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An array of combinations is generated from two arrays,

               indices:  0   1   2   3   4   5
               array 1: [a,  b]      
               array 2: [A,  B,  C]
     repeating array 1:  a,  b,  a,  b,  a,  b      
     repeating array 2:   A,  B,  C,  A,  B,  C      
          combinations: [aA, bB, aC, bA, aB, bC]    length = LCM(2, 3)

INPUT : a, C
OUTPUT: 2, which the index of aC in the generated combinations.

Say, when a and C are given whose indices are already known as 0 and 2 in the original arrays, how to find the index of aC, which is 2? Is there any algorithmic way to do this?

For detailed background information, please check version 1 of this question.

John L.
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Jin Kwon
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2 Answers2

2

(BTW, I think the code in the first edition of the question was the issue -- not the background info with respect to the sexagenary cycle, which is actually good to include to understand the motivation for the problem).

To put the problem a bit differently, you have the sets $\mathbb{Z}/m\mathbb{Z}$ = $\{0,1,...,m-1\}$ and $\mathbb{Z}/n\mathbb{Z} = \{0,1,...,n-1\}$, and you start with the pair (0,0), then (1,1), etc., basically applying the mapping $(x,y)\mapsto(x+1 \mod m,y+1 \mod n)$, until you wrap back around to $(0,0)$. In the easy case where $m=n$ then, you can see that $(i,i)$ was the $i$-th term in the sequence (0-based). In general, if you are asking about the index of $(i,j)$, then you are effectively trying to solve the system of congruences \begin{align*} x &\equiv i \mod m \\ x &\equiv j \mod n \end{align*} for $x$. Here $x$ represents how many times you did the $(+1,+1)$ operation starting with $(0,0)$.

When $m$ and $n$ are coprime, the Chinese remainder theorem gives you an algorithm to compute the solution. When they are not coprime, then that system in general may not have a solution, but in the case where the system was created by the above construction, it does. This math.stackexchange answer shows you how to find the (unique) solution in the range $[0,LCM(m,n))$.

Matthew C
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Yes, there is. Because everything is given, you only want to solve a specific instance of a more general problem. The solution is:

return 2;
catalyst
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