HDTenEightyP/NetNews huggingface dataset:
From: johnl@iecc.cambridge.ma.us (John R. Levine)
Subject: Re: Bias in ACM and IEEE articles on cryptography
Organization: I.E.C.C.
Date: Sat, 15 Aug 1992 18:59:13 GMT

>The ACM articles allowed Rivest, et al., completely free rein to spout off
>with criticism of the NIST DSS, which seemed to boil down to three main
>issues:
>	1) RSA has a bandwagon - you better jump on!
>	2) NSA advised NIST on DSS -- they must have inserted a trap door!
>	3) 512 bits won't work as a maximum key length.

I also read the ACM issue, but my reaction is quite different.  The main
complaint that people had about the proposed NIST scheme is that it had
appeared out of nowhere, with no adequate chance for public review.
Design of cryptographic schemes is very difficult and nobody takes a
scheme seriously until it has withstood attack for quite a while.  One of
the articles pointed out some fairly obvious bugs in the proposed scheme
which makes one wonder whether they'd tested it at all.

The RSA patents are annoying, but RSA does have the advantage that it's
been around for over a decade and has withstood attack.

>In other words, both magazines allowed people with vested *financial* 
>interest in one side of the issue to present their personal views as
>scholarly criticism of the NIST DSS.

In most fast moving technical fields, all the experts have a financial
interest on one side or another.  This doesn't disqualify them from having
useful opinions.  Rivest et al. make no secret of their affiliations, and
even keeping that in mind I found their counterarguments to the NIST
scheme compelling.

Regards,
John Levine, johnl@iecc.cambridge.ma.us, {spdcc|ima|world}!iecc!johnl
Newsgroups: sci.crypt
From: tal@Warren.MENTORG.COM (Tom Limoncelli)
Subject: Re: Secure netnews
Summary: Oh nothing important.  I just have this Perl script that I use to generate this stream of posts whenever someone posts the silly suggestion that netnews could be made secure.
Keywords: silly automated
Organization: Mentor Graphics - IC Group
Date: Wed, 19 Aug 1992 23:15:09 GMT

In <f6@pilot.njin.net> limonce@pilot.njin.net (Tom Limoncelli) writes:

> Of course, that just means that forgers will only need to be a bit
> more careful when crafting their forgeries for them to work.  After
> a few iteration of attempts, they'll have perfected their technique.

At that time, someone will suggest that netnews be reimplemented
from the ground up with security and authenticity as prime
concerns.

Tom
--
There is nothing wrong with talking to yourself...
I do it, and so do I!
-- 
Tom Limoncelli -- tal@warren.mentorg.com (work) -- tal@plts.uucp (play)
"Oh! I thought it was one of those useless demos of everything that
    a GUI builder could do." -Anonymous person watching demo of
         Solaris 2.0's graphical tool for managing NIS+
Newsgroups: sci.crypt
From: graef@rs5.hrz.th-darmstadt.de (Ulrich Graef)
Subject: Re: tabulation of primes
Date: Wed, 12 Aug 1992 15:29:48 GMT
Organization: TU Darmstadt

In article <Bstqr2.Lxq@cs.columbia.edu>, ji@cs.columbia.edu (John Ioannidis) writes:
.
. stuff deleted
.
> OK, so that this is not just a flame, he're your quick proof (is this
> by Euclid?)
.
. proof deleted
.
You must notice, that there exists a small community (even with
mathematicians among them) which does not believe in inductive
proofs. I think, they have a lot of problems with commonly
accepted theorems.

(I do _not_ belong to this group!!!)

Uli


-- 
Ulrich Graef        |  analog (phone) :  +49 6155 62493           
Lichtenbergweg 11   |  digital (Email): graef@hrz.th-darmstadt.de 
D-W-6103 Griesheim  +--------------------------------------------
Germany             |  Nothing is real, look into your mirror!
Newsgroups: sci.crypt
From: pausch@saaf.se (Paul Schlyter)
Subject: Re: Cheap RSA chips? (Was: Re: shift registers??)
Organization: SAAF, Svensk Amat|rAstronomisk F|rening
Distribution: sci
Date: Fri, 21 Aug 92 08:01:53 GMT

