SumArray (A[1..N]): sum = 0 i = 1 for (i < N) : { sum = sum + A[i] i=i+ 1 } return sum
SumArray (A[1..N]): sum = 0 i = 1 for (i < N) : { sum = sum + A[i] i=i+ 1 } return sum
Related questions
Question
I need help question 2
![Problem 2: Formulating a loop invariant for the purpose of proving the correctness of an
iterative algorithm can be a bit of a challenge. However, for the following algorithm, finding a
loop invariant is not difficult. Here is the algorithm:
SumArray (A[1..N]):
sum = 0
i = 1
for (i < N) :
{
sum = sum + A[i]
i = i + 1
}
return sum
(a) State a loop invariant the can be used to prove the correctness of this algorithm (correct
calculation of the sum of all values in A[1..N]).
(b) Show the prove using your invariant in three steps: initialization, maintenance, and
termination.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbb9a233-ccbf-4977-a45b-ea6968028606%2F60a39997-5e0e-43f8-a443-15ec09c5fa18%2Fxd09fe_processed.png&w=3840&q=75)
Transcribed Image Text:Problem 2: Formulating a loop invariant for the purpose of proving the correctness of an
iterative algorithm can be a bit of a challenge. However, for the following algorithm, finding a
loop invariant is not difficult. Here is the algorithm:
SumArray (A[1..N]):
sum = 0
i = 1
for (i < N) :
{
sum = sum + A[i]
i = i + 1
}
return sum
(a) State a loop invariant the can be used to prove the correctness of this algorithm (correct
calculation of the sum of all values in A[1..N]).
(b) Show the prove using your invariant in three steps: initialization, maintenance, and
termination.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps
