def partition(A, l, r):
p = A[l]
stack = A[l]
A[l] = A[r]
A[r] = stack
s = l
for i in range(l, r):
if A[i] <= p:
stack2 = A[i]
A[i] = A[s]
A[s] = stack2
s += 1
stack3 = A[s]
A[s] = A[r]
A[r] = stack3
return s
def quicksort(A, l, r):
if l < r:
q = partition(A, l, r)
quicksort(A, l, q - 1)
quicksort(A, q + 1, r)
return A
I've written "maybe" quicksort algorithm, as I've noticed here the time complexity of partition was O(n) because of the for loop, Also the complexity in quicksort seems to be at least O(n). The question: how is it possible for the entire code to have total time complexity of O(nlogn).
You partition by 2 each level till you get individual elements. Partitioning and comparison what makes the time complexity. You make n comparisons in each level and you will be partioning log2(n) times.
In the worst case, your array is already sorted and you will be partitioning n times and still make n comparisons in each level.
Your sorting function isn't O(nlogn). In the worst case, you are making O(n) recursive calls.
As a simple test:
def test(n):
nums = list(reversed(range(n)))
return sum(quicksort(nums,0,n-1))
Then, for example, test(1100) triggers:
RecursionError: maximum recursion depth exceeded in comparison
Which wouldn't happen if you were calling partition just log(n) times.
On the other hand,
import random
def test2(n):
nums = list(range(n))
random.shuffle(nums)
return sum(quicksort(nums,0,n-1))
works well even for calls like test2(100000), so you do have average case O(nlogn) complexity. This is easy to confirm numerically but difficult to prove. See https://en.wikipedia.org/wiki/Quicksort for a proof.
Related
I am wondering why this brute force approach to a Maximum Sum Subarray of Size K problem is of time complexity nk instead of (n-k)k. Given that we are subtracting K elements from the outer most loop wouldn't the latter be more appropriate? The text solution mentions nk and confuses me slightly.
I have included the short code snippet below!
Thank you
def max_sub_array_of_size_k(k, arr):
max_sum = 0
window_sum = 0
for i in range(len(arr) - k + 1):
window_sum = 0
for j in range(i, i+k):
window_sum += arr[j]
max_sum = max(max_sum, window_sum)
return max_sum
I haven't actually tried to fix this, I just want to understand.
In the calculation of time complexity, O(n)=O(n-1)=O(n-k) ，both represent the complexity of linear growth, thus O(n-k)✖️O(k) = O(n*k). Of course, this question can be optimized to O(n) time complexity by using the sum of prefixes.
def max_sub_array_of_size_k(k, arr):
s = [0]
for i in range(len(arr)):
# sum[i] = sum of arr[0] + ... + arr[i]
s.append(s[-1] + arr[i])
max_sum = float("-inf")
for i in range(1, len(s) + 1 - k):
max_sum = max(max_sum, s[i + k - 1] - s[i - 1])
return max_sum
It's O(k(n-k+1)), actually, and you are right that that is a tighter bound than O(nk).
Big-O gives an upper bound, though, so O(k(n-k+1)) is a subset of O(nk), and saying that the complexity is in O(nk) is also correct.
It's just a question of whether or not the person making the statement about the algorithm's complexity cares about, or cares to communicate, the fact that it can be faster when k is close to n.
I'm trying to find the Master Theorem of this Merge Sort Code, but first I need to find its recurrence relation, but I'm struggling to do and understand both. I already saw some similar questions here, but couldn't understand the explanations, like, first I need to find how many operations the code has? Could someone help me with that?
def mergeSort(alist):
print("Splitting ",alist)
if len(alist)>1:
mid = len(alist)//2
lefthalf = alist[:mid]
righthalf = alist[mid:]
mergeSort(lefthalf)
mergeSort(righthalf)
i=0
j=0
k=0
while i < len(lefthalf) and j < len(righthalf):
if lefthalf[i] < righthalf[j]:
alist[k]=lefthalf[i]
i=i+1
else:
alist[k]=righthalf[j]
j=j+1
k=k+1
while i < len(lefthalf):
alist[k]=lefthalf[i]
i=i+1
k=k+1
while j < len(righthalf):
alist[k]=righthalf[j]
j=j+1
k=k+1
print("Merging ",alist)
alist = [54,26,93,17,77,31,44,55,20]
mergeSort(alist)
print(alist)
To determine the run-time of a divide-and-conquer algorithm using the Master Theorem, you need to express the algorithm's run-time as a recursive function of input size, in the form:
T(n) = aT(n/b) + f(n)
T(n) is how we're expressing the total runtime of the algorithm on an input size n.
a stands for the number of recursive calls the algorithm makes.
