10/13/2006

A student writes:

        Dr. Patt,
        I have a few questions about a few different topics.

        1.) Am I correct in saying that the state variables are s1 and s0, the
        actual wires? In other words they are not the "circles" themselves on 
        the state diagrams, but the actual wires whose different combinations
        represent the states?

That is pretty close.  But, I would prefer you say that the actual wires 
carry the values of the state variables.  You are correct, they are 
s1 and s0.  

Each of the "circles" represent a state.  To do anything useful with these 
states, such as identify a specific output combination with each state, we 
need to represent each state uniquely.  Five states require five unique 
representations, which means three bits to differentiate the five states.  
Therefore, three state variables.  

Recall our example of a danger sign, and the detailed handout I went over
in class.  Each state is specified by the value in all of the master-slave
flipflops.  In that example, there were four states.  Therefore two state
variables are enough.  Each state variable is stored in one master-slave
flipflop, and if I know the value of both of them, then I know the state.  
Note on the handout, in the figure, they are labeled "state var."

The purpose of the sequential circuit is to determine the output as a function
of the current state, and the next state as a function of the current state 
and external inputs.  To do that, I need inputs to the combination logic from 
the current state and from the external inputs.  The current state information 
is completely specified by the values of the state variables, which in the
case of the danger sign is s1 and s0

        2.) When using an immediate value, such as with the AND opcode, 
        where you can AND the last five bits of the sixteen bit value, 
        with a selected register, how would you AND an immediate value 
        longer than five bits? (I hope that is not a run-on sentence). 
        Or is this simply not possible?

In the LC-3, it is "simply not possible."  The microprocessor in your 
laptop (called x86) allows immediate values to have 32 bits.  It does so by
allowing an instruction to consist of as many bits as necessary for the
particular instruction as needed.  Up to a maximum of 120 bits, if I 
remember.  Most ISAs fix the number of bits for all instructions to some fixed
number.  In the case of LC-3 all instructions are 16 bits.  With that number
of bits, the best I could allow was 5 bits for an immediate value.  
Therefore, for the LC-3, immediates must be between -16 and +15.  When 
John Hennessy specified the MIPS ISA at Stanford in 1981, the best he could
do was 4 bits for an immediate.  When Dave Patterson specified the RISC ISA
at Berkeley in 1980, the best he could do was 9 bits for an immediate, if
I remember correctly.

        3.) When talking of the condition codes, are these simply random bits
        inside the computer hardware itself? Or are these imbedded 
        somewhere else? 

The condition codes are stored in one-bit registers.  There are three of
them, an N register, a Z register, and a P register.  As I said in class,
every time a value is written into one of the 8 registers R0 to R7, the
three condition code registers are set to 100, 010, or 001, depending on
whether the value written is Negative, Zero, or Positive.  The next
instruction can then branch (change the PC) based on both the values in these
three one-bit registers and the values of bits [11:9] of that instruction.

        Also, in your example on the board in class on Wednesday, 
        you used the 'z' bit to determine if n=0 had just been loaded 
        to know when to stop our
        program. Say that we wanted to stop when n=1, not n=0. Is there 
        a way to test that using the bits? Or is that some other method 
        all together?

The three one-bit registers N, Z, and P tell you whether the last value
written to a register is negative, zero, or positive.  I suppose I could
have specified a fourth one-bit register (called O) which would be 1 if
the last value written to a register was 0000000000000001, and 0 otherwise.
If I had such a fourth one-bit register, and I change the conditional
branch instruction accordingly, then I could do as you ask.  But I do not
know any computer that has such a condition code.  

However, it is generally possible to do most things, provided you want
to do them badly enough.  So, what you could do, if you want to stop when 
n=1, is to add another instruction to subtract 1 from n and write that
result in some other register that is not being used.  Then branch on that
value being 0, which of course would be the case when n=1.  Got it.

        Muchas Gracias!

De nada.

        <<name withheld to protect the student with three questions>>


Thank you for the questions.  They were all excellent questions, and
showed good insights.  Good luck with the rest of the class.

Yale Patt