In computer science, patience sorting is a sorting algorithm inspired by, and named after, the card game patience. A variant of the algorithm efficiently computes the length of a longest increasing subsequence in a given array.
Overview The algorithm's name derives from a simplified variant of the patience card game. The game begins with a shuffled deck of cards. 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 subsequent card is placed on the leftmost existing pile whose top card has a value greater than or equal to the new card's value, 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. This card game is turned into a two-phase sorting algorithm, as follows. Given an array of n elements from some totally ordered domain, consider this array as a collection of cards and simulate the patience sorting game. When the game is over, recover the sorted sequence by repeatedly picking off the minimum visible card; in other words, perform a k-way merge of the p piles, each of which is internally sorted.
Pseudocode Below is an iterative implementation of Patience Sort, this implementation runs in O ( n 2 ) {\displaystyle O(n^{2})} .
function PatienceSorting(array arr) is n ← length(arr) piles ← empty list of lists
for i = 0 to n - 1 do if length(piles) == 0 then tmp ← new list tmp.append(arr[i]) piles.append(tmp) else placed ← false for j ← 0 to length(piles) - 1 do if arr[i] < last element of piles[j] then piles[j].append(arr[i]) placed ← true break if placed == false then tmp ← new list tmp.append(arr[i]) piles.append(tmp)
return MergePiles(piles)
function MergePiles(list of lists piles) is result ← empty list
while true do minValue ← ∞ minIndex ← -1
for i ← 0 to length(piles) - 1 do if length(piles[i]) > 0 and piles[i][length(piles[i]) - 1] < minValue then minValue ← piles[i][length(piles[i]) - 1] minIndex ← i if minIndex = -1 then break
result.append(minValue) piles[minIndex].remove(piles[minIndex][length(piles[minIndex]) - 1] if length(piles[minIndex]) == 0 then piles.remove(piles[minIndex])
return result
Analysis The first phase of patience sort, the card game simulation, can be implemented to take O(n log n) comparisons in the worst case for an n-element input array: there will be at most n piles, and by construction, the top cards of the piles form an increasing sequence from left to right, so the desired pile can be found by binary search. The second phase, the merging of piles, can be done in O ( n log n ) {\displaystyle O(n\log n)} time as well using a priority queue. When the input data contain natural "runs", i.e., non-decreasing subarrays, then performance can be strictly better. In fact, when the input array is already sorted, all values form a single pile and both phases run in O(n) time. The average-case complexity is still O(n log n): any uniformly random sequence of values will produce an expected number of O ( n ) {\displaystyle O({\sqrt {n}})} piles, which take O ( n log n ) = O ( n log n ) {\displaystyle O(n\log {\sqrt {n}})=O(n\log n)} time to produce and merge. An evaluation of the practical performance of patience sort is given by Chandramouli and Goldstein, who show that a naive version is about ten to twenty times slower than a state-of-the-art quicksort on their benchmark problem. They attribute this to the relatively small amount of research put into patience sort, and develop several optimizations that bring its performance to within a factor of two of that of quicksort. If values of cards are in the range 1, . . . , n, there is an efficient implementation with O ( n log n ) {\displaystyle O(n\log n)} worst-case running time for putting the cards into piles, relying on a Van Emde Boas tree.
Relations to other problems Patience sorting is closely related to a card game called Floyd's game. This game is very similar to the game sketched earlier:
The first card dealt forms a new pile consisting of the single card. Each subsequent card is placed on some existing pile whose top card has a value no less than the new card's value, 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. The difference with the patience sorting algorithm is that there is no requirement to place a new card on the leftmost pile where it is allowed. Patience sorting constitutes a greedy strategy for playing this game. Aldous and Diaconis suggest defining 9 or fewer piles as a winning outcome for n = 52, which happens with approximately 5% probability.
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