Output a googol copies of a string

Question

I am interested in seeing programs which don't ask for any input, print a googol copies of some nonempty string, no less, no more, and then stop. A googol is defined as 10^100, i.e., 1 followed by a hundred 0's in decimal.

Example output:

111111111111111111111111111111111111111111111111111111111111111111111111...

or

Hello world
Hello world
Hello world
Hello world
Hello world
Hello world
...

The string can also be entirely composed of white space or special symbols. The only exception to identical copies of a fixed string is if your language decorates the output in some way that can not be prevented, but could be trivially undone in a wrapper script, like prepending a line number to each line. The wrapper script in such cases need not be provided.

You can assume your computer will never run out of time, but other than that, your program must have a reasonable demand of resources. Also, you must respect any restrictions that the programming language of your choice poses, for example, you can not exceed a maximum value allowed for its integer types, and at no point more than 4 GB of memory must be needed.

In other words, the program should in principle be testable by running it on your computer. But because of the extent of this number you will be expected to prove that the number of copies of the string it outputs is exactly 10^100 and that the program stops afterwards. Stopping can be exiting or halting or even terminating due to an error, but if so, the error must not produce any output that could not easily be separated from the program's output.

This is code-golf, so the solution with the fewest bytes wins.

Example solution (C, ungolfed, 3768 bytes)

#include <stdio.h>

int main() {
  int a00, a01, a02, a03, ..., a99;
  for(a00 = 0; a00 < 10; a00++)
  for(a01 = 0; a01 < 10; a01++)
  for(a02 = 0; a02 < 10; a02++)
  for(a03 = 0; a03 < 10; a03++)
  ...
  for(a99 = 0; a99 < 10; a99++)
    puts("1");
  return 0;
}

Answer 1

Jelly, 6 4 bytes

³Ȯ*¡

This is a niladic link (function w/o arguments) that prints 10²⁰⁰ copies of the string 100, meaning that it prints 10¹⁰⁰ copies of the string that consists of 10¹⁰⁰ copies of the string 100.

Try it online!

Note that the online interpreter cuts the output at 100 KB for practical reasons. The code also works as a full program, but due to implicit output, that program prints one copy too many.

How it works

³Ȯ*¡  Niladic link. No arguments.

³     Set the left argument and initial return value to 100.
 Ȯ    Print the current return value.
  *   Compute 100 ** 100 = 1e200.
   ¡  Call Ȯ 1e200 times. 

Answer 2

Fuzzy Octo Guacamole, 13 12 11 10 bytes

9+ddpp![g] 

Explanation:

9+ddpp![g]
9+           # push 9 and increment, giving 10
  dd         # duplicate, twice. now you have [10,10,10]
    pp       # raise a 10 to the 10th power, then raise that to the 10th again. That ends up being 10^100.
      ![ ]   # for loop, `!` sets the counter to the top of stack
        g    # prints an ASCII art goat. 

Sample of the goat printed:

                  ___.
                 //  \\
                ((   ""
                 \\__,
                /6 (%)\,
               (__/:";,;\--____----_
                ;; :";,:";`;,";,;";`,`_
                  ;:,;;";";,;":,";";,-Y\
                   ;,;,;";";,;":;";"; Z/
                   / ;,";";,;";,;";;"
                  / / |";/~~~~~\";;"
                 ( K  | |      || |
                  \_\ | |      || |
                   \Z | |      || |
                      L_|      LL_|
                      LW/      LLW/

Answer 3

Python, 28 bytes

-1 byte thanks to Jonathan Allan!

Python 2:

i=10**100
while i:print;i-=1

Python 3 (30 bytes):

i=10**100
while i:print();i-=1

Answer 4

Haskell, 28 bytes

main=putStr$[1..10^100]>>"1"

Concatenates 10^100 copies of the string "1" and prints it.

Answer 5

Brainfuck, 480 188 114 106 98 bytes

Just because it needs to be done.

Assumes 8-bit cells with wrapping. Prints 250²⁵⁵ NUL bytes, which is 10¹⁰⁰ times 10¹⁵⁵ times 25²⁵⁵ NUL bytes.

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

Explanation:

>>>>>> is needed to leave a bit of working space.

- produces 255.

[[->>>+<<<]------>>>-] turns this into 255 copies of the value 250, giving a tape that looks like:

0 0 0 0 0 0 250 0 0 250 0 0 ... 250 0 0 [0]

<<<[<<<]+ moves the data pointer back and finishes up the initial data:

0 0 0 [1] 0 0 250 0 0 250 0 0 ...

Then comes the loop: [+...-] initially sets the 1 to a 2, which gets set back to 1 at the end of the loop. The loop terminates when the loop body already set 2 to 1.

Now, the numbers 2 250 250 250 ... 250 represent a counter, in base 250, with each number one greater than the digit it represents.

• [>>>]<<< moves all the way to the right. Since each digit is represented by a non-zero number, this is trivial.

• ->+[<[+>-]>[-<<<<->+>>------>>]<<<<]>>- decreases the counter by 1. Starting with the last digit: the digit gets decremented. If it remains positive, we're done. If it turns to zero, set it to 250, and continue with the digit before.

• [<<<].>>> moves the pointer back before the left-most digit, and this is a nice moment to print a NUL byte. Then re-position to exactly the left-most digit, to see if we're done.

To verify correctness, change the initial - to + to print 250¹ NUL bytes, ++ for 250², etc.

Answer 6

C, 51 bytes

Function g() calls recursive function f() to depth 99.

Excludes unnecessary newline added between f() and g() for clarity.

f(n,i){for(i=10;i--;)n?f(n-1):puts("");}
g(){f(99);}

//call like this
main(){g();}

Prints 1E100 newlines.

Declaration of i as second parameter of f() not guaranteed to work in all versions of C. Tested on my own machine (GCC on CygWin) and on ideone.com (I believe they also run GCC), but not up to f(99) for obvious reasons!

