Encryption and Decryption
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Transcript of Encryption and Decryption
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Encryption and Encryption and Decryption
Speaker:Tsung Ray Wang Advisor:Prof.Li-Chun Wang
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Contents
MODELS,GOALS,AND EARLY CIPHER SYSTEMS
THE SECRECY OF A CIPHER SYSTEM
PRACTICAL SECURITY STREAM ENCRYPTION PUBLIC KEY CRYPTOSYSTEMS
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Model of a cryptographic channel
Plaintext
M
Encipher DecipherPublic channel
Plaintext
Cryptanalyst
Key
)(CDM k)(MEC k
CiphertextKK
Secure channel
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The two primary reasons for using cryptosystems in communications
(1)privacy,to prevent unauthorized persons from exacting information from the channel
(2)authentication,to prevent unauthorized persons from injecting information into the channel
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System GoalsThe major requirements for a cryptosystem
1.To provide an easy and inexpensive means of
encryption and decryption to authorized users in
possession of the appropriate key
2.To ensure that the cryptanalyst’s task of producing an
estimate of the plaintext without benefit of the key is
made difficult and expensive
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Classic Threats
Ciphertext-Only Attack Known-Plaintext Attack Chosen-Text Attack
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Classic Ciphers
Caesar Cipher ex. Plaintext : N O W I S T H E T I M E
: Ciphertext : Q R Z L V W K H W L P H
Polybius square . Plaintext : NOWI S T H E T I M E Ciphertext: 33 43 25 42 34 44 32 51 44 42 23 51
Polyalphabetic cipher . Plaintext: NOWI S T H E T I M E Ciphertext: OQZMXZ O M CS X Q
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Caesar’s alphabet with a shift of 3
Plaintext: ABCDEFGHIJKLMNOPQRSTUVWXYZ
CHIPHERTEXT: DEFGHIJKLMNOPQRSTUVWXYZABC
Polybius square
1 2 3 4 5
1 2 3 4 5
A B C D E
F G H IJ K
L M N O P
Q R S T U
V W X Y Z
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Trithemius progressive key
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THE SECRECY OF A CIPHER SYSTEM
•What is Perfect Secrecy??
• Entropy and Equivocation
• Rate of a language and Redunancy
• Unicity Distance and Ideal Secrecy
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P(Mo)=1/4 M00 C01
C1
C2
2
C3
3
P(M1)=1/4 M1
P(M2)=1/4M2
P(M3)=1/4 M3
Plaintext Ciphertext
Example of perfect secrecy Key
Cs=Tkj(Mi)
S=( ) modulo-Nji
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PRACTICAL SECURITY
Substitution Permutation Product Cipher System The Data Encryption Standard
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Substitution box
n=3
input
2n=8
1
1
0
output
0
1
1
2n=8012
3456
7
01234567
input
output
000
011
001
111
010
000
011
110
100
010
101
100
110
101
111
001
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Permutation box
input output
1
0
0
1
0
0
1
0
0
1
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Individual keying capability
Example of binary key
1 0 1 0 0 0 1 0 1 1 1 1 1 0 1 1 0 1 0 1 1 1 0 1 0
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Initial Permutation (IP)
58 50 42 34 26 18 10 2
60 52 44 36 28 20 12 4 62 54 46 38 30 22 14 6 64 56 48 40 32 24 16 8 57 49 41 33 25 17 9 1 59 51 43 35 27 19 11 3 61 55 45 37 29 21 13 5 63 55 47 39 31 23 15 7
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E-Table Bit Selection
32 1 2 3 4 5 4 5 6 7 8 98 9 10 11 12 1312 13 14 15 16 1716 17 18 19 20 2120 21 22 23 24 2524 25 26 27 28 2928 29 30 31 32 1
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P-Table Permutation
16 7 20 21
29 12 28 17
1 15 23 26
5 18 31 10
2 8 24 14
32 27 3 9
19 13 30 6
22 11 4 25
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Final Permutation (IP-1)
40 8 48 16 56 24 64 3239 7 47 15 55 23 63 3138 6 46 14 54 22 62 3037 5 45 13 53 21 61 2936 4 44 12 52 20 60 2835 3 43 11 51 19 59 2734 2 42 10 50 18 58 2633 1 41 9 49 17 57 25
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Key Permutation PC-1
57 49 41 33 25 17 9
1 58 50 42 34 26 18 10 2 59 51 43 35 27
19 11 3 60 52 44 36
63 55 47 39 31 23 15
7 62 54 46 38 30 22
14 6 61 53 45 37 29
21 13 5 28 20 12 4
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Key Schedule of Left Shifts
Iteration Number of left shifts i
12345678910111213141516
1
21
2222212222221
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Key Permutation PC-2
14 17 11 24 1 5
3 28 15 6 21 10
23 19 12 4 26 8
16 7 27 20 13 2
41 52 31 45 33 48
30 40 51 45 33 48
44 49 39 56 34 53
46 42 50 36 29 32
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STREAM ENCRYPTION
Key Generation Using a Linear Feedback Shift Register
Vulnerabilities of Linear Feedback Shift Registers
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Linear feedback shift register example
feedback
output
x4x3 x2 x1
Modulo-2
adder
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PUBLIC KEY CRYPTOSYSTEMS
Signature Authentication Using a Public Key Cryptosystem
A Trapdoor One-Way Function The Rivest-Shamir-Adelman Scheme The Knapsack Problem A Public Key Cryptosystem Based on a Trapdoor
Knapsack
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The important features of a public key cryptosystem
The encryption algorithm, ,and the decryption algorithm, ,are invertible transformations on the plaintext ,M,or the ciphertext ,C,defined by the key K. That is,for each K and M,
For each K, and are easy to compute. For each K,the computation of from is
computa-tionally intractable.
)()(),( MCMMC EDDE kkkk
E k
Dk
E kDk
E kDk
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Public Key cryptosystem
M Cryptomachine
Subscriber A
DirectoryA-B-C- . . . .
AE
BE
CE
EB
Cryptomachine
Subscriber B
M
DB
)(MC EB
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Signature authenticaton using a public key cryptosystem
MCrypto
machine
A
Date
Cryptomachine
A
Public channel
Directory
)(1
MS E A
))((1
MEC E AB
BEAD
Cryptomachine
B
Cryptomachine
B
M
Directory
Signaturestorage
))((1
MC EE AB
)(1
MS E A
E A
BD
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The Rivest-Shamir-Adelman Scheme
1.Each user chooses his own value of n and another pair of positive integers (e,d) ,and n=pq, =(p-1)(q-1),gcd[ ,d]=1,
ed modulo- =1,and p,q are prime numbers.
2..The user places his encryption key the number pair (n,e),in the public directory.
3. The decryption key consists of the number pair (n,d),of which d is kept secret.
4.messages are first represented as integers in the range (0,n-1)
5.Encryption: modulo-n
Decryption: modulo-neMMEC )()(
dCCDM )()(
)(n
)(n)(n
RSA
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How to Compute e
A variation of Euclid’s algorithm for computing the gcd of and d is to compute e
1.First,compute a series …... where = , =d ,and = modulo- ,until an =0 is found.
than the gcd ( , d )=
2.For each compute numbers and such that
= +
3.If =1,then is the multiplicative inverse of
modulo- .If is a negative number, the solution is
+
,.....,,, 210 xxx
)(n
)(n
0x )(n
1x 1ix 1ix ix
1kx
kx
ix ia ib
ix ia 0xib 1x
1kx1kb
0x 1kb1kb )(n
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The Knapsack problem
1.Let us express the knapsack problem in terms of a knapsack
vector ‘a’ and a data vector ’ x’.
2.The knapsack,S,is the sum of a subset of the components of the
knapsack vector where
= ax
n
n
xxxx
aaaa
,,.........,
,........,,
21
21
n
iii xaS
1
1,0ix
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Super-increasing and how to slove “x”
1.super-increasing is
2.When a is super-incresing,the solution of x is found by starting
with if S (otherwise ) ,and continuing
as follows:
=
where
1
1
i
j ji aa ni ,......,3,2
1nx na 0nx
ix
n
ijijj aaxS
1
0
1 if
otherwise
1..,,.........2,1 nni
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A Public key Cryptosystem Based on a Trapdoor Knapsack
-this scheme,also known as the Merkle-Hellman scheme
method:
1.we form a super-increasing n-tuple a’,and select a prime number
M such that ,also select a random number,W, where 1<W<M,and we form to satisfy the following relationship:
W modulo -M =1,note:the vector a’ and the number M,W,
are all kept hidden.
2.we form a with the elements from a’ as: modulo-M
n
iiaM
11W
1W 1W
ii Waa
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3.When a data vector x is to be transmitted ,we multiply x by a,
yielding the number S,which is sent on the public channel.
n
iii
n
iii xMuloWaxaaxS
11
)mod(
4.The authorized user receives S and converts it to S’ :
= =
n
iii MuloxMuloWaWMuloSWS
1
11 mod)mod(mod
n
iii MuloxMuloWaW
1
1 mod)mod(
n
iii Muloxa
1
mod
n
iii xa
1
5.Since the authorized user knowns the secretly held super-increasing
vector a’ ,he can use S’ to find x.
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CONCLUSION
1.We have presented the basic models and goals of the cryptographic
process,and looked at some early cipher systems.
2.We defined a system that can exhibit perfect secrecy .
3.We outlined the DES algorithm in detail,and we also considered
the use of linear feedback shift registers(LFSR) for stream
encryption systems.
4.RSA scheme ,based on the product of two large prime numbers,
and the Merkle-Hellman scheme,based on the classical knapsack
problem.