December 2003 - cypherpunks-legacy

Re: Save Water... ask me how!
by nobody＠REPLAY.COM 17 Dec '03

17 Dec '03

martian(a)rt66.com wrote: > Ask me how to save water. > http://www.rt66.com/martian Did anyone else go there? What a smell, eh? Whoooeee. I bet that guy saves a *lot* of water. Human-Gus Peter

1 0

SMTP Encryption Extension
by mskala＠langara.csc.UVic.CA 17 Dec '03

17 Dec '03

Lots of people want to encrypt network traffic in large quantities to deter wiretapping, raise awareness, and so on. The usual approach seems to be to encrypt at a very low level, on individual network links. Unfortunately, those plans aren't moving very fast, partly because they require the involvement of people in positions of power at both ends of any given link. If I'm the sysadmin of a major ISP, I've got more important things to worry about than trying to coordinate with my counterpart at the other end of a high-speed link that is ALREADY WORKING FINE, and trying to change the protocol. Well, what are we really concerned about? Not Web traffic - although your web surfing history is private information, the data itself is by nature public and encrypting it is kinda pointless. The same kind of argument applies to news. What we're really concerned about w.r.t. wiretapping is electronic mail. So why not extend SMTP to do strong encryption? When your SMTP mailer connects to another mailer with the encryption extension, they exchange public keys, use those to exchange a session key, and encrypt their traffic. When your smart mailer connects to a stupid mailer, they just speak normal SMTP. This is a win even without any special key management: even if I can't verify that the identity of the remote SMTP server is correct, I'm still protected from passive attacks if I know that only the remote server (whoever that is) can read the traffic. If we're concerned about government surveillance of the populace, passive wiretaps are by far the most significant threat. This has the advantage that it only requires support from the admins of the computers that send and receive email; your ISP doesn't have to cooperate. Also, it doesn't require cooperation between the two ends of a connection: I can put the crypto extension onto my system without worrying about whether any other specific person will cooperate. Finally, it's got good subversive potential; the benefits of having the extension increase exponentially as more people have it. Just imagine if we could get a major Linux distribution to install it by default. Wham! All of a sudden, whenever any two boxes running that distribution connect to each other, they automatically do it securely. The ideal way to implement this would be as a patch to sendmail, but for short term "proof of concept", I'm building a Perl program that will listen on port 25, accept SMTP connections, and pass the data through to a copy of sendmail running on some other port. It also has a client side which gets installed as a "mailer" under sendmail, replacing sendmail's built-in IPC mailer. When two copies of sendmail so equipped connect to each other, they automatically encrypt. I've already got code for a program I call "blackpony", that implements the basic protocol using RSA and Blowfish (extensible to others). It also supports ROTn, in order to provide a cryptosystem people can do in their heads while manually testing the program. I haven't finished testing the integration between it and sendmail, and I'd like to document the protocol more thoroughly, but it does work, and it transparently interoperates with ordinary sendmail. I expect to have it in a form I'll be willing to show people, within the week. If anyone would like a copy or has any comments or suggestions, please let me know. Perl5 is required, as is the Cryptix security library available at http://www.systemics.com/software/. At present the Linux /dev/urandom device (or something like it) is required, but I intend to make that optional eventually. Deploy and enjoy...

1 0

Cato Institute briefing paper on domain names
by Declan McCullagh 17 Dec '03

17 Dec '03

*************** http://www.cato.org/pubs/briefs/bp-033.html INTERNET DOMAIN NAMES Privatization, Competition, and Freedom of Expression by Milton L. Mueller Milton L. Mueller, an associate professor at the Syracuse University School of Information Studies, is the author of many scholarly works on telecommunications policy. Executive Summary There is growing confusion over the administration of Internet top-level domain names (TLDs), the system of suffixes, such as .com, .org, and .edu, that determines a person's e-mail or Web site address on the Internet. We need to define rules and procedures that will permit and encourage competition among administrators of TLDs in response to market demand. Freedom of expression should be a primary concern. Proposals for compulsory national TLDs should be rejected. National TLDs would undermine the international character of the Internet and encourage national governments to enact myriad petty regulations and restrictions on free speech. Domain names should not be equated with trademarks or brand names. We should reject attempts to forge inappropriate links between domain name registration and trademark protection. The U.S. government should encourage the development of property rights and competition by moving the administration of Internet domain names into the private sector. ### ------------------------- Declan McCullagh Time Inc. The Netly News Network Washington Correspondent http://netlynews.com/

