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In psychology we sometimes use balanced latin squares for the order of our tests to counterbalance first-order carry-over effects (fatigue, learning, etc.) .

For our current study we want to pretest 12 stimuli (let's call them A-F) to see whether they're useful for a later study. We don't want to bore our participants, so we wanted to give them only half of all the material we need to test. We're indifferent about the size of the subset of 12 as long as it is anything between 4-8 stimuli per participant.

For a different reason (to achieve sufficient statistical power) we need at least 132 participants (at least 11 runs where each stimulus occurs first), we don't want to exceed this too heavily.

A balanced latin square 6*6 isn't too hard to construct. There is a Matlab script as well.

A   B   F   C   E   D
B   C   A   D   F   E
C   D   B   E   A   F
D   E   C   F   B   A
E   F   D   A   C   B
F   A   E   B   D   C

But is it also possible to construct a balanced (latin) rectangle (6 columns wide), where each letter is followed by another letter an equal amount of times? How many rows (participants) would this yield?

Maybe somebody with a bit more handle on this problem will enjoy the puzzle!

Sorry if my language is too idiosyncratic, if I can clarify with the appropriate jargon I'll duly comply, this is quite outside my field.

Splitting it in the middle and then adding the broken-up orders seemed the wrong approach to me.


Edit: Can I find one computationally? I have no idea how ridiculous that question is, but the sheer number of permutations (479 001 600) does seem daunting.


Edit 2: I didn't want to make this question too much about our experiment, but apparently that made it less clear. I'm sorry. I edited the clarifications into the question.

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This is actually a rather difficult problem to prove in Partition Theory, it's called the Wide Partition Conjecture. Essentially it states that if a partition, $\lambda$ is wide, then there exists a tableau of shape $\lambda$ that is Latin. I believe the conjecture to be true, and it's been proven for shapes up to 10x10. So for you, this means that if the rectangle is longer than it is wide, then there exists a Latin-ization of that tableau –  Nicolas Villanueva Jun 30 '11 at 14:49
    
Woops... I meant that if it's width is greater than it's height. –  Nicolas Villanueva Jun 30 '11 at 15:33
    
@Nic Hm, at least now I don't feel bad I couldn't do this off-the-cuff. I googled a little (mostly to understand the jargon) and found that it holds for all partitions having at most 65 boxes. I would have intuited that there would have to be more than 10 rows in a 6 times X balanced latin rectangle. I am a little confused now - my rectangle will have to be longer than wide, right? Does this mean a Latin tableau exists? –  Ruben Jul 1 '11 at 0:05
    
I don't understand the question. First of all, you write about numbers followed by other numbers, but I don't see any numbers - do you mean letters? Second, what does it mean for one number to follow another? Does "12 stimuli" mean you actually have 12 letters and you're trying to find a something-by-6 array such that each letter occurs equally often, no letter occurring twice in any row or column? Maybe you could give an example with 12 and 6 replaced by, say, 4 and 2 so we could see what you really want. –  Gerry Myerson Jul 1 '11 at 3:05

3 Answers 3

up vote 3 down vote accepted
+50

I need to change your parameters a bit, but it sounded like you would be flexible, so let me suggest the following idea.

The idea only works when the number of stimuli is a prime number $p$. So if you need to test exactly 12 stimuli, then this is useless, but may be you can leave one out, or add a placebo/null test to the mix, and use this idea with $p=11$ or $p=13$.

The scheme has $p(p-1)$ rows and $k$ columns, where $k$ is any number between $2$ and $p$ inclusive.

