In a 2d array of n×n, I am trying to find a number that's larger than its neighbour.
I approached this using the divide and conquer algorithm.
Next, I attempted to prove the correctness of my algorithm by exhibiting a suitable invariant i.e. I approached this by filling up 4x4 grid with random numbers, dividing it and finding the global maxima for each column selected (not sure this is the way to prove the correctness of the algorithm)
The bit I am most confused about is how to analyse the running time for my algorithm i.e. how many elements in the n×n array need to be visited in the worst case and finally if there is a way to show if the algorithm is asymptotically optimal.
You cannot have better complexity than O(n^2), as you need to visit every element of an array. If you did not visit some element, it may turn out that your array does not have any local maxima except global maximum in the element you didn't visit. Brute force algorithm (just checking every element if it is a local maximum) is O(n^2), so there's not much space to do better or worse than that.
Re divide and conquer algorithm -- this seems to be overengineering for the given problem. You gain nothing by it, and you introduce additional complications (how do you handle borders between the parts?)
I'm tutoring a student and one of her assignments is to describe an O(nlogn) algorithm for the closest pair of points in the one-dimensional case. But the restriction is she's not allowed to use a divide-and-conquer approach. I understand the two-dimensional case from a question a user posted some years ago. I'll link it in case someone wants to look at it: For 2-D case (plane) - "Closest pair of points" algorithm.
However, for the 1-D case, I can only think of a solution which involves checking each and every point on the line and comparing it to the closest point to the left and right of it. But this solution isn't O(nlogn) since checking each point will take time proportional to n and the comparisons for each point would take time proportional to 2n. I'm not sure where log(n) would come from without using a divide-and-conquer approach.
For some reason, I can't come up with a solution. Any help would be appreciated.
Hint: If the points were ordered from left to right, what would you do, and what would the complexity be? What is the complexity of ordering the points first?
It seems to me that one could:
Sort the locations into order - O(n log n)
Find the differences between the ordered locations - O(n)
Find the smallest difference - O(n)
The smallest difference defines the two closest points.
The overall result would be O(n log n).
Consider the following numbers:
9,44,32,12,7,45,31,98,35,37,41,8,20,27,83,64,61,28,39,93,29,92,17,13,14,55,21,66,72,23,73,99,1,2,88,77,3,65,83,84,62,5,11,74,68,76,78,67,75,69,70,22,71,24,25,26.
I try to implement an algorithm to remove the least amount of numbers in the list to make the sequence
a) increasing order
b) decreasing order
I already tried with the shortest and longest subsecuence. Dont want the code, only the explanation or a pseudo code,i can't understand how to solve the problem thanks!
This is a lightly camouflaged Longest increasing (decreasing) subsequence problem. The algorithm to solving your problem is as follows:
Find the longest increasing (decreasing) subsequence in the array
Remove all elements that do not belong to the longest increasing subsequence.
Since the increasing/decreasing subsequence is longest, the amount of numbers that you will remove is the smallest.
Wikipedia article has a nice pseudocode for solving the LIS/LDS problem. You can substitute binary search for a linear one unless the original sequence is 1000+ elements long.
Since it has already been mentioned, I will add my 2 cents. Most probably this will be asked in interview in those circumstances, the running time(efficiency) is a major concern. So the same problem can be tackled with many algorithms depending on the time they take to execute.
The best known algorithm is of order O(nlogn). Other important one can like Dynamic programming paradigm can also be applied to yield a solution of O(n^2).
O(n^2) here
O(nlogn) here
Why do divide and conquer algorithms often run faster than brute force? For example, to find closest pair of points. I know you can show me the mathematical proof. But intuitively, why does this happen? Magic?
Theoretically, is it true that "divide and conquer is always better than brute force"? If it is not, is there any counterexample?
For your first question, the intuition behind divide-and-conquer is that in many problems the amount of work that has to be done is based on some combinatorial property of the input that scales more than linearly.
For example, in the closest pair of points problem, the runtime of the brute-force answer is determined by the fact that you have to look at all O(n2) possible pairs of points.
If you take something that grows quadratically and cut it into two pieces, each of which is half the size as before, it takes one quarter of the initial time to solve the problem in each half, so solving the problem in both halves takes time roughly one half the time required for the brute force solution. Cutting it into four pieces would take one fourth of the time, cutting it into eight pieces one eighth the time, etc.
