Download Example 2

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-1Y.
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
W  ( w1
w2
w3
w4 )
C  (c1 c2 w1 c3 w2 w3 w4 )
where c1, c2, and c3 denote the parity check bits.
Encoding
c1  w1  w2  w4
c2  w1  w3  w4
c3  w2  w3  w4
 w1 
 c1   1 1 0 1  
  
  w2 
 c2    1 0 1 1  
 c   0 1 1 1  w3 
 3 
 
 w4 
Example 4
Encode the word W = (1 0 1 1).
Solution
w  1, w  0, w  1, w  1
1
2
3
4
1
 c1   1 1 0 1   1．1  1．0  0．1  1．1  0 
  
 0 
  
 c2    1 0 1 1    1．1  0．0  1．1  1．1   1 
 c   0 1 1 1  1   0．1  1．0  1．1  1．1  0 
 3 
  
  
1
c1  0, c2  1, c3  0
C  (0 1 1 0 0 1 1)
Decoding
S  HR  0
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)
1
Solution
 
(a)
1
 0 0 0 1 1 1 1  0   0 

   
S   0 1 1 0 0 1 1  1    0 
 1 0 1 0 1 0 1  0   0 

   
0
1
 
we conclude that R is a code word. By the check
bits in (1 1 0 1 0 0 1), we get the decoded
Example 5 (2)
(b)
1
 
0
 0 0 0 1 1 1 1  0   0 

   
S   0 1 1 0 0 1 1  1    1 
 1 0 1 0 1 0 1  0   1 

   
1
0
 
Since S  0, the received message R is not a code
word.
E  [e1 e2 e3 e4 e5 e6 e7 ]
1 , if noise changes the ith bit
ei  
0, if noise does ot change the ith bit
R  C  E,
RT  CT  ET
HRT  H(CT  ET )  HCT  HET  0  HET  HET
 e4  e5  e6  e7 


T
HE   e2  e3  e6  e7 
e e e e 
 1 3 5 7
0
0
0
1
1
1
1
 
 
 
 
 
 
 
T
HE  e1  0   e2  1   e3  1   e4  0   e5  0   e6  1   e7 1
1
0
1
0
1
0
1
 
 
 
 
 
 
 
Example 6
R  (1 0 0 1 0 1 0)
0
 
S  1
1
 
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).
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
2a + b = 3
3a + b = 4
4a + b = 6
5a + b = 5
Example 2 (2)
Let
1
 1 1
 


 3
 2 1
Y   4 , A   3 1
 


6
 4 1
5
 5 1
 


 55 15 
A A

 15 5 
T
we have
Example 2 (3)
1

1  2
 55 15  
X
 3
 15 5  
4
5

T
1

1
1

1
1
1
 
 3
 4
 
6
5
 
1
 
 3
1  5  15  1 2 3 4 5   
 

 4
50   15
55  1 1 1 1 1  
6
5
 
1  5  15   68   1.1 
 
  
50   15
55   19   0.5 
Example 2 (4)
We have AX = Y. Then the best solution of X will
be X = (ATA)-1ATY = (1.1, 0.5)T. For this line the
2
sum
E  [1of
 fthe
(1)]2 square
[3  f (2)]error
 [4  is
f (3)]2  [6  f (4)]2  [5  f (5)]2
 [1  1.6]2  [3  2.7]2  [4  3.8]2  [6  4.9]2  [5  6]2  2.7
The fit function is
y = 1.1x + 0.5
Fig 8.15
8.16 Discrete Compartmental Models
• The General Two-Compartment Model
dx
 ( F12  F10 )c1 (t )  F21c2 (t )  I (t )
dt
dy
 F21c1 (t )  ( F21  F20 )c2 (t )
dt
Fig 8.16
Discrete Compartmental Model
 x1 
 
 x2 
X ,

 
 xn 
 y1 
 
 y2 
Y 

 
 yn 
y1  x1  (amount of tracer entering 1)  (amount of tracer leaving 1)
 x1  ( 12 x2   13 x3     1n xn )  ( 21   31     n1 ) x1
 (1   21   31     n1 ) x1   12 x2     1n xn．
y1   11x1   12 x2     1n xn
y2   21x1   22 x2     2 n xn

yn   n1x1   n 2 x2     nn xn
 y1    11  12   1n   x1 
  
 
 y2   21  22   2 n   x2 
   
   
  
  
 yn   n1  n 2   nn   xn 
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
－ 0.05 0.25 


0%Tto C3
then
75%
to
0  C3
 0.2 －
 0

0.3
－ 
Fig 8.18
Example 1 (2)
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
 0.8 0.05 0.25 


T   0.2 0.65
0 
 0

0
.
3
0
.
75


Example 1 (3)
Then one day later,
 0.8 0.05 0.25   100  112.5 


 

Y  TX   0.2 0.65
0   250   182.5 
 0 0.3 0.75   80  135 


 

• Note: m days later, Y = TmX0
X1  TX0 , X 2  TX1 , X3  TX 2 ,  , X n1  TX n
X2  T(TX0 )  T2 X0 , X3  (T2 X0 )  T3X0 , 
X n  T n X0 ,
n  1, 2, 
Example 2
Example 2 (2)
 20 
 
60 

X0   
15
 
 20 
 0.85 0.01 0

0.05 0.98 0.2

T
0.1
0
0.8

0.01 0
 0
0

0
0

1
Example 2 (3)
 0.85 0.01 0

0.05 0.98 0.2

X1  TX0  
0.1
0
0.8

0.01 0
 0
0   20   17.6 
  

0   60   62.8 

0   15   14.0 
   

1   20   20.6 
Para la matriz simétrica:
tenemos  = −9, −9, 9.
 16 4  4 0 


( A  9I 0)   4
1 1 0
  4 1 1 0


  2 4  4 0


( A  9I 0)   4 17  1 0 
  4  1 17 0 


4  4
 7


A   4  8 1
  4 1  8


operacione s
filas

operacione s
filas

 1 1/4  1/ 4 0 


0 0
0 0
0 0

0
0


1 0 4 0


0 0 0 0
0 0 0 0


1
1
k1   k2  k3
4
4
0
 1
 
 
K1   1, K 2    4 
 1
 0
 
 
k1  4k2
  4
 
K 3   1
 0
 
Recuerda que si A es una matriz n × n simétrica, los autovectores
correspondientes a distintos (diferentes) autovalores son ortogonales.
Observa que:
K3  K1 = K3  K2 = 0, K1  K2 = – 4  0
Usando el método de Gram-Schmidt: V1 = K1
 1 
 
 K 2  V1 
V2  K 2  
V1    2 
 V1  V1 
 2 
 
Ahora si que tenemos un conjunto ortogonal y podemos
normalizarlo:








1  4 


0   

3
2
  3 

1   2  1 
,  ,
2   3  3 2 
1   2  1 
   

2  3   3 2



P




0
1
2
1
2
1
4 

3
3 2 
2
1 

3 3 2 
2
1 


3
3 2
Ortogonal