J. Software Engineering & Applications, 2010, 3, 737-745
doi:10.4236/jsea.2010.38085 Published Online August 2010 (http://www.SciRP.org/journal/jsea)
737
A New Tool: Solution Boxes of Inequality
Ferenc Kálovics
Analysis Department, University of Miskolc, Miskolc-Egyetemváros, Hungary.
Received May 4th 2010; revised June 19th 2010; accepted June 23rd 2010.
ABSTRACT
Many numerical methods reduce the solving of a complicated problem to a set of elementary problems. In this way, the
author reduced the finding of solution boxes of a system of inequalities, the computation of integral values with error
bounds, the approximation of global maxima to computing solution boxes of one inequality. These previous papers pre-
sented the mathematical aspects of the problems first of all. In this paper the computational handling of solution boxes
of an inequality (as a new tool of computer methods) is discussed in detail.
Keywords: Inequality, Solution Box, Coded Form of Expression
1. Introduction
Let :RR
m
gD be a continuous multivariate
real function, where





12 12
,,,,...,, m
m
D
xxxxx x
is an open box. Define the box


,, ,Bgc D   where
,R,cD

 as an open box around ,c in which the
relation is the same as between ()
g
c and

(it is sup-
posed that ()gc

). Thus, if () ,gc

 then
()gx

 for all


,, ,xBgc

 if () ,gc

 then
()gx

 for all


,, .xBgc

 Now consider a par-
ticular case. Let 2
12
:RR,(,)gD gxx 12 sin( )xx

2
1
2
,
ln
x
x
where
 
0,,1, 4D   is an open box. Try to
give the box 2
12 12
sin()/ln, (1, 2),1.Bxxxx



Since ( )2.351,gc  an open box around point

1, 2 in which () 1gx  is searched. Use the decom-
position

2
12 12
2
121 2
sin()/ln, (1, 2),1
sin(),(1,2),/ln,(1, 2),.
Bxxxx
BxxBxx









The spontaneous choice 0.5



 is suitable be-
cause sin(1 2)0.5 and 2 1/ln2 0.5, hence the
boxes


12
sin(), (1, 2),0.5Bxx and 2
12
/ln,(1,2),0.5 ,Bx x


around ,c satisfy the properties 12 sin()0.5xx  and
2
12
/ln 0.5,xx respectively. Since the decompositions


 
12
12 12
sin(),(1, 2),0.5
,(1,2),/6,(1,2),5 /6
Bxx
Bxx Bxx 







12
12
,(1,2), /6,(1,2),1
,(1,2),5/12,(1,2),5/3,
Bx Bx
Bx Bx




 


 
where the first row shows that 12 /65 /6,xx

  the
second row shows that 121 /6, 1,
x
xx

5/12,

25/3x
  in the box

12
sin( ),Bxx

(1, 2), 0.5,and the
decomposition

2
12
2
12
/ln,(1,2),0.5
,(1,2),0.5ln,(1,2),1
Bx x
BxB x






where the formula shows that 2
12
0.5,ln 1xx in the
box 2
12
/ln,(1,2),0.5 ,Bx x



 are also fortunate, there-
fore
2
12 12
sin()/ln,(1,2),1Bxxxx













 








/6,,1,4 0,,1,4
0,5/12, 1,40,, 1,5/3
 















 

 
0.5,,1,40, ,1,0.71,1.14,1,2.28.e  
To get the result, spontaneous rules were used for sum
(),

product (),

quotient (/) and composit function
(). In the last step the boxes were computed to some
elementary functions. Naturally, the decomposition can
break down with any spontaneous rule, e.g. the choice

A New Tool: Solution Boxes of Inequality 738
0.95, 0.05

  does not work in the first step, al-
though 1.

  If the guaranteed rules of Theorem 1
of [1] are used, then
 
2
12 12
sin()/ln, (1,2),1
0.84,1.21,1,2.41.
Bxxxx



Our aim is to make a C++ program which can follow
the above decomposition process. Suppose that the con-
tinuous multivariate real function :RR,
m
gD
where



12 12
, , ,,...,,,
m m
Dxxxx xx is built of
the well-known (univariate real) elementary functions:
, ,ln,,sin,cos,tan,cot,
arcsin,arccos, arctan, arccot
rx
x
exx xxxx
xxxx
by the function operations ,,/,.  (It is supposed that
r of r
x
is a positive integer or negative integer or re-
ciprocal of a positive integer. This is only a formal re-
striction because of

1/
/,, N.
n
kn k
xx kn


) The
functions ,log,sinh ,
x
a
axx etc. can be used by the
connections ln,logln/ ln,sinh
xxa
a
aexxax 
()/2,
xx
ee

 etc., respectively, the operation of differ-
ence can be used in form (1).ab ab

 Naturally,
the program can be enlarged by further elementary func-
tions, but we would like to make a program in a publish-
able length. The notations, names, definitions and dis-
cussions are simpler and clearer than they were in the
former papers of the author (experience has slowly be-
come knowledge). Using our new notations, the rules of
Theorem 1 of [1] are as follows. (It is supposed that the
functions ,pq satisfy the restrictions belonging to the
function ,
g
the function e is an allowed elementary
function, furthermore /pq and eq are well-defined
on .D)
Rule of constant addition:




,,,, ,Bp cBpc

  where R, .

 
Rule of constant multiplication:




,,,, ,Bpc Bpc

  where 0R,/.

