Related
(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]
Algorithm:
A sequence of numbers is called a wiggle sequence if the differences
between successive numbers strictly alternate between positive and
negative. The first difference (if one exists) may be either positive
or negative. A sequence with fewer than two elements is trivially a
wiggle sequence.
For example, [1,7,4,9,2,5] is a wiggle sequence because the
differences (6,-3,5,-7,3) are alternately positive and negative. In
contrast, [1,4,7,2,5] and [1,7,4,5,5] are not wiggle sequences, the
first because its first two differences are positive and the second
because its last difference is zero.
Given a sequence of integers, return the length of the longest
subsequence that is a wiggle sequence. A subsequence is obtained by
deleting some number of elements (eventually, also zero) from the
original sequence, leaving the remaining elements in their original
order.
Examples:
Input: [1,7,4,9,2,5]
Output: 6
The entire sequence is a wiggle sequence.
Input: [1,17,5,10,13,15,10,5,16,8]
Output: 7
There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].
Input: [1,2,3,4,5,6,7,8,9]
Output: 2
My soln:
def wiggle_max_length(nums)
[ build_seq(nums, 0, 0, true, -1.0/0.0),
build_seq(nums, 0, 0, false, 1.0/0.0)
].max
end
def build_seq(nums, index, len, wiggle_up, prev)
return len if index >= nums.length
if wiggle_up && nums[index] - prev > 0 || !wiggle_up && nums[index] - prev < 0
build_seq(nums, index + 1, len + 1, !wiggle_up, nums[index])
else
build_seq(nums, index + 1, len, wiggle_up, prev)
end
end
This is working for smaller inputs (e.g [1,1,1,3,2,4,1,6,3,10,8] and for all the sample inputs, but its failing for very large inputs (which is harder to debug) like:
[33,53,12,64,50,41,45,21,97,35,47,92,39,0,93,55,40,46,69,42,6,95,51,68,72,9,32,84,34,64,6,2,26,98,3,43,30,60,3,68,82,9,97,19,27,98,99,4,30,96,37,9,78,43,64,4,65,30,84,90,87,64,18,50,60,1,40,32,48,50,76,100,57,29,63,53,46,57,93,98,42,80,82,9,41,55,69,84,82,79,30,79,18,97,67,23,52,38,74,15]
which should have output: 67 but my soln outputs 57. Does anyone know what is wrong here?
The approach tried is a greedy solution (because it always uses the current element if it satisfies the wiggle condition), but this does not always work.
I will try illustrating this with this simpler counter-example: 1 100 99 6 7 4 5 2 3.
One best sub-sequence is: 1 100 6 7 4 5 2 3, but the two build_seq calls from the algorithm will produce these sequences:
1 100 99
1
Edit: A slightly modified greedy approach does work -- see this link, thanks Peter de Rivaz.
Dynamic Programming can be used to obtain an optimal solution.
Note: I wrote this before seeing the article mentioned by #PeterdeRivaz. While dynamic programming (O(n2)) works, the article presents a superior (O(n)) "greedy" algorithm ("Approach #5"), which is also far easier to code than a dynamic programming solution. I have added a second answer that implements that method.
Code
def longest_wiggle(arr)
best = [{ pos_diff: { length: 0, prev_ndx: nil },
neg_diff: { length: 0, prev_ndx: nil } }]
(1..arr.size-1).each do |i|
calc_best(arr, i, :pos_diff, best)
calc_best(arr, i, :neg_diff, best)
end
unpack_best(best)
end
def calc_best(arr, i, diff, best)
curr = arr[i]
prev_indices = (0..i-1).select { |j|
(diff==:pos_diff) ? (arr[j] < curr) : (arr[j] > curr) }
best[i] = {} if best.size == i
best[i][diff] =
if prev_indices.empty?
