Vector rotation "acrobatics" in "programming Pearl"

If the x length of the vector is L, the first N loops must be shifted to the left.

If the initial position to be moved is I, the acrobatics method is x [I] <-> x [I + N] <-> x [I + 2N]... <-> X [I + kN], when (I + kN) % L = I, it is not shifted, and shift x [I + kN] = x [I] at this time. If the maximum number of M-grams between L and N is d, the first N-bits can be shifted to the left after the d-shift is completed.

To prove the feasibility of this method, that is, to prove whether this method has completed the loop of all BITs to shift N bits left.

1. If the current BIT is set to I, the (I + N) % L shifts the loop left to the I bit. Here, the vector can be understood as a ring connected to the beginning and end, see Figure 1)

2. How can I prove that a total of 1 BITs have been moved?

Proof: after d-shift, the first N-bit cycle can be shifted to the left, considering how many bits are moved each time. If L = α d and N = β d are set, I + kN = mL + I is set because each shift ends at (I + kN) % L = I, then k β d = m α d. Due to α and beta, k = α is the minimum value that makes the equation true, this proves that each shift moves the Alpha bit and ends. Only after the d-shift can all the L = α d-bits be moved to the left.

3. How can we prove that each shift will not affect the previously moved bits?

Proof: There are two proof cases

(1) In a single shift, there is no case that a single position is operated twice, that is, it proves that (I + kN) % L = I can occur twice in a shift. In 2, it has been proved that to make the equation true, at least α needs to be displaced, which proves that in a shift, there is no possibility of a second operation being operated.

(2) During multiple shifts, no operated bits are operated again. The proof is as follows:

Start from 0 and perform the first shift, that is, when I = 0, then x [0] Then x [0 + N], x [0 + N] Then x [(0 + 2N) % L]

...... X [(0 + (α-1) N) % L] ß x [0] In the first shift, it will not affect 1, 2, 3, 4 ...... D-1 and its multiples are proved as follows: Set kN % L = x... R, kN = xL + r, that is, k β d-x α d = r => (k β-x α) d = r

This proves that r can only be a multiple of d, not 1, 2, 3, 4 ...... D-1 and its multiples.

When I = I + 1 = 1 at the end of the first operation, the next cycle will be carried out. Likewise, it can be proved that the shift operation will not affect 0, 2, 3,

4 ...... D-1 and its multiples. When I = D-1 is executed, the shift ends, and the α d bit is removed.

End of proof till now

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Figure 1

Implementation Code 1:

int* vector_loop1(int* array,int len,int loop_num)  {        int i=0,j=0,temp;      int d = gcd(len,loop_num);      while(d)      {          temp = array[i];          while((i+loop_num)%len != j)          {              array[i] = array[(i+loop_num)%len];              i = (i+loop_num)%len;          }            array[i] = temp;          j++;          i=j;          d--;      }        return array;  }

Implementation Code 2

int* vector_loop2(int* array,int len,int loop_num)  {      int i=0,j=0,temp;      temp = array[i];      for(int k=0;k<len;k++)     {          if((i+loop_num)%len == j)          {              array[i] = temp;              j++;             i=j;          }          else          {             array[i] = array[(i+loop_num)%len];              i = (i+loop_num)%len;        }      }     return array; }

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