The recursive version ends up being faster in this case because at each step, we avoid doing a lot of work from dealing with pairs of elements by ensuring that there aren't too many pairs that we actually need to check. Most algorithms that have a divide and conquer solution end up being faster for a similar reason.
For your second question, no, divide and conquer algorithms are not necessarily faster than brute-force algorithms. Consider the problem of finding the maximum value in an array. The brute-force algorithm takes O(n) time and uses O(1) space as it does a linear scan over the data. The divide-and-conquer algorithm is given here:
If the array has just one element, that's the maximum.
Otherwise:
Cut the array in half.
Find the maximum in each half.
Compute the maximum of these two values.
This takes time O(n) as well, but uses O(log n) memory for the stack space. It's actually worse than the simple linear algorithm.
As another example, the maximum single-sell profit problem has a divide-and-conquer solution, but the optimized dynamic programming solution is faster in both time and memory.
Hope this helps!
I recommend you read through the chapter 5 of Algorithm Design, it explains divide-and-conquer very well.
Intuitively, for a problem, if you can divide it into two sub-problems with the same pattern as the origin one, and the time complexity to merge the results of the two sub-problems into the final result is somehow small, then it's faster than solve the orignal complete problem by brute-force.
As said in Algorithm Design, you actually cannot gain too much from divide-and-conquer in terms of time, general you can only reduce time complexity from higher polynomial to lower polynomial(e.g. from O(n^3) to O(n^2)), but hardly from exponential to polynomial(e.g. from O(2^n) to O(n^3)).
I think the most you can gain from divide-and-conquer is the mindset for problem solving. It's always a good attempt to break the original big problem down to smaller and easier sub-problems. Even if you don't get a better running time, it still helps you think through the problem.
I was asked if a Binary Search is a divide and conquer algorithm at an exam. My answer was yes, because you divided the problem into smaller subproblems, until you reached your result.
But the examinators asked where the conquer part in it was, which I was unable to answer. They also disapproved that it actually was a divide and conquer algorithm.
But everywhere I go on the web, it says that it is, so I would like to know why, and where the conquer part of it is?
The book:
Data Structures and Algorithm Analysis in Java (2nd Edition), by Mark Allen Weiss
Says that a D&C algorithm should have two disjoint recursive calls, just like QuickSort does.
Binary Search does not have this, even though it can be implemented recursively.
I think it is not divide and conquer, see first paragraph in http://en.wikipedia.org/wiki/Divide_and_conquer_algorithm
recursively breaking down a problem into two or more sub-problems
which are then combined to give a solution
In binary search there is still only one problem which does just reducing data by half every step, so no conquer (merging) phase of the results is needed.
It isn't.
To complement #Kenci's post, DnC algorithms have a few general/common properties; they:
divide the original problem instance into a set of smaller sub-instances of itself;
independently solve each sub-instance;
combine smaller/independent sub-instance solutions to build a single solution for the larger/original instance
The problem with Binary Search is that it does not really even generate a set of independent sub-instances to be solved, as per step 1; it only simplifies the original problem by permanently discarding sections it's not interested in. In other words, it only reduces the problem's size and that's as far as it ever goes.
A DnC algorithm is supposed to not only identify/solve the smaller sub-instances of the original problem independently of each other, but also use that set of partial independent solutions to "build up" a single solution for the larger problem instance as a whole.
The book Fundamentals of Algorithmics, G. Brassard, P. Bratley says the following (bold my emphasis, italics in original):
It is probably the simplest application of divide-and-conquer, so simple in fact that strictly speaking this is an application of simplification rather than divide-and-conquer: the solution to any sufficiently large instance is reduced to that of a single smaller one, in this case of half size.
Section 7.3 Binary Search on p.226.
In a divide and conquer strategy :
1.Problem is divided into parts;
2.Each of these parts is attacked/solved independently, by applying the algorithm at hand (mostly recursion is used for this purpose) ;
3.And then the solutions of each partition/division and combined/merged together to arrive at the final solution to the problem as a whole (this comes under conquer)
Example, Quick sort, merge sort.
Basically, the binary search algorithm just divides its work space(input (ordered) array of size n) into half in each iteration. Therefore it is definitely deploying the divide strategy and as a result, the time complexity reduces down to O(lg n).So,this covers up the "divide" part of it.
As can be noticed, the final solution is obtained from the last comparison made, that is, when we are left with only one element for comparison.
