Painstaking clear solution for Cipher Map by veky - python coding challenges

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Painstaking solution in Clear category for Cipher Map by veky

def recall_password(grill, cypher):
    password = ""
    for _ in grill:  # must be of len 4
        for grill_row, cypher_row in zip(grill, cypher):
            for grill_letter, cypher_letter in zip(grill_row, cypher_row):
                if grill_letter == 'X':
                    password += cypher_letter
        row1, row2, row3, row4 = grill
        grill = tuple(zip(row4, row3, row2, row1))  # rotate
    return password

Sept. 18, 2014

Comments:

Thenbacker on Oct. 3, 2016, 12:49 a.m.
Fabulous, short and very, very readable!!

tobias.lindblom on Dec. 1, 2016, 4:35 p.m.
3 nested for loops? seems a bit excessive... cannot handle larger NxN grills? here is mine: idx, pwd = range(len(cipher_grille)), [] cidx = [(y,x) for y in idx for x in idx if cipher_grille[y][x]=='X' ] for rot in range(4): # four rotations _pwd.extend([ciphered_password[y][x] for y,x in cidx]) _cidx=sorted([(x,idx[-1]-y) for y,x in cidx]) return ''.join(pwd)

veky on Dec. 2, 2016, 3:15 a.m.
Well, instead of three nested loops, you stuff two of them in a list comprehension. Yes, it's factored out, but in case of n=4 it surely doesn't matter with respect to performance. And yes, my code is specialized for n=4, but I don't view it as a disadvantage in this context. What I dislike about your solution (besides horrible names like cidx:-P) is excessive indexing. Some of it is for factoring, but some is easily eliminable. Every time you use range of len, you should probably use enumerate. Look: <pre class='brush: python'>selected = [(i,j) for i, row in enumerate(grill) for j, cell in enumerate(row) if cell == 'X'] </pre> And almost always it's better to pass a generator expression to a function than a comprehension. Like <pre class='brush: python'>password.extend(cipher[i][j] for i,j in selected) </pre> and sorted too.

tobias.lindblom on Dec. 2, 2016, 6:53 a.m.
hehe... I do give pretty horrible names don't I... and I just dicovered list comprehensions....sorry Your solution is nice and easy to understand. mine have more room for improvement.... and I like space...! ;-)

veky on Dec. 2, 2016, 12:03 p.m.
<blockquote> Your solution is nice and easy to understand. mine have more room for improvement </blockquote> Yes, our solutions are fundamentally different with respect to what they consider important. You value generality and performance, I value readability and education value (Aha!-effect). That's the beauty of CheckiO. :-)

tobias.lindblom on Dec. 2, 2016, 7:01 p.m.
I think my major point is that rotating the whole matrix is a lot of work for no gain except for simplicity. A smart, more efficient (faster, less memory) solution is also a good lesson. ;-) As the matrix is only 4x4, executions times of the two functions are more ore less equal. Slightly better version of my previous code: <pre class='brush: python'>def recall_password(cipher_grille, ciphered_password): # works by rotating only indices where the X:es are # an index of (3,2) (4th row, 3rd column ) becomes (2,1) after rotation # row_index, col_index = col_index, max_index - row_index # max_index is the size of the grille -1 (in this case 3) grille_size = len(cipher_grille) # figure out indices of X:es only X_indices = [divmod(index, grille_size) for index, char in enumerate(''.join(cipher_grille)) if char == 'X'] password ='' for rot in range(4): # four rotations for row_index, col_index in X_indices: # cycle through indices password += ciphered_password[row_index][col_index] if rot==3: return password # no point of doing final rotation... return result # rotates indices clockwise, also does a needed sort X_indices=sorted((col_index,grille_size-1-row_index) for row_index,col_index in X_indices) </pre>

veky on Dec. 2, 2016, 8:46 p.m.
I'm not sure what do you mean by "whole" matrix. You rotate what amounts to a quarter of it anyway. :-) Or do you mean that when the grille is NxN, there is only N Xes?

