Algorithm for Iterating thru a list using pagination? - algorithm

I am trying to come up with an algorithm to iterate thru a list via pagination.
I'm only interested in the initial index and the size of the "page".
For example if my list is 100 items long, and the page length is 10:
1st page: starts at 0, length 10
2nd page: starts at 11, length 10
3rd page: starts at 21, length 10
...
Nth page: starts at 90, length 10
My problem is coming up with an elegant solution that satisfies these cases:
1. list has 9 elements, page length is 10
1st page: starts at 0, length 9
2. list has 84 elements, page length is 10
1st page: starts at 0, length 10
2nd page: starts at 11, length 10
3rd page: starts at 21, length 10
...
Nth page: starts at 80, length 4
I could do this with a bunch of conditionals and the modulo operation, but I was wondering if anyone could offer a better/elegant approach to this problem.
Thanks!


There follows some code doing it the long way in Python which could be used for other languages too; followed by how it could be done in a more maintainable fashion by the intermediate Pythoneer:
>>> from pprint import pprint as pp
>>> n, perpage = 84, 10
>>> mylist = list(range(n))
>>> mylist[:10]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> mylist[-10:] # last ten items
[74, 75, 76, 77, 78, 79, 80, 81, 82, 83]
>>> sublists = []
>>> for i in range(n):
pagenum, offset = divmod(i, perpage)
if offset == 0:
# first in new page so create another sublist
sublists.append([])
# add item to end of last sublist
sublists[pagenum].append(i)
>>> pp(sublists)
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74, 75, 76, 77, 78, 79],
[80, 81, 82, 83]]
>>> # Alternatively
>>> sublists2 = [mylist[i:i+perpage] for i in range(0, n, perpage)]
>>> pp(sublists2)
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74, 75, 76, 77, 78, 79],
[80, 81, 82, 83]]
>>>

Related

Why is there a long list of PolygonVertexIndex without any negatives (In one Fbx file)?

I made a default cylinder using blender and exported it into fbx file. When I looked at the content of the file after converting it into json, I found the field "PolygonVertexIndex" (under "Geometry"-"Objects") contains:"3, 1, 63, 61, 59, 57, 55, 53, 51, 49, 47, 45, 43, 41, 39, 37, 35, 33, 31, 29, 27, 25, 23, 21, 19, 17, 15, 13, 11, 9, 7" and " 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60".
These consistent positive indices series confuse me a lot since I believe fbx
PolygonVertexIndex store indices in "positive, positive, negative" (when the surface is triangular) or "positive, positive, positive, negative" (when the surface is quadrilateral). And I cannot find any information regarding consistent positive indices when I googled.
The whole list looks like:
["PolygonVertexIndex", [[0, 1, 3, -3, 2, 3, 5, -5, 4, 5, 7, -7, 6, 7,
9, -9, 8, 9, 11, -11, 10, 11, 13, -13, 12, 13, 15, -15, 14, 15, 17,
-17, 16, 17, 19, -19, 18, 19, 21, -21, 20, 21, 23, -23, 22, 23, 25, -25, 24, 25, 27, -27, 26, 27, 29, -29, 28, 29, 31, -31, 30, 31, 33, -33, 32, 33, 35, -35, 34, 35, 37, -37, 36, 37, 39, -39, 38, 39, 41, -41, 40, 41, 43, -43, 42, 43, 45, -45, 44, 45, 47, -47, 46, 47, 49, -49, 48, 49, 51, -51, 50, 51, 53, -53, 52, 53, 55, -55, 54, 55, 57, -57, 56, 57, 59, -59, 58, 59, 61, -61, 3, 1, 63, 61, 59, 57, 55, 53, 51, 49, 47, 45, 43, 41, 39, 37, 35, 33, 31, 29, 27, 25, 23, 21, 19,
17, 15, 13, 11, 9, 7, -6, 60, 61, 63, -63, 62, 63, 1, -1, 0, 2, 4, 6,
8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42,
44, 46, 48, 50, 52, 54, 56, 58, 60, -63]], "i", []],
It can be seen that the indices are fine at the start of the list but goes weird later.

The list contains negative numbers to end a polygon and looks consistent. A polygon is not restricted to be a triangle or a quad, it can contain as many vertices as it needs.
One way to create a cylinder mesh is using quads on the sides and circles on both ends. In your case, the two long series of positive numbers are the two circles encoded as two big polygons containing 32 vertices each.

