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have a mode of operation that might increase security and make the ciphertext more random.

You can have the plaintext divided into blocks and have each block pass trough the initialization vector a new initialization vector is generated for each block each block will be XORed with the previous block before being encrypted

this mode has the features of ECB mode because it divides the plaintext into blocks and the features of CBC because an initialization vector is used on each block of plaintext and each block will be XORed with the previous block before being encrypted

you can call it electronic cipher block because the plaintext is divided into blocks and call it cipher block chaining because each block will go through an initalialization vector before being XORed with the previous block before being encrypted and can call it ECBC mode Electronic Code Block Chaining.

But will this new mode of operation idea be secure?

Mathematically, the idea looks like this:

Given, $P_1,P_2,P_3$ and $k_1,k_2,k_3$, generate $IV_1$ randomly, and compute $C_1=E_{k_1}(P_1 \oplus IV_1)$. Then set $IV_2=C_1$ and compute $C_2=E_{k_2}(P_2 \oplus IV_2)$, and set $IV_3=C_2$, and compute $C_3=E_{k_3}(P_3 \oplus IV_3)$.

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this is just the previous question in a more clearer format – Andrew Jul 13 '12 at 23:00
in this mode the attacker will have to reconstruct the message – Andrew Jul 14 '12 at 0:52
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It's not possible to decrypt that, since C1,C2,C3 will all equal IV. $\:$ – Ricky Demer Jul 14 '12 at 4:49
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Or just write some code (pseudo or real, as long as its a normal looking language) – CodesInChaos Jul 14 '12 at 12:00
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@mikeazo, Andrew: The formula given is not actually the same as CBC mode, as it uses a different key for each block (which is quite unusual for a block cipher mode of operation). – Paŭlo Ebermann Jul 15 '12 at 8:51
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closed as not constructive by David Schwartz, CodesInChaos, mikeazo Jul 14 '12 at 17:42

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1 Answer

have a mode of operation that might increase security and make the ciphertext more random.

It does neither as far as I can tell (please post a mathematical representation of your construction).

It definitely does not make the ciphertext more random (compared to a good mode like CBC). I say this because if there were any randomness deficiencies of a block cipher in a strong mode, this would imply that the block cipher itself, and doing your proposed scheme would not be able to fix that.

As to increasing the security, this totally depends on how the new IV is generated for each block and what you mean when you say that "each block pass trough [sic] the initialization vector". But unless you make the system completely impractical, the increase in strength is negligible at best.

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P1 P2 P3 Ek1 Ek2 Ek3 C1 C2 C3 C1= Ek1(P¡⊕C¡_1) , C0=IV C2= Ek2 (P¡⊕C¡_1) , C0=IV C3= Ek3 (P¡⊕C¡_1) , C0=IV – Andrew Jul 14 '12 at 4:25

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