I
nternational Journal “Information
Theories and Applications”, Vol. 17, Number 3, 2010
202
International Journal
INFORMATION THEORIES & APPLICATIONS
Volume 1
7
/ 20
10, Number
3
Editor in chief:
Krassimir Markov
(Bulgaria)
International Editorial Staff
Chairman:
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Alexander Eremeev
(Russia)
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(Armenia)
Alexander Kleshchev
(Russia)
Luis F. de Mingo
(Spain)
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(GB)
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I
nternational Journal
“Information Theories and Applications”, Vol. 17, Number 3, 2010
203
CONSTRAINT
C
ONVEXITY
T
OMOGRAPHY AND LAGRAN
GIAN
A
PPROXIMATIONS
Levon Aslanyan, Artyom Hovsepyan, Hasmik Sahakyan
Abstract
:
This paper considers one particular
problem of general type of discrete tomography problems and
introduces an approximate algorithm for its solution based on Lagrangian relaxation. A software implementation is
given as well.
Keywords:
discrete tomography, lagrangian relaxation
.
ACM Classific
ation Keywords
: F.2.2 Nonnumerical Algorithms and Problems: Computations on discrete
structures.
Introduction
Discrete tomography is a field which deals with problems of reconstructing objects from its projections. Usually in
discrete tomography object
T
, represents a set of points in multidimensional lattice. Some measurements are
performed on
T
, each of which contains projection, which calculates number of points of
T
along parallel
directions. Given finite number of such measurements it is required to
reconstruct object
T
, or if it is not possible
to find unique reconstruction, construct an object which satisfies given projections. The object existence problem
even by given 3 non

parallel projections is NP

complete [1].
In recent years discrete tomograp
hy draws huge attention because of the variety of mathematical formulations
and applications. Theory of discrete tomography is widely used particularly in the field of medical image
processing, which is based on so called computerized tomography.
Lets cons
ider 2

dimensional lattice and horizontal and vertical projections only. Object
T
can be represented as
a
n
m
)
1
,
0
(
matrix, where 1s corresponds to points in
T
. Vector of row sums corresponds to horizontal
projection and vector of column sums to vertical project
ion. So the problem of reconstructing the object by given
horizontal and vertical projections is equivalent to the
)
1
,
0
(

matrix existence problem with given
R
and
S
row
and column sums. The latter problem was solved independently by Gale and Ryser in 1957. The
y gave sufficient
and necessary condition for such a matrix existence and also proposed an algorithm for the matrix construction.
Same problem with condition of rows inequality was investigated in [6].
In many cases orthogonal projections does not contain
enough information for the objects unique reconstruction.
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
204
That's why often we consider different classes of such problems, where we impose additional constraints, for
instance of geometrical nature. Such constraints narrow the solutions set but at the same
time could make the
problem hard to solve. Typical examples of such constraints are convexity and connectivity.
We say that matrix has row (or horizontal) convexity feature if all ones in the row forms a continuous interval.
Same way we define column (or
vertical) convexity. Connectivity is the feature of moving between 1s in
neighboring cells. In our case we consider only vertical and horizontal connectivity (not diagonal).
Existence problem for connected matrices is NP

complete [2]. Existence problems fo
r horizontally or vertically
convex, and for both horizontally and vertically convex matrices are also NP

complete [3].
Different
authors pro
v
ed that horizontally and vertically convex and connected matrices reconstruction problem
can be solved in polynomi
al time. Given description shows how sensitive are this kind of problems to input
conditions. We see that existence problem's complexity changes along with adding new constraints. At the same
ti
m
e there are a lot of other notations of the problem for those
the complexity is not even known. Particularly that
means that they also lacks easy solution algorithms.
So we consider several problems in the field of discrete tomography, propose ways for constructing such
matrices that satisfy constraints (convex or n
early convex, satisfying given parameters or having values near to
given parameters). Further we will formulate the problems as optimization problems and give ways for their
approximation, based on the integer programming relaxation. The question is that i
nteger programming model is
known for being used to reformulate known NP complex optimization problems. This model's (precise or
approximate algorithms construction) investigation is very important and often this model is used to approximate
optimizations
problems [4, 6]. Implemented algorithms and software package based on that algorithms give an
ability to make calculations either for tomography problem or for similar problems, such that those calculations
might guide us or give approximate or precise sol
utions.
In this paper we will consider one problem from the field of discrete tomography, horizontally convex matrix
existence problem.
Horizontally convex matrix existence problem
Since 1's in the horizontally convex matrix are in neighboring position
then if we count the number of 1's in the
matrices rows, that number for convex matrices will be maximum for the ones with same parameters. That's why
problems that are often considered are related to number of neighboring 1's, their constraints and optimi
zation.
)
,
,
(
1
m
r
r
R
,
)
,
,
(
1
n
s
s
S
,
)
,
,
(
'
'
1
'
m
r
r
R
vectors are given. Is there a
n
m
}
{
,
j
i
x
X
matrix such that
R
is row sum vector for that matrix and
S
is column sums vector, and number
of neighboring 1's in row
i
is equal to
'
i
r
.
I
nternational Journal
“Information Theories and Applications”, Vol. 17, Number 3, 2010
205
}
1
,
0
{
∈
,
,
1
,
∑
)
,
min(
,
,
1
,
∑
,
,
1
,
∑
,
'
1
1
1
,
,
1
,
1
,
j
i
i
n
j
j
i
j
i
i
n
j
j
i
j
m
i
j
i
x
m
i
r
x
x
m
i
r
x
n
j
s
x
In other words the problem is following, find the matrix with horizontal convexity in the class of
)
1
,
0
(
matrices
with given row and column sums. This problem is NP

complete, since for the case when
m
i
r
r
i
i
,
,
1
,
1
'
it's equivalent to the horizontally convex matrix existence
problem. Given particular
case just require the matrix to be horizontally convex by neighboring 1s in the rows.
As we already mentioned lot of combinatorial problems are suitable to represent as integer linear optimization
problems. Lets reformulate our
p
roblem
as integer programming problem.
Lets define
}
1
,
0
{
,
j
i
y
variables the way that it provides neighboring 1's in row
i
.
1
,
,
1
;
,
,
1
),
1
(
&
)
1
(
)
1
(
1
,
,
,
n
j
m
i
x
x
y
j
i
j
i
j
i
This can be done by satisfying
conditions
1
1
,
,
,
1
,
,
,
,
j
i
j
i
j
i
j
i
j
i
j
i
j
i
x
x
y
x
y
x
y
So we reformulate the problem
in
the following way.
)
,
,
(
1
m
r
r
R
,
)
,
,
(
1
n
s
s
S
,
)
,
,
(
'
'
1
'
m
r
r
R
vectors are given: Is there a
n
m
}
{
,
j
i
x
X
matrix such that
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
206
}
1
,
0
{
},
1
,
0
{
∈
)
5
(
,
,
1
,
∑
)
4
(
1
,
,
1
,
,
,
1
1
)
3
(
,
,
1
,
∑
)
2
(
,
,
1
,
∑
)
1
(
,
,
'
1
1
,
1
,
,
,
1
,
,
,
,
1
,
1
,
j
i
j
i
i
n
j
j
i
j
i
j
i
j
i
j
i
j
i
j
i
j
i
i
n
j
j
i
j
m
i
j
i
y
x
m
i
r
y
n
j
m
i
x
x
y
x
y
x
y
m
i
r
x
n
j
s
x
Lagrangean relaxation and variable splitting
So we have horizontal row convex matrix existence problem, which is reformulated as linear integer programming
problem
I
. We also know that problem
I
is NP

