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Q. Explain the odd-even transposition algorithm?
The algorithm needs one 'for loop' beginning from I=1 to N it implies that N times and for every value of I, one 'for loop' of J is executed concurrently. So the time complexity of the algorithm is O(n) because there are total n phases and every phase executes either even or odd transposition in O(1) time.
Figure: Example
Q. Describe Independent Loops in fortran? HPF offers extra opportunities for parallel execution by employing the INDEPENDENT directive to declare the iterations of a do-loop is
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composition of two shm in right angles to each other to havingg time period in the ratio 1:2
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