Instead of letters A,B,... I use numbers $0,1,\ldots,p-1$ as table entries. One of the standard constructions for Latin squares is the following: First pick a parameter $m$ that is an integer in the range $1\le m<p$. Then put on row #$i$ and column #$j$ the number that equals the remainder of $i+mj$ when divided by $p$. This gives us a $p\times p$ Latin square. Call it $L(m,p)$. In this Latin square all the pairs of consecutive entries on all rows differ by $m$. Therefore this square alone is the very opposite of balanced. Within this square a zero is always followed by an $m$, a one by $m+1$ et cetera. Note that we count cyclically $\pmod p$, so an entry $p-m$ is always followed by $(p-m)+m=p\equiv 0$, and so forth.

Here comes the trick. We build a larger table with $p(p-1)$ rows by putting all the Latin squares $L(1,p)$, $L(2,p)$, $\ldots$, $L(p-1,p)$ on top of each other. By using all the possible values of $m$ we get a balanced table in the end!

We can take the $k$ first columns of this large table, and be done with it. Each entry occurs on all the columns exactly $p-1$ times. If two distinct entries, say $a$ and $b$, differ by $m=b-a$, the pair $(a,b)$ appears in the rows of $L(m,p)$ exactly $k-1$ times - once per each pair of adjacent columns, and doesn't appear on any other rows. There are no repetitions of stimuli within rows, as the rows are parts of rows of a Latin square.

Why doesn't this work with $p=12$? IOW, why do I insist that $p$ must be a prime? The reason is that the formula $i+m*j \bmod p$ gives a Latin square only, when $m$ is coprime to $p$. For example, if $p=12$ and $m=6$, then the rows of $L(6,12)$ look like 0,6,0,6,...; 1,7,1,7,...

With $p=13$ you get 156 rows, so the table may be too large for you. Another possibly troubling feature of this construction is the following. The rows of the Latin square $L(1,p)$ look like 0,1,2,...; 1,2,3,...; so they have intersections of $k-1$ elements. This may be bad for eliminating secondary correlations from your test. If you do $k=6$ tests per participant, then five participants will see stimulus #1 followed by stimulus #2. That's ok, but I am a bit troubled by the fact that four out of those five will see stimulus #3 next. And the fifth person won't see anything, because his/her day ends after stimulus #2. Similar patterns appear in other component squares $L(m,p)$. If these shortcomings make the construction unusable, then I apologize for wasting your time.

[Edit: Oh boy, I should learn to proofread and not post in such haste. I apologize for the mostly illegible first version :-(]

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1  
Tis a nice construction. –  Willie Wong Jul 2 '11 at 15:52
    
+1 and will accept if no one solves it for 12 items. Thank you so much for your effort. I needed some time to get it (your edit helped!) and it sounds like a clever solution to the puzzle. I'll have to talk to the others about the possibility of changing the number of stimuli (especially when it increases the number of participants needed this might be problematic). By the way you say that m needs to be coprime to p, but because we use all possible values of m (e.g. 1-12) this just means that p needs to be a prime (stronger requirement)? –  Ruben Jul 4 '11 at 7:44
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@Ruben: Correct. That formula gives a $p\times p$ Latin square $L(m,p)$, whenever $m$ and $p$ are coprime. But to get a balanced design, we need to use all $m$ in the range $[1,p-1]$, and as you noticed, this forces $p$ to be a prime. –  Jyrki Lahtonen Jul 4 '11 at 8:48

If I understand the question correctly, you are seeking a $12 \times 6$ Latin rectangle (strictly speaking, this is the transpose of a Latin rectangle) in which each of 12 symbols (each representing a stimulus) occurs exactly once in each column (representing a round of experimentation) and at most once in each row (representing the participants). There is an additional condition that the Latin rectangle be balanced (i.e. each possible pair of symbols occurs in horizontally neighbouring cells).

This cannot be achieved since there will be exactly 6 copies of any symbol in the Latin rectangle, but there will be 11 other symbols. Simply put, they won't fit. The closest you could get is to take a balanced Latin square and chop off the last 6 columns (this is probably not suitable for your experiment).