Answer 7

Commodore VIC 20 machine code (40 bytes)

... here shown as hexadecimal:

1040   A9[64]A2 00 9D 68 10 E8  E0[32]D0 F8 A9 00 9D 68
1050   10 A9[31]20 D2 FF A2 00  A9[64]DE 68 10 30 08 D0
1060   F0 9D 68 10 E8 D0 F3 60

(Started using: SYS 4160)

Meaning of the bytes in brackets

• 0x64 (occurs twice) is the base (100); (values from 2 to 127 should work)
• 0x32 is the exponent (50) (any non-zero value (1-255) should work)
• Note that 100^50 = 10^100; running the program 100^50 times is more RAM efficient than doing it 10^100 times
• 0x31 is the ASCII character to be printed

and at no point no more than 4 GB of memory must be needed.

Is this a typing mistake?

We have the year 1981.

A typical home computer has 1 to 16 KB of RAM! And you will hardly find professional models that have 1 MB or more.

(Ok. Just a joke.)

In other words, the program should in principle be testable by running it on your computer. But because of the extent of this number you will be expected to prove that the number of copies of the string it outputs is exactly 10^100 and that the program stops afterwards.

The program has been tested with other bases and exponents. I have no doubt it will also work with 100 and 50.

At least it does not crash with these numbers (but does not terminate in measurable time either).

The memory size is sufficient for an exponent of 50 and 100 is less than 127 so a base of 100 should not be a problem.

The the basic idea

There is a 50-digit counter that counts in the 100-system. Bytes 0x01-0x64 represent the digits 0-99. The first byte in the counter is the lowest digit. The last byte in the counter (highest digit) is followed by a byte with the value 0x00.

The counter has the initial value 100^50.

An outer loop is writing a byte to the "current channel" ("standard output" on modern systems; typically the screen) and then decrements the counter.

Decrementing is done by an inner loop: It decrements a digit and in the case of an underflow from 1 to 99 it advances to the next digit. If the byte 0x00 at the end of the counter is decremented the program stops.

The assembly code is

    ; Some constants
    base=10
    exponent=100
    character=0x32

    ; Initialize the content of the counter to 100^50
    ; (Fill the first 50 bytes with the value 100)
    lda  #base
    ldx  #0
initLoop:
    sta  counter,x
    inx
    cpx  #exponent
    bne  initLoop
    ; (The terminating 0 at the end of the counter)
    lda  #0
    sta  counter,x

    ; Now do the actual job
outerLoop:
    ; Output a character
    lda  #character
    jsr  (0xFFD2)
    ; Prepare for the inner loop
    ldx  #0
    lda  #base
innerLoop:
    ; Decrement one digit
    dec  counter,x
    ; Result is negative -> must have been the terminating
    ; NUL byte -> Exit
    bmi  progEnd
    ; Result is not zero -> Print the next byte
    bne  outerLoop
    ; Result is zero -> Was 0x01 before -> As 0x01 represents
    ; digit 0 this is an underflow -> set the digit to
    ; "exponent" (100) again (which represents digit 99)
    sta  counter,x
    ; Go to the next digit and...
    inx
    ; ... repeat the inner loop (decrement next digit)
    ; (Note that this conditional branch is ALWAYS taken)
    bne  innerLoop

progEnd:
    rts

counter:
    ; The memory used by the counter is here...

EDIT

The program runs on Commodore C64, too!

Answer 8

Node, 89 bytes

for(i="0".repeat(100)+1;+i;i=i.replace(/0*./,x=>"9".repeat(x.length-1)+~-x))console.log()

Outputs 10¹⁰⁰ newlines. (Theoretically, that is; test by replacing 100 with 1 to output 10¹ newlines instead.)

This works by setting i to the string

00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

(100 zeroes and a 1; a googol reversed), then repeatedly "subtracting 1" with a regex replace and outputting a newline until the string is all zeroes.

A port of the C++ answer would be 49 bytes:

(f=n=>{for(i=10;i--;)n?f(n-1):console.log()})(99)

Answer 9

05AB1E, 6 bytes

Tn°F1?

Explanation

Tn°    # push 10^100
   F   # 10^100 times do
    1? # print 1

Answer 10

Ruby, 20 bytes

(10**100).times{p 1}

Prints 1 followed by a newline 1E100 times.

1E100 does not work as it evaluates to a float, not an arbitrary precision integer.

Answer 11

Befunge 93, 33 bytes

1>01g0`#@!# _01v
d^.1 **52p10-1g<

Unfortunately, Befunge does not have a power function, so almost all of that code is my implementation of a power function. I'm still working on this.

Explanation:

1 > 01g 0` #@!# _ 01g 1- 01p 25** 1. v
d ^                                  <

1: Start off with 1 in the top left so that when we multiply, we don't get 0 every time.

01g: get the character at position (0, 1), which is d, whose ASCII code is 100.

0`: see if the value stored in (0, 1) is greater than 0; this value will change.

#@!# _: Logical not ! to the value we get from the last step (0 or 1), so that if it was 1, now we have 0, and we Note that # means that you skip the next character in the code.

01g 1- 01p: Take the value stored in (0, 1) again, subtract 1 from it, and store this new value at (0, 1)

25**: multiply the top value of the stack by 10

1.: print 1 every time this loops

1 is printed (in theory) googol times, but that quickly runs off of the page that I tested this on.

You can run Befunge 93 code here. For some reason, the top value of the stack is 1.0000000000000006e+100 when it should be 1.0e+100. I don't know where that 6 came from, but I don't think it should be there and that it may be some rounding error or something like that.

Answer 12

///, 36 ASCII characters (4 distinct)

/t./.ttttt//.t/t\..........//t//t...

Outputs the . character 3*10^125 times, meaning that it outputs the string consisting of 3*10^25 repetitions of the . character, 10^100 times.

Explanation:

1. /t./.ttttt/: Replace t. with .ttttt throughout the rest of the program, repeating until no instances of t. remain. This replaces t... with ... followed by 125 ts.
2. /.t/t\........../: Replace .t with t.......... throughout the rest of the program, repeating until no instances of .t remain. This takes the ... followed by 125 ts, and turns it into 125 ts followed by 10^125 occurrences of ....
3. /t//: Remove all remaining ts.
4. t...: This gets replaced with 3*10^125 .s. Output them.

