[2337] in Kerberos
Re: Local password validation (was: kerberizing xlock)
daemon@ATHENA.MIT.EDU (Greg Wohletz)
Thu Nov 5 02:08:51 1992
To: Barry Jaspan <bjaspan@Athena.MIT.EDU>
Cc: hendrick@neptune.ctron.com, kerberos@Athena.MIT.EDU
In-Reply-To: Your message of "Sun, 01 Nov 92 14:23:15 EST."
Date: Wed, 04 Nov 92 22:44:07 -0800
From: Greg Wohletz <greg@duke.cs.unlv.edu>
>In summary, here is a timeline showing the attack:
>
>1. login sends tgt request {U,TGS} to KDC
>2. attacker responds with {U,TGS}_fake
>3. KDC responds with error, login is no longer listening
>4. attacker enters correct fake password for {U,TGS}_fake
>
>5. login sends service ticket {U,V} reques to KDC
>6. attacker responds with {U,V}_fake using session key from {U,TGS}_fake
>7. KDC responds with error, login is no longer listening
>8. login's decryption succeeds
>
>9. login sends mutual auth request to V
>10. attacker responds to mutual auth request
>11. V responds with error, login is no longer listening
>12. login is tricked
OK, now I see what your getting at, and I agree that it is
theoretically possible to accomlish this. I think that V could be
designed and located on the network in such a way as to make this
attact extremely difficult, if not impossible, I'll have to think more
on that. Anyway, In our environment I still feel OK about using this
scheme to accomplish login authentication as I'm not particularly
concerned if an imposter gains access to a workstation cpu since all
he would not actually be able to do anything usefull (nfs, pop, etc.
all require kerberos authentication), and its annoying to have to have
multiple passwords floating around. I don't think anyone will go to
that much trouble just to gain access to a workstation with no usable
disk space, far easier to gain root access through the rom monitor I
should think...
>KERBEROS IS A NETWORK AUTHENTICATION SYSTEM, NOT A PASSWORD VALIDATION
SYSTEM. If you use it as the latter, you must exercise extreme
>caution. In particular, using Kerberos for screensaver/keyboard lock
>programs is a tricky business.
So it would appear.
--Greg
