Question

Multiply $\mathbf{(}\mathbf{1110}{\mathbf{)}}_{\mathbf{2}}$and$\mathbf{(}\mathbf{1010}{\mathbf{)}}_{\mathbf{2}}$using the fast multiplication algorithm.

Short answer

Thus, the answer is$1000110$

Step-by-step solution

Fast Multiplication of integers definition

The algorithm for the fast multiplication of integers is based on the fact that$\mathit{a}\mathit{b}$can be rewritten as:$\mathit{a}\mathit{b}\mathbf{=}\mathbf{(}{\mathbf{2}}^{\mathbf{2}\mathbf{n}}\mathbf{+}{\mathbf{2}}^{\mathbf{n}}\mathbf{)}{\mathit{A}}_{\mathbf{1}}{\mathit{B}}_{\mathbf{1}}\mathbf{+}{\mathbf{2}}^{\mathbf{n}}\mathbf{(}{\mathbf{A}}_{\mathbf{1}}\mathbf{-}{\mathbf{A}}_{\mathbf{0}}\mathbf{)}\mathbf{(}{\mathbf{B}}_{\mathbf{0}}\mathbf{-}{\mathbf{B}}_{\mathbf{1}}\mathbf{)}\mathbf{+}\mathbf{(}{\mathbf{2}}^{\mathbf{n}}\mathbf{+}\mathbf{1}\mathbf{)}{\mathit{A}}_{\mathbf{0}}{\mathit{B}}_{\mathbf{0}}\mathbf{.}$

Use Fast Multiplication

In the documentation of Example 4 (all numerals in base 2), duplicate $\text{a}=1110$by $b=1010$. Take note of that $\text{n}=2$.

In this manner $A_1=11,A_0=10,B_1=10$and $B_0=10$.

Shape $A_1-A_0=11-10=01$and $B_1-B_0=00$.

The accompanying three items: $A_1{B}_1=\left(11\right)\left(10\right),\left(A_1-A_0\right)\left(B_0-B_1\right)=\left(01\right)\left(00\right)$and $A_0{B}_0=\left(10\right)\left(10\right)$.

With a specific end goal to from these items, the calculation would in actuality recurse, accepting rather that we have these answers, in particular

$A_1{B}_1=0110,\left(A_1-A_0\right)\left(B_0-B_1\right)=0000$, also, $A_0{B}_0=0100$

Move these items of different quantities of spots to one side.

Move $A_1{B}_{1}2n=4$places and furthermore $n=2$places, acquiring $01100000$and $011000$.

Move $\left(A_1-A_0\right)\left(B_0-B_1\right)\text{n}=2$ places, acquiring $000000$, and move $A_0{B}_0\text{n}=2$places and furthermore no spots, getting $010000$and$0100$.

Include every one of the five of these paired numbers, getting $1000110$.

Thus, the answer is$1000110$.