AES S-box How and Why?. Notes The general math review slides are taken from a variety of internet...

AES S-box How and Why?

Notes

• The general math review slides are taken from a variety of internet sources. I tried to be diligent in citing, but given the nature of basic math, I may have missed some citation

• This presentation assumes you are comfortable with symmetric ciphers as well as the details of AES.

Basic Math Review: Represent decimals as polynomials

Basic Math Review: Fields, groups, rings• In mathematics, and more specifically in abstract algebra, the term

algebraic structure generally refers to a set with one or more operations defined on it

• A group is an algebraic system consisting of a set, an identity element, one operation and its inverse operation.

• A ring is an algebraic system consisting of a set, an identity element, two operations and the inverse operation of the first operation.

• A field is an algebraic system consisting of a set, an identity element for each operation, two operations and their respective inverse operations.

• GF(p) for any prime, p, this Galois Field has p elements which are the residue classes of integers modulo p.

5

Basic Math Review: Galois Fields

The order of the field is given by pn while p is called the characteristic of the field example: gf(5) = (0; 1; 2; 3; 4)which consists of 5 elements where each of them is a polynomial of degree 0(a constant) whilegf(23) = (0; 1; 2; 2 + 1; 22; 22 + 1; 22 + 2; 22 + 2 + 1) = (0; 1; 2; 3; 4; 5; 6; 7)which consists of 23 = 8 elements where each of them is a polynomial of degree at most 2 evaluated at 2.

Basic Math Review: Galois Field Addition

Basic Math Review: Galois Field Multiplication

Basic Math Review: First refresh your memory on polynomial multiplication

Basic Math Review: Polynomial Division

For additional review of basic polynomial math http://www.doc.ic.ac.uk/~mrh/330tutor/ch04s02.html

http://www.doc.ic.ac.uk/~mrh/330tutor/ch04s02.html

Basic Math Review: Galois Field Multiplication

Basic Math Review: Polynomial Arithmetic

• Can compute using polynomials

• Several alternatives available• ordinary polynomial arithmetic• poly arithmetic with coords mod p


• add or subtract corresponding coefficients

• multiply all terms by each other

• For example:• let f(x) = x3 + x2 + 2 and g(x) = x2 – x + 1f(x) + g(x) = x3 + 2x2 – x + 3f(x) – g(x) = x3 + x + 1f(x) x g(x) = x5 + 3x2 – 2x + 2


• when computing value of each coefficient do calculation modulo some value could be modulo any prime but we are most interested in mod 2

• ie all coefficients are 0 or 1• eg. let f(x) = x3 + x2 and g(x) = x2 + x + 1f(x) + g(x) = x3 + x + 1f(x) x g(x) = x5 + x2


• can write any polynomial in the form:• f(x) = q(x) g(x) + r(x)• can interpret r(x) as being a remainder• r(x) = f(x) mod g(x)

• if have no remainder say g(x) divides f(x)

• if g(x) has no divisors other than itself & 1 say it is irreducible (or prime) polynomial

• arithmetic modulo an irreducible polynomial forms a field


• can compute 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)

• Form a finite field


• Find the results of

(x5+x2+x) * (x7+x4+x3+x2+x )

in GF(28) with irreducible polynomial x8+x4+x3+x+1

Basic Math Review: Polynomial Arithmetic Answer

• Multiply the two polynomials • (x5+x2+x) * x7+x4+x3+x2+x = x5 * (x7+x4+x3+x2+x ) + x2 * (x7+x4+x3+x2+x ) + x *

(x7+x4+x3+x2+x ) = (x12+x7+x2)

• Get the results of• (x12+x7+x2) mod (x8+x4+x3+x+1) = (x5+x3+x2+x+1)

1

1Re1

xxx1xxxx

4

235

348

2458

457812

2712348

x

xxxxmainderxxxx

xxxx

xxxxx

Basic Math Review: Polynomial Arithmetic Polynomial Algebra

• Operation of addition is performed using an XOR operation denoted by . For example, all notations below are equivalent:

(x6 + x4 + x2 + x + 1) + (x7 + x + 1) = x7 + x6 + x4 + x2 + 0 [polynomial notation];

{01010111} {10000011} = {11010100} [binary notation].

