I talked about Rijndael in a couple of previous posts: Generating the Rijndael S-box, Efficient multiplication and division in GF(28), Sieving irreducible monic polynomials over a finite field, Addition and multiplication table for GF(22).

I'm going to talk some more about it today.

The Rijndael non-linear S-box S(x) is a composition of two invertible functions f(g(x)). Last time we showed how to generate these, and their inverses, as 256-entry tables.

As Daemen and Rijmen point out, any function from a finite field to itself can be expressed as a polynomial. In fact, given a tabular form of the function, it is possible to generate the Lagrange polynomial and then simplify.

They also give the polynomial for S(x):

SRD[x] = 05·x255 + 09·x253 + F9·x251 + 25·x247 + F4·x239 + 01·x223 + B5·x191 + 8F·x127 + 63

Well, let's check this.
And while we're at it, let's find the polynomials for g(x), f(x) and even S-1(x) too.
First of all, let's start with the Lagrange polynomial.

Given a table of entries { (00, y00), (01, y01), ..., (ff, yff) }, there is a polynomial L(x) which gives the same output, namely:

L(x) = Σi ∈ { 00, 01, ..., ff } yi pi(x)
where pi(x) = Πj ∈ { 00, 01, ..., ff }, ji (x - j) / (i - j)

Can we simplify this?
Yes. Note that (i - j)-1 term varies over all the non-zero elements of the finite field. Since this is a field, every non-zero element has an inverse, which might or might not be itself.
If the inverse is not itself, we can pair the two inverse elements together, and we get 01, which is the multiplicative identity, so we can ignore it.
What are we left with? The product of those non-zero elements which are their own inverses.
In the case of GF(28), there is only one such element, namely 01.
Great! We can ignore the (i - j)-1 terms altogether:

L(x) = Σi ∈ { 00, 01, ..., ff } yi Πj ∈ { 00, 01, ..., ff }, ji (x - j)

What is this? A sum of 256 different polynomials.
Each of these summands is a product of 255 polynomials of degree 1, and so is a polynomial of degree 255.
But in GF(28), x255 = 01 for all x. So each of the summands is actually a polynomial of degree 254, or less.
So the sum is also a polynomial of degree 254, or less; terms with an exponent >= 255 go away and just contribute to lower-order terms.

Great. We have { (00, y00), (01, y01), ..., (ff, yff) } for f, g, S, and S-1. Let's plug them in and see what happens!

I wrote a Perl script to do this; the script is attached. Here's the output. (It's quite slow; it takes about two and a half minutes to run.)

g(x) = 01 x^254

f(x) =
 63 + 05 x + 09 x^2 + f9 x^4 + 25 x^8 + f4 x^16 + 01 x^32 + b5 x^64 +
 8f x^128

S(x) =
 63 + 8f x^127 + b5 x^191 + 01 x^223 + f4 x^239 + 25 x^247 + f9 x^251 + 09 x^253 +
 05 x^254

