Where are Champernowne's zeroes?

Question

Consider the infinite string of all nonnegative decimal integers concatenated together in order (akin to Champernowne's constant):

0123456789101112131415161718192021222324252627282930...979899100101102103...

Write a program or function that takes in a nonnegative integer that indexes (0-based) into this infinite string. Output a truthy value if the digit indexed is 0, otherwise output a falsy value if the digit is 1-9.

The shortest code in bytes wins.

The first 25 truthy-producing inputs are:

0
11
31
51
71
91
111
131
151
171
191
192
194
197
200
203
206
209
212
215
218
222
252
282
312

Kudos if your program is memory effecient, but this is not a requirement.

Answer 1

Haskell, 25 bytes

(<'1').((show=<<[0..])!!)

Usage example: (<'1').((show=<<[0..])!!) 312 -> True

Answer 2

05AB1E, 5 bytes

Code:

ÝJ¹è_

Explanation:

Ý      # Get the list [0 .. input].
 J     # Join the list.
  ¹    # Get the first input again.
   è   # Get the character on that index.
    _  # Logical negate (0 -> 1, everything else -> 0).

Uses the CP-1252 encoding. Try it online!

Answer 3

Mathematica, 42 40 bytes

(0@@Join@@IntegerDigits@Range@#)[[#]]<1&

Anonymous function. Takes a number as input and returns either True or False as output. A longer, yet more efficient(?) solution:

RealDigits[ChampernowneNumber[],10,1,-#][[1,1]]<1&

Answer 4

CJam, 9 bytes

{_),s=~!}

This is an unnamed block (function) which takes in an integer and returns 0 or 1 accordingly.

Explanation:

{       }        Defines a block
 _               Copy input n
  ),             Increment n and take range
    s            Convert to string - for a list of numbers this concatenates
                 the digits
     =           Index, getting nth digit
      ~          Evaluate the digit character into a number
       !         Logical negation

Online interpreter. Note that ~ evaluates a block. Alternative, you can run this test suite which uses , to filter the first 1000 numbers for truthy values.

Answer 5

MATL, 11 bytes

Qt:qVXzw)U~

Try it Online!

Explanation:

    % Implicitly grab input as an integer (N)
Qt  % Add 1 and duplicate
:q  % Create an array from [0 ... N]
V   % Convert each entry to a string (places spaces between each number)
Xz  % Remove all whitespace
w)  % Get the N+1 element of the string (since MATL uses 1-based indexing natively)
U~  % Convert the result back to a number and negate which yields TRUE if it was '0' and
    % FALSE otherwise

Answer 6

Brachylog, 10 8 bytes

2 bytes thanks to Fatalize.

y@ec:?m0

Try it online!

y@ec:?m0

y         range from 0 to Input, inclusive,
 @e       the digits of every number in that range,
   c      concatenated
    :?m   the Input-th digit
       0  is zero.

Answer 7

Perl 6, 26 25 bytes

{!+map(|*.comb,0..*)[$_]}

A lambda that takes a number as input and returns a True or False.

Memory-efficient.

How it works

1. 0..* – Construct the range from 0 to infinity.
2. map(|*.comb, ) – Lazily iterate the range, replacing each number with the characters of its string representation, and returning a new lazy sequence. The | keeps the new sequence flattened.
3. [$_] – Take the element at the index defined by the (implicitly declared) lambda parameter $_.
4. + – Coerce it to a number. (This step is needed because coercing a string directly to a boolean always gives True unless the string is empty.)
5. ! – Coerce it to a boolean and negate it.

(try it online)

EDIT: -1 byte thanks to b2gills.

Answer 8

Python 3.5, 40 bytes

lambda n:('%d'*-~n%(*range(n),n))[n]<'1'

Test it on repl.it.

How it works

For input n, '%d'*-~n repeats the format string n + 1 times.

(*range(n),n) unpacks the range [0, ..., n - 1] and yields the tuple (0, ..., n).

...%... replaces each occurrence of %d with the corresponding integer in the range, yielding the string 01234567891011...n.

(...)[n]<'1' selects the character at index n and tests if it is less than the character 1.

Answer 9

Jelly, 6 bytes

RDF⁸ị¬

Try it online! or verify all test cases.

How it works

RDF⁸ị¬  Main link. Argument: n

R       Range; yield [1, ..., n].
 D      Decimal; convert all integers in that range to base 10 arrays.
  F     Flatten the result.
   ⁸ị   Extract the digit at index n (1-based).
        This returns 0 if the array is empty (n = 0).
     ¬  Logical NOT; return 1 if the digit is 0, 0 if not.

