multiplication of two numbers in 6 of n/3 bits 6T(n/3) karatsuba

问题

multiplication of two numbers

x*y ----> x =(x0*10^(n/3)+x1*10^(n/3)+x2) and y=(y0*10^(n/3)+y1*10^(n/3)+y2)

It is 9 multiplication of 10^n/3 numbers so 9T(n/3) but it can be reduced to 5 by the following method.

x*y= x0*y0+x1*y1+x2*y2+x0*y1+x0*y2+x1*y0+x1*y3+x2*y0+x2*y1

I am able to reduce the multiplication of two numbers to 5T(n/3) by similar trick like Karatsuba's algorithm

(x0+x1+x2)(y0+y1+y2)-x0*y0-x1*y1-x2*y2= x0*y1+x0*y2+x1*y0+x1*y3+x2*y0+x2*y1

All and all 5 multiplication of n/3 bits number so 5T(n/3)+O(n) but how can I do in 6 multiplication like 6T(n/3)+O(n)

Q1 Can it be reduced to 6 instead of 5 ?

[edit1] correction copied from the reasked question by Spektre

x and y has n bits

x=x0*(10^2n/3)+x1*10^n/3+x2
y=y0*(10^2n/3)+y1*10^n/3+y2
x*y=x2y2+(x2y1+x1y2)10^n/3+(x2y0+x1y1+x0y2)10^2n/3+(x1y0+x0y1)10^n+x0y0*10^4n/3

now 9 multiplication of n/3 bit numbers running time 9T(n/3) which is O(n^2)

with little trick like Karatsuba's multiplication:

first calculate x0y0, x1y1 and x2y2 this is 3 multiplication of n/3 bit numbers
then use x0y0, x1y1 and x2y2 to calculate the others follow:
x2y1+x1y2=(x1+x2)(y1+y2)-x1y1-x2y2 ->> 1 multiplication of n/3 bit number
x2y1+x1y1+x0y2=(x0+x2)(y0+y2)-x0y0-x2y2+x1y1 ->> 1 multiplication of n/3 bit number
x1y0+x0y1=(x0+x1)(y0+y1)-x0y0-x1y1 ->> 1 multiplication of n/3 bit number

Recursively solve the 6 subproblems

and combine them with 7 additions on O(n)-digit numbers.
total it takes now 6 multiplication of n/3 bit numbers running time 6T(n/3)

Q2 how can I reduce this to 5 multiplication instead of 6 ?

Q1 is now obsolete due to mistake in OP

回答1:

Not an answer but this would be unreadable as comment

if you divide digits to 3 groups then greybeard is right

a0=10^(2n/3)
a1=10^(1n/3)
a2=10^(0n/3)
x=x0*a0+x1*a1+x2*a2
y=y0*a0+y1*a1+y2*a2

so multiplication looks like this instead:

x*y=a0(x2*y2+x1*y2+x0*y2)
    +a1(x2*y1+x1*y1+x0*y1)
    +a2(x2*y0+x1*y0+x0*y0)

rewrite to digits:

b0=(x2*y2+x1*y2+x0*y2)
b1=(x2*y1+x1*y1+x0*y1)
b2=(x2*y0+x1*y0+x0*y0)
x*y=a0*b0+a1*b1+a2*b2

and now simplify b0,b1,b2

[hints]

what trick did you exploit?
why is in your result no digit weight a0,a1,a2?

[edit1] if you want to solve this then use standard notation instead of yours

lower significant digits are lower index so:

a0=10^(0n/3)
a1=10^(1n/3)
a2=10^(2n/3) // this is max wieght for operand
a3=10^(3n/3)
a4=10^(4n/3)
a5=10^(5n/3) // this is max weight for multiplication result
x=x0*a0+x1*a1+x2*a2
y=y0*a0+y1*a1+y2*a2

so multiplication looks like this instead:

x*y=a0(x0*y0+x1*y0+x2*y0)
    +a1(x0*y1+x1*y1+x2*y1)
    +a2(x0*y2+x1*y2+x2*y2)

rewrite to digits:

b0=(x0*y0+x1*y0+x2*y0)
b1=(x0*y1+x1*y1+x2*y1)
b2=(x0*y2+x1*y2+x2*y2)
x*y=a0*b0+a1*b1+a2*b2

of coarse b0,b1,b2 are not in range of single digit - instead they are all double digit due to multiplication

so you have to split them into singular digits:

x*y=a0*B0+a1*B1+a2*B2+a3*B3+a4*B4+a5*B5

where:

B0=(b0/a0)%a1
B1=(b0/a1+b1/a0)%a1
B2=(b0/a2+b1/a1+b2)%a1
B3=(b0/a3+b1/a2+b2/a1)%a1
B4=(b0/a4+b1/a3+b2/a2)%a1
B5=(b0/a5+b1/a4+b2/a3)%a1

and also do not forget to handle overflows...

来源：https://stackoverflow.com/questions/26209076/multiplication-of-two-numbers-in-6-of-n-3-bits-6tn-3-karatsuba

标签

algorithm

math

multiplication