-----BEGIN PGP SIGNED MESSAGE----- I have a question I hope someone here might be able to answer: As the method of cryptanalysis of XOR (Ie. index of coincidence) relies on redundancy in the plaintext, would the following be strong: Compress P to get perfect compression (ie. 0 redundancy) Encrypt F (the compressed text) using a repeated key XOR of course this is all rather theoretical as there is no such thing as perfect compression, but I just thought it might be interesting to see if this is indeed strong, superficially it appears so to me... -----BEGIN PGP SIGNATURE----- Version: 2.6.3ia Charset: cp850 iQCVAwUBMj2TvL5OPIRbv66xAQGRpgP+LU0Y8sxzO6rObCYAQdrD8/R/iDJN3m0Z 4ZetS7jcbz7wT6bj2l7Usb0F4h/YMhxtl0y9EQ91ozg35jfRKdy2IwUoMDvqsVSZ wKmaM/DpEt2LDyRQnzIvlNYQp6/eXQoBUb7r9SH/dZbjM7culpjzJLhd07Nx5okE jUmPNBLm9m0= =RHie -----END PGP SIGNATURE----- Datacomms Technologies web authoring and data security Paul Bradley, Paul@fatmans.demon.co.uk Http://www.fatmans.demon.co.uk/crypt/ "Don`t forget to mount a scratch monkey" -----BEGIN PGP PUBLIC KEY BLOCK----- Version: 2.6.3ia mQCNAjH9j+cAAAEEAMBvREiQR0ot9dFCO0TiSCSunAYLv2g1Bc6I3bz8FzKXNH53 6mieJf/W4rD+CxJpT0q9RQaaoRtkHJLwbjfK2il3D7mEahMAyqvF/xRJNqkXfhM3 sRJM0Jh43l+W0M5vwokbEbk25/bxWWGspTsLD3YHbzKnG6pOcL5OPIRbv66xAAUR tCdQYXVsIEJyYWRsZXkgPHBhdWxAZmF0bWFucy5kZW1vbi5jby51az6JAJUDBRAy NwfvNkCBjDT0xHEBATQPA/9TORmN/UjNecj03q4anpvdyCLiez5sKuNbnYK50RiP Jj4QpWWvST3smyQ0A86DrZY/re056MXwQmARESx0rFZxdnD0oORICl5r8dJLIy3b j8rbA5olXwZwKz73/X5s13v/pvHYX4cIsbVK8NHXqh5llSKt6TBAuGgkIGF29z5k C4kAlQMFEDI3B9mdtf/umVkv7QEBcRYD/1FBteLqsUmr81euxqqnnrpLlyHb58B/ 9sdATuua4uSjX46hXDZ264YozspNrzSB4NEdrmXOWVX3fiE0ga6XkSSkIeF23V90 En37Z0BdbFzgF00FRYTFyTq8eezQrdg/+rBPUsZUmG5wpq3e12FKHQsX01i+1mB2 YmqqwCV5e95eiQCVAgUQMh8uSb5OPIRbv66xAQEqJwP/fxQyiCasjFcbDpsFfsYp put5cCC/9pOx6X3DlbKShPMpUOS+A9HsTEmJQN8Iawv1nSwPdtc2cR/GhW6ilVjW LSloGdMVLabm9pGpZZMkRaZlXFUkOv7VhfgsUiL+vIDryBCAwUZCzQiWycjt/cPi mUqFH41Z7NkyO8ZFdi5GGX0= =CMZA -----END PGP PUBLIC KEY BLOCK-----