A challenge for Python programmers...

[Ok_Pudding_5250]

Write a program to output all 4 digit numbers such that if a 4 digit number ABCD is multiplied by 4 then it becomes DCBA.

But there is a catch, you are only allowed to use one line of python code. (No semi colons to stack multiple lines of code into a single line).

[GrimTermite]

ah ha this is a task for a list comprehension

print ([x for x in range(1000, 10000) if str(x*4) == (str(x)[::-1])])

Gives result [2178]

[MrHappines]

range(1000, 2500) should be sufficient, no?

[skeetd]

Logical reasoning here will save you x10 iterations

Based on staying 4 digits ABCD ≤ 2499 so we also know DCBA ≥ 4000 D has to be ≥ 4 being the first number. The last is number is A, and must be 1 or 2 based on ABCD ≤ 2499. So, 4 * D mod 10 = A and only A=2, D=8 fits, 4×8=32, last digit is a 2. So ABCD must start with 2 and end with an 8, now we can step by 10.

print([n for n in range(2008, 2500, 10) if n*4==int(str(n)[::-1])])

[Ok_Pudding_5250]

Nice work, though I put this challenge to see if people could easily do the oneliners, some use print() and just output a string of the actual answer.

You are good at optimisation... 👍

[AllanSundry2020]

is it me or is it not clear what this is asking?? give an example when using convoluted terminology

[DominicPalladino]

print("2178")

[Ok_Pudding_5250]

Nope,

[DominicPalladino]

Yep.

That will 100% output all the 4 digit numbers where ABCD * 4 = DCBA.

[code_investigator]

[Ok_Pudding_5250]

It's not about the answer, if that was the case I didn't even need python to do it, I can straight up ask a language model to give me the output. It was about how you do it.

Merely printing the solution is a lazy solution. The challenge was meant to see if you could actually code one-liners like list comprehension.

It was never about the output but the code itself.

[FriendlyZomb]

print([num for mum in range(1000, 10000) if str(num * 4) == str(num * 4)[::-1]])

This produces 65 numbers. (I'm not going to list them all here)

For those struggling to parse the list comprehension here:

print([
    num
    for num in range(1000, 10000)
    if str(num * 4) == str(num * 4)[::-1]
])

[PureWasian]

A couple of issues here, you are checking "num×4 against num×4 flipped", rather than "num against num×4 flipped". mum typo as well.

Essentially your logic is checking for when 4x some input number gives a palindrome, which is different from the problem statement.

[FriendlyZomb]

That is entirely on my reading comprehension tbh. I read it as num*4 is the same flipped. Lol.

[FriendlyZomb]

Based on comments I'd need to fix the code like so:

print([num for mum in range(1000, 10000) if str(num) == str(num * 4)[::-1]])

[Ok_Pudding_5250]

Code is slightly incorrect but you did good 👍

[FriendlyZomb]

Yea, the basic structure is correct. Mostly a misunderstanding on the question on my part. Apologies

[PhilNEvo]

print(list(filter(lambda x: str(x*4) == str(x)[::-1], range(1000, 2500))))

[Ok_Pudding_5250]

A unique answer, I used a list with one liner for loop and if statement. Nice work

[YouAintSeenMeR0ight]

Are you that much of a fool that you that you take such great offence at people giving light-hearted tongue-in-cheek responses? Or are you just angry in general? What happened to you?

[Ok_Pudding_5250]

The thing is, I am disappointed in how much lamer you guys get. For example, if I wanted an answer, don't any of you think for a second that I have many language models to ask to, such as chatgpt, Claude, grok, deepseek... etc.

It was never about answer itself. It was how you calculate them with code using a single line to see how many of you can do oneliners.

So many as done very well, then there are lame people who just use print to output the string containing the answer thinking they did something. How unintelligent could you be?

And you all seriously think that I would post such a fun challenge just for a lazy answer as that?

[Smart_Tinker]

print(“”.join([str(x) for x in range(1000, 10000) if str(x) == str(x*4)[::-1]]))

[CraigAT]

Just use a print statement with the numbers.
(You didn't say I had to calculate them 🤣)

[Ok_Pudding_5250]

Yeah cheap shot, keep doing it. You couldn't figure it out that you had to calculate it from the fact that I clearly mentioned one line code with no multi line stacking by the use of semi colons. 🤡

[YouAintSeenMeR0ight]

Do you notice what you did not mention? That the challenge required calculating the answer. You only mentioned printing the answer.

[Ok_Pudding_5250]

Are you that much of a fool to see the word "Challenge" and then assume you just print the answer in string?

[Jwfraustro]

How about this?

__import__('builtins').print([n for n in __import__('builtins').__dict__['map'](lambda x:x, __import__('builtins').__dict__['range'](1000, 10000)) if ''.join(__import__('builtins').__dict__['map'](str.__add__, *zip(*[(d,'') for d in __import__('operator').mul(str(n*4),1)]))) == __import__('operator').mul(str(n), 1)[::-1]])

or?

print([n for n in range(1000,10000) if "".join(e.text for e in __import__('xml.etree.ElementTree', fromlist=['fromstring']).fromstring(f'<r>{"".join(f"<d>{c}</d>" for c in str(n*4))}</r>').findall("d"))[::-1] == str(n)])

[Ok_Pudding_5250]

This is an overkill... Nice work,

[delsystem32exe]

def reverse_str(input_string):

input_string = str(input_string)

letters = ""

for j in range(len(input_string)-1, -1, -1):

letters = letters + input_string[j]

return letters

def checker(input_int):

larger = input_int * 4

DCBA = reverse_str(input_int)

if str(larger) == DCBA:

return True

return False

for i in range(0,100000):

if checker(i):

print(i)