Comparing powers.

Question

Compare two numbers N₁ = a^bc, N₂ = d^ef by constructing a function f(a,b,c,d,e,f) that:

• returns 1 if N₁ > N₂
• returns -1 if N₁ < N₂

Note: You are not required to return any value for any other relation between N₁ and N₂. e.g. when they are equal or when their relation is undefined (complex numbers).

other constrains:

• all numbers are integers
• a,b,c,d,e,f may be positive or negative but not zero.
• |a|,|d| < 1000
• |b|,|c|,|e|,|f| < 10¹⁰
• running time less than few seconds

Examples:

f(100,100,100,50,100,100) = 1
f(-100,100,100,50,100,100) = 1
f(-100,99,100,50,100,100) = -1
f(100,-100,-100, -1, 3, 100) = 1
f(535, 10^9, 10^8, 443, 10^9, 10^9) = -1

This is code golf. Shortest code wins.

Answer 1

Ruby 1.9, 280 227 189 171 characters

z=->a,b,c,d,e,f{l=->a{Math.log a}
u=->a,b{[a.abs,a][b&1]}
a=u[a,b=u[b,c]]
d=u[d,e=u[e,f]]
d*a<0?a<=>d :b*e<0?b<=>e :(l[l[a*q=a<=>0]/l[d*q]]<=>f*l[e*r=b<=>0]-c*l[b*r])*q*r}

I know this is a bit longer than the other solutions, but at least this approach should work without calculating a^bc, d^ef, b^c or e^f.

Edit:

• (279 -> 280) Fixed a bug when a**b**c < 0 and d = 1.
• (280 -> 227) Removed an unnecessary check for a special case.
• (227 -> 192) Removed some checks that aren't necessary with the given criteria (non-zero integers, no output necessary for complex values)
• (192 -> 189) Due to all the other checks, I can safely calculate log(log(a)/log(d)) instead of log(log(a))-log(log(d)).
• (189 -> 171) Simplified way to transform equivalent problems into one another.

Testcases:

z[100, 100, 100, 50, 100, 100] == 1
z[-100, 100, 100, 50, 100, 100] == 1
z[-100, 99, 100, 50, 100, 100] == -1
z[100, -100, -100, -1, 3, 100] == 1
z[535, 10**9, 10**8, 443, 10**9, 10**9] == -1
z[-1, -1, 1, 2, 2, 2] == -1
z[1, -5, -9, 2, -1, 2] == -1
z[1, -5, -9, 2, -1, 3] == 1
z[3, -3, 3, -4, 1, 1] == 1
z[-2, 1, 1, 1, 1, 1] == -1
z[1, 1, 1, -1, 1, 1] == 1
z[1, 1, 1, 2, 3, 1] == -1
z[1, 1, 1, 2, -3, 2] == -1
z[1, 1, 1, 2, -3, 1] == 1
z[-1, 1, 1, 1, 1, 1] == -1
z[2, 3, 1, 1, 1, 1] == 1
z[2, -3, 2, 1, 1, 1] == 1
z[2, -3, 1, 1, 1, 1] == -1

Answer 2

Mathematica, 110 chars

z[a_,b_,c_,d_,e_,f_]:=With[{g=Sign[a]^(b^c),h=Sign[d]^(e^f)},If[g!=h,g,g*Sign[Log[Abs[a]]b^c-Log[Abs[d]]e^f]]]

Answer 3

ShortScript, 89 bytes

{CP
$M^ η1 η2
$M^ ζ η3
↑Αζ
$M^ η4 η5
$M^ ζ η6
↔α>ζ↑Ζ1
↔α<ζ↑Ζ-1}

The implementation isn't exactly the described one, but it works.

This answer is non-competing, since ShortScript has been published after this challenge.

Answer 4

Python 2.6 (this doesn't actually work)

import cmath
g=cmath.log
f=lambda a,b,c,d,e,f:-1+2*((c*g(b)+g(g(a))-f*g(e)-g(g(d))).real>0)

today i learnt python has a complex log function. so, blindly double log both sides and look at the real component. works for 4 out of the 5 tests. not sure what's happening with the fourth one.

print f(100,100,100,50,100,100) == 1
print f(-100,100,100,50,100,100) == 1
print f(-100,99,100,50,100,100) == -1
print f(100,-100,-100, -1, 3, 100) == 1 # failure, sadness.
print f(535, 10^9, 10^8, 443, 10^9, 10^9) == -1

Answer 5

Python (99)

from math import*
from numpy import*
l=log
def f(a,b,c,d,e,f):return sign(l(a)*l(b)*c-l(d)*l(e)*f)

Answer 6

Haskell, 44 characters

n True=1
n _=1-2
g a b c d e f=n$a^b^c>d^e^f

Runs under a second for all the test examples on my machine.