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A lottery ticket contains five unique numbers. A set of unique numbers does not contain repeated elements. The winning combinati

RSB [31]

Answer:

Answer explained below

Explanation:

I have given two approaches in implementing the solution.

1. Using the for loop, in which you have to iterate over all the elements in list 1 and check in list 2

2. Use the set intersection method. As intersection will give u the common elements. And we can get there length by using len method.

I have added the code along with the snapshot and inline comment for the ease of you to understand. Please check the methods below. You can use either of them.

METHOD-1:

********** CODE *****************

def matches(tickets,winner):

tickets = set(tickets)

winner = set(winner)

counter = 0 #To Count the common elements

for i in tickets: # Iterate over all the elements in tickets.

if i in winner: # Check the element in the winner list

counter = counter+1

return counter

METHOD -2:

********** CODE ********************

def matches(tickets, winner):

tickets = set(tickets)

winner = set(winner)

return len(tickets.intersection(winner))

6 0

3 years ago

The ____ algorithm was the first public key encryption algorithm developed (in 1977 and published for commercial use.

Whitepunk [10]

The RSA Algorithm was developed by and named after the initials of: Ron Rivist, Adi Shamir, and Leonard Adleman while they were working at MIT. It was developed in 1977 and published in 1978. The algorithm uses the exponentiation modulo in encrypt and decrypt information.

7 0

4 years ago

Alexei was given a key to a substitution cipher. The key showed that the entire alphabet was rotated 13 steps. What type of ciph

larisa [96]

Answer:

ROT13

Explanation:

8 0

4 years ago

) how many bits are used for host number on the child network (subnet) , b) how many usable addresses can exist on this child ne

vovikov84 [41]

Answer:

Machine’s IP=126.127.85.170, Machine’s Netmask=/27, Parent’s Netmask=255.255.240.0 .

a) Machine's Netmask = /27 : therefore no. of remaining bits for hosts =32-27 = 5 bits.

b) No. of usable addresses in the child network = 25 -2 = 32-2 =30 [Since first(network ID of the machine) and last ip (broadcast address of the machine ) addresses are not used ]

c) first usable address is on this child network (subnet) =

First, find out the network id of machine can be found out by doing bitwise AND machine's IP and Machine's subnet mask :

01111110. 01111111.01010101.10101010 (IP)

11111111. 11111111 .11111111 .11100000 (Subnet Mask)

01111110. 01111111. 01010101.10100000 (Network ID )

first usable address is on this child network (subnet) : 01111110. 01111111. 01010101.10100001

: 126.127.85.161

d) what the last usable address is on this child network (subnet) : 01111110. 01111111. 01010101.10111110

: 126.127.85.190

e) what the child network’s (subnet's) broadcast address is :

In the directed broadcast address , all the host bits are 1. Therefore, broadcast address :

01111110. 01111111. 01010101.10111111 (126.127.85.191)

f) what the child network's (subnet's) network number is : Network ID has already been calcuulated in part c .

01111110. 01111111. 01010101.10100000 (126.127.85.160)

6 0

4 years ago

Given the following adjacency matrix, what is the approximate rank vector after one iteration of the power iteration method (use

ValentinkaMS [17]

Note: The matrix referred to in the question is: $M = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]$

Answer:

a) [5/18, 5/18, 4/9]'

Explanation:

The adjacency matrix is $M = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]$

To start the power iteration, let us start with an initial non zero approximation,

$X_o = \left[\begin{array}{ccc}1\\1\\1\end{array}\right]$

To get the rank vector for the first Iteration:

$X_1 = MX_0$

$X_1 = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]\left[\begin{array}{ccc}1\\1\\1\end{array}\right] \\\\X_1 = \left[\begin{array}{ccc}5/6\\5/6\\4/3\end{array}\right]\\$

Multiplying the above matrix by 1/3

$X_1 = \left[\begin{array}{ccc}5/18\\5/18\\4/9\end{array}\right]$

8 0

3 years ago