{ length: 0, prev_ndx: nil }
else
prev_diff = previous_diff(diff)
j = prev_indices.max_by { |j| best[j][prev_diff][:length] }
{ length: (1 + best[j][prev_diff][:length]), prev_ndx: j }
end
end
def previous_diff(diff)
diff==:pos_diff ? :neg_diff : :pos_diff·
end
def unpack_best(best)
last_idx, last_diff =
best.size.times.to_a.product([:pos_diff, :neg_diff]).
max_by { |i,diff| best[i][diff][:length] }
return [0, []] if best[last_idx][last_diff][:length].zero?
best_path = []
loop do
best_path.unshift(last_idx)
prev_index = best[last_idx][last_diff][:prev_ndx]
break if prev_index.nil?
last_idx = prev_index·
last_diff = previous_diff(last_diff)
end
best_path
end
Examples
longest_wiggle([1, 4, 2, 6, 8, 3, 2, 5])
#=> [0, 1, 2, 3, 5, 7]]
The length of the longest wiggle is 6 and consists of the elements at indices 0, 1, 2, 3, 5 and 7, that is, [1, 4, 2, 6, 3, 5].
A second example uses the larger array given in the question.
arr = [33, 53, 12, 64, 50, 41, 45, 21, 97, 35, 47, 92, 39, 0, 93, 55, 40, 46,
69, 42, 6, 95, 51, 68, 72, 9, 32, 84, 34, 64, 6, 2, 26, 98, 3, 43, 30,
60, 3, 68, 82, 9, 97, 19, 27, 98, 99, 4, 30, 96, 37, 9, 78, 43, 64, 4,
65, 30, 84, 90, 87, 64, 18, 50, 60, 1, 40, 32, 48, 50, 76, 100, 57, 29,
arr.size 63, 53, 46, 57, 93, 98, 42, 80, 82, 9, 41, 55, 69, 84, 82, 79, 30, 79,
18, 97, 67, 23, 52, 38, 74, 15]
#=> 100
longest_wiggle(arr).size
#=> 67
longest_wiggle(arr)
#=> [0, 1, 2, 3, 5, 6, 7, 8, 9, 10, 12, 14, 16, 17, 19, 21, 22, 23, 25,
# 27, 28, 29, 30, 32, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 47, 49, 50,
# 52, 53, 54, 55, 56, 57, 58, 62, 63, 65, 66, 67, 70, 72, 74, 75, 77, 80,
# 81, 83, 84, 90, 91, 92, 93, 95, 96, 97, 98, 99]
As indicated, the largest wiggle is comprised of 67 elements of arr. Solution time was essentially instantaneous.
The values of arr at those indices are as follows.
[33, 53, 12, 64, 41, 45, 21, 97, 35, 47, 39, 93, 40, 46, 42, 95, 51, 68, 9,
84, 34, 64, 6, 26, 3, 43, 30, 60, 3, 68, 9, 97, 19, 27, 4, 96, 37, 78, 43,
64, 4, 65, 30, 84, 18, 50, 1, 40, 32, 76, 57, 63, 53, 57, 42, 80, 9, 41, 30,
79, 18, 97, 23, 52, 38, 74, 15]
[33, 53, 12, 64, 41, 45, 21, 97, 35, 92, 0, 93, 40, 69, 6, 95, 51, 72, 9, 84, 34, 64, 2, 98, 3, 43, 30, 60, 3, 82, 9, 97, 19, 99, 4, 96, 9, 78, 43, 64, 4, 65, 30, 90, 18, 60, 1, 40, 32, 100, 29, 63, 46, 98, 42, 82, 9, 84, 30, 79, 18, 97, 23, 52, 38, 74]
Explanation
I had intended to provide an explanation of the algorithm and its implementation, but having since learned there is a superior approach (see my note at the beginning of my answer), I have decided against doing that, but would of course be happy to answer any questions. The link in my note explains, among other things, how dynamic programming can be used here.
Let Wp[i] be the longest wiggle sequence starting at element i, and where the first difference is positive. Let Wn[i] be the same, but where the first difference is negative.
