Make big rocks into small rocks

Question

Welcome to the grinder.

Your task is to make big rocks into small rocks by grinding them.

Take an input of a big rock of size n > 3 and grind it.

Continue to grind the rocks by dumping them into the grinder until the size of all the rocks are 2.

rocks are always grinded into equal even halves. If the result of a grinding is odd take result - 1.

Print the output of each grinding as you proceed.

Examples

input: 5

output: 22

The result are two rocks of size 2

input: 50

output:

2424 //two rocks of size 24
12121212 //four rocks of size 12
66666666 //8 rocks of size 6
2222222222222222

the result is 16 rocks of size 2

input: 30

output:

1414
6666
22222222

the result is 8 rocks of size 2

This is code-golf so the shortest code wins! Have fun and good luck!

Answer 1

TSQL, 61 59 bytes

DECLARE @ int=50

,@x int=1WHILE @>2SELECT @=@/4*2,@x*=2PRINT replicate(@,@x)

Try it out

Answer 2

COW, 297 291 bytes

MoOMoOMoOMoOMoOMoOMoOMoOMoOMoOmoOMoOmoOmoOoommOoMoOMOOmoOMMMmoOMMMmoOOOOMoOmOoMOOMOomoOmoO
MOOMOomOoMOomoOmoomOoMMMOOOMoOmoOMMMmOomOomoomoOmoOMOOMOomOomOomOoMOomoOmoOmoOmoomOomOomOo
mOomOoMMMmoOMMMMOOMOomoOOOMmOomOoMoOmoOmoomOomOoMoomoOmoOmoOMOOMOoMOomoOMoOmOomoomoOMMMOOO
mOoMMMMMMmOoMMMMOomoo

Try it online!

The code prints each number on its own line, and separates iterations with an additional newline. It also prints the first iteration by itself, followed by a newline. So an input of 5 would give an output that looks like 5 2 2 except with newlines instead of spaces. Sample output for 50 is given below.

Explanation tree:

MoOMoOMoOMoOMoOMoOMoOMoOMoOMoOmoOMoOmoOmoOoom ;Store 10 in [0], 1 in [1], and integer input in [3]
mOoMoO                                        ;Store 1 in [2]
MOO                                           ;Loop while [2] is non-zero
   moOMMMmoOMMMmoOOOOMoOmOo                   ;   Copy [3] to [4], clear contents of [5], and store 1 in [5]
   MOO                                        ;   Loop while [4] is non-zero
      MOomoOmoO                               ;      Decrement 4 and move to 6
      MOO                                     ;      Loop while [6] is non-zero
         MOomOoMOomoO                         ;         Decrement [5] and [6]
      moo                                     ;      End loop once [6] is empty
      mOoMMMOOOMoOmoOMMMmOomOo                ;      Copy [5] to [6], and reset [5] to 1, then move back to [4]
   moo                                        ;   End loop now that [4] is empty.  [6] now contains the parity of [3]
   moOmoO                                     ;   Navigate to [6]
   MOO                                        ;   Loop while [6] is non-empty
      MOomOomOomOoMOomoOmoOmoO                ;      Decrememnt [3] and [6]
   moo                                        ;   End loop now that [6] is empty.  [3] now contains the largest even number less than the previous iteration.
   mOomOomOomOomOoMMMmoOMMM                   ;   Copy [1] to [2]
   MOO                                        ;   Loop while [2] is non-empty
      MOomoOOOMmOomOoMoOmoO                   ;      Decrement [2], increment [1], and print the number in [3].
   moo                                        ;   End loop now that [2] is empty
   mOomOoMoo                                  ;   Print a new line
   moOmoOmoO                                  ;   Navigate to [3]
   MOO                                        ;   Loop while [3] is non-empty
      MOoMOomoOMoOmOo                         ;      Decrement [3] twice and increment [4] once
   moo                                        ;   [4] now contains half of [3]
   moOMMMOOOmOoMMM                            ;   Copy [4] to [3] and clear [4]
   MMMmOoMMMMOo                               ;   Copy [3] to [2] and decrement once
moo                                           ;End loop now that [2] is empty

Sample output for input 50:

50

24
24

12
12
12
12

6
6
6
6
6
6
6
6

2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2

Answer 3

05AB1E, 12 11 bytes

¸[4÷·€D=¬<#

Try it online!

