Fault-Tolerant Hello World

At the end of your interview, the Evil Interviewer tells you, "We make all of our applicants take a short coding test, to see if they really know what they are talking about. Don't worry; it's easy. And if you create a working program, I'll offer you the job immediately." He gestures for you to sit down at a nearby computer. "All you have to do is create a working Hello World program. But"--and he grins broadly--"there's a catch. Unfortunately the only compiler we have on this machine has a small bug. It randomly deletes one character from the source code file before compiling. Ok, see you in five minutes!" And he walks out of the room, whistling happily.

Can you guarantee that you will get the job?

Write a program that will print Hello, world! to standard output even after a single character is removed from any position in the file. Or come as close to this as possible.

The Rules

No Extraneous Output - Hello, world! must be the only substantive thing printed to standard output. It is ok to include other characters if they are naturally produced by your language of choice--such as a trailing newline or even something like [1] "Hello, world!" (for example if you were using R), but it must print the exact same thing every time. It cannot print Hello, world!Hello, world! or Hello world!" && x==1 some of the time, for example. Warnings, however, are allowed.

Testing In order to test determine your score, you have to test each possible permutation of the program: test it with each character removed, and see if it produces the correct output. I have included a simple Perl program for this purpose below, which should work for many languages. If it doesn't work for you, please create a test program and include it in your answer.

Scoring Your score is the number of times your program fails. In other words, the number of individual positions in your file where deleting a character prevents your program from working. Lowest score wins. In the event of a tie, the shortest code wins.

Trivial Solutions such as "Hello, world!" in several languages (score of 15) are acceptable, but they aren't going to win.

Happy programming, and may you get the job. But if you fail, you probably didn't want to work for that evil boss anyway.

Perl test script:

use warnings;
use strict;

my $program = 'test.pl'; my$temp_file = 'corrupt.pl';
my $command = "perl -X$temp_file"; #Disabled warnings for cleaner output.
my $expected_result = "Hello, world!"; open my$in,'<',$program or die$!;
local $/; #Undef the line separator my$code = <$in>; #Read the entire file in. my$fails = 0;
for my $omit_pos (0..length($code)-1)
{
my $corrupt =$code;
$corrupt =~ s/^.{$omit_pos}\K.//s;  #Delete a single character

open my $out,'>',$temp_file or die $!; print {$out} $corrupt; #Write the corrupt program to a file close$out;

my $result = $command;    #Execute system command.
if ($result ne$expected_result)
{
$fails++; print "Failure$fails:\nResult: ($result)\n$corrupt";
}
}

print "\n$fails failed out of " . length$code;

• Can the deleted character result in the program not compiling? Is that still counted as not working? – lochok Apr 18 '13 at 7:06
• @lochok, yes, that would count as a failure. Any deleted character which leads to Hello, World! not being printed is a failure. – user7486 Apr 18 '13 at 7:09
• Similar question: codegolf.stackexchange.com/questions/4486/… – mowwwalker Apr 18 '13 at 8:26
• @Walkerneo, thanks! I searched for similar questions and didn't find that one. I think this is significantly different, though. In particular, that question guarantees that only modifications resulting in syntactically valid code have to be handled. – user7486 Apr 18 '13 at 8:33
• The Perl test script should be subject to the same 1-character deletion – xDaizu Jul 3 '17 at 8:45

Befunge, Score 0

I think I cracked it - no single character deletion will change the output.
Deleting any character from line 1 changes nothing - it still goes down at the same place.
Lines 2 and 3 are redundant. Normally line 2 is executed, but if you delete a character from it, the < is missed, and line 3 takes charge.
Deleting newlines doesn't break it either (it did break my previous version).
No test program, sorry.

EDIT: simplified a lot.

                              vv
@,,,,,,,,,,,,,"Hello, world!"<<
@,,,,,,,,,,,,,"Hello, world!"<<


A short explanation of the flow:

1. Befunge starts executing from top-left, moving right. Spaces do nothing.
2. v turns the execution flow downward, so it goes down one line.
3. < turns the execution flow left, so it reads line 2 in reversed order.
4. "Hello, world!" pushes the string to the stack. It's pushed in reversed order, because we're executing right to left.
5. , pops a character and prints it. The last character pushed is printed first, which reverses the string once more.
6. @ terminates the program.
• +1, great solution. I don't think a test program is needed, because it is obvious that there is only a small number of potentially significant deletions. – user7486 Apr 18 '13 at 12:14
• So easy to verify and no discussion involved. Genius! – Karma Fusebox Apr 19 '13 at 22:51
• Congratulations on an excellent solution. – user7486 Apr 29 '13 at 7:39
• I know this isn't code-golf, but here's a version in 66 bytes. – MD XF Feb 28 '18 at 0:26

Perl, Score 0

(147 characters)

Here is my solution, which I managed to get from 4 down to 0:

eval +qq(;$_="Hello, world!";;*a=print()if length==13or!m/./#)|| +eval +qq(;$_="Hello, world!";;print()if*a!~/1/||!m/./#)##)||
print"Hello, world!"


It must appear all on one line to work; line breaks are for "readability" only.

It benefits from Perl's pathologically permissive syntax. Some highlights:

• Barewords that are not recognized are treated as strings. So when eval becomes evl, that is not an error if a string is permissible at that point.
• Unary + operator, which does nothing other than disambiguate syntax in certain situations. This is useful with the above, because function +argument (where + is unary) becomes string + argument (addition) when a function name is mangled and becomes a string.
• Many ways to declare strings: a double quoted string qq( ) can become a single-quoted string q(); a string delimited by parentheses qq(; ... ) can become a string delimited by a semicolon qq; ... ;. # inside strings can eliminate balancing issues by converting things to comments.

The length of this can probably be reduced somewhat, though I doubt ugoren's solution can be beaten.

• +1 for line breaks are for readability only ... – Bakuriu Apr 24 '13 at 14:47

HQ9+

This will never fail to produce the intended result when a character is deleted so gets a score of zero.

HH


When do I start?

• It produces "hello, world" though, not "Hello, world!" as specified. – Paul R Apr 17 '13 at 21:17
• Nonsense, the language was just incorrectly specified on the esolang wiki ;) – skeevey Apr 17 '13 at 21:20
• Good for you if they have an HQ9+ compiler on the interview computer... :) – user7486 Apr 18 '13 at 4:41
• You could write your own buggy compiler. – PyRulez Feb 19 '14 at 20:50
• @mbomb007 This interpreter prints Hello, world! for the H command. – Dennis Dec 14 '16 at 1:28

Befunge-98, score 0, 45 bytes

20020xx""!!ddllrrooww  ,,oolllleeHH""cckk,,@@


Try it online!

Although an optimal solution has already been found (and there's no tie breaker), I thought I'd show that this can be simplified considerably with Befunge 98.

Explanation

The 20020xx reliably sets the delta (the steps of the instruction pointer between ticks) to (2,0) so that starting at the first x, only every other command is executed. See this answer for a detailed explanation of why this works. Afterwards, the code is merely:

"!dlrow ,olleH"ck,@


We first push all the relevant character codes onto the stack with "!dlrow ,olleH". Then ck, means print the top of the stack (,), 13 (c plus 1) times (k). @ terminates the program.

J, 7 points

Selecting every letter with odd position:

   _2{.\'HHeellllo,,  wwoorrlldd!!'
Hello, world!

• The same thing in golfscript would be 4 points: 'HHeelllloo,, wwoorrlldd!!'2% – cardboard_box Apr 18 '13 at 0:07
• @cardboard_box Feel free to submit it. :) – randomra Apr 18 '13 at 10:04
• Upvoting because I understand it, and it doesn't feel like cheating. – Michael Stern Apr 17 '14 at 16:43

Unary, score 0, 74817134662624094502265949627214372599829 bytes

The code is not included due to post length limitations, but it consists of 74817134662624094502265949627214372599829 zeros.

If any of these zeros is removed, the resulting brainfuck program is KSab's hello world program.

Sadly, I didn't get the job because my machine had only 500 GB of hard disk space.

Befunge-93, Score 0 (63 bytes)

I know this isn't a code-golf challenge, but I thought it would be interesting to see if the existing Befunge-93 solution could be improved upon in terms of size. I had originally developed this technique for use in the similar Error 404 challenge, but the need for a wrapping payload in that case made the 3 line solution more optimal.

This isn't quite as good as Martin's Befunge-98 answer, but it's still a fairly significant reduction on the winning Befunge-93 solution.

<>>  "!dlrow ,olleH">:#,_@  vv
^^@_,#!>#:<"Hello, world!"<<>#


Try it online!

Explanation

There are two versions of the payload. For an unaltered program, the first < causes the program to execute right to left, wrapping around to the end of the line, until it reaches a v directing it down to the second line, and a < directing it left into the left-to-right version of the payload.

An error on the second line causes the final < to be shifted left and replaced with a > directing the flow right instead. The # (bridge) command has nothing to jump over, so the code just continues until it wraps around and reaches a ^ at the start of the line, directing it up to the first line, and then a > directing it right into the right-to-left payload.

Most errors on the first line just cause the final v commands to shift across by one, but that doesn't alter the main flow of the code. Deleting the first < is slightly different, though - in that case the execution path just flows directly into the left-to-right payload on the first line.

The other special case is the removal of the line break. When the code wraps around to the end of the line, in this case it's now the end of the what used to be the second line. When it encounters the # command from the right, this jumps over the > and thus continues directly into the right-to-left payload.

In case there's any doubt, I've also tested with the perl script and it confirmed that "0 failed out of 63".

Lost, Score 0, 122 bytes

>>>>>>>>>>>>>%%/>>>>>>>>>>/
\\
>>>>>>>>>>>>>//>>>>>>>>>>>\\
//
/@+-(  "vUvdlrow ,olleH"?<<
///
/@+-("vUvdlrow ,olleH"? <<<


Try it online!

It never stops amazing me that are possible in Lost.

If you haven't seen Lost before it is a 2-D programming language in which instead of starting execution at a fixed location it chooses a random location and direction and starts there. This makes writing any lost program a bit of a radiation hardening challenge normally, since a program must work from every start location. But now we layer on too and every program must work from every start location. With 122 bytes, four directions and a 28 by 6 bounding box, this means there are approximately 80,000 different programs that must be work.

