Cryptography-Digest Digest #370, Volume #10 Wed, 6 Oct 99 21:13:03 EDT
Contents:
Re: rc5-128 cracking $20 per letter ("A Poster")
Re: Does anyone have more information? (John Savard)
RC-5 breaking, $19 per letter (Anonymous)
Re: radioactive random number generator (jjlarkin)
Re: Help: Mobility of the Private Key within PKI (David P Jablon)
Re: classifying algorithms (Doug Stell)
Re: rc5-128 cracking $20 per letter ("Steven Alexander")
----------------------------------------------------------------------------
From: "A Poster" <[EMAIL PROTECTED]>
Subject: Re: rc5-128 cracking $20 per letter
Date: Wed, 6 Oct 1999 19:57:39 -0400
I found these in alt.security.pgp and thought that they might be of interest to this
newsgroup...
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From: "John Croll" <[EMAIL PROTECTED]>
Newsgroups: alt.security.pgp
Subject: RC5-32/12/8 IS DEAD!
Date: Thu, 23 Sep 1999 14:08:28 -0500
Organization: AT&T WorldNet Services
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Xref: tor-nx1.netcom.ca alt.security.pgp:113990
challenge: RC5-32/12/8
solution: B7 5C 0F 56 B9 8C 6A EF D1 5D 3E 56 A1 DF 39 B1
name: richard lee king jr.
address: r l king
p.o.box 236
st. bernice, in.
email: [EMAIL PROTECTED]
phone: 765-832-2916
time: 2.5 days.
method: trivial solution based on ignoring the algorithym and just
sequencing
from 32 to 90 and simultaneously doing and, or, xor, mod, to see if
coincidences showed up. i jotted down suspects which were not many.
then i descrambled the message using my hash table. once the message was
clear i applied the clear text against the cypher text to get the sub keys.
then i anded them 4 at a time. i may have a couple in the wrong order. i
have been awake a long time. please let me know if i failed and i will try
again.
secret message::
RC5-32/12/8 RAPI
D GROWTH IN FAST
COMPUTERS HAS M
ADE THIS OLD HAT
done:
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From: "John Croll" <[EMAIL PROTECTED]>
Newsgroups: alt.security.pgp
Subject: Re: RC5-32/12/8 IS DEAD!
Date: Fri, 24 Sep 1999 13:51:48 -0500
Organization: AT&T WorldNet Services
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Xref: tor-nx1.netcom.ca alt.security.pgp:114095
Contest identifier: RC5-32/12/16
Cipher: RC5-32/12/16 (RC5 with 32-bit wordsize, 12 rounds, and 16*8=128-bit
key)
Start of contest: 28 January 1997, 9 am PST
State of contest: ongoing
IV: a7 8b 00 e8 15 e6 2f 5d
Hexadecimal ciphertext:
i can't make keys but i have demonstrated that i
can decipher any rc5 message regardless of the
length of the key.
there is a trivial solution. it took me 4.5 hours to
crack this code. if i could program then it could
be cracked in a second. rsa is totally ignoring me.
rc5 is dead.
secret message:
rc5-32/12/16
128 bit keys make
finding it hard
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From: "John Croll" <[EMAIL PROTECTED]>
Newsgroups: alt.security.pgp
Subject: rc5-128 trivial solution
Date: Fri, 24 Sep 1999 15:08:53 -0500
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Xref: tor-nx1.netcom.ca alt.security.pgp:114103
Contest identifier: RC5-32/12/16
Cipher: RC5-32/12/16 (RC5 with 32-bit wordsize, 12 rounds, and 16*8=128-bit
key)
Start of contest: 28 January 1997, 9 am PST
State of contest: ongoing
IV: a7 8b 00 e8 15 e6 2f 5d
Hexadecimal ciphertext:
secret message:
rc5-32/12/16
128 bit keys make
finding it hard
done
i can't make keys but i have demonstrated that i
can decipher at least part of any any rc5 message
regardless of the length of the key.
there is a trivial solution. it took me 4.5 hours to
crack this code. if i could program then it could
be cracked in a second. rsa is totally ignoring me.
enclosed is my cpp source for examining the relation
ship of the cypher characters to potential clear characters.
i found that if i input the correct clear character that
the out put looks different. please test my algorythim
and tell me where i am right and where i am wrong.
my code is stupid but if you examine it you see what
it does. it really does work.
i used microsoft c++.
ps.
my theory is based on the idea that each cypher character
can only represent 4 or 5 letters of plain text. after i find the
possibilities then i look at which combinations make sense.
this makes it possible to decypher normal language messages.
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <iostream.h>
#include <conio.h>
#include <ios.h>
void main()
{
void WaitKey(char ASCIIcode);
unsigned int n;
n=32+36;
unsigned int m;
m=95;
unsigned int o;
o=0;
int b[17];
char a;
a=1;
char f;
f=n;
printf( "ASCII %1c \n", f );
char *h[1];
h[1]="5";
f=n;
printf( "ASCII %1d \n", h );
for( int i = 1; i < 255; ++i )
{
f=n;
printf( "ASCII %1c \n", f );
printf( " PLAIN Dec %d tHex: %Xh \n", n , n );
// printf( " SUBTRACT Dec %d tHex: %Xh \n", o, o );
o= _rotl(m,n);
printf( " ROTL Dec %d tHex: %Xh \n", o, o );
o= _rotl(n,m);
printf( " RROTL Dec %d tHex: %Xh \n", o, o );
o= _rotr(m,n);
printf( " ROTR Dec %d tHex: %Xh \n", o, o );
o= _rotr(n,m);
printf( " ROTR Dec %d tHex: %Xh \n", o, o );
o= m & n;
printf( " AND Dec %d tHex: %Xh \n", o, o );
o= n & m;
printf( " AND Dec %d tHex: %Xh \n", o, o );
o= m ^ n;
printf( " XOR Dec %d tHex: %Xh \n", o, o );
o= n ^ m;
printf( " XOR Dec %d tHex: %Xh \n", o, o );
o= m % n;
printf( " MODULA Dec %d tHex: %Xh \n", o, o);
o= n % m;
printf( " MODULA Dec %d tHex: %Xh \n", o, o);
o= m | n;
printf( " OR Dec %d tHex: %Xh \n", o, o);
o= n | m;
printf( " OR Dec %d tHex: %Xh \n", o, o);
// n ++;
b[1]=139^120^18^95;
b[2]=34^184^74^250
b[3]=221 116 55 66 67 71 16 46;
b[5]=164 187 165 117 118 112 89 69;
b[7]=81 193 61 81 27 148 85;
b[9]=64 227 241 251 113 113 191 191;
b[11]=13 42 3 155 108 232 47 27;
printf( "key %Xh %Xh %Xh %Xh \n", b[1], b[3], b[5], b[7]);
printf( "key %Xh %Xh %Xh %Xh \n", b[9], b[11], b[13], b[15]);
WaitKey(0);
}
}
void WaitKey( char ASCIICode )
{
char chTemp;
chTemp = getchar();
while( chTemp != 32 )
{
chTemp = getchar();
}
}
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From: "John Croll" <[EMAIL PROTECTED]>
Newsgroups: sci.crypt
Subject: rc5-128 bit
Date: Sun, 3 Oct 1999 15:46:35 -0500
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Xref: tor-nx1.netcom.ca sci.crypt:116777
this program allows any one who is good at logic problems and word games
to decypher plain text from an rc-5 cypher regardless of key length. the
only
caveat is that the plaintext must conform to standard english grammar. this
method is not well suited to extracting numerical data. as is, the program
only dumps data concerning one cypher hex and one letter at a time. their
is a rem'ed out block where i tried to add scoring for the whole alphabet.
but my programming skills are not up to the task. if you improve it, you
may copyright that improvement.
begin below----------------------
file://microft visual c++ 5.0
#include <stdlib.h >
#include <stddef.h>
#include <stdio.h >
#include <math.h >
#include <ios.h >
#include <iostream.h>
#include <conio.h>
#include <cstring>
#include <ostream>
#include <stdarg.h >
#include <sstream>
void main()
{
int m;
int n;
char ms[3];
int temp1;
int temp2;
int temp3;
int temp4;
unsigned int ROT_IT;
unsigned int ROR_IT;
int ND_IT;
int XOR_IT;
int MOD_IT;
int MOD_IT2;
int OR_IT;
int SCO;
int OUTPA[255][255];
char ns;
n=1;
m=1;
for( int k = 32; k < 128; ++k )
{
for( int j = 32; j < 128; ++j )
{
OUTPA[m][n]=1;
n ++;
}
m ++;
}
cout << "copyright c 1999, Richard Lee King Jr." << endl;
while (m!=999){
cout << "Enter the Hexadecimal Value of the Cypher Byte:(01 to ff)" << endl;
cin >> ms;
temp1=0;
temp2=0;
temp3=0;
temp4=0;
temp1=ms[0];
temp2=ms[1];
temp3=temp1;
temp4=temp2;
if (temp1<64)
{
temp1=(temp1-48)*16;
}
if (temp2<64)
{
temp2=(temp2-48);
}
if (temp3>95)
{
temp1=(temp1-97+10)*16;
}
if (temp4>95)
{
temp2=(temp2-97+10);
}
m=(temp1+temp2);
cout << m << endl;
if (m>128){m=m-128;}
n=32;
SCO=0;
for( int i = 32; i < 128; ++i )
{
ROT_IT= _rotl(m,n);
if ( ROT_IT>255 ){ ROT_IT=ROT_IT%256;}
ROR_IT= _rotr(m,n);
if ( ROR_IT>255 ){ ROR_IT=ROR_IT%256;}
ND_IT= m&n;
XOR_IT= m^n;
MOD_IT= m%n;
MOD_IT2= n%m;
OR_IT=m|n;
if ((MOD_IT2==XOR_IT) && (MOD_IT==m)){SCO ++;}
if ((ROT_IT==192) && (MOD_IT==m)){SCO ++;SCO ++;}
if ((((ROT_IT==0) && (MOD_IT==m)) && (OR_IT==n)) && ((XOR_IT%8)==1)){SCO
++;}
if ((ROT_IT==0) && (MOD_IT==m) && (OR_IT==127) && ((ND_IT%8)==0)){SCO ++;}
if ((ROT_IT==0) && (MOD_IT==MOD_IT2) && (XOR_IT==0) && (ND_IT==n) &&
(OR_IT==n) && (MOD_IT==0)){SCO ++;}
if ((ROT_IT==0) && (MOD_IT2==n) && (ND_IT==n)){SCO ++;}
if ((ROT_IT==0) && (MOD_IT==m) && (OR_IT==121) && (XOR_IT==95)){SCO ++;}
if ((ROT_IT==0) && (MOD_IT==m) && ((ND_IT%8)==0) && (OR_IT==121)){SCO ++;}
if ((ROT_IT==0) && (MOD_IT2==0) && ((ND_IT%8)==0) && (OR_IT==125)){SCO ++;}
if ((ROT_IT==0) && (MOD_IT2==n) && (OR_IT==127) && (ND_IT==98)){SCO ++;}
if ((ROT_IT==0) && (MOD_IT==m) && (OR_IT==126)){SCO ++;}
if ((ROT_IT==128) && (MOD_IT2==n) && (ND_IT==n)){SCO ++;}
if ((MOD_IT==ND_IT) && (XOR_IT==n) && (ROT_IT==0)){SCO ++;}
if ((MOD_IT==ND_IT) && (OR_IT==n) && (ROT_IT==0)){SCO ++;}
file://if (((ND_IT%32)==0) && (M_IT==m)){SCO ++;}
OUTPA[1][n-1]=SCO;
SCO=0;
n ++;
}
cout << "Input Trial Letter to See Logic Output at Bottom: " << endl;
cin >> ns;
n=ns+1;
/* this block does not function correctly.
cout << "A: " << OUTPA[1][65] << " B: " << OUTPA[1][66] << " C: " <<
OUTPA[1][67] << " D: " << OUTPA[1][68] << " E: " << OUTPA[1][69] << " F: "
<< OUTPA[1][70] << " G: " << OUTPA[1][71] << " H: " << OUTPA[1][72] << " I:
" << OUTPA[1][73] << " J: " << OUTPA[1][74] << endl;
cout << "K: " << OUTPA[1][75] << " L: " << OUTPA[1][76] << " M: " <<
OUTPA[1][77] << " N: " << OUTPA[1][78] << " O: " << OUTPA[1][79] << " P: "
<< OUTPA[1][80] << " Q: " << OUTPA[1][81] << " R: " << OUTPA[1][82] << " S:
" << OUTPA[1][83] << " T: " << OUTPA[1][84] << endl;
cout << "u: " << OUTPA[1][85] << " V: " << OUTPA[1][86] << " W: " <<
OUTPA[1][87] << " X: " << OUTPA[1][88] << " Y: " << OUTPA[1][89] << " Z: "
<< OUTPA[1][90]<< endl;
cout << " " << endl;
cout << "a: " << OUTPA[1][97] << " b: " << OUTPA[1][98] << " c: " <<
OUTPA[1][99] << " d: " << OUTPA[1][100] << " e: " << OUTPA[1][101] << "
f: " << OUTPA[1][102] << " g: " << OUTPA[1][103] << " h: " << OUTPA[1][104]
<< " i: " << OUTPA[1][105] << " j: " << OUTPA[1][106] << endl;
cout << "k: " << OUTPA[1][107] << " l: " << OUTPA[1][108] << " m: " <<
OUTPA[1][109] << " n: " << OUTPA[1][110] << " o: " << OUTPA[1][111] << " p:
" << OUTPA[1][112] << " q: " << OUTPA[1][113] << " r: " << OUTPA[1][114] <<
" s: " << OUTPA[1][115] << " t: " << OUTPA[1][116] << endl;
cout << "u: " << OUTPA[1][117] << " v: " << OUTPA[1][118] << " w: " <<
OUTPA[1][119] << " x: " << OUTPA[1][120] << " y: " << OUTPA[1][121] << " z:
" << OUTPA[1][121] << endl;
cout << " " << endl;
cout <<"(0) " << OUTPA[1][48] << " (1) " << OUTPA[1][49] << " (2) " <<
OUTPA[1][48] << " (3) " << OUTPA[1][49] << " (4) " << OUTPA[1][50] << " (5)
" << OUTPA[1][51] << " (6) " << OUTPA[1][52] << " (7) " << OUTPA[1][53] << "
(8) " << OUTPA[1][54] << " (9) " << OUTPA[1][55] << " (10) " << OUTPA[1][56]
<< endl;
cout <<"(/) " << OUTPA[1][47] << " (-) " << OUTPA[1][43] << endl;
*/ end of funky block.
file://if (m>128){m=m-128;}
ROT_IT=_rotl(m,n);
if ( ROT_IT>255 ){ ROT_IT=ROT_IT%256;}
ROR_IT= _rotr(m,n);
if ( ROR_IT>255 ){ ROR_IT=ROR_IT%256;}
ND_IT= m&n;
XOR_IT= m^n;
MOD_IT2= n%m;
MOD_IT= m%n;
OR_IT=m|n;
cout << "ROTL " << ROT_IT << endl;
cout << "ROTR " << ROR_IT << endl;
cout << "AND " << ND_IT << endl;
cout << "XOR " << XOR_IT << endl;
cout << "MOD " << MOD_IT << endl;
cout << "OR " << OR_IT << endl;
cout << "MOD2 " << MOD_IT2 << endl;
cout << " " << endl;
cout << "sub_key= " << (m^n) << endl;
printf ("letter is: %c = %d \n ", n-1, n);
}
}
------------------------------
From: [EMAIL PROTECTED] (John Savard)
Subject: Re: Does anyone have more information?
Date: Wed, 06 Oct 1999 22:23:10 GMT
Michael Pedersen <[EMAIL PROTECTED]> wrote, in part:
>If you go to
>http://www.sunday-times.co.uk/news/pages/tim/99/09/29/timintint02001.html?1341861
>you find an article that a 512-bits RSA have been broken in 12
>microseconds.
>Anyone have some more news on this? Personally I think this is a big
>hoax, but perhaps some of you have more information...
There is no additional information available as yet. The idea that it
could have been broken in 12 microseconds, and by a handheld device,
yet, does appear to be absurd for a number of reasons, as has been
noted by a number of posts in other threads.
However, the report could be a _garbled_ version of something real,
such as a new theoretical approach to quantum computing that could
lead to quantum computers of smaller size, or a way, again
theoretical, to use optical computing techniques to perform quantum
computations in parallel.
John Savard ( teneerf<- )
http://www.ecn.ab.ca/~jsavard/crypto.htm
------------------------------
Date: Thu, 7 Oct 1999 00:35:21 +0200 (CEST)
From: Anonymous <[EMAIL PROTECTED]>
Subject: RC-5 breaking, $19 per letter
I am offering to undercut that other guy's RC-5 decryption charges. I
will decrypt RC-5, RC-4, Twofish, and any other cipher (it does not
matter... I can confidently, positively do them ALL!) for only $19/letter.
Also, if the plaintext character turns out to be Q or Z, I will only
charge half, or $9.50 for that character. WHAT A BARGAIN!!!
As an unrelated service, I can also provide you six guaranteed winning
lottery numbers for only $19 apiece. Just let me know which state or
local lottery you need the numbers for; I'll do the rest.
The only catch is I need all the money in advance, and you must deposit it
directly to my Swiss bank account #102349-A and the funds have to be
verified before I will begin my miraculous work.
Thanks and I look forward to helping you all!!!
K
------------------------------
From: jjlarkin <[EMAIL PROTECTED]>
Subject: Re: radioactive random number generator
Crossposted-To: sci.electronics.design,sci.electronics.equipment
Date: Wed, 06 Oct 1999 15:38:35 -0700
In article <[EMAIL PROTECTED]>, "John E. Kuslich"
<[EMAIL PROTECTED]> wrote:
> The problem is much more difficult than even
> designing a switching power supply for example.
Gosh, I've done 20 or 30 switchers in the last 30 years. I didn't
realize how difficult it is.
> The very first thing you will discover is that there are more ones
> than zeros
As I addressed in a prievious post, this one is actually easy to fix.
Just xor the zener stream with the output of a long cyclic shift
register or, even better, xor it *into* the shift register. That will
not only fix the 1/0 ratio, but wash out nearly all autocorrelations.
Pseudo-random shift registers already pass all randomness tests except
that they repeat (easily fixed: make the repeat period a few million
years) and they have a slight 1/0 bias (repeat last fix). Cyclic codes
are cryptographically crackable precisely because the generator
algorithms are simple hence discoverable. Combining true random noise
with a cyclic generator solves that problem.
A few zener noise generators plus an FPGA full of PRN registers and
xors should create a stream that would take a good chunk of the world
GNP to pick patterns out of, assuming a few terabits of output were
available to work on.
John
* Sent from RemarQ http://www.remarq.com The Internet's Discussion Network *
The fastest and easiest way to search and participate in Usenet - Free!
------------------------------
From: [EMAIL PROTECTED] (David P Jablon)
Subject: Re: Help: Mobility of the Private Key within PKI
Date: Wed, 6 Oct 1999 22:47:14 GMT
One vendor providing PKI security for roaming users in this
model is Entrust. (www.IntegritySciences.com/PKI50.html)
In this product, the user authenticates to a secure key
repository using the SPEKE protocol, and retrieves his private
key.
In article <7tg3u9$6bm$[EMAIL PROTECTED]>, <[EMAIL PROTECTED]> wrote:
>How can I enable the users to be able to encrypt and sign the data from
>any workstation within the network where PKI infrastructure is
>installed?
>
>The user needs his private key in order to encrypt or sign the data.
>However, the private keys are normally stored on either workstation
>directly (encrypted with the hash of the password) or on a smartcard.
>The smart card is not an option.
>
>It would be nice if there was some kind of private key server.
>The user would login through an application that runs on the
>workstation. The application would request user's private key from the
>server. The server would send that private key. Even though the private
>keys are stored on the server only the private key owner can utilize the
>key since the server store the keys in the encrypted fashion:
>TripleDes(hash(user_password),PrivateKey).
>Since only the key owner knows the password the key can only be used by
>the owner.
>
>Do such scenarios currently exist?
>If not then what are the common solutions to the situations where the
>enterprise has a PKI which does not restrict the users to their
>workstations?
>
>--
>Alex Bykov [EMAIL PROTECTED]
======================================================
David P. Jablon
Integrity Sciences, Inc.
[EMAIL PROTECTED]
<http://www.IntegritySciences.com>
------------------------------
From: [EMAIL PROTECTED] (Doug Stell)
Subject: Re: classifying algorithms
Date: Wed, 06 Oct 1999 23:13:50 GMT
On Wed, 06 Oct 1999 17:30:50 GMT, [EMAIL PROTECTED] (jerome) wrote:
>what is KEA ?
The Key Exchange Algoarithm. It is the Type 2 algorithm used in the
FORTEZZA card and is technicallly a key agreement algorithm. It may be
the best authenticated key agreement algorithm that is publically
known, according to people who have worked with "the good stuff" for
two decades.
After much delay, KEA and SKIPJACK were declassified by the NSA on
June 23, 1998 and published together by NIST the following day. See
http://csrc.ncsl.nist.gov/encryption/skipjack-kea.htm for the spec.
KEA is a dual, complementary, semi-ephemeral Diffie-Hellman exchange,
providing both authentication and random contributions by both parties
to the shared secret. The spec has details on a unidirectional varient
for email applications.
KEA is specified to use the DSA parameters, but any Diffie-Hellman
parameters will do.
------------------------------
From: "Steven Alexander" <[EMAIL PROTECTED]>
Subject: Re: rc5-128 cracking $20 per letter
Date: Wed, 6 Oct 1999 16:34:52 -0700
> >> 1602d701fa1ac1ad
>
> Unfortunately, this message is only eight bytes long, and your
> decryption is 16 bytes long.
Oops.
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