In article <1992Aug20.160528.2155@nessie.mcc.ac.uk> segr@nessie.mcc.ac.uk (Simon Read) writes:
>In article <1992Aug19.864.126@ALMAC>, keith.willis@almac.co.uk
>writes:
>
>> You mean that one could implement RSA in hardware for a
>> reasonably low price?  And would such a device still infringe
>> the patents (in the US), or do they only cover software
>> implementations?
>
>Chips that "do RSA" are in fact no more than modular
>multiplication chips (they perform the calculation C=(M**E) mod N,
>for C M E and N very large numbers, typically 512 or 1024 bits).
>
>This can be done quite cheaply. Philips designed but never sold a
>smart card to do it,

Not yet, but Philips will probably sell it in the future.  At least
they are still developing the card (and it does need development,
for instance, it's not yet ISO 7816 compliant, i.e. many smart card
readers cannot handle the Philips RSA card).



-- 
---
Paul Schlyter, Swedish Amateur Astronomer's Society (SAAF)
Nybrogatan 75 A, 6 tr,  S-114 40 Stockholm,  Sweden
InterNet:  pausch@saaf.se
Newsgroups: sci.crypt
From: graef@rs5.hrz.th-darmstadt.de (Ulrich Graef)
Newsgroups: sci.crypt
Subject: Re: tabulation of primes
Date: 12 Aug 92 19:46:03 GMT
Organization: TU Darmstadt

In article <1992Aug12.171656.8755@tessi.com>, george@tessi.com (George Mitchell) writes:
> graef@rs5.hrz.th-darmstadt.de (Ulrich Graef) writes:
> 
> >You must notice, that there exists a small community (even with
> >mathematicians among them) which does not believe in inductive
> >proofs. I think, they have a lot of problems with commonly
> >accepted theorems.
> 
> Whoa!  Euclid's proof does not involve induction!  It is a
> reductio ad absurdum proof, ...

You are right, but I only use the wrong words (english is not
my native language). There is also a group, which worries
about `reductio ad absurdum' proofs.

A friend of mine worry about both!!!

Uli


-- 
Ulrich Graef        |  analog (phone) :  +49 6155 62493           
Lichtenbergweg 11   |  digital (Email): graef@hrz.th-darmstadt.de 
D-W-6103 Griesheim  +--------------------------------------------
Germany             |  Nothing is real, look into your mirror!
From: ray@ole.cdac.com (Ray Berry)
Newsgroups: sci.crypt
Subject: Re: PD public key system
Date: 16 Aug 92 17:41:33 GMT
Organization: Cascade Design Automation

rjones@itchy.dsd.es.com (Ray Jones - Perp) writes:

(description deleted...)

>Any suggestions?  Is it true I can't use RSA?  Is there a scheme that would
>work for the above and that I could also distribute?

	Some time ago Next Inc. announced their intentions to use a scheme termed
"elliptical descent" for public key security.  I'm not qualified to comment
on its applicability or security - perhaps others here can do so.  I don't
recall it being discussed here at the time of the announcement.  My impression
was that it was free of patent liability issues.
-- 
Ray Berry  kb7ht  ray@ole.UUCP  73407.3152@compuserve.com  /* inquire within */
From: Dan Yurman <dyurman@igc.apc.org>
Newsgroups: sci.crypt
Date: 18 Aug 92 06:36 PDT
Subject: Public Key & MS-DOS
Nf-ID: #N:cdp:1465900005:000:480
Nf-From: cdp.UUCP!dyurman    Aug 18 06:36:00 1992


I would appreciate pointers to books which explain how to
understand public key / private key systems.  Also, I would
appreciate pointers to any MS-DOS implementations either as
shareware or commercial products.  As you can tell, I am
new to this field so elementary sources are better. Thanks.

/*---------------------------------------------*/
 * Red Shift is not a brand of chewing tobacco!
 * dyurman@igc.apc.org  43N30 112W01 -7 GMT
 * PO Box 1569 Idaho Falls, ID 83403 USA


Newsgroups: sci.crypt
From: edc323usft@vx24.cc.monash.edu.au
Subject: Determing if a number is Prime
Organization: Computer Centre, Monash University, Australia
Date: Fri, 14 Aug 1992 01:49:42 GMT

Mathmatics not being my strongest field, and time being of concern, I am
in need of an algorithim for determining if a number is indeed prime
(for use in a RSA encryption algorithim).  Also, if anyone has any
advise in handiling 150 - 200 digit numbers in C then I would more than
welcome some advice ...

Travis
From: jhaukioj@sara.cc.utu.fi (Jussi Haukioja)
Newsgroups: sci.logic
Subject: Re: The poor parachutist...
Date: 20 Aug 92 09:28:31 GMT
Organization: University of Turku, Finland


Maybe a group ?.logic.puzzles would be in order?

In article <1992Aug19.090212.1658@gacvx2.gac.edu>, logic@gacvx2.gac.edu writes:
> 
> Now for a riddle of a different flavor..  I first saw this in an interview
> article with Paul Halmos in the Mathematical Intelligencer..  I offer it here
> because it is (1) not as widely known as the riddle above, (2) more difficult
> than the riddle above, and (3) its solution offers some surprises..  Here we
> go.  My wife and I go to a party..  There are 4 other couples there for a total
> of 5 couples.  (No funny business here..  a couple is husband and wife..)  As
> is usual at a social gathering a certain amount of handshaking takes place.. 
> After the handshaking has completed, I interview all of the people at the party
> including my wife.  I ask them how many hands they shook.  (note: spouses don't
> shake hands and no one shakes their own hand)..  Now each person, therefore,
> could have shook as few as 0 hands or as many as 8..  Now, in fact, that range
> of answers is exactly what i get for answers!  I.e., the first person I
> interview responds 0, the second 1, the third 2, ..., the ninth 8..  The riddle
> is "How many hands did my wife shake?"..  I'll even give you the answer..  It
> is 4..  The real riddle, of course, is explaining why this is the correct
> answer..  If "pigeon hole" suddenly occurs to you, you're on the right track..
> Subriddle:  Who is the "meanest" person at this party married to?
> 
> enjoy,
> al hart
> 

First, lets call the nine people you (represented here by U) interviewed by the
letters A-I (A shook 0 hands, B shook 1, ... ,I shook 8 hands).

Now drawing a simple graph helps (actually, the graph could be omitted; it's
just nice to see how things turn out and to be assured that the puzzle is
soluble, that is, that U has not shaken hands with E): we see that the only way
to have the required numbers of handshakes for A-I is:
      Person	    The people with whom s/he shook hands
      	A
	B	    	I
	C		H,I
	D		G,H,I
	E		F,G,H,I
	F		E,G,H,I,U
	G		D,E,F,H,I,U
	H		C,D,E,F,G,I,U
	I		B,C,D,E,F,G,H,U
which shows us:
	U		F,G,H,I

Now we see that I has to be married to A: I has shaken hands with everyone else
(subriddle)
By the same reasoning, H is married to B: H has shaken hands with everyone but A
and B, and A is already married to I
And so on: G&C are married to each other, so U is left with E, who shook 4
hands. Q.E.D.
 
Jussi.

P.S. Sorry for double posting if this happens. My first posting hasn't turned
up, so I sent this again.		
Newsgroups: sci.logic
From: wjcastre@magnus.acs.ohio-state.edu (W.Jose Castrellon G.)
Subject: Re: Non-standard integers.
Organization: The Ohio State University,Math.Dept.(studnt)
Date: Wed, 19 Aug 1992 15:26:30 GMT

In article <1992Aug19.151419.8835@magnus.acs.ohio-state.edu> I write:

>In article <1992Aug19.172922.410@csc.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:

>>
>>...at first it wasn't clear to me that every nonstandard model would have
>>such an  n .  But I suppose this can be seen by taking any odd nonstandard k,
>>and forming  n = k(k-2)(k-4)(k-6).....  .
...
>>
>>Presumably this operation can be made legitimate in some simple way ?

>
>Yes, by Godel coding, but n will necessarily be divisible by some non-standard
>naturals.

