Complexity class of comparison of power towers
Recently I asked a very similar question at Mathematica.SE. I assume you know it, because you participated in the discussion.
Leonid Shifrin suggested an algorithm that solves this problem for the majority of cases, although there were cases when it gave an incorrect answer. But his approach seems correct and it looks like it is possible to fix those defects. Although it was not rigorously proved, his algorithm seems to work in polynomial time. It looks like it would be fair if he got the bounty for this question, but for some reason he didn't want to.
So, this question is not yet settled completely and I am going to look for the complete and correct solution, and will start a new bounty for this question once the current one expires. I do not expect to get a bounty for this answer, but should you choose to award it, I will add it up to the amount of the new bounty so that it passes to whomever eventually solves this question.
For readability I'll write $[a_1,a_2,\ldots,a_n]$ for the tower $a_1^{a_2^{a_3^\cdots}}$.
Let all of the $a_i,b_i$ be in the interval $[2,N]$ where $N=2^K$ (if any $a_i$ or $b_i$ is 1 we can truncate the tower at the previous level, and the inputs must be bounded to talk about complexity).
Then consider two towers of the same height $$ T=[N,N,\ldots,N,x] \quad \mathrm{and} \quad S=[2,2,\dots,2,y] $$ i.e. T is the largest tower in our input range with $x$ at the top, and S is the smallest with $y$ at the top.
With $N, x\ge 2$ and $y>2Kx$ then $$ \begin{aligned} 2^y & > 2^{2Kx} \\ & = N^{2x} \\ & > 2log(N)N^x &\text{ as $x \gt 1 \ge \frac{1+log(log(N))}{log(N)}$} \\ & = 2KN^x \end{aligned} $$ Now write $x'=N^x$ and $y'=2^y>2Kx'$ then $$ [N,N,x]=[N,x']<[2,y']=[2,2,y] $$ Hence by induction $T<S$ when $y>2Kx$.
So we only need to calculate the exponents down from the top until one exceeds the other by a factor of $2K$, then that tower is greater no matter what values fill in the lower ranks.
If the towers have different heights, wlog assume $n>m$, then first we reduce $$ [a_1,a_2,\ldots,a_n] = [a_1,a_2,\ldots,a_{m-1},A] $$ where $A=[a_m,a_{m+1},\ldots,a_n]$. If we can determine that $A>2Kb_m$ then the $a$ tower is larger.
If the towers match on several of the highest exponents, then we can reduce the need for large computations with a shortcut. Assume $n=m$, that $a_j> b_j$ for some $j<m$ and $a_i=b_i$ for $j<i\le m$. Then $$ [a_1,a_2,\ldots,a_m] = [a_1,a_2,\ldots,a_j,X] \\ [b_1,b_2,\ldots,b_m] = [b_1,b_2,\ldots,b_j,X] $$ and the $a$ tower is larger if $(a_j/b_j)^X>2K$. So we don't need to compute $X$ fully if we can determine that it exceeds $\log(2K)/\log(a_j/b_j)$.
These checks need to be combined with a numeric method like the one @ThomasAhle gave. They can solve the problem that method has with deep trees that match at the top, but can't handle $[4,35,15],[20,57,13]$ which are too big to compute but don't allow for one of these shortcuts.
My approach is similarly numeric to Leonid's, but more precise and perhaps easier to analyse. It supports real exponents > 0.
The idea is to represent power towers as a single floating point number with $n$ exponentiations: $(x\mid n) := exp^n(x)$. Normalizing $x\in[0,1)$, this format allows easy comparison between numbers.
What remains is a way to calculate $a^{(x\mid n)}$ for any real, positive $a$. My Python code below is an attempt to do so, while being as numerically stable as possibly, e.g. by using the log-sum trick. My code runs in time proportional to the height of the tower (number of apow calls) and the iterated-log of it's value (number of recursive calls).
I haven't been able to find two towers with values close enough to case my method to fail. At least for integer exponents. With fractional exponents it is possible to create very towers too close for my representation to handle. E.g. $$ 2^{2^{2^{2^0}}} \\ < \\ 2^{2^{2^{2^{(1/2)^{2^{2^{2^2}}}}}}} $$
I would be interested in suggestions to other types of counter examples, especially integer ones.