Binary search does not merge or combine solution.
In short, binary search divides the size of the problem (on which it has to work) into halves but doesn't find the solution in bits and pieces and hence no need of merging the solution occurs!
I know it's a bit too lengthy but i hope it helps :)
Also you can get some idea from : https://www.khanacademy.org/computing/computer-science/algorithms/binary-search/a/running-time-of-binary-search
Also i realised just now that this question was posted long back!
My bad!
Apparently some people consider binary search a divide-and-conquer algorithm, and some are not. I quickly googled three references (all seem related to academia) that call it a D&C algorithm:
http://www.cs.berkeley.edu/~vazirani/algorithms/chap2.pdf
http://homepages.ius.edu/rwisman/C455/html/notes/Chapter2/DivConq.htm
http://www.csc.liv.ac.uk/~ped/teachadmin/algor/d_and_c.html
I think it's common agreement that a D&C algorithm should have at least the first two phases of these three:
divide, i.e. decide how the whole problem is separated into sub-problems;
conquer, i.e. solve each of the sub-problems independently;
[optionally] combine, i.e. merge the results of independent computations together.
The second phase - conquer - should recursively apply the same technique to solve the subproblem by dividing into even smaller sub-sub-problems, and etc. In practice, however, often some threshold is used to limit the recursive approach, as for small size problems a different approach might be faster. For example, quick sort implementations often use e.g. bubble sort when the size of an array portion to sort becomes small.
The third phase might be a no-op, and in my opinion it does not disqualify an algorithm as D&C. A common example is recursive decomposition of a for-loop with all iterations working purely with independent data items (i.e. no reduction of any form). It might look useless at glance, but in fact it's very powerful way to e.g. execute the loop in parallel, and utilized by such frameworks as Cilk and Intel's TBB.
Returning to the original question: let's consider some code that implements the algorithm (I use C++; sorry if this is not the language you are comfortable with):
int search( int value, int* a, int begin, int end ) {
// end is one past the last element, i.e. [begin, end) is a half-open interval.
if (begin < end)
{
int m = (begin+end)/2;
if (value==a[m])
return m;
else if (value<a[m])
return search(value, a, begin, m);
else
return search(value, a, m+1, end);
}
else // begin>=end, i.e. no valid array to search
return -1;
}
Here the divide part is int m = (begin+end)/2; and all the rest is the conquer part. The algorithm is explicitly written in a recursive D&C form, even though only one of the branches is taken. However, it can also be written in a loop form:
int search( int value, int* a, int size ) {
int begin=0, end=size;
while( begin<end ) {
int m = (begin+end)/2;
if (value==a[m])
return m;
else if (value<a[m])
end = m;
else
begin = m+1;
}
return -1;
}
I think it's quite a common way to implement binary search with a loop; I deliberately used the same variable names as in the recursive example, so that commonality is easier to see. Therefore we might say that, again, calculating the midpoint is the divide part, and the rest of the loop body is the conquer part.
But of course if your examiners think differently, it might be hard to convince them it's D&C.
Update: just had a thought that if I were to develop a generic skeleton implementation of a D&C algorithm, I would certainly use binary search as one of API suitability tests to check whether the API is sufficiently powerful while also concise. Of course it does not prove anything :)
The Merge Sort and Quick Sort algorithms use the divide and conquer technique (because there are 2 sub-problems) and Binary Search comes under decrease and conquer (because there is 1 sub-problem).
Therefore, Binary Search actually uses the decrease and conquer technique and not the divide and conquer technique.
Source: https://www.geeksforgeeks.org/decrease-and-conquer/
Binary search is tricky to describe with divide-and-conquer because the conquering step is not explicit. The result of the algorithm is the index of the needle in the haystack, and a pure D&C implementation would return the index of the needle in the smallest haystack (0 in the one-element list) and then recursively add the offsets in the larger haystacks that were divided in the divison step.
Pseudocode to explain:
function binary_search has arguments needle and haystack and returns index
if haystack has size 1
return 0
else
divide haystack into upper and lower half
if needle is smaller than smallest element of upper half
return 0 + binary_search needle, lower half
else
return size of lower half + binary_search needle, upper half
The addition (0 + or size of lower half) is the conquer part. Most people skip it by providing indices into a larger list as arguments, and thus it is often not readily available.
The divide part is of course dividing the set into halves.