T(n/b) represents the recursive calls: The n/b signifies that the input size to the recursive calls is some particular fraction of original input size (the divide part of divide-and-conquer).
f(n) represents the amount of work you need to do to in the main body of the algorithm, generally just to combine solutions from recursive calls into an overall solution (you could say this is the conquer part).
Here's a slightly re-factored definition of mergeSort:
def mergeSort(arr):
if len(arr) <= 1: return # array size 1 or 0 is already sorted
# split the array in half
mid = len(arr)//2
L = arr[:mid]
R = arr[mid:]
mergeSort(L) # sort left half
mergeSort(R) # sort right half
merge(L, R, arr) # merge sorted halves
We need to determine, a, n/b and f(n)
Because each call of mergeSort makes two recursive calls: mergeSort(L) and mergeSort(R), a=2:
T(n) = 2T(n/b) + f(n)
n/b represents the fraction of the current input that recursive calls are made with. Because we are finding the midpoint and splitting the input in half, passing one half the current array to each recursive call, n/b = n/2 and b=2. (if each recursive call instead got 1/4 of the original array b would be 4)
T(n) = 2T(n/2) + f(n)
f(n) represents all the work the algorithm does besides making recursive calls. Every time we call mergeSort, we calculate the midpoint in O(1) time.
We also split the array into L and R, and technically creating these two sub-array copies is O(n). Then, presuming mergeSort(L), sorted the left half of the array, and mergeSort(R) sorted the right half, we still have to merge the sorted sub-arrays together to sort the entire array with the merge function. Together, this makes f(n) = O(1) + O(n) + complexity of merge. Now let's take a look at merge:
def merge(L, R, arr):
i = j = k = 0 # 3 assignments
while i < len(L) and j < len(R): # 2 comparisons
if L[i] < R[j]: # 1 comparison, 2 array idx
arr[k] = L[i] # 1 assignment, 2 array idx
i += 1 # 1 assignment
else:
arr[k] = R[j] # 1 assignment, 2 array idx
j += 1 # 1 assignment
k += 1 # 1 assignment
while i < len(L): # 1 comparison
arr[k] = L[i] # 1 assignment, 2 array idx
i += 1 # 1 assignment
k += 1 # 1 assignment
while j < len(R): # 1 comparison
arr[k] = R[j] # 1 assignment, 2 array idx
j += 1 # 1 assignment
k += 1 # 1 assignment
This function has more going on, but we just need to get it's overall complexity class to be able to apply the Master Theorem accurately. We can count every single operation, that is, every comparison, array index, and assignment, or just reason about it more generally. Generally speaking, you can say that across the three while loops we are going to iterate through every member of L and R and assign them in order to the output array, arr, doing a constant amount of work for each element. Noting that we are processing every element of L and R (n total elements) and doing a constant amount of work for each element would be enough to say that merge is in O(n).