complete. To solve this probl
em we will use a method based on
Lagrangian relaxation.
Obviously if we drop some of the constraints we will get problems relaxation. Assume that we can call one or
several constraints hard in the since that by dropping those constraints we can solve
resulted integer
programming problem more easily. Constraints dropping could be embedded in more common method which is
called Lagrangian relaxation. We can apply Lagrangian relaxation to given method in various ways. One of the
ways, which we will use her
e is following, if the problem can be splitted to subproblems, which have common
variables, first split those variables and then relax their equality constraint.
So, we
take
two set of variables
h
j
i
x
,
and
v
j
i
x
,
by duplicating
j
i
x
,
variables,
a
nd reformulate our
problem as
I
nternational Journal
“Information Theories and Applications”, Vol. 17, Number 3, 2010
207
v
j
i
h
j
i
j
i
v
j
i
h
j
i
i
n
j
j
i
h
j
i
h
j
i
j
i
h
j
i
j
i
h
j
i
j
i
i
n
j
h
j
i
j
m
i
v
j
i
x
x
y
x
x
m
i
r
y
n
j
m
i
x
x
y
x
y
x
y
m
i
r
x
n
j
s
x
,
,
,
,
,
'
1
1
,
1
,
,
,
1
,
,
,
,
1
,
1
,
)
6
(
}
1
,
0
{
},
1
,
0
{
∈
,
)
5
(
,
,
1
,
∑
)
4
(
1
,
,
1
,
,
,
1
1
)
3
(
,
,
1
,
∑
)
2
(
,
,
1
,
∑
)
1
(
We split our original problem using variable splitting to two problems, each of which has its own variable set and
which would be independent without constraint (6). From this point of view constraint
(6) is
the
hardest one. We
will relax co
nstraint (6) using Lagrangian relaxation with coefficients
j
i
λ
,
.
We get following problem
)
(
λ
VSI
, and its optimal value is
)
(
λ
v
VSI
.
}
1
,
0
{
},
1
,
0
{
∈
,
)
5
(
,
,
1
,
∑
)
4
(
1
,
,
1
,
,
,
1
1
)
3
(
,
,
1
,
∑
)
2
(
,
,
1
,
∑
)
1
(
)
)
(
max(
,
,
,
'
1
1
,
1
,
,
,
1
,
,
,
,
1
,
1
,
,
,
,
,
j
i
v
j
i
h
j
i
i
n
j
j
i
h
j
i
h
j
i
j
i
h
j
i
j
i
h
j
i
j
i
i
n
j
h
j
i
j
m
i
v
j
i
j
i
v
j
i
h
j
i
j
i
y
x
x
m
i
r
y
n
j
m
i
x
x
y
x
y
x
y
m
i
r
x
n
j
s
x
x
x
λ
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
208
Then using same method we can further split the problems into subproblems for rows and columns, which itself is
reducing to the finding
of simple path, with given number of edges and biggest weight on directed graph.
We can approach the problem in other way, by relaxing constraint (3) we would split the problem into two
subproblems with
j
i
x
,
and
j
i
y
,
variables. But this paper is limited with fi
rst approach.
Obviously problem
)
(
λ
VSI
is relaxation of problem
I
, hence
)
(
λ
v
VSI
is upper limit for value of
I
. Find
best upper limit means to solve Lagrangian dual problem which is
)
(
min
λ
v
v
VSI
λ
VSD
This is convex non

differential optimization problem: There are different method
s for solving this problem. One of
them is subgradient optimization method. Subgradient optimization on each step calculates the value of
)
(
λ
v
VSI
for given
j
i
λ
,
, in this case that equals to solving following
m
independent problems
(*)
We will try to solve
these problems using algorithm for finding simple path on acyclic directed graph with biggest
cost and given number of edges.
}
1
,
0
{
},
1
,
0
{
∈
∑
1
,
,
1
1
∑
)
max(
'
1
1
1
1
1
1
j
j
n
j
j
j
j
j
j
j
j
j
n
j
j
n
j
j
j
y
x
r
y
n
j
x
x
y
x
y
x
y
r
x
x
c
I
nternational Journal
“Information Theories and Applications”, Vol. 17, Number 3, 2010
209
Decomposed problem on graph and the solution
We consider directed graph
)
,
(
E
V
G
which vertex set consists of vertexes for each
j
x
variab
le plus
s
source and
o
destination. We define edges
in
following way
)
,
(
j
x
s
with weight
j
c
1
),
,
(
j
i
x
x
j
i
with weight
j
c
)
,
(
o
x
j
with weight 0
Consider the paths form
s
to
o
. Only r variables corresponding to
j
x
vertexes, are 1's according to (*) and
among them
'
r
is neighbori
ng 1's. Hence we are interested only in those paths from
s
to
o
that ha
ve
only
r
vertexes and there are only
'
r
with neighboring 1's. We need to find among those paths, the one that has
maximum weight. Now by assigning 1's to variables corresponding to vert
exes we will get solution to
the
problem
(*)
.
Now lets give algorithmic description.
Let
)
,
(
p
j
z
is weight of the longest path from
s
to
j
x
vertex with
p
vertexes on it. Lets
)
,
,
(
q
p
j
w
is
weight of the longest path from
s
to
j
x
which has p vertexes on it and there are
q
neighboring vertexes with
corresponding variables equal to 1. In this case
)
,
(
p
j
z
and
)
,
,
(
q
p
j
w
can be calculated the following
way. First of all consider
)
,
(
p
j
z
s
o
x
j

1
x
j
c
j
c
j
0
c
j
c
j
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
210
)
)
1
,
(
max(
)
,
(
)
1
,
(
j
x
u
j
c
p
u
z
p
j
z
c
j
z
j
And the optimal value we're looking for is
)
,
(
max
)
,
(
r
x
z
r
o
z
j
j
.
This part of the problem is solving
in
the following way.
For
given
n
and
r
]
[
,
,
p
j
p
j
z
Z
array is constructed, where
n
j
,...,
1
and for
j
1
,...,
0
j
p
. In reality for fixed
r
its enough to consider
r
p
,...,
1
layers, but
n
p
,...,
1
will satisfy calculations needed for all
r
.
First of all
1
,
1
z
value is calculated. That's equal to
1
c
. All values of row
1
p
are calculated
in the
same way
j
j
c
z
1
,
. To calculate
p
j
z
,
by our formula we need to know values for
1
p
and for all
1
,...,
1
j
indexes. But in row
1
p
first non

zero value is in
1
p
j
position, which is on diagonal. So calculations
can be done sequentially on
,...
,...,
1
r
p
rows and in rows in order
n
p
j
,...,
. This constructs are
needed for software implementation and these give ability to measure number of operations in calcul
ation. It
doesn't exceed
3
n
, which means polynomial complexity.
Maximal weight paths can be stored in
a
separate array. They can be stored as 0,1 vectors or as indexes of non
1
n
p
r
z
j
,p
j
1
n
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
211
zero elements which however won't significantly decrease number of computations.
Now lets calculate values of
)
,
,
(
q
p
j
w
. First of all lets consider edge values. From
)
,
,
(
q
p
j
w
we have
maximal weight path from
s
to
j
x
which has
p
vertexes and there are
q
pairs with neighboring 1's.
1
p
q
and lets
p
's are decreased up to
1
q
.
)
,
1
,
(
q
q
j
w
's can be non

zero starting from
1
q
j
. For bigger
q
's and smaller
j
's
)
,
,
(
q
p
j
w
's are equal to 0.
Interestingly
q
can't be very small. If
2
1
j
p
then
q
can't be 0 (at least 2 vertexes must have
neighboring indexes).
Let
3
1
j
τ
. In that case
τ
vertex pairs still might not be neighbors, which gives
τ
2
vertexes.
After that
any new vertex addition would add 2 new pairs.
Now lets consider common case. For calculating
)
,
,
(
q
p
j
w
lets consider class where for
j
j
p
and for
p
j
,
pairs
1
p
q
. This class is larger than needed but in reality it doesn't differ much from the minimal
class which is necessary for calculations. For slight transition of edge values class is zeroed before performing
calculations. Lets investigate value of
)
,
,
(
q
p
j
w
. We do cha
in calculations and on each step consider 2
cases
1
1
j
x
and
0
1
j
x
. So we get following values
j
c
q
p
j
w
)
1
,
1
,
1
(
and
)
)
,
1
,
(
(
max
1
j
x
u
c
q
p
u
w
j
We are interested in maxim
um
of these values.
In
j
c
q
p
j
w
)
1
,
1
,
1
(
all indexes are less than preceding and we assume that this value is already
calculated in previous steps. Fo
r calculating
)
)
,
1
,
(
(
max
1
j
x
u
c
q
p
u
w
j
we do next step in chain
calculation.
)
)
,
2
,
(
(
max
)
1
,
2
,
1
(
1
j
u
v
j
c
q
p
v
w
c
q
p
u
w
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
212
And the needed optimal value is
)
'
,
,
(
max
)
'
,
,
(
r
r
x
w
r
r
o
w
j
j
. This problem practically can be solved
in
following way.
For given
n
,
r
,
'
r
we construct the class given above, array
]
[
,
,
,
,
q
p
j
q
p
j
w
W
, where for
n
j
,...,
1
and
j
j
p
,...,
1
and for pair
p
j
,
1
,...,
0
p
q
.
In reality for fixed
r
it's enough to consider
r
p
,...,
1
layers and for
q
all values where
1
r
q
.
But calculations must be done in such sequence to be executable.
First
0
,
1
,
1
w
values are calculated
,
1
0
,
1
,
1
c
w
, all values in row
1
p
are calculated in the same way
j
j
c
w
0
,
1
,
. More,
0
q
values were already considered. To calculate
q
p
j
w
,
,
based on our formula we
need to know values for
1
p
and all
1
,...,
1
j
. But in layer
1
p
with current
q
value is either 0 or
already calculated. Then calculations can be done in layers
,...
,...,
1
r
p
sequentially and in layers in order
of
n
p
j
,...,
. Given constructions are needed for software implementation and give ability to measure
number of calculations. Thos
e are not more than
4
n
which means polynomial complexity.
Maximal weight paths that we're looking for could be stored in separate array as 0,1 vectors or as array of
1
n
p
r
w
j
,p,q
0
n

1
1
j
n
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
213
indexes with non

zero values, which however won't significantly lower number of calculation
s.
Software implementation
Based on given methods a software
system
with
an
UI was implemented, which can be used to solve some
problems from the field of discrete tomography based on Lagrangian relaxation.
There are several fields which are used for data
input. Since we are solving problems in the field of discrete
tomography so input data are projections, in our case row sums and column sums. Also we are giving specific
problem description by additional constraints. So we have special fields for that pur
pose. Then there is special
control which can be used to reformulate given problem as mathematical programming problem. Then we can
choose one or several constraints which we want to relax. Also we
can do variable splitting etc. And there is a
n
output wind
ow which is used for displaying results. For example value of Lagrangian Dual or variables difference
as a result of splitting.
In given example as a problem is considered horizontal row convexity existence problem.
Now lets describe one of the main
classes in the implementation, ProblemBase abstract class. This class is base
for all problem. Class encapsulates problem data. It also has several virtual functions which are used for problem
solution. For example function which reformulates the problem a
s mathematical programming problem, chooses
constraints for relaxation. Important function in ProblemBase is Solve method, which invokes the method for
I
nternational Journal
“Information Theories and Applications”, Vol. 17, Number 3, 2010
214
specific problem. Since most of the problems are reducing to relatively easy problems on graphs and meth
ods for
those solutions can be used for different problems we put those methods in
a
separate library.
Bibliography
1. Gardner R.J., Gritzmann P., Prangenberg D., On the computational complexity of reconstructing lattice sets from their X

rays. Technical R
eport (970

05012), Techn. Univ. Munchen, fak. f. math, 1997.
2. G.J. Woeginger. The reconstruction of polyominoes from their orthogonal projections. Inform. Process. Lett., 77:225

229,
2001.
3. E. Barcucci, A. Del Lungo, M. Nivat, and R. Pinzani. Reconstru
cting convex polyominoes from horizontal and vertical
projections. Theoret. Comput. Sci., 155:321

