Re: A89: Difference between "AMS 2.0x" and "HW 2.00" designation on tica

[Prev][Next][Index][Thread]

Re: A89: Difference between "AMS 2.0x" and "HW 2.00" designation on ticalc.org?


No... he just payed attention in his Stats class.

-Miles Raymond      EML: m_rayman@bigfoot.com
ICQ: 13217756       IRC: Killer2        AIM: kilier2
http://www.bigfoot.com/~m_rayman

----- Original Message -----
From: "Cassady Roop" <croop@oregontrail.net>
To: <assembly-89@lists.ticalc.org>
Sent: Sunday, January 02, 2000 7:54 PM
Subject: Re: A89: Difference between "AMS 2.0x" and "HW 2.00" designation on
ticalc.org?

> Uh...  Were you bored today? :)
>
> Kaus wrote:
>
> > well, close.  there are 16 to be mathematically complete.
> > if each of the following are binary possibilities: where 1 is yes and 0
is
> > no
> > HW1Compatible  (bit 3)
> > HW2Compatible  (bit 2)
> > AMS1Compatible (bit 1)
> > AMS2Compatible (bit 0)
> >
> > And so:
> > 1111
> > 1110
> > 1101
> > 1100
> > 1011
> > 1010
> > 1001
> > 1000
> > 0111
> > 0110
> > 0101
> > 0100
> > 0011
> > 0010
> > 0001
> > 0000
> > or 2^4=16 possible states.  althought some of these are unlikely, such
as:
> >
> > not compatible with anything (0000)
> > not compatible with any AMS, but works on both hardware versions (1100)
> > not compatible with any HW versions, but works on both AMSs (0011)
> > not compatible with any HW versions, but works on AMS1.0x (0001)
> > not compatible with any HW versions, but works on AMS2.0x (0010)
> > not compatible with any AMS, but works on HW1 (0100)
> > not compatible with any AMS, but works on HW2 (1000)
> >
> > others can happen too though:
> >
> > only works on HW1, AMS1.0x (0101)
> > only works on HW2, AMS2.0x (1010)
> > only works on HW1, AMS2.0x (0110)
> > only works on HW2, AMS1.0x (1001)
> >
> > so you see, the answer of 16 is a possible number of states, but since 7
of
> > them are invalid, not accurate of the problem.  so how do we arrive at
the
> > correct answer, 9?
> > well, we combine the two low order bits and the two high order bits.
since
> > at all times, one of the bottom two bits has to be one, and one of the
upper
> > two bits has to be one, we can provide new rangenames:
> > HW:  0=HW1, 1=HW2, 2=both
> > AMS: 0=AMS1.0x, 1=AMS2.0x, 2=both
> > since each range has is a trinary unit, and we have two ranges, we
simply:
> > 3^2=9
> > and that is the correct number of possible combinations for our problem.
> >
> > hehehe
> >
> > --kaus
References: