Welcome to this useless blog, let’s begin!

As my first post, I’m going to present a nice python puzzle, at least it was a puzzle for me. It gave me even the motivation to start a blog…

Recently I was working on a project and accidentally discovered some strange numbers showing up between my sorting results. First I thought it’s a bug, but it’s more like an unintentional featureðŸ˜‰

It was a very simple function which took an array of numbers as parameter and returned another one. It is important that the parameter passed to the function must be a range of numbers starting from 1, without skipping. The actual code just displaced the numbers in the array, placing the number from the original array into the new array. The index in the new array was the actual number from the first one, and the value in the second array was the index from the first. Here is an illustration:

As you see it was nothing fancy…. So again, the value of the first array indicates the index in the second array, and the index of the first array becomes the value in the second array. If you place the numbers following that rule, you’ll get exactly the same number. They got the name mirrored arrays because of that properties. Now the tricky part was that it’s not true for all the permutations of the sequence, and this is why I thought first that there is something wrong with my code. For a range of numbers from 1 to 8, there are 40320 permutations, but only 764 of them can be mirrored.

I looked into the function and tried to figure out if it is a new mathematical discovery I made, or just a silly puzzle. Well the second option seemed more realistic… Back to the topic, the question is how can I generate these numbers?

Luckily the toolbox was already open, so the first implementation was finished quickly. It looks like:

from itertools import permutations def find_mirror_arrays(array_length): ''' First brute-force implementation for finding mirrored arrays Generates all permutations of array_length and test every one of them separately. ''' array = [i for i in range(1, array_length + 1)] array_permutations = permutations(array, array_length) found_arrays = [] for test_array in array_permutations: new_array = [0] * array_length for arr_index, arr_value in enumerate(test_array): new_array[arr_value - 1] = arr_index + 1 if tuple(new_array) == test_array: found_arrays.append(new_array) return found_arrays mirror_arrays = find_mirror_arrays(8)

Now what this snippet does is it generates all the permutations of the required length and then tests every one of them with that simple displacer operation, so after we execute the displacement operation on a sequence we should get the same sequence back. If we get another sequence, of course it means that it’s not a mirrored array.

Now the problem with that implementation is, that it’s a brute force method, primitive enough to cause performance problems above the length of 10 numbers.

The interesting part just started. How to generate those numbers manually to get better performances? Took a little time until I figured out, but finally I got an idea. The simplest way I had in my mind was using a recursive function. Although recursive functions are usually a bit more messy, I don’t like them either, but in theory using them in this case would result with more readable code than doing it by a regular approach.

Here’s the code i got then:

from copy import deepcopy class MirrorCrafter(object): ''' Second implementation, non-bruteforce method, uses a recursive function to generate a tree of solutions and collect all the results. ''' def __init__(self): self._mirrored_arrays = [] def _get_child_branches(self, parent): remaining_numbers = list(parent['remaining_numbers']) for num in remaining_numbers: src_list = list(parent['src_list']) free_cell_index = src_list.index(0) src_list[free_cell_index] = num src_list[num - 1] = free_cell_index + 1 new_remaining_numbers = list(remaining_numbers) try: new_remaining_numbers.remove(num) new_remaining_numbers.remove(free_cell_index + 1) except ValueError: pass try: new_branch except NameError: new_branch = deepcopy(parent) new_branch['child_branches'] = [] new_branch['src_list'] = src_list new_branch['remaining_numbers'] = new_remaining_numbers if new_remaining_numbers: new_branch['child_branches'].append(self._get_child_branches(new_branch)) if 0 not in src_list: self._mirrored_arrays.append(src_list) return new_branch def get_mirrors(self, array_length): tree_branch = { 'src_list': [0] * array_length, 'remaining_numbers': range(1, array_length + 1), 'child_branches': [] } self._get_child_branches(tree_branch) return self._mirrored_arrays mc = MirrorCrafter() mirror_arrays = mc.get_mirrors(8)

This looks much more interesting. It uses a dictionary which has 3 keys. As it calls itself recursively over and over again it creates a tree by appending new branches of solutions. See the following example tree. In a similar way is our tree organized, at least it was the basic idea for it.

