print(a)

Answer 3

class X():
def __init__(self, size):
self.adj = [ [0] * size for i in range(size)]
self.size = size

def add_friend(self, x, y):
if x > self.size or y > self.size or x < 0 or y < 0:
print("Error")
else:
self.adj[x-1][y-1] = 1
self.adj[y-1][x-1] = 1

def remove_friend(self, x, y):
if x > self.size or y > self.size or x < 0 or y < 0:
print("Error")
else:
self.adj[x-1][y-1] = 0
self.adj[y-1][x-1] = 0

x = X(5)

x.add_friend(1, 3)
x.add_friend(1, 5)
x.add_friend(2, 5)
x.add_friend(2, 4)
x.add_friend(4, 5)

n = int(input())
#your code goes here
a =0
for i in range (len(x.adj)):
if i+1==n:
for j in x.adj[i]:
if j==1:
a += 1
print(a)

Answer 4

cory p what part are you not sure about ?
In the 26.1 section 3rd question the details may be helpful if you look at the "try it"

Answer 5

ok, I'm going to look at it again

Answer 6

I think this is the part 26.1 Q3 I need to incorporate, right?:
def display(self):
for row in self.adj:
print()
for val in row:
print('{:4}'.format(val),end="")

Answer 7

Alright, thank you for the guidance. I'll keep working on it :)

Answer 8

Could anyone explain to me what is the input should be? (How does the input number affect the number of connections of the corresponding users?) (because I could not figure out what is the relation of the input with the number of connections of the corresponding users)

Answer 9

cory p Yes something along those lines -