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Presentation Title Your company information. Cryptography2 Syllabus Course Syllabus –أخلاقيات استخدام الانترنت والقوانين المتعلقة بها –مقدمة في أمن المعلومات.

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Presentation on theme: "Presentation Title Your company information. Cryptography2 Syllabus Course Syllabus –أخلاقيات استخدام الانترنت والقوانين المتعلقة بها –مقدمة في أمن المعلومات."— Presentation transcript:

1 Presentation Title Your company information

2 Cryptography2 Syllabus Course Syllabus –أخلاقيات استخدام الانترنت والقوانين المتعلقة بها –مقدمة في أمن المعلومات مفاهيم ومصطلحات التعمية المتناظرة وغير المتناظرة Cryptography التوقيع الرقمي Digital Signature الشهادات الرقمية Digital Certificate –المشاكل الأمنية في بروتوكولات الانترنت: HTTP, SMTP FTP –برتوكولات الحماية SSL, TLS, HTTPS, PGP واستخداماتها في تطبيقات الوب –أنواع الهجوم على الوب: Cross-Site Request Forgery (CSRF)

3 Cryptography3 Syrian Virtual University MWS/MWT Internet & Web Security S2 Dr. Moutasem Shafa’amry

4 Cryptography4 Syllabus Course Syllabus –أخلاقيات استخدام الانترنت والقوانين المتعلقة بها –مقدمة في أمن المعلومات مفاهيم ومصطلحات التعمية المتناظرة وغير المتناظرة Cryptography التوقيع الرقمي Digital Signature الشهادات الرقمية Digital Certificate –المشاكل الأمنية في بروتوكولات الانترنت: HTTP, SMTP FTP –برتوكولات الحماية SSL, TLS, HTTPS, PGP واستخداماتها في تطبيقات الوب –أنواع الهجوم على الوب: Cross-Site Request Forgery (CSRF)

5 Cryptography5 Secrecy Ciphers Stream & Bloc ciphers Secret Key Cryptography Key Exchange Public Key Cryptography Digital Signatures Applications Security What does it say?

6 Why Cryptography? Network information needs to be communicated through insecure channel. Stored information may be accessed without proper authorization. Cryptography is a systematic way to make that harder. Cryptography6

7 Common Security Requirements C onfidentiality :Secrecy(encryption) I ntegrity (signature/encryption) AVAILABILITY CIA (Conf., Integrity, Availability) –Authenticity(signature/encryption) –Non-repudiation (signature) Cryptography7

8 What Cryptography can do? Encryption: only the authorized party can understand the encrypted message. Signature: allow people to verify the authenticity of the message. Cryptography8

9 Classical Cryptography Shift Cipher –(a special case used by Caesar) Substitution Cipher –Affine Cipher –Vigenere Cipher –Hill Cipher Permutation Cipher Cryptography9

10 Crypto-analysis Ciphertext-only attack Known plaintext attack Chosen plaintext attack Adaptive Chosen plaintext attack Cryptography10

11 Crypto-analysis Shift Cipher: English histogram Substitution Cipher: histogram again Affine Cipher: histogram Vigenere Cipher: more complicated stat Hill Cipher: Known plaintext attack Permutation Cipher: histogram + semantics Cryptography11

12 12 C onfidentiality (Secrecy) Scenario: Alice wants to send a message (plaintext p) to Bob. The communication channel is insecure and can be eavesdropped by Trudy. If Alice and Bob have previously agreed on an encryption scheme (cipher), the message can be sent encrypted (ciphertext c) Issues: What is a good cipher? What is the complexity of encrypting/decrypting? What is the size of the ciphertext, relative to the plaintext? If Alice and Bob have never interacted before, how can they agree on a cipher?

13 13 Traditional Cryptography Ciphers were already studied in ancient times Caesar’s cipher: replace a with d replace b with e... replace z with c A more general monoalphabetic substitution cipher maps each letter to some other letter.

14 14 Breaking Traditional Cryptography Armed with simple statistcal knowledge, Trudy can easily break a mono-alphabetic substitution cipher –most frequent letters in English: e, t, o, a, n, i,... –most frequent digrams: th, in, er, re, an,... –most frequent trigrams: the, ing, and, ion,... The first description of the frequency analysis attack appears in a book written in the 9th century by the Arab philosopher al- Kindi

15 Cryptography15 Example (S. Singh, The Code Book, 1999) Ciphertext PCQ VMJYPD LBYK LYSO KBXBJXWXV BXV ZCJPO EYPD KBXBJYUXJ LBJOO KCPK. CP LBO LBCMKXPV XPV IYJKL PYDBL, QBOP KBO BXV OPVOV LBO LXRO CI SX'XJMI, KBO JCKO XPV EYKKOV LBO DJCMPV ZOICJO BYS, KXUYPD: 'DJOXL EYPD, ICJ X LBCMKXPV XPV CPO PYDBLK Y BXNO ZOOP JOACMPLYPD LC UCM LBO IXZROK CI FXKL XDOK XPV LBO RODOPVK CI XPAYOPL EYPDK. SXU Y SXEO KC ZCRV XK LC AJXNO X IXNCMJ CI UCMJ SXGOKLU?' OFYRCDMO, LXROK IJCS LBO LBCMKXPV XPV CPO PYDBLK Any Guesses???

16 16 Frequency Analysis Identyfying comon letters, digrams and trigrams... PCQ VMJYPD LBYK LYSO KBXBJXWXV BXV ZCJPO EYPD KBXBJYUXJ LBJOO KCPK. CP LBO LBCMKXPV XPV IYJKL PYDBL, QBOP KBO BXV OPVOV LBO LXRO CI SX'XJMI, KBO JCKO XPV EYKKOV LBO DJCMPV ZOICJO BYS, KXUYPD: 'DJOXL EYPD, X LBCMKXPV XPV CPO PYDBLK Y BXNO ZOOP JOACMPLYPD LC UCM LBO IXZROK CI FXKL XDOK XPV LBO RODOPVK CI XPAYOPL EYPDK. SXU Y SXEO KC ZCRV XK LC AJXNO X IXNCMJ CI UCMJ SXGOKLU?' OFYRCDMO, LXROK IJCS LBO LBCMKXPV XPV CPO PYDBLK First guess: LBO is THE

17 17 Frequency Analysis Assuming LBO represents THE we replace L with T, B with H, and O with E and get PCQ VMJYPD TH YK TYSE KHXHJXWXV HXV ZCJPE EYPD KHXHJYUXJ THJEE KCPK. CP THE THCMKXPV XPV IYJKT PYDHT, QHEP KHO HXV EPVEV THE LXRE CI SX'XJMI, KHE JCKE XPV EYKKOV THE DJCMPV ZEICJE HYS, KXUYPD: 'DJEXT EYPD, ICJ X LHCMKXPV XPV CPE PYDHLK Y HXNE ZEEP JEACMPTYPD TC UCM THE IXZREK CI FXKL XDEK XPV THE REDEPVK CI XPAYEPT EYPDK. SXU Y SXEE KC ZCRV XK TC AJXNE X IXNCMJ CI UCMJ SXGEKTU?' EFYRCDME, TXREK IJCS THE LHCMKXPV XPV CPE PYDBTK More guesses…?

18 18 Code Plaintext Now during this time Shahrazad had borne King Shahriyar three sons. On the thousand and first night, when she had ended the tale of Ma'aruf, she rose and kissed the ground before him, saying: 'Great King, for a thousand and one nights I have been recounting to you the fables of past ages and the legends of ancient kings. May I make so bold as to crave a favour of your majesty?’ E pilogue, Tales from the Thousand and One Nights X Z A V O I D B Y G E R S P C F H J K L M N Q T U W A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

19 Frequency of Letter Occurrence Cryptography19

20 20 Secret-Key Ciphers A secret-key cipher uses a key to encrypt and decrypt Caesar’s generalized cypher uses modular addition of each character (viewed as an integer) with the key: c i = p i + k mod m p i = c i -k mod m A more secure scheme is to use modular exponentiation to encrypt blocks of characters (viewed as integers): c [i,j] = p [i,j] k mod m where m is a large prime.

21 21 Secret-Key Ciphers made more secure Unlike modular addition, modular exponentiation is considered computationally infeasible (exponential) to invert. Thus, even if Trudy guesses a pair:(c [i,j],p [i,j] ), (for example, she knows the plaintext starts with the words “Dear Bob”) she still cannot compute the key k. Alice and Bob need to share only key k. Bob decrypts using Euler’s Theorem from number theory: p [i,j] = c [i,j] d mod m where d can be easily computed from k and m using Euclid’s gcd algorithm.