Explanation

¸             # wrap input in a list
 [            # start infinite loop
  4÷          # elementwise integer divison by 4
    ·         # elementwise multiplication by 2
     €D       # duplicate each element in the list
       =      # print it
        ¬     # get the first element of the list
         <    # decrease it by 1
          #   # if true: exit loop

Answer 4

Python 2, 55 53 bytes

n=input()
while n[0]>2:n=len(n)*2*[n[0]/4<<1];print n

Divide by 4 and left shift by 1 to get the special division

Answer 5

Haskell, 75 71 60 50 47 bytes

f 0=[]
f n|x<-f$2*div n 4=show n:zipWith(++)x x

Try it online! Edit: As the output is now allowed to be a list including the input, 10 13 bytes can be saved.

Usage:

Prelude> f 50
["50","2424","12121212","66666666","2222222222222222"]

---

Original 60 byte version:

2%x=""
n%x|z<-2*div n 4=([1..x]>>show z)++"\n"++z%(x*2)
(%2)

Try it online! Thanks to Christian Sievers for pointing out the shorter formula.

Usage:

Prelude> (%2)50
"2424\n12121212\n66666666\n2222222222222222\n"

Answer 6

Python 2, 48 47 bytes

s=input()
n=1
while s>3:s=s/4*2;n*=2;print`s`*n

Answer 7

JavaScript (ES6) 64 59 57 Bytes

f=s=>{for(n=1;s>2;)console.log(`${s=s/4<<1}`.repeat(n*=2))}

console.log(f.toString().length); 
f(5);
f(50);
f(30);                                  

Answer 8

C#, 88 86 83 bytes

Saved 3 bytes thanks to Skorm

Saved another byte by changing the while to a for loop which includes variable declarations

Saved 1 bytes thanks to Yodle

n=>{var r="";for(int i,c=2;n>2;c*=2,r+="\n")for(i=0,n=n/4*2;i++<c;)r+=n;return r;};

Anonymous function which returns a string composed of the result of each grinding.

Full program with ungolfed method and test cases [before the last edit!]:

using System;

public class Program
{
    public static void Main()
    {
        Func<int, string> f =
        n =>
        {
            var r = "";
            for (int i, c = 1; n > 2; )  // iterator and counter variable
            {
                    n = n/4 * 2;    // make sure the result if even
                    c *= 2;         // keep track of the number of rocks
                    for (i = 0; i++ < c; )  // store the current line made of [c] rocks of size [n]
                        r += n;
                    r += "\n";      // add a trailing newline to the string resulted from this step
            }
            return r;       // return the entire history
        };

        //test cases:
        Console.WriteLine(f(5));
        Console.WriteLine(f(50));
        Console.WriteLine(f(30));
    }
}

Answer 9

Java, 85 bytes

n->{String s="";for(int q=2,i;n>2;q*=2,s+="\n")for(i=q,n=n/4*2;i-->0;)s+=n;return s;}

Testing and ungolfed

import java.util.function.*;

class Ideone {
  public static void main(String[] args) throws java.lang.Exception {
    Function<Integer, String> f = number -> {
      String result = "";
      for (int quantity = 2, i; number > 2; quantity *= 2) {
        number = number / 4 * 2; // Make sure that the next is half or half - 1 if odd
        for (i = quantity; i > 0; i--) { // copy "quantity" times.
          result += number;
        }
        result += "\n"; // append new line
      }
      return result;
    };
    System.out.println(f.apply(50));
  }
}

Note: I don't know why, Ideone keeps giving internal error, so testing on it is a problem. To test, just copy/paste and run in your standard Java IDE. (It works there, I made sure of it ;) )

Answer 10

CJam, 21 bytes

l~]{{2/-2&_}%_n_2-}g;

Try it online! (As a test suite.)