tobias.lindblom on Dec. 3, 2016, 7:03 a.m.
Not really sure what I am meaning, Yours does operations on all elements of the matrix...X:es and dots. On the other hand you only need one if statement and not very many variables. I did some timing tests and yours turned out about 25% faster. I belive for mine to be faster, there need to be less than 1/4 X:s compared to dots and a much bigger matrix. I think the need for sorting in my function may be the culprit.

veky on Dec. 3, 2016, 7:48 a.m.
Yes, I think that if Xes always make for a quarter of a grille (so you eventually use up all the letters in four steps), my solution is always faster - if for nothing else, because sorting becomes the bottleneck when N gets large. But if NxN grille has only N Xes, then of course, as N gets larger, your solution becomes much better. Even sorting is negligible, since NlogN << N^2. Generalizations are rarely straightforward. :-)

taylormyer on April 21, 2017, 9:55 p.m.
I am aware that this is an old thread but I'm newer to programming so can you explain this line: <pre class='brush: python'> for index, char in enumerate(''.join(cipher_grille)) if char == 'X'] </pre> I tried searching something like it online and I couldn't find anything. Thanks for the help.

veky on April 23, 2017, 4:41 a.m.
Probably because it's just a half of (logical) line. See that unbalanced `]` at the end? :-) The whole thing is (the words to search for are "list comprehension") <pre class='brush: python'>X_indices = [divmod(index, grille_size) for index, char in enumerate(''.join(cipher_grille)) if char == 'X'] </pre> which is moreorless equivalent to <pre class='brush: python'>X_indices = [] for index, char in enumerate(''.join(cipher_grille)): if char == 'X': X_indices.append(divmod(index, grille_size)) </pre>

taylormyer on April 25, 2017, 5:52 a.m.
To be honest, I still don't completely know what it does especially the enumeration method and how that works but thanks! You are obviously quite good at this.

veky on April 25, 2017, 8:52 p.m.
<blockquote> You are obviously quite good at this. </blockquote> Yes, but this is not about me, it's about you. :-) <pre class='brush: python'>for index, char in enumerate('Veky'): print(index, char) </pre> Do you get it now?

taylormyer on April 26, 2017, 12:29 a.m.
Actually yes. This makes it obvious.

senswers on June 13, 2017, 6:34 a.m.
Man, I like your solution so much! It's really great! I've been thinking of a way to calculate index rotation, but gave up for a laziest solution very soon.

Andrij_Petriv on Dec. 20, 2016, 8:01 p.m.
rotated = tuple(zip(*original[::-1]))

veky on Dec. 20, 2016, 8:17 p.m.
Of course. The whole algorithm can be written in about 3 lines. Here I tried to avoid advanced syntax like starargs and slicing.

joshua.m.hublar on Jan. 20, 2017, 2:43 a.m.
I want to understand this form. Its elegance is at once baffling and awe inspiring.

veky on Jan. 20, 2017, 10:22 a.m.
Well, did you understand line 9 in the above solution?

Ruslan_Mintz on Feb. 8, 2017, 12:07 p.m.
I did not. Could you please explain how works <pre class='brush: python'>grill = tuple(zip(row4, row3, row2, row1)) # rotate </pre>

veky on Feb. 8, 2017, 2:04 p.m.
Would you understand zip with two arguments? <pre class='brush: python'>>>> tuple(zip([1,2,3], [4,5,6])) </pre> Try it and see what it does. Then try to generalize. :-) If you get stuck, write again.

taylormyer on April 26, 2017, 12:35 a.m.
I understand the zip part but the line above it: <pre class='brush: python'>row1, row2, row3, row4 = grill </pre> I have never seen this before, does this actually just iterate over the rows and grab the values?

veky on April 26, 2017, 5:25 a.m.
Yes, pretty much. It really names the rows themselves, not their "values" whatever that might be. :-) <pre class='brush: python'>>>> a, b, c, d = range(4) >>> c 2 </pre>

ICTwang on Dec. 23, 2016, 11:26 a.m.
I don't know the meaning of "for _ in grill:"? Would you mind to explain it?

