[rc5] Suggestion for protocol

Remi Guyomarch rguyom at mail.dotcom.fr
Wed Jun 25 00:59:20 EDT 1997

Fedor Kouranov wrote:
> On 06/23/97 Remi Guyomarch <rguyom at mail.dotcom.fr> said:
> >About the network traffic, plain-text + cypher-text + iv + keyspace = 16
> >bytes.
> Ahem... what's iv?

Inversion Value ? Well, I don't know ;-)
The client gets this 64bit value from the proxy each time, and xor the
plaintext with it before encrypting it. I think it's just to 'melt' the
plaintext a bit more... This 'iv' value is given by RSA on their
challenge pages.

> The starting key? 7 bytes. Keyspace - 7 bytes. Plain-
> and ciphertext - 4 bytes each. =3D22 bytes.

Oops, you're right !

Well to be complete on the bandwith subject, here's a fast cut-and-paste
from the v1 source. This packet is send AND received each time the
client request a keyspace, and send another time when the client have
checked the keyspace :

#define PKT_STRLEN 128

typedef struct Packet {
        signed int op;          /* operation code */
        RC5_WORD key[2];        /* the Key starting point */
        RC5_WORD iv[2];         /* the IV */
        RC5_WORD pt[2];         /* the Plaintext */
        RC5_WORD ct[2];         /* the Ciphertext */
        RC5_WORD numkeys;       /* number of iterations */
        char id[PKT_STRLEN];    /* identifier */
} Packet;

So the network bandwith needed is 168*3 =3D 504 bytes for each block
checked (apart from network layer overhead).

Since we have 2^28 blocks to check, the total bandwith needed to
complete the 56bit keyspace is 126 Gb !!
(don't be alarmed, it's just 20 min. (!) of traffic on the MAE-East
see http://www.mfsdatanet.com:80/MAE/east.giga.970623.html)


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