22 Block Cipher Divide input bit stream into n-bit sections, encrypt only that section, no dependency/history between sections Cryptography22 In a good block cipher, each output bit is a function of all n input bits and all k key bits

23 Encryption Mode (ECB) Cryptography23 Electronic Code Book (ECB) mode for block ciphers of a long digital sequence Vulnerable to replay attacks: if an attacker thinks block C 2 corresponds to $ amount, then substitute another C k Attacker can also build a codebook of pairs

24 Encryption Mode (CBC) Cryptography24 Cipher Block Chaining (CBC) mode for block ciphers Inhibits replay attacks and codebook building: identical input plaintext P i =P k won’t result in same output code due to memory-based chaining IV = Initialization Vector – use only once

25 25 How to Establish a Shared Key? What if Alice and Bob have never met and did not agree on a key? The Diffie-Hellman key exchange protocol (1976) allows strangers to establish a secret shared key while communicating over an insecure channel

26 26 The Diffie-Hellman key exchange Alice picks her secret “ half-key” x (a large integer) and two large primes m and g. She sends to Bob: (n, g, g x mod m) Even if Trudy intercepts (n, g, g x mod m), she cannot figure out x because modular logarithms are hard to compute. Bob picks his secret half-key y and sends to Alice: (g y mod m) Again, Trudy cannot figure out y. The shared key is: g xy mod m –Bob computes it as (g x mod m)y mod m –Alice computes it as (g y mod m)x mod m

27 27 Algorithmic Issues (How to do it Fast) How can we efficiently compute modular exponents for large integers? NOTE: It is not efficient to compute q = g x mod m in the obvious way: p = g x q = a mod m

28 28 Repeated Squaring Algorithm represent x in binary: x b-1 x b-2... x 1` x 0 repeat b-1 times g = g 2 mod m This yields p 0 = g mod m p 1 = g 2 mod m p 2 = g 4 mod m … p b-1 = g 2 b-1 mod m for i = 0 to b-1 q = qx i p i mod m The number of arithmetic operations performed is proportional to log x

29 Cryptography29 The Woman-in-the-Middle Attack Trudy can fool Alice and Bob to share a secret key with her How?

30 30 Public Key Ciphers: how to A pair of keys is used (e,d) Key e is made public and is used to encrypt Key d is kept private and is used to decrypt RSA, by Rivest, Shamir, Adleman (1978) is the most popular pubkic key cipher –select a pair of large primes, p and q –let e = pq be the public key –define  (e ) = (p-1)(q-1) –let d be the private key, where 3dmod  (e) = 1 –d is the inverse of 3 mod  (e ) –encrypt x with c = x 3 mod e –decrypt c with x = c d mod e –we have x = x 3d mod e

31 31 Public Key Ciphers: Conclusions RSA is considered secure because the only known way to find d from e is to factor e into p and q, a problem believed to be computationally hard NOTE: The RSA patent expired in September 2000

32 32 Digital Signatures Alice sends a message to Bob encrypting it with Bob’s public key. Bob decrypts the message using his private key. How can Bob determine that the message received was indeed sent by Alice? After all, Trudy also knows Bob’s public key.

33 33 Digital Signatures Alice can provide a digital signature for the message: s = x d mod e If Bob receives both x and s, he computes: –y = s 3 mod e = x d3 mod e = x Thus, if y = x, Bob knows that Alice indeed sent x, since she is the only person who can compute s from x. Also, Alice cannot cheat and deny to have sent message x (nonrepudiation). Using digital signatures, Alice and Bob can authenticate each other and prevent Trudy’s woman-in-the-middle attacks Validating a signed message requires knowledge of the other party’s public key.

34 34 Internet Security Recall that validating a signature requires knowledge of the other party’s public key How do we know other people’s public keys? Certification Authorities (e.g., Verisign) provide certificates that bind identities to public keys A certificate is a pair (id, key) signed by the CA A user needs to know only the public key of the CA

35 35 Internet Security Some secret-key ciphers (triple DES, IDEA, BLOWFISH) are much faster than RSA To communicate securely, a two-phase protocol is adopted: –a shared secret key k is established using RSA –data is transfered between the parties using a secret-key cipher and the shared key k Examples: –SSH (secure shell) for secure host login –SSL (secure socket layer) for secure Web access (https), which uses an additional certification phase

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