Code to be written in python:

Correct answer will be automatically awarded the brainliest!

Since you now have a good understanding of the new situation, write a new num_of_paths function to get the number of ways out. The function should take in a map of maze that Yee Sian sent to you and return the result as an integer. The map is a tuple of n tuples, each with m values. The values inside the tuple are either 0 or 1. So maze[i][j] will tell you what's in cell (i, j) and 0 stands for a bomb in that cell.
For example, this is the maze we saw in the previous question:

((1, 1, 1, 1, 1, 1, 1, 1, 0, 1),
(1, 0, 0, 1, 1, 1, 0, 0, 1, 1),
(0, 1, 1, 1, 0, 0, 1, 1, 1, 0),
(1, 1, 0, 1, 1, 1, 1, 0, 1, 1),
(0, 1, 0, 1, 0, 0, 1, 0, 1, 0),
(1, 0, 1, 1, 1, 1, 0, 1, 1, 1),
(0, 1, 1, 1, 1, 1, 1, 1, 1, 0),
(1, 0, 1, 0, 0, 1, 1, 0, 1, 1),
(1, 0, 1, 1, 1, 0, 1, 0, 1, 0),
(1, 1, 0, 1, 0, 1, 0, 1, 1, 1))
Note: You should be using dynamic programming to pass time limitation test.

Hint: You might find the following algorithm useful:

Initialize an empty table (dictionary), get the number of rows n and number of columns m.
Fill in the first row. For j in range m:
2.1 If maze[0][j] is safe, set table[(0, j)] to be 1 because there's one way to go there.
2.2 If maze[0][j] has a bomb, set table[(0, k)] where k >= j to be 0. Since one cell is broken along the way, all following cells (in the first row) cannot be reached.
Fill in the first column. For i in range n:
3.1 If maze[i][0] is safe, set table[(i, 0)] to be 1 because there's one way to go there.
3.2 If maze[i][0] has a bomb, set table[(i, 0)] and all cells under it to be 0. The reason is same as for the first row.

Main dynamic programming procedure - fill in the rest of the table.
If maze[i][j] has a bomb, set table[(i, j)] = 0.
Otherwise, table[(i, j)] = table[(i - 1, j)] + table[(i, j - 1)]
Return table[(n - 1, m - 1)]

Incomplete code:

def num_of_paths(maze):
# your code here

# Do NOT modify
maze1 = ((1, 1, 1, 1, 1, 1, 1, 1, 0, 1),
(1, 0, 0, 1, 1, 1, 0, 0, 1, 1),
(0, 1, 1, 1, 0, 0, 1, 1, 1, 0),
(1, 1, 0, 1, 1, 1, 1, 0, 1, 1),
(0, 1, 0, 1, 0, 0, 1, 0, 1, 0),
(1, 0, 1, 1, 1, 1, 0, 1, 1, 1),
(1, 1, 0, 1, 0, 1, 0, 0, 1, 1),
(0, 1, 1, 1, 1, 1, 1, 1, 1, 0),
(1, 0, 1, 0, 0, 1, 1, 0, 1, 1),
(1, 0, 1, 1, 1, 0, 1, 0, 1, 0),
(1, 1, 0, 1, 0, 1, 0, 1, 1, 1))


maze2 = ((1, 1, 1, 1, 1, 1, 1, 1, 1),
(1, 1, 1, 1, 1, 1, 1, 1, 1),
(1, 1, 1, 1, 1, 1, 1, 1, 1),
(1, 1, 1, 1, 1, 1, 1, 1, 1),
(1, 1, 1, 1, 1, 1, 1, 1, 1),
(1, 1, 1, 1, 1, 1, 1, 1, 1),
(1, 1, 1, 1, 1, 1, 1, 1, 1))

maze3 = ((1, 0, 1, 1),
(1, 0, 1, 1),
(1, 0, 1, 1),
(1, 0, 1, 1),
(1, 0, 1, 0),
(1, 0, 0, 1))

Test Cases:
num_of_paths(maze1) 2
num_of_paths(maze2) 3003
num_of_paths(maze3) 0

Answer:

def num_of_paths(maze):

   # Initialize the table with zeros

   n = len(maze)

   m = len(maze[0])

   table = [[0 for j in range(m)] for i in range(n)]

   # Fill in the first row

   for j in range(m):

       if maze[0][j] == 0:

           break

       table[0][j] = 1

   # Fill in the first column

   for i in range(n):

       if maze[i][0] == 0:

           break

       table[i][0] = 1

   # Fill in the rest of the table using dynamic programming

   for i in range(1, n):

       for j in range(1, m):

           if maze[i][j] == 0:

               table[i][j] = 0

           else:

               table[i][j] = table[i - 1][j] + table[i][j - 1]

