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" <hrlngrv@aol.com> wrote in message
news:1115163829.327460.196040@g14g2000cwa.googlegroups.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.
>