347, 1996.
4. G. Dahl and T. Flatberg. Lagrangian decomposition for reconstructing hv

convex (0, 1) matrices, Report 303, University of
Oslo, p. 1

13, 2002.
5
. M. Guignard and S. Kim. Lagrangian decomposition: a model yielding stronger lagrangian bounds. Math. Prog., 39:215

228, 1987.
6. H.J. Ryser. Combinatorial properties of matrices of zeros and ones. Canad. J. Math., 9:371

377, 1957.
7. D. Gale. A theorem
on flows in networks. Paci_c J. Math., 7:1073

1082, 1957.
Authors' Information
Levon
Aslanyan
–
Head
of
Department,
Institute
for
Informatics
and
Automation
Problems,
P.Sevak
St.
1,
Yerevan
14,
Armenia,
e

mail:
lasl@.sc
i.am
Artyom Hovsepyan
–
Researcher,
Institute
for
Informatics
and
Automation
Problems,
P.Sevak
St.
1,
Yerevan
14,
Armenia,
e

mail:
artyom.hovsepyan@gmail.com
Hasmik Sahakyan
–
Leading Researcher, Institute for Informatics and Automation Problems, NAS RA, P.Sevak
St. 1, Yerevan 14, Armenia, e

mail:
hasmik@ipia.sci.am
I
nternational Journal
“Information Theories and Applications”, Vol. 17, Number 3, 2010
215
ON
HYPERSIMPLE
wtt

MITOTIC
SETS
, WHICH ARE
NOT
tt

MITOTIC
Arsen H. Mokatsian
Abstract:
A Т

complete wtt

mitotic set is composed, which is not tt

mitotic. A relation is found out between
structure of computably enumerable sets and the density of their unsolvability degrees.
Let us a
dduce some definitions:
A computably enumerable (c.e.) set is
tt

mitotic (
wtt

mitotic) set if it is the disjoint union of two c.e.
sets both of the same
tt

degree (
wtt

degree)
of unsolvability.
Let A be an infinite set.
f
majorize
A if
)
)
(
(
n
z
n
f
n
, where
,
,
1
0
z
z
are the members of A
in strictly increasing order.
A
is hyperimmune
(abbreviated h

immune) if A is infinite and (
f
recursive
) [f does not majorize A].
A is hypersimple if A is c.e. and
A
is hyperimmune.
A
is hype
r
hyperimmune
if A is infinite and
recursive
f
so that
[
]
[
&
]
[
)
(
)
(
)
(
)
(
&
v
f
u
f
u
f
u
f
W
W
v
u
v
u
A
W
W
u
finite
is
].
A is hyperhypersimple if A is computably enumerable and
A
is hyperhyperimmune.
We shall denote T

degrees by small bold Latin letters.
A degree
0
a
is low
if
0
a
(i.e. if the jump
a
has the lowest degree possible).
Theorem (Martin [6]).
a
is the degree of a maximal set
a
is the degree of a hypersimple set
[
a
is c.e. and
0
a
].
Theorem (R. Robinson [8]). Let
b
and
c
be c.e. degrees such that
b
c
and
c
is low. Then there
exist incomparable low c.e. degrees
0
a
and
1
a
, such that
1
0
a
a
b
and
c
a
i
, for
2
i
.
Griffiths ([3]) proved that there is a low c.e. T

degree
u
such that if
v
is a c.e. T

degree and
u
v
then
v
is not completely mitotic.
In this article it is proved the following theorem:
Theorem.
There exists a low c.e. T

degree
u
such that if
v
is a c.e. T

degree and
u
v
then
v
contains
hypersimple
wtt

mitotic set, which is not
tt

mitotic.
From the abovementioned theorems of Martin and
R. Robinson follows that it is impossible to replace
hypersimple by hyperhypersimple.
Keywords
:
computably enumerable (c.e.) set,
mitotic,
wtt

reducibility,
tt

reducibility, hypersimple set, low
degree.
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
216
ACM Classific
ation Keywords
: F. Theory of Computation, F.1.3 Complexity Measures and Classes.
Introduction
We shall use notions and terminology introduced in [9], [10].
The definitions of
tt

and
wtt

reducibilities are from [9].
)
(
x
denotes, that
)
(
x
is defined, and
)
(
x
denotes, that
)
(
x
is undefined.
Definition.
The order pair
<<
k
x
x
,
,
1
>,
>
, where
<
k
x
x
,
,
1
>
is a
k

tuple of integers and
is a
k

ary
Boolean function (
0
k
) is called a
truth

table condition
(or
tt

condition
)
of
norm
k
. The set {
k
x
x
,
,
1
} is
called
the
associated set of the
tt

condition
.
Definition.
The
tt

condition
<<
k
x
x
,
,
1
>,
>
, is
satisfied
by
A
if
1
)
(
,
),
(
1
k
A
A
x
c
x
c
, where
A
c
is characteristic function for
A
.
Each
tt

condition is a finite object; clearly an effective coding can be chosen which maps all
tt

conditions (of
varying norm) onto
, on condition that
x
z
z
x
x
tt
condition

the
of
set
associated
of
member
the
is

max
)
(
.
Assume henceforth that a particular such coding has been chosen. Where we speak of “
tt

condi
tion
x
”, we
shall m
ean the
tt

condition with the code number
x
.
Definition.
A
is
truth

table reducible to
B
(notation:
B
A
tt
) if there is a computable function
f
such that
for all
x
, [
tt
A
x

condition
)
(
x
f
is satisfied by
B
]. We also abbreviate “truth

table reducibility” as
“
tt

reducibility”.
Definition.
A
is weak truth

table reducible to
B
(notation:
B
A
wtt
) if
(
z
)[
B
z
A
c
(
computable
f
)
(
x
)[
)
(
x
f
D
contains all integers whose membership in
B
is used in the computation of
)
(
x
B
z
]].
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
217
Definition.
A c.e. set is
tt

mitotic (
wtt

mitotic) set if it is the disjoint union of two c.e. sets both of the same
tt

degree (
wtt

degree) of unsolvability.
Let
B
A
tt
and
)
(
x
[
tt
A
x

condition
)
(
x
f
is satisfied by
B
] and
f
n
. Then we say that
B
A
tt
by
n
.
Let us modify denotations defined in [
4
] with the purpose to adapt them to our theorem.
We say that
)
,
,
,
(
1
1
o
o
A
A
is
tt

mitotic splitting of
A
if
o
A
and
1
A
are c.e.,
A
A
A
o
1
,
1
A
A
o
,
0
A
A
tt
by
0
and
0
A
A
tt
by
0
1
A
A
tt
by
1
.
Let
h
be a recursive function from
onto
4
.
Define (
i
i
i
i
Z
Y
,
,
,
) to be a qu
adruple (
3
2
1
,
,
,
i
i
i
i
W
W
o
), where
)
,
,
,
(
)
(
3
2
1
0
i
i
i
i
i
h
. If
A
is c.e. then
we say that the
non

tt

mitotic condition
of
i
order is satisfied for
A
, if it is not
the case that (
i
i
i
i
Z
Y
,
,
,
) is
a
tt

mitotic splitting of
A
.
Denotation
.
otherwise
n
if
x
s
n
i
u
s
i
i
k
tt
n
,
0
,
)
(
,
,
)
,
,
(
,
,
where
tt

condition
)
(
n
i
= <<
i
k
i
n
x
x
,
,
1
>,
i
n
> .
We
define two computable functions that will be of use later.
1.
n
m
s
m
Z
i
u
n
m
s
m
Y
i
u
n
s
n
i
k
s
tt
s
tt
:
)
,
,
,
(
:
)
,
,
,
(
,
max
)
,
,
(
3
2
,
2.
],
Z
)
(

