can anybody answer this? I couldn't get the logic

Question

Print pattern like this
Example: Input: 1
Output: 0
Input: 2
Output:
0 0
0 1
1 0
1 1

Answer 1

n - input

Yeah so checking the pattern, you've to print n^2 permutations with 0/1 as numbers in it

Loop for n^2 times and print different arrangments of size "n"

Answer 2

Those are binary number with n digits.
For ex: n=1
Output: 0 or 1
n=2:
a= 0 ,b=0 = 0
0, 1. =1
a= 1, 0. =2
1, 1 =3

n= 3, take
a ,b, c
0 0 0 = 0
0 0 1 = 1
0 1 0 = 2
0 1 1 = 3
1 0 0 = 4
1 0 1 = 5
1 1 0 = 6
1 1 1 = 7

how many number are forming by n digits ..
It's equally n*n binary numbers from 0 to n^2-1 (n squire -1)

for n=4,
you get 0 to 15 (4*4-1=15)
0 0 0 0 (0) to
....
....
1 1 1 1 (15)

can say it as different permutations with values 0,1 filled in n places.
Gayathri hope it helps.....

Answer 3

I think how many permutations can be made by in range(0, n).
What is output if input is 3?
where do you found this?

Answer 4

for Input: 3
Output:
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1

Answer 5

You could have to implement binary number generator function by your own with spaces adding between.. I think it is not difficult....
edit : Gayathri
see this , you can find an idea how to implement it..
https://www.rapidtables.com/convert/number/decimal-to-binary.html

Answer 6

Solve the problem by building a function that return the successor of the inputing result; add +1.
Let B’ becomes the binary successor of the binary number B. Let bn, b(n-1), ..., b0 be the digits in B from left to right. Thus:

Decimal repr of B: B|10|= (bn * 2**n) + ... + (b1 * 2**1) + (b0 * 2**0)

Binary repr: B = bn ... b1 b0

bi = 0 or bi = 1. c0 <= i <= n and n is a non negativ integer in B|10|. b0 is the rightmost digit in B, and bn the leftmost digit. B contains (n + 1) digits.

To get B:s successor B’: Start with the leftmost digit bi in B, where i=0 (pseudo code):

i = 0
while i <= n: # n from B|10|
if bi == 0:
bi = 1
break # Finished!
else:
bi = 0
if i == n: # The leftmost digit.
b(n+1) = 1 # Add a new digit!
i += 1

The successor B’ will contain between (n+1) to (n+2) digits.

Here’s a example of what I meant above:

https://code.sololearn.com/c3mwKO8kCz38/?ref=app

Now try to solve your problem. If you stuck, we help you.

Answer 7

https://www.sololearn.com/post/1132307/?ref=app

this is what I'm asking about👆🙂

Answer 8

i could understand the concept but how to print those binary numbers without using the builtin function

Answer 9

tnq u so much Jayakrishna🇮🇳 🤗i got it

Answer 10

You're welcome Gayathri ..

Answer 11

Python language is logic take that product cross produc. ..diagognal product and print it after that....

Answer 12

0000
0010
0110
0101
1010
0011
1100
1110
0111

Answer 13

For input:3
Output:
0000
0011
01010
1111

Answer 14

print("Hi Binary Printing Algo Here")
inp_num=int(input("Enter The Number : "))
for i in range(0,2**inp_num):
print(bin(i)[2:])

Maybe this should help u ..

Answer 15

Sorry

Answer 16

Gayathri Here's a possible solution:

for i in range(k:=(2**int(input()))): print(bin(i)[2:].zfill(len(bin(k))-3))

# Hope this helps