Re: Internet Privacy Guaranteed ad (POTP Jr.)
At 10:08 PM 2/20/96, IPG Sales wrote:
Unlike Mr. Silvernail, we have a much simplier definition of what we mean by a one time pad - given a message/file of length N, where N is a finite practical number say less than 10 to the 1000th power, that the encrypted ciphertext can be any of the N to the 256th power possibile clear/plain text messages/files. To prove that the IPG system does not work, all you have to do is to prove that is not the case - that our system, without artifically imposed boundary conditions will generate a subset of those possibilities - that is simple and strsight forward - not hyperbole but action - everyone stated how simple it was to break the system, now everyone is back paddling aa fast as they can, like Mr. Metzger and some of the other big bad cyphermouths.
PROOF: Given that N is the length of the message in bits. The number of possible combinations of bits is 2^N. For any message length N > 1, 2^N < N^256. Simple example. Message length is 3 bits. The maximum number of possible combinations of these bits is 8. This is far less than 3^256 (which is more than 10^100, i.e. it overflows the calculator on my Mac). Sorry guys. Try learning some simple math before you try and sell crypto. Clay --------------------------------------------------------------------------- Clay Olbon II | olbon@dynetics.com Systems Engineer | ph: (810) 589-9930 fax 9934 Dynetics, Inc., Ste 302 | http://www.msen.com/~olbon/olbon.html 550 Stephenson Hwy | PGP262 public key: finger olbon@mgr.dynetics.com Troy, MI 48083-1109 | pgp print: B97397AD50233C77523FD058BD1BB7C0 "To escape the evil curse, you must quote a bible verse; thou shalt not ... Doooh" - Homer (Simpson, not the other one) ---------------------------------------------------------------------------
Yes, in trying to answer all the questions you that were posed, we made that mistake, obviously it was a typo - unlike so many of you, we are not perfect - the 10 to the 1000 is correct though On Wed, 21 Feb 1996, Clay Olbon II wrote:
At 10:08 PM 2/20/96, IPG Sales wrote:
Unlike Mr. Silvernail, we have a much simplier definition of what we mean by a one time pad - given a message/file of length N, where N is a finite practical number say less than 10 to the 1000th power, that the encrypted ciphertext can be any of the N to the 256th power possibile clear/plain text messages/files. To prove that the IPG system does not work, all you have to do is to prove that is not the case - that our system, without artifically imposed boundary conditions will generate a subset of those possibilities - that is simple and strsight forward - not hyperbole but action - everyone stated how simple it was to break the system, now everyone is back paddling aa fast as they can, like Mr. Metzger and some of the other big bad cyphermouths.
PROOF:
Given that N is the length of the message in bits. The number of possible combinations of bits is 2^N. For any message length N > 1, 2^N < N^256. Simple example. Message length is 3 bits. The maximum number of possible combinations of these bits is 8. This is far less than 3^256 (which is more than 10^100, i.e. it overflows the calculator on my Mac). Sorry guys. Try learning some simple math before you try and sell crypto.
Clay
--------------------------------------------------------------------------- Clay Olbon II | olbon@dynetics.com Systems Engineer | ph: (810) 589-9930 fax 9934 Dynetics, Inc., Ste 302 | http://www.msen.com/~olbon/olbon.html 550 Stephenson Hwy | PGP262 public key: finger olbon@mgr.dynetics.com Troy, MI 48083-1109 | pgp print: B97397AD50233C77523FD058BD1BB7C0 "To escape the evil curse, you must quote a bible verse; thou shalt not ... Doooh" - Homer (Simpson, not the other one) ---------------------------------------------------------------------------
Clay writes:
Given that N is the length of the message in bits. The number of possible combinations of bits is 2^N. For any message length N > 1, 2^N < N^256.
Uh, nope. 2^N grows asymptotically faster than N^256. Actually, for any constants A and B, A^N grows asymptotically faster than N^B. For A=2, B=256, the crossover happens somewhere before N=4096. 2^4096 = 2^(16*256) > 2^(12*256) = (2^12)^256 = (4096)^256 If the IPG people are using N=5600 (weird choice) then certainly 2^5600 > 5600^256, for what little that's worth. (Ah, my computer science B.S. pays off ;) -Lewis "You're always disappointed, nothing seems to keep you high -- drive your bargains, push your papers, win your medals, fuck your strangers; don't it leave you on the empty side ?" (Joni Mitchell, 1972)
[ IPG Sales ]
... any of the N to the 256th power possibile clear/plain text messages/files.
Excuse me? 256^N.
[ Clay Olbon ] PROOF:
For any message length N > 1, 2^N < N^256.
Excuse me? 2^N is not O(N^k). Cypherpunks used to be a place where I could fairly reliably see high-SNR commentary from real cryptographers/number theorists. Now we have the blind replying to the blind's obfuscatory nonsense, and the useful posts take some effort to find.
Try learning some simple math before you try and sell crypto.
Surely it's not a requirement these days. Peter Monta pmonta@qualcomm.com Qualcomm, Inc./Globalstar
participants (4)
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IPG Sales -
lmccarth＠cs.umass.edu -
olbon＠dynetics.com -
Peter Monta