  
Rule of sum:






,,,,,,,B pqcB pcBqc

  
where (()())/2, .pc qc





 
Rule of quotient:




/,, ,,0Bpqc Bpc   if 0;








/,,,,,, ,Bpqc BpcBqc

  where
2
(()())/(1 ),,pc qc

   if 0.


Rule of product:






,,,,0 ,,0B pqcBpcBqc  
if 0 or ()()0;pcqc











,, ,,,,
,,,, ,
Bpqc BpcBpc
Bqc Bqc





where ,,

 

 if 0,( )0,( )0;pc qc










,,,,,,0,, ,B pqcBpcB pcBqc

 
where 2(),/ ,pc




 if 0,()0,( )0;pcqc










,,,,,, ,,0,BpqcBpcBqc Bqc  
where 2(),/ ,qc




 if 0,( )0,( )0;pc qc










,,,,,,0,, ,BpqcBpcBpc Bqc

 
where sgn( )()/( ),pcpc qc
 
sgn( )( )/(),qcqc pc   if ()() 0.pcqc


Rule of composite function:


,,,,,, ,
cc
lr
BeqcDB qcB qc 

 
 

 

where c
l

is the maximum value for which ()
c
l
e 

,()
c
lqc
  and c
r

is the minimum value for which
() ,().
cc
rr
eqc   If c
l

or c
r

does not exist,
then the actual member is omitted.
After the decomposition process the task is to evalu-
ate the solution boxes belonging to elementary func-
tions. The processes for the 12 cases (which are pub-
lished now for the first time) can be obtained easily
enough by the graphs of these functions. (The starting
box is






1 1,,..., , ,...,, im im
Dxxxx xx in every
case.)
Power function:



112 1
,,
,,...,,,...,,,,,
r
i
im i
im i
Bx c
x
xxxxxbxbx





if Zr


is odd, then 1/
1;
r
b  
if Zr


is even and 0,

 then1/
1,
r
b  
21
;bb


if Zr


is odd and 0,


then 1/
1;
r
b  
if Zr


is even and 0,

 then1/
1,
r
b  
21
;bb


if (0,1), 1/Zrr

 and 0,

 then 1/
1;
r
b  
if


1,, i i
bxc then 1,
i
x
b if

1,, ii
bcx then
1, i
x
b

if


2,, i i
bxc then 2,
i
x
b if

2,, ii
bcx then
2. i
x
b


A New Tool: Solution Boxes of Inequality 739
Exponential function:








11 1
, ,,,...,,,...,,,,
i
x
im im i
Becx xx xxxbx 



if 0,

 then 1ln;b


if

1,, i i
bxc then 1,
i
x
b if

1,, ii
bcx then
1. i
x
b
Logarithm function:


 


12 1 12
ln, ,,,,,...,,,;
im m
Bxcxxxxxxbe

 
if

1,, i i
bxc then 1,
i
x
b if

1,, ii
bcx then
1. i
x
b
Absolute value function:



112 1
,,
,,...,,,...,,,,,
i
im i
im i
Bxc
x
xxxxxbxbx
 


if 0,

 then 12 ,;bb

 
if

1,, i i
bxc then 1,
i
x
b if

1,, ii
bcx then
1, i
x
b
if

2,, i i
bxc then 2,
i
x
b if

2,, ii
bcx then
2. i
x
b
Sinus function:


 

112 1
sin, ,
,,...,,,...,,,,,
i
im i im i
Bxc
x
xxxxxbxbx
 

arcsin ,2xper

 
int (/2),
i
c

if 0,
i
c

then
2,per per


if


0,1 ,   then 12 ,;bxperbxper


if [1,0),

 then 12 ,bxperb

 2
x



;per
if 2,
i
bc then 1122 ,, 2;ybb bby

 
if 1,
i
bc then 22 11 ,, 2;ybb bb y


if

1,, i i
bxc then 1,
i
x
b if

1,, ii
bcx then
1, i
x
b
if


2,, i i
bxc then 2,
i
x
b if


2,, ii
bcx then
2. i
x
b
Cosinus function:


 

112 1
cos, ,
,,...,,,...,,,,,
i
im i imi
Bxc
x
xxxxxbxbx
 

arccos ,2xper

 int (/2),
i
c

if 0,
i
c

then
2,per per


if


0,1 ,   then 12 ,2 ;bxperb xper

 
if [1,0),

 then 12 ,;bxperb xper

 
if 2,
i
bc then 1122 ,, 2;ybb bby

 
if 1,
i
bc then 22 11 ,, 2;ybb bb y


if

1,, i i
bxc then 1,
i
x
b if

1,, ii
bcx then
1, i
x
b

if


2,, i i
bxc then 2,
i
x
b if

2,, ii
bcx then
2. i
x
b

Tangent function:


 

112 1
tan, ,
,,...,,,...,,,,,
i
im i im i
Bxc
x
xxxxxbxbx
 

arctan,2int(/ 2),
i
xperc

 
if 0,
i
c

then
2,per per



if 0,

 then 12 ,;bxperb xper



if 0,


then 12 ,2 ;bxperbxper



 
if 2,
i
bc

then 1221 ,;bbb b



if 1,
i
bc then 2112 ,;bbbb



if


1,, i i
bxc then 1,
i
x
b if

1,, ii
bcx then
1, i
x
b

if


2,, i i
bxc then 2,
i
x
b if

2,, ii
bcx then
2. i
x
b

Cotangent function:


 

