Nonconstructive methods in finite automata

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Nonconstructive methods in finite automata

• Rusin Freivalds
• (University of Latvia)

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To explain the main idea, we consider the
following probabilistic automaton. The initial
probability distribution 1/8, 1/8, 1/8, 1/8,
1/8, 1/8, 1/8, 1/8, 0, 0, 0, 0, 0, 0, 0, 0 The
first 8 states are accepting, the last 8 states
are rejecting.
3
To explain the main idea, we consider the
following probabilistic automaton. The initial
probability distribution 1/8, 1/8, 1/8, 1/8,
1/8, 1/8, 1/8, 1/8, 0, 0, 0, 0, 0, 0, 0, 0 The
first 8 states are accepting, the last 8 states
are rejecting. There are 28 possible input
letters. They interchange the states qi ? qi8
. The input word is in the language if all the
non-zero probabilities have returned to the first
8 states. It is easy to prove that any
deterministic automaton needs 28 states for this
language. Unfortunately, the probabilistic
automaton has non-isolated cut-point.
4
To explain the main idea, we consider the
following probabilistic automaton. The initial
probability distribution 1/8, 1/8, 1/8, 1/8,
1/8, 1/8, 1/8, 1/8, 0, 0, 0, 0, 0, 0, 0, 0 The
first 8 states are accepting, the last 8 states
are rejecting. There are 28 possible input
letters. They interchange the states qi ? qi8
. The input word is in the language if all the
non-zero probabilities have returned to the first
8 states. It is easy to prove that any
deterministic automaton needs 28 states for this
language. Unfortunately, the probabilistic
automaton has non-isolated cut-point.
5
To explain the main idea, we consider the
following probabilistic automaton. The initial
probability distribution 1/8, 1/8, 0, 1/8, 0,
0, 1/8, 1/8, 0, 0, 1/8, 0, 1/8, 1/8, 0, 0 The
first 8 states are accepting, the last 8 states
are rejecting. There are 28 possible input
letters. They interchange the states qi ? qi8
. The input word is in the language if all the
non-zero probabilities have returned to the first
8 states. It is easy to prove that any
deterministic automaton needs 28 states for this
language. Unfortunately, the probabilistic
automaton has non-isolated cut-point.
6
To explain the main idea, we consider the
following probabilistic automaton. The initial
probability distribution 1/8, 1/8, 0, 1/8, 0,
0, 1/8, 1/8, 0, 0, 1/8, 0, 1/8, 1/8, 0, 0 The
first 8 states are accepting, the last 8 states
are rejecting. There are 28 possible input
letters. They interchange the states qi ? qi8
. The input word is in the language if all the
non-zero probabilities have returned to the first
8 states. It is easy to prove that any
deterministic automaton needs 28 states for this
language. Unfortunately, the probabilistic
automaton has non-isolated cut-point.
7
To explain the main idea, we consider the
following probabilistic automaton. The initial
probability distribution 1/8, 1/8, 0, 1/8, 0,
0, 1/8, 1/8, 0, 0, 1/8, 0, 1/8, 1/8, 0, 0 The
first 8 states are accepting, the last 8 states
are rejecting. There are 28 possible input
letters. They interchange the states qi ? qi8
. The input word is in the language if all the
non-zero probabilities have returned to the first
8 states. It is easy to prove that any
deterministic automaton needs 28 states for this
language. Unfortunately, the probabilistic
automaton has non-isolated cut-point.
8
To explain the main idea, we consider the
following probabilistic automaton. The initial
probability distribution 1/8, 1/8, 0, 0, 1/8,
1/8, 1/8, 1/8, 0, 0, 1/8, 1/8, 0, 0, 0, 0 The
first 8 states are accepting, the last 8 states
are rejecting. There are 28 possible input
letters. They interchange the states qi ? qi8
. The input word is in the language if all the
non-zero probabilities have returned to the first
8 states. It is easy to prove that any
deterministic automaton needs 28 states for this
language. Unfortunately, the probabilistic
automaton has non-isolated cut-point.
9
Linear codes is the simplest class of codes. The
alphabet used is a fixed choice of a finite field
GF(q)Fq with q elements. For most of this paper
we consider a special case of GF(2)F2. These
codes are binary codes. A generating matrix G
for a linear n, k code over Fq is a k-by-n
matrix with entries in the finite field Fq, whose
rows are linearly independent. The linear code
corresponding to the matrix G consists of all the
qk possible linear combinations of rows of G.
