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# MOBILE PRGMG WIN 8 CIS 4930

FSU

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This 20 page Class Notes was uploaded by Mrs. Rahul Wuckert on Thursday September 17, 2015. The Class Notes belongs to CIS 4930 at Florida State University taught by Staff in Fall. Since its upload, it has received 98 views. For similar materials see /class/205698/cis-4930-florida-state-university in Comm Sciences and Disorders at Florida State University.

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Date Created: 09/17/15

Modern Ciphers Cryptography after DES Designing a new cipher Competing requirements Code footprint Key agility Efficient softw implementatio Speed Flex bility Security are and hardware n Skipjack Developed by NSA For use with the Clipper chip Caused controversy Secret specification Push for Governmentbased key escrow Being implemented in some constrained environments smack man m m quotM man use dmerem i vamms smack Mes R uund mmmn G Decvypuun mund mquotan Gquot The mmmn F mmmm ms mquot bvxabeaakuv M mmpm N amquot w m macansdemmn sem mg 4pm Imamm mm pm Key scheduhng a gumhm a mwmm mm mm n mm mm mm m mmuancvmaw crwm mm WWW mrmna avv m KEV Increasing key length Product Ciphers and TripleDES Product cipher Consider a cipher E with key length t bits Take two keys k7 and k2 and encrypt a message m by first encrypting it with k7 and then with k2 39 c EuEwm Does the key length increase to 2t imply a corresponding gain in security ManintheMiddle Consider the following strategy to find the keys of a product cipher Obtain some palrs Ph 2 encrypted With the produ c EncK EncK P iff DecK C EncK P Encrypt P7 with all possible keys K Decrypt C7 with all poss ble keys K Select key pairs st D C7 EK C or plaintext and clphertext ct cipher Attack In the example encrypting P1 with K matches decrypting C with the K3 e The pair K1 K3 is a candidate for the double encryption r Thetable has 2 entries no a Security equivalent to u ing a key only one or two bits longeri Timememory tradeoff In order to optimize nding matches Compute A EncK P1 Interpret A as an address location and store the value K there Compute B DecK Ca Interpret B as an address location and store the value K there Ifa key K and a key K try to occupy the same address output K K as a possible key pair Timememory tradeoff 7 Use only part of the block yaiue to address 7 Store only part of the key data TripleDES 3DES is usually implemented as a encryption followed by a decryption with a different key followed by another encryption DESEDE Two and threekey variants to accomplish 112 and 168bit keys DESK1 DES IKZ DESK1 m DESK3 DES IKZ DESK1 m Modes of Operation How to encrypt large messages using block ciphers Large messages In order for encryption to be costef cient a key K should be used to encrypt data of total size much larger than K Typical key sizes are in the range of 80512 bits while blocks are o en 64128 bits Multiple blocks should be encrypted under the same key How to do this ef ciently and securely Dividing a message into chunks Given a message m and blocksize b there should be at least mb blocks to encrypt The last block may not be full ltwill have to be completed with padding bits The padding needs to be uniquely detectable and reversible Example that does NOT work Fill the rest of the block with Us Padding PKCSS Let n be the length ofthe data alter the last full data block 0 s n s b1 where b is blocksize in bitsbytes Fill the rest ofthe block with repeated copies of bn2 the binary representation ofthe number of padding bitsbytes lfn 0 Le ifthe message is an exact mul iple of the blocksize attach a whole new block of b2 s to the end ofthe message Padding data for DES DES has 64bitJBbyte blocksize If he message data is a bytestream take for n the number of bytes ofthe length of data alter last full data block n 0 17 only Use a full byte to encode 8 n239 lfn 3 tan2 000001012 The last block has rst 3 bytes equal to the last three data bytes followed by 5 repetitions of 000001012 lfn o tan2 00001000 and the last block has 8 copies ofthIs Removing the padding For DESPKCSS read last byte B If B does not represent an integer in 1 2 8 report FAILU RE Discard the last B bytes ofthe last block Questions for thinking 7 Why for this padding is it