Benchmarks were made using a pretty old Athlon64 2800+ @1800MHz. Someone with an i7 could post some screenshotsðŸ˜‰ Anyway, take a look at my results:

first implementation on length of 8: time required to run script: 0.239191 time required to run script: 0.238772 time required to run script: 0.235737 second implementation on length of 8: time required to run script: 0.550336 time required to run script: 0.577819 time required to run script: 0.551476 first implementation on length of 9: time required to run script: 2.282437 time required to run script: 2.227491 time required to run script: 2.237375 second implementation on length of 9: time required to run script: 2.481702 time required to run script: 2.500234 time required to run script: 2.568499

As we can see the first primitive algorithm beats our nextgen implementation. But not for so long, it’s just on smaller lengths, look at this:

first implementation on length of 10: time required to run script: 23.976178 time required to run script: 24.317904 time required to run script: 24.385197 second implementation on length of 10: time required to run script: 12.304419 time required to run script: 11.570755 time required to run script: 11.774170 first implementation on length of 11: time required to run script: 286.148981 time required to run script: 295.728056 time required to run script: 298.418401 second implementation on length of 11: time required to run script: 59.683516 time required to run script: 59.883970 time required to run script: 59.646269

And the difference exponentially rises… Not bad.

That would be the result of our discovery, the next thing should be to create a third implementation, hopefully even better than the second one. Let’s see… It would be great if a solution could be found with a regular approach using simple loops… After a little time I came up with this:

from copy import deepcopy def get_mirror_arrays_v3(array_length): ''' Third implementation, non-brute-force method using loops, does not generate any trees, the remaining numbers are calculated every time on the fly. ''' available_numbers = range(1, array_length + 1) num_seq = [[x] + [0] * (array_length - 1) for x in available_numbers] for x in num_seq: x[x[0] - 1] = 1 for count in range(array_length - 2): if count != 0: num_seq = deepcopy(completed_seq) completed_seq = [] for index, seq in enumerate(num_seq): seq = num_seq[index] remaining_numbers = [x for x in available_numbers if x not in seq] if not remaining_numbers: completed_seq.append(seq) for next_num in remaining_numbers: next_seq = list(seq) free_cell_index = next_seq.index(0) next_seq[free_cell_index] = next_num next_seq[next_num - 1] = free_cell_index + 1 completed_seq.append(next_seq) completed_seq[0][array_length - 1] = array_length return completed_seq mirror_arrays = get_mirror_arrays_v3(8)

Now the code does not generate a tree, nor does it store the remaining numbers, everything is calculated on the fly, only the actual sequences are stored and loaded again in every loop. In theory I would say that it should take much more time to calculate everything again every time, but it looks like it’s still faster than working with dictionaries and data storing / loading constantly.

So the long awaited results for the third implementation are awesome. It outperforms both of the solutions mentioned before, look at them:

third implementation on length of 8: time required to run script: 0.065746 time required to run script: 0.065963 time required to run script: 0.068096 third implementation on length of 9: time required to run script: 0.295213 time required to run script: 0.291225 time required to run script: 0.294794 third implementation on length of 10: time required to run script: 1.305196 time required to run script: 1.315555 time required to run script: 1.323389 third implementation on length of 11: time required to run script: 5.836298 time required to run script: 5.799165 time required to run script: 5.847414

Looks like the best solution for now… And our job could be finished here, but there is something more we can do.Â Using the newly created function, it can be easily modified into a recursive function, which guess what, does the job even better! See for yourself:

''' The fourth implementation actually evolved from the third solution but it uses a recursive function again, and the remaining numbers are passed as parameter, so there's no need to calculate them on the fly every time. ''' def recursive_mirror_generator_v2(source, remaining_numbers): if not remaining_numbers: return source completed_seq = [] for next_num in remaining_numbers: next_seq = list(source) free_cell_index = next_seq.index(0) next_seq[free_cell_index] = next_num next_seq[next_num - 1] = free_cell_index + 1 next_remaining_numbers = list(remaining_numbers) try: next_remaining_numbers.remove(next_num) next_remaining_numbers.remove(free_cell_index + 1) except ValueError: pass results = recursive_mirror_generator_v2(next_seq, next_remaining_numbers) if isinstance(results[0], int): completed_seq.append(results) else: completed_seq.extend(results) return completed_seq def get_mirrors_recursively_v2(array_length): available_numbers = range(1, array_length + 1) num_seq = [[x] + [0] * (array_length - 1) for x in available_numbers] completed_seq = [] for x in num_seq: x[x[0] - 1] = 1 remaining_numbers = [y for y in available_numbers if y not in x] completed_seq.extend(recursive_mirror_generator_v2(x, remaining_numbers)) return completed_seq mirror_arrays = get_mirrors_recursively_v2(14)

WARNING! The following results contains elements of humiliating nature, so readers with weak heart or persons under the age of 18 should be advised to skip this content.

fourth implementation on length of 8: time required to run script: 0.013698 time required to run script: 0.016129 time required to run script: 0.015955 fourth implementation on length of 9: time required to run script: 0.049936 time required to run script: 0.052711 time required to run script: 0.061206 fourth implementation on length of 10: time required to run script: 0.190436 time required to run script: 0.195853 time required to run script: 0.201169 fourth implementation on length of 11: time required to run script: 0.730234 time required to run script: 0.771284 time required to run script: 0.804766

At this point the funds devoted to the research of this anomaly have been completely depleted, but anyway I’m satisfied with the results. Feel free to post anyÂ suggestion, comment or even a better implementation, beat my solutions with pleasureðŸ˜‰

**UPDATE: **Thanks for everyone’s help, so now that the actual sequence is identified, I found some very interesting articles about it, along with other algorithms, which outperformed my solutions of course. Special thanks goes toÂ Mikko who created another implementation, which is clearly the winner among all of them.

The first algorithm found is from the combinatorial object server, otherwise known as COSÂ (Â http://theory.cs.uvic.ca/inf/perm/Involutions.htmlÂ ), I just translated it to python, so here it is:

class Involution(object): length = 0 p = list() arrays = list() def store_array(self): self.arrays.append([i + 1 for i in self.p]) def swap(self, i, j): self.p[i], self.p[j] = self.p[j], self.p[i] def cycle(self, n): if n >= self.length: self.store_array() return self.cycle(n + 1) if self.length - n >= 2: self.swap(n, n + 1) self.cycle(n + 2) for i in range(n - 1, -1, -1): if self.p[i] == i: self.swap(i, n) self.swap(n, n + 1) self.cycle(n + 2) self.swap(n, n + 1) self.swap(i, n) else: self.swap(self.p[i], n) self.swap(i, n + 1) self.cycle(n + 2) self.swap(i, n + 1) self.swap(self.p[i], n) self.swap(n, n + 1) def main(self, array_length): self.length = array_length self.p = range(0, self.length) self.cycle(0) return self.arrays inv = Involution() arrays = inv.main(12)

The results are very good:

COS algorithm on length of 8: time required to run script: 0.009012 time required to run script: 0.009840 time required to run script: 0.009145 COS algorithm on length of 9: time required to run script: 0.034899 time required to run script: 0.034381 time required to run script: 0.035642 COS algorithm on length of 10: time required to run script: 0.119194 time required to run script: 0.116180 time required to run script: 0.236970 COS algorithm on length of 11: time required to run script: 0.709367 time required to run script: 0.465431 time required to run script: 0.467473

And finally Mikko’s algorithms:

def invol(n): s1, s2 = ([(1,)], [(1, 2), (2, 1)]) if n < 2: return s1 for k in range(3, n + 1): tmp = [x + (k,) for x in s2] for i in range(k - 1): tmp.extend(aug(s1, i, k)) s1, s2 = s2, tmp return s2 def aug(lst, i, k): for invol in lst: tmp = [p + (p > i) for p in invol] tmp[i:i] = [k] tmp.append(i + 1) yield tuple(tmp) arrays = invol(11)

The results of the first algorithm:

Mikko's first algorithm on length of 8: time required to run script: 0.010087 time required to run script: 0.003362 time required to run script: 0.010010 Mikko's first algorithm on length of 9: time required to run script: 0.018520 time required to run script: 0.016312 time required to run script: 0.014148 Mikko's first algorithm on length of 10: time required to run script: 0.066137 time required to run script: 0.058889 time required to run script: 0.064799 Mikko's first algorithm on length of 11: time required to run script: 0.238079 time required to run script: 0.249475 time required to run script: 0.237163

Here goes the second algorithm:

def invol2(n): s1, s2 = ([[1]], [[1, 2], [2, 1]]) if n < 2: return s1 for k in range(3, n + 1): tmp = [x + [k] for x in s2] for x in s1: tmp.append(x + [k, k - 1]) for i in range(k - 2,0 , -1): x[x.index(i)] += 1 tmp.append(x[:i - 1] + [k] + x[i - 1:] + [i]) s1, s2 = s2, tmp return s2 arrays = invol2(11)

The results for the second:

Mikko's second algorithm on length of 8: time required to run script: 0.005724 time required to run script: 0.010078 time required to run script: 0.004627 Mikko's second algorithm on length of 9: time required to run script: 0.014572 time required to run script: 0.023883 time required to run script: 0.016792 Mikko's second algorithm on length of 10: time required to run script: 0.067655 time required to run script: 0.069951 time required to run script: 0.077429 Mikko's second algorithm on length of 11: time required to run script: 0.247994 time required to run script: 0.319301 time required to run script: 0.262980

And the third algorithm, holding the crown for now:

def invol3(n): s1, s2 = ([[1]], [[1, 2], [2, 1]]) if n < 2: return s1 for k in range(3, n + 1): tmp = [x + [k] for x in s2] for x in s1: tmp.append(x + [k, k - 1]) for i in range(k - 2, 0, -1): a = x[i - 1] x[a - 1 - (a > i)] += 1 w = x[:] w.insert(i - 1, k) w.append(i) tmp.append(w) s1, s2 = s2, tmp return s2 arrays = invol3(11)

With the dominating results:

Mikko's third algorithm on length of 8: time required to run script: 0.002960 time required to run script: 0.003011 time required to run script: 0.003201 Mikko's third algorithm on length of 9: time required to run script: 0.011731 time required to run script: 0.019164 time required to run script: 0.015256 Mikko's third algorithm on length of 10: time required to run script: 0.048309 time required to run script: 0.053429 time required to run script: 0.054088 Mikko's third algorithm on length of 11: time required to run script: 0.186751 time required to run script: 0.279246 time required to run script: 0.207985

Once again thanks, it’s a very nice job. If I find something in the meantime, I’ll post another update…

September 18th, 2011 at 12:19 am

[i for i in range(1, array_length + 1)]

can be written as just

range(1, array_length + 1)

September 18th, 2011 at 11:35 am

In python 2 yes, but in python 3 it would be more like list(range(…))

September 18th, 2011 at 1:08 am

In case you hadn’t found this out already, what you are counting are called permutation involutions. Basically, they are self-inverse permutations, so that if s is in S_8 (the group of permutations of 8 elements), then ss = 1, or equivalently, s = s^{-1}. They can all be decomposed into disjoint transpositions and fixed points. For instance, your initial one is (1 5) (2 8) (3 7) (4 6).

There’s some more information at http://mathworld.wolfram.com/PermutationInvolution.html .

September 18th, 2011 at 6:34 am

Yep, looks like you’re absolutely right, well I obviously didn’t know about that. Thanks for pointing that out!

