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Consider continuous tent map
fc(x)
f(x) = 2 - 2x = 1.111... - 1.b2 b3 b4 ... = 0.u2 u3 u4 ... . where uk = 1 - bk is inversion of the bit bk. Thus after the left shift the upper bit is truncated again but if it is 1 then all the rest bits are inverted. |
| 0 | . | p1 p2 p3 p4 ... |
| s1= p1 | . | (p1 p2 ) (p1 p3 ) (p1 p4 ) ... |
| s2= p1 p2 | . | (p1 p2 p1 p3 ) (p1 p2 p1 p4 ) ... or taking into account that pk2 = 1 |
| . | (p2 p3 ) (p2 p4 ) ... | |
| s3= p2 p3 | . | (p3 p4 ) (p3 p5 ) ... |
| ... | ||
| sn= pn-1 pn | . | (pn pn+1 ) (pn pn+2 ) ... |
| bn = | { |
bn-1
1 - bn-1 |
if sn = 0
if sn = 1 |