But, you can get more particular with counting operations if you want. For the first while loop, every iteration we make 3 comparisons, 5 array indexes, and 2 assignments (constant numbers), and the loop runs until one of L and R is fully processed. Then, one of the next two while loops may run to process any leftover elements from the other array, performing 1 comparison, 2 array indexes, and 3 variable assignments for each of those elements (constant work). Therefore, because each of the n total elements of L and R cause at most a constant number of operations to be performed across the while loops (either 10 or 6, by my count, so at most 10), and the i=j=k=0 statement is only 3 constant assignments, merge is in O(3 + 10*n) = O(n). Returning to the overall problem, this means:
f(n) = O(1) + O(n) + complexity of merge
= O(1) + O(n) + O(n)
= O(2n + 1)
= O(n)
T(n) = 2T(n/2) + n
One final step before we apply the Master Theorem: we want f(n) written as n^c. For f(n) = n = n^1, c=1. (Note: things change very slightly if f(n) = n^c*log^k(n) rather than simply n^c, but we don't need to worry about that here)
You can now apply the Master Theorem, which in its most basic form says to compare a (how quickly the number of recursive calls grows) to b^c (how quickly the amount of work per recursive call shrinks). There are 3 possible cases, the logic of which I try to explain, but you can ignore the parenthetical explanations if they aren't helpful:
a > b^c, T(n) = O(n^log_b(a)). (The total number of recursive calls is growing faster than the work per call is shrinking, so the total work is determined by the number of calls at the bottom level of the recursion tree. The number of calls starts at 1 and is multiplied by a log_b(n) times because log_b(n) is the depth of the recursion tree. Therefore, total work = a^log_b(n) = n^log_b(a))
a = b^c, T(n) = O(f(n)*log(n)). (The growth in number of calls is balanced by the decrease in work per call. The work at each level of the recursion tree is therefore constant, so total work is just f(n)*(depth of tree) = f(n)*log_b(n) = O(f(n)*log(n))
a < b^c, T(n) = O(f(n)). (The work per call shrinks faster than the number of calls increases. Total work is therefore dominated by the work at the top level of the recursion tree, which is just f(n))
For the case of mergeSort, we've seen that a = 2, b = 2, and c = 1. As a = b^c, we apply the 2nd case:
T(n) = O(f(n)*log(n)) = O(n*log(n))
And you're done. This may seem like a lot work, but coming up with a recurrence for T(n) gets easier the more you do it, and once you have a recurrence it's very quick to check which case it falls under, making the Master Theorem quite a useful tool for solving more complicated divide/conquer recurrences.
lately im comparing different types of sort algorithms in python. I noticed that my quicksort isnt handling well inputs where values are repeated.
def compare_asc(a, b):
return a <= b
def partition(a, p, r, compare):
pivot = a[r]
i = p-1
for j in range(p, r):
if compare(a[j], pivot):
i += 1
a[i], a[j] = a[j], a[i]
a[i+1], a[r] = a[r], a[i+1]
return i + 1
def part_quick_sort(a, p, r, compare):
if p < r:
q = partition(a, p, r, compare)
part_quick_sort(a, p, q-1, compare)
part_quick_sort(a, q+1, r, compare)
def quick_sort(a, compare):
part_quick_sort(a, 0, len(a)-1, compare)
return a
Then I test this
import numpy as np
from timeit import default_timer as timer
import sys
test_list1 = np.random.randint(-10000, 10000, size=10000).tolist()
start = timer()
test_list1 = quick_sort(test_list1, compare_asc)
elapsed = timer() - start
print(elapsed)
test_list2 = np.random.randint(0, 2, size=10000).tolist()
start = timer()
test_list2 = quick_sort(test_list2, compare_asc)
elapsed = timer() - start
print(elapsed)
In this example i get RecursionError: maximum recursion depth exceeded in comparison, so i added sys.setrecursionlimit(1000000) and after that i get this output:
0.030029324000224733
5.489867554000284
Can anyone explain why it throws this recursion depth error only during sorting 2nd list ? And why there it is such big time difference ?
Here's a hint: pass a list where all the elements are the same, and watch what it does line by line. It will take time quadratic in the number of elements, and recurse to a level approximately equal to the number of elements.
The usual quicksort partition implementations proceed from both ends, so that in the all-equal case the list slice is approximately cut in half. You can get decent performance in this case for your "only look left-to-right" approach, but the clearest way to do so is to partition into three regions: "less than", "equal", and "greater than".
That can be done in a single left-to-right pass, and is usually called the "Dutch national flag problem". As the text on the linked page says,
The solution to this problem is of interest for designing sorting algorithms; in particular, variants of the quicksort algorithm that must be robust to repeated elements need a three-way partitioning function ...
CODE
For concreteness, here's a complete implementation doing one-pass "left to right" single-pivot 3-way partitioning. It also incorporates other well-known changes needed to make a quicksort robust for production use. Note:
You cannot create a pure quicksort that avoids worst-case quadratic time. The best you can do is average-case O(N*log(N)) time, and (as below, for one way) make worst-case O(N**2) time unlikely.
You can (as below) guarantee worst-case logarithmic recursion depth.
In this approach, a list of all-equal elements is not a bad case, but a very good case: the partitioning routine is called just once total.
The code:
from random import randrange
def partition(a, lo, hi, pivot):
i = L = lo
R = hi
# invariants:
# a[lo:L] < pivot
# a[L:i] == pivot
# a[i:R] unknown
# a[R:hi] > pivot
while i < R:
elt = a[i]
if elt < pivot:
a[L], a[i] = elt, a[L]
L += 1
i += 1
elif elt > pivot:
R -= 1
a[R], a[i] = elt, a[R]
else:
i += 1
return L, R
def qsort(a, lo=0, hi=None):
if hi is None:
hi = len(a)
while True: # sort a[lo:hi] in place
if hi - lo <= 1:
return
# select pivot ar random; else it's easy to construct
# inputs that systematically require quadratic time
L, R = partition(a, lo, hi, a[randrange(lo, hi)])
# must recur on only the shorter chunk to guarantee
# worst-case recursion depth is logarithmic in hi-lo
if L - lo <= hi - R:
qsort(a, lo, L)
# loop to do qsort(a, R, hi)
lo = R
else:
qsort(a, R, hi)
# loop to do qsort(a, lo, L)
hi = L
I've got a function that has two recursive calls and I'm trying to convert it to an iterative function. I've got it figured out where I can do it with one call fairly easily, but I can't figure out how to incorporate the other call.
the function:
def specialMultiplication(n):
if n < 2:
return 1
return n * specialMultiplication(n-1) * specialMultiplication(n-2)
If I just had one of them, it would be really easily:
def specialMult(n, mult = 1):
while n > 1:
(n, mult) = (n-1, n * mult) # Or n-2 for the second one
return mult
I just can't figure out how to add the second call in to get the right answer overall. Thanks!
If you don't mind changing the structure of your algorithm a bit more, you can calculate the values in a bottom-up fashion, starting with the smallest values.
def specialMultiplication(max_n):
a = b = 1
for n in range(1, max_n+1):
a, b = b, a*b*n
return b
Convert the recursion to an iterative function using an auxiliary "todo list":
def specialMultiplication(n):
to_process = []
result = 1
if n >= 2:
to_process.append(n)
while to_process: # while list is not empty
n = to_process.pop()
result *= n
if n >= 3:
to_process.append(n-1)
if n >= 4:
to_process.append(n-2)
return result
create a work list (to_process)
if n >= 2, add n to the list
while to_process is not empty, pop item from list, multiply to result
if n-1 < 2, don't perform "left" operation (don't append to work list)
if n-2 < 2, don't perform "right" operation (don't append to work list)
This method has the advantage of consuming less stack. I've checked the results against recursive version for values from 1 to 25 and they were equal.
Note that it's still slow, since complexity is O(2^n) so it's beginning to be really slow from n=30 (time doubles when n increases by 1). n=28 is computed in 12 seconds on my laptop.
I've successfully used this method to fix a stack overflow problem when performing a flood fill algorithm: Fatal Python error: Cannot recover from stack overflow. During Flood Fill but here Blcknght answer is more adapted because it rethinks the way of computing it from the start.
The OP's function has the same recursive structure as the Fibonacci and Lucas functions, just with different values for f0, f1, and g:
f(0) = f0
f(1) = f1
f(n) = g(f(n-2), f(n-1), n)
This is an example of a recurrence relation. Here is an iterative version of the general solution that calculates f(n) in n steps. It corresponds to a bottom-up tail recursion.
def f(n):
if not isinstance(n, int): # Can be loosened a bit
raise TypeError('Input must be an int') # Can be more informative
if n < 0:
raise ValueError('Input must be non-negative')
if n == 0:
return f0
i, fi_1, fi = 1, f0, f1 # invariant: fi_1, fi = f(i-1), f(i)
while i < n:
i += 1
fi_1, fi = fi, g(fi_1, fi, n) # restore invariant for new i
return fi
Blckknight's answer is a simplified version of this
Is there a fast algorithm to compute the i-th element (0 <= i < n) of the k-th permutation (0 <= k < n!) of the sequence 0..n-1?