Now, outputting 10^100 repetitions of 3*10^25 repetitions of something kind of feels like cheating. This program outputs the . character exactly 10^100 times, using 45 ASCII characters:

/T/tttttttttt//.t/t..........//t//.TTTTTTTTTT

Explanation of this one:

1. /T/tttttttttt/: Replace T with tttttttttt throughout the rest of the program. This replaces TTTTTTTTTT with 100 repetitions of t.
2. /.t/t........../: Replace .t with t.......... throughout the rest of the program. This takes the . followed by 100 ts, and turns it into 100 ts followed by 10^100 .s.
3. /t//: Remove all remaining ts.
4. .TTTTTTTTTT: This gets replaced with 10^100 .s. Output them.

Finally, here's a compromise program, which outputs the . character 2*10^100 times, using 40 characters:

/t./.tttttttttt//.t/t\..........//t//t..

Answer 13

ABCR, 56 bytes

++++++++++AAAAAAAAAA7a*A!(x4bBBBBBBBBBB7b+B@(xa(Ax5b(Box

Turing tarpits are fun, especially when they don't have easy multiplication or exponents. On the other hand, I only needed to use two of the three queues!

Explanation:

++++++++++                                             Set register to 10
          AAAAAAAAAA                                   Queue 10 to queue A 10 times
                    7a*A!(x                            Sum up all of queue A by:
                    7     x                             While the register is truthy:
                     a*                                 Dequeue two elements of A, sum them...
                       A                                ...and queue the result back to A.
                        !(                              If there's only one element left,
                                                        i.e. !(A.len - 1),
                                                        break from the loop.  A now has 100, our exponent.

                             4                        While A's front is truthy:
                              bBBBBBBBBBB              Clone the front of B 10 (our base) times.  (The first iteration fills it up with ten 1s)
                                         7b+B@(x       Sum up all of queue B like we did with A
                                                a(A    Decrement a (so that the loop actually ends. Runs 101 times like it should) x   

                                                       B now contains 10^100 only.

                                                   5b(B x   10^100 times:
                                                       o     print the front of queue A (0)

Answer 14

Batch, 574 242 bytes

@echo off
set t=for /l %%a in (2,1,33554432)do call:
set f=for /l %%a in (2,1,9765625)do call:
%t%a
:a
%t%b
:b
%t%c
:c
%t%d
:d
%f%e
:e
%f%f
:f
%f%g
:g
%f%h
:h
%f%i
:i
%f%j
:j
%f%k
:k
%f%l
:l
%f%m
:m
%f%n
:n
echo

Each loop falls through therefore executing an additional iteration. Loops are limited to ~2³² due to the 32-bit integer limit. The first four loops each count 2²⁵ for a total of 2¹⁰⁰ while the remaining ten loops each count 5¹⁰ for a total of 5¹⁰⁰.

Edit: Saved an unimaginable 58% thanks to @ConorO'Brien.

Answer 15

TI-Basic, 20 bytes

Straightforward. Only eight lines are displayed at once, and previous lines do not stay in memory. Because ᴇ100 is unsupported, we must loop from -ᴇ99 to 9ᴇ99. Then, if I!=0, display the string (which, by the way, is 3). This way, we print it exactly ᴇ100 times.

For(I,-ᴇ99,9ᴇ99:If I:Disp 3:End

Answer 16

PHP, 44 bytes

for($i=bcpow(10,1e2);$i=bcsub($i,print 1););

This snippet will output 1 googol times. It will not run out of memory, but it is terribly slow. I'm using BCMath to be able to handle long integers.

A bit better performing, but not as small (74 bytes):

for($m=bcpow(10,100);$m;$m=bcsub($m,$a))echo str_repeat(a,$a=min(4e9,$m));

Will output the letter a googol times. It will consume almost 4GB of memory, outputting about 4e9 characters at a time.

Answer 17

x86-64 machine code function, 30 bytes.

Uses the same recursion logic as the C answer by @Level River St. (Max recursion depth = 100)

Uses the puts(3) function from libc, which normal executables are linked against anyway. It's callable using the x86-64 System V ABI, i.e. from C on Linux or OS X, and doesn't clobber any registers it's not supposed to.

---

objdump -drwC -Mintel output, commented with explanation

0000000000400340 <g>:  ## wrapper function
  400340:       6a 64                   push   0x64
  400342:       5f                      pop    rdi       ; mov edi, 100  in 3 bytes instead of 5
  ; tailcall f by falling into it.

0000000000400343 <f>:  ## the recursive function
  400343:       ff cf                   dec    edi
  400345:       97                      xchg   edi,eax
  400346:       6a 0a                   push   0xa
  400348:       5f                      pop    rdi       ; mov edi, 10
  400349:       0f 8c d1 ff ff ff       jl     400320 <putchar>   # conditional tailcall
; if we don't tailcall, then eax=--n = arg for next recursion depth, and edi = 10 = '\n'

  40034f:       89 f9                   mov    ecx,edi   ; loop count = the ASCII code for newline; saves us one byte


0000000000400351 <f.loop>:
  400351:       50                      push   rax       ; save local state
  400352:       51                      push   rcx
  400353:       97                      xchg   edi,eax   ; arg goes in rdi
  400354:       e8 ea ff ff ff          call   400343 <f>
  400359:       59                      pop    rcx       ; and restore it after recursing
  40035a:       58                      pop    rax
  40035b:       e2 f4                   loop   400351 <f.loop>
  40035d:       c3                      ret    
# the function ends here

000000000040035e <_start>:

0x040035e - 0x0400340 = 30 bytes

# not counted: a caller that passes argc-1 to f() instead of calling g
000000000040035e <_start>:
  40035e:       8b 3c 24                mov    edi,DWORD PTR [rsp]
  400361:       ff cf                   dec    edi
  400363:       e8 db ff ff ff          call   400343 <f>
  400368:       e8 c3 ff ff ff          call   400330 <exit@plt>    # flush I/O buffers, which the _exit system call (eax=60) doesn't do.

Built with yasm -felf64 -Worphan-labels -gdwarf2 golf-googol.asm && gcc -nostartfiles -o golf-googol golf-googol.o. I can post the original NASM source, but that seemed like clutter since the asm instructions are right there in the disassembly.

putchar@plt is less than 128 bytes away from the jl, so I could have used a 2-byte short jump instead of a 6-byte near jump, but that's only true in a tiny executable, not as part of a larger program. So I don't think I can justify not counting the size of libc's puts implementation if I also take advantage of a short jcc encoding to reach it.

Each level of recursion uses 24B of stack space (2 pushes and the return address pushed by CALL). Every other depth will call putchar with the stack only aligned by 8, not 16, so this does violate the ABI. A stdio implementation that used aligned stores to spill xmm registers to the stack would fault. But glibc's putchar doesn't do that, writing to a pipe with full buffering or writing to a terminal with line buffering. Tested on Ubuntu 15.10. This could be fixed with a dummy push/pop in the .loop, to offset the stack by another 8 before the recursive call.

---

Proof that it prints the right number of newlines:

   # with a version that uses argc-1  (i.e. the shell's $i)  instead of a fixed 100
$ for i in {0..8}; do echo -n "$i: "; ./golf-googol $(seq $i) |wc -c; done
0: 1
1: 10
2: 100
3: 1000
4: 10000
5: 100000
6: 1000000
7: 10000000
8: 100000000
... output = 10^n newlines every time.

---

My first version of this was 43B, and used puts() on a buffer of 9 newlines (and a terminating 0 byte), so puts would append the 10th. That recursion base-case was even closer to the C inspiration.

Factoring 10^100 a different way could maybe have shortened the buffer, maybe down to 4 newlines, saving 5 bytes, but using putchar is better by far. It only needs an integer arg, not a pointer, and no buffer at all. The C standard allows implementations where it's a macro for putc(val, stdout), but in glibc it exists as a real function that you can call from asm.

Printing only one newline per call instead of 10 just means we need to increase the recursion max depth by 1, to get another factor of 10 newlines. Since 99 and 100 can both be represented by a sign-extended 8-bit immediate, push 100 is still only 2 bytes.

Even better, having 10 in a register works as both a newline and a loop counter, saving a byte.

Ideas for saving bytes

A 32-bit version could save a byte for the dec edi, but the stack-args calling convention (for library functions like putchar) makes tail-call work less easily, and would probably require more bytes in more places. I could use a register-arg convention for the private f(), only called by g(), but then I couldn't tail-call putchar (because f() and putchar() would take a different number of stack-args).

It would be possible to have f() preserve the caller's state, instead of doing the save/restore in the caller. That probably sucks, though, because it would probably need to get separately in each side of the branch, and isn't compatible with tailcalling. I tried it but didn't find any savings.

Keeping a loop counter on the stack (instead of push/popping rcx in the loop) didn't help either. It was 1B worse with the version that used puts, and probably even more of a loss with this version that sets up rcx more cheaply.

Answer 18

Haskell, 45 43 bytes

r 0=[]
r i='1':r(i-1)
main=putStr.r$10^100

Answer 19

Pyke, 6 5 bytes

TTX^V

Try it here!

Untested as it crashes my browser. The first 4 chars generate 10^100 and V prints out that many newlines. Test with 100V.

Answer 20

Racket 36 bytes

(for((i(expt 10 100)))(display "1"))

Output:

1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111...

Answer 21

JAISBaL, 4 bytes

˖Q

Chrome can't read all the symbols, and I'm not sure about other browsers, so here's a picture:

Explanation:

# \# enable verbose parsing #\
10^100       \# [0] push 10^100 onto the stack #\
for          \# [1] start for loop #\
    space    \# [2] print a space #\

Pretty simple.... just prints a googol spaces. Three instructions, but the googol constant is two bytes.

(Written in version 3.0.5)

Answer 22

JavaScript ES6, 85 83 bytes

Saved 2 bytes thanks to ETHproductions!

eval([...Array(i=100)].map(_=>`for($${--i}=0;$${i}++<10;)`).join``+"console.log()")

This prints 1e100 newlines.

The inner part generates this program, which is thereafter evaluated.

for($0=0;$0++<10;)for($1=0;$1++<10;)for($2=0;$2++<10;)for($3=0;$3++<10;)for($4=0;$4++<10;)for($5=0;$5++<10;)for($6=0;$6++<10;)for($7=0;$7++<10;)for($8=0;$8++<10;)for($9=0;$9++<10;)for($10=0;$10++<10;)for($11=0;$11++<10;)for($12=0;$12++<10;)for($13=0;$13++<10;)for($14=0;$14++<10;)for($15=0;$15++<10;)for($16=0;$16++<10;)for($17=0;$17++<10;)for($18=0;$18++<10;)for($19=0;$19++<10;)for($20=0;$20++<10;)for($21=0;$21++<10;)for($22=0;$22++<10;)for($23=0;$23++<10;)for($24=0;$24++<10;)for($25=0;$25++<10;)for($26=0;$26++<10;)for($27=0;$27++<10;)for($28=0;$28++<10;)for($29=0;$29++<10;)for($30=0;$30++<10;)for($31=0;$31++<10;)for($32=0;$32++<10;)for($33=0;$33++<10;)for($34=0;$34++<10;)for($35=0;$35++<10;)for($36=0;$36++<10;)for($37=0;$37++<10;)for($38=0;$38++<10;)for($39=0;$39++<10;)for($40=0;$40++<10;)for($41=0;$41++<10;)for($42=0;$42++<10;)for($43=0;$43++<10;)for($44=0;$44++<10;)for($45=0;$45++<10;)for($46=0;$46++<10;)for($47=0;$47++<10;)for($48=0;$48++<10;)for($49=0;$49++<10;)for($50=0;$50++<10;)for($51=0;$51++<10;)for($52=0;$52++<10;)for($53=0;$53++<10;)for($54=0;$54++<10;)for($55=0;$55++<10;)for($56=0;$56++<10;)for($57=0;$57++<10;)for($58=0;$58++<10;)for($59=0;$59++<10;)for($60=0;$60++<10;)for($61=0;$61++<10;)for($62=0;$62++<10;)for($63=0;$63++<10;)for($64=0;$64++<10;)for($65=0;$65++<10;)for($66=0;$66++<10;)for($67=0;$67++<10;)for($68=0;$68++<10;)for($69=0;$69++<10;)for($70=0;$70++<10;)for($71=0;$71++<10;)for($72=0;$72++<10;)for($73=0;$73++<10;)for($74=0;$74++<10;)for($75=0;$75++<10;)for($76=0;$76++<10;)for($77=0;$77++<10;)for($78=0;$78++<10;)for($79=0;$79++<10;)for($80=0;$80++<10;)for($81=0;$81++<10;)for($82=0;$82++<10;)for($83=0;$83++<10;)for($84=0;$84++<10;)for($85=0;$85++<10;)for($86=0;$86++<10;)for($87=0;$87++<10;)for($88=0;$88++<10;)for($89=0;$89++<10;)for($90=0;$90++<10;)for($91=0;$91++<10;)for($92=0;$92++<10;)for($93=0;$93++<10;)for($94=0;$94++<10;)for($95=0;$95++<10;)for($96=0;$96++<10;)for($97=0;$97++<10;)for($98=0;$98++<10;)for($99=0;$99++<10;)console.log()

