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+28 votes

The minimum number of arithmetic operations required to evaluate the polynomial $P(X) = X^5+4X^3+6X+5$ for a given value of $X$, using only one temporary variable is ______.

+62 votes

Best answer

0

to get 6 as answer you will require more than 1 temporary variavle, but in qsn it has mentioned tha only 1 temp variable is available.

+2

i think i get it

x((x^{2}((x^{2})+4))+6)+5 // x^{2} is done 1st

then x((x^{2}((x^{2})+4))+6)+5 now we lost x^{2}

so need to do x^{2} again

is it correct

+5

see, if you even replace x=x^{2}, you have to first store that x in some temp, because x is used in outside last bracket **x**((x^{2}((x^{2})+4))+6)+5 ,

the answer would be 6, if in place of last **x**, it would be **x ^{2}**

try to write in a form of 3 addr code, you will understand what is happening.

+1

Given value of X, using only one temporary variable

Hi @Anu007 ji, I think what ever you are saying looks correct to me.

Answer could be 7 if nothing is given. But explicitly they are saying one temporary variable which means we can store $X^2$ value and avoid recomputation.

ping @kenzou, @joshi_nitish

0

If one could store in memory, it doesn't make sense to mention one temporary variable.Storing in memory is equivalent to having infinite temporaries.One could get answer 6 if storing in memory is allowed.

+21

This will be the code according to answer ...

1. T= x

2. T = T * T [$x^{2}$]

3. T = T + 4 [$x^{2}$ + 4 ]

4. T = T *x [$x^{3} + 4x$]

5. T = T * x [$x^{4} + 4x^{2}$]

6. T = T + 6 [$x^{4} + 4x^{2}+ 6$]

7. T = T * x [$x^{5} + 4x^{3}+ 6x$]

8. T = T + 5 [ $x^{5} + 4x^{3}+ 6x +5$ ]

No of operation = 7

We hav to evaluate that expression using * single temporary variable * .....

0

@Anu007 Yes the answer should be 6. We can take a temporary variable and store 'x^{2' }in it, then this can be done in 6 operations only

+19 votes

Multiplications, 3 Additions]

So answer is 7 arithmetic operations.

+8 votes

^{5}+4x^{3}+6x+5 = x^{3}(x^{2}+4)+6x+5

x^{3} -> 3 multiply operations (also gives x^{2})

So, Totally we need 3 + 1 multiplications and 3 additions = 7 operations

+3 votes

P(X) = X^{5}+4X^{3}+6X+5

RHS

X(X^{4}+4X^{2}+6)+5

X(((X^{2})^{2}+2.2.X^{2}+2^{2})+2)+5

X((X^{2}+2)^{2}+2)+5

Now:

T=X*X

T=T+2

T=T*T

T=T+2

T=T*X

T=T+5

Therefore - (* operator used) =3, (+ operator used) = 3 Total=6

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