• Multiplication in Rijndael is the multiplication of polynomials modulo the irreducible polynomial. For example, in the polynomial notation:

(x6 + x4 + x2 + x + 1) • (x7 + x + 1)

= x13 + x11 + x9 + x8 + x6 + x5 + x4 + x3 + 1,

and

(x13 + x11 + x9 + x8 + x6 + x5 + x4 + x3 + 1) mod (x6 + x4 + x2 + x + 1) = x7 + x + 1.

19

The set of 256 possible byte values, with XOR used as addition, and the multiplication defined as above, has the structure of the finite field GF(28).

Basic Math Review: Matrix multiplication

• Here is a key point: You cannot just multiply each number by the corresponding number in the other matrix. Matrix multiplication is not like addition or subtraction.

From: http://www.freemathhelp.com/matrix-multiplication.html

Basic Math Review: Matrix Multiplication Continued

• The first two steps


Basic Math Review: Matrix Multiplication• Steps 3 & 4


Basic Math Review: Matrix Multiplication• Step 5


Basic Math Review: Matrix Multiplication

Matrix Multiplication is NOT Commutative! Order matters!

You can multiply matrices only if the number of columns in the first matrix equals the number of rows in the second matrix.

2 3

5 6

9 7

2 columns2 rows

1 2 0

3 4 5

Basic Math Review: Matrix Determinants

To find the determinant of a 2 x 2 matrix, multiply diagonal #1 and subtract the product of diagonal #2.

3 1

2 4

Diagonal 1 = 12

Diagonal 2 = -2

12 ( 2) 14

Basic Math Review: Matrix Determinants

To find the determinant of a 3 x 3 matrix, first recopy the first two columns. Then do 6 diagonal products.

5 2 6

2 1 4

3 3 4

5 2

2 1

3 3-20 -24 36

18 60 16

Basic Math Review: Inverse Matrices

When you multiply a matrix and its inverse, you get the identity matrix.

3 1

5 2

2 1

5 3

1 0

0 1


Not all matrices have an inverse! To find the inverse of a 2 x 2 matrix,

first find the determinant.a) If the determinant = 0, the inverse does

not exist! The inverse of a 2 x 2 matrix is the

reciprocal of the determinant times the matrix with the main diagonal swapped and the other terms multiplied by -1.


Example 1: A 3 1

5 2

det(A) 6 (5) 1

A 1 1

1

2 1

5 3

2 1

5 3

Basic Math Review: Scalar Multiplication

To do this, multiply each entry in the matrix by the number outside (called the scalar). This is like distributing a number to a polynomial.

2 4

4 5 0

1 3

8 16

20 0

4 12

Example:

AES – Rijndael mathematics• Rijndael is defined in the Galois field GF (28)

by the irreducible polynomial

• In HEX this is 11B, in binary it is 100011011• Why this one?” The polynomial m(x) (‘11B’) for the

multiplication in GF(28) is the first one of the list of irreducible polynomials of degree 8, given in [LiNi86, p. 378].” page 22 of AES Proposal: Rijndael

• LiNi86 is R. Lidl and H. Niederreiter, Introduction to finite fields and their applications, Cambridge University Press, 1986)

P = x8 + x4 +x3 +x +1

Irreducible PolynomialsA polynomial is irreducible in GF(p) if it does not factor over GF(p). Otherwise it is reducible.

Examples:

The same polynomial is reducible in Z5 but irreducible in Z2.

Are there other irreducible polynomials of power 8? Yes of course, the only list I am aware of is in the book cited by the inventors of Rijndael

Irreducible PolynomialsYou can check the same source that was cited by the inventors of Rijndael. In fact on Amazon, there is the ‘look inside’ option and with a bit of work you can find the table. To save you some time, here are a few irreducible polynomials from that list (in binary form, you may place them in polynomial or hex form if you wish)100101011100111001100111111101001101101011111101110111110001011

OK why degree 8 (9 digits) isn’t that one too many? “Clearly, the result will be a binary polynomial of degree below 8. Unlike for addition, there is no simple operation at byte level.” – page ¾ of the specification

The reason an irreducible but not primitive polynomial is used is that we are trying to make a non-linear permutation function that has diffusion, spreading input bits to output bits in an non-linear way.

The Finite Field GF(28).

• The case in which n is greater than one is much more difficult to describe. In cryptography, one almost always takes p to be 2 in this case. This section just treats the special case of p = 2 and n = 8, that is. GF(28), because this is the field used by Advanced Encryption Standard (AES).

