Petar Minchev's explanation helped clear things up for me, but it was hard for me to parse what everything was, so I made a Python implementation with overly-descriptive variable names and lots of comments. I did a naive recursive solution, the O(n^2) solution, and the O(n log n) solution.
I hope it helps clear up the algorithms!
def recursive_solution(remaining_sequence, bigger_than=None):
"""Finds the longest increasing subsequence of remaining_sequence that is
bigger than bigger_than and returns it. This solution is O(2^n)."""
# Base case: nothing is remaining.
if len(remaining_sequence) == 0:
return remaining_sequence
# Recursive case 1: exclude the current element and process the remaining.
best_sequence = recursive_solution(remaining_sequence[1:], bigger_than)
# Recursive case 2: include the current element if it's big enough.
first = remaining_sequence[0]
if (first > bigger_than) or (bigger_than is None):
sequence_with = [first] + recursive_solution(remaining_sequence[1:], first)
# Choose whichever of case 1 and case 2 were longer.
if len(sequence_with) >= len(best_sequence):
best_sequence = sequence_with
return best_sequence
def dynamic_programming_solution(sequence):
"""Finds the longest increasing subsequence in sequence using dynamic
programming. This solution is O(n^2)."""
longest_subsequence_ending_with = []
backreference_for_subsequence_ending_with = []
current_best_end = 0
for curr_elem in range(len(sequence)):
# It's always possible to have a subsequence of length 1.
longest_subsequence_ending_with.append(1)
# If a subsequence is length 1, it doesn't have a backreference.
backreference_for_subsequence_ending_with.append(None)
for prev_elem in range(curr_elem):
subsequence_length_through_prev = (longest_subsequence_ending_with[prev_elem] + 1)
# If the prev_elem is smaller than the current elem (so it's increasing)
# And if the longest subsequence from prev_elem would yield a better
# subsequence for curr_elem.
if ((sequence[prev_elem] < sequence[curr_elem]) and
(subsequence_length_through_prev >
longest_subsequence_ending_with[curr_elem])):
# Set the candidate best subsequence at curr_elem to go through prev.
longest_subsequence_ending_with[curr_elem] = (subsequence_length_through_prev)
backreference_for_subsequence_ending_with[curr_elem] = prev_elem
# If the new end is the best, update the best.
if (longest_subsequence_ending_with[curr_elem] >
longest_subsequence_ending_with[current_best_end]):
current_best_end = curr_elem
# Output the overall best by following the backreferences.
best_subsequence = []
current_backreference = current_best_end
while current_backreference is not None:
best_subsequence.append(sequence[current_backreference])
current_backreference = (backreference_for_subsequence_ending_with[current_backreference])
best_subsequence.reverse()
return best_subsequence
def find_smallest_elem_as_big_as(sequence, subsequence, elem):
"""Returns the index of the smallest element in subsequence as big as
sequence[elem]. sequence[elem] must not be larger than every element in
subsequence. The elements in subsequence are indices in sequence. Uses
binary search."""
low = 0
high = len(subsequence) - 1
while high > low:
mid = (high + low) / 2
# If the current element is not as big as elem, throw out the low half of
# sequence.
if sequence[subsequence[mid]] < sequence[elem]:
low = mid + 1
# If the current element is as big as elem, throw out everything bigger, but
# keep the current element.
else:
high = mid
return high
def optimized_dynamic_programming_solution(sequence):
"""Finds the longest increasing subsequence in sequence using dynamic
programming and binary search (per
http://en.wikipedia.org/wiki/Longest_increasing_subsequence). This solution
is O(n log n)."""
# Both of these lists hold the indices of elements in sequence and not the
# elements themselves.
# This list will always be sorted.
smallest_end_to_subsequence_of_length = []
# This array goes along with sequence (not
# smallest_end_to_subsequence_of_length). Following the corresponding element
# in this array repeatedly will generate the desired subsequence.
parent = [None for _ in sequence]
for elem in range(len(sequence)):
# We're iterating through sequence in order, so if elem is bigger than the
# end of longest current subsequence, we have a new longest increasing
# subsequence.
if (len(smallest_end_to_subsequence_of_length) == 0 or
sequence[elem] > sequence[smallest_end_to_subsequence_of_length[-1]]):
# If we are adding the first element, it has no parent. Otherwise, we
# need to update the parent to be the previous biggest element.
if len(smallest_end_to_subsequence_of_length) > 0:
parent[elem] = smallest_end_to_subsequence_of_length[-1]
smallest_end_to_subsequence_of_length.append(elem)
else:
# If we can't make a longer subsequence, we might be able to make a
# subsequence of equal size to one of our earlier subsequences with a
# smaller ending number (which makes it easier to find a later number that
# is increasing).
# Thus, we look for the smallest element in
# smallest_end_to_subsequence_of_length that is at least as big as elem
# and replace it with elem.
# This preserves correctness because if there is a subsequence of length n
# that ends with a number smaller than elem, we could add elem on to the
# end of that subsequence to get a subsequence of length n+1.
location_to_replace = find_smallest_elem_as_big_as(sequence, smallest_end_to_subsequence_of_length, elem)
smallest_end_to_subsequence_of_length[location_to_replace] = elem
# If we're replacing the first element, we don't need to update its parent
# because a subsequence of length 1 has no parent. Otherwise, its parent
# is the subsequence one shorter, which we just added onto.
if location_to_replace != 0:
parent[elem] = (smallest_end_to_subsequence_of_length[location_to_replace - 1])
# Generate the longest increasing subsequence by backtracking through parent.
curr_parent = smallest_end_to_subsequence_of_length[-1]
longest_increasing_subsequence = []
while curr_parent is not None:
longest_increasing_subsequence.append(sequence[curr_parent])
curr_parent = parent[curr_parent]
longest_increasing_subsequence.reverse()
return longest_increasing_subsequence