112 1
cot, ,
,,...,,,...,,,,,
i
im i im i
Bxc
x
xxxxxbxbx
 

arccot,2int(/ 2),
i
xperc

  if 0,
i
c

then
2,per per



if 0,

 then 12 ,;bxperb xper



if 0,


then 12 ,2 ;bxperb xper



 
if 2,
i
bc

then 1221 ,;bbb b



if 1,
i
bc then 2112 ,;bbbb



if


1,, i i
bxc then 1,
i
x
b if

1,, ii
bcx then
1, i
x
b

if


2,, i i
bxc then 2,
i
x
b if

2,, ii
bcx then
2. i
x
b

Arcus sinus function:


  
11 1
arcsin, ,
,,...,,,...,,,,
i
im
im i
Bxc
x
xxxxxbx
 

if


/2,/2 ,   then 1sin ;b


if


1,, i i
bxc then 1,
i
x
b if

1,, ii
bcx then
1. i
x
b

Arcus cosinus function:


 

11 1
arccos, ,
,,...,,,...,,,,
i
im im i
Bxc
x
xxxxxbx



if


0, ,

  then 1cos ;b



A New Tool: Solution Boxes of Inequality 740
if

1,, i i
bxc then 1,
i
x
b if

1,, ii
bcx then
1. i
x
b
Arcus tangent function:


 

11 1
arctan, ,
,,...,,,...,,,,
i
im
imi
Bxc
x
xxxxxbx
 

if

/2,/2 ,   then 1tan ;b


if

1,, i i
bxc then 1,
i
x
b if

1,, ii
bcx then
1. i
x
b
Arcus cotangent function:


 

11 1
arc cot,,
,,...,,,...,,,,
i
im
imi
Bxc
x
xxxxxbx
 

if

0, ,

  then 1cot;b


if

1,, i i
bxc then 1,
i
x
b if

1,, ii
bcx then
1. i
x
b
Now we have 6 rules for decomposition and 12 proc-
esses for elementary boxes. Since the application of the
6th rule (rule of composite function) requires the same
investigation as the evaluation of elementary boxes,
therefore the decomposition part of the program contains
17 different cases, and the evaluation part contains 12
cases. At the end of this section, let us emphasize some
facts about solution boxes. 1) If () ,gc

 then
(,,)xBgc

 implies ()gx

 . Consequently,
the box (,,)Bgc

of domain D is assigned to the
interval (,)

 of function values. Similarly, if
()gc

 , then the box (,,)Bgc

of domain D is
assigned to the interval (, )


of function values. The
so-called interval extension functions used in interval
methods (see e.g. in [2]) are inverse type functions, they
assign intervals of function values to boxes of domain.
The handling and application of these two tools require a
highly different mathematical and computational back-
ground. 2) The decomposition of the expression (),
g
x
the determination of the elementary boxes, uses only
the structure of the expression ()
g
x and supposes only
the continuity of
g
on .D 3) The box (,, )Bgc

is
not a symmetrical box around .c Often it has a large
volume, although ()
g
c<

or ()
g
c>

is only just
satisfied.
2. Numerically Coded Form of Expression
When the author introduced the above decomposition of
expressions, the process was followed by a Maple pro-
gram. This code computed a box (,,)Bgc

very simi-
larly as it is done in our numerical example. The program
utilizes strongly the ability of Maple to give the type of
an expression by the function (,)type e t where e is an
expression and t is a type name, furthermore to give
operands from an expression by the function (, )op i e
where i is the position index and e is an expression.
The use of our Maple program is very convenient be-
cause we must only give the function
g
(the expression
()
g
x) in a customary form. Unfortunately, the program
is fairly slow. The conventional program languages (e.g.
C++, Fortran) have no similar functions, but they can
follow the decomposition, if a numerically coded form of
the expression ()
g
x is used. This form is built of triples
of integer or real numbers. The first number in a triple is
the so-called operation code (using the order of rules and
elementary functions):
const. add.1,

const. mult.2, +3, /4,


5,

power func.6,exp7,ln8,


abs. func.9,

sin10, cos11,tan12,


cot 13,arcsin 14,arccos15,arctan 16,


arccot 17.
The second number in the triples is minus i if an
elementary function uses the argument i
x
(the minus
sign identifies the elementary funcions). If the operation
uses a former triple as the first operand, then the ordinal
number of this triple is the second number in the actual
triple. The third number in the triples is the given con-
stant or the exponent if the constant addition, constant
multiplication or the power function is used, respectively.
If the operation uses a former triple as a second operand
(cases of ,/,

 ), then the ordinal number of this triple
is the third number in the actual triple. If the operation is
exp,ln,...,arccot (the first number in the triple is be-
tween 7 and 17), then the third number in the actual triple
is 0. To make the description of expressions by triples
unique, we use the left-right rule in every choice. Exer-
cise the description on the expression 12
(, )gx x

2
12 12
sin()/ln
x
xx x. The four elementary functions used
in the build-up of 12
(, )
g
xx (the elementary boxes be-
long to these functions after the decomposition) are
112
112 212
(),(),,ln.
x
xxx xxTheir descriptions are:
12
2
12
(6, 1,1),(6,2,1),
(6,1,2),ln (8,2,0).
xx
xx



Now we have the sequence of triples:
(6,1,1),(6,2,1),(6,1,2),(8,2,0).


