5. Here are some problems on NP-completeness and intractability. (a) The decisio

Question

5. Here are some problems on NP-completeness and intractability. (a) The decision version of the INTERVAL SCHEDULING problem is the following INTERVAL SCEDULING DECISION (ISD) Input: A set I of intervals, a positive integer k output: Is there is subset I' of pairwise non-overlapping intervals of size at least The decision version of the MAxIMUM INDEPENDENT SET problem is the following. MAXIMUM INDEPENDENT SET DECISION (MISD) Inputs A set G (V, E), a positive integer k. Output: Is there is an independent set V SV of size at least k? Your task is to prove that ISD SP MISD

Explanation / Answer

The Interval Scheduling Decision (ISD) is basically designed to elaborate the reduction in the polynomial time, for seraching and finding the maximum-sized clique.

Let me explain you in step-by-step manner:-

A) prove ISD <p MISD

Step-1:

The initial step is to know the inputs which is consist of intervals on the given line which is represented in the form of pair of points.

Step-2:

The next step is to find the output for the searching the largest intervals, which is not overlap to any two of them at a given point of time.

Step-3:

The next step is to find the output for the searching the largest intervals, which is not overlap to any two of them at a given point of time.

Proof:-

// This function is used to pass the input request as parameter

function IntervalMisd(inputReq)
schedule gets {}
  
   // It uses the While loop statement for checking the highest and lowest finishing time over the period
  
while inputReq is not yet empty
choose a request i_out in inputReq that has the lowest finishing time
schedule gets schedule cup {i_out}
delete all inputReq in inputReq that are not compatible with i_out
end
  
   // Finally, It returns the schedule value
  
return schedule
end

Time Complexity:- O(n log n + n) = O(n log n)

B)

The polynomial time reduction can be obtained using three different ways, i.e, as follow:-

Syntax:- X < mp Y

2. The next step is can be reducted using the truth-table reduction transformation of the following inputs to problem X into a fixed no. of inputs to problem Y, and finally gives the function as the problem Y,

Syntax:- X < ttp Y

3. The last step for the polynomial time reduction is the Turing reduction is can be calculated by the same problem from X and Y, and it will call the subroutine as an output of the program.

Syntax:- X < Tp Y

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