A natural next step would be to modify the scheme so as to incorporate more rows (i.e. participants) and weaken the "Latin" property. I.e., several participants may receive the same stimulus at the same time. In such a modified scheme, we could balance the carry-over effects by ensuring that each ordered pair of symbols (A,B) appear in horizontally adjacent cells an equal number of times (as opposed to exactly once in the Latin square case). If you stick to exactly 6 rounds of testing, then each row contains exactly 5 ordered pairs. Unfortunately, there are $12 \times 11=132$ ordered pairs in total (which is coprime with 5), implying that the smallest balanced such scheme would require at least 132 participants.

If you're flexible on the number of rounds, if you had 7 rounds of testing, it's likely going to be possible to find such a scheme with only 22 participants (since there are 6 ordered pairs per row, and 132 is divisible by 6). That is, it seems likely there will be a $22 \times 7$ matrix containing the symbols $1,2,\ldots,12$ that is balanced (although actually finding one might be a bit tricky).

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Thank you! The columns don't represent rounds of experimentation, a row represents one (or many) run by a participant. The chopping off (what Gerry did) indeed yields the result, that some pairs of don't occur. We need 132 participants anyway (edited my answer to reflect this). We could of course let several participants use the same order (as we usually do with BLS). But if you say the 22x7 matrix will be hard to find, I guess the 12x11 one is even harder?! Thanks for pointing out that the latin property has to be relaxed. –  Ruben Jul 1 '11 at 9:03
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@Douglas: But with a 22 x 7 scheme you will have the problem that some stimuli will tested more frequently than others, as 22 x 7 is not divisible by 12? As they are testing with a larger number of people anyway (for any kind of statistical significance) 132 x 6 might be better, as all the stimuli might then (in theory at least) appear 11 times on each column. I admit that constructing such a beast (if at all possible) may be a lot of fun!! –  Jyrki Lahtonen Jul 1 '11 at 16:29
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@Jyrki: OTOH, if you can construct a $22 \times 7$ scheme, you can permute the symbols and run it 6 times to get it completely balanced. That's still 132 participants, but the rectangle may be easier to construct –  Willie Wong Jul 2 '11 at 14:59
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A good point, @Willie. As you see from my answer, the fact that 12 is not a prime may be problematic. Douglas has a larger database of LS-constructions in his head, so let's wait and see. –  Jyrki Lahtonen Jul 2 '11 at 15:22
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Hum. going back to the 132x6 idea, if it is possible to construct a 11x5 square of {1, ... , 11} such that each number appears 5 times, and in each row, consecutive strings of numbers never sum to 0 mod 12, then we'd also be done. –  Willie Wong Jul 2 '11 at 15:45

I don't understand the question, so this is probably not an answer, but maybe any objection will clarify the question.

A B D G K E  
B C E H L F  
C D F I A G  
D E G J B H  
E F H K C I  
F G I L D J  
G H J A E K  
H I K B F L  
I J L C G A  
J K A D H B  
K L B E I C  
L A C F J D  

There are 12 symbols in 6 columns, each symbol occurs exactly once in each column, no symbol appears twice in any row, no symbol appears with the same symbol immediately to its right more than once.

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Gerry: if you add four blanks in front of a line, this line is inserted in a code block. These look a bit nicer since the font used is monospaced (all characters have the same width). –  t.b. Jul 1 '11 at 7:48
    
+1 Thank you for the objection. I tried to clarify the question and I think Douglas nearly got what I meant. I came up with this chopped BLS myself, but here 6 pairs are missing (E-B, G-D, ...), so it's no longer balanced. –  Ruben Jul 1 '11 at 9:05
    
@Theo, thanks. I'll try (but probably fail) to remember that. –  Gerry Myerson Jul 1 '11 at 13:02
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@Ruben, OK, we're getting somewhere. I'll give it some more thought (but I suspect that if Douglas thinks it's hard, I'll get nowhere fast). –  Gerry Myerson Jul 1 '11 at 13:04

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