1
)
(
)
(

1
)
(
[
)
,
,
(
by
satisfied
condition
by
satisfied
condition
i
i
A
i
i
A
n
tt
n
c
Y
n
tt
n
c
n
s
i
A
L
where
)
,
,
,
(
)
(
3
2
1
0
i
i
i
i
i
h
.
Adduce some information, concerning hypersimple sets.
Definitions
.
Let
A
be an infinite set.
f
majori
z
es A
if
)
)
(
(
n
z
n
f
n
, where
,
,
1
0
z
z
are the members of
A
in
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
218
strictly increasing order.
A
is
hyperimmune
(abbreviated
h

immune
) if
A
is infinite and (
f
recursive
) [
f
does not majori
ze
A
].
A
is
hypersimple
if
A
is computably enumerable and
A
is hyperimmune.
A useful characterization of hyperimmune sets is given in
t
he following theorem.
Theorem
(Kuznecov, Medvedev, Uspenskii [ 7]).
A
is hyperimmune
A
is infinite and
]
[
]
)[
(
&
]
[
)
(
)
(
)
(
)
(
v
f
u
f
u
f
D
D
v
u
v
u
A
D
u
f
recursive
.
Definitions.
A
is
hype
r
hyperrimmune
if
A
is infinit
e and
f
recursive
[
]
[
&
]
[
)
(
)
(
)
(
)
(
&
v
f
u
f
u
f
u
f
W
W
v
u
v
u
A
W
W
u
finite
is
].
A
is
hyperhypersimple
if
A
is computably enumarable and
A
is hyperhyperimmune.
A degree
0
a
is low
if
0
a
(i.e. if the jump
a
has the lowest degree possible).
Theorem
(Martin [6]).
a
is the degree of a maximal set
a
is the degree of a hypersimple set
[
a
is c.e.
and
0
a
].
Theorem
(R. Robinson [8]). Let
b
and
c
be c.e. degrees such that
b
c
and
c
is low. Then there exist
incomparable low c.e. degrees
0
a
and
1
a
, such that
1
0
a
a
b
and
c
a
i
, for
2
i
.
Griffiths ([3]) proved that there is a low c.e. T

degree
u
such that if
v
is a c.e. T

degree and
u
v
then
v
is not
completely mitotic.
Let us prove the following theorem.
Theorem.
There exists a low c.e. T

degree
u
such that if v
is a c.e. T

degree and
u
v
then
v
contains
hypersimple
wtt

mitotic set, which is not
tt

mitotic.
From the abovementioned theorems of Martin and R. Robinson follows that it is impossible to replace
hypersimple
by
hyperhypersimple.
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
219
Proof.
This statement is proved using a finite injury priorit
y argument. We construct a member
U
of
u
in stages
s
,
s
s
U
U
=
. We also construct sets
e
e
V
}
{
to witness that each c.e. T

degree in upper cone of
u
contains a
wtt

mitotic but non

tt

mitotic set.
Denote
0
1
0
\
=
},
)
=
(2
)
(
:
{
=
x
y
y
x
.
Construct
U
,
e
e
V
}
{
to satisfy, for all
e
, the requirements:
)
(
}
{
:
e
e
N
U
e
has a limit in
s
, the stage.
i
e
R
,
: The non

tt

mitotic condition of order
i
is satisfied for
e
V
.
&
]
[
:
)
(
,
&
(
A
D
u
P
u
f
i
i
e
computable
total
is
)
(
)
(
]
)[
(
)
(
)
(
)
(
)
e
z
v
f
u
f
V
D
z
D
D
v
u
v
u
i
.
e
V
e
e
W
P
=
:
~
for some computable functional
.
We also ensure by permitting that
e
T
e
W
U
V
and else
1
0
e
wtt
e
V
V
(where
1
1
0
0
=
&
=
e
e
e
e
V
V
V
V
).
If
e
T
W
U
then the above ensure that
e
T
e
T
e
W
W
U
V
and
e
V
is not
tt

mitotic. Hence,
)
(
e
W
deg
is
not
tt

mitotic but is
wtt

mitotic, and
)
(
=
U
deg
u
is the required degree.
Let
,
be computable bijective pairing function increasing in both coordinates. At each stage
s
place markets
)
,
,
(
s
x
e
on elements of
s
e
V
,
. Values of
will be used both as witnesses to pr
event the
tt

mitoticity of
e
V
sets (by corresponding
i
i
i
i
Z
Y
,
,
,
) and
to ensure that
e
W
is
T

