The base 3 digits are represented as follows
So, if ,
,
,
, and
,
then we are adding the trinary digit 2 to the trinary digit 1, with
no carry coming in. We never have
, nor
,
since those values don't correspond to a trinary digit.
Write a boolean formula in sum of products form, expressing the
carry out ( ) in terms of the five input bits. You
should get a sum of six products, and each product involves two or
three input bits.
I tried to produce bad pipeline behavior. The following loop decrements R1 from 1000 to 0.
There are plenty of instructions after the loop, but we don't care what they are.
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