Chap. 4: Finite Fields Jen-Chang Liu, 2005 Adapted from lecture slides by Lawrie Brown.
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Transcript of Chap. 4: Finite Fields Jen-Chang Liu, 2005 Adapted from lecture slides by Lawrie Brown.
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Chap. 4: Finite Fields
Jen-Chang Liu, 2005
Adapted from lecture slides by Lawrie Brown
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The next morning at daybreak, Star flew indoors, seemingly keen for a lesson. I said, "Tap eight." She did a brilliant exhibition, first tapping it in 4, 4, then giving me a hasty glance and doing it in 2, 2, 2, 2, before coming for her nut. It is astonishing that Star learned to count up to 8 with no difficulty, and of her own accord discovered that each number could be given with various different divisions, this leaving no doubt that she was consciously thinking each number. In fact, she did mental arithmetic, although unable, like humans, to name the numbers. But she learned to recognize their spoken names almost immediately and was able to remember the sounds of the names. Star is unique as a wild bird, who of her own free will pursued the science of numbers with keen interest and astonishing intelligence.
— Living with Birds, Len Howard
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Introduction Finite fields: increasing importance in
cryptography AES, Elliptic Curve, IDEA, Public Key
Goal: mathematics for GF(2n)
01100…0100
Plaintext: 2n
11010…0111
Ciphertext: 2n
T(Plaintext, Key)
1. Arbitrary mapping => large transform table2. Mathematical form, ex. Hill cipher
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Outline Group, rings, and fields Modular arithmetic Finite fields of the form GF(p), p is a
prime Euclid’s algorithm (find GCD) Polynomial arithmetic Finite fields of the form GF(2n)
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Ring+,-,x
Motivation for abstract math
Abstract math.: 數學物件的集合,運算方式
在一個數學物件的集合中,可操作的運算種類? (+, - , x, /)
Group +,-
Field+,-,x,/
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Group Group: {G, •}
G: a set of elements •: binary operation to each pair (a,b) in G
obeys: closure: a•b is also in G associative law: (a•b)•c = a•(b•c) has identity e: e•a = a•e = a has inverses a-1: a•a-1 = e
if commutative a•b = b•a then forms an abelian group
Ex. Integers and +1+2=3
(1+2)+3=1+(2+3)
3+0=0+3=3
2+(-2)=0
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Cyclic Group Def: exponentiation as repeated
application of operator example: a3 = a•a•a
and let identity be: e=a0
cyclic group every element is a power of some fixed
element i.e. b = ak for some a and every b in group a is said to be a generator of the group
EX. integers with addition: 1 as the generator
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Ring Ring: {R, +, x}
R: a set of “numbers” +,x: two operations (addition &
multiplication)
Ring obeys {R,+}: an abelian group multiplication:
has closure is associative: ax(bxc)=(axb)xc distributive over addition: ax(b+c) = axb + axc
if multiplication operation is commutative, it forms a commutative ring
axb = bxa
set of even integers (pos, neg, and 0)
2x4 = 8
2x(4x6) = (2x4)x6
2x(4+6)=2x4+2x6
trivial
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Field Field: {F, +, x}
F: a set of numbers +,x: two operations (addition &
multiplication) Field obeys:
Ring {F, +}: abelian group for addition {F\0, x}: abelian group for multiplication
(ignoring 0) i.e. with multiplication inverse => division is possible
Ex. Real number, {Z, +, x} 不是,無乘法反元素
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Field for n-bit block?
01100…0100
Plaintext: 2n
11010…0111
Ciphertext: 2n
+: additionx: multiplication
?
Problems: 1. the set of plaintext (and ciphertext) is finite 2. how to define +,-,x,/ operations
Ex. 2-bit input 00011011
如何定義 +,-,x,/ 運算?
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Outline Group, rings, and Fields Modular arithmetic Finite fields of the form GF(p) Euclid’s algorithm (find GCD) Polynomial arithmetic Finite fields of the form GF(2n)
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Modulo operator modulo operator: a mod n to be the
remainder (>0) when a is divided by n
) mod (/ nannaa , a and n are integers
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Modulo operator (cont.) Integers a and b are congruence modulo
n a ≡ b mod n when divided by n, a & b have same
remainder eg. 100 ≡ 34 mod 11
b is called the residue of a: b= a mod n integers can always write: a = qn + b usually have 0 <= b <= n-1 -12 mod 7 = 2
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Integers Modulo 7 Example
... -21 -20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 ...
congruence
Z7: residue classmodulo 7
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Question: (mod n) maps all integers into the set Zn={0, 1, 2, …, n-1} Can we perform arithmetic (+,-,x,/)
with this set?
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Modular Arithmetic We can do modular arithmetic with any
set of integers: Zn={0, 1, … , n-1} Under normal arithmetic (+,-,x) Properties:
1. [(a mod n)+(b mod n)] mod n = (a+b) mod n
2. [(a mod n)- (b mod n)] mod n = (a-b) mod n3. [(a mod n)x (b mod n)] mod n = (axb) mod
n
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Modulo 8 Example: add
FindAdditiveinverse
=> group
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Modulo 8 Example: multiply
Not all haveMulti. Inverse Not a field
Does notproduce allelements inZ8
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Modular Arithmetic (cont.) Peculiarities compared with ordinary
arith. if (a+b)≡(a+c) mod n then b≡c mod n Existence of (-a):
if (ab)≡(ac) mod n then b≡c mod n ? Existence of (a-1) ?
((-a)+a+b)≡((-a)+a+c) mod n
((a-1) ab)≡((a-1) ac) mod n
Not always true, eg. n=8Multiplicative inverse exists iff. a is relatively prime to n
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Question Do we have a finite field within Zn ?
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Modulo 7 example: addition
Yes foradditive inverse
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Modulo 7 Example: multiply
Yes formulti. inverse
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Question Do we have a finite field within Zn ? Galois field
GF(p): ZP ,the modulo p is a prime number Under ordinary arithmetic
GF(pn): , p is a prime number 23 does not form a field under normal
arithmetic Under polynomial arithmetic
npZ
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Outline Group, rings, and Fields Modular arithmetic Finite fields of the form GF(p) Euclid’s algorithm (find GCD) Polynomial arithmetic Finite fields of the form GF(2n)
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Galois Fields: GF(p) Theorem: Zn={0,1,…,n-1} is a commutative
ring. Any integer aZn in has a multiplicative inverse iff a is relatively prime to n
In other words, If n is a prime number, then all nonzero
integers in Zn are relatively prime to n
=> all nonzero integers in Zn have multiplicative
inverses=> Zn is a finite field=> GF(p), p is a prime
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Galois Fields: GF(p) GF(p) is the set of integers {0,1, … , p-
1} with arithmetic operations modulo prime p
these form a finite field since have multiplicative inverses
hence arithmetic is “well-behaved” and can do addition, subtraction, multiplication, and division without leaving the field GF(p)
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Example: GF(2)
+ 0 1
01
0 11 0
x 0 1
01
0 00 1
w -w w-1
01
0 x1 1
XOR AND
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Q: How to find the multiplicative inverse?
For GF(7), it is easy to build a table
For GF(1759), how to find the multiplicative inverse of 550?
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Outline Group, rings, and Fields Modular arithmetic Finite fields of the form GF(p) Euclid’s algorithm (find GCD) and
extended Euclid’s algorithm (find multiplicative inverse)
Polynomial arithmetic Finite fields of the form GF(2n)
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Greatest Common Divisor (GCD)
a common problem in number theory GCD (a,b) of a and b is the largest
number that divides evenly into both a and b eg. GCD(60,24) = 12
GCD(.,.)=1: no common factors (except 1) and hence numbers are relatively prime eg. GCD(8,15) = 1 hence 8 & 15 are relatively prime
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Euclid's GCD Algorithm uses theorem that: (a>b>0)
GCD(a,b) = GCD(b, a mod b)
To prove: The set of common divisors of a and b = The set of common divisors of a and (a mod b)
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Example GCD(1970,1066)1970 = 1 x 1066 + 904 gcd(1066, 904)1066 = 1 x 904 + 162 gcd(904, 162)904 = 5 x 162 + 94 gcd(162, 94)162 = 1 x 94 + 68 gcd(94, 68)94 = 1 x 68 + 26 gcd(68, 26)68 = 2 x 26 + 16 gcd(26, 16)26 = 1 x 16 + 10 gcd(16, 10)16 = 1 x 10 + 6 gcd(10, 6)10 = 1 x 6 + 4 gcd(6, 4)6 = 1 x 4 + 2 gcd(4, 2)4 = 2 x 2 + 0 gcd(2, 0)
