Answer:
Initially if x = a = 2437 and y = 875, condition (1) that is x>y is true and output would be x = 1562 with y = 87
Explanation:
According to pseudo code the code will run until x becomes equal to y. Initially if x = a = 2437 and y = 875, condition (1) that is x>y is true. So next step would be x=x-y and output would be x = 1562 with y = 875. With initial values x = a = 2437 and y = 875, x becomes equal to y after 21 iterations and output is x=y=1
X Y
Algorithms are just prototypes of the original program. The output of the algorithm in step 6 is 1.
The given algorithm performs a repeated operation, until the values of x and y are the same
Initially;
[tex]a = 2437 \\b = 875[/tex]
So, the initial values of x and y (as stated in step 2) are:
[tex]x = 2437 \\y = 875[/tex]
When x > y, the following operation is executed
[tex]x = x - y[/tex]
[tex]x = 2437 - 875[/tex]
[tex]x = 1562[/tex]
Next, the iteration checks if x = y.
This is false, so the iteration further checks if x > y
This is true, so the following operation is repeated.
[tex]x = x - y[/tex]
[tex]x = 1562 - 875[/tex]
[tex]x = 687[/tex]
Next, the iteration checks if x = y.
This is false, so the iteration checks if x > y
This is false, so the following operation is executed.
[tex]y = y -x[/tex]
[tex]y = 875-687[/tex]
[tex]y = 188[/tex]
The iteration is repeated until the values of x and y are the same (see attachment for the values of x and y, in each iterating process).
At the end of the iteration, the values of x and y is 1.
Hence, the output of the algorithm in step 6 is 1.
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