Niech jest dodatnia składający się z n cyfry dziesiętne d 1 , d 2 , . . . , d n . Pozwolić$A$$n$$d_1,d_2,...,d_n$$B$ będzie kolejną dodatnią liczbą całkowitą.

Dla celów tego wyzwania, nazywamy się naśladowcę z B , jeśli istnieje co najmniej jedną listę liczb całkowitych dodatnich p 1 , p 2 , . . . , p n takie, że:$A$$B$$p_1,p_2,...,p_n$

$A$ i$B$ nazywane sąwzajemnyminaśladowcami,jeśli$A$ jest naśladowcą$B$ a$B$ jest naśladowcą$A$ .

Przykład

$526$ i$853$ są wzajemnymi naśladowcami, ponieważ:

i:

Wyzwanie

Biorąc pod uwagę dwie dodatnie liczby całkowite $A$ i $B$ , Twoim zadaniem jest wydrukowanie lub zwrócenie prawdziwej wartości, jeśli $A$ i $B$ są naśladowcami lub wartością fałszowania w przeciwnym razie.

Wyjaśnienia i zasady

• Możesz wziąć $A$ i $B$ w dowolnym rozsądnym, jednoznacznym formacie (np. Liczby całkowite, ciągi znaków, listy cyfr, ...)
• $A$ i$B$ mogą być równe. Jeśli liczba jest wzajemną naśladowcą, należy doA007532.
• Zamiast wartości prawda / fałsz, możesz zwrócić dwie wyraźne konsekwencje wartości.
• Dla $1\le A<1000$ i $1\le B<1000$ kod musi zostać wypełniony w mniej niż minutę . Jeśli przyjęcie wyższych wartości zajmuje zbyt dużo czasu, musi jednak być w stanie je rozwiązać teoretycznie.
• To jest golf golfowy .

Przypadki testowe

Truthy:
1 1
12 33
22 64
8 512
23 737
89 89
222 592
526 853
946 961
7 2401
24 4224
3263 9734
86 79424
68995 59227
32028 695345

Falsy:
1 2
3 27
9 24
24 42
33 715
33 732
222 542
935 994
17 2401
8245 4153

Brachylog , 19 bajtów

ẹ{∧ℕ₁;?↔^}ᵐ².+ᵐ↔?∧≜

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Wyjścia true.lubfalse.

Wyjaśnienie

ẹ                     Split the numbers into lists of digits
 {       }ᵐ²          For each digit
  ∧ℕ₁                 Let I be a strictly positive integer
     ;?↔^                Compute the digit to the power I (which is unknown currently)
            .         Call . the list of those new numbers
            .+ᵐ       Their mapped sum results…
               ↔?     …in the reverse of the input
                 ∧≜   Find if there effectively are values for the numbers in . to satisfy
                        these relationships

Łuska , 17 bajtów

Λλ€⁰mΣΠTṪ^ḣ√⁰d)De

Wypróbuj online! Kończy wszystkie przypadki testowe poniżej 1000 w około 11 sekund.

Wyjaśnienie

Λλ€⁰mΣΠTṪ^ḣ√⁰d)De  Implicit inputs, say 12 and 33.
                e  Put into a list: [12,33]
               D   Duplicate: [12,33,12,33]
Λ                  Does this hold for all adjacent pairs:
                    (12,33 is checked twice but it doesn't matter)
                    For example, arguments are 33 and 12.
 λ            )     Anonymous function with arguments 33 (explicit) and 12 (implicit).
             d      Base-10 digits of implicit argument: [1,2]
          ḣ√⁰       Range to square root of explicit argument: [1,2,3,4]
        Ṫ^          Outer product with power: [[1,2],[1,4],[1,8],[1,16],[1,32]]
       T            Transpose: [[1,1,1,1,1],[2,4,8,16,32]]
      Π             Cartesian product: [[1,2],[1,4],...,[1,32]]
    mΣ              Map sum: [3,5,...,33]
  €⁰                Is the explicit argument in this list? Yes.

Dlaczego to działa?

$B = d_1^{p_1} + \cdots + d_n^{p_n}$$d_i$$p_i$$d_i^{p_i} \leq B$$i$$p_i \leq \log_{d_i} B$$d_i \leq 1$, since exponentiating $0$ or $1$ does not change it. In my program the search space is $1 \leq p_i \leq \sqrt{B}$ (to comply with the time restriction; I would use $1 \leq p_i \leq B$ otherwise), so if we have $\lfloor \log_{d_i} B \rfloor \leq \lfloor \sqrt{B} \rfloor$, then everything is fine. If $d_i \geq 3$, this holds for all natural numbers $B$, so the only dangerous case is $d_i = 2$. We have $\lfloor \log_2 B \rfloor > \lfloor \sqrt{B} \rfloor$ only for $B = 8$. In this case $2^3 = 8$, but the search only considers exponents $1$ and $2$. If the other number number $A$ contains the digit $2$, either it has other nonzero digits as well (so the exponent of $2$ cannot be $3$ in the sum), or $A = 2 \cdot 10^k$ for some $k$. In the latter case, $A$ is not a power of $8$, so it cannot be a copycat of $B$ anyway, and the program correctly returns a falsy value regardless of the other computation.

Python 2, 102 bytes

lambda a,b:g(a,b)*g(b,a)
g=lambda a,b,e=1:b==a<1or(b>0<=b-e>=0<a)and(g(a/10,b-(a%10)**e)or g(a,b,e+1))

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05AB1E, 26 22 bytes

εVтLIàgãεYSym}OIyKå}˜P

Takes the input as a list (i.e. [526,853]).

Try it online or verify most test cases in the range [1,999].

Similar as my old answer below, except that the [1,n] list is hardcoded to [1,100], and it creates the cartesian list twice, once for each input-mapping, which is the main bottleneck in terms of performance.

Old 26 bytes answer that's better for performance:

Z©bgL®gãUεVXεYSym}OsN>èå}P

In this version I traded in some bytes to make the performance a lot better so it can run [1,1000] with ease. Test cases containing numbers in the range [1,9999] are done in about a second on TIO. Test cases in the range [10000,99999] in about 10-15 seconds on TIO. Above that it will timeout.

Try it online or verify all test cases with numbers in the range [1,9999].

Explanation:

Z                 # Push the max of the (implicit) input-list (without popping)
                  #  i.e. [526,853] → 853
 ©                # Store it in the register (without popping)
  b               # Convert to binary
                  #  i.e. 853 → 1101010101
   g              # Take its length
                  #  i.e. 1101010101 → 10
    L             # Pop and push a list [1, n]
                  #  i.e. 10 → [1,2,3,4,5,6,7,8,9,10]
     ®            # Push the max from the register
      g           # Take its length
                  #  i.e. 853 → 3
       ã          # Cartesian product the list that many times
                  #  i.e. [1,2,3,4,5,6,7,8,9,10] and 3
                  #   → [[1,1,1],[1,1,2],[1,1,3],...,[10,10,8],[10,10,9],[10,10,10]]
        U         # Pop and store it in variable `X`
ε              }  # Map both values of the input list:
 V                # Store the current value in variable `Y`
  Xε    }         # Map `y` over the numbers of variable `X`
    Y             # Push variable `Y`
     S            # Convert it to a list of digits
                  #  i.e. 526 → [5,2,6]
      ym          # Take each digit to the power of the current cartesian product sublist
                  #  i.e. [5,2,6] and [3,9,3] → [125,512,216]
         O        # Take the sum of each inner list
                  #  i.e. [[5,2,6],[5,2,36],[5,2,216],...,[125,512,216],...]
                  #   → [13,43,223,...,853,...]
          s       # Swap to push the (implicit) input
           N>     # Push the index + 1
                  #  i.e. 0 → 1
             è    # Index into the input-list (with automatic wraparound)
                  #  i.e. [526,853] and 1 → 853
              å   # Check if it's in the list of sums
                  #  i.e. [13,43,223,...,853,...] and 853 → 1
                P # Check if it's truthy for both both (and output implicitly)
                  #  i.e. [1,1] → 1

Haskell, 77 bytes

a#b=a!b&&b!a
a!b|a<1=b==0|b<1=b>1|1<2=or[div a 10!(b-mod a 10^e)|e<-[1..b+1]]

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Perl 6, 87 84 69 bytes

-15 bytes thanks to nwellnhof!

{!grep {!grep $^b,[X+] 0,|map (*X**1..$b.msb+1),$^a.comb},.[0,1,1,0]}

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Anonymous code block that returns True or False.

Explanation:

{!grep {!grep $^b,[X+] 0,|map (*X**1..$b.msb+1),$^a.comb},.[0,1,1,0]}

{                                                                   }  # Anonymous code block
 !grep    # None of:
                                                          .[0,1,1,0]   # The input and the input reverse
       {!grep       # None of
                  [X+]       # All possible sums of
                       0,|   # 0 (this is to prevent single digit numbers being crossed with themself)
                          map                  ,$^a.comb   # Each digit mapped to
                              (*X**           )  # The power of
                                   1..$b.msb+1   # All of 1 to the most significant bit of b plus 1
                                                 # This could just be b+1, but time constraints...
              $^b,  # Is equal to b

JavaScript (Node.js), 116 92 89 86 83 77 bytes

a=>b=>(G=(c,g,f=h=g%10)=>g?c>f&f>1&&G(c,g,h*f)||G(c-f,g/10|0):!c)(a,b)&G(b,a)

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Expect input as (A)(B).

J, 56 bytes

h~*h=.4 :'x e.+/|:>,{x 4 :''<y&*^:(x&>)^:a:y''"+"."+":y'

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Yay, nested explicit definition!

How it works

powers =. 4 :'<y&*^:(x&>)^:a:y'  Explicit aux verb. x = target, y = digit
                             y   Starting from y,
               y&*^:     ^:a:    collect all results of multiplying y
                    (x&>)        until the result is at least x
              <                  Box it.

h=.4 :'x e.+/|:>,{x powers"+"."+":y'  Explicit aux verb. x, y = two input numbers
                            "."+":y   Digits of y
                  x powers"+          Collect powers of digits of y under x
                 {            Cartesian product of each item
           +/|:>,             Format correctly and compute the sums
       x e.                   Does x appear in the list of sums?

h~*h  Tacit main verb. x, y = two input numbers
      Since h tests the condition in only one direction,
      test again the other way around (~) and take the AND.

Python 2, 149 147 143 139 132 118 108 107 106 105 bytes

lambda a,b:g(a,b)*g(b,a)
g=lambda a,b:any(g(a/10,b-(a%10)**-~i)for i in(a*b>0)*range(len(bin(b))))or b==0

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_{-4 bytes, thanks to Vedant Kandoi}

J, 68 bytes

I thought J would perform quite well here, but it ended up being tougher than I expected and would love any suggestions for further golfing...

g=.#@#:@[
1 1-:[:(([:+./[=1#.]^"#.1+g#.inv[:i.g^#@])"."0@":)/"1],:|.

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NOTE: we subtract 3 chars from the TIO count there since f=. on the main function doesn't count

ungolfed

1 1 -: [: (([: +./ [ = 1 #. ] ^"#. 1 + g #.inv [: i. g ^ #@]) "."0@":)/"1 ] ,: |.