Now, for a proof of correctness, we'll use some induction. Let's substitute the initial 100 for other values, generically N. I claim that inserting N will yield 10^N newlines. Let's pipe the result of this to wc -l, which counts the number of newlines in the input. We'll use this modified but equivalent script that takes input N:

eval([...Array(+process.argv[2])].map(_=>`for($${i}=0;$${i++}++<10;)`,i=0).join``+"console.log()")

Now, here's some output:

C:\Users\Conor O'Brien\Documents\Programming
λ node googol.es6 1 | wc -l
10

C:\Users\Conor O'Brien\Documents\Programming
λ node googol.es6 2 | wc -l
100

C:\Users\Conor O'Brien\Documents\Programming
λ node googol.es6 3 | wc -l
1000

C:\Users\Conor O'Brien\Documents\Programming
λ node googol.es6 4 | wc -l
10000

We can see that this transforms the input N for small values to 10^N newlines.

Here is an example output for N = 1:

C:\Users\Conor O'Brien\Documents\Programming
λ node googol.es6 1











C:\Users\Conor O'Brien\Documents\Programming
λ

Answer 23

Mathematica, 48 30 25 bytes

For[n=1,n++<Echo@1*^100,]

Output:

>> 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
>> 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
>> 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
>> 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
>> 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
etc.

Answer 24

Fortran 95, Free-form, Recursive, 117 bytes

Program M
Call R(99)
END
Recursive Subroutine R(L)
IF(L)3,1,1
1 DO 2 I=1,10
2 Call R(L-1)
RETURN
3 PRINT*,0
END

Prints a googol of lines containing

          0

Fortran 90, Recursive, 149 bytes

     CallR(99)
     END
     RecursiveSubroutineR(L)
     IF(L)3,1,1
   1 DO2I=1,10
   2 CallR(L-1)
     RETURN
   3 PRINT*,0
     END     

Recursively calling 100 nested loops, each 10 iterations, makes exactly one googol. N, L, and the loop counters all fit in byte-sized integers.

Tested by replacing 99 with 1, 2, 3, 4, 5 and noting that in each case the resulting line count from "wc" has n+1 zeros.

Fortran II, IV, 66, or 77, 231 bytes:

      N=2*10**9
      DO1I0=1,5**10
      DO1I=1,N
      DO1J=1,N
      DO1K=1,N
      DO1L=1,N
      DO1M=1,N
      DO1I1=1,N
      DO1J1=1,N
      DO1K1=1,N
      DO1L1=1,N
      DO1M1=1,N
1     PRINT2
2     FORMAT(X)
      END

Prints a googol of newlines.

All of these programs will run on 32-bit machines; in fact, the recursive versions would work just fine on a 16-bit machine. One could use fewer loops in the brute-force version by running on an old Cray with its 60-bit integers. Here, ten nested loops of 2*10^9 inside one loop of 5^10 (9765625) equals 10 ^ 100 total iterations.

None of the versions uses any memory to speak of other than the object code itself, the counters, one copy of the output string, and, in the recursive version, a 100-level return stack.

Check the factors by comparing

bc<<<2000000000\^10*5\^10
bc<<<10\^100

Answer 25

Turing machine simulator, 1082 bytes

0 * 6 r 1
1 * E r 2
2 * C r 3
3 * 1 r 4
4 * B r 5
5 * C r 6
6 * F r 7
7 * 4 r 8
8 * 6 r 9
9 * 8 r A
A * 8 r B
B * 3 r C
C * 0 r D
D * 9 r E
E * G r F
F * H r G
G * 8 r H
H * 0 r I
I * 6 r J
J * H r K
K * 9 r L
L * 3 r M
M * 2 r N
N * A r O
O * D r P
P * C r Q
Q * C r R
R * 4 r S
S * 4 r T
T * E r U
U * E r V
V * G r W
W * 6 r X
X * D r Y
Y * E r Z
Z * 0 r a
a * F r b
b * E r c
c * 9 r d
d * F r e
e * A r f
f * H r g
g * D r h
h * E r i
i * 6 r j
j * 6 r k
k * D r l
l * G r m
m * H r n
n * 1 r o
o * 0 r p
p * 8 r q
q * C r r
r * 9 r s
s * G r t
t * 3 r u
u * 6 r v
v * 2 r w
w * 3 r x
x * E r y
y * 0 r z
z * 4 r +
+ * 5 r /
/ * A r =
= * 0 r -
- * H r \
\ * 7 r !
! * A r @
@ * 9 r #
# * 5 r $
$ * A r %
% * B r ^
^ * 5 r &
& * 9 r ?
? * 4 r (
( * C r )
) * E r `
` * 9 r ~
~ * 9 r _
_ * A * .
. 0 I l *
. 1 0 r <
. 2 1 r <
. 3 2 r <
. 4 3 r <
. 5 4 r <
. 6 5 r <
. 7 6 r <
. 8 7 r <
. 9 8 r <
. A 9 r <
. B A r <
. C B r <
. D C r <
. E D r <
. F E r <
. G F r <
. H G r <
. I H r <
. _ * r ]
< _ * r >
< * * r *
> _ = l [
> * * r *
[ _ * l .
[ * * l *
] _ * * halt
] * _ r *

