Staircase numbers

Question

A staircase number is a positive integer x such that its nth digit (one indexed starting with the least significant digit) is equal to x % (n + 1). Thats a bit of a mouthful so lets look at an example. Take 7211311, if we take the modular residues of 7211311 on the range 2-8 we get the following:

7211311 % 2 = 1
7211311 % 3 = 1
7211311 % 4 = 3
7211311 % 5 = 1
7211311 % 6 = 1
7211311 % 7 = 2
7211311 % 8 = 7

These are the digits of 7211311! Thus 7211311 is a staircase number.

Task

Write code that takes when given a positive number as input, will output two distinct values one if the number is a staircase number and the other if it is not.

This is a code-golf competition so your aim should be to minimize the number of bytes in your source code.

Test Cases

Here are the first 13 staircase numbers:

1, 10, 20, 1101, 1121, 11311, 31101, 40210, 340210, 4620020, 5431101, 7211311, 12040210

@Okx the task is just to distinguish positive staircase numbers from positive non-staircase numbers, so the behavior is undefined for 0 and negative numbers. — Paŭlo Ebermann, Jul 2 '17 at 12:40

Silvio Mayolo · Accepted Answer · 2017-07-01 21:08:46Z

11

Haskell, 55 57 bytes

f m|let n#x=n==0||n`mod`10==m`mod`x&&div n 10#(x+1)=m#2

A different approach than the other Haskell solution.

Thanks xnor for saving 2 bytes.

edited Jul 1 '17 at 21:08

answered Jul 1 '17 at 19:22

Silvio Mayolo

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4

\$\begingroup\$ You can use this tip to shorten the let expression. \$\endgroup\$ – xnor Jul 1 '17 at 20:23
\$\begingroup\$ A different, shorter approach. Well done! +1 \$\endgroup\$ – qfwfq Jul 2 '17 at 3:47
\$\begingroup\$ n==0 → n<1 saves a byte. \$\endgroup\$ – Lynn Nov 5 '20 at 15:10

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Erik the Outgolfer · Accepted Answer · 2017-07-01 16:51:12Z

9

Brachylog, 25 21 16 14 bytes

{it+₂;?↔%}ᶠ↔c?

Try it online!

First Brachylog submission :D ~~probably very ungolfed~~...many thanks to Leaky Nun and Fatalize for encouragement and help to golf this from 25 all the way down to just 14. :) :)

edited Jul 1 '17 at 16:51

answered Jul 1 '17 at 15:26

Erik the Outgolfer

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Community · Accepted Answer · 2020-06-17 09:04:33Z

7

05AB1E, 6 bytes

Code:

ā>%JRQ

Uses the 05AB1E encoding. Try it online!

Explanation:

ā        # Get the range [1 .. len(input)]
 >       # Increment by 1
  %      # Vectorized modulo
   J     # Join the array into a single number
    RQ   # Reverse that number and check if it's equal to the original input

edited Jun 17 '20 at 9:04

Community♦

1

answered Jul 1 '17 at 14:48

Adnan

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Herman L · Accepted Answer · 2017-07-01 16:24:10Z

7

Javascript, 42 41 39 38 bytes

-4 bytes thanks to @Shaggy and @ETHProductions

s=>[...s].some(d=>s%i++^d,i=~s.length)

This takes the number as a string and returns false if the number is a staircase number and true otherwise.

Example code snippet:

f=
s=>[...s].some(d=>s%i++^d,i=~s.length)

function update() {
  o.innerText = f(document.getElementById("i").value)
}

<input id="i" type="number">
<button onclick="update()">Test</button>
<p id="o">

edited Jul 1 '17 at 16:24

answered Jul 1 '17 at 15:56

Herman L

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2

\$\begingroup\$ You should be able to drop the ! as the challenge doesn't explicitly specify that you must return true for true and false for false, simply that you must return 2 distinct values. \$\endgroup\$ – Shaggy Jul 1 '17 at 16:01
2

\$\begingroup\$ This is very well golfed, well done. I think you should be able to squeeze out two more bytes if you calculate i yourself: s=>[...s].some(d=>s%i--^d,i=s.length+1) \$\endgroup\$ – ETHproductions Jul 1 '17 at 16:05
2

\$\begingroup\$ Actually, by exploiting the fact that ~x == -(x+1) on integers and x%-y == x%y, I think you can get one more: s=>[...s].some(d=>s%i++^d,i=~s.length) \$\endgroup\$ – ETHproductions Jul 1 '17 at 16:10

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ovs · Accepted Answer · 2017-07-01 17:24:09Z

6

Python 2, 54 bytes

f=lambda n,x=0:10**x>n or(`n%(x+2)`==`n`[~x])*f(n,x+1)

Try it online!

answered Jul 1 '17 at 17:24

ovs

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qfwfq · Accepted Answer · 2017-07-01 18:16:08Z

6

Haskell, 60 bytes

Takes the number as an int

x n|s<-show n=reverse s==(rem n.(+1)<$>[1..length s]>>=show)

edited Jul 1 '17 at 18:16

answered Jul 1 '17 at 18:00

qfwfq

1816 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

5

Mathematica, 60 bytes

FromDigits@Reverse@Mod[#,Range@Length@IntegerDigits@#+1]==#&

Try it online!

@alephalpha golfed it to 48

Mathematica, 48 bytes

FromDigits@Reverse@Mod[#,Range[2,Log10@#+2]]==#&

next one is 24120020

edited Jun 17 '20 at 9:04

Community♦

1

answered Jul 1 '17 at 14:57

J42161217

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Shaggy · Accepted Answer · 2020-11-05 14:43:29Z

5

Japt, 9 7 6 bytes

Takes input as a string.

Ô¶¡%´J

2 bytes saved with help from ETHproductions.

Try it

Ô¶¡%´J     :Implicit input of string U
Ô          :Reverse
 ¶         :Test for equality with 
  ¡        :Map U
   %       :  Modulo
    ´J     :  J (initially -1) prefix decremented

edited Nov 5 '20 at 14:43

answered Jul 1 '17 at 14:55

Shaggy

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\$\begingroup\$ Gosh darn (Y+2, I feel like that could be at least 1 byte shorter... \$\endgroup\$ – ETHproductions Jul 1 '17 at 15:29
1

\$\begingroup\$ ...and it can: ¥£%´JÃw :-) (works because x%y == x%-y in JS) \$\endgroup\$ – ETHproductions Jul 1 '17 at 15:31
\$\begingroup\$ Aha, yes, was trying a few different things to get that calculation down to 2 bytes. \$\endgroup\$ – Shaggy Jul 1 '17 at 15:47

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Okx · Accepted Answer · 2017-07-02 10:40:33Z

4

Neim, 6 bytes

𝐧ᛖ𝕄𝐫𝐣𝔼

Explanation:

𝐧         Get the length of the input, then create an exclusive range
 ᛖ        Add 2 to each element
  𝕄       Modulus
    𝐫      Reverse
     𝐣     Join
      𝔼   Check for equality

Try it online!

edited Jul 2 '17 at 10:40

answered Jul 1 '17 at 15:28

Okx

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\$\begingroup\$ @Thehx Regarding your edit, Neim uses a custom encoding: this one \$\endgroup\$ – Okx Jul 5 '17 at 15:24

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Erik the Outgolfer · Accepted Answer · 2017-07-01 14:48:00Z

2

Jelly, 7 bytes

DJ‘⁸%⁼Ṛ

Try it online!

answered Jul 1 '17 at 14:48

Erik the Outgolfer

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Artyer · Accepted Answer · 2017-07-01 17:09:25Z

2

Python 2, 56 bytes

lambda x:all(`x%(i+2)`==`x`[~i]for i in range(len(`x`)))

Try it online!

answered Jul 1 '17 at 17:09

Artyer

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Sean · Accepted Answer · 2017-07-01 18:28:11Z

Perl 6, 32 bytes

{$_ eq[~] $_ «%«(1+.comb...2)}

Try it online!

.comb is the number of characters in the string representation of the input argument $_ (that is, the number of digits).
1 + .comb ... 2 is the sequence of numbers from one greater than the number of digits down to 2.
«%« is the modulus hyperoperator that gives the remainder when $_, the input argument on its left, is divided by each of the elements of the sequence on its right: $_ % 2, $_ % 3, ....
[~] concatenates those digits into a new number, which is compared with the input argument using the string equality operator eq.

Jörg Hülsermann · Accepted Answer · 2017-07-01 20:23:11Z

2

PHP, 43 bytes

for(;$r<$a=$argn;)$r=$a%~++$i.$r;echo$r>$a;

Try it online!