Here I should have said n will be equal to the product of k's odd predecessors
in  k's  _row_ and other numbers not in  k's  row.
WJCG
From: solovay@math.berkeley.edu (Robert M. Solovay)
Newsgroups: sci.logic
Subject: Re: Non-standard integers.
Date: 17 Aug 1992 08:57:03 GMT
Organization: U.C. Berkeley Math. Department.
Distribution: world
NNTP-Posting-Host: math.berkeley.edu
Summary: Result quoted is false

In article <1992Aug17.141017.373@csc.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:
>I recall once seeing an article by Kemeny, in which he looked at a nonstandard
>model of the integers, in which there would be two (nonstandard) integers,
>n and n+2, both of which would be divisible by all standard integers. The
>hope was to be able to use this model to decide the twin-prime (or related)
>conjectures.
>
If the integer n is divisible by 4, then n + 2 is not. This fact
[obviously true in the standard model] can be proved in PA, and so
holds in non-standard models as well. 

It's a reasonable question (to which I don't know the answer) what the
result of Kemeny is that wft is misremembering.
>
>Thanks,      Bill Taylor.         wft@math.canterbury.ac.nz

Bob Solovay solovay@math.berkeley.edu
From: maj@waikato.ac.nz
Newsgroups: sci.logic
Subject: Re: Presburger arithmetic axiomatization
Date: 16 Aug 92 17:55:40 +1200
Organization: University of Waikato, Hamilton, New Zealand

In article <1992Aug13.201939.6369@cs.sfu.ca>, jamie@cs.sfu.ca (Jamie Andrews) writes:
>      Where can I expect to find easily a precise and elegant
> axiomatization of Presburger arithmetic?  (I know it's
> "arithmetic without multiplication" and I can think of some
> horrible axiomatizations; I wanted to find a nice one.)
> 
> thanks
> --Jamie.

Should we be saying 'Bratislava arithmetic' now ?!
-- 
Murray A. Jorgensen [ maj@waikato.ac.nz ]    University of Waikato
Department of Mathematics and Statistics     Hamilton, New Zealand      
__________________________________________________________________
'Tis the song of the Jubjub! the proof is complete,
if only I've stated it thrice.'
From: jhaukioj@sara.cc.utu.fi (Jussi Haukioja)
Newsgroups: sci.logic
Subject: Re: The poor parachutist...
Date: 19 Aug 92 08:09:44 GMT
Organization: University of Turku, Finland

In article <1992Aug18.131259.14667@wisipc.weizmann.ac.il>, jhsegal@wiscon.weizmann.ac.il (Livy) writes:
> Ok..Now let's see you,logic gguys resolving this question:
> Let's assume that somewhere(on other planet??) there are 2 cities:a city with
> honest people ,who NEVER lie,and a city with crooks,who NEVER TELL the TRUTH.
> One day per year they are going to shoping to the other city,so they are all
> mixed together in both cities.On the very same crazy day,a poor parachutist
> jumped exactly in one of the two cities.The guy know he is in one of those
> cities,but he doesn't know in which of them.Now THE BIG QUESTION is:What
> question should he ask one of the people he sees,so he could deduce in which
> city he is.Remember:The man he is asking could be a liar or a honest man.HE (the
> parchutist) doesn't know it.
> 
> NOW,let's see your brains working!!
> Livy. (I know the answer!!)
 
Easy. Just ask someone: "Do you live here?". If the answer is "yes", you're in
the honest city, if it's "no", you're in the liars' city. The parachutist won't
know whether he's talking to a liar or not, but he'll know which city he's in.

Jussi.
Newsgroups: sci.logic
From: dabrowsa@zuni.ucs.indiana.edu (Andrew Dabrowski)
Subject: Re: ZFC+~Con(ZFC)
Summary: Nonstandard arithmetic
Organization: UCS, Indiana University, Bloomington, IN
Date: Sun, 16 Aug 1992 01:04:25 GMT

If a model satisfies ~Con(ZFC) then it must actually contain some object
which it thinks is a natural number coding a proof of 1=0, say.  This
object cannot be a standard natural number, or it would mean ZFC really is
inconsistent, so it must be a nonstandard number.  Hence such a structure
models nonstandard arithmetic.  I don't know of any systematic treatment
of models obtained this way, though.

Andrew Dabrowski			Try again.  Fail again.  Fail better.
dabrowsa@ucs.indiana.edu		-Worstward Ho
Indiana U/Bloomington			 Samuel Beckett
From: jhaukioj@sara.cc.utu.fi (Jussi Haukioja)
Newsgroups: sci.logic
Subject: Re: The poor parachutist...
Date: 20 Aug 92 08:56:53 GMT
Organization: University of Turku, Finland

Maybe a group ?.logic.puzzles would be in order?