Turing machine simulator

I do not know if this counts as the correct output, since it has 82 leading spaces.

I do not know if this respects the 4 GB limit, so, if it doesn't, then it's non-competitive and just for showcase. The output is 1e100 bytes, so that should be deducted from the memory byte count. The final byte count is 82 bytes.

Here is an explanation:

The first 80 lines of code are 80 different states that generate the base-19¹ loop count 6EC1BCF4688309GH806H932ADCC44EEG6DE0FE9FAHDE66DGH108C9G3623E045A0H7A95AB594CE99A.

The next 19 lines of code are the counter state, which decrements the count every time a character is printed.

The next 6 lines are the printer state, which appends an =.

Finally, the last 2 lines are the cleaner state, which are needed to make sure the only output is =====...=====. Leading/trailing spaces do not count as output, since they are unavoidable side-effects.

The program then halts.

^{1I did the math for that.}

Answer 26

Brainfuck, 45 bytes

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

It outputs 254×250²⁵⁵ null bytes and exits with an error.

Brainfuck, 50 48 bytes, without errors

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

It outputs 254×250²⁵⁵ times \xfa\xfb\xfc\xfd\xfe\xff, and 1524×250²⁵⁵ bytes in total.

Answer 27

Pyth, 8 7 bytes

V^T100G

Link

Solution is tested with small output, but it should print abcdefghijklmnopqrstuvwxyz 1e100 times.

For some reason, the p was unneeded, as 31343 (Maltysen) said.

Answer 28

Pyth, 7 Bytes

New (Competing)

V^T*TTG

Explanation

G=The alphabet
Repeat 10^(10*10) times
    print(G)

---

Old (Non Competing) 7 Bytes

*G^T*TT

Explanation

G=The alphabet
G*(10^(10*10))==G*10^100

Answer 29

Java, 198 179 155 bytes

import java.math.*;class a{void A(a[]x){for(BigInteger b=BigInteger.ZERO;!(b=b.add(BigInteger.ONE)).equals(BigInteger.TEN.pow(100));)System.out.print(x);}}

Prints (x == null ? null : a string that starts with [La;@ or something like that) 10¹⁰⁰ times in O(forever) time.

Answer 30

Python 3, 32 bytes

for i in range(10**100):print()

Alternate solution, 33 bytes:

[print()for i in range(10**100)]

Answer 31

Java, 153 bytes

import java.math.*;interface M{static void main(String[]a){for(BigInteger i=BigInteger.ZERO;!i.add(i.ONE).equals(i.TEN.pow(100));)System.out.print(1);}}

Output: 1e100 1s

I know there is another Java answer which is also pretty close. Mine's got a main and is still shorter though.

This is my first code-golf entry. Tips appreciated.

Answer 32

PHP, 39 38 bytes

for($a=str_pad(e,99,0);T^$a++;)echo!0;

Run like this:

php -r 'for($a=str_pad(e,99,0);T^$a++;)echo!0;' 2>/dev/null

STDERR (34 bytes)

If a variant with output to STDERR is valid, then this outputs PHP Notice: Use of undefined constant T - assumed 'T' in Command line code on line 1 a googol times.

php -r 'for($a=str_pad(@e,99,0);T^++$a;);'

Explanation

for (
    $a = str_pad("e", 99, "0"); // Creates a string "e000" (98 zeroes).
    "T" ^ $a++;                 // Increment the string, binary XOR with
                                // "T": when the alphabetic part of the 
                                // string becomes "da" this results in 
                                // falsy "0", ending the loop.
)
    echo!0;                     // Print a single "1".

Note that the alphabetic part iterates from e..z (22), aa..az (26), ba..bz (26), ca..cz (26) for a total of 100, or a 1 with 2 zeroes. So only 98 zeroes need to be added to arrive at 1E100. The loop stops only when the alphabetic part is 2 digits and the second digit becomes d. The d is conveniently skipped when it's only 1 digit (since it starts at e).

Perf

This doesn't consume a ridiculous amount of resources (in fact very little) and runs fast.

Tweaks

• Saved a byte by using 2 alphabetic chars so less padding characters (99 vs 101)
• Added a variant using STDERR, saving 5 bytes

Answer 33

LOLCODE, 224 bytes

HAI 1.2
I HAS A n ITZ A NUMBR
n R 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
IM IN YR loop UPPIN YR v TIL BOTH SAEM v AN n
BTW THIS WHERE I PRINT
VISIBLE "HAI!!1"
IM OUTTA YR loop
KTHXBYE

I had absolutely no idea how to calculate 10100, so why not just write it out? Apparently, there's no ^ operator or power function in LOLCODE.

I didn't count the bytes for the BTW part, since it's a comment for explaining the code. "explain"

Also, my ressource didn't quite explain how datatypes work in-depth, so we can only assume that NUMBR can store this much.

Answer 34

Perl 6, 18 bytes

say 1 for ^10**100

Notes:

• Perl 6 integers have no upper limit, so 10**100 calculates the googol just fine.
• The ^n syntax is short for 0 ..^ n, which constructs a range from 0 (inclusive) to n (not inclusive), which conceptually contains n integers but is stored as a memory-efficient Range object.
• The for loop iterates over it without keeping already iterated elements in memory, so the program's RAM usage will remain pretty stable - and far below 4 GB - throughout its runtime.