• The AES works primarily with bytes (8 bits), represented from the right as:

• b7b6b5b4b3b2b1b0.The 8-bit elements of the field are regarded as polynomials with coefficients in the field Z2:

• b7x7 + b6x6 + b5x5 + b4x4 + b3x3 + b2x2 + b1x1 + b0.The field elements will be denoted by their sequence of bits, using two hex digits.

The multiplicative inverse

• Multiplication in Galois Field, however, requires more tedious work. Suppose f(p) and g(p) are polynomials in gf(pn) and let m(p) be an irreducible polynomial (or a polynomial that cannot be factored) of degree at least n in gf(pn). We want m(p) to be a polynomial of degree at least n so that the product of two f(p) and g(p) does not exceed 11111111 = 255 as the product needs to be stored as a byte. If h(p) denotes the resulting product then

• h(p) = (f(p) * g(p)) (mod m(p))• On the other hand, the multiplicative inverse of f(p) is given by a(p) such• that• (f(p) * a(p)) (mod m(p)) = 1

The multiplicative inverseNote that calculating the product of two polynomials and the multiplicative inverse of a polynomial requires both reducing coeficients modulo p and reducing polynomials modulo m(p). The reduced polynomial can be calculated easily with long division while the best way to compute the multiplicative inverse is by using Extended Euclidean Algorithm. The details on the calculations in gf(28) is best explained in the following example.

Princeton University offers this calculator for multiplicative inverse http://www.cs.princeton.edu/~dsri/modular-inversion-answer.php?n=9&p=8 We will look at two methods for calculating multiplicative inverse later in this lesson, but you can also use this calculator if you prefer, or the multiplicative inverse table I have provided.There is also a calculator online that does all modular arithmetic http://ptrow.com/perl/calculator.pl

http://www.cs.princeton.edu/~dsri/modular-inversion-answer.php?n=9&p=8

http://ptrow.com/perl/calculator.pl

More on Galois Fields

• Galois Field sizes can be defined with various field sizes like GF(16) or GF(256). AES is not the only cryptographic algorithm to use these finite fields. Quad also uses finite fields under a polynomial modulus

Implementing GF(p^k) arithmetic

Theorem: Let f(x) be an irreducible polynomialof degree k over Zp.

The finite field GF(pk) can be realized as the set

of degree k-1 polynomials over Zp, with additionand multiplication done modulo f(x).

Example: Implementing GF(2^k)

By the theorem the finite field GF(25) can be realized as

the set of degree 4 polynomials over Z2, with additionand multiplication done modulo the irreducible polynomialf(x)=x5+x4+x3+x+1.

The coefficients of polynomials over Z2 are 0 or 1.So a degree k polynomial can be written down by k+1 bits.For example, with k=4:

x3+x+1 (0,1,0,1,1) x4+ x3+x+1 (1,1,0,1,1)

Implementing GF(2^k)

Addition: bit-wise XOR (since 1+1=0)

x3+x+1 (0,1,0,1,1) + x4+ x3+x (1,1,0,1,0)------------------------------- x4 +1 (1,0,0,0,1)

Multiplication: Polynomial multiplication, and then remainder modulo the defining polynomial f(x):

Implementing GF(2^k)

For small size finite field, a lookup table is the most efficientmethod for implementing multiplication.

(1,1,0,1,1) *(0,1,0,1,1)

= (1,1,0,0,1)

Mathematical background(Cont.)

• Polynomials with coefficients in GF(28)• The operation consisting of multiplication by a fixed polynomial a( x )

can be written as matrix multiplication where the matrix is a circulant matrix. We have:

What is a multiplicative inverse?•In mathematics, the reciprocal, or multiplicative inverse, of a number x is the number which, when multiplied by x, yields 1. The multiplicative inverse for the real numbers, for example, is 1/x. To avoid confusion by writing the inverse using set specific notation, we generally write x-1.

•Zero does not have a reciprocal, as division by 0 is undefined

Multiplication of polynomials

• Finite field multiplication is more difficult than addition and is achieved by multiplying the polynomials for the two elements concerned and collecting like powers of x in the result. Since each polynomial can have powers of x up to 7, the result can have powers of x up to 14 and will no longer fit within a single byte.

• This situation is handled by replacing the result with the remainder polynomial after division by a special eighth order irreducible polynomial, which, for Rijndael, is:

• m(x) = x8 + x4 + x3 + x +1Note: we will discuss why this irreducible polynomial in just a little

while.

Multiplication Polynomials by Repeated Shifts• The finite field element {00000010} is the polynomial x, which means that

multiplying another element by this value increases all it’s powers of x by 1. This is equivalent to shifting its byte representation up by one bit so that the bit at position i moves to position i+1. If the top bit is set prior to this move it will overflow to create an x8 term, in which

• case the modular polynomial is added to cancel this additional bit, leaving a result that fits within a single byte.

• For example, multiplying {11001000} by x, that is {00000010}, the initial result is• 1{10010000}. The ‘overflow’ bit is then removed by adding 1{00011011}, the

modular• polynomial, using an exclusive-or operation to give a final result of {10001011}.NOTE: This why the implementation of creating the S-boxes includes four steps of

shifting. This is in essence, multiplying.

Affine Transformation What is it

This concept originates in graphics, and is also used in transforming graphics. Moving pixels in one direction or another is very similar to moving a value in a matrix, so the concept gets applied to matrices (as in AES)

In geometry, an affine transformation or affine map or an affinity (from the Latin, affinis, "connected with") between two vector spaces (strictly speaking, two affine spaces) consists of a linear transformation followed by a translation:

In general, an affine transform is composed of linear transformations (rotation, scaling or shear) and a translation (or "shift"). Several linear transformations can be combined into a single one, so that the general formula given above is still applicable. For our purposes it is just a word for a linear transformation. This video also gives a good explanation:

http://www.youtube.com/watch?v=4vrYNxlkrpI

Now what are linear transformation? Well for more detail

http://www.samiam.org/galois.html





Rijndael Galois

• The Rijndael Block Chiper uses finite field arithmetic in the field of GF( 28 ) or the Galois

• Field. For Rijndael we show this eld as a polynomial where each b is a bit in a byte that can contain the binary value of either 1 or 0.

• b7x7 + b6x6 + b5x5 + b4x4 + b3x3 + b2x2 + b1x + b0• The values used in Rijndael are displayed in hexadecimal value form and

each hexadecimal value correspond to a polynomial representation. Fore example: Here is a byte containing the hexadecimal value of 57 and it corresponding binary and polynomial representations

• Binary = 01010111 Polynomial = x6 + x4 + x2 + x + 1

Rijndael Galois• The s-box is generated by determining the multiplicative inverse for a given number

in the Galois led, zero would be set to zero. The multiplicative inverse is transformed using the following affine transformation

• where [x0, ..., x7] is the multiplicative inverse as a vector.• Now you can do this with matrix mathematics, first multiplying the matrix by the

vectors, then xoring with the values in the final column. This affine transformation is the sum of multiple rotations of the byte as a vector, where addition is the XOR operation. NOTE This is what we are duplicating in the function to create this transform that we did earlier this week

Why these choices?• We don’t know all the why’s of these choices, but we do

know some.• The last vector that is xord in is meant to prevent the generation

of a fixed point In other words S-box(a)=a• The matrix obviously has meaning. Now they did not state it, but

looking at it even briefly reveals a pattern.• Look at the next slide to analyze the choices

AES S-Box Design

• The S-box is generated by determining the multiplicative inverse for a given number in Rijndael's Galois field. The multiplicitive inverse is then transformed using the following affine transformation matrix

• GF(28) = GF(2)[x]/(x8 + x4 + x3 + x + 1)

Now you can do this with matrix math we already described, or you can do it with the function we provided to you yesterday (and is also at the end of this presentation)

Why these choices?• In the submission of Rijndael Joan Daemen & Vincent Rijmen stated reasons for

many things, such as the particular modulus to use. But questions still may be in your mind?

• Why this matrix for the affine transform and not some other?• Why this particular vector for the affine transform and not some other?

• Here are the only answers they gave, and the following is a quote directly from their submission.

How to calculate this yourself

• Unfortunately, no, it is not the same as computing the inverse of a matrix. Also, there are at least two issues involved here. One is what are elements of this finite field GF(2^8)? The other is how are they represented on the computer (i.e. in binary)? And then we can get to the question of how to invert them.

• First of all, an element of GF(2^8) is a polynomial of degree less than 8 over GF(2) (with coefficients in GF(2)) modulo an irreducible polynomial of degree 8. So the first thing you have to do is pick a polynomial of degree 8, and use this as your field modulus. The AES standard says what this modulus is.

