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Julli [10]
3 years ago
15

Suppose that instead of swapping element A[i] with a random element from the subarray A[i..n], we swapped it with a random eleme

nt from anywhere in the array: PERMUTE-WITH-ALL (A) 1, n = A.length 2, for i = 1 to n 3, swap A[i] with A[RANDOM(1, n)] Does this code produce a uniform random permutation? Why or why not?
Computers and Technology
1 answer:
Anastaziya [24]3 years ago
7 0

Answer:

The answer to this question can be defined as follows:

Explanation:

Its Permute-with-all method, which doesn't result in a consistent randomized permutation. It takes into account this same permutation, which occurs while n=3. There's many 3 of each other, when the random calls, with each one of three different values returned and so, the value is=  27. Allow-with-all trying to call possible outcomes as of 3! = 6  

Permutations, when a random initial permutation has been made, there will now be any possible combination 1/6 times, that is an integer number m times, where each permutation will have to occur m/27= 1/6. this condition is not fulfilled by the Integer m.

Yes, if you've got the permutation of < 1,2,3 > as well as how to find out design, in which often get the following with permute-with-all  chances, which can be defined as follows:

\bold{PERMUTATION \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \                           PROBABILITY}

\bold{  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4/27= 0.14 }\\\bold{ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5/27=0.18}\\

\bold{            \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \               5/27=0.18}\\\bold{\ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \                       5/27=0.18}

\bold{    \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \                                                  4/27=0.14}

Although these ADD to 1 none are equal to 1/6.

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Explanation:

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The Average time complexity = 1/6 * (n-1)/2 + (n-1)/3 + n/2

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