Ruby FFI read_array_of_uint8 returns different results for the same read_string

I'm trying to pinpoint a bug we're having and this is as far as I could come.
We're creating a memory pointer which is then filled by a C library:
hash_result_pointer = FFI::MemoryPointer.new(:uint8, 144)
CLibrary(hash_result_pointer)
We're testing this on a macOS High Sierra and on a Ubuntu Xenial via Docker. Here are the outcomes for both:
macOS High Sierra
p hash_result_pointer.read_string
# "\x0F\xE1\x11%\v\xC2)a#"
p hash_result_pointer.read_array_of_uint8(144)
# [15, 225, 17, 37, 11, 194, 41, 97, 64, 0, 32, 156, 191, 7, 38, 254, 119, 123, 104, 145, 2, 160, 137, 112, 10, 193, 71, 15, 3, 134, 245, 5, 6, 24, 234, 1, 100, 4, 188, 45, 158, 28, 120, 130, 42, 62, 152, 15, 1, 97, 48, 16, 0, 41, 125, 5, 4, 20, 142, 3, 10, 3, 156, 1, 65, 10, 122, 15, 52, 67, 52, 109, 38, 13, 18, 128, 47, 44, 53, 90, 23, 72, 32, 82, 56, 17, 42, 126, 57, 71, 71, 80, 9, 180, **65**, 80, 18, 9, 5, 56, 26, 17, 2, 121, 21, 52, 21, 92, 74, 19, 92, 83, 84, 45, 89, 123, 34, 117, 75, 89, 8, 7, 19, 12, 8, 5, 8, 8, 25, 1, 20, 14, 48, 3, 24, 57, 12, 32, 18, 90, 3, 35, 15, 34]
Ubuntu Xenial via Docker
p hash_result_pointer.read_string
# "\x0F\xE1\x11%\v\xC2)a#"
p hash_result_pointer.read_array_of_uint8(144)
# [15, 225, 17, 37, 11, 194, 41, 97, 64, 0, 32, 156, 191, 7, 38, 254, 119, 123, 104, 145, 2, 160, 137, 112, 10, 193, 71, 15, 3, 134, 245, 5, 6, 24, 234, 1, 100, 4, 188, 45, 158, 28, 120, 130, 42, 62, 152, 15, 1, 97, 48, 16, 0, 41, 125, 5, 4, 20, 142, 3, 10, 3, 156, 1, 65, 10, 122, 15, 52, 67, 52, 109, 38, 13, 18, 128, 47, 44, 53, 90, 23, 72, 32, 82, 56, 17, 42, 126, 57, 71, 71, 80, 9, 180, **66**, 80, 18, 9, 5, 56, 26, 17, 2, 121, 21, 52, 21, 92, 74, 19, 92, 83, 84, 45, 89, 123, 34, 117, 75, 89, 8, 7, 19, 12, 8, 5, 8, 8, 25, 1, 20, 14, 48, 3, 24, 57, 12, 32, 18, 90, 3, 35, 15, 34]
As you can see, there is one array element that differs by 1 (65 vs 66) between the two environments. I have no clue why this is happening and I also don't know what could cause this and where shall I look further. Would appreciate any tip!

Ruby: using `.each` or `.step`, step forward a random amount for each iteration

(Also open to other similar non-Rails methods)
Given (0..99), return entries that are randomly picked in-order.
Example results:
0, 5, 11, 13, 34..
3, 12, 45, 67, 87
0, 1, 2, 3, 4, 5.. (very unlikely, of course)
Current thought:
(0..99).step(rand(0..99)).each do |subindex|
array.push(subindex)
end
However, this sets a single random value for all the steps whereas I'm looking for each step to be random.

Get a random value for the number of elements to pick, randomly get this number of elements, sort.
(0..99).to_a.sample((0..99).to_a.sample).sort
#⇒ [7, 20, 22, 29, 45, 48, 57, 61, 62, 76, 80, 82]
Or, shorter (credits to #Stefan):
(0..99).to_a.sample(rand(0..99)).sort
#⇒ [7, 20, 22, 29, 45, 48, 57, 61, 62, 76, 80, 82]
Or, in more functional manner:
λ = (0..99).to_a.method(:sample)
λ.(λ.()).sort
To feed exactly N numbers:
N = 10
(0..99).to_a.sample(N).sort
#⇒ [1, 5, 8, 12, 45, 54, 60, 65, 71, 91]

There're many ways to achieve it.
For example here's slow yet simple one:
# given `array`
random_indexes = (0...array.size).to_a.sample(rand(array.size))
random_indexes.sort.each { |i| puts array[i] }
Or why don't you just:
array.each do |value|
next if rand(2).zero?
puts value
end
Or you could use Enumerator#next random number of times.

Below example returns a sorted array with random entries from given range based on randomly picked true or false from array [true, false]:
(0..99).select { [true, false].sample }
=> [0, 3, 12, 13, 14, 17, 20, 24, 26, 28, 30, 32, 34, 35, 36, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 53, 54, 55, 56, 58, 59, 60, 61, 62, 65, 67, 69, 70, 71, 79, 81, 84, 86, 91, 93, 94, 95, 98, 99]
To reduce the chances of a bigger array being returned, you can modify your true/false array to include more falsey values:
(0..99).select { ([true] + [false] * 9).sample }
=> [21, 22, 28, 33, 37, 58, 59, 63, 77, 85, 86]

Associating values from one array to another in Ruby

I'm building a simple program after learning a little bit of Ruby. I'm trying to associate values from one array to another here's what I've got so far.
ColorValues = ["Black", "Brown", "Red", "Orange", "Yellow", "Green", "Cyan", "Blue", "Violet", "Pink", "Grey"]
(0..127).each_slice(12) {|i| p i}
.each_slice returns these arrays
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
[12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]
[24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35]
[36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47]
[48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59]
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71]
[72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83]
[84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95]
[96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107]
[108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119]
[120, 121, 122, 123, 124, 125, 126, 127]
What I'm attempting to do is then take the returned arrays and associate them with each color in the ColorValues[] array
i.e.
"Black" = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
"Brown" = [12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]
I'm sure there's a simple way of doing this I'm just not sure how to go about doing it.