reducible to
e
V
. Initially
2)
1)
,
4(
=
,0)
,
(
x
e
x
e
for all
x
e
,
.
Also define a function
)
,
,
(
s
i
e
for all
i
e
,
(at each stage
s
),
i
i
e
=
,0)
,
(
for all
i
e
,
. We use
to
ensure that only members of sufficiently large magnitude enter
U
at stage
s
, so we can satisfy the lowness
requirements
e
N
.
According to the theorem
(Kuznecov, Medvedev, Uspenskii)
the satis
faction of
i
e
P
,
(for all
i
) ensure the
hypersimplicity of
e
V
.
Order the requirements in the following priority ranking:
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
220
,
,
,
,
,
,
,
,
,
2
2
2
1
1
1
0
0
0
P
R
N
P
R
N
P
R
N
The
e
e
P
}
~
{
do not appear in this ranking.
e
N
requires attention
if it is not
satisfied
and
]
)[
(
}
{
s
e
e
U
.
i
e
R
,
requires attention
if it is not
satisfied
and
)
(
&
)
(
)
(
x
x
x
s
i
s
i
y
,
where
)
)
,
,
(
,
(
=
s
s
i
e
e
y
.
)
,
,
,
(
i
i
i
i
Z
Y
is
threatening A through x at stage
s
if it is partially satisfied and all the following hold:
i)
s
i
,
ii)
)
,
,
(
s
i
A
L
x
,
iii)
s
i
s
i
Z
Y
,
iv)
)
)(
(
)
(
m
Z
Y
m
c
s
i
s
i
s
A
for all
)
,
,
(
s
x
i
k
m
.
(Note, that actually
i
e
R
,
is partially satisfied, if
i
e
R
,
requires attention (via some
)
,
,
(
s
y
) and
corresponding
2
,
1
y
y
belong to
e
V
,
1
y
belongs to
1
s
U
. See Construction, Part A , a) ).
We will build
s
s
U
U
=
and
s
e
s
s
V
V
,
=
for all
e
. Initially all requirements
e
N
,
i
e
R
,
are declared
unsatisfied
.
Construction
Stage
0
=
s
. Let
=
0
U
,
=
,0
e
V
for all
e
.
Stage
1
s
.
Part A
.
Act on the highest priority requirement which requires attention, if such a requirement exists.
a) If
e
N
requires attention
then set
s
s
i
e
s
i
e
,
ˆ
,
ˆ
=
1
,
ˆ
,
ˆ
for each
e
i
e
ˆ
,
ˆ
. This action prevents
injury to
e
N
by lower priority requirements as we assume that
s
bounds the use of the halting computation.
Define
)
1
,
,
(
s
not specified in Part A to be the same as
)
,
,
(
s
.
Declare
e
N
satisfied; declare all lower priority
R
,
N
unsatisfied.
I
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“Information Theories and Applications”, Vol. 17, Number 3, 2010
221
If
i
e
R
,
require attention
via
s
s
i
e
e
y
1),
,
,
(
,
=
then set
2}
1,
{
=
~
,
1
,
y
y
V
V
s
e
s
e
and
1}
{
=
~
1
y
U
U
s
s
. (Note that such
i
e
,
cannot be
e
, so
)
,
,
(
1)
,
,
(
s
i
e
s
i
e
). Declare
i
e
R
,
partially satisfied.
Define
1
,
1
,
~
,
~
s
e
s
e
U
V
not specified in Part A, a) to be the same as
1
,
1
,
,
s
e
s
e
U
V
respectively.
b)
If
(
i
i
i
i
Z
Y
,
,
,
)
is threatening
1
,
~
s
e
V
through
y
at stage
1
s
(so
i
e
R
,
is partially satisfied via
)
,
,
,
,
(
s
s
i
e
e
y
),
then
set
}
{
~
~
~
1
,
1
,
y
V
V
s
e
s
e
and
}
{
~
=
~
~
1
1
y
U
U
s
s
.
If
i
e
R
,
is partially satisfied, via
)
,
,
,
,
(
s
s
i
e
e
y
,
whether (
i
i
i
i
Z
Y
,
,
,
) is threatening
1
,
~
s
e
V
through
y
at stage
1
s
or not define
=
1)
,
,
,
,
(
1
s
s
i
e
e
s
s
s
i
e
e
),
,
,
(
,
.
Define
)
1
,
,
(
,
~
~
,
~
~
1
1
,
1
,
s
U
V
s
e
s
e
not specified in Part A, b) to be the same as
)
,
,
(
,
~
,
~
1
,
1
,
s
U
V
s
e
s
e
respectively.
Such definition of
1
allow us to satisfy
i
e
R
,
requirement (after Part A) whether
(
i
i
i
i
Z
Y
,
,
,
) is
threatening
1
,
~
s
e
V
through
y
or
not
(if don't take into consideration higher priority requirements).
Declare
i
e
R
,
satisfied; declare all
lower priority
N
R
,
unsatisfied
.
Part B
. If
s
e
s
e
W
W
x
,
1
,
\
then set
and
1)
,
,
(
~
~
=
1
1
,
*
1
,
s
x
e
V
V
s
e
s
e
.
j
all
for
ω
1)
1),
1,
,
(
,
(
=
1)
,
,
(
1
2
s
s
j
x
e
e
s
j
x
e
Find all
i
ˆ
such that
1)
,
,
(
)
1),
,
ˆ
,
(
,
(
1
s
x
e
s
s
i
e
e
and declare
i
e
R
ˆ
,
unsatisfied
for each such
i
ˆ
.
Define
)
1
,
,
(
,
2
*
1
,
s
V
s
e
not specified in Part B to be the same as
)
1
,
,
(
,
~
~
1
1
,
s
V
s
e
respectively.
Note that for all
)
,
,
(
,
s
i
e
s
is increasing in both
e
and
i
.
Part C.
Let
}
ˆ
ˆ
ˆ
ˆ

1
,
ˆ
,
ˆ
{
max
{
1
e,i
i
,
e
i
,
e
s
i
e
m
s
with
all
for
,
I
nternational Journal
“Information Theories and Applications”, Vol. 17, Number 3, 2010
222
}
}
ˆ
ˆ
ˆ

1
,
ˆ
,
ˆ
{
2
e,i
,i
e
i
,
e
s
i
e
with
all
for
.
If
)
(
)
(
1
,
z
z
s
i
, denote
)
(
0
z
D
y
y
z
i
. Then, if
0
0
m
z
and
i
e
P
,
is not satisfied,
set
}
{
~
~
&
}
1
,
{
)
(
)
(
0
1
1
*
1
,
1
,
z
U
U
y
y
V
V
z
D
y
y
s
s
s
e
s
e
i
.
Set
.
all
for
i
i
s
s
s
i
e
e
s
s
i
e
e
ˆ
,
1)
1),
,
ˆ
,
(
,
(
=
)
1
,
1)
,
ˆ
,
(
,
(
2
Define
)
1
,
,
(
,
,
1
,
1
,
s
U
V
s
e
s
e
not specified in Part C, to be the same as
)
1
,
,
(
,
~
~
,
2
1
,
*
1
,
s
U
V
s
e
s
e
respectively.
Declare
i
e
P
,
satisfied
, declare all lower priority
N
R
,
unsatisfied.
Verification
Lemma 1.
For all
e
,
i
:
1.
e
N
is met,
)
,
(
=
)
,
,
(
lim
i
e
s
i
e
s
exists.
2
i
e
R
,
is met,
s
s
i
e
e
s
),
,
,
(
,
lim
exists.
Proof.
By induction on
i
e
j
,
.
Suppose there exists a stage
0
s
such that for all
i
e
ˆ
,
ˆ
with
j
i
e
ˆ
,
ˆ
:
1.
i
e
N
ˆ
,
ˆ
is met and never acts after stage
0
s
,
)
ˆ
,
ˆ
(
)
,
ˆ
,
ˆ
(
lim
i
e
s
i
e
s
exists and is attained by
0
s
.
2.
i
e
R
ˆ
,
ˆ
is met and never acts after stage
0
s
,
)
,
ˆ
(
),
,
ˆ
,
ˆ
(
lim
s
s
i
e
e
s
exists and is attained by
0
s
.
1). The proof of point 1
is similar to Lemma 1 of Theorem 2.2.2 [3].
After stage
0
s
the requirements
i
e
N
ˆ
,
ˆ
,
i
e
R
ˆ
,
ˆ
(for all
i
e
ˆ
,
ˆ
with
j
i
e
ˆ
,
ˆ
) do not injury
j
N
. Positive
requirements
i
e
P
ˆ
,
ˆ
for all
i
e
ˆ
,
ˆ
with
j
i
e
ˆ
,
ˆ
, can injury
j
N
only finitely. So there is stage
1
s
after which if
j
N
receives attention,
then it is met and never injured, so there is a
1
2
s
s
after which
j
N
does not receive
attention. (Else set
1
2
s
s
). Thus
)
,
(
)
1
,
,
(
2
i
e
s
i
e
, because
)
1
,
,
(
2
s
i
e
is not changed after.
2). Now consider point 2. Note, that positive requirements
i
e
P
,
,
e
P
~
injury each of the requirements
I
nternational Journal
“Information Theories and Applications”, Vol. 17, Number 3, 2010
223
with lower priority only finitely.
Let the stage
1
s
is such stage, that
0
1
s
s
and
i
e
N
ˆ
,
ˆ
,
i
e
R
ˆ
,
ˆ
,
i
e
P
ˆ
,
ˆ
,
i
e
P
ˆ
,
ˆ
~
(for all
i
e
ˆ
,
ˆ
with
j
i
e
ˆ
,
ˆ
) are met and never acts after stage
1
s
.
The following Lemma is used (in [4
], [ 5]) for building the non