Answer 10

Python 3, 44 bytes

lambda n:''.join(map(str,range(n+1)))[n]<'1'

An anonymous function that takes input via argument and returns True or False as appropriate.

How it works

lambda n      Anonymous function with input n
range(n+1)    Yield the range [0, n]...
map(str,...)  ...convert all elements to string...
''.join(..)   ...concatenate...
...[n]        ...yield nth character...
:...<'1'      ...return True if int(character)==0 else return False

Try it on Ideone

Answer 11

Pyth, 8 7 bytes

Thanks to @LeakyNun for -1 byte

!s@jkUh

This is my first attempt at golfing in Pyth.

A full program that prints True or False as appropriate.

Try it online

First 25 truthy inputs

How it works

!s@jkUh    Program. Input: Q
      hQ   Head. Yield Q+1
     U     Unary range. Yield [0, Q]
   jk      Join. Join on empty string
  @     Q  Index. Yield string[Q]
 s         Integer. Convert to integer
!          Logical negation. 0 -> True, all other digits -> False
           Print. Print result implicitly

Answer 12

S.I.L.O.S, 141 bytes

readIO
i+1
lblL
c=C
p=1
lblc
p*10
c/10
if c c
p/10
lbln
m=C
m/p
m%10
p/10
i-1
if i C
GOTO H
lblC
if p n
C+1
GOTO L
lblH
m/m
m-1
m|
printInt m

Try it online!

Uses only 5 integers, maximum memory efficiency \o/

Explanation

We generate as many digits as the input in the Champernowne's constant.

In the main loop, we do the following:

• Find the length of the current number by floor_dividing it by 10 repeatedly until it reaches 0, and then count the number of divisions used.
• Instead of storing the number of divisions, we store 10 to that number power instead.
• Iterate through each digit as such: the 100s digit of 1234 is obtained by (1234/10)%10 where / is floor division.
• For each digit generated, take 1 from the input, while checking if the input reached zero.
• If the input reaches zero, check if the current digit is 0 and then halts.

Answer 13

JavaScript (ES6), 45 bytes + Kudos

f=(n,s='0')=>s[n]?!+s[n]:f(n-s.length,-~s+'')

My best non-Kudos version was 34 bytes:

n=>!+(g=s=>s[n]||g(s+i++))('',i=0)

Answer 14

JavaScript (ES6), 47 bytes

n=>[...Array(n+1)].reduce((a,_,i)=>a+i,'')[n]<1

Answer 15

Javascript (ES6), 42 33 bytes

n=>!+(r=i=>i>n?'':i+r(i+1))(0)[n]

Example:

let f =
n=>!+(r=i=>i>n?'':i+r(i+1))(0)[n]

// test all integers in [0, 312]
for(var n = 0, list = []; n <= 312; n++) {
  f(n) && list.push(n);
}
console.log(list.join(','));

Answer 16

Groovy, 56 bytes

def f(n){def s=''<<'';(0..n).each{s<<it};!(s[n] as int)}

Nothing fancy, but I'm trying out some new things.

def f(n) {
  def s = ''<<''           // declare a StringBuffer
  (0..n).each { s << it }
  !(s[n] as int)           // Groovy considers a non-null char truthy, so we have to cast 
}

Answer 17

Perl, 24 bytes

Includes +1 for -p

Run with input on STDIN:

zero.pl <<< 31

print 1 for zero, nothing otherwise

zero.pl

$_=!(map/./g,0..$_)[$_]

Answer 18

Ruby, 35 23 bytes

This is an anonymous function that concatenates [0..n], takes the nth index and checks if that char is "0" (less than "1"). Golfing suggestions welcome.

->n{([*0..n]*'')[n]<?1}

Ungolfing

->n{...}   # Create an anonymous function with parameter n.
[*0..n]    # Create an array of the range [0..n].
[...]*''   # Join the array using the empty string.
(...)[n]   # Take the char at the n-th index of the string.
<?1        # Check if the char is < "1" (that is, "0").

Answer 19

Actually, 9 8 bytes

This answer concatenates the range [0..n], takes the nth index and checks if that char is "0". Golfing suggestions welcome. Try it online!

;urεjE≈Y

Ungolfing

;          Duplicate n
 ur        Increment the duplicate and create range [0..n].
   εj      Join the range with an empty string. Stack: <string> n
     E     Take the char at the n-th index.
      ≈    int(a)
       Y   Logical NOT. If the digit is 0, then return 1, else return 0.

Answer 20

PHP, 36 bytes

<?=!join(range(0,$a=$argv[1]))[$a];

Print 1 if Champernowne argument-th decimal is 0, else print '' (empty string).