Then:
Wp[k] = max(1+Wn[k'] for k<k'<n, where A[k'] > A[k]) (or 1 if no such k' exists)
Wn[k] = max(1+Wp[k'] for k<k'<n, where A[k'] < A[k]) (or 1 if no such k' exists)
This gives an O(n^2) dynamic programming solution, here in pseudocode
Wp = [1, 1, ..., 1] -- length n
Wn = [1, 1, ..., 1] -- length n
for k = n-1, n-2, ..., 0
for k' = k+1, k+2, ..., n-1
if A[k'] > A[k]
Wp[k] = max(Wp[k], Wn[k']+1)
else if A[k'] < A[k]
Wn[k] = max(Wn[k], Wp[k']+1)
result = max(max(Wp[i], Wn[i]) for i = 0, 1, ..., n-1)
In a comment on #quertyman's answer, #PeterdeRivaz provided a link to an article that considers various approaches to solving the "longest wiggle subsequence" problem. I have implemented "Approach #5", which has a time-complexity of O(n).
The algorithm is simple as well as fast. The first step is to remove one element from each pair of consecutive elements that are equal, and continue to do so until there are no consecutive elements that are equal. For example, [1,2,2,2,3,4,4] would be converted to [1,2,3,4]. The longest wiggle subsequence includes the first and last elements of the resulting array, a, and every element a[i], 0 < i < a.size-1 for which a[i-1] < a[i] > a[i+1] ora[i-1] > a[i] > a[i+1]. In other words, it includes the first and last elements and all peaks and valley bottoms. Those elements are A, D, E, G, H, I in the graph below (taken from the above-referenced article, with permission).
Code
def longest_wiggle(arr)
arr.each_cons(2).
reject { |a,b| a==b }.
map(&:first).
push(arr.last).
each_cons(3).
select { |triple| [triple.min, triple.max].include? triple[1] }.
map { |_,n,_| n }.
unshift(arr.first).
push(arr.last)
end
Example
arr = [33, 53, 12, 64, 50, 41, 45, 21, 97, 35, 47, 92, 39, 0, 93, 55, 40,
46, 69, 42, 6, 95, 51, 68, 72, 9, 32, 84, 34, 64, 6, 2, 26, 98, 3,
43, 30, 60, 3, 68, 82, 9, 97, 19, 27, 98, 99, 4, 30, 96, 37, 9, 78,
43, 64, 4, 65, 30, 84, 90, 87, 64, 18, 50, 60, 1, 40, 32, 48, 50, 76,
100, 57, 29, 63, 53, 46, 57, 93, 98, 42, 80, 82, 9, 41, 55, 69, 84,
82, 79, 30, 79, 18, 97, 67, 23, 52, 38, 74, 15]
a = longest_wiggle(arr)
#=> [33, 53, 12, 64, 41, 45, 21, 97, 35, 92, 0, 93, 40, 69, 6, 95, 51, 72,
# 9, 84, 34, 64, 2, 98, 3, 43, 30, 60, 3, 82, 9, 97, 19, 99, 4, 96, 9,
# 78, 43, 64, 4, 65, 30, 90, 18, 60, 1, 40, 32, 100, 29, 63, 46, 98, 42,
# 82, 9, 84, 30, 79, 18, 97, 23, 52, 38, 74, 15]
a.size
#=> 67
Explanation
The steps are as follows.
arr = [3, 4, 4, 5, 2, 3, 7, 4]
enum1 = arr.each_cons(2)
#=> #<Enumerator: [3, 4, 4, 5, 2, 3, 7, 4]:each_cons(2)>
We can see the elements that will be generated by this enumerator by converting it to an array.
enum1.to_a
#=> [[3, 4], [4, 4], [4, 5], [5, 2], [2, 3], [3, 7], [7, 4]]
Continuing, remove all but one of each group of successive equal elements.
