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Extended Euclidean algorithm

Encyclopedia : E : EX : EXT : Extended Euclidean algorithm

The extended Euclidean algorithm is an algorithm used to calculate the greatest common divisor (gcd, or also highest common factor, HCF) of two integers a and b, as well as integers x and y such that

[ax + by = \gcd(a, b). \,]
(This equation is known as Bezout's identity.) The algorithm is essentially the same as the Euclidean algorithm, but keeps track of the quotient during the computation in order to find x and y as above.

The extended Euclidean algorithm is particularly useful when a and b are coprime, since x is the multiplicative inverse of a modulo b.

Contents

Informal formulation of the algorithm

Dividend Divisor Quotient Remainder
120 23 5 5
23 5 4 3
5 3 1 2
3 2 1 1
2 1 2 0

It's assumed that the reader is already familiar with Euclid's algorithm.

To illustrate the extension of the Euclid's algorithm, consider the computation of gcd(120, 23), which is shown on the table on the left. Notice that the quotient in each division is recorded as well alongside the remainder.

In this case, the remainder in the last line(which is 0) indicates that the gcd is 1; that is, 120 and 23 are coprime (also called relatively prime). For the sake of simplicity, the example chosen is a coprime pair; but the more general case of gcd other than 1 also works similarly.

There are two methods to proceed, both using the division algorithm, which will be discussed separately.

The iterative method

This method attempts to solve the required equation with the sum replaced by the remainders in each step of the algorithm, which is larger than their gcd, but are decreasing in magnitude, and so will eventually become the required equation. To start with, express each line from the last table with the division algorithm, focusing on the remainder:

Remainder = Dividend - Quotient × Divisor
5 = 120
5 × 23
3 = 23
4 × 5
2 = 5

1 × 3
1 = 3

1 × 2
0 = 2
2 × 1

The first line fits the pattern of the original equation, with 120 and 23 as the terms. However from the second line onwards, the terms are decreasing. To proceed, observe that by nature of the Euclidean algorithm:

Thus the terms can be substituted with the previous two line. By mathematical induction, we can then represent each remainder as sum of multiples of the two original numbers, one after another.

Corresponding to this example, notice the divisor on the third line, 3, is the same as the remainder on the second line. Further, the dividend on the third line, 5, is the same as the divisor on the second line, and the remainder on the first line.

Carrying out this induction on the example given:

5 = 120 - 5 × 23 = = 1 × 120 - 5 × 23
3 = 23 - 4 × 5 = 1×23 - 4 × (1×120 - 5×23) =
× 120 + 21 × 23
2 = 5 - 1 × 3 = (1×120 - 5×23) - 1 × (-4×120 + 21×23) = 5 × 120 - 26 × 23
1 = 3 - 1 × 2 = (-4×120 + 21×23) - 1 × (5×120 - 26×23) =
× 120 + 47 × 23

  1. The first line is correct.
  2. In the second line, 23 is okay, but 5 has to be substituted with the results of first line.
  3. In the third line, substitute 5 and 3 with the first and second line.
  4. In the fourth line, substitute 3 and 2 with the second and third line.
The last line read 1 = −9×120 + 47×23, which is the required solution: x = −9 and y = 47.

This also means that −9 is the multiplicative inverse of 120 modulo 23, and that 47 is the multiplicative inverse of 23 modulo 120. Restating it using mathematical expression:

−9 × 120 ≡ 1 (mod 23) and also 47 × 23 ≡ 1 (mod 120).
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The recursive method

This attempt to solve the original equation directly, by reducing the dividend and divisor gradually, from the first line to the last line, which can then be substituted with trivial value and work backward to obtain the solution.

Consider the original equation:

120 x
23 y = 1
(5×23+5) x
23 y = 1
23 (5x+y)
5 x = 1
...
1 a
0 b = 1

Notice that the equation remains unchanged after decomposing the original dividend in terms of the divisor plus a remainder, and then regroup terms. If we have solution to the equation in the second line, then we can work backward to find x and y as required. Although we don't have the solution yet to the second line, notice how the magnitude of the terms decreased(120 and 23 to 23 and 5). Hence, if we keep applying this, eventually we'll reach the last line, which obviously has (1,0) as a trivial solution. Then we can work backward and gradually find out x and y.

Dividend = Quotient x Divisor + Remainder
120 = 5 x 23
5
23 = 4 x 5
3
...

For the purpose of explaining this method, the full working will not be shown. Instead some of the repeating steps will be described to demostrate the principle behind this method.

Start by rewriting each line from the first table with division algorithm, focusing on the dividend this time(because we'll be substituting the dividend).

120 x0
23 y0 = 1
(5×23+5) x0
23 y0 = 1
23 (5x0+y0)
5 x0 = 1
23 x1
5 y1 = 1
(4×5+3) x1
5 y1 = 1
5 (4x1+y1)
3 x1 = 1
5 x2
3 y2 = 1

  1. Assume that we were given x2=2 and y2=-3 already, which is indeed a valid solution.
  2. x1=y2=-3
  3. Solve 4x1+y1=x2 by substituting x1=-3, which gives y1=2-4(-3)=14
  4. x0=y1=14
  5. Solve 5x0+y0=x1 by substituting x0=14, so y0=-3-5(14)=-73

The table method

The table method is probably the simplest method to carry out with a pencil and paper. It is similar to the recursive method, although it does not directly require algebra to use and only requires working in one direction. The main idea is to think of the equation chain [\gcd(x, y), \gcd(y, x\mod y), \dots, \gcd(z, 1)] as a sequence of divisors [x, y, x\mod y, \dots, 1]. In the running example we have the sequence 120, 23, 5, 3, 2, 1. Any element in this chain can be written as a linear combination of the original [x] and [y], most notably, the last element, [\gcd(x, y)], can be written in this way. The table method involves keeping a table of each divisor, written as a linear combination. The algorithm starts with the table as follows:

a b d
1 0 120
0 1 23

The elements in the [d] column of the table will be the divisors in the sequence. Each [d_i] can be represented as the linear combination [d_i = a_i \cdot x + b_i \cdot y]. The [a] and [b] values are obvious for the first two rows of the table, which represent [x] and [y] themselves. To compute [d_i] for any [i > 2], notice that [d_i = d_ \mod d_]. Suppose [d_i = d_ - k \cdot d_]. Then it must be that [a_i = a_ - k \cdot a_] and [b_i = b_ - k \cdot b_]. This is easy to verify algebraicly with a simple substitution.