September 18th, 2011 at 5:09 am

I found your introduction a little vague and I had a hard time understanding what you were trying to do. It wasn’t clear if you wanted to find these mirrored arrays, or if you wanted to avoid them, and were looking for arrays that weren’t mirrored…

Anyway, I just wanted to say try to make your intro as clear as possible. But, hey, maybe I’m just a bit dullðŸ˜‰

September 18th, 2011 at 6:35 am

I’ll try to be more clear if I post anything else next time

September 19th, 2011 at 5:06 pm

Here’s a shorter solution, just generating the involutions

from the recurrence: http://pastebin.com/U6qz7P6V

Timings:

length = 8, best of 5 runs of invol: 0.005 s

length = 9, best of 5 runs of invol: 0.020 s

length = 10, best of 5 runs of invol: 0.081 s

length = 11, best of 5 runs of invol: 0.331 s

length = 8, best of 5 runs of get_mirrors_recursively_v2: 0.028 s

length = 9, best of 5 runs of get_mirrors_recursively_v2: 0.101 s

length = 10, best of 5 runs of get_mirrors_recursively_v2: 0.371 s

length = 11, best of 5 runs of get_mirrors_recursively_v2: 1.421 s

September 20th, 2011 at 6:50 am

Your code is very nice, and congratulations, it really beats mine!

From the comment above when I saw finally that these numbers really exists in math, I searched for other solutions too. I will update the post with your solution and that other one I found, although your code > the one I found > my code. So my solution is the slowest.

September 20th, 2011 at 7:53 am

Actually my solution is still pretty unoptimized. Last night I wrote a version that’s 25% faster. It’s basically the same algorithm with some useless work removed (e.g. not creating tuples for output, and doing only one ‘+’ operation in the innermost loop). If you like I’ll send the code when I get home.

September 20th, 2011 at 10:33 am

Sounds interesting, yes it would be nice to post the optimized code. Send it please if itâ€™s not a problem. Thanks!

September 20th, 2011 at 6:25 pm

Here it is:

def invol2(n):

s1, s2 = ([[1]], [[1,2], [2,1]])

if n < 2:

return s1

for k in range(3,n+1):

tmp = [x+[k] for x in s2]

for x in s1:

tmp.append(x + [k,k-1])

for i in range(k-2,0,-1):

x[x.index(i)] += 1

tmp.append(x[:i-1] + [k] + x[i-1:] + [i])

s1, s2 = s2, tmp

return s2

In case something eats the whitespace, here's a link: http://pastebin.com/tbrDjWKg

September 21st, 2011 at 9:34 pm

Nice work, the code is shorter, but as I ran it, it seems that your first version is faster, now I profiled the new one and it looks like that the index method takes much time. I tried to fix it, but I have no idea how it is working at all

If you can have a look at it, it would be great. Thanks!

September 22nd, 2011 at 8:41 pm

That’s odd, for me the 2nd version is clearly faster. Here are my timings,

measured with timeit.py (fastest of 3 repeats):

Python 2.7.1:

n 8 9 10 11 12 13 14

invol1 0.003 0.010 0.040 0.165 0.673 2.844 12.321

invol2 0.002 0.009 0.034 0.130 0.516 2.114 9.116

invol3 0.002 0.007 0.027 0.102 0.405 1.633 7.016

Python 3.1:

n 8 9 10 11 12 13 14

invol1 0.003 0.012 0.048 0.198 0.798 3.338 14.470

invol2 0.003 0.009 0.034 0.131 0.514 2.107 9.070

invol3 0.002 0.007 0.026 0.100 0.399 1.624 7.029

“invol1” and “invol2” are the 2 versions that I posted above. “invol3”

is a further optimization to invol2: I replaced the inner loop body

x[x.index(i)] += 1

tmp.append(x[:i-1] + [k] + x[i-1:] + [i])

by

a = x[i-1]; x[a-1-(a>i)] += 1

w = x[:]

w.insert(i-1, k)

w.append(i)

tmp.append(w)

How does it work: on each iteration of the outer loop s1 contains all the involutions of length k-2, and s2 contains the involutions of length k-1. The involutions of length k are obtained from these:

– take each element of s2 and stick k to the end: that’s an involution of length k that leaves k at its own position.

– the other involutions of length k have the value k swapped with some earlier value i. The remaining k-2 elements are an involution of length k-2 (and we happen to have all those in s1), except the elements must be re-labeled since the value i was taken. The “+=1” or “+(p>i)” stuff in the code handles the re-labeling (try with pencil and paper e.g. with k=5).