Does it match the digit pattern?

Question

Recently I had a Maths test and noticed that a certain number on the test matched an interesting pattern. The number (28384) matched a generic digit sequence that looks like this

(n)(x)(n+1)(x)(n+2)(x)(n+3) etc...

where n and x are single digit integers. The sequence can begin with either x or n and end with either x or n+y.

Your task is, given a multi digit positive integer, output a truthy or falsey value, depending on whether the input matches the pattern. The input will be between 4 and 18 digits long. You may take input as a string representation of the integer. The input will not begin with a 0 but can contain or end with 0s.

n+y will always be a single digit number (hence why the length limit is 18).

Test Cases

These should output a truthy value

182838485868788898
4344
85868
12223242526

And these should be falsey

12345
6724013635
36842478324836
1222232425
5859510511

As with all code golfs, shortest code wins! Good luck and may the odds, be ever in your favour!

Answer 1

Python 2, 84 81 80 79 bytes

-1 byte thanks to ovs

lambda x:g(x,1)|g(x,0)
g=lambda x,a:len(set(x[a::2]))==(x[a<1::2]in"123456789")

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---

Python 3, 82 79 78 77 bytes

lambda x:g(x,1)|g(x,0)
g=lambda x,a:len({*x[a::2]})==(x[a<1::2]in"123456789")

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Slightly shorter in Python 3, but I didn't think it deserved its own answer.

---

Explanation

We set up a function g that takes string and an index (either 1 or 0). g then returns whether or not len(set(x[a::2])), that is the number of unique digits in every other position, is equal to (x[a==0::2]in"123456789"), whether or not the other digits are in ascending order. If the digits are in ascending order this returns whether or not they are all the same, if not it will ask if the set is empty, which it cannot be, thus always returning false.

Answer 2

Jelly, 13 11 bytes

Ds2ZI’M¦Ẏ¬Ạ

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Explanation:

Ds2ZI’M¦Ẏ¬Ạ Accepts an integer
D           Get individual digits
  2         2
 s          Split into chunks of specific length
   Z        Zip
    I       Take deltas
     ’      Decrement
      M     Take maximal indices
       ¦    Apply at specific indices
        Ẏ   Reduce dimensionality
         ¬  Vectorized NOT
          Ạ Check if all are truthy

Answer 3

05AB1E, 15 bytes

TG9LNýN.øŒ})˜Iå

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Explanation

TG9LNýN.øŒ})˜Iå
TG        }     # For 1 to 9...
  9L             # Push [1 .. 9]
    Ný           # Join with current value 
      N.ø        # surround with current value
         Π      # Push substrings
           )    # Wrap stack to array
            ˜   # Deep flatten the array
             I  # Push input
              å # Is the input in the array?
                # Implicit print

It should work (test cases did) but if you find any flaws please let me know.

14 Bytes if no output counts as falsy:

TG9LNýN.øŒIåi1

Answer 4

D, 117 bytes

int f(string n){int o=n[0]!=n[2],x=n[o];foreach(i,c;n){if(i%2!=o&&i>1&&c!=n[i-2]+1||i%2==o&&c!=x)return 0;}return 1;}

Definitely suboptimal, but it works fine

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Answer 5

Haskell, 108 113 97 95 bytes

d(x:_:r)=x:d r
d r=r
c(x:n:r)=and$all(==x)(d r):zipWith(==)(d$n:r)[n..]
f s@(_:n:_)=c s||c(n:s)

Sample call: f "182838485868788898" yields True

Ungolfed version with explanations:

-- Take every other element from the string, starting with the first
d (x:_:r) = x : d r
d r       = r
c (x:n:r) = and $ all (== x) (d r)              -- Every other char is equal to the first
                : zipWith (==) (d $ n:r) [n..]  -- The remaining chars are a prefix of n(n+1)(n+2)...
f s@(_:n:_) = c s      -- The case where s already starts with x
           || c (n:s)  -- If not, prepend a dummy x and try again

Answer 6

JavaScript (ES6), 66 63 60 bytes

Takes input as a string.

s=>[...s].every((d,i)=>d-s[j^(k=i+j&1)]==k*i>>1,j=s[2]-s[0])

Test cases

let f =

s=>[...s].every((d,i)=>d-s[j^(k=i+j&1)]==k*i>>1,j=s[2]-s[0])

console.log('[Truthy]');
console.log(f("182838485868788898"))
console.log(f("4344"))
console.log(f("85868"))
console.log(f("12223242526"))

console.log('[Falsy]');
console.log(f("12345"))
console.log(f("6724013635"))
console.log(f("36842478324836"))
console.log(f("1222232425"))

Answer 7

C (gcc), 123 bytes

#define X*s&&*s++-++n&&(r=0)
t,n,x,r;f(char*s){t=*s==s[2];for(r=1,n=s[t],x=s[!t],s+=2;*s;)t||X,*s&&*s++-x&&(r=0),t&&X;n=r;}

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Answer 8

Python 3, 99 96 89 bytes

• Saved 3 bytes: use of all() function
• @WheatWizard saved 7 bytes:shorthanding & | and replace extra variable by k<1

lambda x,g=lambda x,k:(x[k<1::2]in'123456789')&all(j==x[k]for j in x[k::2]):g(x,0)|g(x,1)

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Explanation:

First split the string into two lists: one with odd-indexed and other with even-indexed elements. The two lists A and B are supposed be such that either:

1. A contains same number and B contains consecutive numbers in ascending order.

OR just the opposite

2. B contains same number and A contains consecutive numbers in ascending order.

The consecutive condition is checked by: a in '123456789'

The same-number condition is checked by: all(i=a[x] for i in a)