veky on Dec. 26, 2016, 5:47 a.m.
I tried to write this solution minimizing references to the outside world (yes, even natural numbers:), using just basic Python constructs. It occured to me that I can even do without `range` (which Guido considers a necessary evil) if I use the fact that grill must have exactly 4 elements. :-)

danny.luker_6f2b7a7a6c6340b297 on Jan. 10, 2017, 2:24 p.m.
Very cool way to rotate! Wish I'd of thought of that.

veky on April 23, 2017, 5:32 a.m.
It's the "standard" (`zip/reversed`) way, just written in painstaking detail. :-)

abehad701 on June 6, 2017, 12:15 a.m.
i like yow you rotate

potter777 on May 10, 2018, 3:52 a.m.
just awesome, i hope someday having better solutions like this

veky on May 11, 2018, 8:50 a.m.
It's just a matter of time. And effort. :-)

yanhua.dev on July 6, 2018, 3:43 a.m.
I like your code's excellent readability very much. While I am using Python, I am not clear on how to balance the readability and applying Python advantaged language features ( map, list comprehension, join ect...) Below is my code, I feel I am a little bit overusing Python advantaged language features, it may not friendly for reader. May I know your suggestions? Thanks. <pre class='brush: python'>def recall_password(cipher_grille, ciphered_password): def x_positions(lst): return [i for i, x in enumerate(lst) if x == 'X'] def decode_row(positions, row): return ''.join([row[i] for i in positions]) def rotate_90(grille): return tuple(zip(*reversed(grille))) password = '' for _ in range(4): positions = map(x_positions,cipher_grille) password += ''.join(map(decode_row, positions,ciphered_password)) cipher_grille = rotate_90(cipher_grille) return password </pre>

veky on July 6, 2018, 2:17 p.m.
Well, it all depends on the intended audience. Who are you writing for? As you've noticed above, my style varies greatly, mostly because I want to showcase various not-so-well-known corners of Python. Here I tried to present the fact that even with using just the basic constructs of Python, the solution can be written that is short and easily understandable. Of course, that just means that CiO missions are much easier than most real-world tasks. :-] In many other solutions, I tried to present the opposite. :-D At the meta-level, of course, I'm trying to show that Python is a fantastically expressive language. ;-) You above obviously tried to emphasize naming of various pieces of calculation, which, while laudable in a much more complicated task, here seems forced. You just make the reader jump up and down trying to match the name with the definition, without much benefit.

yanhua.dev on July 6, 2018, 3:27 p.m.
From your comments and show cases, I learned a lot :-). Thanks very much.

Splitter on Aug. 4, 2019, 4:21 p.m.
Why painstaking? Because of loops? Your implementation of the rotation is great. Did you use `tuple` for the sake of speed?

veky on Aug. 4, 2019, 8:13 p.m.
The name, motivation and rotation technique I've explained somewhere above. About the tuple: I never do such things for speed. It's simply that I view the things of constant length, especially when I unpack them, as tuples.

bsquare on Sept. 6, 2019, 2:39 p.m.
Very interesting solution!

piter239 on April 26, 2020, 5:19 p.m.
"Painstaking" is exactly how I felt about my solution for this. So many things to do by hand... For some reason I thoght I have to do it with list comprehension. But now I see - there is a time and a place for each tool! Thanks again!

kdim on Jan. 28, 2021, 4:33 p.m.
<pre class='brush: python'>for _ in grill </pre> tricky combination

viktor.chyrkin on Dec. 29, 2021, 1:33 p.m.
Good, but only for perfect squares.

veky on Dec. 29, 2021, 1:51 p.m.
What do you mean by "perfect" here? As far as I know, it only works for 4^2 (as required by task). 😇

viktor.chyrkin on Dec. 29, 2021, 1:55 p.m.
Yeah. that's what I meant. Just if it said it could be a not perfect square you'd be in trouble.

veky on Dec. 29, 2021, 2:12 p.m.
Ok, now I know what you mean, but I'm still curious about your terminology. In which sense is 4^2 perfect? — as opposed to, say, 6^2 — which literally could be called "perfect square" since 6 is a perfect number 😉

viktor.chyrkin on Jan. 1, 2022, 1:51 p.m.
I guess so.^_^

List, list and list again

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