1.
A.) Start with a guess for the square root, x0. We can choose x0 = n/2.
Compute x1 = (x0 + n/x0)/2.
If x1 is close enough to x0 then we can stop and return x1 as the result. Otherwise, we
can set x0 = x1 and go back to the previous step
It is possible to compute the square root of a positive integer n by beginning with a
guess for the square root, which is denoted by x0. We can go with x0 = n/2 as our
option. After that, you compute x1 as the following: x1 = (x0 + n/x0)/2. If the distance
between x1 and x0 is small enough (for instance, if the difference between them is less
than 0.001), then we can call x1 the outcome. We can simply reset x0 to equal x1 and
continue the process.
B.)
Set x0 = n/2.
Set x1 = (x0 + n/x0)/2.
If the difference between x1 and x0 is less than 0.001, return x1.
Otherwise, set x0 = x1 and go back to step 2.
Algorithm begins by making a best guess for the value of x0, which is the square root.
Then, it takes the average of x0 and n/x0 to come up with a new guess called x1, which
it computes. If this new guess is sufficiently similar to the one that came before it, then
we will use that one as the result. If this is not the case, we will set x0 to equal x1 and
continue the operation
2.
A).
Euclid 2 ( Mn )
algorithm
data
substations rather
uses
integer
than
)
Algorithm Euclid ( Mn
compute god Cnn )
Input
:
nonnegative
The
greatest
while
it
by
Two
output !
code !
based
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n
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m
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of two integers ( numbers) , the largest number that divides them both without a
reminder.
board
of
to
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