In problem $$$\overbrace{(((a ^ n) ^ {{}^n}) ^ {{}^{{}^{...}}})}^{b\ times\ n} \mod m$$$
My approach is calculating each part $$$((a ^ n) \mod m)$$$ then $$$(((a ^ n) ^ n) \mod m) = ((t ^ n) \mod m)$$$ ... all together in $$$O(\log n)$$$ each part and $$$O(b \times \log n)$$$ total time complexity
Can I improve it somehow faster where $$$b$$$ is large ($$$b \leq 10^{16}$$$) and ($$$m$$$ can be composite number)
In problem $$$\overbrace{a ^ {(n ^ {(n ^ {(...)})})}}^{b\ times\ n} \mod m$$$
I only know the way to use bignum to calculate $$$n ^ n$$$ then $$$n ^ {n ^ n}$$$ all together then I just have to calculate the modulo of $$$a ^ {...} \mod m$$$ but the total complexity will be very huge (since the number of digits of bignum will raise very fast)
Can I apply these formula from phi-function:
- Can I improve it somehow faster where $$$n$$$ is large ($$$n \leq 10^{16}$$$) and ($$$m$$$ can be composite number)