{VERSION 5 0 "SUN SPARC SOLARIS" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 2 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 2 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 276 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 258 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE " " 0 261 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 262 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 263 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 256 "" 0 "" {TEXT -1 0 "" }{TEXT 256 0 "" }{TEXT 257 17 "ACCESS 2003 - RSA" }}{PARA 257 "" 0 "" {TEXT -1 0 "" }{TEXT 258 0 "" }{TEXT 259 16 "Thursday June 19" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 260 15 "Our Plan Today:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 592 "To help clarify the RSA \+ public-key encryption method that Jim introduced yesterday, we will do the example worked out in the Tom Davis \"Cryptography\" notes, page \+ 13-14. These are great notes - you can think of them as the Cliff Not es for \"The Code Book.\" Davis' example uses small numbers and a one- letter message, and Maple will do all of our computations. I believe \+ this will make the algorithm more clear to you. The explanation of th e Algorithm on page 6-7 of the Rivest-Shamir-Adleman paper is also a v ery concise outline which will make more and more sense as you play wi th examples." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 589 "After we digest Davis' example we will try somewhat larger pr ime numbers, to help prepare you for the part of your group project in which you send yourselves (and me) encoded messages . In your actua l project, which will be posted later today, you will impliment a medi um-sized version of an RSA cipher system (big enough to send short mes sages, but not really big enough to be secure). You will also incorpo rate the \"secure signature\" feature, which we will discuss today. Y ou can read about this in section 8.2 of Davis' notes, or in section I V of the Rivest-Shamir-Adleman paper. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 287 "We will stop our lab work by 10:10 today so that we have plenty of time to make our way over to JTB 120 \+ for the lecture by Professor Seger at 10:30; we will back here all o f tomorrow morning to tie up loose ends we don't finish today, and to \+ get a good start on your week 1 projects. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 626 "The number theory we've talked about, and the relation to RSA cryptography, is usually taught in ou r number theory course, Math 4400. This is a senior level course, so \+ I would expect that some of what we've talked about has been challengi ng. Still, math majors can take Math 4400 fairly early in their under graduate careers since it does not have very many prerequisites beyond algebra and the ability to reason mathematically. As far as ACCESS g oes, we hope that you're enjoying the magic hidden in modular arithm etic, and that you are pleasantly surprised that this \"abstract\" mat hematics turns out to be so practical." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 238 "I will make you do a lot of your own typing in Part I below, so that you can begin learning MAPLE and comm on errors which users make. Therefore, in the file you download, many of the Maple commands which you see in the hardcopy are gone." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 267 7 "Part I " }}{PARA 258 "" 0 "" {TEXT -1 18 "The Davis Example:" }}{PARA 0 "" 0 " " {TEXT -1 196 "In this example Bob is going to send a message to Alic e. I will follow Davis' numbering of the steps on pages 13-14. We a re also going to use his table on page 9 to convert letters to numbers .\n" }}{PARA 0 "" 0 "" {TEXT -1 5 "1) " }{TEXT 261 5 "Alice" }{TEXT -1 173 " must create her public key, for Bob to use when he encripts h is message to her. So she picks two prime numbers, see page 13: (p=2 3 and q=41). Define p and q using Maple." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 4 "2) " }{TEXT 268 5 "Alice" }{TEXT -1 100 " defines her modulus to be the product of p and q. This will be the \+ first piece of her public key. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "3a) " } {TEXT 269 5 "Alice" }{TEXT -1 1 " " }{TEXT 271 9 "privately" }{TEXT -1 459 " computes the auxillary modulus N2:=(p-1)*(q-1), which is rela ted to Euler's theorem, with which Alice will pick her encoding and de coding powers. No one else will ever see or use this number. First s he finds a number e which is relatively prime to N2; this will be the public encoding power and she will tell it to the world. A good e mu st be relatively prime to (p-1)*(q-1), so that Alice will be able to f ind a decoding power d. So we check the gcd: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "7 ) " }{TEXT 270 5 "Alice" }{TEXT -1 1 " " }{TEXT 280 10 "priva tely " }{TEXT -1 540 "finds her \"secret\" decoding power. Since she \+ does this step sooner than Davis says, we will too. Since e is relati vely prime to (p-1)*(q-1) it has a multiplicative inverse d, mod (p-1) *(q-1). By Euler's Theorem, which Jim talked about yesterday, this d \+ will be the decoding power. I wouldn't be surprised if you're still a mazed and confused by this fact, but like I tell my students in every \+ class, confusion is the first step to understanding. I'll be happy to talk with anyone who wants to understand this part of the math better ...." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 361 "isolve(e*z+y*N2=1); #isolve stands for \"integer solve\", \n #i.e. find integer solu tions. Here the unknowns \n #are z and y, since we already set e= 7 \n #and N2=880 We don't really care about the y-value, \n # we just want z. this is because when we interpret \n #e*z+y*N2=1 \+ in N2 clock arithmetic, N2 is congruent to 0, so\n #so ez=1 mod N2 " }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 197 "d:=-377 mod N2;\n #using mod N2 will get d back \+ in usual residue range.\n #We get DAvis' solution from step 7! Fo r \"fun\" you could\n #try getting this number from the Euclidean \+ algorithm." }}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "3b) Now " }{TEXT 272 5 "Alice" }{TEXT -1 171 " is ready to rece ive messages. She yells from the rooftops: My modulus is N=943. My \+ encryption power is e=7. If you want to send me a message use the enc rypt function:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 232 "encrypt:= x->x^e mod N;\n #if you defined e and N above, \n #then their values will be used for the encrypt function\n #this command is Maplese for encrypt(x)=x^e mod N.\n #So this is the syntax for defining \n #ele mentary functions." }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 " " {TEXT 262 5 "Note:" }{TEXT -1 97 " In class and in the picture note s I made for you we use the letter E for the encrypt function. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "4) The m essage which " }{TEXT 273 3 "Bob" }{TEXT -1 117 " wishes to send Alice is the letter Y. He consults Davis' table on page 9. The number tha t corresponds to Y is 35. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "M:=35;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 3 "5) " }{TEXT 274 3 "Bob" }{TEXT -1 48 " encrypts the message using Alice's public key: " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "C:=encrypt(M); #either of these should w ork\nC:=M^e mod N;\n" }}}{PARA 0 "" 0 "" {TEXT -1 2 "6)" }{TEXT 276 5 " Bob" }{TEXT -1 31 " sends the number 545 to Alice." }}{PARA 0 "" 0 "" {TEXT -1 4 "8) " }{TEXT 275 5 "Alice" }{TEXT -1 88 " decodes the m essage using her decoding power d, which she found in step 7, a while \+ ago." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "decrypt:=y->y^d mod N;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "decrypt(C); #both o f these should work\nC^d mod N;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}} {PARA 0 "" 0 "" {TEXT 263 6 "Alice " }{TEXT 277 97 "consults the table , sees that 35 corresponds to Y, and understands what Bob has sent. W E DID IT!" }{TEXT 278 2 " " }{TEXT 279 146 "Except with Alice's puny \+ primes our message pieces can only be one letter long, so what we've g ot is really no better than a substitution cipher. " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 259 "" 0 "" {TEXT -1 9 "Part II: " }}{PARA 260 "" 0 "" {TEXT -1 22 "A more practical size." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 359 " In your project eve ryone will pick primes bigger than 10^50, so that your moduli will be \+ bigger than 10^(100). This is still not big enough to be secure, but \+ you will be able to send messages with up to 50 characters per packet. (And so that decoding doesn't get too tedious for all groups, you w ill be limited to a total message at most 2 packets.) " }}{PARA 0 "" 0 "" {TEXT -1 239 " For now we will pick primes bigger than 10^6, \+ and use message packets of 6 characters. This means our message packe ts will have up to 12 digits per packet, which will be less than our m odulus N, since N will be greater than 10^12. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 271 "restart: \+ #this will clear all old definitions.\n #It's a good idea to restart when you begin\n #new work - of course you might ne ed to \n #go back and re-enter some old commands that\n \+ #you need again. (Repetition is good pedagogy.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 385 "randomize();#this will tell the \" random\" number generator \n #where to start. the \"seed\" i t generates is based\n #on the system clock, so if you all h it this at\n #the same time you might get the same \"random \" numbers.\n #That would be bad. See help windows to see h ow to\n #make your random numbers unlike your classmates'.