Solving palindromic 'Triangle Quest' puzzle in Python

Question

I'm trying to solve this programming puzzle:

You are given a positive integer N (0 < N < 10). Your task is to print a palindromic triangle of size N.

For example, a palindromic triangle of size 5 is:
1
121
12321
1234321
123454321
You can't take more than two lines. You have to complete the code using exactly one print statement.

Note: Using anything related to strings will give a score of 0. Using more than one for-statement will give a score of 0.

I can think only of 'dumb' way to do this:

for i in range(1, N+1):
    print([0, 1, 121, 12321, 1234321, 123454321, 12345654321, 1234567654321, 123456787654321, 12345678987654321][i])

Is there a more elegant solution?

Somehow, there is now way to get a non-zero score, as printing is related to strings. — DainDwarf, Dec 28 '15 at 13:21

score 10 · Accepted Answer · answered Dec 28 '15 at 13:40

10

I ended up doing the following (thanks @raina77ow for the idea):

for i in range(1, N+1):
    print((111111111//(10**(9-i)))**2)

answered Dec 28 '15 at 13:40

planetp

12,388
16
74
140

6

Slightly better way: `print((10**i//9)**2)`. – simonzack Dec 28 '15 at 13:45

Alex Aza · Answer 2 · 2018-03-02T07:18:27.720

10

for i in range(1,int(input())+1):
   print(int((10**i-1)/9)**2)

1 -> (   10 - 1) / 9 =    1,    1 *    1 = 1
2 -> (  100 - 1) / 9 =   11,   11 *   11 = 121
3 -> ( 1000 - 1) / 9 =  111,  111 *  111 = 12321
4 -> (10000 - 1) / 9 = 1111, 1111 * 1111 = 1234321

edited Mar 02 '18 at 07:18

answered Mar 02 '18 at 07:12

Alex Aza

73,821
25
151
132

score 6 · Answer 3 · answered Oct 06 '17 at 02:53

6

def palindrome(N):
    for i in range(1, N + 1):
        print(int('1' * i)**2)

palindrome(int(input()))

1 * 1 = 1
11 * 11 = 121
111 * 111 = 12321

answered Oct 06 '17 at 02:53

noobninja

840
12
14

If the coding challenge rules exclude strings, then this solution will fail. – noobninja Oct 06 '17 at 18:51

vesche · Answer 4 · 2015-12-28T14:02:45.687

5

for i in range(1,6):
    print (((10 ** i - 1) // 9) ** 2)

Here's a wtf one liner:

f=lambda n:n and[f(n-1),print((10**n//9)**2),range(1,n+1)];f(5)

edited Dec 28 '15 at 14:02

answered Dec 28 '15 at 13:46

vesche

1,742
1
11
19

score 1 · Answer 5 · edited Jul 13 '19 at 09:54

1

I think the following code should work. I have used the most basic method, so that most people are going to understand it:

  N = int(input())
  arr = []
  for i in range(1,N+1):
         arr.append(i)
         print(arr+arr[-2: :-1])

edited Jul 13 '19 at 09:54

double-beep

4,567
13
30
40

answered Jul 13 '19 at 07:18

Taalib Zama

61
3

score 1 · Answer 6 · answered Jul 08 '20 at 13:03

1

  for i in range(1,int(input())+1): #More than 2 lines will result in 0 score. Do not leave a blank line also
    print(''.join(list(map(lambda x:str(x),list(range(i+1))[1:]))+list(map(lambda x:str(x),list(reversed(list(range(i))[1:]))))))

answered Jul 08 '20 at 13:03

Pravin Kumar

47
5

score 1 · Answer 7 · answered Jul 18 '20 at 19:22

1

This is a simple string-less version:

for i in range(1, int(input()) + 1):
    print(sum(list(map(lambda x: 10 ** x, range(i)))) ** 2)

answered Jul 18 '20 at 19:22

Thomas

11
1

PascalVKooten · Answer 8 · 2015-12-28T14:40:33.460

0

Code golfing and taking advice of simon and rain:

set(map(lambda x:print((10**x//9)**2),range(1,N+1)))

edited Dec 28 '15 at 14:40

answered Dec 28 '15 at 13:48

PascalVKooten

19,461
15
93
149

score 0 · Answer 9 · answered Dec 28 '15 at 13:50

0

for i in range(1, N + 1):
    print(*list(range(1, i + 1)) + list(range(i - 1, 0, -1)), sep = None)

answered Dec 28 '15 at 13:50

Keith Hall

14,474
3
54
68

This separates the numbers by space (and using `sep=""` is not allowed) – planetp Dec 28 '15 at 14:34
@planetp, you are correct that it separates the numbers by space, but the OP didn't specify that separating the numbers by space isn't allowed - after all, the answer is still a palindromic triangle, just with spaces :) however, even if it is not valid, I thought it was worth including because it is potentially more understandable than the math answers :) – Keith Hall Dec 28 '15 at 14:37
Way late to the party, but... ```python reduce(lambda x, y: x*10 + y, list(range(1, i + 1)) + list(range(i - 1, 0, -1))) ``` – chris.guethle Dec 14 '16 at 19:52

score 0 · Answer 10 · answered Jan 15 '16 at 07:47

0

Just because every solution offered so far involves range() which I feel is overused in Python code:

from math import log10

i = 1
while (N > log10(i)): print(i**2); i = i * 10 + 1

answered Jan 15 '16 at 07:47

cdlane

37,186
5
26
72

score 0 · Answer 11 · edited Dec 26 '20 at 14:37

0

I am able to print in list format using below:

    for i in range(1,5):
        print [j for j in range(1,i+1) ], [j for j in range(i-1,0,-1) ]

Result:

[1] []
[1, 2] [1]
[1, 2, 3] [2, 1]
[1, 2, 3, 4] [3, 2, 1]
[1, 2, 3, 4, 5] [4, 3, 2, 1]

edited Dec 26 '20 at 14:37

Ann Zen

25,080
7
31
51

answered Sep 26 '17 at 16:12

Vaibhav jain

31
2

score 0 · Answer 12 · answered Jun 17 '18 at 10:08

    for i in range(2,int(raw_input())+2): 
        print ''.join(([unicode(k) for k in range(1,i)]))+""+''.join(([unicode(k) for k in range(i-2,0,-1)]))
        print ''.join(map(unicode,range(1,i)))+""+''.join(map(unicode,range(i-2,0,-1)))

I hope it will help.

score 0 · Answer 13 · edited Aug 13 '19 at 06:50

0

Use this code:

prefix = ''
suffix = ''

for i in range(1,n):
    middle = str(i)
    string = prefix + middle + suffix
    print(string)

    prefix = prefix + str(i)
    suffix = ''.join(reversed(prefix))

edited Aug 13 '19 at 06:50

jkdev

10,404
15
54
77

answered Aug 13 '19 at 04:40

QOPS

623
5
15

@jkdev what was wrong with my code, why the down vote – QOPS Aug 13 '19 at 08:31
I only added “Use this code:” before the code block. No edits to the code block itself. – jkdev Aug 13 '19 at 08:35

TheRealG's Channel · Answer 14 · 2020-05-02T12:32:14.230

for i in range(1,int(input())+1):
    print(int(str('1'*i))**2)

Basically on each iteration, the number '1' in string is multiplied by i times then converted to an integer then squared. So e.g.

On the 3rd iteration --> output = 111^2 = 12321

Edit: Noticed constraints was to answer with str() function

So we have a sequence 1, 11, 111, 1111 nth = an + (a(r^(n-1) - 1)) / (r - 1) where |r > 1|

therefore, solution;

for i in range(1,int(input())+1):
    print(pow((((10**i - 10))//9) + 1, 2))

Solving palindromic 'Triangle Quest' puzzle in Python

14 Answers14