The conquer part is determining whether and on what position in the processed part there is a searched element.
Dichotomic in computer science refers to choosing between two antithetical choices, between two distinct alternatives. A dichotomy is any splitting of a whole into exactly two non-overlapping parts, meaning it is a procedure in which a whole is divided into two parts. It is a partition of a whole (or a set) into two parts (subsets) that are:
1. Jointly Exhaustive: everything must belong to one part or the other, and
2. Mutually Exclusive: nothing can belong simultaneously to both parts.
Divide and conquer works by recursively breaking down a problem into two or more sub-problems of the same type, until these become simple enough to be solved directly.
So the binary search halves the number of items to check with each iteration and determines if it has a chance of locating the "key" item in that half or moving on to the other half if it is able to determine keys absence. As the algorithm is dichotomic in nature so the binary search will believe that the "key" has to be in one part until it reaches the exit condition where it returns that the key is missing.
Divide and Conquer algorithm is based on 3 step as follows:
Divide
Conquer
Combine
Binary Search problem can be defined as finding x in the sorted array A[n].
According to this information:
Divide: compare x with middle
Conquer: Recurse in one sub array. (Finding x in this array)
Combine: it is not necessary.
A proper divide and conquer algorithm will require both parts to be processed.
Therefore, many people will not call binary-search a divide and conquer algorithm, it does divide the problem, but discards the other half.
But most likely, your examiners just wanted to see how you argue. (Good) exams aren't about the facts, but about how you react when the challenge goes beyond the original material.
So IMHO the proper answer would have been:
Well, technically, it consists only of a divide step, but needs to conquer only half of the original task then, the other half is trivially done already.
BTW: there is a nice variation of QuickSort, called QuickSelect, which actually exploits this difference to obtain an on average O(n) median search algorithm. It's like QuickSort - but descends only into the half it is interested in.
Binary Search is not a divide and conquer approach. It is a decrease and conquer approach.
In divide and conquer approach, each subproblem must contribute to the solution but in binary search, all subdivision does not contribute to the solution. we divide into two parts and discard one part because we know that the solution does not exist in this part and look for the solution only in one part.
The informal definition is more or less: Divide the problem into small problems. Then solve them and put them together (conquer). Solving is in fact deciding where to go next (left, right, element found).
Here a quote from wikipedia:
The name "divide and conquer" is sometimes applied also to algorithms that reduce each problem to only one subproblem, such as the binary search algorithm for finding a record in a sorted list.
This states, it's NOT [update: misread this phrase:)] only one part of divide and conquer.
Update:
This article made it clear for me. I was confused since the definition says you have to solve every sub problem. But you solved the sub problem if you know you don't have to keep on searching..
The Binary Search is a divide and conquer algorithm:
1) In Divide and Conquer algorithms, we try to solve a problem by solving a smaller sub problem (Divide part) and use the solution to build the solution for our bigger problem(Conquer).
2) Here our problem is to find an element in the sorted array. We can solve this by solving a similar sub problem. (We are creating sub problems here based on a decision that the element being searched is smaller or bigger than the middle element). Thus once we know that the element can not exist surely in one half, we solve a similar sub-problem in the the other half.
3) This way we recurse.
4) The conquer part here is just returning the value returned by the sub problem to the top the recursive tree
I think it is Decrease and Conquer.
Here is a quote from wikipedia.
"The name decrease and conquer has been proposed instead for the
single-subproblem class"
http://en.wikipedia.org/wiki/Divide_and_conquer_algorithms#Decrease_and_conquer
According to my understanding, "Conquer" part is at the end when you find the target element of the Binary search. The "Decrease" part is reducing the search space.
Binary Search and Ternary Search Algorithms are based on Decrease and Conquer technique. Because, you do not divide the problem, you actually decrease the problem by dividing by 2(3 in ternary search).
Merge Sort and Quick Sort Algorithms can be given as examples of Divide and Conquer technique. You divide the problem into two subproblems and use the algorithm for these subproblems again to sort an array. But, you discard the half of array in binary search. It means you DECREASE the size of array, not divide.
No, binary search is not divide and conquer. Yes, binary search is decrease and conquer. I believe divide and conquer algorithms have an efficiency of O(n log(n)) while decrease and conquer algorithms have an efficiency of O(log(n)). The difference being whether or not you need to evaluate both parts of the split in data or not.