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Anvisha [2.4K]
2 years ago
7

A DTP firm has published and printed flyers for an upcoming fundraising event. In which section of the flyer would you find the

contact details of the DTP firm?
A.)body
B.)credits
C.)navigation
D.)title
Computers and Technology
1 answer:
QveST [7]2 years ago
6 0
The Contact Details of the DTP firm would usually be found at the bottom of the flyer, to give way to the event to be advertised on the top and the middle of the flyer.
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Explanation:

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Exercise 9.1.6: Checkerboard, v1 Spoint
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Answer:

In order to get following pattern, we have to use numpy package.

Following code with work perfectly fine and will print the pattern.

Python code starts as below

*********************************************

<em># Python program to print 8 X 8 alternative 1 and 0's. 3rd and 4th row with all 0's </em>

<em># checkerboard pattern using numpy </em>

<em># We need Following pattern </em>

<em># 0 1 0 1 0 1 0 1 </em>

<em># 1 0 1 0 1 0 1 0 </em>

<em># 0 1 0 1 0 1 0 1 </em>

<em># 0 0 0 0 0 0 0 0 </em>

<em># 0 0 0 0 0 0 0 0 </em>

<em># 1 0 1 0 1 0 1 0 </em>

<em># 0 1 0 1 0 1 0 1 </em>

<em># 1 0 1 0 1 0 1 0 </em>

<em> </em>

<em>import numpy as np </em>

<em> </em>

<em># function to print Checkerboard pattern </em>

<em>def printcheckboard(n): </em>

<em> </em>

<em>        print(" Customized Checkerboard pattern:") </em>

<em>        # create a n * n matrix   </em>

<em>        x = np.zeros((n, n), dtype = int) </em>

<em>        y = np.zeros((n, n), dtype = int) </em>

<em>        # fill with 1 the alternate rows and columns </em>

<em>        x[1::2, ::2] = 1 </em>

<em>        x[::2, 1::2] = 1 </em>

<em>       # fill with 0 the alternate rows and columns </em>

<em>        y[1::2, ::2] = 0 </em>

<em>        y[::2, 1::2] = 0 </em>

<em> </em>

<em>        # print the pattern  for first 3 rows</em>

<em>        for i in range(0,3): </em>

<em>                for j in range(n): </em>

<em>                        print(x[i][j], end =" ") </em>

<em>                print() </em>

<em>        # print the pattern   for next two rows with all 0's</em>

<em>        for k in range(3,5): </em>

<em>                for l in range(n): </em>

<em>                        print(y[k][l], end =" ") </em>

<em>                print() </em>

<em>         # print the pattern  for last 3 rows with alternative 1 and 0.        </em>

<em>        for i in range(5,8): </em>

<em>                for j in range(n): </em>

<em>                        print(x[i][j], end =" ") </em>

<em>                print() </em>

<em> </em>

<em># Calling the function code </em>

<em>n = 8 </em>

<em>printcheckboard(n)</em>

**************************************

End of the Python Code.

Explanation:

In this you have to use Python3.7 and numpy should be installed as well in order to execute the code successfully.

2 D matrix will be created using Python numpy library and checkboard pattern is generated using array slicing.

Here n=8 and it will generate the checkerboard pattern of alternative 0 and 1. However, we need row 4th and 5th as all 0. So we have initialized two arrays matrix as x and y.

Comments in the code is self explanatory.

PS: Please make sure code is edited in IDE so that tabs or space issues can be taken care.

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2 years ago
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