Answer 9

PHP, 68 bytes

for(;$v<10;)$s.=strstr(+$v.join(+$v,range(1,9)).+$v++,$argn);echo$s;

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Output part of search string starting from and including the first occurrence of input to the end of search string as truthy value and nothing for falsy

for 2 Bytes more you can replace echo$s; with !!echo$s; to get 1 as truthy value

Find the occurrence of the input in one of the following strings in the array

Array
(
    [0] => 0102030405060708090
    [1] => 1112131415161718191
    [2] => 2122232425262728292
    [3] => 3132333435363738393
    [4] => 4142434445464748494
    [5] => 5152535455565758595
    [6] => 6162636465666768696
    [7] => 7172737475767778797
    [8] => 8182838485868788898
    [9] => 9192939495969798999
)

Answer 10

MATL, 15 bytes

2L&),duw]hSFTX=

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With some help of @LuisMendo in chat. Note that, if empty output + error is also considered 'falsy', the X can be left out, bringing the score to 14 bytes.

2L&)     % Split the input into odd and even-indexed elements
    ,   ] % Do twice (new feature since MATL v20.0), i.e., on both halves of the input
     d     % Pairwise differences of the array. Results in [0,0,...] for the 'constant' part,
            %  and [1,1,...] for the 'increasing' part.
      u      % Get unique elements. Results in [0] for the constant part, [1] for the increasing part.
       w      % Swap the stack to do the same for the other half of the input.
         hS    % Horizontal concatenation followed by sort. Results in [0,1] for the desired string.
           FTX= % Check if the result is indeed [0,1]. Implicit display.

Answer 11

JavaScript (ES6), 54 bytes

f=
s=>[...s].every((e,i)=>i<2|e-s[i-2]==(s[2]!=s[0]^i%2))

<input oninput=o.textContent=this.value[3]?f(this.value):``><pre id=o>

Takes input as a string.

Answer 12

Julia, 57 56 bytes

~s=(!x=occursin(s,join(x*k for k='0':'9')x);!s[1]|!s[2])

• replace || by | by @MarcMush

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• input: the number as decimal string

Answer 13

Mathematica, 121 bytes

(m[x_]:=Take[s=IntegerDigits@#,{x,Length@s,2}];w[b_,n_]:=Union@Differences@m@b=={1}&&Length@Union@m@n==1;w[1,2]||w[2,1])&

Answer 14

Pyth, 20 bytes

ASm.+sMd.Tcz2&-GZ-H1

Output [] when the number matches the digit pattern, anything else otherwise.

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Explanations (example with input 85868)

ASm.+sMd.Tcz2&-GZ-H1

          cz2           # Chop the input in pairs: ['85', '86', '8']
        .T              # Transpose, ignore absences: ['888', '56']
     sM                 # Convert to integers: [[8, 8, 8], [5, 6]]
  m.+  d                # Compute deltas: [[0, 0], [1]]
 S                      # Sort: [[0, 0], [1]]
A                       # Assign first list to G and second list to H
              -GZ       # Filter 0 on G (on absence): [0, 0] -> []
                 -H1    # Filter 1 on H (on absence): [1] -> []
             &          # Check both lists are empty (logical and): [] & [] -> []

Answer 15

Pyth, 17 bytes

qU2Ssm{.+d.TcjQT2

Try it here

Same algorithm as my Jelly answer.

Explanation:

qU2Ssm{.+d.TcjQT2 Accepts an integer
             jQT  Take digits of input
            c   2 Split in pairs
          .T      Transpose
     m            Map the following on each of the two resulting lists:
       .+d          Take deltas
      {             Deduplicate
    s             The list is now in [[a, b, ...], [A, B, ...]] format, convert it to [a, b, ..., A, B, ...]
   S              Sort
qU2               Check if equal to [0, 1]

Answer 16

Python 3, 167 161 157 131 106 bytes

-55 bytes thanks to @WheatWizard's suggestions

def g(t):k,c,f,j=t[::2],t[1::2],'123456789',''.join;return(len({*k})and j(c)in f)or(len({*c})and j(k)in f)

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Answer 17

Java (OpenJDK 8), 128 119 118 108 107 104 bytes

s->{int i=3,a=s[2]-s[0],b=s[3]-s[1];for(;++i<s.length;)a+=b-(b=s[i]-s[i-2]==a?a:2);return(a|b)==1&a!=b;}

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Explanation:

s->{                             // lambda
  int i=3,                       //  iterating index
      a=s[2]-s[0],               //  diff of even-indexed characters
      b=s[3]-s[1];               //  diff of odd-indexed characters
  for(;++i<s.length;)            //  iterate
    a+=b-(b=                     //   swap a and b
        s[i]-s[i-2]==a?a:2       //    or set b to 2 if the diffs don't match
      ));                        //
  return (a|b)==1                //  return true if both a and b are in (0,1)
        &a!=b;                   //         but different
}

Answer 18

Retina, 47 bytes

.
$*11;
(1+)(?<=\1;1+;\1)

^1+;1+

^;?(;1;)+;?$

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Outputs 1 if it matches the pattern, 0 if it does not

Explanation

.
$*11;

Convert each digit n to n+1 in unary, separated by semicolons

(1+)(?<=\1;1+;\1)

(Trailing newline) converts each digit to the difference between itself and the one 2 spots before it

^1+;1+

(Trailing newline) removes the first 2 digits

^;?(;1;)+;?$

Counts the number of matches of this pattern, which checks for alternating 0s and 1s

Answer 19

Husk, 19 bytes

S¤|o←Ẋ~&Λȯ¬→-E↔TC2d

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Answer 20

Vyxal, 11 bytes

y"v¯vUfsk≈⁼

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