d = enum1.reject { |a,b| a==b }
#=> [[3, 4], [4, 5], [5, 2], [2, 3], [3, 7], [7, 4]]
e = d.map(&:first)
#=> [3, 4, 5, 2, 3, 7]
Add the last element.
f = e.push(arr.last)
#=> [3, 4, 5, 2, 3, 7, 4]
Next, find the peaks and valley bottoms.
enum2 = f.each_cons(3)
#=> #<Enumerator: [3, 4, 5, 2, 3, 7, 4]:each_cons(3)>
enum2.to_a
#=> [[3, 4, 5], [4, 5, 2], [5, 2, 3], [2, 3, 7], [3, 7, 4]]
g = enum2.select { |triple| [triple.min, triple.max].include? triple[1] }
#=> [[4, 5, 2], [5, 2, 3], [3, 7, 4]]
h = g.map { |_,n,_| n }
#=> [5, 2, 7]
Lastly, add the first and last values of arr.
i = h.unshift(arr.first)
#=> [3, 5, 2, 7]
i.push(arr.last)
#=> [3, 5, 2, 7, 4]
I watched the RubyConf 2013 talk on Pry and I have decided I ought to give it a good try.
I am working with some large arrays. It would be easier to work with my code if Pry would display returned arrays the way IRB does. What seems odd is that pry will not add newlines if the number of chars in the displayed array is small but it will add them when the number of chars in the displayed array surpasses some threshold (appears to be 26 chars in my case). Does anybody know how to make Pry stop doing this?
IRB:
main 001(0) > a = [] #=> []
main 002(0) > (1..100).each{|i| a << i} #=> 1..100
main 003(0) > a
=> [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]
Pry:
[1] pry(main)> a = []
=> []
[2] pry(main)> (1..26).each{ a << 1 }
=> 1..26
[3] pry(main)> a
=> [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[4] pry(main)> a << 1
=> [1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1,
1]
Edit your .pryrc file to include
Pry.config.print = proc { |output, value| output.puts "=> #{value.inspect}" }
To get pry to print inline like robertjlooby's answer, but retain the pretty-printing:
# in your .pryrc
Pry.config.print = lambda do |output, value, _pry_|
_pry_.pager.open do |pager|
pager.print _pry_.config.output_prefix
Pry::ColorPrinter.pp(value, pager, 9e99)
end
end
adapted from pry source
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]]
>>>
I'm looking form a programatic way to take an integer sequence and spit out a closed form function. Something like:
Given: 1,3,6,10,15
Return: n(n+1)/2
Samples could be useful; the language is unimportant.
This touches an extremely deep, sophisticated and active area of mathematics. The solution is damn near trivial in some cases (linear recurrences) and damn near impossible in others (think 2, 3, 5, 7, 11, 13, ....) You could start by looking at generating functions for example and looking at Herb Wilf's incredible book (cf. page 1 (2e)) on the subject but that will only get you so far.
But I think your best bet is to give up, query Sloane's comprehensive Encyclopedia of Integer Sequences when you need to know the answer, and instead spend your time reading the opinions of one of the most eccentric personalities in this deep subject.
Anyone who tells you this problem is solvable is selling you snake oil (cf. page 118 of the Wilf book (2e).)
There is no one function in general.
For the sequence you specified, The On-Line Encyclopedia of Integer Sequences finds 133 matches in its database of interesting integer sequences. I've copied the first 5 here.
A000217 Triangular numbers: a(n) = C(n+1,2) = n(n+1)/2 = 0+1+2+...+n.
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431
A130484 Sum {0<=k<=n, k mod 6} (Partial sums of A010875).
0, 1, 3, 6, 10, 15, 15, 16, 18, 21, 25, 30, 30, 31, 33, 36, 40, 45, 45, 46, 48, 51, 55, 60, 60, 61, 63, 66, 70, 75, 75, 76, 78, 81, 85, 90, 90, 91, 93, 96, 100, 105, 105, 106, 108, 111, 115, 120, 120, 121, 123, 126, 130, 135, 135, 136, 138, 141, 145, 150, 150, 151, 153
A130485 Sum {0<=k<=n, k mod 7} (Partial sums of A010876).
