Which big number is bigger?

Question

Input

Integers a1, a2, a3, b1, b2, b3 each in the range 1 to 20.

Output

True if a1^(a2^a3) > b1^(b2^b3) and False otherwise.

^ is exponentiation in this question.

Rules

This is code-golf. Your code must terminate correctly within 10 seconds for any valid input on a standard desktop PC.

You can output anything Truthy for True and anything Falsey for False.

You can assume any input order you like as long as its specified in the answer and always the same.

For this question your code should always be correct. That is it should not fail because of floating point inaccuracies. Due to the limited range of the input this should not be too hard to achieve.

Test cases

3^(4^5) > 5^(4^3)
1^(2^3) < 3^(2^1)
3^(6^5) < 5^(20^3)
20^(20^20) > 20^(20^19)
20^(20^20) == 20^(20^20)
2^2^20 > 2^20^2
2^3^12 == 8^3^11
1^20^20 == 1^1^1
1^1^1 == 1^20^20

Answer 1

Perl 6, 31 29 bytes

-2 bytes thanks to Grimy

*.log10* * ***>*.log10* * ***

Try it online!

Believe it or not, this is not an esolang, even if it is composed of mostly asterisks. This uses Arnauld's formula, with log10 instead of ln.

Answer 2

R, 39 bytes

function(x,y,z)rank(log2(x)*(y^z))[1]<2

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Returns FALSE when a > b and TRUE if b < a

Answer 3

05AB1E, 11 9 11 7 bytes

.²Šm*`›

Port of @Arnauld's JavaScript and @digEmAll's R approaches (I saw them post around the same time)
-2 bytes thanks to @Emigna
+2 bytes as bug-fix after @Arnauld's and @digEmAll's answers contained an error
-4 bytes now that a different input order is allowed after @LuisMendo's comments

Input as [a1,b1], [a3,b3], [a2,b2] as three separated inputs.

Try it online or verify all test cases.

Explanation:

.²       # Take the logarithm with base 2 of the implicit [a1,b1]-input
  Š      # Triple-swap a,b,c to c,a,b with the implicit inputs
         #  The stack order is now: [log2(a1),log2(b1)], [a2,b2], [a3,b3]
   m     # Take the power, resulting in [a2**a3,b2**b3]
    *    # Multiply it with the log2-list, resulting in [log2(a1)*a2**a3,log2(b1)*b2**b3]
     `   # Push both values separated to the stack
      ›  # And check if log2(a1)*a2**a3 is larger than log2(b1)*b2**b3
         # (after which the result is output implicitly)

Answer 4

Java (JDK), 56 bytes

(a,b,c,d,e,f)->a>Math.pow(d,Math.pow(e,f)/Math.pow(b,c))

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Credits

• @Ørjan Johansen for finding a bug in my solution.
• Saved 10 bytes by reusing @tsh's advantageous operands agencing.

Answer 5

Wolfram Language (Mathematica), 23 bytes

#2^#3Log@#>#5^#6Log@#4&

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Answer 6

J, ¹¹ 9 bytes

>&(^.@^/)

Try it online!

Arguments given as lists.

• > is the left one bigger?
• &(...) but first, transform each argument thusly:
• ^.@^/ reduce it from the right to the left with exponention. But because ordinary exponentiation will limit error even for extended numbers, we take the logs of both sides

Answer 7

Python 3, 68 bytes

lambda a,b,c,d,e,f:log(a,2)*(b**c)>log(d,2)*(e**f)
from math import*

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Port of @Arnualds answer, but with the base for log changed.

Answer 8

Clean, 44 bytes

import StdEnv
$a b c d e f=b^c/e^f>ln d/ln a

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Uses an adaptation of Arnauld's formula.

Answer 9

05AB1E, 13 bytes

Uses the method from Arnauld's JS answer

2F.²IIm*ˆ}¯`›

Try it online!

Answer 10

Excel, 28 bytes

=B1^C1*LOG(A1)>E1^F1*LOG(D1)

Excel implementation of the same formula already used.

Answer 11

JavaScript, 51 bytes

f=(a,b,c,h,i,j)=>(l=Math.log)(a)*b**c-l(h)*i**j>1e-8

Surprisingly, the test cases doesn't show any floating-point error. I don't know if it ever does at this size.