\n " }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 90 "rand(); #random number generator,\n \+ #default range is between 0 and 12 digits " }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "bigger:=ran d(1..10^51): #much bigger\nbigger();" }}{PARA 11 "" 1 "" {TEXT -1 0 " " }}}{PARA 0 "" 0 "" {TEXT -1 8 "For now," }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "good:=rand(1..10^7):\ngood();" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 213 "for i from 1 to 100 do\n x1:=good():\n if (x1>10^6 #check number is b ig enough\n and\n isprime(x1)=true) #and check if number is p rime\n then print(x1); #if it is, let's see it\n fi;\nod: \n \+ " }}}{PARA 0 "" 0 "" {TEXT -1 259 "It's unlikely your numbers agree w ith mine. (Well, in truth if you all start the generator at the same \+ place, you'll all get the same so-called random numbers.) But let's a ll use two of mine. We are repeating the process we worked out in the tiny example.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 230 "p:= ; #I got these with my mouse, by \n \+ #highlighting with left mouse,\n #clicking curs or, and pasting with\n #middle mouse (at least on our syst em) \nq:= ;\nN:= ; #our modulus" }}}{PARA 0 "" 0 " " {TEXT -1 227 "To see that a system of this size is not secure, try t he following command. This is the command that would fail if we had c hosen primes of length 200 instead of 12, and that's the reason RSA is secure when you use huge primes." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "ifactor(N);" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 28 "3) Find an encoding power e" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "N2:=(p-1)*(q-1);" }}{PARA 11 "" 1 " " {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 86 "for i from 1 to 10 do\n x2:=good();\n if g cd(x2,N2)=1\n then print(x2);\n fi\nod:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "e:= ;\ngcd( ); #check relative prime" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "7) Get de coding power" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "isolve(e*z + y*N2 =1);\n #find decryption power" }}{PARA 11 "" 1 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "d:= mod N2;\n #use t he mouse to copy and paste big numbers" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT 264 16 "Technical Point:" }{TEXT -1 644 " \+ When we get to step 6, or certainly step 8 Maple will complain when w e try to compute large powers of large numbers, so we have to lead it \+ through this modular computation in smaller steps. The procedure below does an analgous computation to what Jim did yesterday, and what Davi s also outlines, except using powers of 10 rather than powers of 2. I t's fine with me if you just use this procedure, but it's even finer i f you can understand it. The encryption algorithm makes use of the di git procedure which picks off the coefficients of powers of 10 in the \+ decimal expansion of a number. So make sure to load digit before you \+ load encrypt." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "digit:=(x,n )->trunc(x/10^n)-10*trunc(x/10^(n+1));" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "digit(123.56,-1);\ndi git(123.56,2);\ndigit(123.56,0);\n #check how digit picks off the di gits corresponding\n #to powers of 10" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 454 "encrypt:= proc(M1,E,N3) #message, encipher power,mo dulus\n #we assume all M1's, E's have at most 105 digit s\n local i,j, #indices\n L1, #list of succesive 10th p owers of M1\n ans; #answer\n L1[1]:=M1 mod N3;\n for i fr om 2 to 105 do\n L1[i]:=L1[i-1]^10 mod N3;\n od:\n ans:=1: #i nitialize answer\n for j from 1 to 105 do\n ans:=ans*(L1[j]^digit (E,j-1)) mod N3;\n od:\n\nRETURN(ans);\nend:\n \n \+ " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 14 "Let's check!!!" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "M:=123 45678910;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 23 "secret:=encrypt(M,e,N);" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "encrypt(secret,d, N);\n #decryption is just encryption with a different power" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 6 "YES!!!" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 26 "Part III\nAn Actual Message" }}{PARA 263 "" 0 "" {TEXT -1 0 "" }}{PARA 262 "" 0 "" {TEXT -1 0 "" }{TEXT 265 44 "We Use Davis' Table on page 9 to encrypt \" " }{TEXT -1 9 "I'm Dizzy" } {TEXT 266 27 "\". We will need two chunks" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 31 "M1:=196749101445;\nM2:=62626163;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%#M1G\"-X95\\n>" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %#M2G\")jhii" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "C1:=encrypt (M1,e,N);\nC2:=encrypt(M2,e,N);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "encrypt(C1,d,N);\nencrypt(C2,d,N);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}{MARK "46 1" 49 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }