Magic popcount numbers

Question

There’s a famous tricky algorithm for counting the number of set bits in a 32-bit unsigned integer:

int popcount(unsigned x) {
   x = (x & 0x55555555) + ((x >> 1) & 0x55555555);
   x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
   x = (x & 0x0F0F0F0F) + ((x >> 4) & 0x0F0F0F0F);
   x = (x & 0x00FF00FF) + ((x >> 8) & 0x00FF00FF);
   x = (x & 0x0000FFFF) + ((x >>16) & 0x0000FFFF);
   return x;
}

I won’t explain it here. But imagine similar code for 512-bit integers! The hexadecimal constants would be huge, and form a pretty pattern. Your task is simply to print this exact output:

0x55555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555
0x33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333
0x0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f
0x00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff
0x0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff
0x00000000ffffffff00000000ffffffff00000000ffffffff00000000ffffffff00000000ffffffff00000000ffffffff00000000ffffffff00000000ffffffff
0x0000000000000000ffffffffffffffff0000000000000000ffffffffffffffff0000000000000000ffffffffffffffff0000000000000000ffffffffffffffff
0x00000000000000000000000000000000ffffffffffffffffffffffffffffffff00000000000000000000000000000000ffffffffffffffffffffffffffffffff
0x0000000000000000000000000000000000000000000000000000000000000000ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

No trailing spaces, please — though a single trailing newline is optional.

This is code-golf, so the shortest answer (in bytes) wins.

Answer 1

Python 2, 52 49 46 bytes

The kth number is given by 2**512/(2**2**k + 1). This is for a 512 bit number, so it's trivial to extend the pattern to different widths.

l=2;exec"print'0x%0128x'%(2**512/-~l);l*=l;"*9

3 bytes saved thanks to Dennis.
3 bytes saved thanks to xnor.

Answer 2

PHP, 111 110 108 bytes

One byte saved thanks to @user59178.

<?=($p=str_pad)("0x",130,5).$p($s="\n0x",131,3);for($x=1;$x<65;$x*=2)echo($p($s,131,$p(0,$x,0).$p(f,$x,f)));

What´s the pattern for 1024 bits? :D

Answer 3

Vim, 32 bytes

i5<CR>3<Esc>qqYpVrf$<C-V>{yPG1vr0q6@q<C-V>{I0x<Esc>

I only need to manually write the first 5 and 3, and the macro takes care of the rest, "doubling the bit count" each time it runs. The order of the steps in the macro is a bit weird (make a new f line, block-wise copy, reuse the visual block size to put 0s in the f line), but it's the fastest variant I've found.

Answer 4

Retina, 43 bytes

:`
0x128$*5
:`5
3
;{:`33
0f
0(f+)(0+)
0$2$1

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Explanation

This makes a lot of use of the generally underused : option which lets you print intermediate results, because it's a lot shorter to modify a single line than to build up the whole output.

:`
0x128$*5

This replaces the empty input with 0x followed by 128 5s and prints it to generate the first line.

:`5
3

This one replaces the 5s with 3s to generate the second line and prints it as well.

;{:`33
0f

This is the last special-cased line and it turns every two 3s into 0f to generate the third line. This also starts a loop through the last two stages ({). However, this stage won't do anything after the first iteration except print the current state. The ; suppresses the output at the very end of the program to avoid duplicating the last line.

0(f+)(0+)
0$2$1

This substitution now transforms each line into the next, by swapping every other pair of fs and 0s. The "every other pair" condition is enforced by matching a zero in front of the f, which makes it impossible to match consecutive pairs since matches cannot overlap.

Answer 5

J, 46 34 bytes

I'm working on golfing this, but this baby likes to stay at 46 bytes... Not anymore! -12 bytes thanks to miles!

