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Nat Resources in American Hist

by: Myrl Mills

Nat Resources in American Hist FOR 204

Marketplace > Syracuse University > College Of Forest Resources > FOR 204 > Nat Resources in American Hist
Myrl Mills
GPA 3.82


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This 6 page Class Notes was uploaded by Myrl Mills on Wednesday October 21, 2015. The Class Notes belongs to FOR 204 at Syracuse University taught by Staff in Fall. Since its upload, it has received 18 views. For similar materials see /class/226266/for-204-syracuse-university in College Of Forest Resources at Syracuse University.

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Date Created: 10/21/15
introduction to cryptography by konrad LiIntroduction to cryptography ICryptography is the process of encryptingdecrypting data streams using some EMDM methods encryption function ID decryption function M essage C cipher text encrypted Enc ion The root fungiquot AHHJHHJSJ password for HHJSDHSA 2042234 is HE8U3278 jas3jjae DEFDSASA 7 Little history about crypto ISubstitution cipher A substitution cipher is one in which the units of the plaintext usually letters or numbers are replaced with other symbols or groups of symbols The actual order of the uniis of the plaintext is not changed The most known is Caesar cipherand ROT13 Example GREEN FLIES 9 HSFFO GLJFI39 LMore transposition Transposition ciphers rearrange the letters of the plaintext without changing the letters themselves For exam e a very simple transposition cipher is the rail fence in which the plaintext is staggered between two rows and then read off to give the ciphertext In a two row rail fence the message MERCHANT TAYLORS39 SCHOOL becomes M R H N T Y 0 S C 0 L E C A T A L R S H 0 Which is read out as MRHNTYOSCOLECATALRSHO E And more polyalphabetic IPolyalphabetic substitution cipher Block cipher Oligenere cipher t 3 which means each letter is mapped to three positions to its right in the alphabet then to seven positions and t en ten positions M THI SCI PHE RIS CER TAI NLY NOT SEC URE C WOS VJS SOO UPC FLB WHS QSI QVD VLM XYO And more product ciphers It39s a combination of substitution and transposition cipher Ekm m xor k Em m4msm5m1m2ma OxorO00xor111xor011xor1 0 Example K 255m DXFF 111111112 BDODOB Ascn 60m w1111002x0r111111112 110000112 Ox63 10011112 Xor 111111112 101100002 oxao quot And last stream cipher Basically block ciphers with each encryption transformation change for each symbol of plaintext being encrypted Vernam Cipher Ekm mxor k 1 lt i ltt If the key string is randomly chosen and never used again the cipher is called onetime system or onetime pad 7 Symmetric ciphers All of those we just discussed Enigma was one of them polyalphabetic with rotors Problems with them Key distribution 0 and Q must meet in secrecy IWhat ifT wants to communicate too Q must create another key Comes out that for five members we have 10 keys Keys n n12 ISolution Public keys Public keys Public keys crypto is based on the idea that a user can have two keys a private key and a public key Public is used only to encrypt the plaintext while the private is only used for decrypting We will assume that there is no way to derive the private key from the ciphertext encrypted using the public key Unsecured enennei public key C ciphertext E public key Need for authentication Have a third party giving out public keys not good Have a repository called public le This public le would have the public keys for other entities man int 7 RSA generating key RSA RivestShamirAdleman RSA Algorithm 5 a pair of prime numoes so large that factoring is beyond all corrpuung capabiimes Here is algorithm Take two primes p and q take mer product n pq N is called madeMS User chooses a nurrber e we no common factors except 1 and less mm pe1xqe1 Then anome nurrber is found using me atmded Euclidean algorithm edrl mod p71q71 o e is called pubit women d is called private ypu e t The public Key is the par n and e The private key is the pair n and d RSA encrypting 1 The receiver M distribuhes hisher public key 2 The sender F composes the ciphertext c from In c m 3 mod n e and n are M public keys 3 F sends the ciphertext c 4 The receiver M decrypls the message mcdmodn RSA example p5q11n55 The least common multiple of p1q1 is 20 Let e 7 ed 1 mod p1q1 0 d 3 Let m message 2 co c m 3modn27 mod55 To decrypt mcdmodn183mod552 Summary pplied Cryptographyquot The PostScript version of the book is at httpwwwcacrmathuwaterloocahac and ii FREE There is a lot more but needless to say it would require a whole semester


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