Using the elements of this sequence, the functions
1212
,sin( )
x
xxx and 2
12
/ln
x
x can be obtained in the
form:

A New Tool: Solution Boxes of Inequality 741
12 12
2
12
(5,1, 2),sin()(10,5,0),
/ln (4,3,4).
xx xx
xx


Since the last operation can be written in form (3,6,7),
the complete description of 12 (, )
g
xx is:
{(6, 1,1),(6, 2,1),(6, 1,2),(8, 2,0),
(5,1,2),(10,5, 0),(4, 3,4),(3,6,7)}.

It can be seen that the preparation of numerically
coded forms of expressions is easy enough, nevertheless
the author has a short Maple program also for this
one-off work. (But a numerically coded form can be used
a thousand or a million times in an application.) The con-
struction of our Maple program can be illustrated (on the
former expression) by the Table 1.
The 1st column shows the index (serial number) of the
expression appearing in the 2nd column. The 4th column
contains the temporary triples which use the indexes of
the 1st column for identification. Going downwards the
temporary triples, the elementary tripes receive their new
indexes, and going backwards, the other triples receive
their new indexes (6th column). Using these new indexes
the final triples (7th column) are obtained. Ordering them
by the indexes of the 6th column the result is ready (8th
column). This result differs from our previous one forma-
Table 1. Creation of numerically coded form
iCompo-
nents Triples ii New
triples Result
112
2
12
sin( )
/ln
xx
x
x

(3,2,3) 8 (3,7,6)(6, 1,2)

212
sin( )
x
x(10, 4, 0) 7 (10, 5, 0)(8,2, 0)

32
12
/ln
x
x(4,5,6)  6 (4,1,2) (6, 1,1)

412
x
x (5,7,8)  5 (5,3, 4) (6, 2,1)

52
1
x
(6,1, 2)

 1 √ (5,3, 4)
62
ln
x
(8,2,0)

 2 √ (4,1,2)
71
x
(6, 1,1)

 3 √ (10, 5,0)
82
x
(6, 2,1)

 4 √ (3,7,6)
lly (here the left-right rule is omitted), but they are
equivalent in essence. The arrays of Maple program
resi
i
tr
i
com
x
,,,, belong to vector x components of
(),
g
x triples, new indexes and result, respectively. The
array v is an auxiliary vector. The complete code is as
follows.
DECLARATIONS
> x:=array(1..10): com:=array(1..10): tri:=array(1..10,1..3):
> ii:=array(1..10): res:=array(1..10,1..3): v:=array(1..3):
INITIAL VALUES
> com[1]:=sin(x[1]*x[2])+x[1]^2 / ln(x[2]):
> i:=0: imax:=1:
COMPONENTS and TRIPLES
> while i <= imax do
> i:=i+1: g:=com[i]:
> if type(g,name) then v:=[0,op(1,g),0]: fi:
> if type(g,`+`) then
> if type(op(1,g),numeric) then v:=[1,g-op(1,g),op(1,g)]:
> elif type(g-op(1,g),numeric) then v:=[1,op(1,g),g-op(1,g)]:
> else v:=[3,op(1,g),g-op(1,g)]: fi: fi:
> if type(g,`*`) then
> if type(op(1,g),numeric) then v:=[2,g/op(1,g),op(1,g)]:
> elif type(g/op(1,g),numeric) then v:=[2,op(1,g),u/op(1,g)]:
> elif type(op(2,g),`^`) and op(2,op(2,g)) < 0 then v:=[4,op(1,g),op(2,g)^(-1)]:
> else v:=[5,op(1,g),g/op(1,g)]: fi: fi:
> if type(g,`^`) then v:=[6,op(1,g),op(2,g)]: fi:
> if type(g,function) then
> if op(0,g)=exp then v:=[7,op(1,g),0]: fi: if op(0,g)=ln then v:=[8,op(1,g),0]: fi:
> if op(0,g)=abs then v:=[9,op(1,g),0]: fi: if op(0,g)=sin then v:=[10,op(1,g),0]: fi:
> if op(0,g)=cos then v:=[11,op(1,g),0]: fi: if op(0,g)=tan then v:=[12,op(1,g),0]: fi:
> if op(0,g)=cot then v:=[13,op(1,g),0]: fi: if op(0,g)=arcsin then v:=[14,op(1,g),0]: fi:
> if op(0,g)=arccos then v:=[15,op(1,g),0]: fi:if op(0,g)=arctan then v:=[16,op(1,g),0]: fi:
> if op(0,g)=arccot then v:=[17,op(1,g),0]: fi: fi:
> if v[1]=0 then tri[i,1]:=6: tri[i,2]:=-v[2]: tri[i,3]:=1: fi:
> if v[1] > = 6 and v[1] < = 17 then
> if type(v[2],name) then tri[i,1]:=v[1]: tri[i,2]:=-op(1,v[2]): tri[i,3]:=v[3]:
> else imax:=imax+1: com[imax]:=v[2]: tri[i,1]:=v[1]: tri[i,2]:=imax: tri[i,3]:=v[3]: fi: fi:
> if v[1]=1 or v[1]=2 then
> imax:=imax+1: com[imax]:=v[2]: tri[i,1]:=v[1]: tri[i,2]:=imax: tri[i,3]:=v[3]: fi:
> if v[1] > = 3 and v[1] < = 5 then
> imax:=imax+1: com[imax]:=v[2]: imax:=imax+1: com[imax]:=v[3]:
> tri[i,1]:=v[1]: tri[i,2]:=imax-1: tri[i,3]:=imax: fi:

A New Tool: Solution Boxes of Inequality 742
> od:
NEW INDEXES
> ind:=1:
> for i from 1 to imax do if tri[i,2] < 0 then ii[i]:=ind: ind:=ind+1: fi: od:
> for i from imax by -1 to 1 do if tri[i,2] > 0 then ii[i]:=ind: ind:=ind+1: fi: od:
RESULT
> for ind from 1 to imax do
> for i from 1 to imax do
> if ii[i]=ind then res[ind,1]:=tri[i,1]:
> if tri[i,2] < 0 then res[ind,2]:=tri[i,2]: res[ind,3]:=tri[i,3]: fi:
> if tri[i,2] > 0 then res[ind,2]:=ii[tri[i,2]]:
> if tri[i,1]=3 or tri[i,1]=4 or tri[i,1]=5 then res[ind,3]:=ii[tri[i,3]]: else res[ind,3]:=tri[i,3]:
> fi: fi: fi:
> od: od:
> print(`The result`, res):
Now we have a suitable form of expressions to make a
fast program for our original problem.
3. A Complete C++ Program for
Computation of Solution Boxes
Here the function segment
s
olbox is made for the
computation of solution box B[g,c,α] where
 

1 1
:RR,
, ,...,,,,R
m
m m
gDD
xxx xc D 
 

have the former definitions. The headline of this segment
contains the input parameters , (in tripleDDGg
form),,,,number of triplesccalp mmnt

 .
The result


,,Bgc

(as output parameter) is placed
into the array .B The segment contains 3 essential parts
(great loops). In the first one the function values of
components of ()
g
x at point c are computed by using
the triples in succession. These function values are
placed into the array fv and will be used in the de-
composition steps. In the second loop the decomposition
(making elementary boxes) goes on. In the normal case
(1),nt first the last triple (expression ()
g
x) and

are placed into the array of non-elementary expressions
().ne At this time the number of elements in this array is
1 (1).ine

Applying the suitable rule for this element,
new elements of the array ne or (and) those of the array
of elementary expressions ()el are created. The values
,,

 etc. appearing in the rules are denoted by
,,,,begade ep and
s
shows the number of new boxes.
The indexes ine and iel show the numbers of ele-
ments in the arrays ne and . el The decomposition
procedure is finished when there is no element for ex-
amination in the array ,ne that is the necessary ele-
mentary expressions are all in the array . el In the third
great loop of segment
s
olbox the (iel pieces of) ele-
mentary boxes are evaluated and the array B is made.
As it was mentioned earlier, the application of the com-
posite function rule requires the same investigation as the
evaluation of elementary boxes. Therefore the second
and third loops of the segment contain similar lines (only
some parameters are different) between case 6 and case
17.
#include <iostream.h>
#include <math.h>
double B[10][3];
/*DECLARATIONS*/
void solbox(double D[][3],double G[][4],double c[],double alp,int m,int nt)
{double fv[100],ne[100][5],el[100][5],al,be,ga,de,ep,x,y,w,alf,per;
int i,j,k,l,ii,jj,kk,ine,iel,s; const double Pi=3.14159265;
/*FUNCTION VALUES TO COMPONENTS OF EXPRESSION g(x)*/
for (i=1; i<=nt; i++)
{j=(int)G[i][1]; k=(int)G[i][2]; l=(int)G[i][3]; w=G[i][3];
if (k<0) x=c[-k]; else x=fv[k]; if (j>=3 && j<=5) y=fv[l];
switch (j)
{case 1:fv[i]=x+w;break; case 2:fv[i]=x*w;break; case 3:fv[i]=x+y;break;
case 4:fv[i]=x/y;break; case 5:fv[i]=x*y;break; case 6:fv[i]=pow(x,w);break;
case 7:fv[i]=exp(x);break; case 8:fv[i]=log(x);break; case 9:fv[i]=fabs(x);break;
case 10:fv[i]=sin(x);break; case 11:fv[i]=cos(x);break; case 12:fv[i]=tan(x);break;
case 13:fv[i]=1/tan(x);break; case 14:fv[i]=asin(x);break; case 15:fv[i]=acos(x);break;
case 16:fv[i]=atan(x);break; case 17:fv[i]=Pi/2-atan(x);break;}}
/*DECOMPOSITION BY USING THE 6 RULES*/
for (i=1; i<=3; i++) ne[1][i]=G[nt][i]; ne[1][4]=alp; ine=1; iel=0; i=0;
while (i<ine && nt>1)
{i++; j=(int)ne[i][1]; k=(int)ne[i][2]; l=(int)ne[i][3]; w=ne[i][3]; al=ne[i][4]; s=0;