Using Enumerable#zip and Hash::[]
colorValues = ["Black", "Brown", "Red", "Orange", "Yellow", "Green", "Cyan", "Blue", "Violet", "Pink", "Grey"]
Hash[colorValues.zip((0..127).each_slice(12))]
# => {"Black"=>[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
# "Brown"=>[12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23],
# "Red"=>[24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35],
# "Orange"=>[36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47],
# "Yellow"=>[48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
# "Green"=>[60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71],
# "Cyan"=>[72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83],
# "Blue"=>[84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95],
# "Violet"=>[96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107],
# "Pink"=>[108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119],
# "Grey"=>[120, 121, 122, 123, 124, 125, 126, 127]}

You can use zip for that:
ColorValues.zip( (0..127).each_slice(12) )

Method #zip is convenient, but it turns out that your problem is so frequent in Ruby, that I have overloaded #>> operator on Array class to perform zipping into a hash. First, install y_support gem. Then,
require 'y_support/core_ext/array' # or require 'y_support/all'
h = ColorValues >> ( 0..127 ).each_slice( 12 ) # returns a hash like in falsetru's answer
h["Black"] #=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

How to get a specific sequence like this?

There are 100 numbers:
1, 1, 2, 2, 3, 3,.. 50, 50.
How can I get a sequence which has one number between the two 1s, two numbers between the two 2s, three numbers between the two 3s,.. and fifty numbers between the two 50s using the hundred numbers?
Does anyone have a better idea than brute force? Or prove it there's no solution when n = 50.