In article <1992Aug19.090212.1658@gacvx2.gac.edu>, logic@gacvx2.gac.edu writes:
> 
> Now for a riddle of a different flavor..  I first saw this in an interview
> article with Paul Halmos in the Mathematical Intelligencer..  I offer it here
> because it is (1) not as widely known as the riddle above, (2) more difficult
> than the riddle above, and (3) its solution offers some surprises..  Here we
> go.  My wife and I go to a party..  There are 4 other couples there for a total
> of 5 couples.  (No funny business here..  a couple is husband and wife..)  As
> is usual at a social gathering a certain amount of handshaking takes place.. 
> After the handshaking has completed, I interview all of the people at the party
> including my wife.  I ask them how many hands they shook.  (note: spouses don't
> shake hands and no one shakes their own hand)..  Now each person, therefore,
> could have shook as few as 0 hands or as many as 8..  Now, in fact, that range
> of answers is exactly what i get for answers!  I.e., the first person I
> interview responds 0, the second 1, the third 2, ..., the ninth 8..  The riddle
> is "How many hands did my wife shake?"..  I'll even give you the answer..  It
> is 4..  The real riddle, of course, is explaining why this is the correct
> answer..  If "pigeon hole" suddenly occurs to you, you're on the right track..
> Subriddle:  Who is the "meanest" person at this party married to?
> 
> enjoy,
> al hart
> 

First, lets call the nine people you (represented here by U) interviewed by the
letters A-I (A shook 0 hands, B shook 1, ... ,I shook 8 hands).

Now drawing a simple graph helps (actually, the graph could be omitted; it's
just nice to see how things turn out and to be assured that the puzzle is
soluble, that is, that U has not shaken hands with E): we see that the only way
to have the required numbers of handshakes for A-I is:
      Person	    The people with whom s/he shook hands
      	A
	B	    	I
	C		H,I
	D		G,H,I
	E		F,G,H,I
	F		E,G,H,I,U
	G		D,E,F,H,I,U
	H		C,D,E,F,G,I,U
	I		B,C,D,E,F,G,H,U
which shows us:
	U		F,G,H,I

Now we see that I has to be married to A: I has shaken hands with everyone else
(subriddle)
By the same reasoning, H is married to B: H has shaken hands with everyone but A
and B, and A is already married to I
And so on: G&C are married to each other, so U is left with E, who shook 4
hands. Q.E.D.
 
Jussi.
From: mab@wdl39.wdl.loral.com (Mark A Biggar)
Subject: Re: The poor parachutist...
Organization: Loral Western Development Labs
Date: Fri, 21 Aug 1992 17:40:13 GMT

The simplest question to ask someone is "Do you live here?".
If you are in the dsihonest city anyone you ask will say "no", but
if you are in the honest city anyone will say "yes".

--
Mark Biggar
mab@wdl1.wdl.loral.com
From: zeleny@husc9.harvard.edu (Michael Zeleny)
Newsgroups: sci.logic
Subject: Value Judgments
Date: 17 Aug 92 20:23:43 GMT
Article-I.D.: husc3.1992Aug17.162344.14825
Organization: The Phallogocentric Cabal

I am interested in finding some treatments of modal logic of value
judgments, as distinct from decision theory.  I guess it is a subspecies of
deontic logic, but a proper Greek term would be appreciated.  All I've seen
so far is a rather uninspired Russian development by Ivin.

cordially,
mikhail zeleny@husc.harvard.edu
"Un de mes plus grands plaisirs est de jurer Dieu quand je bande."
From: zeleny@husc9.harvard.edu (Michael Zeleny)
Newsgroups: sci.logic
Subject: Re: The poor parachutist...
Date: 21 Aug 92 01:52:58 GMT
Article-I.D.: husc3.1992Aug20.215259.14963
Organization: The Phallogocentric Cabal

In article <1185@kepler1.rentec.com> 
andrew@rentec.com (Andrew Mullhaupt) writes:

>In article <1992Aug18.131259.14667@wisipc.weizmann.ac.il>
> jhsegal@wiscon.weizmann.ac.il (Livy) writes:

L
>>Remember:The man he is asking could be a liar or a honest man.HE (the
>>parchutist) doesn't know it.

AM:
>Usually, you encourage the parachutist to ask "What name would a liar give to
>the other city?". However, I am not convinced that this "works". It makes some
>assumptions about lying which are not clear. For example, it assumes (which is
>reasonable) that the honest people have noticed the "liars" and have come to
>some understanding of what is involved. However, what lying means to the liars
>is difficult to say, especially since they can't agree on it. Since the truthful
>citizens know this as well, they give you an answer which a liar would give you,
>but this is no use unless the liar's answer is well defined.
>
>Right?

I think that your worries could be summarized in an observation that the
above approach will not work in any first-order extensional (i.e. purely
truth-functional) language, regardless of any additional assumptions about
the knowledge of the liars and the truth-tellers concerning each other's
behavior.  At least so it would seem at first blush, unless we can
eliminate the strong conditional by replacing it with some sort of
dispositional idiom.  Would anyone care to try?

>Later,
>Andrew Mullhaupt

cordially,
mikhail zeleny@husc.harvard.edu
"Un de mes plus grands plaisirs est de jurer Dieu quand je bande."

P.S. The Phallogocentric Cabal has concluded its activities for this
summer.  The next meeting will occur during the week of September 18th,
with Tal Kubo speaking on infinite combinatorics.  T-shirts, syllabi, and
other paraphernalia may be requested by responding to this message.
Newsgroups: sci.logic
From: jamie@cs.sfu.ca (Jamie Andrews)
Subject: Presburger arithmetic axiomatization
Organization: CSS, Simon Fraser University, Burnaby, B.C., Canada
Date: Thu, 13 Aug 1992 20:19:39 GMT

     Where can I expect to find easily a precise and elegant
axiomatization of Presburger arithmetic?  (I know it's
"arithmetic without multiplication" and I can think of some
horrible axiomatizations; I wanted to find a nice one.)

thanks
--Jamie.
Newsgroups: sci.logic
Subject: Re: The poor parachutist...
From: jhsegal@wiscon.weizmann.ac.il (Livy)
Date: Wed, 19 Aug 1992 19:43:06 GMT
Organization: Weizmann Institute of Science, Computation Center

You are damn right! It took me ages to think about that....
Livy
Newsgroups: sci.logic
Subject: Re: Non-standard integers.
From: wft@math.canterbury.ac.nz (Bill Taylor)
Date: 18 Aug 92 13:33:42 +1200
Distribution: world
Organization: Department of Mathematics, University of Canterbury

>>model of the integers, in which there would be two (nonstandard) integers,
>>n and n+2, both of which would be divisible by all standard integers. 

>It's a reasonable question (to which I don't know the answer) what the
>result of Kemeny is that wft is misremembering.

Sorreeee; bit of a boo-boo here. Brain damage my only excuse.

Should be:  there is a nonstandard integer n, which is divisible by all
standard integers. Thus  (n+1) and (n-1) are not divisible by any of them,
and are thus "standardly" twin primes. Of course, they may still be
composites in the whole nonstandard system. So this doesn't really help out
on the twin prime conjecture.

However; as I recall, Kemeny observed that (n-1), (n+1) are the *only* possible
primes in the "row" of numbers   .... , n-2 , n-1 , n , n+1 , n+2 , n+3 , ... 
Hence there are no 3-tuples of primes of type (p, p+2 , p+4) in the nonstandard
part of the model, (or any n-tuples with n>2). 

Thus, the n-tuple-primes conjectures of Hardy would all be false. The only
problem with this whole operation is showing that there can be a model with
*all* the "rows" having such an `n' in them. Kemeny reluctantly left this 
loop-hole unplugged.

Has anyone heard of the paper ? Or more precisely, followups to it ?

--------------------------------------------------------------------------
            Bill Taylor              wft@math.canterbury.ac.nz 
--------------------------------------------------------------------------
 Stop press ! I have just scaveged up a copy of Kemeny's paper.

It is "Undecidable problems of elementary number theory",
       Math Annalen, Bd. 135, (1958), pp 160-169.

I would still appreciate hearing of any follow-ups, though !
Newsgroups: sci.logic