There are 12 poker chips and one of them is different in weight than the others. Using only a calibrated balance scale, how can you determine the odd chip in only 3 weighings?
- Step 1: Balance 2 piles of 4 chips
- Case 1: Both Piles Level
o Step 2: Take 3 of the 4 unknown chips and weigh against 3 of the known normals
 Case 1: Both Piles Level
• Step 3: Take the remaining chip and weigh against a known normal to get answer
 Case 2: Piles not level
• Take the 3 potentially light or heavy chips (you know which one they are now) and balance two of them to see which is the lighter or heaver one or if it’s the one in your hand.
- Case 2 (Worst Case): Piles not Level
o Identify the 4 potentially heavy and 4 potentially light chips from step 1
o Step 2
 Side 1 of scale: 3 potentially light chips and 1 of the potentially heavy chips
 Side 2 of scale: 3 of the 4 known normal chips from step 1 and the 1 leftover potentially light chip
 Case 1: Both Piles Level
• Step 3: Take the remaining 3 potentially heavy chips and balance 2 of them, seeing which is the heaver one or if it’s the 3rd one in your hand
 Case 2: Side 1 is higher
• Step 3: As none of the chips on side 2 can be heavy, it must be one of the 3 potentially light chips on side 1, so do like step 3 above for them to determine the light chip
 Case 3: Side 2 is higher
• Step 3: Since there is only one heavy chip on side 1 and one light chip on side 2, we know that one of them is the culprit, so take one of them and weigh it against a known normal chip and get your answer!
I came to this conclusion by thinking about how I wanted to have 3 chips at the end that were either all potentially heavy or potentially light. This made me want to eliminate 5 of the 8 unknowns in step 2. The key was having only one potentially heavy chip on side 1 and only one potentially light chip on side 2 so that if side 2 becomes higher, it must be one of those two chips. I kinda lucked out picking groups of 4 for step one, but I spent considerable time examining other groupings for step1 and 3 groups of 4 seemed to be getting the closest.
-Chaz Morantz

Thanks Chaz.

If you don’t want to use INDEX() and MATCH() you might like to try:

=SUMPRODUCT(($A2=dataName)*(B$1=dataID)*(dataResult)) in B2. It’s not an array formula (so no Ctrl + Shift + Enter) and it’s a lot more intuitive.

Where there’s no match, you’ll see a 0. You can hide these in a number of ways:

1. =IF(SUMPRODUCT(($A2=dataName)*(B$1=dataID)*(dataResult))=0,”",SUMPRODUCT(($A2=dataName)*(B$1=dataID)*(dataResult)))

2. Select the cells, right-click and select “Format Cells”. From the “Custom” menu, append two semi-colons to the format. e.g. “General” becomes “General;;”

Plus others.

I had a look at the email of your post (that maintained the correct formatting for some reason) and think I have an idea as to what you are after. My interpretation is that you want to match two criteria; the name in the leftmost column of Sheet1 and the number in the top row Sheet1 – if there’s a match in Sheet2, then return the value in the rightmost column of Sheet2.

Lets start with Sheet2 because this is where your raw data is kept. In fact, if it’s ok, I’ll rename the worksheet to Data

Data

At this point, I’d always recommend using named ranges on these columns, but to keep the solution in cell reference terms, we can substitute this in later.

Now, turning our attention to Sheet1 (which I’ll call Output)…

Output

To reiterate, what I think you want to do is populate the coloured cells (containing ???) when there is a match to the column and row. As mentioned, there are more ways to do this than I care to mention, but the first that comes to mind involves using INDEX() and MATCH() (thank goodness, being the title of the blog entry and all

Let’s start by clicking on the first cell we want to calculate on the Output worksheet: B2.

From earlier, INDEX() can be described as:

=INDEX(array,row_num,column_num)

Here, the array we are after is the rightmost column in the Data worksheet (Data!$C$1:$C$8). Being only one column wide, the only column_num we can have is 1. But how do we work out the row_num when there are two criteria? One way to do this is to concatenate the two search terms. You do this by using an ampersand (&). You then concatenate the columns you want to search so you’re comparing apples to apples (so to speak.)