The Problem: Langford Pairing
The problem is known as Langford Pairing. From Wikipedia:
These sequences are named after C. Dudley Langford, who posed the problem of constructing them in 1958. As Knuth1 describes, the problem of listing ALL Langford pairings for a given N can be solved as an instance of the exact cover problem (one of Karp's 21 NP-complete problem), but for large N the number of solutions can be calculated more efficiently by algebraic methods.
1: The Art of Computer Programming, IV, Fascicle 0: Introduction to Combinatorial Algorithms and Boolean Functions
It's worth noting that there is no solution for N = 50. Solutions only exist for N = 4k or N = 4k - 1. This is proven in a paper by Roy A. Davies (On Langford's Problem II, Math. Gaz. 43, 253-255, 1959), which also gives the pattern to construct a single solution for any feasible N.
Related links
John Miller's page on Langford's Problem
A Perl program based on Roy A. Davies' pattern
Dave Moore's pattern
mathworld.wolfram - Langford's Problem
A CSP model for Microsoft Solver Foundation
Brute force in Java
Here are some solutions that my quick and dirty brute force program was able to find for N < 100. Nearly all of these were found in under a second.
3 [3, 1, 2, 1, 3, 2]
4 [4, 1, 3, 1, 2, 4, 3, 2]
7 [7, 3, 6, 2, 5, 3, 2, 4, 7, 6, 5, 1, 4, 1]
8 [8, 3, 7, 2, 6, 3, 2, 4, 5, 8, 7, 6, 4, 1, 5, 1]
11 [11, 6, 10, 2, 9, 3, 2, 8, 6, 3, 7, 5, 11, 10, 9, 4, 8, 5, 7, 1, 4, 1]
12 [12, 10, 11, 6, 4, 5, 9, 7, 8, 4, 6, 5, 10, 12, 11, 7, 9, 8, 3, 1, 2, 1, 3, 2]
15 [15, 13, 14, 8, 5, 12, 7, 11, 4, 10, 5, 9, 8, 4, 7, 13, 15, 14, 12, 11, 10, 9, 6, 3, 1, 2, 1, 3, 2, 6]
16 [16, 14, 15, 9, 7, 13, 3, 12, 6, 11, 3, 10, 7, 9, 8, 6, 14, 16, 15, 13, 12, 11, 10, 8, 5, 2, 4, 1, 2, 1, 5, 4]
19 [19, 17, 18, 14, 8, 16, 9, 15, 6, 1, 13, 1, 12, 8, 11, 6, 9, 10, 14, 17, 19, 18, 16, 15, 13, 12, 11, 7, 10, 3, 5, 2, 4, 3, 2, 7, 5, 4]
20 [20, 18, 19, 15, 11, 17, 10, 16, 9, 5, 14, 1, 13, 1, 12, 5, 11, 10, 9, 15, 18, 20, 19, 17, 16, 14, 13, 12, 8, 4, 7, 3, 6, 2, 4, 3, 2, 8, 7, 6]
23 [23, 21, 22, 18, 16, 20, 12, 19, 11, 8, 17, 4, 1, 15, 1, 14, 4, 13, 8, 12, 11, 16, 18, 21, 23, 22, 20, 19, 17, 15, 14, 13, 10, 7, 9, 3, 5, 2, 6, 3, 2, 7, 5, 10, 9, 6]
24 [24, 22, 23, 19, 17, 21, 13, 20, 10, 8, 18, 4, 1, 16, 1, 15, 4, 14, 8, 10, 13, 12, 17, 19, 22, 24, 23, 21, 20, 18, 16, 15, 14, 11, 12, 7, 9, 3, 5, 2, 6, 3, 2, 7, 5, 11, 9, 6]
27 [27, 25, 26, 22, 20, 24, 17, 23, 12, 13, 21, 7, 4, 19, 1, 18, 1, 4, 16, 7, 15, 12, 14, 13, 17, 20, 22, 25, 27, 26, 24, 23, 21, 19, 18, 16, 15, 14, 11, 9, 10, 5, 2, 8, 3, 2, 6, 5, 3, 9, 11, 10, 8, 6]
28 [28, 26, 27, 23, 21, 25, 18, 24, 15, 13, 22, 10, 6, 20, 1, 19, 1, 3, 17, 6, 16, 3, 10, 13, 15, 18, 21, 23, 26, 28, 27, 25, 24, 22, 20, 19, 17, 16, 14, 12, 9, 7, 11, 4, 2, 5, 8, 2, 4, 7, 9, 5, 12, 14, 11, 8]
31 [31, 29, 30, 26, 24, 28, 21, 27, 18, 16, 25, 13, 11, 23, 6, 22, 5, 1, 20, 1, 19, 6, 5, 17, 11, 13, 16, 18, 21, 24, 26, 29, 31, 30, 28, 27, 25, 23, 22, 20, 19, 17, 15, 12, 14, 9, 10, 2, 3, 4, 2, 8, 3, 7, 4, 9, 12, 10, 15, 14, 8, 7]
32 [32, 30, 31, 27, 25, 29, 22, 28, 19, 17, 26, 13, 11, 24, 6, 23, 5, 1, 21, 1, 20, 6, 5, 18, 11, 13, 16, 17, 19, 22, 25, 27, 30, 32, 31, 29, 28, 26, 24, 23, 21, 20, 18, 16, 15, 12, 14, 9, 10, 2, 3, 4, 2, 8, 3, 7, 4, 9, 12, 10, 15, 14, 8, 7]
35 [35, 33, 34, 30, 28, 32, 25, 31, 22, 20, 29, 17, 14, 27, 10, 26, 5, 6, 24, 1, 23, 1, 5, 21, 6, 10, 19, 14, 18, 17, 20, 22, 25, 28, 30, 33, 35, 34, 32, 31, 29, 27, 26, 24, 23, 21, 19, 18, 16, 13, 15, 12, 9, 4, 2, 11, 3, 2, 4, 8, 3, 7, 9, 13, 12, 16, 15, 11, 8, 7]
36 [36, 34, 35, 31, 29, 33, 26, 32, 23, 21, 30, 17, 14, 28, 10, 27, 5, 6, 25, 1, 24, 1, 5, 22, 6, 10, 20, 14, 19, 17, 18, 21, 23, 26, 29, 31, 34, 36, 35, 33, 32, 30, 28, 27, 25, 24, 22, 20, 19, 18, 16, 13, 15, 12, 9, 4, 2, 11, 3, 2, 4, 8, 3, 7, 9, 13, 12, 16, 15, 11, 8, 7]
40 [40, 38, 39, 35, 33, 37, 30, 36, 27, 25, 34, 22, 20, 32, 17, 31, 13, 11, 29, 7, 28, 3, 1, 26, 1, 3, 24, 7, 23, 11, 13, 21, 17, 20, 22, 25, 27, 30, 33, 35, 38, 40, 39, 37, 36, 34, 32, 31, 29, 28, 26, 24, 23, 21, 19, 16, 18, 15, 12, 4, 6, 14, 2, 5, 4, 2, 10, 6, 9, 5, 8, 12, 16, 15, 19, 18, 14, 10, 9, 8]
43 [43, 41, 42, 38, 36, 40, 33, 39, 30, 28, 37, 25, 23, 35, 20, 34, 16, 14, 32, 5, 31, 8, 6, 29, 2, 5, 27, 2, 26, 6, 8, 24, 14, 16, 22, 20, 23, 25, 28, 30, 33, 36, 38, 41, 43, 42, 40, 39, 37, 35, 34, 32, 31, 29, 27, 26, 24, 22, 21, 19, 17, 15, 18, 12, 7, 1, 3, 1, 13, 4, 3, 11, 7, 10, 4, 9, 12, 15, 17, 19, 21, 18, 13, 11, 10, 9]
44 [44, 42, 43, 39, 37, 41, 34, 40, 31, 29, 38, 26, 24, 36, 20, 35, 16, 14, 33, 5, 32, 8, 6, 30, 2, 5, 28, 2, 27, 6, 8, 25, 14, 16, 23, 20, 22, 24, 26, 29, 31, 34, 37, 39, 42, 44, 43, 41, 40, 38, 36, 35, 33, 32, 30, 28, 27, 25, 23, 22, 21, 19, 17, 15, 18, 12, 7, 1, 3, 1, 13, 4, 3, 11, 7, 10, 4, 9, 12, 15, 17, 19, 21, 18, 13, 11, 10, 9]
52 [52, 50, 51, 47, 45, 49, 42, 48, 39, 37, 46, 34, 32, 44, 29, 43, 26, 24, 41, 20, 40, 16, 14, 38, 7, 9, 36, 2, 35, 3, 2, 33, 7, 3, 31, 9, 30, 14, 16, 28, 20, 27, 24, 26, 29, 32, 34, 37, 39, 42, 45, 47, 50, 52, 51, 49, 48, 46, 44, 43, 41, 40, 38, 36, 35, 33, 31, 30, 28, 27, 25, 23, 21, 19, 22, 15, 8, 6, 1, 18, 1, 17, 4, 5, 6, 8, 13, 4, 12, 5, 11, 15, 10, 19, 21, 23, 25, 22, 18, 17, 13, 12, 11, 10]
55 [55, 53, 54, 50, 48, 52, 45, 51, 42, 40, 49, 37, 35, 47, 32, 46, 29, 27, 44, 23, 43, 20, 17, 41, 10, 11, 39, 4, 38, 8, 2, 36, 4, 2, 34, 10, 33, 11, 8, 31, 17, 30, 20, 23, 28, 27, 29, 32, 35, 37, 40, 42, 45, 48, 50, 53, 55, 54, 52, 51, 49, 47, 46, 44, 43, 41, 39, 38, 36, 34, 33, 31, 30, 28, 26, 24, 25, 21, 19, 16, 22, 9, 14, 6, 3, 18, 7, 5, 3, 15, 6, 9, 13, 5, 7, 12, 16, 14, 19, 21, 24, 26, 25, 22, 18, 15, 13, 1, 12, 1]
63 [63, 61, 62, 58, 56, 60, 53, 59, 50, 48, 57, 45, 43, 55, 40, 54, 37, 35, 52, 32, 51, 29, 27, 49, 23, 20, 47, 17, 46, 12, 9, 44, 10, 3, 42, 2, 41, 3, 2, 39, 9, 38, 12, 10, 36, 17, 20, 34, 23, 33, 27, 29, 32, 35, 37, 40, 43, 45, 48, 50, 53, 56, 58, 61, 63, 62, 60, 59, 57, 55, 54, 52, 51, 49, 47, 46, 44, 42, 41, 39, 38, 36, 34, 33, 31, 28, 30, 25, 26, 22, 19, 8, 18, 24, 11, 6, 4, 21, 5, 7, 8, 4, 6, 16, 5, 15, 11, 7, 13, 14, 19, 18, 22, 25, 28, 26, 31, 30, 24, 21, 16, 15, 13, 1, 14, 1]
64 [64, 62, 63, 59, 57, 61, 54, 60, 51, 49, 58, 46, 44, 56, 41, 55, 38, 36, 53, 33, 52, 29, 27, 50, 23, 20, 48, 17, 47, 12, 9, 45, 10, 3, 43, 2, 42, 3, 2, 40, 9, 39, 12, 10, 37, 17, 20, 35, 23, 34, 27, 29, 32, 33, 36, 38, 41, 44, 46, 49, 51, 54, 57, 59, 62, 64, 63, 61, 60, 58, 56, 55, 53, 52, 50, 48, 47, 45, 43, 42, 40, 39, 37, 35, 34, 32, 31, 28, 30, 25, 26, 22, 19, 8, 18, 24, 11, 6, 4, 21, 5, 7, 8, 4, 6, 16, 5, 15, 11, 7, 13, 14, 19, 18, 22, 25, 28, 26, 31, 30, 24, 21, 16, 15, 13, 1, 14, 1]
67 [67, 65, 66, 62, 60, 64, 57, 63, 54, 52, 61, 49, 47, 59, 44, 58, 41, 39, 56, 36, 55, 33, 30, 53, 26, 24, 51, 20, 50, 13, 11, 48, 12, 3, 46, 4, 45, 3, 7, 43, 4, 42, 11, 13, 40, 12, 7, 38, 20, 37, 24, 26, 35, 30, 34, 33, 36, 39, 41, 44, 47, 49, 52, 54, 57, 60, 62, 65, 67, 66, 64, 63, 61, 59, 58, 56, 55, 53, 51, 50, 48, 46, 45, 43, 42, 40, 38, 37, 35, 34, 32, 29, 31, 28, 25, 23, 21, 27, 18, 9, 10, 2, 5, 22, 2, 8, 6, 19, 5, 9, 17, 10, 16, 6, 8, 14, 15, 18, 21, 23, 25, 29, 28, 32, 31, 27, 22, 19, 17, 16, 14, 1, 15, 1]