Answer 35

Actually, 9 bytes

2╤╤DW;Y.D

Try it online!

This takes a really long time to run. Here is a faster version that only prints 100 zeroes.

Explanation:

2╤╤DW;Y.D
2╤╤         push 10**(10**2)
   D        subtract 1
    W       while the value on top of the stack is truthy:
     ;Y.      duplicate it, boolean negate it (0 if truthy, 1 if falsey), pop it and print it
        D     decrement the other copy
            (after the while loop finished, (10**100)-1 zeroes will have been printed, separated by newlines)
            implicitly print the only stack value (0), followed by a newline

Answer 36

Pushy, 5 bytes

Non-competing as the language postdates the challenge.

THe:"

Prints 10¹⁰⁰ newlines. You can Try it online, but output is truncated at 100KB, and the site becomes very unresponsive when you run it.

T      \ Push 10
 H     \ Push 100
  e    \ Exponentiate - pops both and computes 10 ^ 100
   :   \ Pop the googol and that many times do:
    "  \   Print the whole stack - just a newline, as it's now empty.

You can verify this generates a googol by running THe# in the online interpreter. The reference implementation is in Python and therefore has no upper limit on integer size. The googol itself does not take too much data to store, in fact you only need 333 bits to store an integer of that size (although as Python integer this seems to take up 58 bytes) - nowhere near the memory restriction on this challenge.

Answer 37

Bash, 72 71 bytes

Prints 10^100 at signs.
Saved 1 byte thanks to @FlipTack.

printf '@%.0s' $(eval echo $(seq 1 $(echo 10^100 | bc | tr "\n" "\0"))) 

WARNING: This is probably extremely slow.

---

This is my first code golf, so I hope I did well.

Answer 38

TrumpScript 189 bytes

I is 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000A is,I -I;We is 1000001 -1000000as long as,I >A;:say,A;I is,I -We;!America is great

Try it online!

Unfortunately 100 bytes taken up by the number as TrumpScript has no exponentials

Explanation

I is googol (using googol for clarity) sets the variable I to 10^100

A is I -I sets A to 0. 0 in the code as a number is invalid as all number must be > 1000000

We is 1000001 -1000000 assigns we as 1

as long as, I > A;: starts while loop checking that I > 0

say A prints out 0

I is,I -We subtracts 1 from I

America is great ends the program

Answer 39

SmileBASIC, 20 bytes

FOR I=1TO 1E100?NEXT

Answer 40

.COM, 21 bytes

0000h: 83 FC 9A 74 0B B9 10 27 51 E8 F4 FF 59 E2 F9 C3
0010h: B4 02 CD 21 C3                                 

    org 100h
f:  cmp sp, -2-4*25
    jz g
    mov cx, 10000
r:  push cx
    call f
    pop cx
    loop r
    ret
g:  mov ah, 2
    int 33
    ret       

Answer 41

C# - 127 123 98 bytes

using System;using b=Numerics.BigInteger;void f(){for(b n=b.Pow(10,100);n-->0;)Console.Write(0);}}

• 1 byte from Kevin Cruijssen!
• 24 bytes from Phaeze!

Prints 10^100 copies of 0.

Answer 42

Python 2, 35 bytes

Under more realistic programming circumstances, Python allows you to multiply strings by a number, as with 'x' * 100, which would produce a string that's a hundred characters long and composed of nothing but the letter x.

However, it appears that Python string objects don't appreciate having indices whose values exceed the maximum of int. Python is content to re-cast large int values to a special bignum sort of thing, but the longest string that Python can handle is on the order of several gigabytes, and assuming each character in Python is 1 byte, it would take much, much more memory to contain the string that would unfold from 'x'*10**100.

As such, Python's built-in arbitrary arithmetic and the humble while loop allow us to produce a more practical program, which spits out one x per line for a googol lines. Thankfully, this proves to be shorter than the admittedly more readable for x in xrange(10**100):, as range() and xrange() are both limited to the largest possible value for a C long type, which is Python's native non-bignum int.

i=0
while i<10**100:
    print'x';i+=1

Answer 43

Common Lisp, 33

(dotimes(i(expt 10 100))(terpri))

Print newlines.

Answer 44

Swift 3, 55 bytes

Dependency: Foundation for pow(73 bytes if you count import statement)

stride(from:0,to:pow(10,100),by:1).map{_ in print(1)}

I know there is another answer above. However, pow(10,100.0) generates error message: fatal error: Double value cannot be converted to Int because the result would be greater than Int.max, and it also requires Foundation as well

Answer 45

Clojure, 36 bytes

(doseq[i(range(.pow 100M 50))](prn))

Using java interlop cuts it down from original (reduce *'(repeat 10 100)) since clojure core does not have exponent. dotimes can only work with longs as was pointed out in comment, sadly; I had to resort to lazy seq (+5 bytes).

Answer 46

Common Lisp, 37 bytes

(loop repeat(expt 10 100)do(print'x))

Nothing fancy. Print X followed by newline 10^100 times.

Answer 47

Sed + bash, 107 bytes

sed "s/Z/\;done/g;s/\([A-L]\)/\1=n\;while((\1--))\;do /g"<<<"n=10;An=10**9;BCDEFGHIJKLpwdZZZZZZZZZZZZ"|bash

By removing the "|bash" at the end, you can see that the bash script generated (with newlines added for readability) is

n=10;A=n;while((A--));do n=10**9;B=n;while((B--));do C=n;while((C--));do
D=n;while((D--));do E=n;while((E--));do F=n;while((F--));do G=n;while((G--));do
H=n;while((H--));do I=n;while((I--));do J=n;while((J--));do K=n;while((K--));do
L=n;while((L--));do pwd;done;done;done;done;done;done;done;done;done;done;done;
done

which is similar to my original bash solution, below, except that it runs 11 loops of 10^9 nested in one loop of 10, for a total of 10^100 iterations. It never uses a value greater than 10^9, which fits in 32 bits. Each output line is generated with the shell builtin "pwd" instead of "echo", to save a byte.

