问题
First, I know
- lower bound is O(nlogn)
- and how to prove it
And I agree the lower bound should be O(nlogn).
What I don't quite understand is:
For some special cases, the # of comparisons could actually be even lower than the lower bound. For example, use bubble sort to sort an already sorted array. The # of comparisons is O(n).
So how to actually understand the idea of lower bound?
The classical definition on Wikipedial: http://en.wikipedia.org/wiki/Upper_and_lower_bounds does not help much.
My current understanding of this is:
lower bound of the comparison-based sorting is actually the upper bound for the worst case.
namely, how best you could in the worst case.
Is this correct? Thanks.
回答1:
lower bound of the comparison-based sorting is actually the upper bound for the best case.
No.
The function that you are bounding is the worst-case running time of the best possible sorting algorithm.
Imagine the following game:
- We choose some number n.
- You pick your favorite sorting algorithm.
- After looking at your algorithm, I pick some input sequence of length n.
- We run your algorithm on my input, and you give me a dollar for every executed instruction.
The O(n log n) upper bound means you can limit your cost to at most O(n log n) dollars, no matter what input sequence I choose.
The Ω(n log n) lower bound means that I can force you to pay at least Ω(n log n) dollars, no matter what sorting algorithm you choose.
Also: "The lower bound is O(n log n)" doesn't make any sense. O(f(n)) means "at most a constant times f(n)". But "lower bound" means "at least ...". So saying "a lower bound of O(n log n)" is exactly like saying "You can save up to 50% or more!" — it's completely meaningless! The correct notation for lower bounds is Ω(...).
回答2:
The problem of sorting can be viewed as following.
Input: A sequence of n numbers . Output: A permutation (reordering) of the input sequence such that a‘1 <= a‘2 ….. <= a‘n.
A sorting algorithm is comparison based if it uses comparison operators to find the order between two numbers. Comparison sorts can be viewed abstractly in terms of decision trees. A decision tree is a full binary tree that represents the comparisons between elements that are performed by a particular sorting algorithm operating on an input of a given size. The execution of the sorting algorithm corresponds to tracing a path from the root of the decision tree to a leaf. At each internal node, a comparison ai aj is made. The left subtree then dictates subsequent comparisons for ai aj, and the right subtree dictates subsequent comparisons for ai > aj. When we come to a leaf, the sorting algorithm has established the ordering. So we can say following about the decison tree.
1) Each of the n! permutations on n elements must appear as one of the leaves of the decision tree for the sorting algorithm to sort properly.
2) Let x be the maximum number of comparisons in a sorting algorithm. The maximum height of the decison tree would be x. A tree with maximum height x has at most 2^x leaves.
After combining the above two facts, we get following relation.
n! <= 2^x
Taking Log on both sides. \log_2n! <= x
Since \log_2n! = \Theta(nLogn), we can say x = \Omega(nLog_2n) Therefore, any comparison based sorting algorithm must make at least \Omega(nLog_2n) comparisons to sort the input array, and Heapsort and merge sort are asymptotically optimal comparison sorts.
回答3:
When you do asymptotic analysis you derive an O or Θ or Ω for all input.
But you can also make analysis on whether properties of the input affect the runtime.
For example algorithms that take as input something almost sorted have better performance than the formal asymptotic formula due to the input characteristics and the structure of the algorithm. Examples are bubblesort and quicksort.
It is not that you can go bellow the lower boundaries. It only behavior of the implementation on specific input.
回答4:
Imagine all the possible arrays of things that could be sorted. Lets say they are arrays of length 'n' and ignore stuff like arrays with one element (which, of course, are always already sorted.
Imagine a long list of all possible value combinations for that array. Notice that we can simplify this a bit since the values in the array always have some sort of ordering. So if we replace the smallest one with the number 1, the next one with 1 or 2 (depending on whether its equal or greater) and so forth, we end up with the same sorting problem as if we allowed any value at all. (This means an array of length n will need, at most, the numbers 1-n. Maybe less if some are equal.)
Then put a number beside each one telling how much work it takes to sort that array with those values in it. You could put several numbers. For example, you could put the number of comparisons it takes. Or you could put the number of element moves or swaps it takes. Whatever number you put there indicates how many operations it takes. You could put the sum of them.
One thing you have to do is ignore any special information. For example, you can't know ahead of time that the arrangement of values in the array are already sorted. Your algorithm has to do the same steps with that array as with any other. (But the first step could be to check if its sorted. Usually that doesn't help in sorting, though.)
So. The largest number, measured by comparisons, is the typical number of comparisons when the values are arranged in a pathologically bad way. The smallest number, similarly, is the number of comparisons needed when the values are arranged in a really good way.
For a bubble sort, the best case (shortest or fastest) is if the values are in order already. But that's only if you use a flag to tell whether you swapped any values. In that best case, you look at each adjacent pair of elements one time and find they are already sorted and when you get to the end, you find you haven't swapped anything so you are done. that's n-1 comparisons total and forms the lowest number of comparisons you could ever do.
It would take me a while to figure out the worst case. I haven't looked at a bubble sort in decades. But I would guess its a case where they are reverse ordered. You do the 1st comparison and find the 1st element needs to move. You slide up to the top comparing to each one and finally swap it with the last element. So you did n-1 comparisons in that pass. The 2nd pass starts at the 2nd element and does n-2 comparisons and so forth. So you do (n-1)+(n-2)+(n-3)+...+1 comparisons in this case which is about (n**2)/2.
Maybe your variation on bubble sort is better than the one I described. No matter.
For bubble sort then, the lower bound is n-1 and the upper bound is (n**2)/2
Other sort algorithms have better performance.
You might want to remember that there are other operations that cost besides comparisons. We use comparisons because much sorting is done with strings and a string comparison is costly in compute time.
You could use element swaps to count (or the sum of swaps and elements swaps) but they are typically shorter than comparisons with strings. If you have numbers, they are similar.
You could also use more esoteric things like branch prediction failure or memory cache misses or for measuring.
来源：https://stackoverflow.com/questions/15625150/understanding-of-lower-bound-for-comparison-based-sorting-algorithm