72 [72, 70, 71, 67, 65, 69, 62, 68, 59, 57, 66, 54, 52, 64, 49, 63, 46, 44, 61, 41, 60, 38, 36, 58, 33, 30, 56, 27, 55, 23, 20, 53, 17, 14, 51, 10, 50, 5, 3, 48, 4, 47, 3, 5, 45, 4, 10, 43, 14, 42, 17, 20, 40, 23, 39, 27, 30, 37, 33, 36, 38, 41, 44, 46, 49, 52, 54, 57, 59, 62, 65, 67, 70, 72, 71, 69, 68, 66, 64, 63, 61, 60, 58, 56, 55, 53, 51, 50, 48, 47, 45, 43, 42, 40, 39, 37, 35, 32, 34, 31, 28, 26, 24, 22, 29, 15, 13, 8, 6, 25, 7, 12, 9, 11, 21, 6, 8, 19, 7, 18, 13, 15, 9, 16, 12, 11, 22, 24, 26, 28, 32, 31, 35, 34, 29, 25, 21, 19, 18, 2, 16, 1, 2, 1]
75 [75, 73, 74, 70, 68, 72, 65, 71, 62, 60, 69, 57, 55, 67, 52, 66, 49, 47, 64, 44, 63, 41, 39, 61, 36, 33, 59, 30, 58, 26, 24, 56, 20, 17, 54, 14, 53, 5, 6, 51, 7, 50, 3, 5, 48, 6, 3, 46, 7, 45, 14, 17, 43, 20, 42, 24, 26, 40, 30, 33, 38, 36, 39, 41, 44, 47, 49, 52, 55, 57, 60, 62, 65, 68, 70, 73, 75, 74, 72, 71, 69, 67, 66, 64, 63, 61, 59, 58, 56, 54, 53, 51, 50, 48, 46, 45, 43, 42, 40, 38, 37, 35, 32, 29, 34, 28, 25, 23, 31, 12, 15, 9, 11, 27, 21, 4, 13, 10, 8, 22, 4, 9, 12, 19, 11, 18, 15, 8, 10, 16, 13, 23, 25, 29, 28, 32, 21, 35, 37, 34, 31, 27, 22, 19, 18, 2, 16, 1, 2, 1]
76 [76, 74, 75, 71, 69, 73, 66, 72, 63, 61, 70, 58, 56, 68, 53, 67, 50, 48, 65, 45, 64, 42, 40, 62, 36, 33, 60, 30, 59, 26, 24, 57, 20, 17, 55, 14, 54, 5, 6, 52, 7, 51, 3, 5, 49, 6, 3, 47, 7, 46, 14, 17, 44, 20, 43, 24, 26, 41, 30, 33, 39, 36, 38, 40, 42, 45, 48, 50, 53, 56, 58, 61, 63, 66, 69, 71, 74, 76, 75, 73, 72, 70, 68, 67, 65, 64, 62, 60, 59, 57, 55, 54, 52, 51, 49, 47, 46, 44, 43, 41, 39, 38, 37, 35, 32, 29, 34, 28, 25, 23, 31, 12, 15, 9, 11, 27, 21, 4, 13, 10, 8, 22, 4, 9, 12, 19, 11, 18, 15, 8, 10, 16, 13, 23, 25, 29, 28, 32, 21, 35, 37, 34, 31, 27, 22, 19, 18, 2, 16, 1, 2, 1]
83 [83, 81, 82, 78, 76, 80, 73, 79, 70, 68, 77, 65, 63, 75, 60, 74, 57, 55, 72, 52, 71, 49, 47, 69, 44, 42, 67, 39, 66, 36, 33, 64, 30, 27, 62, 23, 61, 17, 14, 59, 15, 58, 7, 4, 56, 5, 11, 54, 4, 53, 7, 5, 51, 14, 50, 17, 15, 48, 11, 23, 46, 27, 45, 30, 33, 43, 36, 39, 42, 44, 47, 49, 52, 55, 57, 60, 63, 65, 68, 70, 73, 76, 78, 81, 83, 82, 80, 79, 77, 75, 74, 72, 71, 69, 67, 66, 64, 62, 61, 59, 58, 56, 54, 53, 51, 50, 48, 46, 45, 43, 41, 38, 40, 37, 34, 32, 29, 26, 35, 25, 16, 13, 24, 31, 8, 6, 3, 28, 12, 9, 3, 10, 6, 8, 22, 13, 21, 16, 20, 9, 19, 12, 10, 18, 26, 25, 29, 24, 32, 34, 38, 37, 41, 40, 35, 31, 28, 22, 21, 20, 19, 2, 18, 1, 2, 1]
84 [84, 82, 83, 79, 77, 81, 74, 80, 71, 69, 78, 66, 64, 76, 61, 75, 58, 56, 73, 53, 72, 50, 48, 70, 45, 43, 68, 39, 67, 36, 33, 65, 30, 27, 63, 23, 62, 17, 14, 60, 15, 59, 7, 4, 57, 5, 11, 55, 4, 54, 7, 5, 52, 14, 51, 17, 15, 49, 11, 23, 47, 27, 46, 30, 33, 44, 36, 39, 42, 43, 45, 48, 50, 53, 56, 58, 61, 64, 66, 69, 71, 74, 77, 79, 82, 84, 83, 81, 80, 78, 76, 75, 73, 72, 70, 68, 67, 65, 63, 62, 60, 59, 57, 55, 54, 52, 51, 49, 47, 46, 44, 42, 41, 38, 40, 37, 34, 32, 29, 26, 35, 25, 16, 13, 24, 31, 8, 6, 3, 28, 12, 9, 3, 10, 6, 8, 22, 13, 21, 16, 20, 9, 19, 12, 10, 18, 26, 25, 29, 24, 32, 34, 38, 37, 41, 40, 35, 31, 28, 22, 21, 20, 19, 2, 18, 1, 2, 1]
87 [87, 85, 86, 82, 80, 84, 77, 83, 74, 72, 81, 69, 67, 79, 64, 78, 61, 59, 76, 56, 75, 53, 51, 73, 48, 46, 71, 43, 70, 39, 36, 68, 33, 30, 66, 27, 65, 23, 17, 63, 14, 62, 16, 7, 60, 12, 3, 58, 4, 57, 3, 7, 55, 4, 54, 14, 17, 52, 12, 16, 50, 23, 49, 27, 30, 47, 33, 36, 45, 39, 44, 43, 46, 48, 51, 53, 56, 59, 61, 64, 67, 69, 72, 74, 77, 80, 82, 85, 87, 86, 84, 83, 81, 79, 78, 76, 75, 73, 71, 70, 68, 66, 65, 63, 62, 60, 58, 57, 55, 54, 52, 50, 49, 47, 45, 44, 42, 40, 41, 37, 35, 32, 38, 31, 28, 26, 24, 34, 8, 15, 9, 11, 6, 29, 13, 5, 10, 8, 25, 6, 9, 5, 22, 11, 21, 15, 20, 10, 13, 18, 19, 24, 26, 28, 32, 31, 35, 37, 40, 42, 41, 38, 34, 29, 25, 22, 21, 20, 18, 2, 19, 1, 2, 1]
