In a list of distinct numbers a1,…an , we say that the elements ai and aj are inverted if i<j but ai>aj. Suppose that all orderings of a1,…an are equally probable under some probability distribution (In other words we shuffled a set of numbers perfectly to obtain the order a1,…,an.) What is the expected number of inversions?
Consider the following greedy algorithm for finding a matching.
Initialize S to an empty set of edges
While there is an edge where both vertices are unmatched by S
Add the above edge to S
What is the running time of this algorithm?
Give an example of a graph where the algorithm does not find the maximum matching.
Show that the matching found by the algorithm always has at least half as many edges as a maximum matching.
(Bonus) Now consider a similar algorithm for finding a maximum weight matching in an edge-weighted graph: Greedily add the heaviest edge possible to the current matching; stop when no further edge can be added. Show that this algorithm finds a matching whose weight is at least half the optimum.
Consider the following factor 2 approximation algorithm for the minimum cardinality vertex cover problem. Find a depth first search tree in the given graph, G, and output the set, say S, of all the non-leaf vertices of this tree. Show that S is indeed a vertex cover for G and |S|≤2⋅OPT.
Hint: Use the tree to construct a matching that includes all the vertices in S.
These solutions may offer step-by-step problem-solving explanations or good writing examples that include modern styles of formatting and construction of bibliographies out of text citations and references. Students may use these solutions for personal skill-building and practice. Unethical use is strictly forbidden.For any pair (i,j) of two elements from the array a1,a2,….an are two possibilities: either i<j and ai<aj or i<j and ai>aj (hence inversion). So the probability for each pair to be an inversion is ½.
But the number of pairs in the array is nC2= n(n-1)/2. So intuitively the total amount of expected inversions would be equal to ½* n(n-1)/2= n(n-1)/4.
However it is necessary to attempt giving a more rigorous proof based on the random probabilities of the inversions after shuffling the values.
We randomly take a pair (i,j) with i<j such that ai>aj (hence it is an inversion).
The indicator for the random variable of the event of being an inversion is ½ because there are only two equally-likely possibilities (as we proved above). So the expected value E[Indicator_i,j]=1/2.
The requested expected number of inversions from the entire array will be equal to the sum of all Indicator_i,j....
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