Computer science problem
In computer science, the longest increasing subsequence problem aims to find a subsequence of a given sequence in which the subsequence's elements are sorted in an ascending order and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous or unique. The longest increasing subsequences are studied in the context of various disciplines related to mathematics, including algorithmics, random matrix theory, representation theory, and physics. The longest increasing subsequence problem is solvable in time formula_1 where formula_2 denotes the length of the input sequence.
Example.
In the first 16 terms of the binary Van der Corput sequence
0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15
one of the longest increasing subsequences is
0, 2, 6, 9, 11, 15.
This subsequence has length six; the input sequence has no seven-member increasing subsequences. The longest increasing subsequence in this example is not the only solution: for instance,
0, 4, 6, 9, 11, 15
0, 2, 6, 9, 13, 15
0, 4, 6, 9, 13, 15
are other increasing subsequences of equal length in the same input sequence.
Relations to other algorithmic problems.
The longest increasing subsequence problem is closely related to the longest common subsequence problem, which has a quadratic time dynamic programming solution: the longest increasing subsequence of a sequence formula_3 is the longest common subsequence of formula_3 and formula_5 where formula_6 is the result of sorting formula_7 However, for the special case in which the input is a permutation of the integers formula_8 this approach can be made much more efficient, leading to time bounds of the form formula_9
The largest clique in a permutation graph corresponds to the longest decreasing subsequence of the permutation that defines the graph (assuming the original non-permuted sequence is sorted from lowest value to highest). Similarly, the maximum independent set in a permutation graph corresponds to the longest non-decreasing subsequence. Therefore, longest increasing subsequence algorithms can be used to solve the clique problem efficiently in permutation graphs.
In the Robinson–Schensted correspondence between permutations and Young tableaux, the length of the first row of the tableau corresponding to a permutation equals the length of the longest increasing subsequence of the permutation, and the length of the first column equals the length of the longest decreasing subsequence.
Efficient algorithms.
The algorithm outlined below solves the longest increasing subsequence problem efficiently with arrays and binary searching. 
It processes the sequence elements in order, maintaining the longest increasing subsequence found so far. Denote the sequence values as formula_10 etc. Then, after processing formula_11 the algorithm will have stored an integer formula_12 and values in two arrays:
Because the algorithm below uses zero-based numbering, for clarity formula_43 is padded with formula_66 which goes unused so that formula_14 corresponds to a subsequence of length formula_68 A real implementation can skip formula_52 and adjust the indices accordingly.
Note that, at any point in the algorithm, the sequence
formula_70
is increasing. For, if there is an increasing subsequence of length formula_71 ending at formula_72 then there is also a subsequence of length formula_73 ending at a smaller value: namely the one ending at formula_74 Thus, we may do binary searches in this sequence in logarithmic time.
The algorithm, then, proceeds as follows:
 P = array of length N
 M = array of length N + 1
 M[0] = -1 // undefined so can be set to any value
 L = 0
 for i in range 0 to N-1: //N-1 included
 // Binary search for the smallest positive l ≤ L
 // such that X[M[l]] ;= X[i]
 lo = 1
 hi = L + 1
 while lo ; hi:
 mid = lo + floor((hi-lo)/2) // lo ;= mid ; hi
 if X[M[mid]] ;= X[i]
 hi = mid
 else: // if X[M[mid]] ; X[i]
 lo = mid + 1
 // After searching, lo == hi is 1 greater than the
 // length of the longest prefix of X[i]
 newL = lo
 // The predecessor of X[i] is the last index of 
 // the subsequence of length newL-1
 P[i] = M[newL-1]
 M[newL] = i
 if newL ; L:
 // If we found a subsequence longer than any we've
 // found yet, update L
 L = newL
 // Reconstruct the longest increasing subsequence
 // It consists of the values of X at the L indices:
 // ..., P[P[M[L]]], P[M[L]], M[L]
 S = array of length L
 k = M[L]
 for j in range L-1 to 0: //0 included
 S[j] = X[k]
 k = P[k]
 return S
Because the algorithm performs a single binary search per sequence element, its total time can be expressed using Big O notation as formula_75 discusses a variant of this algorithm, which he credits to Donald Knuth; in the variant that he studies, the algorithm tests whether each value formula_76 can be used to extend the current longest increasing sequence, in constant time, prior to doing the binary search. With this modification, the algorithm uses at most formula_77 comparisons in the worst case, which is optimal for a comparison-based algorithm up to the constant factor in the formula_78 term.
Example run
Length bounds.
According to the Erdős–Szekeres theorem, any sequence of formula_79 distinct integers has an increasing or a decreasing subsequence of length formula_80 For inputs in which each permutation of the input is equally likely, the expected length of the longest increasing subsequence is approximately formula_81 
In the limit as formula_2 approaches infinity, the Baik-Deift-Johansson theorem says, that the length of the longest increasing subsequence of a randomly permuted sequence of formula_2 items has a distribution approaching the Tracy–Widom distribution, the distribution of the largest eigenvalue of a random matrix in the Gaussian unitary ensemble.
Online algorithms.
The longest increasing subsequence has also been studied in the setting of online algorithms, in which the elements of a sequence of independent random variables with continuous distribution formula_84 – or alternatively the elements of a random permutation – are presented one at a time to an algorithm that must decide whether to include or exclude each element, without knowledge of the later elements. In this variant of the problem, which allows for interesting applications in several contexts, it is possible to devise an optimal selection procedure that, given a random sample of size formula_2 as input, will generate an increasing sequence with maximal expected length of size approximately formula_86
The length of the increasing subsequence selected by this optimal procedure has variance approximately equal to formula_87 and its limiting distribution is asymptotically normal after the usual centering and scaling.
The same asymptotic results hold with more precise bounds for the corresponding problem in the setting of a Poisson arrival process.
A further refinement in the Poisson process setting is given through the proof of a central limit theorem for the optimal selection process
which holds, with a suitable normalization, in a more complete sense than one would expect. The proof yields not only the "correct" functional limit theorem
but also the (singular) covariance matrix of the three-dimensional process summarizing all interacting processes.
References.
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