Answer 21

R, 61 57 bytes

Thanks to @plannapus for 4 bytes.

n=scan();strsplit(paste(0:n,collapse=""),"")[[1]][n+1]==0

Creates a vector of numbers 0:n (for 0 indexing), creates a string of them, pulls the nth value from string (adjusting for 0 indexing). Converts to numeric and tests if it is 0.

Answer 22

Bash, 31 28 bytes

seq -s "" 0 $1|egrep ^.{$1}0

Output is non-empty (truthy) or empty (falsy). Test it on Ideone.

Answer 23

Julia, 21 20 bytes

!n=join(0:n)[n+1]<49

Thanks to @LuisMendo for golfing off 1 byte!

Try it online!

Answer 24

GolfScript, 12 bytes

~.),""*\=48=

Explanation:

~             Evaluate the input.
 .            Duplicate it
  )           Increment the duplicate.
   ,          Create an array from 0 to it.
    ""*       Join it with an empty string.
       \=     Get the n-th index of this string, where n is the input
         48=  Is it equal to 0?

Try it online or verify all test cases!

Answer 25

Javascript (ES5): 61 60 bytes

function(b){for(s="";s.length<b;)s+=s.length;return 0==s[b]}

Ungolfed:

function a(b){
  for(var s="";s.length<b;)
    s+=s.length;
  }
  return (s[b]==0);
}

Old:

function(n){s="";while(s.length<n)s+=s.length;return s[n]==0}

Ungolfed old:

function a(n){
  var str="";
  while(str.length<n)str+=str.length; //Create String as long as needed
  return str[n]==0 //Check for 0 and return
}

Answer 26

C, 154 bytes

s(n,v,k,z){for(k=1;(z=n%10,n/=10)&&!v||k<v;++k); return v?z:k;}
f(n,i,j,c){for(i=0,j=0;;++i){c=s(i,0,0,0);j+=c;if(j>n){c=s(i,j-n,c,0);break;}}return !c;}

the function that calculate the value is f(n,0,0,0) where n is the input index. it can calculate from index changing "return !c" in "return c" the value of the array in that index... i don't understand how but it seems to work ok....

main()
{int   i,r;
 char  a[]="0123456789101112131415161718192021222324252627282930313233343536";

 for(i=0; i<1000; ++i) 
    if(r=f(i,0,0,0))  
        printf("%u|",i);
}
/*
 154
 0|11|31|51|71|91|111|131|151|171|191|192|194|197|200|203|206|209|212|215|218|222
|252|282|312|342|372|402|432|462|491|492|494|497|500|503|506|509|512|515|518|522|552
|582|612|642|672|702|732|762|791|792|794|797|800|803|806|809|812|815|818|822|852
|882|912|942|972|
*/

Answer 27

CoffeeScript, 56 bytes

a=(b)->
 s=""
 while s.length<b #loop for building string with required length
  s+=s.length     #add number
 "0"==s[b]        #return, if the number at the position equals zero

Answer 28

zsh, 31 bytes

exit ${${(j..):-{0..$1}}[$1+1]}

exit 0 is true in zsh

Answer 29

C#, 71 bytes

And I thought it was short at first, but then I had to add n+=11 to prevent it from throwing a System.IndexOutOfRangeException when numbers below 11 were input

return String.Join("",Enumerable.Range(0,n+=11).ToArray())[n]=='0'?1:0;

Answer 30

Pyke, 7 bytes

m`sQ@b!

Try it here!

Answer 31

PowerShell v2+, 32 bytes

param($n)(-join(0..$n))[$n]-eq48

Does the same as other answers -- takes the range from 0 to input $n, -joins together into a string, takes the [$n] index, checks if it's -equal to 48 (i.e., ASCII for 0).

Note this will only work up to input 49999, as the PowerShell range operator .. only supports ranges of 50,000 elements or fewer. For larger inputs, you'll need to use the following, based on my previous Champernowne answer, at 58 bytes:

param($n)$j=$n;for($c='';$j-ge0;$j--){$c+=$i++}$c[$n]-eq48

Answer 32

Octave, 41 29 bytes

Credit to Dennis, whose Julia answer made me realize I wasn't treating input 0 correctly.

@(n)sprintf('%i',0:n)(n+1)<49

This is an anonymous function.

Try at Ideone.

Explanation

For input n, this generates the Champernowne digits corresponding to the first n+1 numbers starting at 0 (which is more than enough). sprintf is used for this so that no blank spaces appear. The result is true if the n+1-th digit (1-based) is zero, that is, if its ASCII code is less than 49.

Answer 33

R, 57 bytes

grepl("0$",substr(paste(0:(n=scan()),collapse=""),1,n+1))

Build a string corresponding to 0:n, keeps the first n+1 characters and check if the last character is a zero.