The requirement of linear independence is
equivalent to saying that all the qk linear
combinations are distinct.
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The linear combinations of the rows in G are
called codewords. However we are interested in
something more. We need to have the codewords
not merely distinct but also to be removed as far
as possible each from another in terms of
Hamming distance. Hamming distance between two
vectors v(v1, ..., vn) and w(w1, ..., wn) is
the number of indices i such that vi ? wi.
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The textbook P.Garret The Mathematics of
Coding Theory(2004) contains Theorem A. For
any integer n 4 there is a 2n, n binary code
with a minimum distance between the codewords at
least n/10. However the proof of this theorem
has a serious defect. It is non-constructive. It
means that we cannot find these codes or describe
them in a useful manner. This is why P.Garret
calls them mirage codes.
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Definition. A generating matrix G of a linear
code is called cyclic if along with an arbitrary
row (v1, v2, v3, ..., vn) the matrix G contains
also a row (v2, v3,...,vn,v1). We would wish to
prove a reasonable counterpart of Theorem A for
cyclic mirage codes, but this attempt
fails. Instead we construct a slightly more
complicated structure of mirage codes for which a
counterpart of Theorem A can be proved.
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We consider binary generating matrices. Let p be
an odd prime number, and x be a binary word of
the length p. The generating matrix G(p, x) has p
rows and 2p columns. Let x x1 x2 x3 ... xp.
The first p columns (and all p rows) make a unit
matrix with elements 1 on the main diagonal and 0
in all the other positions. The last p columns
(and all p rows) make a cyclic matrix with x x1
x2 x3 ... xp as the first row, xp x1 x2 x3 ...
xp-1 as the second row, and so on. 1 0 0 0 ... 0
x1 x2 x3 x4... xp 0 1 0 0 ... 0 xp x1 x2
x3 ... xp-1 0 0 1 0 ... 0 xp-1 xp x1 x2 ...
xp-2 .............................................
... 0 0 0 0 ... 1 x2 x3 x4 x5... x1
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Lemma. For arbitrary x, if h1 h2 h3 ... hp hp1
hp2 h p3 ... h2p is a codeword in the linear
code corresponding to G(p, x), then hp h1 h2 ...
h p-1 h p-1 hp-2 h2p hp1 hp2... h2p-1 is also
a codeword.
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There are 2p codewords of the length 2p. If the
codeword is obtained as a linear combination with
the coefficients c1, c2, ..., cp then the first
p components of the codeword equal c1 c2 ... cp.
We denote by R(x, c1 c2 ... cp) the subword
containing the last p components of this
codeword. Lemma. If c1 c2 ...cp 0 0 0...0, then
R(x, c1 c2 ... cp)0 00...0, for arbitrary x.
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Definition. We will call a word trivial if all
its symbols are equal. Otherwise we call the word
nontrivial. Lemma. If c1 c2 ...cp is trivial,
then R(x, c1 c2 ... cp) is trivial for arbitrary
x. Proof. Every symbol of R(x, c1 c2 ... cp)
equals x1 x2 ... xp mod 2. Lemma. If x is
trivial, then R(x, c1 c2 ... cp) is trivial for
arbitrary c1 c2 ...cp. .
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Definition. Word x x1 x2 x3 x4... xp is
called a cyclic shift of the word y y1 y2 ...
yp if there exists i such that x1 yi, x2
yi1, ..., xp yip where the addition is modulo
p. If (i, p) 1, then we say that this cyclic
shift is nontrivial. Lemma. If x is a cyclic
shift of y, then R(x, c1 c2 ... cp) is a cyclic
shift of R(y, c1 c2 ... cp). Lemma. If p is an
odd prime, x is a nontrivial word and y is a
nontrivial cyclic shift of x, then x ?
y. Lemma. If p is an odd prime and c1 c2 ...cp
is nontrivial, then the set T c1 c2 ... cp
R(x, c1 c2 ... cp) x is in 0, 1p and R(x, c1
c2 ... cp is nontrivial has a cardinality
which is a multiple of p.
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If q is a prime number, the set of the codewords
with the operation "component-wise addition" is
a group. Finite groups have useful properties.
We single out Lagrange's Theorem. The order of a
finite group is the number of elements in
it. Lagrange's Theorem. Let GR be a finite
group. Let H be a subgroup of GR. Then the order
of H divides the order of GR.
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For arbitrary fixed c1 c2 ...cp , the set R(x,
c1 c2 ... cp) x is in 0, 1p with algebraic
operation "component-wise addition modulo p" is a
group. We denote this group by B. By D we denote
the group of all 2p binary words of the length p
with the same operation. Lemma. For arbitrary c1
c2 ...cp, x and y, R(x, c1 c2 ... cp) R(y, c1
c2 ... cp) R(xy, c1 c2 ... cp). In other
words, the map D ?B defined by x ? R(x, c1 c2 ...
cp) is a group homomorphism. The kernel of the
group homomorphism is the set ker_0 x
R(x, c1 c2 ... cp) 0 0 0 ...0. The image of
the group homomorphism is the set B. For
arbitrary z in B, by kerz we denote the set
kerz x R(x, c1 c2 ... cp) z.
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Lemma. For arbitrary z in B, card(kerz)
card(ker0). Lemma. For arbitrary z in B,
card(kerz) card(D)/card(B). Lemma. If x
contains (p - 1) zeroes and 1 one, and c1 c2 ...
cp is nontrivial, then R(x, c1 c2 ... cp) is
nontrivial. Proof. For such an x, the number of
ones in R(x, c1 c2 ... cp) is the same as the
number of ones in c1 c2 ...cp.
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Consider the sequence 2, 2¹, 2², , 2p-2,
2p-1, 2p, and the corresponding sequence of
the remainders of these numbers modulo
p r0,r1,r2,,rp-2,rp-1,rp, Little Fermat
theorem asserts that if p is prime exceeding 2,
then is rp-1 congruent to 1 modulo p. If p-1 is
the smallest element of this sequence congruent
to 1 modulo p then we say that 2 is a primitive
root modulo p.
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Emil Artin made in 1927 a famous conjecture
the validity of which is still an open
problem If r is neither -1 nor a square,
then r is a primitive root for infinitely
many primes.
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Lemma. If p is an odd prime such that 2 is a
primitive root modulo p and c1 c2 ...cp is
nontrivial, then the set R(x, c1 c2 ... cp) x
is in 0, 1p is either of cardinality 1 or of
cardinality 2. Proof. By Lagrange's Theorem the
order 2p of the group B divides the order of the
group D. Hence the order of B is 2p for some
integer b. The neutral element of these groups is
the word 000...0. It belongs to every
subgroup. There are two possible cases 1)
111...1 is in B, 2) 111...1 is not in B.
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Case 1) 111...1 is in B. T c1 c2 ... cp
R(x, c1 c2 ... cp) x is in 0, 1p and R(x, c1
c2 ... cp is nontrivial Then card(T)
card(B) - 2, and card(T) is a multiple of p.
Hence 2b card(B) 2 mod p and 2b-1 1 mod p.
Since 2 is a primitive root modulo p, either 2b-1
2p-1 or 2b-1 20. If 2b-1 2p-1, then 2b
2pand for this fixed c1 c2 ... Cp the map x
?R(x, c1 c2 ... cp) takes distinct x'es into
distinct R(x, c1 c2 ... cp) 's. If 2b-1 20,
then 2b 2 and B 000...0, 111...1 , but this
is impossible.
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Case 2) 111...1 is not in B. T c1 c2 ... cp
R(x, c1 c2 ... cp) x is in 0, 1p and R(x,
c1 c2 ... cp is nontrivial Then card(T)
card(B) - 1, and card(T) is a multiple of
p. Hence 2b 1 mod p. Since 2 is a primitive
root modulo p, either 2b 2p-1 or 2b 20 . If
2b 2p-1 then card(B) 2p-1 .Hence for
arbitrary z in T c1 c2 ... cp , card(kerz) 2.
If 2b 20 , then B 000...0 but this is
impossible.
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Definition. We say that the numbering
? ?0(x), ?1(x), ?2(x), ... of
1-argument partial recursive functions is
computable if the 2-argument function U(n, x)
?n(x) is partial recursive. Definition. We say
that a numbering ? is reducible to the
numbering ? if there exists a total recursive
function f(n) such that, for all n and x,
?n(x) ? f(n)(x). Definition. We say that
a computable numbering f of all 1-argument
partial recursive functions is a Gödel numbering
if every computable numbering (of any class of
1-argument partial recursive functions) is
reducible to f.