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Finding Inverses can extend Euclid’s algorithm:
EXTENDED EUCLID(m, b)1.(A1, A2, A3)=(1, 0, m);
(B1, B2, B3)=(0, 1, b)2. if B3 = 0
return A3 = gcd(m, b); no inverse3. if B3 = 1
return B3 = gcd(m, b); B2 = b–1 mod m4. Q = A3 div B35. (T1, T2, T3)=(A1 – Q B1, A2 – Q B2, A3 – Q B3)6. (A1, A2, A3)=(B1, B2, B3)7. (B1, B2, B3)=(T1, T2, T3)8. goto 2
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Inverse of 550 in GF(1759)
( 商 )
4 1 0
B2=A2-Q*B2=0-3*1=-3
inverse
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Recall: inverse matrix in Hill cipher
Find inverse matrix for
Note that 26 is not a prime, not every element in {0,1,2,…,25} has multiplicative inverse
Inverse formula:
75
49K
921
227
43
1
95
47
4579
11K
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Inverse of Hill cipher
Q A3 B3 26 171 17 91 9 81 8 18 1 0
A2 B2 0 1 1 -1-1 2 2 -3
2515
125
207483
506161
921
22723
921
227
17
11K
Multiplicative inverse
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Outline Group, rings, and Fields Modular arithmetic Finite fields of the form GF(p) Euclid’s algorithm (find GCD) Polynomial arithmetic Finite fields of the form GF(2n)
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Motivation for GF(2n) For a 8-bit block
Z256 ={0,1,…,255} is not a field => no division
The largest prime < 256 is 251 Z251 ={0,1,…,250} is a field => 251,
…,255 are wasted Is that possible to find a field for Z256 ?
Yes. Define new arithmetic operations for Z256
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Mapping from GF(2n) to polynomials
Ex. 8-bit block
10010111
01234567 11101001 xxxxxxxx =>
• To build the field for {0,1,2,…,2n-1}=> Require modular polynomial arithmetic
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Polynomial Arithmetic Polynomials
3 polynomial arithmetic available: ordinary polynomial arithmetic Polynomial with coefficients in Zp; arithmetic
on the coefficients is performed modulo p Polynomial with coefficients in Zp, and the
polynomials are defined modulo a polynomial m(x); arithmetic on the coefficients is performed modulo p
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1. Ordinary Polynomial Arithmetic
Add(+) or subtract(-) corresponding coefficients
Multiply(x) all terms by each other eg. 1101, 0111
f(x) = x3 + x2 + 1 and g(x) = x2 + x + 1f(x) + g(x) = x3 + 2x2 + x + 2f(x) – g(x) = x3 - x f(x) x g(x) = x5+2x4+2x3+2x2+x+1
GF(24) ?1. Not in {0,1} 2. degree>3
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2. Polynomial Arithmetic with Modulo Coefficients
Polynomial coefficients are in a field F={p,+,x}
We are most interested in coefficients in GF(2) eg. 1101, 0111 let f(x) = x3 + x2 + 1 and g(x) = x2 + x + 1f(x) + g(x) = x3 + x f(x) x g(x) = x5 + x + 1GF(24) ?
degree > 3
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3. Modular Polynomial Arithmetic for GF(2n)
Polynomial coefficients are in a field F={p,+,x} Ex. coefficients are in GF(2)
If multiplication results in poly. of degree > n-1 Reduce it by modulo some irreducible poly.
m(x) f(x) = q(x) m(x) + r(x)=> r(x) = f(x) mod m(x)
if m(x) has no divisors other than itself & 1 say it is irreducible (or prime) polynomial
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GF(23) =m(x)
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Modular Polynomial Arithmetic
Computation in field GF(2n) polynomials with coefficients modulo 2 whose degree is less than n hence must reduce modulo an irreducible
poly of degree n (for multiplication only) Can always find an inverse
Extended Euclid’s Inverse algorithm to find
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Computational Considerations
since coefficients are 0 or 1, can represent any such polynomial as a bit string
addition becomes XOR of these bit strings
multiplication is shift & XOR modulo reduction done by repeatedly
substituting highest power with remainder of irreducible poly (also shift & XOR) – see textbook for details
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Summary
01100…0100
Plaintext: 2n
11010…0111
Ciphertext: 2n
Mathematicswithin GF(2n)