Inter
national
J
our
nal
of
Electrical
and
Computer
Engineering
(IJECE)
V
ol.
8,
No.
6,
December
2018,
pp.
4577
–
4583
ISSN:
2088-8708,
DOI:
10.11591/ijece.v8i6.pp4577-4583
4577
I
ns
t
it
u
t
e
o
f
A
d
v
a
nce
d
Eng
ine
e
r
i
ng
a
nd
S
cie
nce
w
w
w
.
i
a
e
s
j
o
u
r
n
a
l
.
c
o
m
Domain
Examination
of
Chaos
Logistics
Function
As
A
K
ey
Generator
in
Cryptograph
y
Alz
Danny
W
o
w
or
1
and
V
ania
Beatrice
Liwandouw
2
1
Department
of
Informatics
Engineering,
Satya
W
acana
Christian
Uni
v
ersity
,
Salatig
a,
Indonesia
2
Department
of
Computing
Science,
Radboud
Uni
v
ersity
,
Nijme
gen,
The
Netherland
Article
Inf
o
Article
history:
Recei
v
ed
September
28,
2017
Re
vised
May
09,
2018
Accepted
May
22,
2018
K
eyw
ord:
Domain
Examination
Logistic
Function
Chaos
Cryptograph
y
ABSTRA
CT
The
use
of
logistics
functions
as
a
random
number
generator
in
a
cryptograph
y
algo-
rithm
is
capable
of
accommodating
the
dif
fusion
properties
of
the
Shannon
principle.
The
problem
that
occurs
is
initialization
x
0
w
as
static
and
w
as
not
af
fected
by
changes
in
the
k
e
y
,
so
that
the
algorithm
will
generate
a
random
number
that
is
al
w
ays
the
same.
This
study
design
three
schemes
that
can
pro
viding
the
fle
xibility
of
the
input
k
e
ys
in
conducting
the
e
xamination
of
the
v
alue
of
the
domain
logistics
function.
The
results
of
each
schemes
do
not
sho
w
a
pattern
that
is
directly
proportional
or
in
v
erse
with
the
v
alue
of
x
0
and
relati
v
e
error
x
0
and
successfully
fulfill
the
properties
of
the
b
utterfly
ef
fect.
Thus,
the
e
xistence
of
logistics
functions
in
generating
chaos
numbers
can
be
accommodated
based
on
k
e
y
inputs.
In
addition,
the
resulting
random
numbers
are
dis-
trib
uted
e
v
enly
o
v
er
the
chaos
range,
thus
reinforcing
the
algorithm
when
used
as
a
k
e
y
in
cryptograph
y
.
Copyright
c
2018
Institute
of
Advanced
Engineering
and
Science
.
All
rights
r
eserved.
Corresponding
A
uthor:
Alz
Dann
y
W
o
w
or
Af
filiation
F
aculty
of
Information
T
echnology
,
Satya
W
acana
Christian
Uni
v
ersity
,
Jl.
Dipone
goro
52-60,
Salatig
a
50711,
Indonesia
+62852
0071
0079
alzdann
y
.w
o
w
or@staf
f.uksw
.edu
1.
INTR
ODUCTION
The
logisti
c
function
f
(
x
)
=
r
x
(1
x
)
or
in
the
iterati
v
e
form
x
i
+1
=
r
x
i
(1
x
i
)
w
as
usually
used
as
a
generator
in
the
generation
of
chaos-based
random
numbers
[1].
This
function
is
able
to
accommodate
the
dif
fusion
properties
of
Shannon’
s
principle
on
cryptograph
y
algorithms,
since
the
y
ha
v
e
a
sensiti
vity
to
initial
v
alues.
Research
[2],
[3],
[4],
uses
numbers
as
static
initialization
on
logistics
functions
in
cryptograph
y
algo-
rithms.
This
means
that
changes
to
the
k
e
y
don’
t
ha
v
e
an
ef
fect
on
random
numbers,
thus
the
algorithm
will
still
acquire
a
random
number
that
is
al
w
ays
the
same.
Research
[5],
also
uses
numbers
as
input
v
alues
that
directly
fill
in
by
user
,
ne
v
erthele
ss
becomes
inef
ficient
since
the
user
needs
more
input
other
than
the
k
e
y
and
plainte
xt.
This
research
not
only
combines
algorithms
such
as
[6],
[7],
[8],
b
ut
designs
a
ne
w
algorithm
to
obtain
an
unique
k
e
y
.
Additionally
,
the
function
will
generate
a
random
number
,
if
the
initialization
domain
r
and
x
0
as
seed
is
limited
to
a
certain
v
alue,
0
<
x
0
<
1
,
and
r
=
4
.
The
v
alue
of
r
is
constant,
making
the
strength
of
the
algorithm
rests
on
the
v
alue
of
x
0
.
Moreo
v
er
it
needs
an
e
xamination
process
by
using
an
y
scheme
that
can
increase
x
0
comple
xity
space,
on
the
contrary
remain
in
logistics
function
domain.
This
study
pro
vides
the
fle
xibility
of
k
e
y
inputs
that
are
ef
ficient
and
can
generate
dif
ferent
random
numbers
of
logistics
functions.
J
ournal
Homepage:
http://iaesjournal.com/online/inde
x.php/IJECE
I
ns
t
it
u
t
e
o
f
A
d
v
a
nce
d
Eng
ine
e
r
i
ng
a
nd
S
cie
nce
w
w
w
.
i
a
e
s
j
o
u
r
n
a
l
.
c
o
m
Evaluation Warning : The document was created with Spire.PDF for Python.
4578
ISSN:
2088-8708
2.
PR
OPOSED
EXAMIN
A
TION
SCHEME
The
e
xamination
process
of
the
logistics
function
domain
conducted
to
a
wide
v
ariety
of
inputs
that
allo
w
it
to
be
used
as
a
k
e
y
.
Suppose
the
k
e
y
k
1
;
k
2
;
:
:
:
;
k
8
is
the
result
of
a
con
v
ersion
of
eight
ASCII
characters.
K
e
y
input
is
set
up
to
eight
characters,
considering
the
user’
s
ability
to
recall
k
e
y
and
also
consider
the
comple
xity
of
the
k
e
y
guessing
space
when
the
use
of
the
characters
too
little.
Each
k
e
y
input
of
l
ess
than
eight
characters,
then
do
the
padding
process
with
character
”
x
”
that
equi
v
alent
to
167
in
ASCII.
This
study
pro
vides
three
algorithms
that
are
used
for
the
e
xamination
process
of
initialization
v
alue
in
domain
0
<
x
0
<
1
,
as
sho
wn
in
the
general
scheme
in
Figure
1.
The
ratio
of
r
a
is
designed
as
a
comparison
of
tw
o
v
alues
to
accommodate
the
domain
from
initialization.
Let
r
a
=
p
a
=q
a
where
q
a
>
p
a
,
for
a
=
1
;
2
;
3
and
p
a
;
q
a
2
R
.
x
0
A
SC
I
I
e
n
c
o
d
e
,
k
i
,
i = 1,2,…, 8.
examination
scheme
r
a
= p
a
/q
a
a = 1,2,3.
m
a
x
key = 8
p
a
d
d
i
n
g
c
h
a
r
a
c
t
e
r
Ҥ
”
Y
N
Figure
1.
General
Scheme
of
x
0
Initialization
V
alue
Determination.
2.1.
The
First
Scheme
Gi
v
en
each
k
i
2
Z
256
for
i
=
1
;
2
;
:
:
:
;
8
is
the
decimal
base
number
of
ASCII
con
v
ers
ion
results.
T
o
be
able
to
mak
e
changes
at
e
v
ery
turn
of
the
input,
gi
v
en
inde
x
v
alue
d
i
2
Z
256
for
j
=
1
;
2
;
:
:
:
;
8
as
a
constant
v
alue
which
is
multiplied
by
the
v
alue
k
i
.
Scheme-1
is
determined
by
using
r
1
=
p
1
=q
1
where,
p
1
=
(
k
1
)
2
+
k
7
d
7
+
k
3
d
3
+
k
4
d
4
+
k
5
d
1
+
k
6
d
6
+
k
8
d
8
+
k
1
d
5
(1)
q
1
=
8
X
i
=1
k
i
=
8
(
k
7
+
d
2
)
k
1
d
3
+
k
4
d
5
+
k
6
d
8
+
k
3
d
1
+
k
5
d
6
+
k
8
d
7
(
k
2
+
d
4
)
(2)
Determination
of
p
1
v
alue
is
obtained
from
the
sum
of
multiplication
k
i
and
d
i
which
is
performed
based
on
position,
it
is
only
for
k
1
and
d
5
there
is
crossing
position
and
at
k
1
squared.
Multiplication
of
position
dif
ference
is
also
done
to
acquire
the
q
1
v
alue,
ho
we
v
er
in
sum
with
the
a
v
erage
v
alue
of
each
k
i
.
The
multi-
plication
combination
is
done
as
a
v
ariation
to
g
ain
a
unique
rat
io
v
alue,
taking
into
account
the
requirement
q
1
>
p
1
.
2.2.
Second
scheme
Scheme-2
also
uses
the
same
decimal
number
and
inde
x
v
alue,
where
each
k
i
,
d
i
2
Z
256
for
i;
j
=
1
;
2
;
:
:
:
;
8
.
The
ratio
v
alue
is
determined
by
the
equation
r
2
=
p
2
=q
2
,
p
2
=
(
k
1
+
k
2
+
k
3
+
k
4
+
k
5
+
k
6
)
=
6
=
6
X
i
=1
k
i
=
6
(3)
q
2
=
k
1
d
1
+
k
2
d
2
+
k
3
d
3
+
k
4
d
4
+
k
5
d
5
+
k
6
d
6
+
k
7
d
7
+
k
8
d
8
=
8
X
i
=1
k
i
d
i
(4)
The
p
2
v
alue
is
obtained
using
the
a
v
erage
of
the
first
six
v
alues
of
k
i
,
whereas
to
earn
the
q
2
v
alue
is
the
sum
of
the
multiplication
of
k
i
and
d
i
v
alues
according
to
the
order
of
each
v
alue.
2.3.
Third
scheme
Ev
ery
k
i
;
d
j
2
Z
256
for
i;
j
=
1
;
2
;
:
:
:
;
8
:
Scheme-3
is
obtained
based
on
r
3
=
p
3
=q
3
,
where
the
determination
of
the
numerator
and
denominator
is
gi
v
en
in
Equation
(5)
and
Equation
(6).
The
p
3
v
alue
is
the
IJECE
V
ol.
8,
No.
6,
December
2018:
4577
–
4583
Evaluation Warning : The document was created with Spire.PDF for Python.
IJECE
I
SSN:
2088-8708
4579
a
v
erage
of
the
multiplication
k
i
with
d
i
only
at
the
first
v
alue,
the
third
v
alue,
and
the
sixth
v
alue
with
each
inde
x.
p
3
=
(
k
1
d
1
+
k
2
+
k
3
d
3
+
k
4
+
k
5
+
k
6
d
6
)
=
6
(5)
q
3
=
3(
k
1
d
1
)
+
7(
k
2
d
2
)
+
11(
k
3
d
3
)
+
13(
k
4
d
4
)
+
17(
k
5
d
5
)
+
23(
k
6
d
6
)
+
6(
k
7
d
7
)
+
27(
k
8
d
8
)
(6)
The
v
a
lue
of
q
3
is
t
he
sum
of
the
k
i
and
d
i
multiplications
based
on
the
inde
x
with
the
constants
selected
by
dif
ferent
increments.
3.
RESUL
T
AND
DISCUSSION
3.1.
Analysis
of
Examination
Pr
ocess
Domain
e
xamination
of
the
logistics
function
is
performed
based
on
the
three
schemes
gi
v
en
in
t
he
pre
vious
section.
Referring
to
the
general
scheme
in
Figure
1,
then
testing
related
v
ariations
of
inputs
that
allo
w
to
be
used
as
a
k
e
y
.
Each
possible
k
e
y
is
an
ASCII
character
whose
decimal
basis
is
in
the
range
0
to
255.
The
problem
that
occurs
is
not
all
numbers
ha
v
e
a
character
.
Consequently
the
test
for
the
lo
west
number
can’
t
start
from
decimal
0
instead
in
decimal
32
which
is
proportional
to
the
space
character
,
whereas
testing
for
the
lar
gest
number
at
decimal
255
is
equi
v
alent
to
the
character
.
In
addition
to
character
testing
for
minimum
and
maximum
decimals,
k
e
y
tests
are
also
tested
with
one
bit
dif
ference,
so
that
it
can
be
seen
ho
w
sensiti
v
e
each
scheme
is
to
generating
initialization
v
alues.
T
able
1
is
the
simulation
result
of
each
scheme
in
obtaining
the
v
alue
of
x
0
.
T
able
1.
K
e
y
V
ariations
T
est
on
Each
Scheme.
T
est
Input
K
e
ys
Schem
e-1
Scheme-2
Scheme-3
1
ZZZZZZ
0.002416239247
0.020477815700
0.082448573324
2
ZZZZZY
0.002417771534
0.020467836257
0.082481509183
3
fti
0.002441029054
0.024268127467
0.082049869712
4
ftj
0.002439355077
0.024284705051
0.082201209246
5
$4LaT1g4
0.006268369980
0.025188594809
0.095909955612
6
$4LaT1g3
0.006272557467
0.025266127990
0.096168154762
7
(space
8
character)
0.018665158371
0.027777777778
0.096054888500
8
0.003244514742
0.027777777778
0.096054888508
9
•
y
•
y
•
y
•
y
•
y
•
y
•
y
•
y
0.001456374283
0.027777777778
0.096054888508
T
esting
with
one
bit
dif
ference
(numbers
1
through
6)
sho
ws
the
changes
in
v
alues
that
be
gin
to
occ
ur
in
the
4th
or
5th
mantissa
of
the
x
0
v
alue
in
each
schema.
Accordingly
,
the
process
of
domain
e
xamination
for
each
scheme
succeeds
to
generate
dif
ferent
initialization.
This
condition
corresponds
to
the
need
of
logistics
functions
in
obtaining
random
numbers
based
on
chaos.
T
ests
with
the
same
eight-character
input
based
on
the
smallest
decimal,
the
medium
decimal,
and
the
lar
gest
decimal
are
gi
v
en
successi
v
ely
in
the
numbers
7,
8,
and
9
in
T
able
1.
The
initialization
v
alues
in
scheme-2
and
scheme-3
obtain
the
same
v
alue,
although
the
input
is
v
ery
dif
ferent.
This
condition
occurs
since
the
determination
ratio
of
r
2
and
r
3
using
the
a
v
erage
process
to
get
the
numerator
,
while
for
the
denominator
using
the
addition
and
multiplication
of
characters
with
the
inde
x
v
alue
with
the
same
position.
On
the
other
hand,
scheme-1
uses
a
combination
of
multi
plicity
of
position
dif
ference,
squared
and
mean
process.
In
that
case,
scheme-1
k
eeps
generating
dif
ferent
initialization
v
alues.
3.2.
Relati
v
e
Err
or
T
est
Relati
v
e
error
testing
[9]
w
as
conducted
to
see
whether
linear
character
reduction
w
ould
yield
also
linear
results
on
the
x
r
v
alues
with
proportional
or
in
v
ersely
proportional.
Use
E
R
=
j
c
a
p
a
j
=c
a
100%
,
the
k
e
y
•
y
•
y
•
y
•
y
•
y
•
y
•
y
•
y
is
chosen
as
the
reference
v
alue
c
a
(number
1
in
T
able
2),
and
the
approximation
v
alue
as
the
k
e
y
input
p
a
less
than
eight
characters
•
y
(iterated
from
number
2
to
number
8).
Domain
Examination
of
Chaos
Lo
gistics
Function
As
A
K
e
y
Gener
ator
in
Crypto
gr
aphy
(Alz
Danny
W
owor)
Evaluation Warning : The document was created with Spire.PDF for Python.
4580
ISSN:
2088-8708
T
able
2.
Relati
v
e
error
test
on
k
e
y
v
ariation
dif
ferences.
T
est
Input
k
e
ys
Scheme-1
Scheme-2
Scheme-3
Relati
v
e
Error
Scheme-1
Scheme
2
Scheme-3
1
•
y
•
y
•
y
•
y
•
y
•
y
•
y
•
y
0.00145637428
0.027777777778
0.09605488850
2
•
y
•
y
•
y
•
y
•
y
•
y
•
y
0.00151184278
0.030084945729
0.10267089320
3.81
8.31
6.89
3
•
y
•
y
•
y
•
y
•
y
•
y
0.00237830973
0.032442748091
0.10498487627
63.30
16.79
9.30
4
•
y
•
y
•
y
•
y
•
y
0.00247676219
0.032778687034
0.10922797121
70.06
18.00
13.71
5
•
y
•
y
•
y
•
y
0.00260130457
0.032743277230
0.11525621373
78.62
17.88
19.99
6
•
y
•
y
•
y
0.00272929162
0.032262996942
0.11525043975
87.40
16.15
19.98
7
•
y
•
y
0.00287551307
0.031283195241
0.10350303974
97.44
12.62
7.75
8
•
y
0.00146481070
0.029781420765
0.09849935979
0.58
7.21
2.54
The
relati
v
e
error
results
in
each
scheme
do
not
sho
w
a
proportional
or
in
v
ersely
proportional
pattern.
So,
it
will
complicate
cryptanalyst
to
see
the
pattern
of
input
changes
on
the
same
character
.
3.3.
Linear
Regr
ession
T
est
One-to-one
correspondence
between
the
input-output
is
also
important
so
that
cryptanalyst
dif
ficult
to
reconstructing
scheme
and
mak
e
prediction
of
t
he
k
e
y
used
as
input.
Thi
s
relationship
can
be
seen
through
a
linear
re
gression
test
based
on
the
rate
of
change
on
the
resulting
v
alue
of
x
0
.
Figure
2,
is
the
result
of
each
scheme
visualized
using
the
Scatter
plot.
The
diagram
of
each
scheme
has
no
linear
relationship,
because
when
the
curv
e
matching
process
is
used,
the
coef
ficient
of
determination
(
R
2
)
is
close
to
zero.
This
test
illustrates
that
an
y
changes
to
the
k
e
y
characters
is
done
patterned,
will
not
pro
vide
an
initial
v
alue
x
0
linearly
patterned,
either
proportional
or
in
v
ersely
.
scheme 1
0.000
0.001
0.002
0.003
0.004
0
1
2
3
4
5
6
7
8
y = 9.589E-5x + 0.0018
R
²
= 0.1502
scheme 2
0.026
0.028
0.030
0.032
0.034
0
1
2
3
4
5
6
7
8
y = 0.0002x + 0.0301
R
²
= 0.1
scheme 3
0.080
0.091
0.103
0.114
0.125
0
1
2
3
4
5
6
7
8
y = 0.0007x + 0.1026
R
²
= 0.0567
Figure
2.
Diagram
Scatter
Initialization
V
alue
x
0
based
on
K
e
y
P
attern
Changes.
3.4.
Butterfly
Effect
T
est
The
b
utterfly
ef
fect
test
is
used
to
indicate
the
change
in
bits
in
the
k
e
y
input,
whether
it
gi
v
es
a
lar
ge
change
to
the
output.
Suppose
that
as
a
comparator
k
e
y
ZZZZZZ
and
ZZZZZY
are
selected
which
has
a
dif-
ference
of
one
bit.
The
result
of
the
tw
o
k
e
ys
with
scheme-1
is
obtained
by
a
random
number
of
the
first
500
iterations,
sho
wn
in
Figure
3.
ZZZZZZ
0
0.25
0.5
0.75
1
i
0
125
250
375
500
ZZZZZY
0
0.25
0.5
0.75
1
i
0
125
250
375
500
x
i
x
i
Figure
3.
Results
of
Random
Numbers
with
Dif
ferent
Inputs
1
Bit
for
Scheme-1.
V
isually
,
the
random
number
generated
is
v
ery
dif
ferent
although
the
dif
ference
in
initialization
v
alue
x
0
is
only
0
:
0000151538
1
:
538
10
6
or
in
an
absolute
relati
v
e
of
0.063%.
A
minor
change
in
input
and
a
major
change
in
output
pro
v
es
that
the
1st
scheme
has
successfully
fulfilled
the
b
utterfly
ef
fect.
IJECE
V
ol.
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No.
6,
December
2018:
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–
4583
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IJECE
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SSN:
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4581
The
initial
v
alue
dif
ference
of
x
0
is
0.0487%
in
scheme-2.
But
the
rate
of
change
occurs
v
ery
signif-
icantly
that
appears
in
the
Scatter
diagram
in
Figure
4.
So,
the
scheme-2
has
also
fulfilled
the
b
utterfly
ef
fect
test.
ZZZZZZ
0
0.25
0.5
0.75
1
i
0
125
250
375
500
ZZZZZY
0
0.25
0.5
0.75
1
i
0
125
250
375
500
x
i
x
i
Figure
4.
Results
of
Random
Numbers
with
Dif
ferent
Inputs
1
Bit
for
Scheme-2.
Significant
changes
also
occur
at
random
v
alues
with
scheme-3,
as
sho
wn
in
Figure
5,
although
the
dif
ference
is
0.0399%
at
input
v
alue
x
0
.
So,
the
scheme-3
also
meets
the
properties
of
the
b
utterfly
ef
fect.
ZZZZZZ
0
0.25
0.5
0.75
1
i
0
125
250
375
500
ZZZZZY
0
0.25
0.5
0.75
1
i
0
125
250
375
500
x
i
x
i
Figure
5.
Results
of
Random
Numbers
with
Dif
ferent
Inputs
1
Bit
for
Scheme-3.
The
three
schemes
ha
v
e
succeeded
in
fulfilling
t
he
properties
of
the
b
utterfly
ef
fect,
thus
the
e
xistence
of
the
logistics
function
in
generating
chaos
numbers
can
be
accommodated
based
on
k
e
y
inputs.
Each
scheme
can
be
used
as
a
complement
to
a
cryptograph
y
algorithm
to
meet
the
dif
fusion
properties
of
Shannon’
s
principle.
3.5.
Analysis
Of
The
Algorithm
Ability
Each
scheme
is
tested
in
correlation
[10],
to
detect
the
connectedness
of
the
random
number
generated
based
on
the
input.
While
MAPE
[11],
it
is
used
to
find
out
ho
w
massi
v
e
the
dif
ference
of
random
v
alue
of
k
e
y
change.
Used
three
v
ariations
of
the
input
[12],
the
firs
t
is
the
same
character
input,
a
second
input
alphabetic
characters
that
re
v
olv
e
around
the
26-character
alphabet.
While
the
last
test,
used
alphabet
,
symbols,
and
numbers.
Calculation
of
correlation
in
T
able
3
sho
w
that
there
are
tw
o
v
alues
on
scheme-1
and
scheme-3
which
correlation
is
ne
g
ati
v
e,
besides
the
rest
is
positi
v
e.
Cryptographically
,
the
ne
g
ati
v
e
v
alue
is
not
too
influential,
hence
it
seen
ho
w
close
the
v
alue
to
zero
indicating
the
unrelated
tw
o
random
numbers
are
generated.
In
the
conte
xt
of
the
relations,
this
same
analogy
can
be
used
to
test
the
dif
ference
of
tw
o
random
numbers
generated.
T
able
3.
1
Bit
dif
ferences
test
with
k
e
y
v
ariations.
T
est
Input
k
e
ys
Correlation
v
alue
MAPE
Scheme-1
Scheme
2
Scheme-3
Sc
heme-1
Scheme-2
Scheme-3
1
ZZZZZZ
-
ZZZZZY
0.1739823
0.04831424
0
:
0685288
23.9966
3486.022
13.192
2
fti
-
ftj
0.0220125
0.05469541
0.00900384
612.172
210.2993
46.6942
3
$4LaT1g4
-
$4LaT1g3
0
:
011209
0.02911591
0.00971799
64.9543
38.1335
380.158
Ov
erall
the
correlation
v
alue
generated
by
each
scheme
is
within
the
range
of
0.00
-
2.99.
Based
on
[13],
the
interv
al
sho
ws
the
strength
of
a
v
ery
weak
relationship.
This
condition
pro
vides
information
t
hat
1
bit
k
e
y
input
dif
ference,
can
generate
dif
ferent
random
numbers
on
each
scheme.
Domain
Examination
of
Chaos
Lo
gistics
Function
As
A
K
e
y
Gener
ator
in
Crypto
gr
aphy
(Alz
Danny
W
owor)
Evaluation Warning : The document was created with Spire.PDF for Python.
4582
ISSN:
2088-8708
In
additi
on
to
the
correlation
and
MAPE
analysis,
we
also
tested
the
distrib
ution
of
random
number
data
using
box-plot
diagrams.
-
0.
2
0
0.
2
0.
4
0.
6
0.
8
1
1.
2
Figure
6.
Box
Plot
Random
Number
of
Each
Scheme
Output
with
V
arious
Input
V
ariations
Based
on
Figure
6,
each
box
has
almost
the
same
size
in
which
the
upper
and
lo
wer
whisk
er
lines
v
ary
slightly
,
b
ut
the
maximum
v
alue
is
al
w
ays
close
to
one
and
the
minimum
is
near
zero.
The
distrib
ution
of
data
in
the
chaos
range
will
strengthen
the
cryptograph
y
algorithm
if
used
as
a
k
e
y
,
this
condition
will
certainly
complicate
cryptanalyst
to
be
able
to
search
for
infinitely
man
y
numbers
although
limited.
4.
CONCLUSION
Each
designed
scheme
is
capable
of
pro
viding
k
e
y
input
fle
xibility
that
can
e
x
ecute
domai
n
logistics
function
v
alues.
A
1
bit
dif
ference
in
the
k
e
y
character
af
fects
e
v
ery
random
number
generation,
so
each
k
e
y
will
generate
a
dif
ferent
random
number
sequence.
Under
the
k
e
y
input
conditions
of
the
sam
e
eight
characters,
the
1st
scheme
is
better
at
generating
dif
ferent
initialization
v
alues
than
the
scheme-2
and
schema-3.
In
addition,
the
one
character
reduction
of
the
eight
identical
characters
in
the
k
e
y
input
does
not
sho
w
a
proportional
or
re
v
erse
pattern
with
the
initial
x
0
v
alues
and
the
relati
v
e
error
x
0
.
The
resulting
random
numbers
distrib
uted
e
v
enly
o
v
er
the
chaos
range
will
amplify
the
algorithm
when
used
as
a
k
e
y
in
cryptograph
y
.
This
condition
will
certainly
complicate
cryptanalyst
to
be
able
to
search
for
infinitely
man
y
numbers
although
limited.
The
three
schemes
ha
v
e
succeeded
in
fulfilling
the
nature
of
the
b
utterfly
ef
fect,
thus
the
e
xistence
of
the
logistics
function
in
generating
chaos
numbers
can
be
accommodated
based
on
k
e
y
inputs.
Each
scheme
can
be
used
as
a
complement
to
a
cryptograph
y
algorithm
to
satisfy
the
dif
fusion
properties
of
the
Shannon
principle.
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ane
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2017,
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Montgomery
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ile
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Makridaki
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Liw
andouw
,
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&
W
o
w
or
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A.D.,
2015,
Desain
Algoritma
Berbasis
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ub
us
Rubik
dalam
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an
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eknik
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dan
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Sarw
ono,
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ualitatif
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Ilmu.
IJECE
V
ol.
8,
No.
6,
December
2018:
4577
–
4583
Evaluation Warning : The document was created with Spire.PDF for Python.
IJECE
I
SSN:
2088-8708
4583
A
CKNO
WLEDGMENT
Thank
you
to
Satya
W
acana
Christian
Uni
v
ersity
Research
and
Community
Service
Center
(BP3M)
for
the
research
funding
support
through
the
Internal
Oblig
atory
Research
scheme
in
the
2016
fiscal
year
.
BIOGRAPHIES
OF
A
UTHORS
Alz
Danny
W
o
w
or
is
currently
a
lecturer
at
the
F
aculty
of
Informa
tion
T
echnology
,
Satya
W
acana
Christian
Uni
v
ersity
in
Salatig
a,
Indonesia.
He
recei
v
ed
bachelor
and
master
de
gree
in
mathemat-
ics
and
informatics
from
Satya
W
acana
Christian
Uni
v
ersity
,
in
2005
and
2011
respecti
v
ely
.
His
researches
are
in
fields
of
Primiti
v
e
Crypt
ograph
y
,
Symmetric
Cryptograph
y:
Bloc
k
Cipher
and
Pseudorandom.
V
ania
Beatrice
Liwandouw
is
a
Master
of
Cyber
Security
student
at
F
aculty
of
Science,
Radboud
Uni
v
ersity
,
Nijme
gen,
The
Netherland.
She
recei
v
ed
her
Bachelor
de
gree
in
engineering
informatics
at
the
F
aculty
of
Information
T
echnology
,
Satya
W
acana
Christian
Uni
v
ersity
,
Salatig
a,
Indonesia.
Her
research
interests
are
in
the
Design
and
Implem
entation
of
Symmetric
Cryptographic
Algo-
rithms.
Domain
Examination
of
Chaos
Lo
gistics
Function
As
A
K
e
y
Gener
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