![]() ![]() In the brief period while all are changing, the switches will read some spurious position. In the transition between the two states shown above, all three switches change state. The problem with natural binary codes is that physical switches are not ideal: it is very unlikely that physical switches will change states exactly in synchrony. If that device uses natural binary codes, positions 3 and 4 are next to each other but all three bits of the binary representation differ: ![]() Many devices indicate position by closing and opening switches. The use of Gray code in these devices helps simplify logic operations and reduce errors in practice. Gray codes are widely used to prevent spurious output from electromechanical switches and to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems. That way, incrementing a value from 1 to 2 requires only one bit to change, instead of two. In Gray code, these values are represented as " 001" and " 011". The reflected binary code ( RBC), also known as reflected binary ( RB) or Gray code after Frank Gray, is an ordering of the binary numeral system such that two successive values differ in only one bit (binary digit).įor example, the representation of the decimal value "1" in binary would normally be " 001" and "2" would be " 010". ![]() Thus, for our given situation, we can only obtain 9 unique sets of groups.Ordering of binary values, used for positioning and error correction Given M initial players, there are M - 1 unique rounds. One simply fixes a position in a matrix and rotates the other indices clockwise. From the wikipedia page, the algorithm is straightforward. That solution mentions Round-robin tournament. I see that the OP has provided a solution from the linked math.so solution, but I would like to provide a working solution of the other answer on that page that gets to the heart of this problem. Unless I'm failing to see how they can be applied, I don't think any of these solutions quite solves this problem. For example, in the case of my class, I want to split the 10 students into 5 pairs of 2 on 13 different occasions, and I want pairs to be unique until they no longer can be (i.e., after 9 occasions have passed). But my problem is actually that I want to split the set into M/N subsets of size N. The solutions suggested below work well when I only want to split the set into a single subset of size N, each time. I realize my original question wasn't exactly clear. I don't understand Python, and so I'm looking for an easy R solution.)Įdit: Thanks to all for suggestions. (Someone seems to have posed this same question, but. How can I do this? I started by, but the trouble I am having is that these unique permutations don't necessarily yield unique groups. (The actual problem here: I have a class of 10 people and I want to split them into 5 pairs for a duration of 13 weeks, but I don't want anyone to be in a repeat pairing until they have been in a pairing with everyone in the class.) I'd like no element in the large set to be in a repeating group until they have been in a small group with every one else. I have a large set of size M (let's say 10), and I want to, repeatedly for a certain number of occasions (let's say 13), randomly split it into M/N smaller groups of size N (let's say 2). ![]()
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