A New Tool: Solution Boxes of Inequality 743
switch (j)
{case 1: be=al-w;s=1;break;
case 2: be=al/w;s=1;break;
case 3: be=(al+fv[k]-fv[l])/2;ga=al-be;s=2;break;
case 4: if (al==0) {be=0;s=1;}; if (al !=0) {ga=(al*fv[k]+fv[l])/(1+al*al);be=al*ga;s=2;};break;
case 5: if (al==0 || fv[k]*fv[l]*al < 0) {be=0;ga=0;s=2;}
if (al!=0 && fv[k]==0 && fv[l]==0) {be=sqrt(fabs(al));ga=be;de=-be;ep=de;s=4;}
if (al!=0 && fv[k] !=0 && fv[l]==0) {be=2*fv[k];ga=al/be;de=0;s=3;}
if (al!=0 && fv[k]==0 && fv[l]!=0) {ga=2*fv[l];be=al/ga;de=be;ep=0;s=4;}
if (fv[k]*fv[l]*al>0) {be=fv[k]/fabs(fv[k])*sqrt(al*fv[k]/fv[l]);
ga=fv[l]/fabs(fv[l])*sqrt(al*fv[l]/fv[k]);de=0;s=3;};break;
case 6: if (w>0 && (l+1)/2*2==l+1) {be=pow(al,1/w); s=1;}
if (w>0 && l/2*2==l && al>=0) {be=pow(al,1/w);ga=-be;s=2;}
if (w<0 && (l-1)/2*2==l-1 && al !=0) {be=pow(al,1/w);s=1;}
if (w<0 && l/2*2==l && al>0) {be=pow(al,1/w);ga=-be;s=2;}
if (w>0 && w<1 && al>=0) {be=pow(al,1/w);s=1;};break;
case 7: if (al > 0) {be=log(al);s=1;};break;
case 8: be=exp(al);s=1;break;
case 9: if (al >= 0) {be=al;ga=-al;s=2;};break;
case 10:if (fabs(al)<=1)
{x=asin(fabs(al));per=2*Pi*(int)(fv[k]/2/Pi); if(fv[k]<0) per=per-2*Pi;
if (al>=0) {be=x+per;ga=Pi-x+per;}; if (al<0) {be=Pi+x+per;ga=2*Pi-x+per;}
if (ga<fv[k]) {y=be;be=ga;ga=y+2*Pi;}; if (be>fv[k]) {y=ga;ga=be;be=y-2*Pi;}; s=2;};break;
case 11:if (fabs(al)<=1)
{x=acos(fabs(al));per=2*Pi*(int)(fv[k]/2/Pi); if(fv[k]<0) per=per-2*Pi;
if (al>=0) {be=x+per;ga=2*Pi-x+per;}; if (al<0) {be=Pi-x+per;ga=Pi+x+per;}
if (ga<fv[k]) {y=be;be=ga;ga=y+2*Pi;}; if (be>fv[k]) {y=ga;ga=be;be=y-2*Pi;}; s=2;};break;
case 12:x=atan(fabs(al));per=2*Pi*(int)(fv[k]/2/Pi); if(fv[k]<0) per=per-2*Pi;
if (al>=0) {be=x+per;ga=Pi+x+per;}; if (al<0) {be=Pi-x+per;ga=2*Pi-x+per;}
if (ga<fv[k]) {be=ga;ga=be+Pi;}; if (be>fv[k]) {ga=be;be=ga-Pi;};s=2;break;
case 13:x=Pi/2-atan(fabs(al));per=2*Pi*(int)(fv[k]/2/Pi); if(fv[k]<0) per=per-2*Pi;
if (al>=0) {be=x+per;ga=Pi+x+per;}; if (al<0) {be=Pi-x+per;ga=2*Pi-x+per;}
if (ga<fv[k]) {be=ga;ga=be+Pi;}; if (be>fv[k]) {ga=be;be=ga-Pi;};s=2;break;
case 14:if (fabs(al)<=Pi/2) {be=sin(al);s=1;};break;
case 15:if (al>=0 && al<=Pi) {be=cos(al);s=1;};break;
case 16:if (fabs(al)<Pi/2) {be=tan(al);s=1;};break;
case 17:if (al>0 && al<Pi) {be=1/tan(al);s=1;};break;}
for (ii=1; ii<=s; ii++)
{switch (ii)
{case 1:alf=be;kk=k;break; case 2:alf=ga;if (j>=3 && j<=5) kk=l;break;
case 3:alf=de;kk=k;break; case 4:alf=ep;if (j>=3 && j<=5) kk=l;break;}
if (G[kk][2]<0) {iel++; for (jj=1; jj<=3; jj++) el[iel][jj]=G[kk][jj];el[iel][4]=alf;}
if (G[kk][2]>0) {ine++; for (jj=1; jj<=3; jj++) ne[ine][jj]=G[kk][jj];ne[ine][4]=alf;}}}
if (nt==1) {for (ii=1; ii<=4; ii++) el[1][ii]=ne[1][ii];iel=1;}
/*EVALUATING THE ELEMENTARY BOXES BY USING THE 12 PROCESSES*/
for (i=1; i<=m; i++) {B[i][1]=D[i][1]; B[i][2]=D[i][2];}
for (i=1; i<=iel; i++)
{j=(int)el[i][1]; k=(int)-el[i][2]; l=(int)el[i][3]; w=el[i][3]; al=el[i][4]; s=0;
switch (j)
{case 6: if (w>0 && (l+1)/2*2==l+1) {be=pow(al,1/w);s=1;}
if (w>0 && l/2*2==l && al>=0) {be=pow(al,1/w);ga=-be;s=2;}
if (w<0 && (l-1)/2*2==l-1 && al!=0) {be=pow(al,1/w);s=1;}
if (w<0 && l/2*2==l && al>0) {be=pow(al,1/w);ga=-be;s=2;}
if (w>0 && w<1 && al>=0) {be=pow(al,1/w);s=1;};break;
case 7: if (al > 0) {be=log(al);s=1;};break;
case 8: be=exp(al);s=1;break;
case 9: if (al >= 0) {be=al;ga=-al;s=2;};break;
case 10:if (fabs(al)<=1)
{x=asin(fabs(al));per=2*Pi*(int)(c[k]/2/Pi);if(c[k]<0) per=per-2*Pi;
if (al>=0) {be=x+per;ga=Pi-x+per;}; if (al<0) {be=Pi+x+per;ga=2*Pi-x+per;}
if (ga<c[k]) {y=be;be=ga;ga=y+2*Pi;}; if (be>c[k]) {y=ga;ga=be;be=y-2*Pi;}; s=2;};break;
case 11:if (fabs(al)<=1)
{x=acos(fabs(al)); per=2*Pi*(int)(c[k]/2/Pi);if(c[k]<0) per=per-2*Pi;
if (al>=0) {be=x+per;ga=2*Pi-x+per;}; if (al<0) {be=Pi-x+per;ga=Pi+x+per;}
if (ga<c[k]) {y=be;be=ga;ga=y+2*Pi;}; if (be>c[k]) {y=ga;ga=be;be=y-2*Pi;}; s=2;};break;
case 12:x=atan(fabs(al));per=2*Pi*(int)(c[k]/2/Pi);if(c[k]<0) per=per-2*Pi;
if (al>=0) {be=x+per;ga=Pi+x+per;}; if (al<0) {be=Pi-x+per;ga=2*Pi-x+per;}
if (ga<c[k]) {be=ga;ga=be+Pi;}; if (be>c[k]) {ga=be;be=ga-Pi;};s=2;break;
case 13:x=Pi/2-atan(fabs(al));per=2*Pi*(int)(c[k]/2/Pi);if(c[k]<0) per=per-2*Pi;

