Re: Short Signature required
From: David Cross [MS] (dcross_at_online.microsoft.com)
Date: 11/06/03
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Date: Thu, 6 Nov 2003 05:23:00 -0800
No, ECC is not supported by Windows or CryptoAPI at this time.
-- David B. Cross [MS] -- This posting is provided "AS IS" with no warranties, and confers no rights. http://support.microsoft.com "Ken Cattanach" <randd@no_spam.opdicom.com> wrote in message news:O%23wSCOEpDHA.2868@TK2MSFTNGP09.phx.gbl... > Hi, > > We are developing an application for Win 98, Me, 2000, XP and 2003. Part of > the application's functionality is required to be activated by entering a > special code - similar to MS Windows XP Activation... > > I was wanting to utilize assymetric cryptography to generate and verify the > activation codes, however algorithms like RSA generate digital signatures > that are the same size as their key. Not everyone is connected to the > internet and manual entry of the code is a requirement. Obviously with a > 512 bit or 1024 bit key this is too long for a user to type in. > > There seem to be a lot of products that require activation out there, many > of which claim to using public key cryptography - how do they generate such > short activation codes? > > One thing that I am looking at is using Elliptic Curve Cryptography (ECC). > Because of its small key size it generates much smaller signatures. Does > anyone know if this is a good idea? If so, then does anyone know of any > good commercial or free ECC libraries? I'd prefer ECDSA compliance and if > it were a windows compatible CSP then that would be even better... > > Oddly enough, I noticed that in the CAPICOM documentation in MSDN Library > there is mention of ECDSA in the form of a CSP type definition: > > CAPICOM_PROV_EC_ECDSA_SIG The CSP that supports the Elliptic Curve > Digital Signature Algorithm (ECDSA) functions and algorithms required for > digital signatures. > > > However, I couldn't see a Cryptographic Service Provider of this type. Does > this mean that ECDSA is actually going to be supported by Windows soon? > > Thanks. > > Ken. > >
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