1 0

DNA encrypted messages (fwd)
by Jim Choate 17 Dec '03

17 Dec '03

Forwarded message: >From owner-traveller(a)Phaser.ShowCase.MPGN.COM Mon Oct 20 18:05:46 1997 Message-Id: <199710202034.WAA19474(a)vip.cybercity.dk> From: "Mark Seemann" <dko3835(a)vip.cybercity.dk> To: <traveller(a)MPGN.COM> Subject: DNA encrypted messages Date: Mon, 20 Oct 1997 22:28:39 +0200 X-MSMail-Priority: Normal X-Priority: 3 X-Mailer: Microsoft Internet Mail 4.70.1155 MIME-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 7bit Sender: owner-traveller(a)Phaser.ShowCase.MPGN.COM Reply-To: traveller(a)MPGN.COM Tue, 21 Oct 1997 00:08:58 +1300 Andrew Moffatt-Vallance wrote: > To crib a plot line from ST:TNG (who cribed it from Niven I think). Assuming > that the aliens have a reasonable biotech, why not encode it in the DNA > sequences of some local life forms. Sharks would be a good bet, or jellyfish. > Remember something like 90% of any DNA sequence is just waste space. I actually used this idea maybe ten adventures back in my current campaign, although I got the idea myself and thought it was pretty original. Well, that just goes to show... The setup was a little different. A psionic noble (yes, I'm playing off the Psionic Knights campaign) was persued during the psionic suppressions, and encoded the whereabouts of his (psionic) noble's order's tresuary in plasmid rings in his blood, leaving behind him a subtle clue for friendly successors to pick up. The player characters found his buried body, dug it up and analyzed what was left of his deteriorated blood. The party's (NPC) doctor found the following obviously artificial encoding as part of more 'natural' DNA: AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAACCCGGGTTTAAACCCGGGTTTAAACCCGGGTTTAAACCCGGGTTT AAACCCGGGTTTAAACCCGGGTTTAAACCCGGGTTTAAACCCGGGTTT AAACCCGGGTTTAAACCCGGGTTTAAACCCGGGTTTAAACCCGGGTTT AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT AAAAACAAGAATACAACCACGACTAGAAGCAGGAGTATAATCATGATT CAACACCAGCATCCACCCCCGCCTCGACGCCGGCGTCTACTCCTGCTT GAAGACGAGGATGCAGCCGCGAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ATAAGGCGTCTAATGGCAAAGAAGAAGACAGCACTACGCCAGCCTGAA AGGCGTATCGCACCGAGGCGTATACACATGCTAGCCGCAGAAAGGCGT CGTAGGCCTCTCGCAAGGATAATACGACGTATCCAGCCTCAAGCACTC CGAGCACAGCCGCGCATGCGTCAGAGGCCCGCAATGATCCAGATACTC GCAAGCACTGCCGCACGCAGGCTGCCCAGGCTAGCACGAATTATTCAG ATACAGAGGCCCGCACGTATGATACGACAACCTCAGCTCCAGCGACCT GCACGAATTGCACTCCACATGGCACGACGTATCATGCGTGCACGAATT GCACTCCACATGGCAGAACACCAGCTCATGGCACTACTCAGGCGTGCC GCAAGGCTCGCACTCCACATGGCAATCATGAGGATCGCACCGAGGCCT CTAGCAATTCAGCCTCAAATGCGTGCCGCAAGTCCCAGGATACCAGCA AGTCGAGACGCACTCCACCGTCGAGAACCTGCACAGCCTGCACTCCAC ATGGCAGAAATGCCCCCCGCCAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA I explained to the players that 'A' was for Adenine, 'C' for Cytocine, 'G' for Guanine, and 'T' for Tymine (the for possible DNA bases) and let them loose without any further help at all. They actually broke the code, and it was very satisfying for both them and me. Here's a little challenge for you all: Can any of you guys break the code? Mark Seemann mark(a)dk-online.dk http://www2.dk-online.dk/users/mark_seemann

1 0

Re: SMTP Encryption Extension
by Ariel Glenn 17 Dec '03

17 Dec '03

>Well, what are we really concerned about? Not Web traffic - although your >web surfing history is private information, the data itself is by nature >public and encrypting it is kinda pointless. Not at all! Most of my time is spent trying to keep 'restricted data' from being scooped by the wrong people... from the Web. We have licensed data, students' private data, surveys and voting (the results of which should be available only in certain formats and then only to certain people), etc. etc... So data kept on the filesystem has to be protected and it has to be encrypted when it is sent out to the browsing (legitimate) user. Nice try... Ariel Glenn ariel(a)columbia.edu

2 1

Re: A Beehive in their Bonnet, from The Netly News
by Jim Ray 17 Dec '03

17 Dec '03

-----BEGIN PGP SIGNED MESSAGE----- At 12:41 PM 10/20/97 -0400, Declan McCullagh wrote: ... >A Beehive in Their Bonnet >by Declan McCullagh (declan(a)well.com) > > John Caldeira's remark in a beekeeper's discussion > area on Prodigy seemed innocuous enough: It > described how to rid honeybees of parasites. "Two > ounces of Mavrik mixed with one gallon of water and > sprayed on the bottom boards of one's hives several > times a year. It is reportedly very effective," > Caldeira wrote. > > Honey may attract flies, but this discussion about its > production drew swarms of bureaucrats. The Texas > Department of Agriculture charged Caldeira with > violating a state law banning unapproved discussions > of bee medicines and fined the 42-year-old hobbyist > $600. A draft statement from the department > complains that Caldeira was "criticizing > regulations" and improbably argues that "the honey > market in the U.S. could become depressed because > of widespread concerns about its safety and due to > Mr. Caldeira's suggestion that commercial beekeepers > are using Mavrik in their hives." Effectively conceding that Mr. Caldeira has more reputation- capital than they do. Gee, I wonder what an enterprising bee- keeper would need to do to keep these bureaucrats at bay? ;^) Suppose that instead of "Mavrik" the solution were instead the regular burning of hemp byproducts nearby? If nothing else, this would certainly lead to docile bees... JMR -----BEGIN PGP SIGNATURE----- Version: 2.6.2 Comment: Freedom isn't Freeh. iQEPAwUBNEvklzUhsGSn1j2pAQGqqQfPV4xHkB2JIySmL7mdcNSbu9D79EVPeGPu lk448L9pQw6UgDQy6cIn6+xY4/0zwEGItf0gEPwkT7upTBKpoOJEIk7zlJPgw0H1 7WZZDARj1scqM7YVIIq5ce1c/3jMUfqTZ3HcXhxJ9rPtHmU4+PW5Y6uJisSE2fKY f3JrGhcfOfkdhGPSBNZTmtfwnBZRORXbv1aM+ibqgLcipWr7SrfyrLoE8VQr4pEd BJ7aBgoQK0cNw4f9vLcETTrPCrEsJqbRoFfl97apehp1zmSoKYn1AeriRK4qE0mD v7dY2sclypzpH59wF45mddMUO7uYdaZ7EzxDV3ll6qqEpA== =krnc -----END PGP SIGNATURE-----