0, 1, 3, 6, 10, 15, 21, 21, 22, 24, 27, 31, 36, 42, 42, 43, 45, 48, 52, 57, 63, 63, 64, 66, 69, 73, 78, 84, 84, 85, 87, 90, 94, 99, 105, 105, 106, 108, 111, 115, 120, 126, 126, 127, 129, 132, 136, 141, 147, 147, 148, 150, 153, 157, 162, 168, 168, 169, 171, 174, 178, 183
A104619 Write the natural numbers in base 16 in a triangle with k digits in the k-th row, as shown below. Sequence gives the leading diagonal.
1, 3, 6, 10, 15, 2, 1, 1, 14, 3, 2, 2, 5, 12, 4, 4, 4, 13, 6, 7, 11, 6, 9, 9, 10, 7, 12, 13, 1, 0, 1, 10, 5, 1, 12, 8, 1, 1, 14, 1, 9, 7, 1, 4, 3, 1, 2, 2, 1, 3, 4, 2, 7, 9, 2, 14, 1, 2, 8, 12, 2, 5, 10, 3, 5, 11, 3, 8, 15, 3, 14, 6, 3, 7, 0, 4, 3, 13, 4, 2, 13, 4, 4, 0, 5, 9, 6, 5, 1, 15, 5, 12, 11, 6
A037123 a(n) = a(n-1) + Sum of digits of n.
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 46, 48, 51, 55, 60, 66, 73, 81, 90, 100, 102, 105, 109, 114, 120, 127, 135, 144, 154, 165, 168, 172, 177, 183, 190, 198, 207, 217, 228, 240, 244, 249, 255, 262, 270, 279, 289, 300, 312, 325, 330, 336, 343, 351, 360, 370, 381
If you restrict yourself to polynomial functions, this is easy to code up, and only mildly tedious to solve by hand.
Let , for some unknown
Now solve the equations
…
which simply a system of linear equations.
If your data is guaranteed to be expressible as a polynomial, I think you would be able to use R (or any suite that offers regression fitting of data). If your correlation is exactly 1, then the line is a perfect fit to describe the series.
There's a lot of statistics that goes into regression analysis, and I am not familiar enough with even the basics of calculation to give you much detail.
But, this link to regression analysis in R might be of assistance
The Axiom computer algebra system includes a package for this purpose. You can read its documentation here.
Here's the output for your example sequence in FriCAS (a fork of Axiom):
(3) -> guess([1, 3, 6, 10, 15])
2
n + 3n + 2
(3) [[function= -----------,order= 0]]
2
Type: List(Record(function: Expression(Integer),order: NonNegativeInteger))
I think your problem is ill-posed. Given any finite number of integers in a sequence with
no generating function, the next element can be anything.
You need to assume something about the sequence. Is it geometric? Arithmetic?
If your sequence comes from a polynomial then divided differences will find that polynomial expressed in terms of the Newton basis or binomial basis. See this.
There is no general answers; a simple method can be implemented bu using Pade approximants; in two words, assume your sequence is a sequence of coefficients of the Taylor expansion of an unknown function, then apply an algorithm (similar to the continued-fraction algorithm) in order to "simplify" this Taylor-expansion (more precisely: find a rational function very close to the initial (and truncated) function. The Maxima program can do it: look at "pade" on the page: http://maxima.sourceforge.net/docs/manual/maxima_28.html
Another answer tells about the "guess" package in the FriCAS fork of Axiom (see previous answer by jmbr). If I am not wrong; this package is itself inspired from the Rate program by Christian Krattenthaler; you can find it here: http://www.mat.univie.ac.at/~kratt/rate/rate.html Maybe looking at its source could tell you about other methods.