This just compares the logarithm of the numbers.

Equality tolerance is equal to 1e-8.

Answer 12

TI-BASIC, 27 31 bytes

ln(Ans(1))Ans(2)^Ans(3)>Ans(5)^Ans(6)(ln(Ans(4

Input is a list of length \$6\$ in Ans.
Outputs true if the first big number is greater than the second big number. Outputs false otherwise.

Examples:

{3,4,5,5,4,3
   {3 4 5 5 4 3}
prgmCDGF16
               1
{20,20,20,20,20,19       ;these two lines go off-screen
{20 20 20 20 20 19}
prgmCDGF16
               1
{3,6,5,5,20,3
  {3 6 5 5 20 3}
prgmCDGF16
               0

Explanation:

ln(Ans(1))Ans(2)^Ans(3)>Ans(5)^Ans(6)(ln(Ans(4   ;full program
                                                 ;elements of input denoted as:
                                                 ; {#1 #2 #3 #4 #5 #6}

ln(Ans(1))Ans(2)^Ans(3)                          ;calculate ln(#1)*(#2^#3)
                        Ans(5)^Ans(6)(ln(Ans(4   ;calculate (#5^#6)*ln(#4)
                       >                         ;is the first result greater than the
                                                 ; second result?
                                                 ; leave answer in "Ans"
                                                 ;implicit print of "Ans"

---

Note: TI-BASIC is a tokenized language. Character count does not equal byte count.

Answer 13

Jelly, 8 bytes

l⁵×*/}>/

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Based on Arnauld’s JS answer. Expects as input [a1, b1] as left argument and [[a2, b2], [a3, b3]] as right argument.

Now changed to use log to the base 10 which as far as correctly handles all the possible inputs in the range specified. Thanks to Ørjan Johansen for finding the original problem!

Answer 14

bc -l, 47 bytes

l(read())*read()^read()>l(read())*read()^read()

with the input read from STDIN, one integer per line.

bc is pretty fast; it handles a=b=c=d=e=f=1,000,000 in a little over a second on my laptop.

Answer 15

C++ (gcc), 86 bytes

Thanks to @ØrjanJohansen for pointing out a flaw in this and @Ourous for giving a fix.

#import<cmath>
int a(int i[]){return pow(i[1],i[2])/pow(i[4],i[5])>log(i[3])/log(*i);}

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Takes input as a 6-integer array. Returns 1 if \$a^{b^c} > d^{e^f}\$, 0 otherwise.

Answer 16

APL(NARS), chars 36, bytes 72

{>/{(a b c)←⍵⋄a=1:¯1⋄(⍟⍟a)+c×⍟b}¨⍺⍵}

Here below the function z in (a b c )z(x y t) would return 1 if a^(b^c)>x^(y^t) else would return 0; test

  z←{>/{(a b c)←⍵⋄a=1:¯1⋄(⍟⍟a)+c×⍟b}¨⍺⍵}
  3 4 5 z 5 4 3
1
  1 2 3 z 3 2 1
0
  3 6 5 z 5 20 3
0
  20 20 20 z 20 20 19
1
  20 20 20 z 20 20 20
0
  2 2 20 z 2 20 2
1
  2 3 12 z 8 3 11
0
  1 20 20 z 1 1 1
0
  1 1 1 z 1 20 20
0
  1 4 5 z 2 1 1
0

{(a b c)←⍵⋄a=1:¯1⋄(⍟⍟a)+c×⍟b} is the function p(a,b,c)=log(log(a))+c*log(b)=log(log(a^b^c)) and if aa=a^(b^c) with a,b,c >0 and a>1 bb=x^(y^t) with x,y,t >0 and x>1 than

aa>bb <=> log(log(a^b^c))>log(log(x^y^t))  <=>  p(a,b,c)>p(x,y,t)

There is a problem with the function p: When a is 1, log log 1 not exist so I choose to represent that with the number -1; when a=2 so log log a is a negative number but > -1 .

PS. Seen the function in its bigger set in which is defined

p(a,b,c)=log(log(a))+c*log(b)

appear range for a,b,c in 1..20 is too few... If one see when it overflow with log base 10, the range for a,b,c could be 1..10000000 or bigger for a 64 bit float type.