T

mitotic set :
Lemma.
If
i
i
i
i
Z
Y
,
,
,
is threating
A
through
x
at stage
s
A
A
x
s
,
and for all
x
m
such that
)
,
,
(
s
x
i
k
m
we have
)
(
m
A
A
s
m
, then the non

T

mitotic condition of order
i
is satisfied for
A
.
Similar lemma is thrue for
tt

reduci
bility.
Let
0
s
is such stage that
i
e
N
ˆ
,
ˆ
,
i
e
R
ˆ
,
ˆ
i
e
P
ˆ
,
ˆ
are met and never acts after stage
1
s
.
If there isn’t such
)
,
1)
,
,
(
,
(
s
s
i
e
e
y
(where
1
s
s
), that
i
e
R
,
is partially satisfied (via
y
), then
i
e
R
,
is met.
If there exists such
y
and
i
i
i
i
Z
Y
,
,
,
never threatens
e
V
through
y
after stage
s
, then certainly the
condition is satisfied. On the other hand, if
i
i
i
i
Z
Y
,
,
,
threatens
e
V
through
y
at time
s
t
, then put
y
into
e
V
at time
1
t
, and never put any other number
)
,
,
(
t
y
i
k
into
e
V
after stage
t
, so
i
e
R
,
is met.
Lemma 2.
i
e
P
,
is met.
According to Lemma1
)
ˆ
,
ˆ
(
)
,
ˆ
,
ˆ
(
)
(
0
0
i
e
s
i
e
s
[
, for all
with
i
e
ˆ
,
ˆ
)
,
(
)
ˆ
,
ˆ
(
i
e
i
e
&
)
ˆ
,
ˆ
(
,
ˆ
),
,
ˆ
,
ˆ
(
,
ˆ
&
0
0
i
e
e
s
s
i
e
e
for all
]
with
)
,
(
)
ˆ
,
ˆ
(
ˆ
,
ˆ
i
e
i
e
i
e
.
Denote
)}
,
(
)
ˆ
,
ˆ
(
ˆ
,
ˆ

)
ˆ
,
ˆ
(
,
ˆ
{
)},
,
(
)
ˆ
,
ˆ
(

)
ˆ
,
ˆ
(
{
max
0
i
e
i
e
i
e
i
e
e
i
e
i
e
i
e
m
with
all
for
all
for
.
Then if
&
]
[
)
(
&
(
A
D
u
u
f
i
computable
total
is
)
(
0
,
)
(
)
(
}
,
,
1
,
0
{
&
)
(
)
(
)
(
]
)[
(
)
(
0
)
z
s
i
s
v
f
u
f
i
D
m
z
z
s
D
D
v
u
v
u
.
Thus, according to constraction
i
e
P
,
is met, because
)
(
z
i
D
enters
e
V
.
I
nternational Journal
“Information Theories and Applications”, Vol. 17, Number 3, 2010
224
Lemma 3
.
e
T
e
W
U
V
.
Proof.
By permitting: in the construction a number
k
enters
e
V
only if a number less than or equal to
k
enters
U
or enters
e
W
.
Lemma 4.
For all
e
,
e
P
~
is satisfied, that is
e
V
e
W
=
.
Proof.
To determine whether
e
W
z
we need to find a stage such that
)
,
,
(
s
z
e
has attained its limit.
e
V
computably determines
)
,
(
,
,0),
(
z
e
e
(note that
)
,
,
(
s
y
e
changes only if a number
)
,
,
(
s
y
e
enters
e
V
).
Find a stage
z
s
such that
z
s
e
V
,
¹
e
z
V
=
1
¹
,
1
z
where
)}
,
(
,
,0),
(
{
max
=
z
e
e
z
. Then
e
W
z
iff
z
s
e
W
z
,
.
Lemma 5.
e
V
is
wtt

mitotic.
Proof. 1) Prove
1
0
e
wtt
e
V
V
(and hence
1
e
wtt
e
V
V
).
To determine whether
0
e
V
x
find such stage
s
, that
1
,
s
e
V
¹
1
=
2
e
V
x
¹
2
x
. Then
0
,
0
s
e
e
V
x
V
x
,
because
i)
if
s
i
ˆ
,
, such that
(
i
i
i
i
Z
Y
,
,
,
) is threatening
1
,
~
s
e
V
through
x
at stage
s
ˆ
, then
0
e
V
x
, only if a
number less than or equal to
1
x
enters,
ii)
otherwise, then find a stage
s
(this stage
s
obligatory is
s
) such that
1
x
1
e
V
. If after the stage
s
such changes happen in
1
,
s
e
V
¹
2
x
, which lead to displacement of marker
)
,
,
(
s
x
e
and
we have
0
,
s
e
V
x
, then
0
e
V
x
. Thus
0
,
0
s
e
e
V
x
V
x
.
2) Prove
0
1
e
wtt
e
V
V
.
The proof is similar to abovementioned in item 1), only without point
ii).
I
nternational Journal
“Information Theories and Applications”, Vol. 17, Number 3, 2010
225
Bibliography
1.
R. G. Downey and T. A. Slaman. Completely mitotic r. e. degrees. In: Ann. Pure Appl. Logic, pp.119
–
152, 41, 1989.
2.
R. G. Downey and M. Stob. Splitting theorems in recursion theory. In: Ann. Pure Appl. Logic, n. 1, v.
65, 106 p., 1993.
3.
E. J. Griffiths. Co
mpletely Mitotic Turing degrees, Jump Classes and Enumeration Degrees. In: Ph. D.
Thesis, University of Wisconsin

