Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Mystery(arr, lb, ub) if(arr[lb]>arr[up]) swap...
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Mystery(arr, lb, ub)
if(arr[lb]>arr[up])
swap(arr[lb], ar[ub])
if(lb+1>=ub)
then return
t=floor((ub-lb+1)/3)
Mystery(arr, lb, ub-t)
Mystery(arr, lb+t, ub)
Mystery(arr, lb, ub-t)
Considering the recursive algorithm provided:
Derive a recurrence relation for the worst-case running time of this algorithm and calculate its time complexity using Master's Theorem.
Given: lb is lower bound and ub is upper bound of array arr.?
if(arr[lb]>arr[up])
swap(arr[lb], ar[ub])
if(lb+1>=ub)
then return
t=floor((ub-lb+1)/3)
Mystery(arr, lb, ub-t)
Mystery(arr, lb+t, ub)
Mystery(arr, lb, ub-t)
Considering the recursive algorithm provided:
Derive a recurrence relation for the worst-case running time of this algorithm and calculate its time complexity using Master's Theorem.
Given: lb is lower bound and ub is upper bound of array arr.?
Most Upvoted Answer
Mystery(arr, lb, ub) if(arr[lb]>arr[up]) swap(arr[lb], ar[ub]) if(lb+1...
Recurrence Relation Derivation
To analyze the recursive algorithm, we define T(n) as the worst-case running time, where n is the number of elements in the array defined by the bounds lb and ub.
1. Base Case:
- If lb + 1 >= ub, the algorithm terminates. Thus, T(1) = O(1) for a single element.
2. Recursive Case:
- The algorithm performs a swap and then makes three recursive calls:
- Mystery(arr, lb, ub-t)
- Mystery(arr, lb+t, ub)
- Mystery(arr, lb, ub-t)
- The size of the problem reduces by approximately t (floor((ub-lb+1)/3)), where t is a third of the current size.
Thus, the recurrence relation can be expressed as:
T(n) = T(n - t) + T(n - t) + T(t) + O(1)
This simplifies to:
T(n) = 2T(n - t) + T(t) + O(1)
Considering t is approximately n/3, we can rewrite this as:
T(n) = 2T(2n/3) + O(1)
Time Complexity Calculation using Master’s Theorem
To apply Master’s Theorem:
- We have:
- a = 2
- b = 3/2
- f(n) = O(1)
We need to evaluate n^(log_b(a)):
- log_b(a) = log_(3/2)(2)
- Since f(n) is polynomially smaller than n^(log_b(a)), we fall into case 1 of the Master’s Theorem.
Thus, we conclude:
T(n) = Θ(n^(log_(3/2)(2)))
This implies that the time complexity is approximately O(n^0.585), indicating a sub-linear growth rate relative to n.
Conclusion
The recursive algorithm's worst-case time complexity is O(n^0.585), showcasing its efficiency in processing the array within defined bounds.
To analyze the recursive algorithm, we define T(n) as the worst-case running time, where n is the number of elements in the array defined by the bounds lb and ub.
1. Base Case:
- If lb + 1 >= ub, the algorithm terminates. Thus, T(1) = O(1) for a single element.
2. Recursive Case:
- The algorithm performs a swap and then makes three recursive calls:
- Mystery(arr, lb, ub-t)
- Mystery(arr, lb+t, ub)
- Mystery(arr, lb, ub-t)
- The size of the problem reduces by approximately t (floor((ub-lb+1)/3)), where t is a third of the current size.
Thus, the recurrence relation can be expressed as:
T(n) = T(n - t) + T(n - t) + T(t) + O(1)
This simplifies to:
T(n) = 2T(n - t) + T(t) + O(1)
Considering t is approximately n/3, we can rewrite this as:
T(n) = 2T(2n/3) + O(1)
Time Complexity Calculation using Master’s Theorem
To apply Master’s Theorem:
- We have:
- a = 2
- b = 3/2
- f(n) = O(1)
We need to evaluate n^(log_b(a)):
- log_b(a) = log_(3/2)(2)
- Since f(n) is polynomially smaller than n^(log_b(a)), we fall into case 1 of the Master’s Theorem.
Thus, we conclude:
T(n) = Θ(n^(log_(3/2)(2)))
This implies that the time complexity is approximately O(n^0.585), indicating a sub-linear growth rate relative to n.
Conclusion
The recursive algorithm's worst-case time complexity is O(n^0.585), showcasing its efficiency in processing the array within defined bounds.
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