Napisz program lub funkcję, która przyjmuje dodatnią liczbę całkowitą. Możesz założyć, że dane wejściowe są prawidłowe i możesz je traktować jako ciąg. Jeśli liczba jest dowolna z

123
234
345
456
567
678
789

następnie wypisz prawdziwą wartość. W przeciwnym razie wypisz wartość fałszowania . Na przykład dane wejściowe

1
2
3
12
122
124
132
321
457
777
890
900
1011
1230
1234

wszystkie muszą skutkować wynikiem fałszowania. (Dane wejściowe nie będą miały zer wiodących, więc nie musisz się martwić takimi rzeczami 012.)

Najkrótszy kod w bajtach wygrywa.

Python, 24 bajty

range(123,790,111).count

Anonimowa funkcja, która wyprowadza 0 lub 1. Tworzy listę [123, 234, 345, 456, 567, 678, 789]i zlicza, ile razy pojawi się wejście.

f=range(123,790,111).count

f(123)
=> 1
f(258)
=> 0

Python, 24 bajty

lambda n:n%111==12<n<900

Po prostu wiele łańcuchów warunków.

Haskell, 22 bajty

(`elem`[123,234..789])

Anonimowa funkcja. Generuje listę o równych odstępach [123, 234, 345, 456, 567, 678, 789]i sprawdza, czy dane wejściowe są elementem.

Brachylog , 9 bajtów

h:2j:12+?

Wypróbuj online! lub Zweryfikuj wszystkie przypadki testowe.

Kredyty dla Dennisa za algorytm .

W języku angielskim „(udowodnij, że) pierwsza cyfra wejścia, połączona ze sobą dwa razy, dodaj 12, wciąż jest wejściem”.

Python 2, 25 bajtów

lambda n:`n-12`==`n`[0]*3

Przetestuj na Ideone .

Brain-Flak 76 + 3 = 79 bajtów

Ta odpowiedź jest golfem tej odpowiedzi. Właściwie nie wiem dokładnie, jak działa moja odpowiedź, ale DJMcMayhem daje dobre wyjaśnienie w swojej oryginalnej odpowiedzi, a moja odpowiedź jest modyfikacją jego.

([]<>)<>({}[({})]<>)<>(({}[({})()()()()()]<>{}{}<(())>)){{}{}(((<{}>)))}{}{}

Jest uruchamiany z flagą -a ascii, która dodaje 3 bajty.

Wyjaśnienie (swego rodzaju)

Począwszy od oryginalnego działającego rozwiązania:

([]<>)<>({}[({})]<>)<>({}[({})]<>)({}{}[()()])({}<({}[()()()])>)(({}{}<(())>)){{}{}(((<{}>)))}{}{}

Przeprowadzam to przez prosty algorytm golfowy, który napisałem i otrzymuję:

([]<>)<>({}[({})]<>)<>(({}[({})]<>{}[()()]<({}[()()()])>{}<(())>)){{}{}(((<{}>)))}{}{}

Stąd widzę sekcję, która <({}[()()()])>{}zasadniczo mnoży się przez jedną, co sprawia, że ​​równa się {}[()()()]zmniejszeniu całego kodu do:

([]<>)<>({}[({})]<>)<>(({}[({})]<>{}[()()]{}[()()()]<(())>)){{}{}(((<{}>)))}{}{}

Wreszcie negatywy można łączyć:

([]<>)<>({}[({})]<>)<>(({}[({})()()()()()]<>{}{}<(())>)){{}{}(((<{}>)))}{}{}

Brainfuck, 32 bajty

+>,+>,>,-[-<-<->>],[<]<[<]<[<]<.

Wypróbuj online!

Kredyty dla Lynn za rdzeń algorytmu.

Galaretka , 6 bajtów

DI⁼1,1

Wypróbuj online! lub zweryfikuj wszystkie przypadki testowe .

Jak to działa

DI⁼1,1  Main link. Argument: n (integer)

D       Decimal; convert n to base 10.
 I      Increments; compute the differences of all pairs of adjacent digits.
   1,1  Yield [1, 1].
  ⁼     Test the results to both sides for equality.

05AB1E , 5 bajtów

¥XX‚Q

Wyjaśnienie

¥      # deltas
    Q  # are equal to
 XX‚   # [1,1]

Wypróbuj online

MATL , 8 bajtów

d1=tn2=*

Wypróbuj online!

To zostanie wydrukowane 1 1 dla prawdziwych danych wejściowych i tablicy z 0nią dla wartości fałszowania, ponieważ jest to fałsz w MATL.

Wyjaśnienie:

d           % Calculate the difference between consecutive digits
 1=         % Push an array of which elements equal one
   t        % Duplicate this array
    n       % Push the length of this array
     2=     % Push a one if the length is 2, and a zero otherwise
            % Now, if we have a truthy input, the stack looks like:
            %   [1 1]
            %   1
            % And if we have a falsy input, the stack looks something like this:
            %   [1 0]
            %   1
            % Or this:
            %   [1 1]
            %   0
       *    % Multiply the top two elements

Java 7, 46 bytes

boolean f(int a){return a>12&a<790&a%111==12;}

After trying several things with Leaky Nun in chat, this seems to be the shortest. Sometimes you just have to do things the straightforward way :/

Explanation:

boolean f(int a){
    return a>12         Is it more than 12? (stupid edge case)
           &
           a<790        Is it in range the other way? 
           &
           a%111==12;   Is it 12 more than a multiple of 111? 
}

Perl 6,  35 29 24 21  19 bytes

{.chars==3&&'0123456789'.index: $_}
{$_ (elem) (123,*+111...789)}
{$_∈(123,*+111...789)}
*∈(123,*+111...789)
*∈(123,234...789)

Explanation:

# Whatever lambda ( the parameter is ｢*｣ )
*

∈ # is it an element of:

# this sequence
(
  123,
  234,

  # deduce rest of sequence
  ...

  # stop when you generate this value
  789
)

Usage:

my &code = *∈(123,234...789);

say code 123; # True
say code 472; # False

say (  *∈(123,234...789)  )( 789 ); # True

Retina, 26

.
$*: 
^(:+ ):\1::\1$

Outputs 1 for truthy and 0 for falsey.

Try it online (First line added to allow multiple testcases to be run).

Ruby, 32 30 25 + 2 = 27 bytes

+2 bytes for -nl flags.

Takes input on STDIN and prints true or false.

p"123456789"[$_]&.size==3

See it on repl.it: https://repl.it/DBn2/2 (Click on ▶️ and then type input into the console below.)

Brain-Flak, 99 bytes

([{}]({})<>)<>([{}]{}<>)(({})<([{}]{})((){[()](<{}>)}{})>)([{}]{})((){[()](<{}>)}{})<>{{{}}<>{}}<>

Try it online!

This is 98 bytes of code +1 for the -a flag.

This prints 1 for truthy, and either 0 or nothing (which is equivalent to 0) for falsy

Brain-Flak, 114 bytes

([((()()()){}){}]{})(((()()()){}())<>)<>{({}<(({}[(((((()()()){}()){}){}){}){}]())){(<{}>)<>({}[()])<>}{}>[()])}<>

Try it online!

Correct version (in the spirit of the question): takes the integer as input, output 0 for falsey and 1 for truthy.

This is not stack clean.

Algorithm

Let the input be n.

The output is truthy iff (n-123)(n-234)(n-345)(n-456)(n-567)(n-678)(n-789)=0.

I computed those seven numbers by first subtracting 12 and then subtract 111 7 times, and then computed the logical double-NOT of those seven numbers and added them up.

For truthy results, the sum is 6; for falsey results, the sum is 7.

Then, I subtract the sum from 7 and output the answer.

R, 30 22 bytes

scan()%in%(12+1:7*111)

Not particularly exciting; check if input is in the sequence given by 12 + 111k, where k is each of 1 to 7. Note that : precedes * so the multiplication happens after the sequence is generated.

C# (Visual C# Interactive Compiler), 41 30 23 bytes

First code-golf submission, be gentle :)

a=>{return a>12&&a<790?a%111==12:false;};
a=>a>12&&a<790?a%111==12:false
a=>a>12&a<790&a%111==12

Try it online!

• -11 bytes thanks to Kirill L.
• Another -7 bytes thanks to ASCII-only.

Brainfuck, 43 bytes

,>,>,>,[>>]<[[-<-<->>]<+[>>]<++[>>->]<+<]>.

Bah, I'm no good at this. Outputs \x01 if the output is one of the strings 123, 234, …, 789; outputs \x00 otherwise.

(I beat Java 7, though…)

Try it online!

JavaScript ES6, 26 bytes

n=>1>(n-12)%111&n>99&n<790

This takes advantage of the fact that I'm using bit-wise logic operators on what are essentially booleans (which are bit-based!)

Thanks to Titus for saving 2.

Excel - 62 57 35 31 bytes

Based on Anastasiya-Romanova's answer, but returning Excel's TRUE/FALSE values.

=AND(LEN(N)=3,MID(N,2,1)-MID(N,1,1)=1,MID(N,3,1)-MID(N,2,1)=1)

Further, we can get to

=AND(LEN(N)=3,MID(N,2,1)-LEFT(N)=1,RIGHT(N)-MID(N,2,1)=1)

since both RIGHT and LEFT return a single character by default.

And, inspired by some of the Python solutions:

=AND(LEN(N)=3,MOD(N,111)=12,N<>900)

Thanks to Neil for 4 more bytes...

=AND(N>99,MOD(N,111)=12,N<900)

Brachylog (2), 7 bytes

ẹ~⟦₂-_2

Try it online!

Explanation

ẹ~⟦₂-_2
ẹ        Split into digits
 ~⟦₂     Assert that this is an increasing range; take its endpoints
    -_2  Assert that the starting minus ending endpoint is -2

As a full program, we get a truthy return if all assertions hold, a falsey return if any fail.

CJam, 13 9 bytes

A,s3ewqe=

Try it online!

Explanation

A,s        e# Push "0123456789".
   3ew     e# Split it into contiguous length-3 chunks: ["012" "123" "234" ... "789"].
      q    e# Push the input.
       e=  e# Count the number of times the input appears in the array.

Excel - 104 bytes

=IF(LEN(N)<3,"Falsy",IF(AND(LEN(N)=3,MID(N,2,1)-MID(N,1,1)=1,MID(N,3,1)-MID(N,2,1)=1),"Truthy","Falsy"))

Explanation:

The syntax for the IF formula in Excel is:

IF( condition, [value_if_true], [value_if_false] )

If the length of input N, where it's a name of the reference cell, is less than 3, then it will return Falsy. Else, if the length of input N is 3 and both of the difference of second digit and first digit and the difference of third digit and second digit are equal to 1, then it will return Truthy.

Dyalog APL, 10 bytes

Takes string argument.

1 1≡¯2-/⍎¨

1 1≡ Is {1, 1} identical to

¯2-/ the reversed pair-wise difference of

⍎¨ each character taken as a number?

TryAPL online! (⍎ has been emulated with e for security reasons.)

Perl, 18 bytes

Includes +1 for -p

Run with the input on STDIN

123.pl <<< 123

123.pl:

#!/usr/bin/perl -p
$_=$_=/./.2==$_-$&x3

PHP, 31 bytes

<?=($n=$_GET[n])-12==$n[0]*111;

Check if first digit of (number minus 12) is multiple of 111

PowerShell v3+, 24 bytes

($args[0]-12)/111-in1..7

Uses the same "multiple of 111 plus 12" trick as some other answers, but goes the other direction. Takes input $args[0], subtracts 12, divides by 111, and checks whether that's -in the range 1..7. Outputs a Boolean true/false value. Requires v3+ for the -in operator.

Test Cases

PS C:\Tools\Scripts\golfing> 123,234,345,456,567,678,789|%{.\easy-as-one-two-three.ps1 $_}
True
True
True
True
True
True
True

PS C:\Tools\Scripts\golfing> 1,2,3,12,122,124,132,321,457,777,890,900,1011,1230,1234|%{.\easy-as-one-two-three.ps1 $_}
False
False
False
False
False
False
False
False
False
False
False
False
False
False
False

ARM Machine Code, 18 bytes

Hex dump (little endian):

3803 d105 6808 ebc0 2010 b280 f2a0 1001 4770

This is a function that takes in a length, pointer pair for the string. The output is bash-style, it outputs 0 for true and a non-zero value for false. In C the function would be declared int oneTwoThree(size_t length, char* string). The instruction encoding is thumb-2, which has 2 and 4 byte instructions. Tested on a Raspberry Pi 3.

Ungolfed assembly:

.syntax unified
.text
.global oneTwoThree
.thumb_func
oneTwoThree:
    @Input: r0 - the number of characters in the string
    @r1 - A pointer to the (not necessarily NUL-terminated)
    @string representation of the number (char*)
    @Output: r1 - 0 if the number is in 123,234,...,789, else non-zero (bash-style)
    subs r0,r0,#3
    bne end @Return non-zero if r0!=3
    ldr r0,[r1] @Remember that this is little endian
    @So the first digit is the most siginificant byte
    @I.e. if the input was 123 then r0 contains 0xXY010203 where XY is garbage

    rsb r0,r0,r0,lsr #8 @r0=(r0>>8)-r0 (rsb is reverse subtract)
    uxth r0,r0 @r0&=((1<<16)-1) (mask off top half)
    @Now r0 is 0x0101 iff we have a matching number
    sub r0,r0,#0x101
    @Now r0 is 0 iff the string fit the specification

    end:
    bx lr @return

Testing script (also assembly):

.syntax unified
.text
.global main
.thumb_func
main:
    push {r4,lr}
    ldr r4,[r1,#4] @r0=argv[1]
    mov r0,r4
    bl strlen
    @Now r0 is the length of the string argv[1]
    mov r1,r4
    bl oneTwoThree @oneTwoThree(strlen(argv[1]),argv[1])
    cmp r0,#0
    it ne
    movne r0,#1 @Output through return code, 1 if false
    pop {r4,pc}

JavaScript (ES6), 34 bytes

And one more option in JS. Takes input as a string and outputs 0 for false and 1 for true.

n=>++n[0]==n[1]&++n[1]==n[2]&!n[3]

See my other solutions here and here

Try it

f=
n=>++n[0]==n[1]&++n[1]==n[2]&!n[3]
i.addEventListener("input",_=>o.innerText=f(i.value))

<input id=i type=number><pre id=o>