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MaRussiya [10]
3 years ago
11

2.2-2 Consider sorting numbers stored in array by first finding the smallest element n A of and exchanging it with the element i

n . Then find the second smallest A AOE1 element of A, and exchange it with . Continue in this manner for the firs AOE2 1 elements of A. Write pseudocode for this algorithm, which is known as selection sort. What loop invariant does this algorithm maintain? Why does it need to run for only the firs elements, rather than for all elements? Give the best-case n 1 n and worst-case running times of selection sort in ‚-notation.
Computers and Technology
1 answer:
True [87]3 years ago
7 0

Answer:

Follows are the explanation of the choices:

Explanation:

Following are the Pseudocode for selection sort:                      

for j = 0 to k-1 do:

      SS = i

      For l = i + 1 to k-1 do:

        If X(l) < X(SS)

          SS= l

        End-If

      End-For

      T = X(j)

      X(j) = X(SS)

      X(SS) = T

    End-For

Following are the description of Loop invariants:

The subarray  A[1..j−1] includes the lowest of the j−1 components, ordered into a non-decreasing order, only at beginning of the iteration of its outer for loop.  

A[min] is the least amount in subarray A[j.. l−1] only at beginning of the each loop-inner iterations.                      

Following are the explanation for third question:

Throughout the final step, two elements were left to evaluate their algorithm. Its smaller in A[k-1] would be placed as well as the larger in A[k]. One last is the large and medium component of its sequence because most and the last two components an outer loop invariant has been filtered by the previous version. When we do this n times, its end is a repetitive, one element-sorting phase.

Following is the description of choosing best-case and worst-case in run- time:

The body the if has never been activated whenever the best case time is the list is resolved. This number of transactions are especially in comparison also as a procedure, that will be (n-1)(((n+2)/2)+4).    

A structure iterator at every point in the worst case that array is reversed, that doubles its sequence of iterations in the inner loop, that is:(n−1)(n+6) Since both of them take timeΘ(n2).

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Suppose that the format for license plates in a certain state is two letters followed by four numbers. (a) How many different pl
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Answer:

(a) 6,760,000 plates

(b) 3,407,040 plates

(c) 6,084,000 plates

Explanation:

The very first thing to note about this question is the number of characters involved in the license plate format (6 characters in this case; 2 letters and 4 numbers). The letters come first and then the numbers follow.

There are a total of 26 possible letters (A-Z) and 10 possible numbers (0 - 9) that can be chosen. We can then proceed to the first question;

(a) Here, the total number of possible plates is to be determined. This is done as follows:

Character 1 (Letter): There are 26 possible letters

Character 2 (Letter): There are 26 possible letters

Character 3 (number): There are 10 possible numbers

Character 4 (number): There are 10 possible numbers

Character 5 (number): There are 10 possible numbers

Character 6 (number): There are 10 possible numbers

So, total number of different plates will be obtained by multiplying all the possibilities: 26 × 26 × 10 × 10 × 10 × 10 = 6,760,000 plates

(b) This second part puts a constraint on the usage of the numbers, unlike the question (a), where there was no constraint at all.

Since there is no constraint on the letters, we can write that:

Character 1 (Letter): There are 26 possible letters

Character 2 (Letter): There are 26 possible letters

For the first number as well, we can write:

Character 3 (number): There are 10 possible numbers

However, for the remaining characters, the possibilities will continually reduce by a value of 1, since we can not use a number that has been used before. So,

Character 4 (number): There are 9 possible numbers

Character 5 (number): There are 8 possible numbers

Character 6 (number): There are 7 possible numbers

So, total number of different plates will be: 26 × 26 × 10 × 9 × 8 × 7 = 3,407,040 plates

(c) Here, repetitions are allowed as in questions (a), but there can not be four zeros. This implies that the maximum number of zeros in any plate will be three. Thus, there will be maximum possibilities on all characters until the last one which will be constrained.

Character 1 (Letter): There are 26 possible letters

Character 2 (Letter): There are 26 possible letters

Character 3 (number): There are 10 possible numbers

Character 4 (number): There are 10 possible numbers

Character 5 (number): There are 10 possible numbers

Character 6 (number): There are 9 possible numbers

Total number of plates will therefore be: 26 × 26 × 10 × 10 × 10 × 9 = 6,084,000 plates.

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Answer:

true I know this cuz I smart

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