Introduction
The Extended Euclidean algorithm is an extension to the Euclidean algorithm and computes, in addition to the greatest common divisor of integers a and b, also the coefficients of Bezout's identity which are integers x and y such that
The Extended Euclidean algorithm is an extension to the Euclidean algorithm and computes, in addition to the greatest common divisor of integers a and b, also the coefficients of Bezout's identity which are integers x and y such that
ax + by = gcd(a,b).
The gcd of and b is mentioned in https://kali-mathlessions.blogspot.com/2016/09/section-002-gcd.html
In case of Euclidean algorithm we are interested in only the remainder and not the quotients. Because we are interested in only the value of gcd(a,b) and not the quotients a and b, But in Extended Euclidean algorithm as we are interested in a and b so we need to keep the quotients and these quotients are used to compute the a and b. This is illustrated as below for the same example for which we had computed the gcd in the gcd chapter.
In both cases we proceed by succession of divisions.
Outline of solution
We provide the outline of the solution here
Llet $r_0 = a$ and $r_1=b$ and we have values computed upto $r_n$ then we compute
$r_{n+1} = r_{n-1} - m_{n+1} * r_n\cdots(1)$ and
We repeat the steps until we get $r_{n+1} = 0$
Then $r_n$ is the gcd
Then we repeat using (1) and replacing $r_i$ by combination of $r_{i-1}$ and $r_{i-2}$ for i for i going From n down to 3. and thus we get the result
We wind it by providing an example
Problem 1
Compute a ,b for GCD(1001,104)
Solution 1
We have
The gcd of and b is mentioned in https://kali-mathlessions.blogspot.com/2016/09/section-002-gcd.html
In case of Euclidean algorithm we are interested in only the remainder and not the quotients. Because we are interested in only the value of gcd(a,b) and not the quotients a and b, But in Extended Euclidean algorithm as we are interested in a and b so we need to keep the quotients and these quotients are used to compute the a and b. This is illustrated as below for the same example for which we had computed the gcd in the gcd chapter.
In both cases we proceed by succession of divisions.
Outline of solution
We provide the outline of the solution here
Llet $r_0 = a$ and $r_1=b$ and we have values computed upto $r_n$ then we compute
$r_{n+1} = r_{n-1} - m_{n+1} * r_n\cdots(1)$ and
We repeat the steps until we get $r_{n+1} = 0$
Then $r_n$ is the gcd
Then we repeat using (1) and replacing $r_i$ by combination of $r_{i-1}$ and $r_{i-2}$ for i for i going From n down to 3. and thus we get the result
We wind it by providing an example
Problem 1
Compute a ,b for GCD(1001,104)
Solution 1
We have
- 1001 = 104*9 + 65 so $65 = 1001 - 104 *9 \cdots (1)$
- 104 = 65*1 + 39 so $39 = 104 -65 \cdots (2)$
65 = 39*1 + 26 so $26 = 65 - 39 \cdots (3)$
39 = 26*1 + 13 so so $13 = 39 - 26 \cdots (4)$
26 = 13*2 + 0 - As the remainder is zero so we have the GCD 13.
- Now $13 = 39-26$ (using (4))
- $= 39 - (65-39)$ (using (3))
- $= 2 * 39 - 65$
- $= 2 * (104-65) - 65 $ (using (2))
- $= 2 * 104 - 3 * 65$
- $= 2 * 104 - 3 * (1001 - 104 *9)$ using (1)
- $= 29 * 104 - 3 * 1001$
- $= 1001 * (-3) + 104 * 29 $ (to put in form of ax + by)
- So we get $GCD(1001,104)=13= -3 * 1001 + 29 * 1001$
- So we have for a = 1001 and b = 104 the value of gcd(a,b) = 13, x = -3 and y = 29
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