necessary to add a whole padding lockto messages that have iength an exact multiple of the blocksize e is it possipie to design a uniquely decodable paddingtnat oes not need to pad an exactr ttlng message ECB The simplest way to encrypt data using a block cipher is to encrypt each data block separately Electronic Code Book ECB mode Not secure for large messages f plaintext blocks ever repeat their corresponding ciphertext blocks are equal facilitates given and chosenplaintext ciphertext attacks reasonable for small amounts of random data such as an initialization vector used by other data encryption modes other keys etc CBC Cipher Block Chaining mode Requires an IV initialization vector 005 IV 6i1 EPi1 69 Cr PM DCi1 69 Ci 0 encryption ofthe ith plaintext block p E D39 basic ECB encryption decryption CBC properties PM Dci1 69 of Secure for large messages within limits ECB is selfsynchronizing If block 0 is lost during transmission the following block will not decrypt correctly 0 is needed to decrypt CM However c2 will decrypt correctly if CM is received Error propagation rate 19 b139 If jth bit of c is received incorrectly the whole of p decrypts incorrectly as well as the jth bit of pm Later blocks are not affected w mm Cipher Feed Back CFB mode Notation S selects s leftmost bits of data 3 shift left by s discarding s leftmost bits 0 Ems 99 P Hms C9 0 m r 8llcl r0 IV CFB properties pam m m r S0 Secure for large messages within limits CFB decryption reuses ECB encryption for decryption Shorter code CFB errorpropagation 1 gt 1 R lfjth bit of block 0 counting from right is received incorrectly the jth bit of p is corrupted A further s k bits will be corrupted where k is the smallest number such that j 3k 2 R the register buffer size In the worst case 0 and SK R Assuming R is a multiple of s a common case More CFB properties The value 3 can be tuned to eliminate the need for padding For instance s 8 for a data bytestream Smaller amounts of data can be independently encrypted s8 allows for encryption of single bytes Adequate method for streaming data no need to buffer data until blocklength bits of data are available for transmission However thatmpies an added cost one ECB encryption per 5 bits instead of per block CFB is selfsynchronizing m w u l Output Feed Back OFB mode 0 Ems pp p Ems 0 39 m r SllErls r0 IV OFB properties p EmSW n1rlltltSIIErls Secure for large messages within limits OFB decryption reuses ECB encryption for decryption Shorter code OFB errorpropagation 1 gt 1 OFB does not feed the received ciphertext in its register so a wrong ciphertext bit only affects the same bit of the plaintext OFB does NOT selfsynchronize Instead it requires synchronization it is a synchronous mode The register at the receiverwill be permanently ahead of the register at the sender if a ciphertext block is lost in transmission More OFB properties As in CFB OFB can be used with streaming data by tuning the value 8 Also ifs is as large as the smallest data unit it does not require padding Counter mode Encryption of ith I plaintext block His ci Ei Pi r Decryption of ith A plaintext block Di Ei Ci Counter mode properties Error propagation 1 gt 1 Each corrupted bit of ciphertext results in the same bit corrupted after decryption Synchronous requires synchronization Enables precomputation of keystream Fully parallelizable Random access mode Security as good as other modes I Notes of caution An pair key IV should never be used to encrypt more than one message Some modes like counter require only that IV be not reusable Other modes require that IV be unpredictable CBC Encryption does not provide integrity protection This is particularly problematic with OFBcounter Why E uw sh Awme v 5m mm mm E uw sh v edbvamceschnew v as H vs AanmaMNa3vwas an AES mam mam Mummy 3mm ngdevendem s Bans E uw sh Fmste sumqu u a may mmmn E uw sh vuund U quotWWW If mm mm Kay schedme mm m kevsmedme mm mums 7 mum Wampum Gummy mm m mm p 3mm 5 baxes m m mmmmmm mm m emansmn m x m mmm Kay schedme 2 Key schedule algorithm Initialize P S withfrac7 6 XOR P S with cyclically extended key For P S PP2 84254 84 255 Do Replace P P2 by EncPS 0 Replace P3 P by EncPs PPz Replace P7 Pm by EncPS Pl llpl Replace s u s by EI39ICP S P17 P18 Replace 30250502531337 EquotCPSSo252 s6253 Notes The S boxes are read as simple