Question: What is the drawback of Double DES algorithm? How is it overcome by Triple DES?

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- Consider a message 'n' Which is to the encrypted
- The corresponding block cipher for one application of the DES applications is represented by E
_{k}where K is the corresponding DES key - The output of 2-DES is C = E
_{k2}( E_{ }(m)) to decrypt similarly, m = Dk_{1}(Dk_{2}(C)) - The cipher 2-DES should offer additional security, equivalent to both K
_{1}& K_{2}. - The cipher 2-DES obtained by the repeated application of DES is called 2DES = DES X DES. This is called a product cipher obtained by the composition of two cipher such an idea can similarly by extended to multiple ciphers
- The composition of two ciphers with two different keys cannot be obtained by a single application of DES with a key
- This 2DES is encrypted to provide security equivalent to 56 X 12 = 112 bits
- But such a cipher can be attached by an attach method which is called meet in the middle ( MIM ) attach
- As double DES , DES has a problem of the MIM attach, triple DES was developed
- The expected security of 3-DES is 112 bits

- The corresponding block cipher for one application of the DES applications is represented by E

**Triple DES**

- There are in general two flavors of 3-DES & two ways for implementation of 3-DES
- The first implementation uses three keys K
_{1},K_{2},K_{3}the ciphertent of 'm' is thus obtained by

C = DES_{K1}[DESk_{2} ( DESK_{3} (m))]

The second way to implement 3-DES is using two keys, thus

C = DES_{K1}[DES^{-1}k_{2} ( DESK_{3} (m))]

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