'0x',"1'5','3','0f'(128$#)"{~2^i.7

Try it online! :D

Result

   '0x',"1'5','3','0f'(128$#)"{~2^i.7
0x55555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555
0x33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333
0x0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f0f
0x00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff00ff
0x0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff0000ffff
0x00000000ffffffff00000000ffffffff00000000ffffffff00000000ffffffff00000000ffffffff00000000ffffffff00000000ffffffff00000000ffffffff
0x0000000000000000ffffffffffffffff0000000000000000ffffffffffffffff0000000000000000ffffffffffffffff0000000000000000ffffffffffffffff
0x00000000000000000000000000000000ffffffffffffffffffffffffffffffff00000000000000000000000000000000ffffffffffffffffffffffffffffffff
0x0000000000000000000000000000000000000000000000000000000000000000ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

---

For this answer, I needed (ideally) a verb with rank 0 1 so as to use it in the u"v definition of rank; however, miles observed that 0 _ was sufficient for the task at hand.

┌──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┐
│= │< │<.│<:│> │>.│>:│+ │+.│+:│* │*.│*:│_ 0 0│_ 0 0│0 0 0│0 0 0│0 0 0│0 0 0│0 0 0│0 0 0│0 0 0│0 0 0│0 0 0│0 0 0│0 0 0│
├──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┤
│- │-.│-:│% │%.│%:│^ │^.│$ │$.│$:│~.│~:│0 0 0│0 _ _│0 _ _│0 0 0│2 _ 2│0 0 0│0 0 0│0 0 0│_ 1 _│_ _ _│_ _ _│_ 0 0│_ 0 0│
├──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┤
│| │|.│, │,.│,:│; │;:│# │#.│#:│! │/:│\:│0 0 0│_ 1 _│_ _ _│_ _ _│_ _ _│_ _ _│1 _ _│_ 1 _│1 1 1│_ 1 0│0 0 0│_ _ _│_ _ _│
├──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┤
│[ │[:│] │{ │{.│{:│}.│}:│".│":│? │?.│a │_ _ _│_ _ _│_ _ _│1 0 _│_ 1 _│_ 0 0│_ 1 _│_ 0 0│1 _ _│_ 1 _│0 0 0│_ 0 0│_ _ _│
├──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┤
│A │A.│b │C │C.│d │D │e │e.│E │E.│f │H │_ _ _│1 0 _│_ _ _│_ _ _│1 1 _│_ _ _│_ _ _│_ _ _│_ _ _│_ _ _│0 _ _│_ _ _│_ _ _│
├──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┤
│i │i.│i:│I │I.│j │j.│L │L.│M │o │o.│p │_ _ _│1 _ _│0 _ _│_ _ _│1 _ _│_ _ _│0 0 0│_ _ _│_ 0 0│_ _ _│_ _ _│0 0 0│_ _ _│
├──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼──┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┤
│p.│p:│q │q:│r │r.│s │s:│S │t │T │u:│x:│1 1 0│0 _ _│_ _ _│0 0 0│_ _ _│0 0 0│_ _ _│_ _ _│_ _ _│_ _ _│_ _ _│_ _ _│_ _ _│
└──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┘

Here you see a bunch of verbs' string representations with their respective ranks. This is the script I used to generate it.

Answer 6

Pyth, 26 bytes

V9%"0x%0128x"/^2 512h^2^2N

Port of my Python answer.

Answer 7

Actually, 25 bytes

9r`╙╙u9╙╙\#"0x%0128x"%`Mi

Try it online!

This solution uses the fact that f(n) = 2**512//(2**2**n + 1) (where // is floored division) to compure the values.

Explanation:

9r`╙╙u9╙╙\#"0x%0128x"%`Mi
9r`╙╙u9╙╙\#"0x%0128x"%`M   for n in range(1, 10):
      9╙╙\                   2**2**9//
   ╙╙u                                (2**2**n + 1)
          #"0x%0128x"%       pad with zeroes to 128 digits, prefix with "0x"
                        i  flatten and implicitly print

Answer 8

JavaScript (Firefox 30+), 139 113 112 92 83 80 bytes

_=>[for(x of"970123456")(f=y=>y--?f(y)+(x>6?x-4:y>>x&1&&'f'):'0x')(128)].join`
`

Finally hit the recursive sweet spot :-) Uses a handy-dandy string comprehension to save 3 bytes over .map:

