Consider continuous tent map
f_{c}(x)
f(x) = 2 - 2x = 1.111... - 1.b_{2 }b_{3 }b_{4 } ... = 0.u_{2 }u_{3 }u_{4 } ... . where u_{k} = 1 - b_{k} is inversion of the bit b_{k}. Thus after the left shift the upper bit is truncated again but if it is 1 then all the rest bits are inverted. |
0 | . | p_{1} p_{2} p_{3} p_{4} ... |
s_{1}= p_{1} | . | (p_{1 }p_{2 }) (p_{1 }p_{3 }) (p_{1 }p_{4 }) ... |
s_{2}= p_{1 }p_{2} | . | (p_{1 }p_{2 }p_{1 }p_{3 }) (p_{1 }p_{2 }p_{1 }p_{4 }) ... or taking into account that p_{k}^{2} = 1 |
. | (p_{2 }p_{3 }) (p_{2 }p_{4 }) ... | |
s_{3}= p_{2 }p_{3} | . | (p_{3 }p_{4 }) (p_{3 }p_{5 }) ... |
... | ||
s_{n}= p_{n-1 }p_{n} | . | (p_{n }p_{n+1 }) (p_{n }p_{n+2 }) ... |
b_{n} = | { |
b_{n-1}
1 - b_{n-1} |
if s_{n} = 0
if s_{n} = 1 |