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• Definition. We say that a Gödel numbering ? is
  a Kolmogorov numbering if for arbitrary
  computable numbering ? (of any class of
  1-argument partial recursive functions) there
  exist constants c gt 0, d gt 0, and a total
  recursive function f(n) such that
• for all n and x, ?n(x) ?f(n)(x),
• - for all n, f(n) cn d.

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There exist many distinct Kolmogorov numberings.
We now fix one of them and denote it by ? .
Since Kolmogorov numberings give indices for all
partial recursive functions, for arbitrary x and
p, there is an i such that ?i(p)x . Let i(x,
p) be the minimal i such that ?i(p) x . It is
easy to see that if x1?x2, then i(x1,p) ?
i(x2,p) . We consider all binary words x of
the length p and denote by x(p) the word x
such that i(x,p) exceeds i(y,p) for all
binary words y of the length p different from x.
It is obvious that i 2p-1.
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We introduce a partial recursive function µ(z, e,
p) defined as follows. Above when defining
G(p,x) we considered auxiliary function R(x,
c1 c2 ... cp) . To define µ(z, e, p) we
consider all 2p binary words x of the length p.
If z is not a binary word of the length 2p, then
µ(z, e, p) is not defined. If e is not in
0,1, then µ(z, e, p) is not defined. If z
is a binary word of the length 2p and e is in
0,1, then we consider all x in 0,1p such
that R(x,d(z))e(z). If there are no such x,
then µ(z, e, p) is not defined. If there is
only one such x , then µ(z, e, p) x. If
there are two such x, then the
first such x in the lexicographical order, for
e1 µ(z, e, p) the second such
x in the lexicographical order, for e0 If there
are more than two such x, then µ(z, e, p) is not
defined.
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Now we introduce a computable numbering of some
partial recursive functions. This numbering is
independent of p. For each p (independently from
other values of p) we order the set of all the
22p binary words z of the length 2p z0, z1, z2,
... , z22p-1. We define z0 as the word 0 0 0 ...
0. The words z1, z2, ..., z2p are words with
exactly one symbol 1. We strictly follow a rule
"if the word zi contains less symbols 1 than the
word zj, then i lt j". Words with equal number of
the symbols 1 are ordered lexicographically.
Hence z22p-11 1 1...1.
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For each p, we define ?0(p) µ (z0, 0, p) ?1(p
) µ (z0, 1, p) ?2(p) µ (z1, 0, p) ?3 (p) µ
(z1, 1, p) ?4 (p) µ (z2, 0, p) ?5 (p) µ (z2, 1,
p) ... ?22p-2(p) µ (z22p-1, 0, p) ?22p-1(p) µ
(z22p-1, 1, p) For j 22p1, ?j (p) is
undefined.
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We have fixed a Kolmogorov numbering ? and we
have just constructed a computable numbering ?
of some partial recursive functions. Lemma.
There exist constants c gt 0 and d gt 0
(independent of p) such that for arbitrary i
there is a j such that ?i (t) ? j(t)
for all t , and j cid. We consider
generating matrices G(p,x(p)) for linear codes
where p is an odd prime such that 2 is a
primitive root modulo p, and, as defined above,
x(p) is an arbitrary binary word of the length
p such that ?i(p) x(p) implies i 2p-1. We
denote the corresponding linear code by
LC2(p). We prove several lemmas showing that, if
p is sufficiently large, then Hamming distances
between arbitrary two codewords are no less than
4p/19.
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A Hermitian matrix (or self-adjoint matrix) is a
square matrix with complex entries which is equal
to its own conjugate transpose that is, the
element in the ith row and jth column is equal to
the complex conjugate of the element in the jth
row and ith column, for all indices i and j
or written with the conjugate transpose For
example, is Hermitian. If the eigenvalues of a
Hermitian matrix are all positive, then the
matrix is positive definite if they are all
non-negative, then the matrix is positive
semidefinite.
