8.13 Cryptography

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Nov 21, 2013 (3 years and 8 months ago)

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8.13 Cryptography


Introduction

Secret writing

means
code
.


A simple code

Let the letters a, b, c,

., z be represented by the
numbers 1, 2, 3,

, 26. A sequence of letters cab
then be a sequence of numbers. Arrange these
numbers into an
m


n
matrix
M
. Then we select a
nonsingular
m


m
matrix
A
. The new sent message
becomes
Y
=
AM
, then
M
=
A
-
1
Y
.

8.14 An Error Correcting Code


Parity Encoding

Add an extra bit to make the number of one is
even

Example 2

(a)
W

= (1 0 0 0 1 1) (b)
W

= (1 1 1 0 0 1)

Solution

(a) The extra bit will be 1 to make the number of
one is 4 (even).

The code word is then
C

= (1 0 0 0 1 1 1).

(b) The extra bit will be 0 to make the number of
one is 4 (even).

So the edcoded word is C = (1 1 1 0 0 1 0).

Fig 8.12

Example 3

Decoding the following

(a)
R

= (1 1 0 0 1 0 1)

(b)
R

= (1 0 1 1 0 0 0)

Solution

(a) The number of one is 4 (even), we just drop the
last bit to get (1 1 0 0 1 0).

(b) The number of one is 3 (odd). It is a parity error.


Hamming Code

where
c
1
,
c
2
, and
c
3

denote the parity check bits.

)
(
4
3
2
3
1
2
1
w
w
w
c
w
c
c

C
)
(
4
3
2
1
w
w
w
w

W
Encoding

4
3
2
3
4
3
1
2
4
2
1
1
w
w
w
c
w
w
w
c
w
w
w
c












































4
3
2
1
3
2
1
1
1
1
0
1
1
0
1
1
0
1
1
w
w
w
w
c
c
c
Example 4

Encode the word W = (1 0 1 1).

Solution


,
1
,
0
,
1
3
2
1



w
w
w
1
4

w


































































0
1
0
1
1
1
1
0
1
1
0
1
1
1
1
0
0
1
1
1
1
1
0
0
1
1
1
1
1
0
1
1
1
1
0
1
1
0
1
1
0
1
1
3
2
1












c
c
c
0
,
1
,
0
3
2
1



c
c
c
)
1
1
0
0
1
1
0
(

C
Decoding

0
HR
S


T
Example 5

Compute the syndrome of (a) R = (1 1 0 1 0 0
1) and (b) R = (1 0 0 1 0 1 0)

Solution


(a)







we conclude that R is a code word. By the check
bits in (
1

1

0
1

0 0 1), we get the decoded
message (0 0 0 1).













































0
0
0
1
0
0
1
0
1
1
1
0
1
0
1
0
1
1
1
0
0
1
1
0
1
1
1
1
0
0
0
S
Example 5 (2)

(b)







Since
S



0
, the received message
R

is not a code
word.













































1
1
0
0
1
0
1
0
0
1
1
0
1
0
1
0
1
1
1
0
0
1
1
0
1
1
1
1
0
0
0
S
]
[
7
6
5
4
3
2
1
e
e
e
e
e
e
e

E
bit
th

the
change
ot

does

noise

if
bit
th

the
changes

noise

if
,
0
,
1
i
i
e
i




T
T
T
T
T
T
T
HE
HE
0
HE
HC
E
C
H
HR







)
(
,
E
C
R


T
T
T
E
C
R






















7
5
3
1
7
6
3
2
7
6
5
4
e
e
e
e
e
e
e
e
e
e
e
e
T
HE













































































1
1
1
0
1
1
1
0
1
0
0
1
1
1
0
0
1
0
1
0
0
7
6
5
4
3
2
1
e
e
e
e
e
e
e
T
HE
Example 6

Changing zero to one gives the code word
C

= (1
0 1 1 0 1 0). Hence the first, second, and
fourth bits from
C

we arrive at the decoded
message (1 0 1 0).

)
0
1
0
1
0
0
1
(

R











1
1
0
S
8.15 Method of Least Squares


Example 2

If we have the data (1, 1), (2, 3), (3, 4), (4, 6),
(5,5), we want to fit the function
f
(
x
)

=ax + b.
Then





a + b =
1



2
a + b =
3



3
a + b =
4




4
a + b =
6




5
a + b =
5


Example 2 (2)

Let







we have

,
5
6
4
3
1

















Y

















1
5
1
4
1
3
1
2
1
1
A







5
15
15
55
A
A
T
Example 2 (3)






























































































5
.
0
1
.
1
19
68
55
15
15
5
50
1
5
6
4
3
1
1
1
1
1
1
5
4
3
2
1
55
15
15
5
50
1
5
6
4
3
1
1
5
1
4
1
3
1
2
1
1
5
15
15
55
1
T
X
Example 2 (4)

We have
AX

=
Y
. Then the best solution of
X
will
be
X

= (
A
T
A
)
-
1
A
T
Y
= (1.1, 0.5)
T
.
For this line the
sum of the square error is





The fit function is




y =
1.1
x +
0.5

7
.
2
]
6
5
[
]
9
.
4
6
[
]
8
.
3
4
[
]
7
.
2
3
[
]
6
.
1
1
[
)]
5
(
5
[
)]
4
(
6
[
)]
3
(
4
[
)]
2
(
3
[
)]
1
(
1
[
2
2
2
2
2
2
2
2
2
2





















f
f
f
f
f
E
Fig 8.15

8.16 Discrete Compartmental Models


The General Two
-
Compartment Model

)
(
)
(
)
(
)
(
)
(
)
(
)
(
2
20
21
1
21
2
21
1
10
12
t
c
F
F
t
c
F
dt
dy
t
I
t
c
F
t
c
F
F
dt
dx








Fig 8.16

Discrete Compartmental Model































n
n
y
y
y
x
x
x


2
1
2
1
,
Y
X

n
n
n
n
n
n
x
x
x
x
x
x
x
x
x
y
1
2
12
1
1
31
21
1
1
31
21
1
3
13
2
12
1
1
1
)
1
(
)
(
)
(
)
1

leaving
tracer
of
amount
(
)
1

entering
tracer
of
amount
(



































n
nn
n
n
n
n
n
n
n
x
x
x
y
x
x
x
y
x
x
x
y

























2
2
1
1
2
2
22
1
21
2
1
2
12
1
11
1











































n
nn
n
n
n
n
n
x
x
x
y
y
y







2
1
2
1
2
22
21
1
12
11
2
1









Fig 8.17

Example 1


See Fig 8.18. The initial amount is 100, 250, 80
for these three compartment.

For Compartment 1 (C1):




20% to C2



0% to C3

then

80% to C1



For C2:

5% to C1



30% to C3

then

65% to C2

For C3:

25% to C1



0% to C3

then

75% to C3
















3
.
0
0
0
2
.
0
25
.
0
05
.
0
T
Fig 8.18


That is,

New C1 = 0.8C1 + 0.05C2 + 0.25C3

New C2 = 0.2C1 + 0.65C2 + 0C3

New C3 = 0C1 + 0.3C2 + 0.75C3

We get the transfer matrix as
















75
.
0
3
.
0
0
0
65
.
0
2
.
0
25
.
0
05
.
0
8
.
0
T
Example 1 (2)

Example 1 (3)

Then one day later,









































135
5
.
182
5
.
112
80
250
100
75
.
0
3
.
0
0
0
65
.
0
2
.
0
25
.
0
05
.
0
8
.
0
TX
Y

Note
:
m
days later,
Y

=
T
m
X
0

n
n
TX
X
TX
X
TX
X
TX
X





1
2
3
1
2
0
1
,
,
,
,


,
)
(
,
)
(
0
3
0
2
3
0
2
0
2
X
T
X
T
X
X
T
TX
T
X





,
2
,
1
,
0


n
n
n
X
T
X
Example 2

Example 2 (2)
















20
15
60
20
0
X















1
0
01
.
0
0
0
8
.
0
0
1
.
0
0
2
.
0
98
.
0
05
.
0
0
0
01
.
0
85
.
0
T
Example 2 (3)














































6
.
20
0
.
14
8
.
62
6
.
17
20
15
60
20
1
0
01
.
0
0
0
8
.
0
0
1
.
0
0
2
.
0
98
.
0
05
.
0
0
0
01
.
0
85
.
0
0
1
TX
X
Para la matriz sim
é
trica:



tenemos


=

9, −9, 9.




















8
1
4
1
8
4
4
4
7
A




























0
0
0
0
0
0
0
0
0
4
1/
1/4
1
filas
s
operacione
0
0
0
1
1
4
1
1
4
4
4
16
)
0
9
(
I
A























0
4
1
,
1
1
0
2
1
K
K
3
2
1
4
1
4
1
k
k
k































0
0
0
0
0
0
0
0
0
4
0
1
filas
s
operacione
0
0
0
17
1
4
1
17
4
4
4
2
)
0
9
(
I
A












0
1
4
3
K
2
1
4
k
k


Recuerda que si
A

es una matriz
n

×

n

simétrica, los autovectores
correspondientes a
distintos
(diferentes)

autovalores
son ortogonales.

Observa que:


K
3



K
1

=
K
3



K
2

= 0,
K
1



K
2

=


4


0

Usando el m
é
todo de Gram
-
Schmidt:
V
1

=
K
1





Ahora si que tenemos un conjunto ortogonal y podemos
normalizarlo:


























2
2
1
1
1
1
1
2
2
2
V
V
V
V
K
K
V


















































2
3
1
2
3
1
2
3
4
,
3
2
3
2
3
1
,
2
1
2
1
0



















2
3
1
3
2
2
1
2
3
1
3
2
2
1
2
3
4
3
1
0
P
Ortogonal