Any order of the permutations may be chosen, it does not have to be lexicographical. There are algorithms that construct the k-th permutation in O(n) (see below). But here the complete permutation is not needed, just its i-th element. Are there algorithms that can do better than O(n)?
Is there an algorithm that has a space complexity less than O(n)?
There are algorithms that construct the k-th permutation by working on an array of size n (see below), but the space requirements might be undesirable for large n. Is there an algorithm that needs less space, especially when only the i-th element is needed?
Algorithm that constructs the k-th permutation of the sequence 0..n-1 with a time and space complexity of O(n):
def kth_permutation(n, k):
p = range(n)
while n > 0:
p[n - 1], p[k % n] = p[k % n], p[n - 1]
k /= n
n -= 1
return p
Source: http://webhome.cs.uvic.ca/~ruskey/Publications/RankPerm/MyrvoldRuskey.pdf
What jkff said. You could modify an algorithm like the one you posted to just return the i-th element of the k-th permutation, but you won't save much time (or space), and you certainly won't reduce the Big-O complexity of the basic algorithm.
The unordered permutation code that you posted isn't really amenable to modification because it has to loop over all the elements performing its swaps, and it's painful to determine if it's possible to break out of the loop early.
However, there's a similar algorithm which produces ordered permutations, and it is possible to break out of that one early, but you still need to perform i inner loops to get the i-th element of the k-th permutation.
I've implemented this algorithm as a class, just to keep the various constants it uses tidy. The code below produces full permutations, but it should be easy to modify to just return the i-th element.
#!/usr/bin/env python
''' Ordered permutations using factorial base counting
Written by PM 2Ring 2015.02.15
Derived from C code written 2003.02.13
'''
from math import factorial
class Permuter(object):
''' A class for making ordered permutations, one by one '''
def __init__(self, seq):
self.seq = list(seq)
self.size = len(seq)
self.base = factorial(self.size - 1)
self.fac = self.size * self.base
def perm(self, k):
''' Build kth ordered permutation of seq '''
seq = self.seq[:]
p = []
base = self.base
for j in xrange(self.size - 1, 0, -1):
q, k = divmod(k, base)
p.append(seq.pop(q))
base //= j
p.append(seq[0])
return p
def test(seq):
permuter = Permuter(seq)
for k in xrange(permuter.fac):
print '%2d: %s' % (k, ''.join(permuter.perm(k)))
if __name__ == '__main__':
test('abcd')
This algorithm has a little more overhead than the unordered permutation maker: it requires factorial to be calculated in advance, and of course factorial gets very large very quickly. Also, it requires one extra division per inner loop. So the time savings in bailing out of the inner loop once you've found the i-th element may be offset by these overheads.
FWIW, the code in your question has room for improvement. In particular, k /= n should be written as k //= n to ensure that integer division is used; your code works ok on Python 2 but not on Python 3. However, since we need both the quotient and remainder, it makes sense to use the built-in divmod() function. Also, by reorganizing things a little we can avoid the multiple calculations of n - 1
#!/usr/bin/env python
def kth_permutation(n, k):
p = range(n)
while n:
k, j = divmod(k, n)
n -= 1
p[n], p[j] = p[j], p[n]
return p
def test(n):
last = range(n)
k = 0
while True:
p = kth_permutation(n, k)
print k, p
if p == last:
break
k += 1
test(3)
output
0 [1, 2, 0]
1 [2, 0, 1]
2 [1, 0, 2]
3 [2, 1, 0]
4 [0, 2, 1]
5 [0, 1, 2]
You probably cannot get the i'th digit of the k'th permutation of n elements in O(n) time or space, because representing the number k itself requires O(log(n!)) = O(n log n) bits, and any manipulations with it have corresponding time complexity.