Map sequence of integers into repeating sequence

Question

I have a continuous, ordered sequence of integers that start less than zero and go above zero; for instance ...,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7 . I need to map these into a repeating sequence 3,6,9,12 , with the initial condition that f(0)=3 . So for instance f(-2)=9, f(-1)=12, f(0)=3, f(1)=6, f(2)=9, f(3)=12, f(4)=3 and so on. Is there a compact way to express this operation in Java, so that we are able to just use the remainder (%) operator and avoid any if/else statements?

Answer 1

This would server your purpose:

    int repeatInt =  12 - ((11 - seqInt)*3) % 12 ;

where repeatInt = 3,9,6 or 12, the output 
  and seqInt = a number from your sequence

( forgive me for any syntax mistakes, I tried the code in PHP, and changed it to Java )

Answer 2

int repeatingSequence[] = { 3,6,9,12 };
int fZero = 0;
int len = repeatingSequence.length;

public int f(int n) {
  return repeatingSequence[ (n % len + fZero + len) % len ];
}

Works for positives, negatives, and every fZero as the initial mapping condition.

Answer 3

I would suggest the following to handle negative numbers.

public int f(int n, int ... sequence) {
  int idx = n % sequence.length;
  if (idx < 0) idx += sequence.length;  
  return sequence[idx];
}


int x = f(n, 3,6,9,12);

however in this specific example there is a much simpler solution ;)

public int f(int n) {
    return (n & 3) * 3 + 3;
}