1	2	3	4
1
2	B	C	A	D
3	C	D	B	A
4	D	A	C	B

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What is it?

Wikipedia defines a latin square as "an n × n array filled with n different symbols, each occurring exactly once in each row and exactly once in each column.". Latin squares are useful to reduce order-effects when designing experiments with multiple conditions. For example, in an experiment comparing a technique A vs B vs C, if all participants test A first, then B, then C, we might observe poor results for C not because C is worse than A or B, but because of participants' fatigue. Instead, if we order conditions based on a randomly generated latin square, we could obtain the following order.

By ordering conditions using a latin square, we reduce the order-effect since conditions appear the same number of times in each position. However, there is still an issue caused by the repeated sequences: B is often preceded by A, C by B, and A by C. This could cause a carry-over effect.

Balanced Latin Squares (the ones generated above) are special cases of latin square that remove immediate carry-over effects: A condition will precede another exactly once (or twice, if the number of conditions is odd). This page is a simple generator of balanced latin square. The generator uses the method proposed and mathematically proved by James V. Bradley in "Complete Counterbalancing of Immediate Sequential Effects in a Latin Square Design".

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Source code

The code of this generator is open source. Feel free to copy the following function to generate balanced latin squares directly from your code.

Javascript

// How to use: // var conditions = ["A", "B", "C", "D"]; // balancedLatinSquare(conditions, 0) //=> ["A", "B", "D", "C"] // balancedLatinSquare(conditions, 1) //=> ["B", "C", "A", "D"] // balancedLatinSquare(conditions, 2) //=> ["C", "D", "B", "A"] // ... function balancedLatinSquare(array, participantId) { result = []; // Based on "Bradley, J. V. Complete counterbalancing of immediate sequential effects in a Latin square design. J. Amer. Statist. Ass.,.1958, 53, 525-528. " for (var i = 0, j = 0, h = 0; i < array.length; ++i) { var val = 0; if (i < 2 || i % 2 != 0) { val = j++; } else { val = array.length - h - 1; ++h; } var idx = (val + participantId) % array.length; result.push(array[idx]); } if (array.length % 2 != 0 && participantId % 2 != 0) { result = result.reverse(); } return result; }