The same without `all` clear solution for All the Same by Tinus_Trotyl - python coding challenges

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The same without `all` solution in Clear category for All the Same by Tinus_Trotyl

def all_the_same(elements):
   return elements == elements[1:] + elements[:1]

Jan. 31, 2018

Comments:

tom-tom_f993b3abc4814d4cb5476e on Jan. 31, 2018, 8 p.m.
Nice!

veky on Feb. 1, 2018, 4:50 p.m.
LOL, this is actually an interesting one. I'd even give it a few thumbs if you change the category. :-P

Tinus_Trotyl on Feb. 1, 2018, 5:42 p.m.
Well, at first I'd like hear to your motives why you think it doesn't apply for clear.

veky on Feb. 14, 2018, 6:56 a.m.
You're right that it's different this time. This has nothing to do with Python. But logically, the fact that "sequence is a fixpoint of rotation" => "sequence is constant" in my eyes requires a proof, which is not exactly trivial. Maybe clear/creative distinction isn't meant for these kind of creative processes, but then again, since bryukh left, noone knows what they are meant for. :-] As another example, what do you say about <a href="https://py.checkio.org/mission/all-the-same/publications/veky/python-3/my-hovercraft-is-full-of-ells/?">https://py.checkio.org/mission/all-the-same/publications/veky/python-3/my-hovercraft-is-full-of-ells/?</a> (Also on the logic level; please disregard other Python-level obfuscations.) Is it obvious to you that this is logically right solution?

Vasily__Chibilyaev on April 12, 2018, 11:32 a.m.
brilliant!

tigerhu3180 on April 13, 2018, 2:49 a.m.
createive,i like it

gnosis on Oct. 23, 2018, 3:26 a.m.
Is this equivalent to the following? <pre class='brush: python'>def all_the_same(elements): return elements == elements[-1:] + elements[1:] </pre> Am I wrong in reasoning that these are the only two such wrap-around shifts that would work?

mdhirsch on March 14, 2019, 9:34 p.m.
If elements is a large list, this is a fairly expensive operations. Three new arrays are made, one is small, but the other one almost the size of the original and one the same size. But bonus points for creativity.

Tinus_Trotyl on March 23, 2019, 8:23 p.m.
You're partly right I think (thanks for pointing on this tho, but please have a look at the following). To me, there only are 2 new lists made (together they have the same amount of items as the original one). And here is on what I believe why : <pre class='brush: python'>>>> def y(x): x[3] = 0 >>> a = [1,2,3,4,5] >>> a <<< [1, 2, 3, 4, 5] >>> y(a) # Will python process list 'a' in function 'y' locally as a new list 'x' ? >>> a # As you will see, apparently not !!! <<< [1, 2, 3, 0, 5] </pre> "NOW, WHAT THE HELL IS THIS. . . ?" Yeah, that was what I was thinking too when I noticed the first time . . . :-)

mdhirsch on March 28, 2019, 11:04 p.m.
The problem is that a slice of an array is yet another array: <pre class='brush: python'>>>> a = [1,2,3,4,5] >>> b = a[1:] >>> b [2, 3, 4, 5] >>> a[2]=0 >>> a [1, 2, 0, 4, 5] >>> b [2, 3, 4, 5] >>> </pre> so <pre class='brush: python'>elements == elements[1:] + elements[:1] </pre> creates 2 arrays, elements[1:] and elements[:1] (the second of those is small), and then appends them together to create a 3rd array. This has the memory overhead of holding the arrays, and the overhead of copying the values twice. In your example you didn't slice the array, just replace one of its entries. This, as you pointed out, does not create an new array.

Tinus_Trotyl on March 29, 2019, 8:39 a.m.
okay :-)

snow_7 on March 20, 2020, 6:32 a.m.
Wow! It's cool!

mr_vads on July 20, 2021, 8:39 p.m.
this is not obvious one. can anybody explain, why length of starting list is the same as the result of concatenating two lists?

Tinus_Trotyl on Aug. 1, 2021, 8:38 a.m.
Imagine list x = [a, b, c, d, e] then x[:1] == [a] and [x:] == [b, c, d, e] So x[1:] + x[:1] == [b, c, d, e] + [a] == [b, c, d, e, a] though if this is not the answer you were looking for, feel free to ask (more explicite).

mr_vads on Aug. 1, 2021, 5:10 p.m.
thx for reply! now I get it, and this solution is genious for this task, imho! p.s. not sure about efficiency for this solution, I used set() to solve this one

dangu on March 11, 2023, 6:52 a.m.
This is absolute genius!, I would call it the mirror effect solution!

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