Probabilistic verification
Mario Szegedy, Rutgers
www/cs.rutgers.edu/~szegedy/07540
Lecture 5
C= (n,k,d)q codes
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n = Block length
k = Information length
d = Distance
k/n = Information rate
d/n = Distance rate
q = Alphabet size
Linear Codes
• C = {x | x Є L}
(L is a linear subspace of Fqn)
• Δ(x,y) = Δ(x-z,y-z)
• min Δ(0,x) = min Δ(x,y)
x,y Є L
• k = dimension of the message space = dim L
• n = dimension of the whole space (in which the code words lie)
• Generator matrix: {xG | x
• “Parity” check matrix: {y
Fqk}
n
Є Fq | yH = 0}
Є
Reed-Solomon codes
• The Reed–Solomon code is an error-correcting code that works by
oversampling a polynomial constructed from the data.
• C is a [n, k, n-k+1] code; in other words, it is a linear code of length n
(over F) with dimension k and minimum distance n-k+1.
Welch-Berlekamp decoding
algorithm for RS codes
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Length = n (n ≤ |F|)
Degree = k
# of errors =e
We assume: k+2e < n.
Received code word = y
E = non-zero polynomial that is zero at the bad “bits” of y
deg E ≤ e
y E = B.
deg B ≤ e + k
NOTATION:
I.
II.
v = w (exc e) means “word v is the same as word w except at e places)
v = w (exc H) means “word v is the same as word w except on set H)
We have
P = y (exc Err), where P is the encoded word, and |Err| ≤ e.
We can find E and B such that y E = B by solving a system of linear equations:
E0 + E1 fj + E2 fj2 + … + Ee fje = yj B0 + yj B1 fj + yj B2 fj2 + … + yj Be fje+k
(1 ≤ j ≤ n)
Welch-Berlekamp decoding
algorithm for RS codes
Since P = y (exc Err), it follows that
PE = yE = B (exc Err),
But deg PE = k + e, and deg B = k + e, and since
k+e < n–e, we have
PE = B.
Now we can easily recover P by simply dividing polynomial B by E.
Generalization of the WB algorithm
• C is a general code (replacing RS)
α.
β.
γ.
δ.
A * C is a subset of B
dim A > e
d(B) > e
d(A) + d(C) ≥ n
solve
y a = b (we need β. for such non-zero a to exist)
let c be the code word we are looking for
ca = ya = b (exc Err) α.,γ.
→
ca = b, since ca Є B. →
→
b/a = c (exc zeros(a))
→
δ.
b/a = c (exc n – d(A))
→
we can recover c
Multi-variate polynomials
Two variate polynomial of multi-degree (2,2):
General n variate polynomial:
P(x1,x2,…,xn) = P(x) = Σα aα xα
Multi-degree
deg y
x3y5
x2y3
deg x
Multi-degree (3,5)
Multi-degree
deg y
deg x
General multi-degree pattern. Multi-degree polynomials
of a fix pattern form a vector space over F.
Multi-degree
deg y
P→P+P
deg x
Under multiplication degree patterns add like vectors
Total degree d
deg y
leading
terms
total degree 5
deg x
for each coeff vector α in P(x) we have
Σi α i ≤ d
LEMMA: Let S be a subset of a field F, |S|=d+1 ≥ 1.
Then any function from Sn to F has a unique
extension to a multi-degree (d,…,d) polynomial.
PROOF:
1. there exists a low degree extension:
By linearity it is enough to show that any function that takes a single
non-zero value has an extension. Assume the non-zero value is
taken over (a1,a2,…,an). Then over S it can be expressed as:
const ∙ ΠiΠs (xi - s)
s runs through S, but leaves out ai
LEMMA: Let S be a subset of a field F, |S|=d+1 ≥ 1.
Then any function from Sn to F has a unique
extension to a multi-degree (d,…,d) polynomial.
2. The low degree extension is unique:
Because of linearity it is enough to show that the identically zero
function on Sn has a unique extension (which is the 0 polynomial).
In 1 dimension this follows from the fact that a degree d polynomial
cannot have more than d roots. Assume that we have proven the
statement for d-1. →
If we fix xi to any constant s in S, we get an identically zero
polynomial.
→
Now fix all the other variables (anyhow), except xi, we get a
polynomial taking zeros on all points of S. → All such uni-variate
polynomial has to take zero everywhere. → P(x)=0.
(d+1)n-1
xi
d+1
All other variables
0
Zippel’s lemma
THEOREM (Schwartz-Zippel). Let
be a polynomial of degree d over a field, F. Let S be a finite subset
of F and let
be selected randomly from S. Then
PROOF: Induction on n. For n = 1, P can have at most d roots. This gives us the
base case. Now, assume that the theorem holds for all polynomials in n − 1
variables. We can then consider P to be a polynomial in x1 by writing it as
Since P is not identically 0, there is some i such that Pi is not identically 0.
Take the largest such i. Then
.
PROOF: Induction on n. For n = 1, P can have at most d roots. This gives us the
base case. Now, assume that the theorem holds for all polynomials in n − 1
variables. We can then consider P to be a polynomial in x1 by writing it as
Since P is not identically 0, there is some i such that Pi is not identically 0.
Take the largest such i. Then
. Now we randomly pick
.
from S. By the induction hypothesis,
PROOF: Induction on n. For n = 1, P can have at most d roots. This gives us the
base case. Now, assume that the theorem holds for all polynomials in n − 1
variables. We can then consider P to be a polynomial in x1 by writing it as
Since P is not identically 0, there is some i such that Pi is not identically 0.
Take the largest such i. Then
. Now we randomly pick
.
from S. By the induction hypothesis,
If
then
is of degree i.
Event A:
Event B:
Maximal number of zeros of a
multi-degree (d,…,d) polynomial
qn – (q – d)n
Homework