A New Tool: Solution Boxes of Inequality 744
if (al>=0) {be=x+per;ga=Pi+x+per;}; if (al<0) {be=Pi-x+per;ga=2*Pi-x+per;}
if (ga<c[k]) {be=ga;ga=be+Pi;}; if (be>c[k]) {ga=be;be=ga-Pi;};s=2;break;
case 14:if (fabs(al)<=Pi/2) {be=sin(al);s=1;};break;
case 15:if (al>=0 && al<=Pi) {be=cos(al);s=1;};break;
case 16:if (fabs(al)<Pi/2) {be=tan(al);s=1;};break;
case 17:if (al>0 && al<Pi) {be=1/tan(al);s=1;};break;}
for (ii=1; ii<=s; ii++) { alf=be; if (ii==2) alf=ga;
if (alf < c[k] && alf > B[k][1]) B[k][1]=alf;
if (alf > c[k] && alf < B[k][2]) B[k][2]=alf;}}}
/*THE CALLING SEGMENT WITH THE SAMPLE EXPRESSION*/
void main()
{double D[10][3],G[100][4],c[10],alp; int m,nt,i,j;
double g[][3]={{6,-1,1},{6,-2,1},{6,-1,2},{8,-2,0},{5,1,2},{10,5,0},{4,3,4},{3,6,7}};
D[1][1]=0.;D[1][2]=3.141593;D[2][1]=1.;D[2][2]=4.; c[1]=1.;c[2]=2.; alp=1.; m=2; nt=8;
for (i=1; i<=nt; i++) {for (j=1; j<=3; j++) G[i][j]=g[i-1][j-1];}; solbox(D,G,c,alp,m,nt);
for (i=1; i<=m; i++) {cout << B[i][1] << " " << B[i][2] << endl;}}
The calling (main) segment uses a simple trick (push
down of indexes) for the easy handling of triple forms.
Using the form (produced by the Maple program)
{(6,1,2), (8,2,0), (6,1,1),(6,2,1),
(5,3,4),(4,1,2),(10,5,0),(3,7,6)}

and the form (produced by us)
{(6, 1,1),(6, 2,1),(6, 1,2),(8, 2,0),
(5,1,2), (10, 5,0), (4,3, 4), (3,6,7)}

of our sample expression 12
(, )
g
xx , the result is the
same:


2
12 12
sin()/ln, (1, 2),1
0.839584,1.20543, 1,2.41086,
Bxxxx  

which was given (with rounded values) in the first sec-
tion.
4. A numerical Example for Illustration
The surfaces given by formulas
22
312312
2sin()2, 2xxxxxx
have their intersection over the two dimensional interval


3, 3,3, 3. Maple 11 can produce Figure 1 with
these data. Our aim is to give an approximating value for
the volume of the section set S determined by the sur-
faces.
Since the set S is the solution set of the system of
inequalities:
Figure 1. Intersection
22
1233 12
2sin()20, 20,xxxx xx


one of methods of [3] can be used. This method searches
for disjunct open solution boxes of an inequality system
and it is based on the following four principles. 1) If


1,,0Bfc is a solution box to the inequality 1() 0fx
and


2,,0Bfc is a solution box to the inequality
2() 0,fx then the box




12
,,0 ,,0BfcBf c is a
solution box to the system of the two inequalities. 2) If
1
B and 2 B are m-dimensional boxes, then the set
12
BB

can be divided into (at most) 2m boxes easily.
In this way a sequence of solution boxes and another
sequence of boxes waiting for examination can be gener-
ated. 3) A boundary region is used to prevent the ap-
pearance of too small boxes. Here this region is defined
by the inequality 3 () 10fx 
 and a cube with edges
of 3 10

is excluded at a region point .c 4) The most
promising box (the box with maximum volume) is cho-
sen from boxes waiting for examination in every step.
The essential input data are:
1123
( )2sin()2
{(6,1,1),(6,2,1),..., (3,6,7),(1,8, 2.)},
fxxx x