Explanation

l~]      e# Read input, evaluate and wrap it in a singleton list.
{        e# Do while...
  {      e#   Map this block over the list of rocks.
    2/   e#   Halve the rock.
    -2&  e#   Bitwise AND with -2, clearing the least-significant bit and
         e#   rounding down to an even integer.
    _    e#   Duplicate.
  }%
  _n     e# Print a copy of the current list of rocks.
  _2-    e# Continue if the current list of rocks contains values that aren't 2.
}g
;        e# Discard the final result to prevent printing it again.

Answer 11

Jelly, 13 12 11 bytes

:4Ḥx2µȦпṖY

TryItOnline!

Note: the OP stated that the input may also be in the output.

How?

:4Ḥx2µȦпṖY - Main link: rockSize
     µ      - monadic separation
       п   - loop collect intermediate results while
      Ȧ     - any
:4          -     integer division (vectorises) by 4
  Ḥ         -     double (vectorises)
   x2       -     repeat the elements 2 times
         Ṗ  - pop (remove the trailing list of zeros)
          Y - join with line feeds

Version without the input displayed for 12 bytes: :4Ḥḟ0x2µÐĿḊG

Answer 12

Pyth, 18 16 13 bytes

WhQ=Q*2my/d4\n

^{* \n is a new line}
Explanation :

W              # While
 hQ            # first element of Q - 0 is falsy
   =Q          # assign to Q
     *2        # the double of the (list) that is returned
       m       # form this \/ map
         /d4   # divide every element by 4
        y      # and double
            \n # print Q

Try here

Answer 13

PHP, 72 67 64 bytes

for($n=$argv[$k=1];$n>2;)echo str_repeat($n=$n/2&~1,$k*=2),"\n";

Takes argument from command line. Run with -r.

Answer 14

MATL, 13 bytes

`K/kEthttH>]x

Try it online!

`       % Do...while
  K/k   %   Divide by 4 and round down. Takes input implicitly in the first iteration
  E     %   Multiply by 2
  th    %   Attach a copy of itself (creates a longer array)
  t     %   Duplicate. This copy will be used for further grinding, keeping the original
  tH>   %   Duplicate. True if the values exceed 2. Used as loop condition
]       % End. The loop exits if the latest array contains 2
x       % Delete last copy. Implicitly display the entire stack

Answer 15

Perl, 40 35 30 + 1 = 31 bytes

Run with the -n flag

-4 bytes thanks to @Dada

say$_ x($.*=2)while$_=$_>>1&~1

Try it online!

Perl automatically reads the input into the variable $_ when -n is set. $. is a special variable set to 1 at the beginning of the program by the interpreter, so I can use it as a base for doubling. Every iteration of the while loop, it bit-shifts $_ down and performs a logical AND against the negative of itself minus one to cancel out the ones bit.

Answer 16

PowerShell 3+, 58 54 bytes

for($s=$input;$s;$s=($s-shr2)*2){"$s"*(2-shl($i++)-1)}

Thanks TimmyD for saving me 4 bytes!

Slightly Ungolfed (formatting)

for ( $s = $input ; $s ; $s = ( $s -shr 2 ) * 2 ) {
    "$s" * (2 -shl ($i++)-1)
}

Explanation

I'm using the same divide by 4 multiply by 2 trick as a lot of the other answers, but I ran into a problem. PowerShell converts numbers to floating point if needed during division, and for golfing that's annoying because $v/4*2 becomes something unsighlty like [int]($v/4)*2. I got around that using bitshifting for the division with -shr.

For calculating how many times to print an iteration I just take (2^$i)-1 which works nicely and has the added effect of leaving out the input value. Trying to just multiply by 2 was problematic because starting from 0 makes it hard to increase the value with just $i*=2 and starting from 1 requires too much correction to get the number right.

Since PowerShell doesn't have an operator for it, and I wanted to avoid [Math]::Pow(), I relied on bitshifting again for my powers of 2.

