#=======================================================================
# Author: Isai Damier
# Title: Longest Increasing Subsequence
# Project: geekviewpoint
# Package: algorithms
#
# Statement:
# Given a sequence of numbers, find a longest increasing subsequence.
#
# Time Complexity: O(n^2)
#
# Sample Input: [8,1,2,3,0,5]
# Sample Output: [1,2,3,5]
#
# DEFINITION OF SUBSEQUENCE:
# A sequence is a particular order in which related objects follow
# each other (e.g. DNA, Fibonacci). A sub-sequence is a sequence
# obtained by omitting some of the elements of a larger sequence.
#
# SEQUENCE SUBSEQUENCE OMISSION
# [3,1,2,5,4] [1,2] 3,5,4
# [3,1,2,5,4] [3,1,4] 2,5
#
# SEQUENCE NOT SUBSEQUENCE REASON
# [3,1,2,5,4] [4,2,5] 4 should follow 5
#
# STRATEGY:
# Illustrating by finding a longest increasing subsequence
# of [8,1,2,3,0,5]:
#
# - Start by finding all subsequences of size 1: [8],[1],[2],[3],[0],[5];
# each element is its own increasing subsequence.
#
# - Since we already have the solutions for the size 1 subsequences,
# we can use them in solving for the size two subsequences. For
# instance, we already know that 0 is the smallest element of an
# increasing subsequence of size 1, i.e. the subsequence [0].
# Therefore, all we need to get a subsequence of size 2 is add an
# element greater than 0 to [0]: [0,5]. The other size 2
# subsequences are: [1,2], [1,3], [1,5], [2,3], [2,5], [3,5].
#
# - Now we use the size 2 subsequences to get the size 3 subsequences:
# [1,2,3], [1,2,5], [1,3,5], [2,3,5]
#
# - Then we use the size 3 subsequences to get the size 4 subsequences:
# [1,2,3,5]. Since there are no size 5 solutions, we are done.
#
# SUMMARY:
# Instead of directly solving the big problem, we solved a smaller
# version and then 'copied and pasted' the solution of the subproblem
# to find the solution to the big problem. To make the 'copy and paste'
# part easy, we use a table (i.e. array) to track the subproblems
# and their solutions. This strategy as a whole is called Dynamic
# Programming. The tabling part is known as memoization, which means
# writing memo.
#
# To recognize whether you can use dynamic programming on a problem,
# look for the following two traits: optimal substructures and
# overlapping subproblems.
#
# Optimal Substructures: the ability to 'copy and paste' the solution
# of a subproblem plus an additional trivial amount of work so to
# solve a larger problem. For example, we were able to use [1,2]
# itself an optimal solution to the problem [8,1,2] to get [1,2,3]
# as an optimal solution to the problem [8,1,2,3].
#
# Overlapping Subproblems: Okay. So in our approach the solution grew
# from left to right: [1] to [1,2] to [1,2,3] etc. But in reality
# we could have solved the problem using recursion trees so that
# for example [1,2] could be reached either from [1] or from [2].
# That wouldn't really be a problem except we would be solving for
# [1,2] more than once. Anytime a recursive solution would lead to
# such overlaps, the bet is dynamic programming is the way to go.
#
# [1] [2]
# / | \ / | \
# / | \ / | \
# / | \ / | \
# [1,2] [1,3] [1,5] [1,2] [2,3] [2,5]
#
# NOTE:
# Dynamic Programming = Optimal Substructures + Overlapping Subproblems
# Divide and Conquer = Optimal Substructures - Overlapping Subproblems
# see merge sort: http://www.geekviewpoint.com/python/sorting/mergesort
#
# Alternate coding: Not really much difference here, just another code
# that some readers will find more intuitive:
#
# m = [1]*len(A)
#
# for x in range(len(A)):
# for y in range(x):
# if m[y] >= m[x] and A[y] < A[x]:
# m[x]+=1
#
# max_value = max(m)
#
# result = []
# for i in range(m-1,-1,-1):
# if max == m[i]:
# result.append(A[i])
# max-=1
#
# result.reverse()
# return result
#=======================================================================
def LIS( A ):
m = [0] * len( A ) # m = [1]*len(A) not important here
for x in range( len( A ) - 2, -1, -1 ):
for y in range( len( A ) - 1, x, -1 ):
if A[x] < A[y] and m[x] <= m[y]:
m[x] += 1 # or use m[x] = m[y] + 1
#===================================================================
# Use the following snippet or the one line below to get max_value
# max_value=m[0]
# for i in range(m):
# if max_value < m[i]:
# max_value = m[i]
#===================================================================
max_value = max( m )
result = []
for i in range( len( m ) ):
if max_value == m[i]:
result.append( A[i] )
max_value -= 1
return result
import unittest
from dynamic_programming import LongestIncreasingSubsequence as subsequence
class Test( unittest.TestCase ):
def testLIS( self ):
A = [1, 2, 3, 4, 5, 6, 7, 8, 9]
B = [9, 8, 7, 6, 5, 4, 3, 2, 1]
C = [50, 2, 45, 3, 5, 44, 1, 8, 40, 12, 37, 15]
D = [2, 3, 5, 8, 12, 37]
actual = subsequence.LIS( A )
self.assertEquals( A, actual )
self.assertEquals( 1, len( subsequence.LIS( B ) ) )
actual = subsequence.LIS( C )
self.assertEquals( D, actual )