1 0

Re: A Math problem related to cryptography
by phelix＠vallnet.com 17 Dec '03

17 Dec '03

On 19 Oct 1997 18:36:26 -0500, Jeffery Walker <jeff(a)beol.net> wrote: > > I have been working on some math and C++ code to implement it. The >problem relates to a possible new means of factoring. However, I have run >into a problem which I cannot solve. Indeed it maybe impossible. The >problem is this: > > Find and equation using only multiplication, division, GCD, LCM and >powers (not addition or subtraction) in terms of a and b which is equal to >a+b for all Integers. If so desired it can be assumed that a and b are >relative primes with a>b. Note that equivalent problems include but aren't >limited to finding a similar equation for a+1. If other functions such as >logs are required they may be allowed but I would have to look into it. > > To encourage people to answer I am offering $20 for a solution and >$10 for a proof that it is impossible. I will judge all answers as to the >validity both in terms of actually being correct (or in the case of a proof >a valid argument) and in that they do not use functions such as addition or >subtraction which are not allowed. > (why do I get the feeling that I'm doing somebody's homework assignment) Claim: In the realm of Integers, There does not exist a formula using only multplication,division,GCD,LCM,and powers such that f(a,b)=a+b for all integers a,b Proof (by contradiction): I assume that a,b are primes > 2. Lets also define x,y,z,w,m,n and any integer >=0 and C,D,E as any integer > 0. Now let's look at the operations available to us: multiplication and powers: C=C*(a^0)*(b^0) CD=E(a^0)*(b^0) where E=CD a=1*(a^1)*(b^0) a^2=1(a^2)*(b^0) Cab^2=C*(a^1)*(b^2) well , all results can be expressed in the following form: C*(a^x)*(b*y) division: a !| b and b !| a (here !| means "does not divide") (From here on out, lets assume that a !| C,D,E and b !| C,D,E. it makes things simpler) Ca/a=C*(a^0)*(b*0) Ca/D=E*(a^1)*(b^0) assuming that E | C I won't go through all this. It should be obvious that for all allowed divisions, the result can be expressed as C*(a^x)*(b*y) GCD: gcd(a,b)=1 gcd(a,C)=1 gcd(Ca,Da)=max(a,gcd(C,D)) gcd(Ca,Db)=gcd(C,D) gcd(C(a^x),C(a^y))=C(a^(min(x,y))) gcd(C(a^x)*(b^y),D(a^z)*(b^w))=gcd(C,D)*(a^(min(x,y)))*(b^(min(z,w)))) =E(a^m)*(b^n) for E=gcd(C,D) m=min(x,z) and n=min(y,w) so, all results of GCD can be expressed in the form: C(a^x)(b^y) LCM: LCM(C,D)=CD/gcd(C,D) LCM(C(a^x)*(b^y),D(a^z)*(b^w)) =CD*(a^(x+z))*(b^(y+w))/gcd(C(a^x)*(b^y),D(a^z)*(b^w)) =( CD/gcd(C,D) * (a^(|x+z-min(xz)|)) * (b^(y+w-min(yw)) = E(a^m)*(b^n) where E=CD/gcd(C,D)=lcm(C,D), m=x+z-min(x,z) , n=y+w-min(y,w) so, all results of LCM can be expressed in the form: C(a^x)(b^y) Now, as you can probably tell by now, I don't remember a damn thing about groups, rings, and fields. If i did, I could have done all of the above in one paragraph. SO, what I'm trying to say is that all the numbers you can operate on here will be of the form C(a^x)(b^y). SO you really have to prove at least the following C(a^x)(b^y)=a+b for some C>0, x,y>=0 integers Now, you could probably say that C=a+b, x,y=0, but that's a cop out and is probably not allowed. Now, watch while I cop out and provide an easy counterexample: assume some formula exists such that f(a,b)=a+b with f only using *,/,gcd,lcm,^ operators let a=5, b=7 Since I know that ultimately anything I do with constants and the operators 8,/,gcd,lcm,^ will result in something in the form of C(a^x)(b^y), I won't even mess with the formula there must exist some C>0, x,y>=0 that makes: C(5^x)(7^y)=5+7=12 of x>=1 and y>=1 then C(5^x)(7^y)>12 for all C>0 if x=0 and y=0, C=12 (this works) if x=1,y=0 then 5C=12. there is no C which satisfies this. if x=0,y=1 then 7C=12. there is no C which satisfies this. so, the only solution is C=12, or more generally, C=a+b oops, that can't be the formula since we're not allowed to do that. QED. Well, that was pretty sloppy, so lets be more general about this: C(a^x)(b^y)=a+b Now, I'm sure that somewhere in the history of mathematics somebody has proven that for primes a and b where a,b>2 that a*b>a+b. So, for any x,y,C>0, C(a^x)(b^y)>a+b. C, of course, cannot be negative, or zero, so all were left with is x or y being zero. Let y=0. we have: C*a^x=a+b C*a^x - a = b a*(C*a^(x-1) - 1 ) = b a*E=b Now, a and b are primes and a!=b. From the fundamental theorem of algebra, every number can be represented by a series of primes such as: p1^a1 * p2^a2 * p3^a3 * ... In a*E=b, if we factor each side down to its primes, we find that the right hand side does not contain the prime 'a' which is on the left hand side, so a*E != b. A similar argument can be used for the case x=0. This leaves the only solution to C(a^x)(b^y)=a+b being x=0,y=0 C(a^0)(b^0)=C=a+b which isn't allowed. QED Now, If you're like me, you've probably read the above and don't have a f*cking clue what I meant (I sure wouldn't if I read all that crap). So here's the executive summary. With the operators you've defined on the number set you want to work with, no matter how complicated or crazy your formula is, I can express the result at C(a^x)(b^y). And hopefully, I've shown that C(a^x)(b^y) cannot equal a+b except in the trivial case of C=a+b (with x,y=0) which is what you don't want. >Thanks, >Jeff >P.S. I am very interested in joining the coderpunks list as I do program >and am interested in cryptography as related to programming. If someone >would be kind enough to recommend me I thank you. > I wasn't aware that you had to be invited to the coderpunks list. Just send a message to majordomo(a)toad.com with the following in the body: subscribe coderpunks -- Phelix, who really should have been able to give a better proof (especially since one of my degrees has the word 'Mathematics' on it)

2 1

Re: A Math problem related to cryptography
by phelix＠vallnet.com 17 Dec '03

17 Dec '03

On 19 Oct 1997 18:36:26 -0500, Jeffery Walker <jeff(a)beol.net> wrote: > > I have been working on some math and C++ code to implement it. The >problem relates to a possible new means of factoring. However, I have run >into a problem which I cannot solve. Indeed it maybe impossible. The >problem is this: > > Find and equation using only multiplication, division, GCD, LCM and >powers (not addition or subtraction) in terms of a and b which is equal to >a+b for all Integers. If so desired it can be assumed that a and b are >relative primes with a>b. Note that equivalent problems include but aren't >limited to finding a similar equation for a+1. If other functions such as >logs are required they may be allowed but I would have to look into it. > > To encourage people to answer I am offering $20 for a solution and >$10 for a proof that it is impossible. I will judge all answers as to the >validity both in terms of actually being correct (or in the case of a proof >a valid argument) and in that they do not use functions such as addition or >subtraction which are not allowed. > (why do I get the feeling that I'm doing somebody's homework assignment) Claim: In the realm of Integers, There does not exist a formula using only multplication,division,GCD,LCM,and powers such that f(a,b)=a+b for all integers a,b Proof (by contradiction): I assume that a,b are primes > 2. Lets also define x,y,z,w,m,n and any integer >=0 and C,D,E as any integer > 0. Now let's look at the operations available to us: multiplication and powers: C=C*(a^0)*(b^0) CD=E(a^0)*(b^0) where E=CD a=1*(a^1)*(b^0) a^2=1(a^2)*(b^0) Cab^2=C*(a^1)*(b^2) well , all results can be expressed in the following form: C*(a^x)*(b*y) division: a !| b and b !| a (here !| means "does not divide") (From here on out, lets assume that a !| C,D,E and b !| C,D,E. it makes things simpler) Ca/a=C*(a^0)*(b*0) Ca/D=E*(a^1)*(b^0) assuming that E | C I won't go through all this. It should be obvious that for all allowed divisions, the result can be expressed as C*(a^x)*(b*y) GCD: gcd(a,b)=1 gcd(a,C)=1 gcd(Ca,Da)=max(a,gcd(C,D)) gcd(Ca,Db)=gcd(C,D) gcd(C(a^x),C(a^y))=C(a^(min(x,y))) gcd(C(a^x)*(b^y),D(a^z)*(b^w))=gcd(C,D)*(a^(min(x,y)))*(b^(min(z,w)))) =E(a^m)*(b^n) for E=gcd(C,D) m=min(x,z) and n=min(y,w) so, all results of GCD can be expressed in the form: C(a^x)(b^y) LCM: LCM(C,D)=CD/gcd(C,D) LCM(C(a^x)*(b^y),D(a^z)*(b^w)) =CD*(a^(x+z))*(b^(y+w))/gcd(C(a^x)*(b^y),D(a^z)*(b^w)) =( CD/gcd(C,D) * (a^(|x+z-min(xz)|)) * (b^(y+w-min(yw)) = E(a^m)*(b^n) where E=CD/gcd(C,D)=lcm(C,D), m=x+z-min(x,z) , n=y+w-min(y,w) so, all results of LCM can be expressed in the form: C(a^x)(b^y) Now, as you can probably tell by now, I don't remember a damn thing about groups, rings, and fields. If i did, I could have done all of the above in one paragraph. SO, what I'm trying to say is that all the numbers you can operate on here will be of the form C(a^x)(b^y). SO you really have to prove at least the following C(a^x)(b^y)=a+b for some C>0, x,y>=0 integers Now, you could probably say that C=a+b, x,y=0, but that's a cop out and is probably not allowed. Now, watch while I cop out and provide an easy counterexample: assume some formula exists such that f(a,b)=a+b with f only using *,/,gcd,lcm,^ operators let a=5, b=7 Since I know that ultimately anything I do with constants and the operators 8,/,gcd,lcm,^ will result in something in the form of C(a^x)(b^y), I won't even mess with the formula there must exist some C>0, x,y>=0 that makes: C(5^x)(7^y)=5+7=12 of x>=1 and y>=1 then C(5^x)(7^y)>12 for all C>0 if x=0 and y=0, C=12 (this works) if x=1,y=0 then 5C=12. there is no C which satisfies this. if x=0,y=1 then 7C=12. there is no C which satisfies this. so, the only solution is C=12, or more generally, C=a+b oops, that can't be the formula since we're not allowed to do that. QED. Well, that was pretty sloppy, so lets be more general about this: C(a^x)(b^y)=a+b Now, I'm sure that somewhere in the history of mathematics somebody has proven that for primes a and b where a,b>2 that a*b>a+b. So, for any x,y,C>0, C(a^x)(b^y)>a+b. C, of course, cannot be negative, or zero, so all were left with is x or y being zero. Let y=0. we have: C*a^x=a+b C*a^x - a = b a*(C*a^(x-1) - 1 ) = b a*E=b Now, a and b are primes and a!=b. From the fundamental theorem of algebra, every number can be represented by a series of primes such as: p1^a1 * p2^a2 * p3^a3 * ... In a*E=b, if we factor each side down to its primes, we find that the right hand side does not contain the prime 'a' which is on the left hand side, so a*E != b. A similar argument can be used for the case x=0. This leaves the only solution to C(a^x)(b^y)=a+b being x=0,y=0 C(a^0)(b^0)=C=a+b which isn't allowed. QED Now, If you're like me, you've probably read the above and don't have a f*cking clue what I meant (I sure wouldn't if I read all that crap). So here's the executive summary. With the operators you've defined on the number set you want to work with, no matter how complicated or crazy your formula is, I can express the result at C(a^x)(b^y). And hopefully, I've shown that C(a^x)(b^y) cannot equal a+b except in the trivial case of C=a+b (with x,y=0) which is what you don't want. >Thanks, >Jeff >P.S. I am very interested in joining the coderpunks list as I do program >and am interested in cryptography as related to programming. If someone >would be kind enough to recommend me I thank you. > I wasn't aware that you had to be invited to the coderpunks list. Just send a message to majordomo(a)toad.com with the following in the body: subscribe coderpunks -- Phelix, who really should have been able to give a better proof (especially since one of my degrees has the word 'Mathematics' on it)

1 0

RC5 encryption
by vcarlos35＠juno.com 17 Dec '03

17 Dec '03

Forgive this seemingly naive question, but.... Where does RC5 derive its security from? I know about the ROTL stuff and the key expansion methods (I have the RSA Cryptobytes, HAC, AC, etc.) However, I personally can't see how it would work. I know its supposedly secure because of the data-dependent rotations, and other features. Correct me if I'm wrong, but a ROTL assembler command shifts all bits to the LEFT one and moves the "overflow" bits back into the right end of the register: 0100111011 to 0011101101 if it rotates left by 2 How does this make it hard to decrypt? The round function seems MUCH weaker than IDEA's, CAST's, or even GOST and Blowfish. Vincent Carlos

1 0

Re: consensus on pgp? can we consolidate for action?
by nobody＠REPLAY.COM 17 Dec '03

17 Dec '03

Lucky Green wrote: > I encrypt both my SMTP and POP links using SafePassage Secure Tunnel. > 128 bit SSL v3 with client certs. You can download a free eval copy of > SPST at http://www.c2.net/products/spst/ Without the certs, it shouldn't > take longer than 15 min. to set this up at your site. > > Disclaimer: I work for C2. Will these constant lies from C2Nut employees never stop? I've been at C2 for three months, now, and I haven't seen Lucky do any work, yet. CorporateSpyMonger

1 0