Australian Gov't Offers $560k Cryptographic Protocol For Free

mask.of.sanity writes "Australia's national welfare agency will release its 'unbreakable' AU$560,000 smart card identification protocol for free. The government agency wants other departments and commercial businesses to adopt the Protocol for Lightweight Authentication of ID (PLAID), which withstood three years of design and testing by Australian and American security agencies. The agency has one of Australia's most advanced physical and logical converged security systems: staff can access doors and computers with a single centrally-managed identity card, and user identities can be automatically updated as employees leave, are recruited or move to new departments. PLAID, which will be available soon, is to be used in the agency's incoming fleet of contact-less smartcards that are currently under trial by staff. It will replace existing identity cards that operate on PKI encryption."

Comments

Re:contactless smart cards are the way to go

[Burkin]

Until you lose your wallet and the person who finds it has complete control to ruin every aspect of your life connected to said card... ...

Yes, because clearly they would have no system to revoke lost cards.

Re:I laugh ...

[mark-t]

Meh.... unbreakable encryption is easy, or so close to it that the difference is largely irrellevant:

1. Find any two nice and large prime numbers and publish them. Call them A and B. Call their product C. Let n = one less than the number of bits in C.
2. Both the source and destination can pick any number that is coprime to (A-1)*(B-1), call them Xs and Xd. They do not share this information.
3. The source and destination then compute Ys and Yd, respectively, such that their own X*Y is congruent to 1 mod (A*B). They do not share this information.
4. The source takes n bits from the data, D, and applies the following transform: D = D ^ Xs mod C. This data is transmitted.
5. The destination then applies the transform D = D ^ Xd mod C and transmits that back to the source.
6. The source applies the transform D = D ^ Ys mod C and transmits that to the destination
7. The destination finally applies D = D ^ Yd mod C, and in this final transform retrieves the unencrypted data.

This allows one to completely securely transmit up to n bits of data from a source stream, and because the source and destination can pick new X and Y values with every transmission, and unencrypted data is never found on any transmitted data stream. The likelihood of breaking it is genuinely 1 in 2^n and can only be broken by brute force attack. Factoring methods will not break the encryption because what would normally be associated as a public/private key pair (X,Y) in some other encryption protocols is never shared with the other party.

Re:A little more info

[owlstead]

The protocol looks unremarkable. They pass some entropy and IDs back and forth, using conventional standards based encryption and hash algorithms.

That's a good thing.

Their problem is keeping the cards secure and they state clearly that they are using commercially available smart cards.

Which is also a good thing, as long as these cards have been analyzed well. I would be worried if they were using cards with "military grade" security meaning that they were only analyzed by few, without any standardized security level like FIPS or CC.

There are secrets in the cards, an RSA private key and an AES master key. The bigger problem is keeping these secrets in the cards and distributing the keys to cards. The PLAID protocol has no bearing on these matters.

Sorry, but you are wrong on both matters.

The RSA private key and AES master keys are not on the card. It contains the RSA public key and the AES derived key (one that is specific to the card).

There are many interesting things about this protocol. Lets have a list so I can get a few mod points on this old discussion:

• No ID before authentication (card ID is encrypted with public RSA key, standard RSA encryption uses random padding)
• No RSA private key encryption for the authentication (vulnerable to attack)
• Uses standardized, up to date algorithms (SHA-1 is only used in a secure way as far as I can see)
• Uses RSA public key on the card, which is *faster* than ECC because the public exponent will likely be small (010001h normally)

Ok, for some disadvantages

• Requires contact-less processor card with AES and hardware RSA support
• Access is much slower than with AES only authentication
• Time and power usage of RSA calculations may make it more difficult to do a successful authentication
• Unremarkable (probably has been invented earlier)
• Requires terminal that performs RSA private key encryption
• Requires RSA private key to be present on reader side, key cannot be revoked
• Still requires a single master key (hopefully it will never be leaked)

All in all, this protocol is very interesting for mutual authentication. I'll have to look into it further (e.g. how much the private key needs to stay private).