ISO/IEC 9797-1 MAC Algorithm 3 how to?

13 Apr 2005 2:41 PM

Frank Fritze

Hi,

I need to calculate the Cryptographic checksum using ISO/IEC 9797-1 MAC
algorithm 3 with block cipher DES, zero IV (8bytes), and ISO 9797-1
padding method 2. The MAC length must be 8 bytes.

Any ideas on how to do it?

As a sample, the data I am working on now is:
Data =
'72C29C2371CC9BDB65B779B8E8D37B29ECC154AA56A8799FAE2F498F76ED92F2'
Kmac = '7962D9ECE03D1ACD4C76089DCE131543'

Resultant MAC[Kmac](Data)='5F1448EEA8AD90A7' (I want to know how to get
to this value).

13 Apr 2005 3:11 PM

Valery Pryamikov

Frank,
I don't have 9797 handy, so I'm working from memory here:
Padding method is to append single a 1 bit to the end of your message and as
many 0 bit as it necessary to get padded message length to be multiple of 64
bit. Kmac shows that you are have two keys;
the first key (first 64 bits of Kmac) is used to DES CBC encrypt padded
message using IV = {0};
the last cipher block should be encrypted with second key (last 64 bits of
Kmac) using DES ECB;
and you need to encrypt result of previous encryption with the fist key
using DES ECB;
(its analogy of using 3DES EEE with two keys on the last block).

that should be give you required results.

-Valery.
http://www.harper.no/valery

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"Frank Fritze" <frank.fri***@bdr.de> wrote in message
news:dfcb9d01.0504130641.6b27b0db@posting.google.com...
> Hi,
>
>
> I need to calculate the Cryptographic checksum using ISO/IEC 9797-1 MAC
> algorithm 3 with block cipher DES, zero IV (8bytes), and ISO 9797-1
> padding method 2. The MAC length must be 8 bytes.
>
>
> Any ideas on how to do it?
>
>
> As a sample, the data I am working on now is:
> Data =
> '72C29C2371CC9BDB65B779B8E8D37B29ECC154AA56A8799FAE2F498F76ED92F2'
> Kmac = '7962D9ECE03D1ACD4C76089DCE131543'
>
>
> Resultant MAC[Kmac](Data)='5F1448EEA8AD90A7' (I want to know how to get
> to this value).

13 Apr 2005 3:12 PM

Valery Pryamikov

And just as a side note:
this question is better fit for microsoft.public.security.crypto group.

-Valery.
http://www.harper.no/valery

Show quoteHide quote

"Frank Fritze" <frank.fri***@bdr.de> wrote in message
news:dfcb9d01.0504130641.6b27b0db@posting.google.com...
> Hi,
>
>
> I need to calculate the Cryptographic checksum using ISO/IEC 9797-1 MAC
> algorithm 3 with block cipher DES, zero IV (8bytes), and ISO 9797-1
> padding method 2. The MAC length must be 8 bytes.
>
>
> Any ideas on how to do it?
>
>
> As a sample, the data I am working on now is:
> Data =
> '72C29C2371CC9BDB65B779B8E8D37B29ECC154AA56A8799FAE2F498F76ED92F2'
> Kmac = '7962D9ECE03D1ACD4C76089DCE131543'
>
>
> Resultant MAC[Kmac](Data)='5F1448EEA8AD90A7' (I want to know how to get
> to this value).

13 Apr 2005 8:41 PM

Frank Fritze

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frank.fri***@bdr.de (Frank Fritze) wrote in message news:<dfcb9d01.0504130641.6b27b0db@posting.google.com>...
> Hi,
>
>
> I need to calculate the Cryptographic checksum using ISO/IEC 9797-1 MAC
> algorithm 3 with block cipher DES, zero IV (8bytes), and ISO 9797-1
> padding method 2. The MAC length must be 8 bytes.
>
>
> Any ideas on how to do it?
>
>
> As a sample, the data I am working on now is:
> Data =
> '72C29C2371CC9BDB65B779B8E8D37B29ECC154AA56A8799FAE2F498F76ED92F2'
> Kmac = '7962D9ECE03D1ACD4C76089DCE131543'
>
>
> Resultant MAC[Kmac](Data)='5F1448EEA8AD90A7' (I want to know how to get
> to this value).

In addition to my question:

I've already tried to calculate it over MACTripledes ... but with no success..
And I'm sure, it can be done with .NET, isn't it ?

Regards

Frank

14 Apr 2005 5:22 AM

Valery Pryamikov

Did you see my post where I answered you how to do it? It could be done with
..Net if you simply follow my perscription. and right - MACTripleDES is a
different thing.

-Valery.
http://www.harper.no/valery

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"Frank Fritze" <frank.fri***@bdr.de> wrote in message
news:dfcb9d01.0504131241.1cf7e340@posting.google.com...
> frank.fri***@bdr.de (Frank Fritze) wrote in message
> news:<dfcb9d01.0504130641.6b27b0db@posting.google.com>...
>> Hi,
>>
>>
>> I need to calculate the Cryptographic checksum using ISO/IEC 9797-1 MAC
>> algorithm 3 with block cipher DES, zero IV (8bytes), and ISO 9797-1
>> padding method 2. The MAC length must be 8 bytes.
>>
>>
>> Any ideas on how to do it?
>>
>>
>> As a sample, the data I am working on now is:
>> Data =
>> '72C29C2371CC9BDB65B779B8E8D37B29ECC154AA56A8799FAE2F498F76ED92F2'
>> Kmac = '7962D9ECE03D1ACD4C76089DCE131543'
>>
>>
>> Resultant MAC[Kmac](Data)='5F1448EEA8AD90A7' (I want to know how to get
>> to this value).
>
>
> In addition to my question:
>
> I've already tried to calculate it over MACTripledes ... but with no
> success..
> And I'm sure, it can be done with .NET, isn't it ?
>
> Regards
>
> Frank