How to calculate the Multiplicative inverse• The multiplicative inverse for an element a of a finite field can be

calculated a number of different ways:• By multiplying a by every number in the field until the product is one. This is a

Brute-force search.• By using the Extended Euclidean algorithm• By making a logarithm table of the finite field, and performing subtraction in

the table. Subtraction of logarithms is the same as division.

Compute multiplicative inverse

• Remember that in modular arithmetic, the modular multiplicative inverse of x is the number a such that ax ≡ 1 (mod n). This multiplicative inverse exists if and only if a and n are coprime. For example, the inverse of 3 modulo 11 is 4 because it is the solution to 3x ≡ 1 (mod 11).

• Put another way you are trying to find a number that is coprime to x mod n.

• The extended Euclidean algorithm may be used to compute it. In fact this is probably the most common way to compute it.

Extended EuclideanThe extended Euclidean algorithm is a version of the Euclidean algorithm; its input are two integers a and b and the algorithm computes their greatest common divisor (GCD) as well as integers x and y such that ax + by = gcd(a,b). This works because the steps of Euclid's algorithm always deal with sums of multiples of a and b.

We are going to look first at a video demonstration of this, then at an example here in this presentation, then at a second and different video demonstration. The hope is that seeing it done three different times, should assist you in understanding it. Here is the first video demohttp://www.youtube.com/watch?v=fz1vxq5ts5I very clear, easy to follow

http://www.youtube.com/watch?v=fz1vxq5ts5I

Extended Euclidean

Consider as an example the computation of gcd(120,23) with Euclid's algorithm:

•120 / 23 = 5 remainder 5•23 / 5 = 4 remainder 3•5 / 3 = 1 remainder 2•3 / 2 = 1 remainder 1•2 / 1 = 2 remainder 0

• In this case, the remainder in the second-to-last line indicates that the gcd is 1; that is, 120 and 23 are coprime.

Extended Euclidean step 2Now do a little algebra on each of the above lines:120 / 23 = 5 r 5 => 5 = 120 - 5*2323 / 5 = 4 r 3 => 3 = 23 - 4*55 / 3 = 1 r 2 => 2 = 5 - 1*33 / 2 = 1 r 1 => 1 = 3 - 1*22 / 1 = 2 r 0 => 0 = 2 - 2*1Now observe that the first line contains multiples of 120 and 23. Also, the rightmost values are in each case the remainders listed on the previous line, and the left side of the differences are the residues from two lines up. We can thus progressively calculate each successive remainder as sums of products of our two original values. Here we rewrite the second equations in the above table: 5 = 120 - 5*23 = 1*120 - 5*23 3 = 23 - 4*5 = 1*23 - 4*(1*120 - 5*23) = -4*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) = -9*120 + 47*23

Notice that the last line says that 1 = -9*120 + 47*23, which is what we wanted: x = -9 and y = 47.

This means that -9 is the multiplicative inverse of 120 modulo 23, because 1 = -9 * 120 (mod 23).

Extended Euclidean step 3So remember the original formula? ax + by = gcd(a,b). That means in this case b is the modulo, a is the number you are trying to find the multiplicative inverse for. Here is an example ( too see this entire example online http://www.fact-index.com/e/ex/extended_euclidean_algorithm.html

http://www.fact-index.com/e/ex/extended_euclidean_algorithm.html

Euclid’s Greatest Common Divisor Algorithm

• Since m mod n = m - m/nn, We can now apply statement 4 and statement 3 to see that gcd(m,n) = gcd(m mod n, n) = gcd(n, m mod n)

5. For integers m and n with n > 0, gcd(m,n) = gcd(n, m mod n).• We can apply statement 5 to derive an algorithm for

computing - greatest common divisors• Euclid’s Algorithm:• Input: integers m and n, not both zero

Output: d = gcd(m,n)• If n == 0

d = |m|else

while n 0c = nn = m mod nm = c

d = mreturn d

Euclid’s Greatest Common Divisor Algorithm• One of the special properties of the greatest common

divisor of two numbers is that it can be written as an integer linear combination of the numbers

• Example 2: gcd(32,24) = 8• 32 = 124 + 8• Thus 8 = 132 + (-1)24• Example 3: gcd(54,42)

54 = 142 + 1242 = 312 + 612 = 26 + 0, so 6 = gcd(54,42)Using back-substitution:6 = 142 - 312 and 12 = 154 - 142 Thus 6 = 142 - 3(154 - 142 ) = 442 + (-3)54