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1/8 0 0 1/8 0 0 0 1/8 0
0 0 0 0 0 0 0 0 1/8 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1/8 0 0
1/8 0 0 0 0 0 0 0 0 0
0 1/8 0 0 1/8 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 1/8 0 0 1/8 0 0
0 0 0 0 0 0 0 0 1/8 0
0 1/8 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
1/8 0 0 0 0 0 0 0 0
0 1/8 0 0 0 1/8 0 0 1/8
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
41
A density matrix is a self-adjoint (or Hermitian)
positive-semidefinite matrix, (possibly infinite
dimensional), of trace one, that describes the
statistical state of a quantum system. The
formalism was introduced by John von Neumann
(according to other sources independently by Lev
Landau and Felix Bloch) in 1927.
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A finite quantum automaton with mixed states is
described by the inititial density matrix, the
unitary matrices corresponding to the input
letters, and the measurements.
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1/8 1/8 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1/8 1/8 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1/8 1/8 0
0 0 0 0 0 0 0 0 0 0
0 0 0 1/8 1/8 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 1/8 1/8 0 0 0 0
0 0 0 0 0 0 0 0 0 0
1/8 1/8 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
1/8 1/8 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1/8 1/8 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0
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1/8 1/8 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1/8 1/8 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1/8 1/8 0
0 0 0 0 0 0 0 0 0 0
0 0 0 1/8 1/8 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 1/8 1/8 0 0 0 0
0 0 0 0 0 0 0 0 0 0
1/8 1/8 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
1/8 1/8 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1/8 1/8 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0
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1/8 0 0 0 0 0 0 0
0 1/8 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1/8 0 0
0 0 0 0 0 0 1/8 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 1/8 1/8 0 0 0
0 0 0 0 0 0 0 0 0 0
0 1/8 1/8 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 1/8 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1/8 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1/8 0 0
0 0 0 0 0 0 1/8 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 1/8 0 0 0 0 0
0 0 0 1/8 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1/8
0 0 0 0 0 0 0 1/8
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1/8 0 0 0 0 0 0 0
0 1/8 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1/8 0 0
0 0 0 0 0 0 1/8 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 1/8 1/8 0 0 0
0 0 0 0 0 0 0 0 0 0
0 1/8 1/8 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 1/8 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1/8 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1/8 0 0
0 0 0 0 0 0 1/8 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 1/8 0 0 0 0 0
0 0 0 1/8 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1/8
0 0 0 0 0 0 0 1/8
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1/8 1/8 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1/8 1/8 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1/8 1/8 0
0 0 0 0 0 0 0 0 0 0
0 0 0 1/8 1/8 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 1/8 1/8 0 0 0 0
0 0 0 0 0 0 0 0 0 0
1/8 1/8 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
1/8 1/8 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1/8 1/8 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0
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1/8 1/8 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1/8 1/8 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1/8 1/8 0
0 0 0 0 0 0 0 0 0 0
0 0 0 1/8 1/8 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 1/8 1/8 0 0 0 0
0 0 0 0 0 0 0 0 0 0
1/8 1/8 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
1/8 1/8 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1/8 1/8 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0
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p1 p2 d1 p3 d2 d3 d4
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When studying Hamming distances and Hamming codes
for binary strings we use the independence of
distinct symbols in the string. There is no such
independence in permutations. Hamming distance
between any two n-permutations either equals 0 or
is no less than 2.
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We wish to construct to construct a set of
n-permutations containing as many as possible
elements such that Hamming distance between any
two of them is as large as possible.
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There are n! n-permutations. We now wish to
construct a subset of them such that Hamming
distance between any two of them is at least
3 . How large the subset may be?
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Even and Odd Permutations
A permutation is even if and only if the diagram
has an even number of crossings.
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There are n! n-permutations. Half of them are
even, half of them are odd. The even permutations
are a subgroup, the odd permutations are a
co-set. The number of these permutations is
large enough. Unfortunately, the Hamming distance
is too small.
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Permutahedron
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How many automorphisms Fano plane has?   Take any
line (e.g. the line 6,1,5 ) and consider what
permutations of vertices are possible that
conserve the line 6,1,5. Every permutation of
this kind is described by positions of the
vertices 0,2,3,4. This gives us 4! 24
possibilities. Since there are 7 lines, the
number of automorphisms is 24 times 7 168.
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The automorphism group of Fano plane is a group
of order 168 generated by
These are the permutation matrices associated
with the coordinate permutations
, ,
, and
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• A finite projective plane of order n is a set of
  n2 n 1 points and a set of
• n2 n 1 subsets of points called lines such
  that
• Any two distinct points determine a unique line
• Any two distinct lines determine a unique point
• Every point is contained in exactly n 1 lines
• Every line contains exactly n 1 points
• The finite projective plane of order n 2, the
  Fano plane, is shown above.