PHP, 44 bytes

prints 1 for true and nothing for false

for(;$r<$a=$argn;)$r=$a%~++$i.$r;echo$r==$a;

Try it online!

edited Jul 1 '17 at 20:23

answered Jul 1 '17 at 17:35

Jörg Hülsermann

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Jim · Accepted Answer · 2017-07-02 19:57:46Z

2

Pyth, 13 bytes

-1 bytes thanks to Okx.

qsjk_m%QhdSl`

Try it online!

Explanation

             QQ    # Implicit input
          Sl`Q     # Generate [1, len(str(Q))]
     m%Qhd         # For digit d in above range, perform Q % (d + 1)
 sjk_              # Reverse, then convert to number
q             Q    # Test equality with input

Alternate solution, still 13 bytes (thanks to karlkastor)

qi_.e%Q+2k`QT

Try it online! That's essentially the same as the first solution, excepted that it uses i to convert from array of numbers to a number, and that the range is generated differently.

edited Jul 2 '17 at 19:57

answered Jul 1 '17 at 16:48

Jim

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1

\$\begingroup\$ You can replace ss`M_ with jk_ to save 2 bytes. \$\endgroup\$ – Okx Jul 1 '17 at 17:13
\$\begingroup\$ @Okx I need it because j outputs a string whereas I need a number to compare with the input (which is a number). \$\endgroup\$ – Jim Jul 1 '17 at 17:59
1

\$\begingroup\$ Another 13 byte solution would be: qi_.e%Q+2k`QT using enumerated map (.e) instead of map. And converting the remainders to a base 10 int from the list instead of using join. \$\endgroup\$ – KarlKastor Jul 1 '17 at 18:13

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Dada · Accepted Answer · 2017-07-10 13:58:16Z

2

C++,104 bytes

1) original version:

int main(){int N,T,R=1;cin>>N;T=N;for(int i=1;i<=log10(N)+1;i++){if(N%(i+1)!=T%10){R=0;}T/=10;}cout<<R;}

2) in a readable form:

int main()
{
    int N, T, R = 1;

    cin >> N;
    T = N;

    for (int i = 1; i <= log10(N) + 1; i++)
    {
        if (N % (i + 1) != T % 10)
        {
            R = 0;
        }

        T /= 10;
    }

    cout << R;
}

Try it Online!

edited Jul 10 '17 at 13:58

Dada

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answered Jul 5 '17 at 9:37

koita_pisw_sou

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Razetime · Accepted Answer · 2020-12-21 17:07:53Z

APL (Dyalog Extended), ²¹ 19 bytes

{⍵=10⊥⌽⍵|⍨1↓⍳1+≢⍕⍵}

Try it online!

Creates two lists of digits and check if they are equal.

Explanation

{⍵=10⊥⌽⍵|⍨1↓⍳1+≢⍕⍵} ⍵ → input
               ≢⍕⍵  length of ⍵ as a string
             1+     +1
            ⍳       range from 1 to that
          1↓        Drop first element 
       ⍵|⍨          ⍵ mod each element in range
      ⌽             reversed
   10⊥              converted from base 10
 ⍵=                 equals the input?

alephalpha · Accepted Answer · 2017-07-01 14:59:05Z

1

Pari/GP, 42 bytes

n->Vecrev(s=digits(n))==[n%d|d<-[2..#s+1]]

Try it online!

edited Jul 1 '17 at 14:59

answered Jul 1 '17 at 14:53

alephalpha

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officialaimm · Accepted Answer · 2017-07-01 15:40:50Z

1

Python 2, 66 60 58 57 bytes

Thanks to @Leaky nun for 6 bytes: remove unneeded x and (shouldnot check for 0)
Thanks to @Einkorn Enchanter for 1 byte: use of enumerate

lambda x:all(a==`x%(i+2)`for i,a in enumerate(`x`[::-1]))

Try it online!

edited Jul 1 '17 at 15:40

answered Jul 1 '17 at 15:04

officialaimm

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1

\$\begingroup\$ Shorter if you enumerate \$\endgroup\$ – Wheat Wizard♦ Jul 1 '17 at 15:09
\$\begingroup\$ Thank you. And hey, I just realised you were wheat wizard.. :D \$\endgroup\$ – officialaimm Jul 1 '17 at 15:42
\$\begingroup\$ This is a very late comment, but enumerate can take a second argument "number to start counting from", so x%i and enumerate(…,2) saves two bytes. \$\endgroup\$ – Lynn Nov 5 '20 at 15:06

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Python 3: 63 Bytes

lambda m:all(int(x)==m%(n+2)for n,x in enumerate(str(m)[::-1]))

If I could count the number of times I wished 'enumerate' were shorter...