Madison, 1998.
4.
H. Lachlan. The priority method. Zeitschrift für mathematische Logik und Grundlagen der Mathematik,
vol.13, pp.1

10, 1967.
5.
R. Ladner. Mitotic Enumerable Sets. In: The Journal of Symbolic Logic, pp. 199

211, v. 38, n. 2, June
1973.
6.
D. A. Martin. A Theorem on Hyperhypersimple Sets. In: The journal of symbolic logic. vol.28, pp.273

278,1963.
7.
Yu.T. Medvedev. On Nonisomorphic Recu
rsively Enumerable Sets (Russian). In: Doklady Akademii
Nauk SSSR, n.s., vol. 102, pp. 211

214, 1955.
8.
R. W. Robinson. Interpolation and embedding in the recursively enumerable degrees. In: Ann. Of
Math. (2), 93, pp. 285

314, 1971.
9.
H. Rogers. Theory of Recu
rsive Functions and Effective Computability. In: McGraw

Hill Book Company,
1967.
10.
R. I. Soare. Recursively Enumerable Sets and Degrees. In: Springer

Verlag, 1987.
11.
Authors' Information
Arsen Mokatsian
–
Leading Researcher at the
Institute for Informatic
s and Automation Problems
of the
National Academy of Sciences of
Republic of
Armenia
, P.Sevak street, no. 1, 0014 Yerevan,
Armenia;
e

mail: arsenmokats
ia
n@gmail.com
I
nternational Journal
“Information Theories and Applications”, Vol. 17, Number 3, 2010
226
A N
EW
A
LGORITM
FOR
THE L
ONGEST
C
OMMON
S
UBSEQUENCE
P
ROBLEM
Vahagn Minasyan
Abstract
:
This paper discusses the problem of determining the longest common subsequence (LCS) of two
sequences. Here we view this problem in the background of the well known algorithm for the longest increasing
subsequence (LIS). This new approach leads us to a new
online algorithm which runs in
(
)
time and in
(
)
space where
is the length of the input sequences and
is the number of minimal matches between them.
Using an advanced technique of van Emde Boas trees the time complexity bound can be reduce
d to
(
)
preserving the space bound of
(
)
.
Keywords
: longest common subsequence, longest increasing subsequence, online algorithm.
ACM Classification Keywords
:
G.2.1 Discrete mathematics: Combinatorics
Introduction
Let
and
,
, be two sequences over some alphabet
of size
,
. A sequence
,
, over
is called a
subsequence
of
, if
can be obtained from
by deleting some of its elements, that is if exists a set of
indices
{
}
such that
and
for
.
is said to be a
common subsequence
of
and
, if it is a
subsequence of both sequences
and
;
is said to be a
longest common subsequence
(LCS)
of
and
, if it
has
the maximum length among all common subsequences of
and
; that length is called the
LCS length
of
and
. In general the longest common subsequence is not unique.
The
Longest Common Subsequence Problem
(LCS Problem)
is to determine a LCS of
an
d
. Oftenthe
problem of determining theLCS length is also referred to as LCS Problem. This is due to the fact that most of
algorithms intended to find the LCS length can easily be modified to determine a LCS [Bergroth
,
2000]. In this
paper we will concen
trate on determining the LCS length rather thandetermining an actual LCS.The first known
solution of the LCS Problem is based on dynamic programming [Cormen
,
2009].F
or
and
denote by
the LCS length of sequences
and
; thus
is the LCS length of
and
. Note
that thefollowing recursion holds for
:
I
nternational Journal
“Information Theories and Applications”, Vol. 17, Number 3, 2010
227
{
if
or
(1)
if
{
}
if
Based on this relation it is easy to construct an algorithm which fills an array of size
, where
(
)

th cell
contains the value of
. As it follows form (1) such algorithm has to fill the rest of array before obtaining the
value of
(
)

th ce
ll, so it will determine the LCS length of sequences
and
in
(
)
time and
(
)
space (
(
)
time for filling each cell and
(
)
space for holding each cell).A simple trick can be used to make
this algorithm require only
(
)
space to obtain the value o
f the
(
)

th cell [
Cormen
,
2009
]. Here we
give some definitions which will be used later in the paper. For
and
the pair
(
)
is
called
matching
between sequences
and
if
; it is called
minimal (or dominant) matching
if for
every
other matching
(
)
such that
it holds
and
or
and
.Note that if
and
are two integers such that
and
, then theLCS Problem for two sequences of size
and
is
asymptotically not harder than the LCS Problem for two sequences of size
and
. Indeed, given two
sequences of size
and
and an algorithm which solves the LCS Problem for two sequences of size
and
, we can lengthen the given sequences
(by appending to themsymbols which don’t occur in the initial
sequences) up to size
and
respectively and pass the resulting two sequences to the given algorithm. It is
easy to see that such algorithm will solve the LCS Problem for two sequences of
size
and
in asymptotically
the same time and space bounds as the given algorithm solves the LCS Problem for two sequences of size
and
. This means that each lower bound for the LCS Problem for two sequences of size
andeachupper
bound for t
he LCS Problem for two sequences of size
are respectively lower and upper bounds for the LCS
Problem for two sequences of size
and
(recall that
). At [
Aho
,
1976
] the LCS Problem is examined
using the decision tree model of computation where t
he decision tree vertices represent “equal

unequal”
comparisons. There it is shown that each algorithm solving the LCS Problem and fitting this model has time
complexity lower bound of
(
)
, where
is the number of distinct symbols occurring in the sequ
ences (i.e. the
alphabet size). This means that the LCS Problem with unrestricted size of the alphabet has time complexity lower
bound of
(
)
, as such LCS Problem can be viewed as an LCS Problem with restricted alphabet of size
(
)
.In practice the unde
rlying encoding scheme for the symbols of the alphabet implies a topological order
between them. Algorithms which take into account this fact don’t fit the decision tree model with “equal

unequal”
comparisons examined at [
Aho
,
1976
]. At [
Masek
,
1980]
it is
presented an algorithm which applies the “Four
Russians” trick to the dynamic programming approach, thusit doesn’t fit the model examined at [
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