95 [95, 93, 94, 90, 88, 92, 85, 91, 82, 80, 89, 77, 75, 87, 72, 86, 69, 67, 84, 64, 83, 61, 59, 81, 56, 54, 79, 51, 78, 48, 46, 76, 43, 40, 74, 36, 73, 33, 30, 71, 26, 70, 18, 15, 68, 17, 9, 66, 6, 65, 13, 14, 63, 4, 62, 6, 9, 60, 4, 15, 58, 18, 57, 17, 13, 55, 14, 26, 53, 30, 52, 33, 36, 50, 40, 49, 43, 46, 48, 51, 54, 56, 59, 61, 64, 67, 69, 72, 75, 77, 80, 82, 85, 88, 90, 93, 95, 94, 92, 91, 89, 87, 86, 84, 83, 81, 79, 78, 76, 74, 73, 71, 70, 68, 66, 65, 63, 62, 60, 58, 57, 55, 53, 52, 50, 49, 47, 45, 42, 39, 44, 38, 35, 32, 41, 31, 28, 34, 25, 37, 16, 12, 7, 11, 8, 3, 5, 10, 29, 3, 7, 27, 5, 8, 12, 11, 24, 16, 10, 23, 19, 20, 21, 22, 25, 28, 32, 31, 35, 39, 38, 42, 34, 45, 47, 44, 41, 37, 29, 27, 19, 24, 20, 23, 21, 2, 22, 1, 2, 1]
96 [96, 94, 95, 91, 89, 93, 86, 92, 83, 81, 90, 78, 76, 88, 73, 87, 70, 68, 85, 65, 84, 62, 60, 82, 57, 55, 80, 52, 79, 49, 46, 77, 43, 40, 75, 36, 74, 33, 30, 72, 26, 71, 18, 15, 69, 17, 9, 67, 6, 66, 13, 14, 64, 4, 63, 6, 9, 61, 4, 15, 59, 18, 58, 17, 13, 56, 14, 26, 54, 30, 53, 33, 36, 51, 40, 50, 43, 46, 48, 49, 52, 55, 57, 60, 62, 65, 68, 70, 73, 76, 78, 81, 83, 86, 89, 91, 94, 96, 95, 93, 92, 90, 88, 87, 85, 84, 82, 80, 79, 77, 75, 74, 72, 71, 69, 67, 66, 64, 63, 61, 59, 58, 56, 54, 53, 51, 50, 48, 47, 45, 42, 39, 44, 38, 35, 32, 41, 31, 28, 34, 25, 37, 16, 12, 7, 11, 8, 3, 5, 10, 29, 3, 7, 27, 5, 8, 12, 11, 24, 16, 10, 23, 19, 20, 21, 22, 25, 28, 32, 31, 35, 39, 38, 42, 34, 45, 47, 44, 41, 37, 29, 27, 19, 24, 20, 23, 21, 2, 22, 1, 2, 1]
For instructional purposes, here's the source code:
import java.util.*;
public class LangfordPairing {
static void langford(int N) {
BitSet bs = new BitSet();
bs.set(N * 2);
put(bs, N, new int[2 * N]);
}
static void put(BitSet bs, int n, int[] arr) {
if (n == 0) {
System.out.println(Arrays.toString(arr));
System.exit(0); // one is enough!
}
for (int i = -1, L = bs.length() - n - 1;
(i = bs.nextClearBit(i + 1)) < L ;) {
final int j = i + n + 1;
if (!bs.get(j)) {
arr[i] = n;
arr[j] = n;
bs.flip(i);
bs.flip(j);
put(bs, n - 1, arr);
bs.flip(i);
bs.flip(j);
}
}
}
public static void main(String[] args) {
langford(87);
}
}
Solutions for some N values are missing; they are known to require an abnormally large number of operations just to find the first solution by brute force.
Note that as mentioned, there are generally many solutions for any given N. For N = 7, there are 26 solutions:
[7, 3, 6, 2, 5, 3, 2, 4, 7, 6, 5, 1, 4, 1]
[7, 2, 6, 3, 2, 4, 5, 3, 7, 6, 4, 1, 5, 1]
[7, 2, 4, 6, 2, 3, 5, 4, 7, 3, 6, 1, 5, 1]
[7, 3, 1, 6, 1, 3, 4, 5, 7, 2, 6, 4, 2, 5]
[7, 1, 4, 1, 6, 3, 5, 4, 7, 3, 2, 6, 5, 2]
[7, 1, 3, 1, 6, 4, 3, 5, 7, 2, 4, 6, 2, 5]
[7, 4, 1, 5, 1, 6, 4, 3, 7, 5, 2, 3, 6, 2]
[7, 2, 4, 5, 2, 6, 3, 4, 7, 5, 3, 1, 6, 1]
[5, 7, 2, 6, 3, 2, 5, 4, 3, 7, 6, 1, 4, 1]
[3, 7, 4, 6, 3, 2, 5, 4, 2, 7, 6, 1, 5, 1]
[5, 7, 4, 1, 6, 1, 5, 4, 3, 7, 2, 6, 3, 2]
[5, 7, 2, 3, 6, 2, 5, 3, 4, 7, 1, 6, 1, 4]
[1, 7, 1, 2, 6, 4, 2, 5, 3, 7, 4, 6, 3, 5]
[5, 7, 1, 4, 1, 6, 5, 3, 4, 7, 2, 3, 6, 2]
[1, 7, 1, 2, 5, 6, 2, 3, 4, 7, 5, 3, 6, 4]
[2, 7, 4, 2, 3, 5, 6, 4, 3, 7, 1, 5, 1, 6]
[6, 2, 7, 4, 2, 3, 5, 6, 4, 3, 7, 1, 5, 1]
[2, 6, 7, 2, 1, 5, 1, 4, 6, 3, 7, 5, 4, 3]
[3, 6, 7, 1, 3, 1, 4, 5, 6, 2, 7, 4, 2, 5]
[5, 1, 7, 1, 6, 2, 5, 4, 2, 3, 7, 6, 4, 3]
[2, 3, 7, 2, 6, 3, 5, 1, 4, 1, 7, 6, 5, 4]
[4, 1, 7, 1, 6, 4, 2, 5, 3, 2, 7, 6, 3, 5]
[5, 2, 7, 3, 2, 6, 5, 3, 4, 1, 7, 1, 6, 4]
[3, 5, 7, 4, 3, 6, 2, 5, 4, 2, 7, 1, 6, 1]
[3, 5, 7, 2, 3, 6, 2, 5, 4, 1, 7, 1, 6, 4]
[2, 4, 7, 2, 3, 6, 4, 5, 3, 1, 7, 1, 6, 5]
Related links
John Miller - Langford's Problem - Oddity for some N values
OEIS A014552 - Number of solutions to Langford problem
Attachments
Source code and output on ideone.com