Answer 17

Python 2, 47 Bytes

Since the OP said that a 1D array which included the input was fine I have come up with this recursive function, which unfortunately only ties with the current Python winner.

f=lambda s,n=1:[s]*n+(f(s/4*2,n*2)if s>3else[])

Answer 18

Vim 61 54 bytes

qqYpPJ0yw:s;\d*;="/2
;g
:let @t=(">7)+1
@tkjjG@qq@q

TryItOnline!

Unprintables:

qqYpPJ0yw:s;\d*;^R=^R"/2
;g
:let @t=(^R">7)+1
@tkjjG@qq@q

Luckily vim automatically truncates on x/2.

Answer 19

Perl, 47 bytes

$a=<>>>1;say 2*(($a>>=1)||die)x(1<<$_)for 1..$a

No command-line options, this time (unusually for Perl). The basic idea is that as all the rocks at any given step are the same size, we just record the size (in $a) and the number (in $_), rather than recording the whole list. I couldn't find a way to get rid of the space (or +) after say; you can move the 2* but it won't parse correctly if it's followed by an opening parenthesis.

I can't help but shake the feeling that this is improvable, but I can't see how.

Answer 20

JavaScript, 71 63 59 58 Bytes

Well, I came up with this javascript solution. Totally new at golfing, but I foudn this a fun challenge

Saved 4 bytes thanks to Titus suggestion using a for loop.

ungolfed basis:

for(o = i = 30; i > 1; i= i/4<<1) {
   console.log(`${i}`.repeat(o / i));
}

Golfed version

for(o=i=30;i>1;i=i/4<<1){console.log(`${i}`.repeat(o/i));}

I'm open for suggestions how to improve it / learn golfing

input tester

function test(a){
var el = document.getElementById('output');
var text = []; 
for(o = i = parseInt(a); i > 1; i= i/4<<1) {
   text.push(`${i}`.repeat(o / i));
}
  el.innerText = text.join('\n');
}

<input type="input" onchange="test(this.value)"><button>run</button>
<pre id="output">
</pre>

Answer 21

BASH, 81 bytes

n=$1
h=1
while [ ${n//2} ];do
printf -v s %$[h=h*2]s
echo ${s// /$[n=n/4*2]}
done

Answer 22

Swift, 84 Bytes

func g(n:Int){var n=n,i=2;while n>2{n=n/4*2;print(Array(repeating:n,count:i));i*=2}}

---

Ungolfed

func grind(rockSize: Int) {
    var rockSize = rockSize
    var rockCount = 1

    while rockSize > 2 {
        rockSize = rockSize / 4 * 2
        rockCount *= 2

        let output = Array(repeating: rockSize, count: rockCount)
        print(output)
    }
}

Answer 23

Befunge, 45 bytes

&1vg0_\:.\v
:\<  ^!:-1<p00:*2\.:*2/4,+55_@#`2

Try it online!

Explanation

&           read the rock size
1           initialise the count
<           start of main loop going right to left

  \         swap the size to the top of the stack
  :2`#@_    if size is not > 2 then exit
  55+,      output a line break
  4/2*      size = size/4*2, i.e. split into even halves
  :.        output the size
  \         swap the count to the top of the stack
  2*        count = count*2
  :00p      save count for later

  <         start of inner loop
    1-      decrement the count
    :!^_    break out of the loop if the count is zero
    \       swap the size to the top of the stack
    :.      output the size
    \       swap the count to the top of the stack
    v       back to the start of the inner loop    

  0g        restore the saved count
  v         back to the start of the main loop

Answer 24

Javascript, 106 bytes

First code golf, thought I'd have a go. (It's not very good).

for(;i[i.length-1]>3;){for(var x=0;x<i.length;x++)i[x]/=2,i[x]%2===1&&i[x]--;i=i.concat(i),console.log(i)}

Unminified:

while (input[input.length - 1] > 3) {
    for (var x = 0; x < input.length; x++) {
        input[x] /= 2;
        if (input[x] % 2 === 1) input[x]--;
    }
    input = input.concat(input);
    console.log(input);
}