Euclidean VIDEO Tutorials

• http://www.youtube.com/watch?v=fz1vxq5ts5I very clear, easy to follow

• http://www.youtube.com/watch?v=shaQZg8bqUM This is a little longer, but a pretty good one.

http://www.youtube.com/watch?v=fz1vxq5ts5I

http://www.youtube.com/watch?v=shaQZg8bqUM

Here is another method method One divided by a given number is the multiplicative inverse of that number. To find the multiplicative inverse of the number a: •Find the logarithm for a •Subtract 255 by a's logarithm •Take the anti-log of the resulting number •This is the multiplicative inverse (In other words, 1 / a) Here is some code which uses the above log and anti-log tables to calculate the multiplicative inverse: unsigned char gmul_inverse(unsigned char in) {

/* 0 is self inverting */if(in == 0)

return 0;else

return atable[(255 - ltable[in])];}This example comes from http://www.samiam.org/galois.html

Rijendael step by step- The s-box

• This is the standard s-box

Multiplicative inverse

• The S-box is generated by determining the multiplicative inverse for a given number in GF(28) = GF(2)[x]/(x8 + x4 + x3 + x + 1), Rijndael's finite field (zero, which has no inverse, is set to zero). Rijndael uses a characteristic 2 finite field with 256 elements, which can also be called the Galois field GF(28). It employs the following reducing polynomial for multiplication: x8 + x4 + x3 + x + 1.

• The multiplicative inverse is then transformed using the following affine transformation:

S-box Derivation

The S-box maps byte x to byte z via the function z = Ax-1+b:

Input byte x: x7x6x5x4x3x2x1x0

Compute the inverse in GF(28): y7y6y5y4y3y2y1y0 (non-linear, vs. attacks)(use 0 as inverse of 0)

Compute this linear function z in GF(28): (to complicate attacks) (A is simple to implement) b chosen so

xzandxz

Calculate the multiplicative inverse• Recall that the multiplicative inverse in a modulo n world is defined as being the

number, a-1, such that• (a)(a-1) ≡ 1 (mod n)• Thus, for example, 3 and 7 are multiplicative inverses of each other in a mod 10 world

while 5 and 11 are multiplicative inverses of each other in a mod 27 world.• For very small moduli, finding the multiplicative inverse of a number is not terribly

difficult to do via exhaustive search since, by definition, (a)(a-1) = kn + 1• So it becomes a simple matter of writing this as (a-1) = (kn + 1)/a• and trying successive values of k until one is found that results in the right hand side

being an integer. All that is then left is to reduce the resulting integer modulo n. But as the modulus grows, this process quickly becomes unwieldy and impractical. What is needed is a more systematic means of finding a-1.

Calculate the multiplicative inverse

We have to find x such that ax ≡ 1 (mod m) Above equivalence can be stated alternatively as:ax = pm + 1 or ax + my = 1 known as Bézout’s identity

Table of Multiplicative inverses

Generating the standard S-Box

• The next slides provide this. However if you wish to alter the s-box you should attempt the following:

• Find a different irreducible polynomial to use as the mod operation.• Find a different affine transform, that simply means a different matrix to

multiple the multiplicative inverse by. This changes the shifts/operations you see in the following slides.

Rijendael step by step- The s-box

• This will generate the Rijndael (AES) S-box, which is represented here with hexadecimal notation:

Rijendael step by step- step one

• Take a number for GF(28) lets pick 2. Looking at the multiplicative inverse table that gives us 8d in hex or 10001101 in binary

• Now we need to do four iterations of the process of affine transformation. We start by putting the multiplicative inverse into two variables s and x:

• s = 10001101• x = 10001101

Rijendael step by step- step two (first iteration)• Rotate s(10001101) to the left = 00011010• If the high bit is one make the low bit 1• else low bit is 0 Now in this case the high bit WAS 1 so we change the low bit• so s is 00011011

xor with x so 00011011 xor 10001101= 10010110s = 00011011 x = 10010110

Rijendael step by step- step three (second iteration)• rotate s (00011011 )to the left = 00110110• If then(shown above) still gives us 00110110• xor with x so 00110110 xor 10010110 = 10100000• s = 00110110 x = 10100000

Rijendael step by step- step four (third iteration)• rotate s(00110110) to the left = 01101100• if then (shown above ) still gives us 01101100• xor with x so 01101100 xor 10100000= 11001100• so s = 01101100 x = 11001100

Rijendael step by step- step five(fourth iteration)• rotate s(01101100) to the left = 11011000• if then (shown above) still gives us 11011000• xor with x so 01101100 xor 11001100= 00010100• so s = 11011000 x = 00010100

Rijendael step by step- step six

• Now x (00010100) gets xor'd with decimal 99 (hex x63 binary 1100011) = 1110111 or 77

And if we check the s-box table for what 2 should give us we get

So what are we seeing?