Try it online!

edited Jun 17 '20 at 9:04

Community♦

1

answered Jul 5 '17 at 14:38

bendl

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\$\begingroup\$ Yep, and I just realized its exactly the same as the answer @officialaimm gave... Should I remove? \$\endgroup\$ – bendl Jul 5 '17 at 14:41
\$\begingroup\$ Theirs is in python 2 and you came up with it independently so I would leave it. \$\endgroup\$ – Wheat Wizard♦ Jul 5 '17 at 14:44
\$\begingroup\$ Can save two bytes by starting your enumeration at 2 and rearranging the logical: lambda m:all(m%n==int(x)for n,x in enumerate(str(m)[::-1],2)) \$\endgroup\$ – nocturama Jul 5 '17 at 15:48

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Olivier Grégoire · Accepted Answer · 2017-07-05 16:03:17Z

1

Java (OpenJDK 8), 60 bytes

n->{int m=2,r=0,t=1;for(;n>=t;t*=10)r+=n%m++*t;return r==n;}

Try it online!

A non-string version.

edited Jul 5 '17 at 16:03

answered Jul 5 '17 at 15:48

Olivier Grégoire

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Alex Ferretti · Accepted Answer · 2017-07-06 12:49:03Z

1

Java 8, 156 149 bytes

interface B{static void main(String[]s){String f="";for(int i=1;i<=s[0].length();)f=new Long(s[0])%++i+f;System.out.print(f.equals(s[0]));}}

Ungolfed :

interface B {
    static void main(String[] s) {
        String f = "";
        for (int i = 1; i <= s[0].length();)
            f = new Long(s[0]) % ++i + f;
        System.out.print(f.equals(s[0]));
    }
}

Try it Online !

UPDATE :
-7 bytes : removed useless {} and replaced Integer.parseInt(...) by new Integer(...)
-9 bytes : thanks to Kevin Cruijssen, removed a bunch of useless (), used Long instead of Integer and print instead of println. Thanks Kévin !

edited Jul 6 '17 at 12:49

answered Jul 3 '17 at 12:06

Alex Ferretti

714 bronze badges

1

\$\begingroup\$ Nice answer, +1 from me. Btw, some small things to golf: new Integer can be new Long (-3 bytes); println can be print (-2 bytes); and you can remove the parenthesis surrounding new Long(s[0])%i+f; (-4 bytes). \$\endgroup\$ – Kevin Cruijssen Jul 5 '17 at 12:29
\$\begingroup\$ Really nice ! Thanks, i'll update this ! \$\endgroup\$ – Alex Ferretti Jul 6 '17 at 12:45

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Neil · Accepted Answer · 2017-09-16 00:41:09Z

1

Charcoal, 20 15 bytes

⌊Ｅθ⁼ιＩ﹪Ｉθ⁻⁺¹Ｌθκ

Try it online! Outputs - for a staircase number, nothing otherwise. Link is to verbose version of code.

edited Sep 16 '17 at 0:41

answered Jul 1 '17 at 18:58

Neil

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Lynn · Accepted Answer · 2020-11-05 15:21:45Z

1

Haskell, 53 bytes

f n=foldr(\(k,_)a->10*a+mod n k)0(zip[2..]$show n)==n

Try it online!

answered Nov 5 '20 at 15:21

Lynn

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Daniel · Accepted Answer · 2017-07-01 15:11:00Z

0

Python 2, 61 bytes

lambda x:[`x%(n+2)`for n in range(len(`x`))][::-1]==list(`x`)

answered Jul 1 '17 at 15:11

Daniel

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\$\begingroup\$ Nope, your new golf is a byte shorter. :) \$\endgroup\$ – Wheat Wizard♦ Jul 1 '17 at 15:15

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streetster · Accepted Answer · 2017-07-01 21:35:35Z

q/kdb+, 34 bytes

Solution:

{s~raze($)x mod'2+(|)(!)(#)s:($)x}

Example:

q){s~raze($)x mod'2+(|)(!)(#)s:($)x}7211311 / this is a staircase number (true)
1b
q){s~raze($)x mod'2+(|)(!)(#)s:($)x}7211312 / this is not (false)
0b
q)t(&){s~raze($)x mod'2+(|)(!)(#)s:($)x}each t:1 + til 1000000 / up to a million
1 10 20 1101 1121 11311 31101 40210 340210

Explanation:

Cast the input number to a string, count from 0..length of string, add 2 to all, reverse it and feed each number into mod along with the original input. Cast the result of the mod to a string and reduce the list, check if it is equal to the string of the input number:

{s~raze string x mod'2 + reverse til count s:string x} / ungolfed solution
{                                                    } / lambda function
                                           s:string x  / convert input to string, save as s
                                     count             / return length of this string
                                 til                   / like python's range() function
                         reverse                       / reverses the list
                     2 +                               / adds two to each element in the list
               x mod'                                  / ' is each both, so feeds x, and each element of the list to modulo function
        string                                         / converts output list to string list ("7";"2";"1"..etc)
   raze                                                / reduce list ("721...")
 s~                                                    / is s equal to this reduced list, returns boolean

Notes:

Most of the solution is for generating the 2,3,4.. list, I have another solution that does less stuff, but winds up being 37 bytes after golfing:

{s~x mod'reverse 2 + til count s:("J"$) each string x} / ungolfed
{s~x mod'(|)2+til(#)s:("J"$)each($)x}                  / golfed

NikoNyrh · Accepted Answer · 2017-07-05 11:47:32Z

0

Clojure, 75 bytes

#(=(sort %)(sort(map(fn[i c](char(+(mod(Integer. %)(+ i 2))48)))(range)%)))

Input is a string, using map and the trailing % ended up being shorter than for[i(range(count %))] approach.

answered Jul 5 '17 at 11:47

NikoNyrh

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archaephyrryx · Accepted Answer · 2017-07-05 16:51:18Z

0

Haskell, 62 bytes

f x=and$zipWith(==)(reverse$show x)$map(head.show.mod x)[2..]

Instead of reversing the (infinite) list of moduli, it truncates the list by zipping it with the reversed string-respresentation of the integral x, which it then ensures is equal element-wise.

answered Jul 5 '17 at 16:51

archaephyrryx

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Xcali · Accepted Answer · 2017-09-16 05:10:14Z

0

Perl 5, 41 bytes

39 bytes of code + 2 flags -pa

map{$\||=$_!=$F[0]%++$n}0,reverse/./g}{

Try it online!

Outputs nothing (undef) for staircase numbers, 1 for anything else

answered Sep 16 '17 at 5:10

Xcali

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Kevin Cruijssen · Accepted Answer · 2020-11-05 15:23:42Z

MathGolf, 7 bytes

mækî)%Ñ

Input as a string.

Try it online.

Explanation:

m        # Map over each character of the (implicit) input-string,
 æ       # using the following four characters as inner code-block:
         #  (implicitly push the current character of the map)
  k      #  Push the input as an integer
   î     #  Push the 1-based map-index
    )    #  Increase it by 1
     %   #  Take the input-integer modulo (1-based) index+1
         # After the map, the digits are implicitly joined together to a string
      Ñ  # Check that this string is a palindrome
         # (after which the entire stack joined together is output implicitly as result)

Staircase numbers

Task

Test Cases

29 Answers 29

Haskell, 55 57 bytes

Brachylog, 25 21 16 14 bytes

05AB1E, 6 bytes

Code:

Explanation:

Javascript, 42 41 39 38 bytes

Python 2, 54 bytes

Haskell, 60 bytes

Mathematica, 60 bytes

Mathematica, 48 bytes

Japt, 9 7 6 bytes

Neim, 6 bytes

Jelly, 7 bytes

Python 2, 56 bytes

Perl 6, 32 bytes

PHP, 43 bytes

PHP, 44 bytes

Pyth, 13 bytes

C++,104 bytes

APL (Dyalog Extended), 21 19 bytes

Explanation

Pari/GP, 42 bytes

Python 2, 66 60 58 57 bytes

Python 3: 63 Bytes

Java (OpenJDK 8), 60 bytes

Java 8, 156 149 bytes

Charcoal, 20 15 bytes

Haskell, 53 bytes

Python 2, 61 bytes

q/kdb+, 34 bytes

Clojure, 75 bytes

Haskell, 62 bytes

Perl 5, 41 bytes

MathGolf, 7 bytes

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