That's a lot more efficient than what I came up with and I got the same
results. However, according to the OP, the correct output should be:

_ | 8 | 8
1 | _ | 5
_ | _ | _
8 | 6 | _
9 | 0 | 2
_ | _ | _
_ | _ | _
4 | 5 | _
_ | _ | _
_ | _ | _
_ | _ | _
_ | _ | _
_ | _ | _
_ | _ | _

The entries that are 4|5|5 and _|4|5 are kicking my butt! Specifcally, and
this is where I'm stuck, the last 5 should not appear in column H.

Biff

"Harlan Grove" wrote in message
oups.com...
Vasant Nanavati wrote...
...
. . . And you still say there's nothing original in these NGs?
...

Nope. The ABS(x-m)<w/2 idiom has been mentioned before for testing
whether x falls between m-w/2 and m+w/2 without having to calculate m
twice, and the ratio term is akin to the standard way of counting
distinct entries in a range containing duplicates.

The gist is that each of C, D or E must account for some but not all of
the total count of all C:E in I:K.