• The output is actually the multiplicative invers of the input then put through this affine transform.

• Still not convinced lets see it again?

Rijendael step by step- step one

• Take a number for GF(28) lets pick 7. Looking at the multiplicative inverse table that gives us d1 in hex or 11010001 in binary

• Now we need to do four iterations of the process of affine transformation. We start by putting the multiplicative inverse into two variables s and x:

• s = 11010001• x = 11010001

Rijendael step by step- step two (first iteration)• Rotate s(11010001) to the left = 10100010• If the high bit is one make the low bit 1• else low bit is 0 Now in this case the high bit WAS 1 so we change the low bit• so s is 10100011

xor with x so 10100011 xor 11010001 = 1110010s = 10100011 x = 1110010

Rijendael step by step- step three (second iteration)• rotate s(10100011) to the left = 01000110• If then(shown above) • gives us 01000111• xor with x so 01000111 xor 1110010 = 00110101• s = 01000111 x = 00110101

Rijendael step by step- step four (third iteration)• rotate s(01000111) to the left = 10001110 • if then (shown above ) still gives us 10001110• xor with x so 10001110 xor 00110101 = 10111011• so s = 10001110 x = 10111011

Rijendael step by step- step five(fourth iteration)• rotate s(10001110) to the left = 00011100 • if then (shown above) • gives us 00011101• xor with x so 00011101 xor 10111011 = 10100110• so s = 00011101 x = 10100110

Rijendael step by step- step six

• Now x (10100110) gets xor'd with decimal 99 (hex x63 binary 1100011) = 11000101 or c5

And if we check the s-box table for what 7 should give us we get

That is IT!!

• That is how you generate the entire Rijndael s box!• Is this clearer? Well lets reiterate the basics:1.Use GF(28) and look at each possible value in that filed one at a time.2.Find the multiplicative inverse, then compute the transform. 3.That takes four steps, and gives us the output for the s box

But wait…

• What was the matrix we saw earlier? Well that is an alternative way to do this. You can put all the GF(28) values in a matrix, and use the matrix we saw to produce the multiplicative inverse of the GF(28) matrix…OR you can use the formula we just did!

• What does this mean? Well either method works, but I think you will find the formula method easier to code in any programming language and easier to alter.

Alter it? How

• Some obvious solutions come into play• What about different shifts? Left or right, or by more than one? Will that

affect the output?• Of course a different value of GF(28) perhaps GF(29) for 512 bits?• Perhaps building this output table (this s box) then xoring it with the round

key so there is a different s-box each round? And it is semi key dependent?• Of course you could construct an entirely different affine transform.

How to change an AES s-box

• Remember you want the entire cipher to meet• Strict Avalanche Criteria• Bit independence criteria• Balance

• As much as possible the s-box should fulfill this. So when making any change, we know that we must make certain our change meets these criteria

So what changes can we make• Isomorphic fields to the underlying field can be generated by using different

irreducible polynomials of the same degree. There are a total of 30 irreducible polynomials of degree 8 to choose from. This gives you 29 alternatives to the traditional s-box for AES, each with well tested security. (note for more details you can look into Rabin’s test for irreducibility).

• Change the affine transform. This is a little more tricky but safe if you simply alter parameters within the existing transform. Section 5.2 of “ Algebraic Construction and Cryptographic Properties of Rijndael Substitution” discusses this in detail.

• Change the translation vector (the final number you xor with). Obviously there are 255 possible variations.

Comments on changing the affine transform• “We propose to increase the complexity and security of AES S-box by

modifying the affine transformation and adding an affine transformation”

• -AN IMPROVED AES S-BOX AND ITS PERFORMANCE ANALYSIS “improved AES s-box performance.pdf”

AES S-box How and Why?. Notes The general math review slides are taken from a variety of internet...

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