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My for loop prints all the consecutive subsequence of a list. For example, suppose a list contains [0, 1,2,3,4,5,6,7,8,9]. It prints,
0
0,1
0,1,2
0,1,2,3
........
0,1,2,3,4,5,6,7,8,9
1
1,2
1,2,3
1,2,3,4,5,6,7,8,9
........
8
8,9
9
for i in range(10)
for j in range(i, 10):
subseq = []
for k in range(i, j+1):
subseq.append(k)
print(subseq)
The current algorithmic complexity of this for loop is O(N^3). Is there any way to make this algorithm any faster?
I don't know Python (this is Python, right?), but something like this will be a little faster version of O(N^3) (see comments below):
for i in range(10):
subseq = []
for j in range(i, 10):
subseq.append(j)
print(subseq)
Yes, that works:
[0]
[0, 1]
[0, 1, 2]
[0, 1, 2, 3]
[0, 1, 2, 3, 4]
[0, 1, 2, 3, 4, 5]
[0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7, 8]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[1]
[1, 2]
...
[7, 8]
[7, 8, 9]
[8]
[8, 9]
[9]
It’s not possible to do this in less than O(n3) time because you’re printing a total of O(n3) items. Specifically, split the array in quarters and look at the middle two quarters of the array. Pick any element there - say, the one at position k. That will be printed in at least n2 / 4 different subarrays: pick any element in the first quarter, any element in the last quarter, and the subarray between those elements will contain the element at position k.
This means that any of the n / 2 items in the middle two quarters gets printed at least n2 / 4 times, so you print at least n3 / 8 total values. There’s no way to do that in better than O(n3) time.
I need for given N create N*N matrix which does not have repetitions in rows, cells, minor and major diagonals and values are 1, 2 , 3, ...., N.
For N = 4 one of matrices is the following:
1 2 3 4
3 4 1 2
4 3 2 1
2 1 4 3
Problem overview
The math structure you described is Diagonal Latin Square. Constructing them is the more mathematical problem than the algorithmic or programmatic.
To correctly understand what it is and how to create you should read following articles:
Latin squares definition
Magic squares definition
Diagonal Latin square construction <-- p.2 is answer to your question with proof and with other interesting properties
Short answer
One of the possible ways to construct Diagonal Latin Square:
Let N is the power of required matrix L.
If there are exist numbers A and B from range [0; N-1] which satisfy properties:
A relativly prime to N
B relatively prime to N
(A + B) relatively prime to N
(A - B) relatively prime to N
Then you can create required matrix with the following rule:
L[i][j] = (A * i + B * j) mod N
It would be nice to do this mathematically, but I'll propose the simplest algorithm that I can think of - brute force.
At a high level
we can represent a matrix as an array of arrays
for a given N, construct S a set of arrays, which contains every combination of [1..N]. There will be N! of these.
using an recursive & iterative selection process (e.g. a search tree), search through all orders of these arrays until one of the 'uniqueness' rules is broken
For example, in your N = 4 problem, I'd construct
S = [
[1,2,3,4], [1,2,4,3]
[1,3,2,4], [1,3,4,2]
[1,4,2,3], [1,4,3,2]
[2,1,3,4], [2,1,4,3]
[2,3,1,4], [2,3,4,1]
[2,4,1,3], [2,4,3,1]
[3,1,2,4], [3,1,4,2]
// etc
]
R = new int[4][4]
Then the algorithm is something like
If R is 'full', you're done
Evaluate does the next row from S fit into R,
if yes, insert it into R, reset the iterator on S, and go to 1.
if no, increment the iterator on S
If there are more rows to check in S, go to 2.
Else you've iterated across S and none of the rows fit, so remove the most recent row added to R and go to 1. In other words, explore another branch.
To improve the efficiency of this algorithm, implement a better data structure. Rather than a flat array of all combinations, use a prefix tree / Trie of some sort to both reduce the storage size of the 'options' and reduce the search area within each iteration.
Here's a method which is fast for N <= 9 : (python)
import random
def generate(n):
a = [[0] * n for _ in range(n)]
def rec(i, j):
if i == n - 1 and j == n:
return True
if j == n:
return rec(i + 1, 0)
candidate = set(range(1, n + 1))
for k in range(i):
candidate.discard(a[k][j])
for k in range(j):
candidate.discard(a[i][k])
if i == j:
for k in range(i):
candidate.discard(a[k][k])
if i + j == n - 1:
for k in range(i):
candidate.discard(a[k][n - 1 - k])
candidate_list = list(candidate)
random.shuffle(candidate_list)
for e in candidate_list:
a[i][j] = e
if rec(i, j + 1):
return True
a[i][j] = 0
return False
rec(0, 0)
return a
for row in generate(9):
print(row)
Output:
[8, 5, 4, 7, 1, 6, 2, 9, 3]
[2, 7, 5, 8, 4, 1, 3, 6, 9]
[9, 1, 2, 3, 6, 4, 8, 7, 5]
[3, 9, 7, 6, 2, 5, 1, 4, 8]
[5, 8, 3, 1, 9, 7, 6, 2, 4]
[4, 6, 9, 2, 8, 3, 5, 1, 7]
[6, 3, 1, 5, 7, 9, 4, 8, 2]
[1, 4, 8, 9, 3, 2, 7, 5, 6]
[7, 2, 6, 4, 5, 8, 9, 3, 1]
You are given 2 lists, the first with a elements and the second with b elements, with a < b.
For each element e in list a, you want to take a element f in list b, and replace e with e-f. You cannot use a element twice unless it appears in list b twice.
The problem is to find the minimum value of the largest element of list a.
For example, say list a is [1, 2, 3, 4], and list b is [5, 6, 7, 8, 9, 10, 11, 12]. We would take the e's to be 5, 6, 7, 8, so that list a becomes [5-1, 6-2, 7-3, 8-4], with the largest element being 4. So 4 is the answer.
Another example: if list a is [1, 4, 7] and list b is [-1, 3, 4, 5, 6, 7, 8], we would take the e's to be -1, 4, 7, so that list a becomes [2, 0, 0], and the answer is 2. So 2 is the answer.
I know this is poorly worded, if I could do anything to better word it, please let me know. I tried first sorting list a and list b, then did not know what to do.
If you could help, please do.
Thanks!
calculate the values of the list:
(java)
List listA = ...;
List listb = ...;
for(int i = 0; i < listA.size(); i++){
listA.set(i, listA.get(i) - listb.get(i));
}
find the highest value in listA:
iHighestValue = listA.get(0); //setting it to 0 would not work with lists containing only negative integers
for(int j = 1; j < listA.size(); j++){
if(listA.get(j) > iHighestValue)
iHighestValue = listA.get(i);
}
[Edit]: sorry, it doesn't show as code (don't know why)
Given I have an array such as follows:
arr = [8, 13, 14, 10, 6, 7, 8, 14, 5, 3, 5, 2, 6, 7, 4]
I would like to count the number of consecutive number sequences. Eg in the above array the consecutive number sequences (or array-slices) are:
[13,14]
[6,7,8]
[6,7]
And hence we have 3 such slices. What is an efficient Algorithm to count this? I know how I can do it O(N^2) but I'm looking for something which is better than that.
arr = [8, 13, 14, 10, 6, 7, 8, 14, 5, 3, 5, 2, 6, 7, 4]
p arr.each_cons(2).chunk{|a,b| a.succ == b || nil}.count #=> 3
nilhas a special meaning to the chunk-method: it causes items to be dropped.
arr = [8, 13, 14, 10, 6, 7, 8, 14, 5, 3, 5, 2, 6, 7, 4]
result = []
stage = []
for i in arr:
if len(stage) > 0 and i != stage[-1]+1:
if len(stage) > 1:
result.append(stage)
stage = []
stage.append(i)
print result
Output:
[[13, 14], [6, 7, 8], [6, 7]]
The time complexity of this code is O(n). (There's only one for loop. And it's not hard to see that each iteration in the loop is O(1).)
