Multiple Memory Structure Bit Reversal Algorithm Based on Recursive Patterns of Bit Reversal Permutation

Table 1

Relation between elements of 16-element BRP (𝑛=16) for split = 1 and split = 2.

Normal order	Reverse order	Split (𝑆) = 1				Split (𝑆=2)
		Blocks (1,…,2^𝑠)	Index of the block 𝑖	Reverse order calculation	Calculation method; 𝑚=0,…,𝑠−1(𝑚=0); 𝑗=1,…,2^𝑚; 𝑚=0: 𝑗=1; 𝑖=1,…,𝑛/2^𝑠−1; 𝑖=1,…,7	Blocks (1,…,2^𝑠)	Index of the block 𝑖	Reverse order calculation	Calculation method; 𝑚=0,…,𝑠−1(𝑚=0,1); 𝑗=1,…,2^𝑚; 𝑚=0: 𝑗=1,𝑚=1: 𝑗=1,2; 𝑖=1,…,𝑛/2^𝑠−1; 𝑖=1,2,3
0	0	1	0	0	Initialized		0	0	Initialized
1	8		1	8 = 8 + 0	Use Elster’s Linear Bit Reversal method	1	1	8 = 8 + 0	Use Elster’s Linear Bit Reversal method
2	4		2	4 = 4 + 0			2	4 = 4 + 0
3	12		3	12 = 12 + 0			3	12 = 12 + 0
4	2		4	2 = 2 + 0			0	2	Initialized
5	10		5	10 = 10 + 0		2	1	10 = 8 + 2	𝐵(2^𝑚+𝑗)[𝑖]=𝐵(𝑗)[𝑖]+2^𝑠−𝑚−1
6	6		6	6 = 6 + 0			2	6 = 4 + 2	For 𝑚=0, 𝑗=1, 𝑖=1,2,3; 𝐵(2^0+1)[𝑖]=𝐵(1)[𝑖]+2^2−0−1
7	14		7	14 = 14 + 0			3	14 = 12 + 2	𝐵(2)[𝑖]=𝐵(1)[𝑖]+2
8	1	2	0	1	Initialized		0	1	Initialized
9	9		1	9 = 8 + 1	𝐵(2^𝑚+𝑗)[𝑖]=𝐵(𝑗)[𝑖]+2^𝑠−𝑚−1; For 𝑚=0,𝑗=1,𝑖=1,…,7; 𝐵(2^0+1)[𝑖]=𝐵(1)[𝑖]+2^2−0−1; 𝐵(2)[𝑖]=𝐵(1)[𝑖]+1	3	1	9 = 8 + 1	𝐵(2^𝑚+𝑗)[𝑖]=𝐵(𝑗)[𝑖]+2^𝑠−𝑚−1
10	5		2	5 = 4 + 1			2	5 = 4 + 1	For 𝑚=1,𝑗=1,𝑖=1,2,3; 𝐵(2^1+1)[𝑖]=𝐵(1)[𝑖]+2^2−1−1
11	13		3	13 = 12 + 1			3	13 = 12 + 1	𝐵(3)[𝑖]=𝐵(1)[𝑖]+1
12	3		4	3 = 2 + 1			0	3	Initialized
13	11		5	11 = 10 + 1		4	1	11 = 10 + 1	𝐵(2^𝑚+𝑗)[𝑖]=𝐵(𝑗)[𝑖]+2^𝑠−𝑚−1
14	7		6	7 = 6 + 1			2	7 = 6 + 1	For 𝑚=1,𝑗=2,𝑖=1,2,3; 𝐵(2^1+2)[𝑖]=𝐵(2)[𝑖]+2^2−1−1
15	15		7	15 = 14 + 1			3	15 = 14 + 1	𝐵(4)[𝑖]=𝐵(2)[𝑖]+2