_=>[..."970123456"].map(x=>(f=y=>y--?f(y)+(x>6?x-4:y>>x&1&&'f'):'0x')(128)).join`
`

.replace is also 83 bytes:

_=>"970123456".replace(/./g,x=>(f=y=>y--?f(y)+(x>6?x-4:y>>x&1&&'f'):'0x')(128)+`
`)

If a leading newline were allowed, this would also be 80 bytes:

_=>"970123456".replace(/./g,x=>(f=y=>y--?f(y)+(x>6?x-4:y>>x&1&&'f'):`
0x`)(128))

Answer 9

05AB1E, 26 22 21 bytes

05AB1E uses the CP-1252 encoding.

9F„0x0NÍo×9ooNoo>÷hJ,

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Explanation

9F                      # for N in [0 ... 8]
  „0x                   # push the string "0x"
     0NÍo×              # push 2**(N-2) zeroes
          9oo           # 2**2**9
                 ÷      # //
             Noo>       # (2**2**N+1)
                  h     # converted to base-16
                   J    # join everything to string
                    ,   # print with a newline

Other versions that might be improved

9F9ooNoo>÷h¾6o×J7o£R…0xÿ,
9F9ooNoo>÷h0žy×ìR7o£R…0xÿ,
9FX0No×1No×JCh7o×1K7o£…0xÿ,
8ÝovX0y×1y×JCh7o×1K7o£…0xÿ,
9F1NoÅ0D>)˜JCh1K7o×7o£…0xÿ,

Answer 10

Python 2, 60 bytes

k=1
exec"print'0x%0128x'%int(256/k*('0'*k+'1'*k),2);k*=2;"*9

Try it online!

Answer 11

Bubblegum, 65 bytes

00000000: c5cb 4501 0441 1043 d17b d4fc 254b d110  ..E..A.C.{..%K..
00000010: f7cb 9761 9e7a 8d45 e451 4ce4 564c 04d7  ...a.z.E.QL.VL..
00000020: 2e11 b02b 8f08 80df aa5e 11fe fc77 762c  ...+.....^...wv,
00000030: 428b 5b8e ae8b 30c1 13b6 ce8b b091 377a  B.[...0.......7z
00000040: 01                                       .

Obligatory Bubblegum answer.

Answer 12

Haskell, 84 72 bytes

Porting @orlp's answer:

import Text.Printf
mapM(\n->printf"0x%0128x\n"$div(2^2^9)$2^2^n+1)[0..8]

94 byte alternatives without the power of Text.Printf:

import Data.List
mapM(putStrLn.("0x"++))$transpose$("53"++).reverse<$>sequence(["0f"]<*[1..7])

r=[0..127]
mapM(putStrLn.("0x"++))$('5'<$r):('3'<$r):[["0f"!!mod(div x(2^y))2|x<-r]|y<-[0..6]]

Answer 13

Batch, 216 bytes

@echo off
set s=5
call:c
set s=3
call:c
set a=0
set b=f
for /l %%i in (1,1,7)do call:l %%i
exit/b
:l
set s=%a%%b%
:c
for /l %%j in (0%1,1,6)do call set s=%%s%%%%s%%
echo 0x%s%
set a=%a%%a%
set b=%b%%b%

Answer 14

PowerShell v2+, 68 bytes

5,3|%{"0x"+"$_"*128}
($a=1)..7|%{"0x"+('0'*$a+'f'*$a)*(128/($a*=2))}

PowerShell doesn't have arbitrary precision integers without using [bigint] calls, and those can't be easily converted to hexadecimal, so we're instead treating this as a string-based challenge.

The first line handles the repeating 5 and 3 by just doing a string multiplication out to 128 characters and tacking a 0x on the front.

The next line loops from $a=1 to 7, each iteration outputting another string. Again we have the 0x tacked onto the front, and we're doing string multiplication in the middle to construct the appropriate number of 0 and f concatenated together, and then doing string multiplication of that out to the appropriate number of characters. Note we're using variable $a here, and not the loop counter $_, so we can properly scale (otherwise we'd need to loop like 1,2,4,8,16,32,64|%{...}, which is longer).

The resulting strings are left on the pipeline, and output via implicit Write-Output happens at program completion, with a newline between elements.

Answer 15

V, 43 bytes

64i0fòYpÓ¨¨0«©¨f«©©û2}/²²³³òdd{3ÄÒ5jÒ3Îi0x

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This uses one of the longest compressed regexes I've ever needed in a V answer. Here is the more readable version, where I added a byte for readable regexes, and changed the unprintable escape character to <esc>

64i0f<esc>òYpÓö((0+)(f+)){2}/²²³³òdd{3ÄÒ5jÒ3Îi0x

Explanation (using the readable version):

64i0f<esc>                                          " Insert 64 "0f"s and escape to normal mode
          ò                      ò                  " Recursively:
           Yp                                       "   Duplicate this line
             Ó                                      "   Substitute:
              ö                                     "     (Optionally Turn the readable version on)
               ((0+)(f+))                           "     One or more '0's followed by one or more 'f's
                         {2}                        "     Repeated twice
                            /                       "   With:
                             ²²                     "     The second capture group twice (the '0's)
                               ³³                   "     Followed by the third capture group twice (the 'f's)
                                                    "   Once the search is not found, the loop will break
                                  dd                " Delete a line (because we have one too many)
                                    {               " Move to the first line
                                     3Ä             " Make three copies of this line
                                       Ò5           " Replace the first one with '5's
                                         jÒ3        " Move down a line and replace the second with '3's
                                            Î       " On every line:
                                             i0x    "   Insert a '0x'

Answer 16

JavaScript (ES6), 74 72 70 bytes

Includes the optional trailing newline.

f=(i=1152)=>i--?f(i)+(i&127?'':`
0x`)+('53'[y=i>>7]||i&1<<y-2&&'f'):''

f=(i=1152)=>i--?f(i)+(i&127?'':`
0x`)+('53'[y=i>>7]||i&1<<y-2&&'f'):''

console.log(f())

Answer 17

Ruby, 66 60 45 bytes

x=2;9.times{puts"0x%0128x"%(2**512/-~x);x*=x}

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Idea stolen from orlp

Answer 18

Pyth - 31 30 bytes

To get pattern except for the 3's and 5's it cumulative reduces, each time doubling the chunks.

jm+"0x".[dd128+`3+`5.u.iNN6"0f

Try it online here.

Answer 19

Vim 72 bytes

i0x128a5Ypll128r3o0x64a0fa0Ypqqffdt0fft0p@qq@qqwYp@qq@w@w@w@w:%s/0$

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Unprintables:

i0x^[128a5^[Ypll128r3o0x^[64a0f^[a0^[Ypqqffdt0fft0p@qq@qqwYp@qq@w@w@w@w:%s/0$

The 4 @ws at the end are bugging me, but because I was relying on the @q to fail out at the end of a line, it fails out the @w as well. I might try just running q 32 times and seeing if it messes up the later lines.

Answer 20

C, 146 bytes

#define F for(i=0;++i<129;)s[i+1]=
#define P ;puts(s);
i,j;f(){char s[131]={'0','x'};F'5'P F'3'P for(j=1;(j*=2)<129;){F(i-1)%j<j/2?'0':'f'P}}

Ungolfed:

#define F for(i=0;++i<129;)s[i+1]=
#define P ;puts(s);
i,j;f(){
  char s[131]={'0','x'};
  F'5'P
  F'3'P
  for(j=1;(j*=2)<129;){
    F(i-1)%j<j/2?'0':'f'P 
  }
}

Answer 21

brainfuck, 211 bytes

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

Try it online!

Answer 22

GNU sed 4.2.2, 77

s/^/0x5/
:
s/x5{,127}$/&5/
t
p
y/5/3/
p
s/33/0f/gp
:a
s/(f+)(0+f+)/\2\1/gp
ta

Try it online.

