In Section 1.2 .3, we studied Euclid's algorithm for...

In Section $1.2 .3,$ we studied Euclid's algorithm for computing the greatest common divisor (ged) of two positive integers: the largest integer which divides them both. Here we will look at an alternative algorithm based on divide-and-conquer. (a) Show that the following rule is true. \[ \operatorname{gcd}(a, b)=\left\{\begin{array}{ll} 2 \operatorname{gcd}(a / 2, b / 2) & \text { if } a, b \text { are even } \\ \operatorname{gcd}(a, b / 2) & \text { if } a \text { is odd }, b \text { is even } \\ \operatorname{gcd}((a-b) / 2, b) & \text { if } a, b \text { are odd } \end{array}\right. \] (b) Give an efficient divide-and-conquer algorithm for greatest common divisor. (c) How does the efficiency of your algorithm compare to Euclid's algorithm if $a$ and $b$ are $n$ -bit integers? (In particular, since $n$ might be large you cannot assume that basic arithmetic operations like addition take constant time.)

In Section $1.2 .3,$ we studied Euclid's algorithm for computing the greatest common divisor (ged) of two positive integers: the largest integer which divides them both. Here we will look at an alternative algorithm based on divide-and-conquer.
(a) Show that the following rule is true.
\[
\operatorname{gcd}(a, b)=\left\{\begin{array}{ll}
2 \operatorname{gcd}(a / 2, b / 2) & \text { if } a, b \text { are even } \\
\operatorname{gcd}(a, b / 2) & \text { if } a \text { is odd }, b \text { is even } \\
\operatorname{gcd}((a-b) / 2, b) & \text { if } a, b \text { are odd }
\end{array}\right.
\]
(b) Give an efficient divide-and-conquer algorithm for greatest common divisor.
(c) How does the efficiency of your algorithm compare to Euclid's algorithm if $a$ and $b$ are $n$ -bit integers? (In particular, since $n$ might be large you cannot assume that basic arithmetic operations like addition take constant time.)

Transcript

00:01 All right, so here we have that a and b are positive integers.

00:38 And i'm just going to go ahead and assume that a is greater than or equal to b.

00:42 One of them should be greater than or equal to the other.

00:45 And we want to prove that the greatest common divisor of 2a to the power of a minus 1 and 2 to the power of b minus 1 is equal to 2 to the...

01:05 All right, so here we have that a and b are positive integers.

01:12 And i'm just going to go ahead and assume that a is greater than or equal to b.

01:16 And one of them should be greater than or equal to the other.

01:20 And we want to prove that the greatest common device, so that r is going to be equal to a mod b.

01:53 Okay, so that means that the gcd of a and b, being equal to the gcd of b and o.

02:00 R is also equal to the gcd of b, a, such that a is equal to b times q plus r.

02:16 Okay, and by the euler division method, so we get that the euclidean method, i apologize, the gcd of a and b is equal to the gcd of b and r.

02:37 We also know that r is going to be equal to modus b.

02:44 So we're just going to rewriting that there.

02:48 Well, so we can use that in our equation, right? so if we're looking at the gcd of 2 to the power of a minus 1, and 2 to the power of b minus 1, that is going to be equal to the greatest common advisor of 2 to the power of b minus 1, a mod b...

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