There is only a solution to this problem for pairs of numbers from 1-n where n = 4m or n = 4m-1 for any positive integer m.
Update:
For any solution the odd number pairs must occupy two odd-numbered or two even-numbered positions. The even number pairs must occupy one of each. When there are an odd number of odd pairs (eg. 1 1 2 2 3 3 4 4 5 5 - 3 odd pairs) there is no solution. There's no way to place the first number of each pair, without resulting in a clash when you try to place the second.
See http://en.wikipedia.org/wiki/Langford_pairing
Another update:
My answer was basically from Knuth. I've been thinking it through, though, and came up with the following on my own.
For any sequence {1 1 2 2 ... n n} there are, say, m odd pairs, n-m even pairs, and 2n positions in which to place them (i.e. n positions of each parity).
If you place the even pairs first then you use n-m even positions and n-m odd positions, thus you have m positions of each parity left in which to place the odd pairs.
The odd pairs must be placed in positions of the same parity. If m is even there is no problem because half the pairs will will be placed in odd positions, and half in even positions.
If m is odd, however, you can only place m-1 of the odd pairs, at which point you'll have one odd pair left to place, and one position of each parity. As the odd pair requires positions of the same parity there is no solution when m (the number of odd pairs) is odd.

I adopted an elimination approach with paper and pencil for L(2,7).
Place the two 7s. Just three possibilities
Place the two 6s and so on.
It is obvious when placing the 1s, 2s, and 3s simultaneously whether there is a solution or not.
I found 18 of the 26 solutions thus.
When time and paper permits I will tackle L(2,8).

I'm pretty sure I've seen this in Knuth's "The Art of Computer Programming". I'll look it up when I get home tonight.

I think a good heuristic would be to start with the biggest numbers. You'll still need backtracking though.
The problem with the problem is that having a solution for n-1 usually doesn't help you much to find a solution for n.

Is it always possible? I can see the pattern is probable and not definite. It doesn't work for 1s or 1s and 2s.
If it is probablistic, the best idea would be to start with brute force and keep on eliminating wrong options as an when you encounter them.

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