- Patience sorting
**Patience sorting**is asorting algorithm , based on asolitaire card game , that has the property of being able to efficiently compute the length of thelongest increasing subsequence in a givenarray .**The card game**The game begins with a

shuffle d deck of cards, labeled $1,\; 2,\; ldots,\; n$.The cards are dealt one by one into a sequence of piles on the table, according to the following rules.

# Initially, there are no piles. The first card dealt forms a new pile consisting of the single card.

# Each new card may be placed either on an existing pile that has a higher value card on the top (if any such pile exists), thus increasing the number of cards in that pile, or to the right of all of the existing piles, thus forming a new pile.

# When there are no more cards remaining to deal, the game ends.The object of the game is to finish with as few piles as possible. D. Aldous and P. Diaconis

David Aldous andPersi Diaconis . [*http://www-stat.stanford.edu/~cgates/PERSI/year.html#99 Longest increasing subsequences: from patience sorting to the Baik-Deift-Johansson theorem.*] "Bull. (new series) of the Amer. Math. Society, Volume 36", number 4, pages 413–432, p.414] suggest defining 9 or fewer piles as a winning outcome for $n\; =\; 52$, which has approximately 5% chance to happen.**Algorithm for sorting**Given an $n$-element array with an ordering relation as an input for the sorting, consider it as a collection of cards, with the (unknown in the beginning) statistical ordering of each element serving as its index. Note that the game never uses the actual value of the card, except for comparison between two cards, and the relative ordering of any two array elements is known.

Now simulate the patience sorting game, played with the "greedy strategy", i.e., placing each new card on the leftmost pile that is legally possible to use. At each stage of the game, under this strategy, the labels on the top cards of the piles are increasing from left to right. To recover the sorted sequence, collect the minimum card from the top row each time.

**Complexity**If values of cards are in the range $1,\; ldots,\; n$, there is an efficient implementation with $O(n\; cdot\; log\; log\; n)$

worst-case running time for putting the cards into piles (which makes the worst case for a complete sort $O(n\; log\; n)$) and $O(n)$ space spent for supporting structures, relying on avan Emde Boas tree . A description is given in the work by S. Bespamyatnikh and M. Segal.Sergei Bespamyatnikh and Michael Segal. [*http://www.pims.math.ca/publications/preprints/bespamyatnikh-segal_6-12-99.ps.gz Enumerating Longest Increasing Subsequences and Patience Sorting.*] "Pacific Inst. for the Math. Sci. Preprints", PIMS-99-3., pp.7–8]When no assumption is made about values, the greedy strategy can be implemented in $O(n\; log\; n)$ comparisons in worst case. In fact, one can implement it with an array of stacks ordered by values of top cards and, for inserting a new card, use a binary search, which is $O(log\; p)$ comparisons in worst case, where $p$ is the number of piles. But there is no known efficient way (aka $O(n\; log\; n)$ worst case) of using the structure obtained by the greedy strategy to complete the sorting. The card with the least value is at top of leftmost pile but when this card is removed this is no longer the case in the new configuration, and some work has to be done. Finding the next card by searching it among all tops of piles, as in the wikibooks implementation, gives a $O(n^2)$ worst case. Inserting the first pile in the remaining list of piles, with respect to the top cards ordering, can cost $O(n)$ each time and so gives also a $O(n^2)$ worst case (take $k+1,\; ldots,\; 2k,\; 1,\; ldots,\; k$ as starting list of values).

**Algorithm for finding the longest increasing subsequence**First, execute the sorting algorithm as described above. The resulting number of piles will equal the length of the longest increasing subsequence. Whenever a card is placed on top of a pile, put a back-

pointer to the top card in the previous pile (that, by assumption, has a lower value than the new card has). In the end, follow the back-pointers from the top card in the last pile to recover a decreasing subsequence of the longest length; its reverse is an answer to the longest increasing subsequence algorithm.**History**According to D. Aldous and P. Diaconis, patience sorting was first recognized as an algorithm to compute the longest increasing subsequence length by Hammersley, [

*J.M. Hammersley. A few seedlings of research. In "Proc. Sixth Berkeley Symp. Math. Statist. and Probability, Volume 1", pages 345–394. University of California Press, 1972. MR*] and by A.S.C. Ross and independently**53**:9457, p.362Bob Floyd as a sorting algorithm. Initial analysis was done by Mallows. [*C.L. Mallows. Patience sorting. "Bull. Inst. Math. Appl.", 9:216–224, 1973.*]**ee also****External links***The Bazaar version control system — uses the patience sorting algorithm for merge resolution [

*http://revctrl.org/PreciseCodevilleMerge*] .**References**

*Wikimedia Foundation.
2010.*