   # Return the value in the bottom right corner of the table

   return table[n - 1][m - 1]

maze1 = ((1, 1, 1, 1, 1, 1, 1, 1, 0, 1),

(1, 0, 0, 1, 1, 1, 0, 0, 1, 1),

(0, 1, 1, 1, 0, 0, 1, 1, 1, 0),

(1, 1, 0, 1, 1, 1, 1, 0, 1, 1),

(0, 1, 0, 1, 0, 0, 1, 0, 1, 0),

(1, 0, 1, 1, 1, 1, 0, 1, 1, 1),

(1, 1, 0, 1, 0, 1, 0, 0, 1, 1),

(0, 1, 1, 1, 1, 1, 1, 1, 1, 0),

(1, 0, 1, 0, 0, 1, 1, 0, 1, 1),

(1, 0, 1, 1, 1, 0, 1, 0, 1, 0),

(1, 1, 0, 1, 0, 1, 0, 1, 1, 1))

maze2 = ((1, 1, 1, 1, 1, 1, 1, 1, 1),

(1, 1, 1, 1, 1, 1, 1, 1, 1),

(1, 1, 1, 1, 1, 1, 1, 1, 1),

(1, 1, 1, 1, 1, 1, 1, 1, 1),

(1, 1, 1, 1, 1, 1, 1, 1, 1),

(1, 1, 1, 1, 1, 1, 1, 1, 1),

(1, 1, 1, 1, 1, 1, 1, 1, 1))

maze3 = ((1, 0, 1, 1),

(1, 0, 1, 1),

(1, 0, 1, 1),

(1, 0, 1, 1),

(1, 0, 1, 0),

(1, 0, 0, 1))

print(num_of_paths(maze1))

print(num_of_paths(maze2))

print(num_of_paths(maze3))

What is wrong with the following code?
int name = "Steve":
if (name = "Steve") {
System.out.prſntln("Hi Steve!);
}

In this problem, you will derive the efficiency of a CSMA/CD-like multiple access protocol. In this protocol, time is slotted and all adapters are synchronized to the slots. Unlike slotted ALOHA, however, the length of a slot (in seconds) is much less than a frame time (the time to transmit a frame). Let S be the length of a slot. Suppose all frames are of constant length L = kRS, where R is the transmission rate of the channel and k is a large integer. Suppose there are N nodes, each with an infinite number of frames to send. We also assume that dprop < s,="" so="" that="" all="" nodes="" can="" detect="" a="" collision="" before="" the="" end="" of="" a="" slot="" time.="" the="" protocol="" is="" as="">

⢠If, for a given slot, no node has possession of the channel, all nodes contend for the channel; in particular, each node transmits in the slot with probability p. If exactly one node transmits in the slot, that node takes possession of the channel for the subsequent k â 1 slots and transmits its entire frame.

If some node has possession of the channel, all other nodes refrain from transmitting until the node that possesses the channel has finished transmitting its frame. Once this node has transmitted its frame, all nodes contend for the channel.

Note that the channel alternates between two states: the productive state, which lasts exactly k slots, and the nonproductive state, which lasts for a random number of slots. Clearly, the channel efficiency is the ratio of k/(k + x), where x is the expected number of consecutive unproductive slots

a. For fixed N and p, determine the efficiency of this protocol.
b. For fixed N, determine the p that maximizes the efficiency.
c. Using the p (which is a function of N) found in (b), determine the efficiency as N approaches infinity
d. Show that this efficiency approaches 1 as the frame length becomes large

Hi, I have a CodeHS Python assignment called "Exclamation Points"


the assignment is as follows and I have no idea where to start with this if someone could help me please:


"Words are way more edgy when you replace the letter i with an exclamation point!


Write the function exclamations that takes a string and then returns the same string with every lowercase i replaced with an exclamation point. Your function should:


Convert the initial string to a list

Use a for loop to go through your list element by element

Whenever you see a lowercase i, replace it with an exclamation point in the list

Return the stringified version of the list when your for loop is finished"

In a network, servers receive requests from which of the following?
clients, which are the networked computers that request data.
O other networked computers, which use encrypted messages.
O ISPs, which control the type of data that can be sent on the network.
O routers, which direct the data to the correct destination.

HELP WILL MARK BRAINLEST

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This image shows a web designer's grids for different pages on a website. The uppermost box on each page is the website's identity. Which important
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A simplicity
B.
consistency
Ос
clarity
D. harmony

The largest form of a computer network is called
कम्प्युटर नेटवर्कको सबैभन्दा ठूलो रूप
aekar.xe.​

Match each characteristic to its operating system.

[A] This system can be ported to more computer platforms than any of its counterparts.
[B] The operating system functions require many system resources.
[C] The operating system works only with specific hardware.

[1] Microsoft Windows Operating System
[2] Linux Operating System
[3] Apple Operating System

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A) They enable the transmission of videos.
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C) They allow files to be shared by sending attachments.
D) They provide increased security.

which type of computer is used to process large amount of data​

Hey tell me more about your service

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The purpose of a good web page design is to make it
and

plz i need help what is wrong on line 14

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Linux

Microsoft

Apple

IBM

Design and implement a program that reads a series of 10 integers from the user and prints their average. Read each input value as a string, and then attempt to convert it to an integer using the Integer.parseInt method. If this process throws a NumberFormatException (meaning that the input is not a valid number), print an appropriate error message and prompt for the number again. Continue reading values until 10 invalid integers have been entered.

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