Describing MATCH():

=MATCH(Lookup_value,Lookup_array,Match_type)

So our Lookup_value for this first cell is $A2&B$1. We use the dollar ($) symbol in front of the A in $A1 to confine the search queries to the first column (i.e. the ones with name1…name8 in them). Similarly, we use the dollar ($) symbol in from of the 1 in B$1 to confine the searches to the first row. That way, when we copy the formula to the other cells, we’re always matching the leftmost/top column that corresponds the cell we’re in.

As mentioned, our Lookup_array is also a concatenation, however this time it is Data!$B$1:$B$8&Data!$A$1:$A$8. We have dollar ($) symbols in front of of all row and column references here because the raw data doesn’t move (i.e. its position is not relative the cells).

This results in:
=MATCH(Output!$A2&Output!B$1,Data!$B$1:$B$8&Data!$A$1:$A$8,0)

So…putting it altogether we have:

=INDEX(Data!$C$1:$C$8,MATCH(Output!$A2&Output!B$1,Data!$B$1:$B$8&Data!$A$1:$A$8,0),1)

but pressing return results in #VALUE. This is because concatenating strings in a function such as this returns an array. To formulate cells like this instead of press Enter after entering the formula, you need to press Shift and Enter at the same time. This will put curly brackets ({ and }) around the formula. In doing so, for this example, you should now see the number 5 in B2 which corresponds to the two criteria in the raw data.

You only need to do this once, as we’re going to now place the cursor on the bottom right-hand corner of B2 until the cross appears and will drag the formula to the other cells in the table. You can’t copy and paste the formula to more than one cell as Excel will complain that you “cannot change part of an array.”

The only problem now is that there are #N/As where there isn’t a corresponding value in the raw data What’s a quick way to get rid of this. What I do is use a simple if statement to return nothing if the result is an #N/A. What I do in this instance is use a combination of the IF() and ISNA() functions.

To describe this in broad terms:
=IF(ISNA(cell value),nothing,something)

i.e. if the cell value is #N/A then return nothing; otherwise, return something. In this case, it’s the MATCH() function that is returning the #N/A, so we only need to use that as the in the above.

Our formula for B2 now becomes:

={IF(ISNA(MATCH(Output!$A2&Output!B$1,Data!$B$1:$B$8&Data!$A$1:$A$8,0)),””,INDEX(Data!$C$1:$C$8,MATCH(Output!$A2&Output!B$1,Data!$B$1:$B$8&Data!$A$1:$A$8,0),1))}

(without the curly brackets – these are put in the cells formula after pressing Shift + Enter.

As you can see, this formula is ridiculously long. Named ranges will improve this. For instance defining:

dataID => Data!$A$1:$A$8

dataName => Data!$B$1:$B$8
dataResult => Data!$C$1:$C$8

reduces the formula to:

={IF(ISNA(MATCH($A2&B$1,dataName&dataID,0)),””,INDEX(dataResult,MATCH($A2&B$1,dataName&dataID,0),1))}

Output

Hopefully this is what you were after

Hi Arvin,

I’m not too sure what you’re after as the table has become distorted in Wordpress, making it difficult to work out what values are in B1 & A3.

Could you email me your spreadsheet to ryan@kirgs.com.

You can certainly use lookup functions across worksheets and you can also match multiple criteria. As with most things, there are many ways to go about doing this, but with a bit more information I can hopefully provide you with a suitable approach – perhaps one that uses INDEX() and MATCH() rather than VLOOKUP()

Ryan

SHEET2
A B C
1 name1 5
2 name2 6
3 name3 7
4 name4 8
5 name5 9
6 name6 10
7 name7 11
8 name8 12

Back to the drawing board

Your very articulate response is much appreciated and I look forward to trying the method out!

Thanks.

The idea is then that we do the weighings, and the results tell us the label of the odd ball. So if the weighings were: left pan heavy, right pan heavy, balance, then the odd ball is 120

However, since we don’t know whether the ball is heavy or light, each ball has to get two labels which are opposites of each other (so 120 would also have 210).

There are twenty-seven possible labels, but we leave out 000, 111, and 222. This leaves 24, which when paired up gives us exactly twelve sets. These sets have the property that the balls are split up 4-4-4 for each weighing.

What I’ve written isn’t a proof, but it should get you started. At the very least, it is an easy way to remember the solution to this problem.