thus, we deduce, we *actually* need breakpoints at these positions,
and that unused portions common to **all** cases can be deduced
-and marked "x"
+and marked "x" by looking at the columns above which all contain "x":
| |28|26|24| |20|16| |12|10|8| |4 |
x x
Also in order to save gates, in the example above there are only three
cases (32 bit, 16 bit, 8 bit) therefore only three sets of logic
are required to construct the larger overall computational result
-from the "smaller chunks", rather than at first glance, with there
+from the "smaller chunks". At first glance, with there
being 9 actual partitions (28, 26, 24, 20, 16, 12, 10, 8, 4), it
would appear that 2^9 (512!) cases were required, where in fact
there are only three.