S^(-1)(x) =
 52 + f3 x + 7e x^2 + 1e x^3 + 90 x^4 + bb x^5 + 2c x^6 + 8a x^7 +
 1c x^8 + 85 x^9 + 6d x^10 + c0 x^11 + b2 x^12 + 1b x^13 + 40 x^14 + 23 x^15 +
 f6 x^16 + 73 x^17 + 29 x^18 + d9 x^19 + 39 x^20 + 21 x^21 + cf x^22 + 3d x^23 +
 9a x^24 + 8a x^25 + 2f x^26 + cf x^27 + 7b x^28 + 04 x^29 + e8 x^30 + c8 x^31 +
 85 x^32 + 7b x^33 + 7c x^34 + af x^35 + 86 x^36 + 2f x^37 + 13 x^38 + 65 x^39 +
 75 x^40 + d3 x^41 + 6d x^42 + d4 x^43 + 89 x^44 + 8e x^45 + 65 x^46 + 05 x^47 +
 ea x^48 + 77 x^49 + 50 x^50 + a3 x^51 + c5 x^52 + 01 x^53 + 0b x^54 + 46 x^55 +
 bf x^56 + a7 x^57 + 0c x^58 + c7 x^59 + 8e x^60 + f2 x^61 + b1 x^62 + cb x^63 +
 e5 x^64 + e2 x^65 + 10 x^66 + d1 x^67 + 05 x^68 + b0 x^69 + f5 x^70 + 86 x^71 +
 e4 x^72 + 03 x^73 + 71 x^74 + a6 x^75 + 56 x^76 + 03 x^77 + 9e x^78 + 3e x^79 +
 19 x^80 + 18 x^81 + 52 x^82 + 16 x^83 + b9 x^84 + d3 x^85 + 38 x^86 + d9 x^87 +
 04 x^88 + e3 x^89 + 72 x^90 + 6b x^91 + ba x^92 + e8 x^93 + bf x^94 + 9d x^95 +
 1d x^96 + 5a x^97 + 55 x^98 + ff x^99 + 71 x^100 + e1 x^101 + a8 x^102 + 8e x^103 +
 fe x^104 + a2 x^105 + a7 x^106 + 1f x^107 + df x^108 + b0 x^109 + 03 x^110 + cb x^111 +
 08 x^112 + 53 x^113 + 6f x^114 + b0 x^115 + 7f x^116 + 87 x^117 + 8b x^118 + 02 x^119 +
 b1 x^120 + 92 x^121 + 81 x^122 + 27 x^123 + 40 x^124 + 2e x^125 + 1a x^126 + ee x^127 +
 10 x^128 + ca x^129 + 82 x^130 + 4f x^131 + 09 x^132 + aa x^133 + c7 x^134 + 55 x^135 +
 24 x^136 + 6c x^137 + e2 x^138 + 58 x^139 + bc x^140 + e0 x^141 + 26 x^142 + 37 x^143 +
 ed x^144 + 8d x^145 + 2a x^146 + d5 x^147 + ed x^148 + 45 x^149 + c3 x^150 + ec x^151 +
 1c x^152 + 3e x^153 + 2a x^154 + b3 x^155 + 9e x^156 + b7 x^157 + 38 x^158 + 82 x^159 +
 23 x^160 + 2d x^161 + 87 x^162 + ea x^163 + da x^164 + 45 x^165 + 24 x^166 + 03 x^167 +
 e7 x^168 + 19 x^169 + e3 x^170 + d3 x^171 + 4e x^172 + dd x^173 + 11 x^174 + 4e x^175 +
 81 x^176 + 91 x^177 + 91 x^178 + 59 x^179 + a3 x^180 + 80 x^181 + 92 x^182 + 7e x^183 +
 db x^184 + c4 x^185 + 20 x^186 + ec x^187 + db x^188 + 55 x^189 + 7f x^190 + a8 x^191 +
 c1 x^192 + 64 x^193 + ab x^194 + 1b x^195 + fd x^196 + 60 x^197 + 05 x^198 + 13 x^199 +
 2c x^200 + a9 x^201 + 76 x^202 + a5 x^203 + 1d x^204 + 32 x^205 + 8e x^206 + 1e x^207 +
 c0 x^208 + 65 x^209 + cb x^210 + 8b x^211 + 93 x^212 + e4 x^213 + ae x^214 + be x^215 +
 5f x^216 + 2c x^217 + 3b x^218 + d2 x^219 + 0f x^220 + 9f x^221 + 42 x^222 + cc x^223 +
 6c x^224 + 80 x^225 + 68 x^226 + 43 x^227 + 09 x^228 + 23 x^229 + c5 x^230 + 6d x^231 +
 1d x^232 + 18 x^233 + bd x^234 + 5e x^235 + 1b x^236 + b4 x^237 + 85 x^238 + 49 x^239 +
 bc x^240 + 0d x^241 + 1f x^242 + a6 x^243 + 6b x^244 + d8 x^245 + 22 x^246 + 01 x^247 +
 7a x^248 + c0 x^249 + 55 x^250 + 16 x^251 + b3 x^252 + cf x^253 + 05 x^254

Daemen and Rijmen's polynomial rendition of S(x) is confirmed.

Also note how simple g(x) is, and how complex S-1(x) is.

Finally, I must confess that none of this is actually useful. The tabular form is much more convenient for applications. This is just a fun theoretical exercise.

(Well, I had fun.)