Answer 23

C#, 168 bytes

()={string R="",o="0",f="f";for(int i=0,j;i<9;i++){R+="0x";if(i>2){o+=o;f+=f;}for(j=0;j<128;){R+=i<1?"5":i<2?"3":o+f;j+=i>1?(int)Math.Pow(2,i-1):1;}R+="\n";}return R;};

Answer 24

Stax, 19 bytes

⌡hÅék╝94"ºé♪╛#V┐5í╒

Run and debug it

Unpacked, ungolfed, and commented, it looks like this.

512r        [0..511]
{:Brm       convert each to bits and reverse each
M           transpose matrix, filling missing elements in rectangle with zero
m           map over each element of array, using the rest of the program.  outputs implicitly.
  4/        split bits into groups of 4
  {:b|Hm    convert each 4-bit binary number to single digit hex string
  .0xp      print "0x" without newline

Run this one

Answer 25

///, 193 bytes

/*/\/\///X/
0x*F/5555*G/FFFF*T/3333*U/TTTT*o/0f0f*1/oooo*t/00ff*2/tttt*g/0000*h/ffff*4/ghgh*6/gghh*7/gggg*8/hhhh*9/7788/0xGGGGGGGGXUUUUUUUUX11111111X22222222X44444444X66666666X78787878X99X77988

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Answer 26

JavaScript (Node.js), 72 bytes

n=1n<<512n;f=(i=2n)=>i-n?`
0x`+(n/-~i+n).toString(16).slice(1)+f(i*i):''

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JavaScript (Node.js), 69 bytes from Arnauld's

f=(i=4608)=>i?f(i-=4)+(i&508?'':`
0x`)+('53'[y=i>>9]||i&1<<y&&'f'):''

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Answer 27

Vyxal j, 136 bits^v1, 17 bytes

‛0x53f‛0f6ʀE•J₇vẎ+

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Math? No we don't do that here. We do string manipulation instead.

Explained

‛0x53f‛0f6ʀE•J₇vẎ+­⁡​‎‎⁡⁠⁢⁣‏⁠‎⁡⁠⁢⁤‏⁠‎⁡⁠⁣⁡‏‏​⁡⁠⁡‌⁢​‎⁠⁠⁠‎⁡⁠⁤⁡‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁣⁢‏⁠‎⁡⁠⁣⁣‏⁠‎⁡⁠⁣⁤‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁤⁢‏‏​⁡⁠⁡‌⁢⁡​‎‎⁡⁠⁤‏⁠‎⁡⁠⁢⁡‏⁠‎⁡⁠⁢⁢‏‏​⁡⁠⁡‌⁢⁢​‎‏​⁢⁠⁡‌⁢⁣​‎‎⁡⁠⁤⁣‏⁠‎⁡⁠⁤⁤‏⁠‎⁡⁠⁢⁡⁡‏‏​⁡⁠⁡‌⁢⁤​‎‎⁡⁠⁡‏⁠‎⁡⁠⁢‏⁠‎⁡⁠⁣‏⁠‎⁡⁠⁢⁡⁢‏‏​⁡⁠⁡‌⁣⁡​‎‏​⁢⁠⁡‌­
      ‛0f           # ‎⁡The string "0f"...
            •       # ‎⁢with each character repeated n times...
         6ʀE        # ‎⁣for each n in [2**x for x in range(0, 7)].
             J      # ‎⁤Appended to...
   53f              # ‎⁢⁡the list [5, 3].
# ‎⁢⁢The top of the stack is [5, 3, "0f", "00ff", "0000ffff", ...]
              ₇vẎ   # ‎⁢⁣Take the first 128 items of each item.
                    # ‎⁢⁣This reuses previous digit/chars if the item isn't long enough
‛0x              +  # ‎⁢⁤Add the string "0x" to each item
# ‎⁣⁡The j flag joins on newlines before printing

💎 Created with the help of Luminespire at https://vyxal.github.io/Luminespire