what is the divide and conquer recurrence that would arise for the problem

Question

suppose you are given n bit integers asuming for common sense n as power of 2 .it is required to multiply them using divide and conquer method .what is the divide and conquer recurrence that would arise for the problem

a) T(n)=4T(n/2)+c      

b) a) T(n)=2T(n/2)+n              

c) a) T(n)=4T(n/2)+n2          

d) a) T(n)=4T(n)+n

Answer 1

According to Gauss method answer should be T(n) = T(n/2) + n

Otherwise T(n) = 4T(n/2) + n

Comments

why you are dividing problem in to 4 parts.can you please explain a bit more

Answer 2

A=(a_n,a_n-1,a_{n-2................}a₀)₂  and B=(b_n,b_n-1,.....................b₀)  be 2 n bit no

then A can be written as A=2^n/2A_msb+A_lsb  

                                    B=2^n/2B_msb+Blsb

 A_msb is the n/2 most significant bit of A ie n/2 leftmost bit of the no.

A_lsb   is the n/2  least significant bit of A ie is n/2 rightmost bit of the no.

so A*B=(2^n/2A_{msb    +}      A_lsb  )(   2^n/2B_{msb     +    Blsb} )   =   2^n*A_msb*B_{msb  +}    2^n/2(A_msb * _Blsb   + A_lsb *B_msb)  +    A_{lsb    Blsb} 

this equation says that mul of 2 n bit no can be carried out using mul of 4 n/2 digit no +some shift operatin 

so the required corresponding recurrence if T(n)= time complexity to mul 2 n bit no

T(n)  =  4*T(n/2)   +Cn

but it can also be solved in T(n)=  3*T(n/2) +cn

for more optimiation ---https://en.wikipedia.org/wiki/Karatsuba_algorithm