Memasukkan

Integer a1, a2, a3, b1, b2, b3 masing-masing dalam kisaran 1 hingga 20.

Keluaran

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

^ adalah eksponensial dalam pertanyaan ini.

Aturan

Ini adalah kode-golf. Kode Anda harus berakhir dengan benar dalam waktu 10 detik untuk setiap input yang valid pada PC desktop standar.

Anda dapat menampilkan apa pun yang Sejati untuk Benar dan apa pun yang Falsey untuk False.

Anda dapat mengasumsikan setiap input yang Anda suka asalkan ditentukan dalam jawaban dan selalu sama.

Untuk pertanyaan ini, kode Anda harus selalu benar. Itu seharusnya tidak gagal karena ketidakakuratan floating point. Karena jangkauan input yang terbatas ini seharusnya tidak terlalu sulit untuk dicapai.

Uji kasus

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

Perl 6 , 31 29 byte

-2 byte terima kasih kepada Grimy

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

Cobalah online!

Percaya atau tidak, ini bukan esolang, bahkan jika itu sebagian besar terdiri dari tanda bintang. Ini menggunakan rumus Arnauld , dengan log10 bukan ln.

R , 39 byte

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

Cobalah online!

Mengembalikan FALSE kapan a > bdan BENAR jikab < a

05AB1E , 11 9 11 7 byte

.²Šm*`›

Pelabuhan @Arnauld 's JavaScript dan @digEmAll ' s R pendekatan (aku melihat mereka mengirim sekitar waktu yang sama)
-2 byte berkat @Emigna
2 byte sebagai bug-fix setelah @Arnauld 's dan @digEmAll jawaban' s terkandung kesalahan
-4 byte sekarang karena urutan input yang berbeda diizinkan setelah komentar @LuisMendo

Input [a1,b1], [a3,b3], [a2,b2]tiga dipisahkan input.

Cobalah secara online atau verifikasi semua kasus uji .

Penjelasan:

.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)

Java (JDK) , 56 byte

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

Cobalah online!

Kredit

• @ Ørjan Johansen karena menemukan bug dalam solusi saya.
• Disimpan 10 byte dengan menggunakan kembali operan menguntungkan @ tsh .

Bahasa Wolfram (Mathematica) , 23 byte

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

Cobalah online!

J , ¹¹ 9 byte

>&(^.@^/)

Cobalah online!

Argumen diberikan sebagai daftar.

• > apakah yang kiri lebih besar?
• &(...) tetapi pertama-tama, ubah setiap argumen dengan demikian:
• ^.@^/kurangi dari kanan ke kiri dengan eksponensial. Tetapi karena eksponensial biasa akan membatasi kesalahan bahkan untuk nomor yang diperluas, kami mengambil log dari kedua sisi

Clean, 44 bytes

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

Try it online!

Uses an adaptation of Arnauld's formula.

Python 3, 68 bytes

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

Try it online!

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

05AB1E, 13 bytes

Uses the method from Arnauld's JS answer

2F.²IIm*ˆ}¯`›

Try it online!

Excel, 28 bytes

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

Excel implementation of the same formula already used.

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.

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.

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);}

Try it online!

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

Jelly, 8 bytes

l⁵×*/}>/

Try it online!

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!

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.

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.