I would do as below using Enumerable#slice_before:
a = [8, 13, 14, 10, 6, 7, 8, 14, 5, 3, 5, 2, 6, 7, 4]
prev = a[0]
hash = Hash[a.slice_before do |e|
prev, prev2 = e, prev
prev2 + 1 != e
end.map{|e| [e,e.size] if e.size > 1}]
hash # => {[13, 14]=>2, [6, 7, 8]=>3, [6, 7]=>2}
hash.size # => 3
I think this can be done in O(N) time. If you just want the count,
Iterate through the array. Initialize counter to 0.
If next element is one more or one less than current element, increment the counter.
Continue iterating till the next element is not one more or one less than current element.
Repeat steps 2 and 3 until you reach the end.
If you want sections of continuously increasing consecutive elements (not clear from your question)
Iterate through the array. Initialize counter to 0.
If next element is one more than current element, increment the counter.
Continue iterating till the next element is not one more than current element.
Repeat steps 2 and 3 until you reach the end.
Say we have a 0-indexed sequence S, take S[0] and insert it in a place in S where the next value is higher than S[0] and the previous value is lower than S[0]. Formally, S[i] should be placed in such a place where S[i-1] < S[i] < S[i+1]. Continue in order on the list doing the same with every item. Remove the element from the list before putting it in the correct place. After one iteration over the list the list should be ordered. I recently had an exam and I forgot insertion sort (don't laugh) and I did it like this. However, my professor marked it wrong. The algorithm, as far as I know, does produce a sorted list.
Works like this on a list:
Sorting [2, 8, 5, 4, 7, 0, 6, 1, 10, 3, 9]
[2, 8, 5, 4, 7, 0, 6, 1, 10, 3, 9]
[2, 8, 5, 4, 7, 0, 6, 1, 10, 3, 9]
[2, 5, 4, 7, 0, 6, 1, 8, 10, 3, 9]
[2, 4, 5, 7, 0, 6, 1, 8, 10, 3, 9]
[2, 4, 5, 7, 0, 6, 1, 8, 10, 3, 9]
[2, 4, 5, 0, 6, 1, 7, 8, 10, 3, 9]
[0, 2, 4, 5, 6, 1, 7, 8, 10, 3, 9]
[0, 2, 4, 5, 1, 6, 7, 8, 10, 3, 9]
[0, 1, 2, 4, 5, 6, 7, 8, 10, 3, 9]
[0, 1, 2, 4, 5, 6, 7, 8, 3, 9, 10]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Got [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Since every time an element is inserted into the list up to (n-1) numbers in the list may be moved and we must do this n times the algorithm should run in O(n^2) time.
I had a Python implementation but I misplaced it somehow. I'll try to write it again in a bit, but it's kinda tricky to implement. Any ideas?
The Python implementation is here: http://dpaste.com/hold/522232/. It was written by busy_beaver from reddit.com when it was discussed here http://www.reddit.com/r/compsci/comments/ejaaz/is_this_equivalent_to_insertion_sort/
It's a while since this was asked, but none of the other answers contains a proof that this bizarre algorithm does in fact sort the list. So here goes.
Suppose that the original list is v1, v2, ..., vn. Then after i steps of the algorithm, I claim that the list looks like this:
w1,1, w1,2, ..., w1,r(1), vσ(1), w2,1, ... w2,r(2), vσ(2), w3,1 ... ... wi,r(i), vσ(i), ...
Where σ is the sorted permutation of v1 to vi and the w are elements vj with j > i. In other words, v1 to vi are found in sorted order, possibly interleaved with other elements. And moreover, wj,k ≤ vj for every j and k. So each of the correctly sorted elements is preceded by a (possibly empty) block of elements less than or equal to it.
Here's a run of the algorithm, with the sorted elements in bold, and the preceding blocks of elements in italics (where non-empty). You can see that each block of italicised elements is less than the bold element that follows it.
