Biff,
Works 98% with visible exceptions:
A B C D E F G H I J K
1
2 2 8 8 8 8 8 8
3 1 0 5 1 5 9 5 1
4 8 3 4 5 2 4
5 8 6 7 8 6 6 2 8
6 9 0 2 9 0 2 2 9 0
Most work great as above but then,
7 1 1 0 1 1 1
8 0 0 6 0 0 6 0
9 9 9 0 9 9 0 0 9
10 5 5 7 5 5 5 7
11 0 0 7 0 0 0
12 0 8 8 8 8
13 7 5 5 7 5 5 7 5 7
14 7 7 1 7 7 1 7
15 0 0 7 0 0 0
16 8 8 5 8 8 8 5
17 5 5 6 5 5 5 5
18 6 6 0 6 6 6
I'm adding this one because it worked as well:
19 9 9 9 9 9 9 9

Notice the way the doubles are typically showing in G & H and the 755
situation in line 13 (above example)... courious. These are the only type
situations that failed through hundreds of samples I ran.

This is great! Not sure if this is the response you could live without and
I'm not sure if I understand the formula you wrote but it is very very close
to done.
I would understand if you had enough but if you're into the challenge I'll
await your best shot... You guys rock!
Luke


"Biff" wrote:

If you only knew! <lol

OK, I think I have it but I hope Harlan comes back with a more efficient
solution.

It's a 2 formula approach.

Enter this formula in F2 and copy to G2:

=IF(AND(SUMPRODUCT(--(ISNUMBER(MATCH($I2:$K2,$C2:$E2,0))))=2,SUMPRODUC T(--(ISNUMBER(MATCH($C2:$E2,$I2:$K2,0))))=2,COUNTIF($ I2:$K2,C2)0),C2,"")

Enter this formula in H2:

=IF(AND(SUMPRODUCT(--(ISNUMBER(MATCH($I2:$K2,$C2:$E2,0))))=2,SUMPRODUC T(--(ISNUMBER(MATCH(C2:E2,I2:K2,0))))=2,COUNTIF($I2:$ K2,E2)0,COUNT(F2,G2)<COUNT(I2:K2)),E2,"")

Now, select the range F2, G2, H2 and copy down as needed.

Biff

"Luke" wrote in message
...
Stirred things up a bit it appears.
A question arises in that
in F2
if C2,D2 & C2,E2 are = to either I2,J2 OR J2,I2 OR J2,K2 OR K2,I2 OR K2,J2
Shouldn't it also, in the same sentance say:
but only if I2"" or J2"" OR K2"", OTHERWISE display the contents of
I2,J2 OR J2,I2 OR J2,K2 OR K2,I2 OR K2,J2...?
Like wise in G2 and H2.
Just a thought here, but perhaps a way out of a good but kickin when it
comes to 455 444 senarios :)

Luke

"Biff" wrote:

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.