Actually carrying out the table method though is simpler than the above equations would indicate. To find the third row of the table in the example, just notice that 23 divides 5 times into 120. Now, each value in the table is value two rows above it, minus 5 times the value immediately above it. This correctly leads to [a_3 = 1 - 5 \cdot 0 = 1], [b_3 = 0 - 5 \cdot 1 = -5], and [d_3 = 1 \cdot 120 - 5 \cdot 23 = 5]. After repeating this method to find each line of the table, the final values for [a] and [b] will solve [ax + by = \gcd(x, y)\,]:

a b d
1 0 120
0 1 23
1 -5 5
-4 21 3
5 -26 2
-9 47 1

This method is simple, requiring only the repeated application of one rule, and leaves the answer in the final row of the table with no backtracking.

Formal description of the algorithm

Iterative method

By routine algebra of expanding and grouping like terms(refer to last section), the following algorithm for iterative method is obtained:
  1. Apply Euclidean algorithm, and let qn(n starts from 1) be a finite list of quotients in the division.
  2. Initialize x0, x1 as 1, 0, and y0, y1 as 0,1 respectively.
  3. # Then for each i so long as qi is defined,
  4. # Compute xi+1= xi-1- qixi
  5. # Compute yi+1= yi-1- qiyi
  6. # Repeat the above after incrementing i by 1.
  7. The answer are the second-to-last of xn and yn.
Pseudocode for this method is shown below.

function extended_gcd(a, b)
x := 0    lastx := 1
y := 1    lasty := 0
while b ≠ 0
temp := b
quotient := a div b
b := a mod b
a := temp


temp := x
x := lastx-quotient*x
lastx := temp


temp := y
y := lasty-quotient*y
lasty := temp
return

Recursive method

Solving the general case of the equation in the last corresponding section, the following algorithm results:
  1. If a is divisible by b, the algorithm ends and return the trivial solution x=0, y=1.
  2. Otherwise, repeat the algorithm with b and a modulus b, storing the solution as x' and y'.
  3. Then, the solution to the current equation is x=y', and y = x' minus y' times quotient of a divided by b
Which can be directly translated to this pseudocode: 
function extended_gcd(a, b)
if a mod b = 0
return 
else
temp := extended_gcd(b, a mod b)
x := first(temp)
y := last(temp)
return

Proof of correctness

Let d be the gcd of a and b. We wish to prove that a*x + b*y = d.

See the Euclidean algorithm for the proof that the gcd(a,b) = gcd(b,a mod b) which this proof depends on in the recursive call step.

Computing a multiplicative inverse in a finite field

The extended Euclidean algorithm can also be used to calculate the multiplicative inverse in a finite field.

Pseudocode

Given the irreducible polynomial f(x) used to define the finite field, and the element a(x) whose inverse is desired, then a form of the algorithm suitable for determining the inverse is given by the following. NOTE: remainder() and quotient() are functions different from the arrays remainder[ ] and quotient[ ]. remainder() refers to the remainder when two numbers are divided, and quotient() refers to the integer quotient when two numbers are divided. For example, remainder(5/3) = 2 and quotient(5/3) = 1. Equivalent operators in the C language are % and / respectively.

pseudocode:

remainder[1] := f(x)
remainder[2] := a(x)
auxiliary[1] := 0
auxiliary[2] := 1
i := 2
do while remainder[i] > 1
i := i + 1
remainder[i] := remainder(remainder[i-2] / remainder[i-1])
quotient[i] := quotient(remainder[i-2] / remainder[i-1])
auxiliary[i] := -quotient[i] * auxiliary[i-1] + auxiliary[i-2]
inverse := auxiliary[i]

Note

The minus sign is not necessary for some finite fields in the step.

auxiliary[i] := -quotient[i] * auxiliary[i-1] + auxiliary[i-2]

This is true since in the finite field GF(28), for instance, addition and subtraction are the same. In other words, 1 is its own additive inverse in GF(28).

Example

For example, if the polynomial used to define the finite field GF(28) is f(x) = x8 + x4 + x3 + x + 1, and x6 + x4 + x + 1 =  in big-endian hexadecimal notation, is the element whose inverse is desired, then performing the algorithm results in the following:

i remainder[i]  quotient[i]  auxiliary[i]
 1   x8 + x4 + x3 + x + 1     0
2  x6 + x4 + x + 1    1
3  x2  x2 + 1  x2 + 1
4  x + 1  x4 + x2  x6 + x4 + x4 + x2 + 1
5  1  x + 1  x7 + x6 + x3 + x2 + x2 + x + 1 + 1 

Note: Addition in a binary finite field is XOR.

Thus, the inverse is x7 + x6 + x3 + x = , as can be confirmed by multiplying the two elements together.

References

External links

 

From Wikipedia, the Free Encyclopedia. Original article here. Support Wikipedia by contributing or donating.
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