[4, 8, 6, 1, 2, 7, 5, 0, 3, 9]
[4, 8, 6, 1, 2, 7, 5, 0, 3, 9]
[4, 6, 1, 2, 7, 5, 0, 3, 8, 9]
[4, 1, 2, 6, 7, 5, 0, 3, 8, 9]
[1, 4, 2, 6, 7, 5, 0, 3, 8, 9]
[1, 2, 4, 6, 7, 5, 0, 3, 8, 9]
[1, 2, 4, 6, 5, 0, 3, 7, 8, 9]
[1, 2, 4, 5, 6, 0, 3, 7, 8, 9]
[0, 1, 2, 4, 5, 6, 3, 7, 8, 9]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
If my claim is true, then the algorithm sorts, because after n steps all the vi are in order, and there are no remaining elements to be interleaved. But is the claim really true?
Well, let's prove it by induction. It's certainly true when i = 0. Suppose it's true for i. Then when we run the (i + 1)st step, we pick vi+1 and move it into the first position where it fits. It certainly passes over all vj with j ≤ i and vj < vi+1 (since these are sorted by hypothesis, and each is preceded only by smaller-or-equal elements). It cannot pass over any vj with j ≤ i and vj ≥ vi+1, because there's some position in the block before vj where it will fit. So vi+1 ends up sorted with respect to all vj with j ≤ i. So it ends up somewhere in the block of elements before the next vj, and since it ends up in the first such position, the condition on the blocks is preserved. QED.
However, I don't blame your professor for marking it wrong. If you're going to invent an algorithm that no-one's seen before, it's up to you to prove it correct!
(The algorithm needs a name, so I propose fitsort, because we put each element in the first place where it fits.)
Your algorithm seems to me very different from insertion sort. In particular, it's very easy to prove that insertion sort works correctly (at each stage, the first however-many elements in the array are correctly sorted; proof by induction; done), whereas for your algorithm it seems much more difficult to prove this and it's not obvious exactly what partially-sorted-ness property it guarantees at any given point in its processing.
Similarly, it's very easy to prove that insertion sort always does at most n steps (where by a "step" I mean putting one element in the right place), whereas if I've understood your algorithm correctly it doesn't advance the which-element-to-process-next pointer if it's just moved an element to the right (or, to put it differently, it may sometimes have to process an element more than once) so it's not so clear that your algorithm really does take O(n^2) time in the worst case.
Insertion sort maintains the invariant that elements to the left of the current pointer are sorted. Progress is made by moving the element at the pointer to the left into its correct place and advancing the pointer.
Your algorithm does this, but sometimes it also does an additional step of moving the element at the pointer to the right without advancing the pointer. This makes the algorithm as a whole not an insertion sort, though you could call it a modified insertion sort due to the resemblance.
This algorithm runs in O(n²) on average like insertion sort (also like bubble sort). The best case for an insertion sort is O(n) on an already sorted list, for this algorithm it is O(n) but for a reverse-sorted list since you find the correct position for every element in a single comparison (but only if you leave the first, largest, element in place at the beginning when you can't find a good position for it).
A lot of professors are notorious for having the "that's not the answer I'm looking for" bug. Even if it's correct, they'll say it doesn't meet their criteria.
What you're doing seems like insertion sort, although using removes and inserts seems like it would only add unnecessary complexity.
What he might be saying is you're essentially "pulling out" the value and "dropping it back in" the correct spot. Your prof was probably looking for "swapping the value up (or down) until you found it's correct location."
They have the same result but they're different in implementation. Swapping would be faster, but not significantly so.
I have a hard time seeing that this is insert sort. Using insert sort, at each iteration, one more element would be placed correctly in the array. In your solution I do not see an element being "fully sorted" upon each iteration.
The insert sort algorithm begin:
let pos = 0
if pos == arraysize then return
find the smallest element in the remaining array from pos and swap it with the element at position pos
pos++
goto 2