lookup tables For instance if 32 is given the 8bit input which is the binary expansion ofthe integer 127 then 32 27 is returned 521 applications of Blow sh are required to install a new 39 There are 18 Parray entries and 4x256 SBox entries total of1042 entries Each application of Blow sh replaces two ofthese entries Blowfish facts Low keyeaglllty andor nlgn Blowflsh s speed makes lt m man sma es n oodc orceror Blow sh lmpractlcal ln appllcatlonstnat encrypt constr lned erlvlrorlrnerlts ln errnedla e amounts of Small 647m blockslze data such as typlcal of makes tlnsec r appllcatlonstnaten r pt mal rfllellan lel ou otner securlty sultes verSlons Wlth up to 47 rounds mmdummn m prhevs Madman prhenypes C assma mphevs Keyword continued 0 Substitute characters according to the rule easeaerghtjklmnamessavwx ekey asserghtjlmnpqssavxz 0 Encrypting llThe magic words are squeamish ossifrage chemaglcwordsaresqueam shoaslfrage qaohkdbyu nwpknopmSokhbpa ppbrnkdo Monoalphabetic ciphers 0 A monoealphabetic cipher operates by 7 Substituting one charader for another s me substitution is applied irrespective of the character position in the piainte The keyword cipher is an example of a mono alphabetic cipher The Caesar cipher is another exam le apeaerghajktnnapqxssuvwxyz aergha k1quot o q stuvwx zabc Alphabet permutations 0 The English alphabet has 26 characters T e num h er of di erent monoealphabetic ciphers equals the number of different permutations of the alpha et 7 26 I 4 03291461127e26 7 This number is 89 bits long or about 11 bytes 0 Could one assume that if the specifi permutation is not known that such ciphers are 5 Exhaustive search times Table 21 Average Time Required for Exlluuslive Key Smirrh I Inc required I mt umlier 0 lkrunlhe t Time rtquimd m t tncl39nlllunps nun plums11gt Kc Silt um H m e 35 3 muka 1 l5 iiilllhnuuds 5 it in m m itmm tzx 54x mlmm m1 59x inh unr 1w 4 x min 2x mm 64 x lull 3mm 5 4 x 1mm Figure from textbook Stallings Cryptography and Network Security Cryptanalysis Unlike suggested by exhaustive search times mono alphabetic substitution ciphers are not hard to cryptanalyze recover the keyplaintext The reason is that a natural language message contains a lot of structure that is not present in a random string of characters 7 Character frequency 7 Sequence of character frequency English character frequency Figurc 25 Rvkluu Fnquknc n men in Engim Tm Figure from textbook Stallings Cryptography and Network Security Multiletter ciphers 0 Multialetter ciphers work by substituting a group of letters 2 3 or more at a time by another group of letters usually the same length The Playfair cipher uses square diagrams to substitute digrams of the plaintext e The Hill Cipher uses matrix operations to substitute letter sequences n at a time where n is a parameter of the cipher Playfair cipher A Keyword is chosen without repeated characters say we have chosen Cryptoquiz To encrypt split the word into digrams Use ll letter for repeated characters in he same digram say x IMonkey gt MO NK EY Collect gt CO LX LE CT CO encrypts as OA CT encrypts as RC MO encrypts as GZ Hill Cipher 0 Takes n successive letters Each letter of the English alphabet is assigned avalue ma0b 152 y24 225 0 A set of linear equations is used to define the encryption using modular arithmetic 0 Structure 3 characters at a time 7 C K P K P2 K P5 mod26 7 C2 K2 P K2 P2 K15 P5 mod 26 7 C5 K5 P K5 P2 ijj P5 mod26 Example of Hill Cipher 17175 K 211821 2 219 Plaintext paymoremoney First 3 letters of plaintext p a y 15 0 24 First 3 letters of ciphertext l n s 11 13 18 111715170524mod26 1321151802124mod26 18215 20 1924mod26 Hill Cipher Characteristics 0 Using several characters at a time makes it difficult to collect statistics of ciphertext distribution 7 More ciphertext is required 0 There is a large number of keys to chose from 7 All the invertible matrices of dimension n x n and entries in 0 7 2 25 How to decrypt I The receiver knows the key and can solve the system of equations for unknowns P 7 C K P K2 P2 K3 P3 mod 26 7 C2 K2 P K22 P2 K23 P3 mod 26 7 C3 K3 P K32 P2 K33 P3 mod 26 This system has solutions ff the matrix K is invertible Let M be its inverse Then 7 PMCM2 C2M3 C3 mod26 7 P2M2 C M22 C2MZ3 C3 mod 26 7 P3M3 C M32 C2M33 C3 mod 26

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