22
2312
() 2
{(6,1,2),(6,2, 2),..., (3,3,5),(1,6, 2.)},
fxx xx 

the initial box


(3,3),(3,3),(2,4)I which in-
cludes the set , S and the number b N of solution
boxes looked for, as stop criteria. For the values
445
10,510,10
b
N the results are
27.0008,28.1484,28.4231,
respectively. The running times are (with the original
Lahey Fortran 90 version 4.5 code on a PC of two 2 GHz
processor):
2sec,26sec,107sec,
respectively. The fast increase of running time shows that
neither the 3rd nor the 4th principle work well if b N is

A New Tool: Solution Boxes of Inequality 745
large. (Only a few hundred solution boxes were searched
for in [3].) Now consider another possibility. Since
()1 ,
S
vol Sdxdydz
and [4,5] contain methods for computation of integrals,
perhaps better results can be obtained. The method of [5]
is based on the following three principles. 1) The value
of a (simple, double, triple, etc.) integral can be given by
the volume of a suitable set T. If the box
I
contains
this set, then a good scanning of T gives an approxima-
tion of the integral value and the scanning of the set
I
T also facilitates computation of an error bound.
Here the set T can be described by the system of ine-
qualities
22
1233 124
2sin()20, 20,10,xxxxxxx
and the scanning works by computing solution boxes.
The scanning of the set
I
T is based on the examina-
tion of “the worst inequality” at points .cIT 2) If
1
B and 2 B are m-dimensional boxes, then the set
12
BB can be divided into (at most) 2m boxes easily.
3) The too small boxes are filtered by the simple condi-
tion ().vol B
  Naturally, the value  has a strong
influence on the available error bound. The program is
stopped when there is no more box waiting for examina-
tion. Now the essential input data are: the expressions
123
(),(), ()
f
xfxfx in triple form, the initial box

(3,3), (3,3),I (2,4),

(0,1) which includes the
set ,
T and the filter parameter .  For the values
5
10 , 

 6 10 ,
 7
10, the results are
28.99531.0169, 29.04990.4840, 29.07050.2294,
respectively. The running times are (with the original
Visual C++ version 6.0 code on the former PC of two 2
GHz processor):
4sec,16sec,79sec,
respectively. Since the difference between the third and
first results is 0.0752 only, the third error bound sug-
gests that all three of the results are good enough ap-
proximations. The extra work of the second and third
cases was spent mostly on improving the error bound.
Now make a goal-oriented transformation of the method
of [3]. If the third principle (boundary region) and the
fourth principle (most promising box) are omitted and
principles of method of [5] belonging to error bound and
filter parameter  are used, then the new method will
be competitive. For the values 5 10 , 

 6
10 ,
 7
10

,
the results are
29.08620.6765,29.08290.3124, 29.08260.1447,
respectively. The running times are (with the new Lahey
Fortran code):
1sec,5sec,24sec,
respectively. These results are better from all points of
view than the previous ones were. Since now the differ-
ence between the first and third results is 0.0036 only,
the 3 values suggest much more accurate approximation
than the guaranteed error bounds do. On the other hand,
by the geometrical meaning, here the solution (section)
set S (and its complementary set) was covered with
cuboids, while the solution set T (and its complemen-
tary set) was covered with four-dimensional (abstract)
boxes at the triple integral. It is not surprising that the
computational effort increases with dimension. Finally
some remarks are given.
1) The computation efforts (the evaluation times) be-
longing to (,,)Bgc

and ()
g
c can be characterized
well enough by the formula: effort

(,, )10Bgc 


effort


(),
g
c with respect to all three programming
languages mentioned and problems of [1,3-5].
2) The speeds belonging to the evaluation of
(,,)Bgc

satisfy the relations: speed(C++)

200

speed(Maple), speed(Fortran)  300 speed(Maple)
for the examples of [1]. Nevertheless, the Fortran codes
are hardly faster than the C++ ones ( differences of only
0-10 percent in running time) for the examples of [5].
3) We always used floating point arithmetic for com-
puting solution boxes (zone function values by former
name). Since these boxes are computed by lower esti-
mates (see proofs of [1]), we have never happened to
obtain a faulty result because of the effect of rounding
errors.
4) The author would gladly send the C++, Fortran,
Maple codes mentioned to interested readers in e-mail as
an attached file.
REFERENCES
[1] F. Kálovics, “Creating and Handling Box Valued Func-
tions Used in Numerical Methods,” Journal of Computa-
tional and Applied Mathematics, Vol. 147, No. 2, 2002,
pp. 333-348.
[2] R. Hammer, M. Hocks, U. Kulisch and D. Ratz, “Nu-
merical Toolbox for Verified Computing,” Springer-
Verlag, Berlin, 1993.
[3] F. Kálovics and G. Mészáros, “Box Valued Functions in
Solving Systems of Equations and Inequalities,” Numeri-
cal Algorithms, Vol. 36, No. 1, 2004, pp. 1-12.
[4] F. Kálovics, “Zones and Integrals,” Journal of Computa-
tional and Applied Mathematics, Vol. 182, No. 2, 2005,
pp. 243-251.
[5] F. Kálovics, “Two Improved Zone Methods,” Miskolc
Mathematical Notes, Vol. 8, No. 2, 2007, pp. 169-179.