HOMEWORK If didn't assign 4-31 --4-35, assign it now

NOTE:  I want to say more about CLA but don't want to break-up
       shifter/multiplier accross two lectures

There are other fast adders (e.g. carry save)

Shifter

    Just a string of D-flops; output of one is input of next

        Input to first is the serial input

        Output of last is the serial output         

    But want more.

        Left and right shifting (with serial input/output)
        
        Parallel load

        Parallel Output
        
        Don't shift every cycle

    Parallel output is just wires.

    Shifter has 4 modes (left, right, nop, load) so

        4-1 mux inside

        2 control lines must come in

    Handout (also on web) shows diagram (Fig AA in my notes)

    Our shifters are slow for big shifts; barrel shifters are better

HOMEWORK:  A 4-bit shift register initially contains 1101.  It is
           shifted six times to the right with the serial input being
           101101.  What is the contents of the register after each
           shift.

HOMEWORK:  Same register, same init condition.  For
           the first 6 cycles the opcodes are left, left, right, nop,
           left, right and the serial input is 101101.  The next cycle
           the register is loaded (in parallel) with 1011.  The final
           6 cycles are the same as the first 6.  What is the contents
           of the register after each cycle?

Multipliers

Recall how to do multiplication.

    Multiplicand times multiplier gives product

    Multiply multiplicand by each digit of multiplier

    Put the result in the right column

    Then add the partial products just produced

We will do it the same way ...

... BUT differently

    We have binary so each "digit" of the multiplier is 1 or zero.
    Hence "multiplying" by the digit means either

        Getting the multiplicand

        Getting zero

    Use an "if appropriate bit of multiplier is 1" stmt

    To get "appropriate bit"

        Use LOB

        Shift right (so next bit is LOB)
    
    Putting in the right column is done by shifting the multiplicand
    left one bit each time (even if the multiplier bit is zero)

    Instead of adding partial products at end, keep a running sum

        Don't add zero if multiplier bit is zero just skip step

This results in the following algorithm

    product <- 0
    for i = 0 to 31
        if LOB of multiplier = 1
            product = product + multiplicand
        shift multiplicand left 1 bit
        shift multiplier right 1 bit

Do on board 4-bit addition (8-bit registers) 1100 x 1101

The circuit diagram is figure 4.20 (handout)

What about the control?

    Always give the ALU the ADD operation

    Always send a 1 to the multiplicand to shift left     

    Always send a 1 to the multiplier to shift right

    Pretty boring so far but

        Send a 1 to write line in product if and only if
        LOB multiplier is a 1

        I.e. send LOB to write line

        I.e. it really is pretty boring

This works! ...

... but is wasteful of resourses and hence is
    slower
    hotter
    bigger
    all these are bad

    Product register must be 64 bits since the product is 64 bits

    Why is multiplicand register 64 bits?

        So that we can shift it left

        I.e. for our convenience

    Why is ALU 64-bits?

        Because the product is 64 bits

        But we are only adding a 32-bit quantity to the
        product at any one step.

        Hmmm.

        Maybe we can just pull out the correct bits from the product.

        Would be tricky to pull out bits in the middle
        because which bits to pull changes each step

    POOF!!

Solving both problems at once

    DON'T shift the multiplicand left

        Hence register is 32-bits and not a shifter

    Instead shift the product right!

    Add the HO 32-bits of prod reg to Multiplicand
    and place result back into HO 32-bits

        Only do this if the current multiplier bit is one.

        Do NOT need the carry-out from the adder; why?

            Because the HOB of the product BEFORE the add is zero

This results in the following algorithm

    product <- 0
    for i = 0 to 31
        if LOB of multiplier = 1
            product[32-63] = product[32-63] + multiplicand
        shift product right 1 bit
        shift multiplier right 1 bit

The circuit diagram is figure 4.21 (handout)

What about control
    
    Just as boring as before

    Send ADD, 1, 1  to ALU, multiplier, Product (shift right)

    Send LOB to Product (write)

Redo same example on board

A final trick ("gate bumming", like code bumming of 60s)

    There is sort of a waste of registers.

        Not for the multiplicand since we always need all 32 bits  

        But once we use a multiplier bit, we can toss it

        And the product is half unused at beginning and only slowly ...

        POOF!!

    "Timeshare" the LO half of the "product register"

        In the beginning LO half contains the multiplier

        Each step we shift right and more goes to product
        less to multiplier

    The algorithm changes to

    product <- multiplier
    for i = 0 to 31
        if LOB of product = 1
            product[32-63] = product[32-63] + multiplicand
        shift product-multiplier-register right 1 bit

Redo same example on board

The above was for unsigned 32-bit multiplication

For signed multiplication

    Save the signs of the multiplier and multiplicand

    Convert multiplier and multiplicand to non-neg numbers

    Use above algorithm

        Don't need the 32nd iteration since multiplier HOB of
        multiplier is zero (sign bit)

    Compliment product if original signs were different
        
There are faster multipliers.

Skip Division

Skip Floating Point

HOMEWORK Read 4.11 "Historical Perspective"
         Start Reading Chapter 5