How can I write this pattern in Python?

Question

1.44/5 (2 votes)

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Print the following pattern in python for the given number of rows.
Pattern for N = 5
1 2 3 4 5
11 12 13 14 15
21 22 23 24 25
16 17 18 19 20
6 7 8 9 10

What I have tried:

Python

n = int(input())
upper = int(n/2)+1
lower = n - upper
start_i_lower = 0

for i in range(0, upper):
  if i != 0:
    i = i + i
  start = n*i + 1
  end = start + n

  for j in range(start, end):
    print(j, end =" ")
  start_i_lower = i
  print()

start_i_lower -= 1

for i in range(0, lower):
  if(n%2 == 0):
    start_i_lower -= 1

  start = n*(start_i_lower-1) + 1
  end = start + n

  for j in range(1, n+1):
        print(j+n, end =" ")

  print()

Posted 30-Sep-20 5:42am

Member 14952571

Updated 13-May-23 8:23am

Richard MacCutchan

v2

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Comments

Richard MacCutchan 30-Sep-20 11:52am

What is the question?

OriginalGriff 30-Sep-20 12:28pm

And?
What does it do that you didn't expect, or not do that you did?
What have you tried to do to find out why?
Are there any error messages, and if so, where and when? What did you do to make them happen?

This is not a good question - we cannot work out from that little what you are trying to do.
Remember that we can't see your screen, access your HDD, or read your mind - we only get exactly what you type to work with.
Use the "Improve question" widget to edit your question and provide better information.

5 solutions

Solution 8

n = int(input())

if n%2 == 0:
m = n//2
else:
m = n//2+1

for i in range(1, m+1, 1):
for j in range(1, n+1, 1):
print(2*n*(i-1)+j, end ='')
print(' ',end = '')
print()

r = n - m
for i in range(r, 0, -1):
for j in range(1, n+1, 1):
print((2*i-1)*n + j, end ='')
print(' ', end ='')
print()

Posted 13-May-23 8:23am

Member 16003557

Solution 7

n = int(input())
if n % 2 == 0:
    n1 = int(n/2)
    n2 = n-n1
else:
    n1 = int(n/2)+1
    n2 = n-n1

x = 1
i = 1
i1 = 1

#Upper pattern
while i <= n1:
    for j in range(1, n+1):
        print(x, end=" ")
        x = x+1
    print()
    if i == n1:
        break
    x = x+n
    i = i+1


#Lowe Pattern
if n % 2 == 0:
    while i1 <= n2:
        for k in range(1, n+1):
            print(x, end=" ")
            x = x+1
        x=x-(2*n)
        print()
        i1 = i1+1
        x = x-n
        y=y+1

else:
    while i1 <= n2:
        x = x-(2*n)
        for k in range(1, n+1):
            print(x, end=" ")
            x = x+1
        x=x-n
        print()
        i1 = i1+1

Posted 21-Jan-23 20:03pm

Member 15901317

Updated 21-Jan-23 20:10pm

v2

Solution 6

n = int(input( ))
P = 1
for i in range(1, n + 1):
    for j in range(P, P + n):
        print(j, end=" ")
    print()
    if i == ((n + 1)//2):
        if (n % 2) != 0:
            P = n*(n - 2) + 1
        else:
            P = n * (n - 1) + 1
    elif i > ((n + 1) // 2):
        P = P - 2 * n
    else:
        P = P + 2 * n

It's a pretty simple and straightforward question. Sure a little tricky one

Posted 24-Jun-22 1:08am

x_cud_x

Updated 24-Jun-22 1:28am

v2

Comments

CHill60 24-Jun-22 7:33am

A word to the wise - answering "old" questions that already have solutions is often perceived as "rep point hunting" and in the past has led to accounts being suspended. My advice is to stick to posting solutions to more recent posts, and if there is already a solution posted, make sure you add commentary to explain why yours is better/different/an alternative approach.

Solution 5

The more optimized version will be:

Python

n = 5
h = 0
d = int(n/2)+1
for r in range(0, n):
	s=''
	if r<d:
	    h = r*2
	elif r==d:
		h = d
	else:
	    h = int(d/3)
	for c in range(1, n+1):
		s+=str(h*n+c) + ' '
	print(s)

Posted 27-Dec-21 6:08am

Member 15481083

Updated 27-Dec-21 6:10am

v2

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This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Maciej Los · Accepted Answer · 2020-09-30T21:19:00

Solution 1

Try this:

Python

n = 5
h = 0
d = int(n/2)+1
for r in range(0, n):
	s=''
	for c in range(1, n+1):
		if r<d:
			h = r*2
		elif r==d:
			h = d
		else:
			h = int(d/3)
		s+=str(h*n+c) + ' '
	print(s)

Result:

1 2 3 4 5 
11 12 13 14 15 
21 22 23 24 25 
16 17 18 19 20 
6 7 8 9 10

Posted 30-Sep-20 21:19pm

Maciej Los

Comments

CPallini 1-Oct-20 3:31am

5.

Maciej Los 1-Oct-20 3:51am

Thank you, Carlo. I'm exploring new programming area and i'm truly fascinated ;)

CPallini 1-Oct-20 3:58am

:thumbsup:

How can I write this pattern in Python?

5 solutions

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Solution 1

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How can I write this pattern in Python?

5 solutions

Solution 8

Solution 7

Solution 6

Solution 5

Solution 1

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