• It is easy to show that
• For every prime power q pk, there exists a
  projective plane of order q. 

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From László Babai's Home Page http//people.cs
.uchicago.edu/laci/
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There are n! n-permutations. We now wish to
construct a subset of them such that Hamming
distance between any two of them is the maximal
possible, i.e. n . How large the subset
may be?

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There are n! n-permutations. We now wish to
construct a subset of them such that Hamming
distance between any two of them is the maximal
possible, i.e. n . How large the subset
may be?
The answer is n . Indeed, such a subset
exists
12345
23451
34512
45123
51234 and it is maximal.
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There are n! n-permutations. We now wish to
construct a subset of them such that Hamming
distance between any two of them is at least
n-1 . How large the subset may be?
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There are n! n-permutations. We now wish to
construct a subset of them such that Hamming
distance between any two of them is at least
n-1 . How large the subset may be?
The answer is n(n-1). For rather many n
such a subset exists 12345 23451
34512 45123 51234 13524
24135 35241 41352 52413
14253 25314 31425 42531
53142 15432 21543 32154
43215 54321

and it is maximal.

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How to construct such a set of permutations for
arbitrary n ?
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How to construct such a set of permutations for
arbitrary n ?
12345 23451 34512
45123 51234 13524 24135
35241 41352 52413 14253
25314 31425 42531 53142
15432 21543 32154 43215
54321 Let us re-write this set as
01234 12340 23401 34012
40123 02413 13024 24135
30241 41302 03142 14203
20314 31420 42031 04321
10432 21043 32104
43210
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How to construct such a set of permutations for
arbitrary n ?
12345 23451 34512
45123 51234 13524 24135
35241 41352 52413 14253
25314 31425 42531 53142
15432 21543 32154 43215
54321 Let us re-write this set as
01234 12340 23401 34012
40123 02413 13024 24135
30241 41302 03142 14203
20314 31420 42031 04321
10432 21043 32104
43210 Have we improved anything?
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01234 12340 23401
34012 40123 02413 13024
24135 30241 41302 03142
14203 20314 31420 42031
04321 10432 21043 32104
43210 is the set of all linear functions
modulo 5 x x1
x2 x3 x4 2x
2x1 2x2 2x3
2x4 3x 3x1 3x2
3x3 3x4 4x
4x1 4x2 4x3
4x4 Since 5 is a prime number, they are
distinct permutations.
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Lemma. If for a pair (n.d) the equality
G(n,d) n(n-1)(n-2)...(d1)d holds, then
G(n-1,d) (n-1)(n-2)...(d1)d.
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It would be very nice, had we proven that
G(n,d) n(n-1)(n-2)...(d1)d. This would
imply
However, to prove this theorem we need much less.
It suffices to
prove that G(n, const.n) superexponential (n)
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On the other hand, we are interested in the
Hamming set of permutations being an algebraic
group with a small number of generating elements.
This would allow us to have a small alphabet for
the languages recognized by quantum and
deterministic automata.
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Lagrange's theorem. For any finite group G, the
order (number of elements) of every subgroup H of
G divides the order of G.
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There are n! n-permutations. We now wish to
construct a subset of them such that Hamming
distance between any two of them is n-2
. How large the subset may be?

89
There are n! n-permutations. We now wish to
construct a subset of them such that Hamming
distance between any two of them is n-2
. How large the subset may be? We expect
n(n-1)(n-2). If n5, then n(n-1)(n-2)60.
Indeed, there are 60 even 5-permutations. If
n6, then n(n-1)(n-2)120. How to construct such
a set?

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This is a group with two generating
elements g1(x) 6x5 and g2(x) x4 3x 1
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Sharply 2-transitive groups were constructed
using linear functions. If n is a prime number,
this gives us the needed group. What we can do
to construct a sharply 3-transitive group?
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Cayley transform In mathematics, the Cayley
transform, named after Arthur Cayley (1821 -
1895), has a cluster of related meanings. As
originally described by Cayley (1846), the Cayley
transform is a mapping between skew-symmetric
matrices and special orthogonal matrices.
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where K  w2  x2  y2  z2
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Thank you
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Rezerve
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