It seems to me that for the problem to be in P, we need to non-numerical methods. It doesn't seem unlikely at all, that certain analytical cases are harder than P.
from math import log, exp
def normalize((x,n)):
""" Adjusts n to put x in the range [0,1) """
if x >= 1: return normalize((log(x),n+1))
if x < 0: return normalize((exp(x),n-1))
return (x,n)
normdec = lambda f: lambda *a: normalize(f(*a))
@normdec
def apow(a,(x,n)):
""" Calculates a^(x|n) """
if a == 1: return (1,0)
if a < 1: return (rpow(x,n)*log(a), 1)
if n == 0: return (x*log(a), 1)
if n >= 1:
y, k = cpow(log(log(a)), x, n-1)
return (y, k+2)
@normdec
def cpow(c,x,n):
""" Calculates (x|n) + c """
if c == 0: return (x, n)
if n == 0: return (x + c, 0)
z = rpow(x,n-1)
if z <= log(abs(c)):
return (exp(z)+c, 0)
if c < 0: y, k = cpow(log(1-exp(log(-c)-z)), x, n-1)
if c > 0: y, k = cpow(log(1+exp(log(c)-z)), x, n-1)
return (y, k+1)
def rpow(x,n):
""" Calculates (x|n) as a float
Returs Infinty if the value is out of range"""
try:
for _ in range(n):
x = exp(x)
except OverflowError:
# We get into this case in two situations
# 1) We are calculating log((x|n) + c) and c contributes very little
# 2) We are calculating a^(x|n) for a < 1 and (x|n) is so small it doesn't
# fit into a float
return float('inf')
return x
def powtow(bs):
""" Calculates b[0]**b[1]**b[2]**...**b[m-1] in the (x|n) form.
Equivalent to `foldr apow (1,0) bs'
e.g. apow(b[0], apow(b[1], apow(b[2], (1,0)))) """
if not bs: return (1,0)
return apow(bs[0], powtow(bs[1:]))
if __name__ == "__main__":
print powtow([1,2,3,4,5])
print powtow([2,3,4,5])
print powtow([5,4,3,2])
print powtow([4,4,3,3,3])
print powtow([3,3,3,3,3])
print powtow([4,6,8,8,9])
print powtow([2,2,5,2,7,4,9,3,7,6,9,9,9,9,3,2])
print powtow([3,3,6,3,9,4,2,3,2,2,2,2,2,3,3,3])
print powtow([2,3,2,3,5,8])
print powtow([3,2,2,7,6,7])
print powtow([2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,2,2,2])
print powtow([2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,16])
print powtow([9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9])
print powtow([1.54090919967, 1.46228204461, 1.78555495826, 1.75545819035, 2.21730941808, 1.0797564499, 7.90630125423, 0.881978093585, 1.75085618709, 2.23911325176, 1.39697337886, 1.16053659586, 1.5939192079, 6.11401961748, 0.844860266481, 1.92758094038, 4.64573316954, 0.870819420274, 1.49026447511, 1.77839910981, 1.46208378213, 2.29956158055, 1.00884903003, 1.77521724246, 2])
print powtow([2.32185478602, 1.88198918762, 2.27614315145, 1.77518235487, 1.4841479727, 0.563158971798, 0.732132856919, 0.669957968262, 2.16345101714, 2.23185963501, 0.824885385628, 0.873101580546, 1.45714899023, 2.3973000247, 0.507154709525, 1.94022843601, 1.29982267606, 0.578058713016, 1.58207655843, 1.79417433851, 1.18630377782, 1.37314328673, 0.655551609076, 1.57569812897, 1])
print powtow([2, 2, 2, 1])
print powtow([2, 2, 2, 2, .5, 2, 2, 2, 2])
flip = lambda (a,b): (b,a)
snd = lambda (a,b): b
import itertools
f = lambda xs: [1./x for x in xs]
# Print all power towers of permutations [2..6], sorted
print list(reversed(sorted((flip(powtow(p)),p) for p in itertools.permutations(range(2,7)))))
# Print all power towers of permutations [1/2..1/6] sorted
print list(reversed(sorted((flip(powtow(f(p))),p) for p in itertools.permutations(range(2,7)))))
Examples:
powtow([2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,2,2,2]) = (0.1184590219613409, 18)
powtow([9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9]) = (0.10111176550354063, 18)
powtow([2,2,5,2,7,4,9,3,7,6,9,9,9,9,3,2]) = (0.10111176550354042, 17)
powtow([3,3,6,3,9,4,2,3,2,2,2,2,2,3,3,3]) = (0.19648862015624008, 17)
Counter examples:
powtow([2,2,2,2,2,2,2]) = (0.8639310719129168, 6)
powtow([3,2,2,2,2,2,2]) = (0.8639310719129168, 6)