Tested by changing 10**9 to "2" or "3" and verifying that the script generates 10 * N^11 lines.

Bash, 312 bytes

n=2*10**9
a=9765625;while((a--));do
b=n;while((b--));do c=n;while((c--));do d=n;while((d--));do e=n;while((e--));do
f=n;while((f--));do g=n;while((g--));do h=n;while((h--));do i=n;while((i--));do
j=n;while((j--));do k=n;while((k--));do echo "Hello, World!"
done;done;done;done;done;done;done;done;done;done;done

(2*10^9)^10 * 9765725 is exactly one googol.

This needs no memory because it doesn't store anything other than the 11 counters and one copy of the "Hello, World!" string. The counters are integers that never exceed 32 bits.

Unfortunately I couldn't find a way to do this with a recursive function.

Answer 48

LiveCode, 83 bytes

on g c
   if c=100 then
      put 1 after fld 1
   else
      repeat 10
         g c+1
      end repeat
   end if
end g

Logically pretty straightforward, but validated by changing the test value for c. Set it to 1, you get ten 1's. Set it to 2, you get 100. Set it to 3 you get 1,000, etc.

LiveCode automatically formats with leading spaces, but it's not required for the code to work.

The argument c will default to empty, and in LiveCode it's perfectly fine to add 1 to empty (and results in 1).

LiveCode can write to stdout, but it's a visual environment and putting text into a field is more common.

Answer 49

Groovy, 31 bytes

{Eval.me(it*"1${'0'*10**100}")}

Answer 50

Chip, 3815+3 = 3818 bytes

+3 bytes for flag -w

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Try it online!

Oof.

It's a beeping massive binary counter: 333 bits are needed to express 1 googol, so there are 333 half-adders @ and 333 registers Z. (Actually only 100/log₁₀(2) = ~332.2 bits are required, but this solution can't take advantage of those massive savings). There are also a few periodic caches K, which are required for this to even run correctly.

Note about the TIO: instead of using -w (which is a fake stdin consisting of infinite null bytes), it uses regular input. The result is that it terminates at the end of input, rather than attempting to outlive the universe by a factor of 10⁸⁰.

The termination condition is encoded by the presence or absence of ~ near the Zs. If you worked from the bottom up, writing a '1' when you see a ~, and '0' when you don't, you'll have the binary representation for 1 googol.

Here is a TIO that only prints 100 times. This one does use -w, showing off the termination condition.

Also scattered around are *a, *e, and *f. These work together to produce an ASCII '1' for the output. If they were absent, null characters would be printed instead.

Answer 51

cQuents, 11 bytes

"#10^100::0

Given enough time (and memory, although it times out on TIO instead of running out of output space), will print 10^100 copies of whatever is to the right of 0.

Explanation

"             No join on output
 #10^100      Set n to 10^100
        ::    Print the sequence from 1 to n
          0   Each item in the sequence is 0

Try it online!

Answer 52

HadesLang, 3153 bytes

I know this is really bad...but at least I tried, right?

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end

Answer 53

ForceLang, 58 bytes

set k 10.pow 100
label a
set k k+-1
io.write 1
if k
goto a

Answer 54

Charcoal, 9 bytes

Note that Charcoal uses a custom codepage.

ＦＸ¹⁰¦¹⁰⁰Ｄ

Prints one googol newlines. Takes around 3.16887646×10⁹⁰ years to finish running.

Explanation

Ｆ        For
 Ｘ       Exponentiate
  ¹⁰¦¹⁰⁰  10, 100
Ｄ        Print (Dump) contents of canvas ("") plus newline

This will not freeze your computer (luckily!) because Charcoal has a 10ms minimum delay between prints by default. Charcoal used range as of this challenge's posting, therefore using more than 4GB of memory, but this has been fixed.

Answer 55

Dip, 8 bytes

TT*T^(¹p

Prints googol newlines.

Explanation:

TT       # Pushes 10 twice
             # Stack: [10, 10]
  *      # Multiply 10 by 10
             # Stack: [100]
   T     # Push 10
             # Stack: [100, 10]
    ^    # Raise 10 to 100
             # Stack: [Googol]
     (   # Do Googol times
             # Stack: []
      ¹  # Push ""
             # Stack: [""]
       p # Print
             # Stack: []

Better version that puts to use some features released after this challenge: (4 bytes)

UX(q

Explanation:

U    # Push 100
         # Stack: [100]
 X   # Raise 10 to 100
         # Stack: [Googol]
  (  # Repeat Googol times
         # Stack: []
   q # Print a newline
         # Stack: []

Answer 56

05AB1E, 8 bytes

TTT*mFõ,

Prints googol newlines.

Try it online! (Not recommended)

Explanation:

TTT      # Push 10 three times
             # Stack: [10, 10, 10]
   *     # Multiply 10 by 10
             # Stack: [10, 100]
    m    # Raise 10 to 100
             # Stack: [Googol]
     F   # Do Googol times
             # Stack: []
      õ  # Push ""
             # Stack: [""]
       , # Print
             # Stack: []

Answer 57

PHP7 (64-bit), 36 bytes

<?=bcsub(bcpow(10,bcpow(10,100)),1);

Should output 10^(10^100)-1 so 10^100 times the digit 9.

As of PHP 7.0.0, there are no particular restrictions regarding the length of a string on 64-bit builds. Source

Answer 58

VBA 117 bytes

Sub q():Dim x(100):While x(100)=0:Debug.Print:x(0)=x(0)+1:i=0:While x(i)>9:x(i)=0:i=i+1:x(i)=x(i)+1:Wend:Wend:End Sub

Invoke with q. Prints a googol newlines to immediate window. Easy to test by changing the "While x(100)" to say "While x(2)", which prints a hundred newlines.

I could use "MsgBox 0" in place of "Debug.Print" (saving 3 characters) but that would involve pressing return on the output more times than feasible, countering the whole idea of "program".