This program is built with a mix of standard lost methods and standard radiation hardening methods. To start here is the normal Hello World for lost:

v<<<<<<<<<<>>>>>>>>>>>
>%?"Hello, worldvU"-+@


It is comprised of a "collector" which gathers all the stray instruction pointers and points them to the beginning of the actual program.

v<<<<<<<<<<>>>>>>>>>>>
>               v


A more detailed explanation of how this program works can be found here.

Now we want to modify this program to use the duplicate alignment method of radiation hardening. This method works by having two copies of the desired code and entering them from the end of the line.

>>>>>>>>>vv
program <
program <


This way when on of the programs is edited it becomes misaligned and the other one is executed

>>>>>>>>>vv
progam <
program <


To do this with our lost program is pretty simple:

>>>>>>>>>>>>>>>>>>>>>\\
@+-"Uvdlrow ,olleH"?%<
@+-"Uvdlrow ,olleH"?%<


However there are some problems with this. The alignment of the "s is an issue because it causes infinite loops (in the example above there are basically two strings ">" lain out vertically, so if you start on the quotes going in the right direction you just push the > character endlessly). So you need to make sure the quotes don't line up even when a character has been deleted.

>>>>>>>>>>>>>>>>>>>>>>>\\
@+-  "Uvdlrow ,olleH"?%<
@+-"Uvdlrow ,olleH"?%  <


Also since v is needed to escape the inside of the string it's deletion is potentially a problem. So we need a backup.

>>>>>>>>>>>>>>>>>>>>>>>>>\\
@+-(  "vUvdlrow ,olleH"?%<
@+-("vUvdlrow ,olleH"?%  <


Now while alignment prevents most pointers from entering the wrong program, if we delete a non-critical character from a program, e.g.

>>>>>>>>>>>>>>>>>>>>>>>>>\\
@+-(  "vUvdlrow ,olle"?%<
@+-("vUvdlrow ,olleH"?%  <


Then if execution starts on the % it doesn't need to care about the alignment at all and it will just execute the bad program. So we need to move % before the alignment

>>>>>>>>>>>>>>>>>>>>>>%%\\
@+-(  "vUvdlrow ,olleH"?<
@+-("vUvdlrow ,olleH"?  <


Now another line of collector is needed to fill the gap we made by adding non-redirecting characters to our collector (%s).

>>>>>>>>>>>>>>%%/>>>>>>>/
>>>>>>>>>>>>>>//>>>>>>>>\\
@+-(  "vUvdlrow ,olleH"?<
@+-("vUvdlrow ,olleH"?  <


(Some small changes are glossed over here mainly just edge casey nonsense)

Now we have to worry about what happens when a newline is deleted. This would normally be ok because the alignment doubling strategy works for these, however for us it opens the door to a lot of potential infinite loops for example in the program:

>>>>>>>>>>>>>>%%/>>>>>>>/
>>>>>>>>>>>>>>//>>>>>>>>\\@+-(  "vUvdlrow ,olleH"?<
@+-("vUvdlrow ,olleH"?  <


we now have a bunch of columns with no redirecting characters on them, so pointers starting there can just loop up or down endlessly.

The trick we use here is to add (nearly) blank lines so that deleting a newline doesn't smash two big lines together.

>>>>>>>>>>>>>>%%/>>>>>>>/

>>>>>>>>>>>>>>//>>>>>>>>\\

@+-(  "vUvdlrow ,olleH"?<

@+-("vUvdlrow ,olleH"?  <


So if we delete a newline:

>>>>>>>>>>>>>>%%/>>>>>>>/

>>>>>>>>>>>>>>//>>>>>>>>\\
@+-(  "vUvdlrow ,olleH"?<

@+-("vUvdlrow ,olleH"?  <


Nothing fundamentally changed about our program. However these lines can't be completely blank as it leaves the door open for infinite loops. In fact they need to have at least two (because one could be deleted) redirecting characters.

>>>>>>>>>>>>>>%%/>>>>>>>/
//
>>>>>>>>>>>>>>//>>>>>>>>\\
//
@+-(  "vUvdlrow ,olleH"?<
//
@+-("vUvdlrow ,olleH"?  <


Now if we delete the last newline our quotes align again:

>>>>>>>>>>>>>>%%/>>>>>>>/
//
>>>>>>>>>>>>>>//>>>>>>>>\\
//
@+-(  "vUvdlrow ,olleH"?<
//@+-("vUvdlrow ,olleH"?  <


Which is no good so we use 3 characters on the 3rd filler line:

>>>>>>>>>>>>>>%%/>>>>>>>/
//
>>>>>>>>>>>>>>//>>>>>>>>\\
//
@+-(  "vUvdlrow ,olleH"?<
///
@+-("vUvdlrow ,olleH"?  <


From here we arrive at the final program with just some minor fiddling. Iteratively fixing and finding mistakes. I've covered everything I think is interesting, but if you are interested in exactly why this or that just leave a commend and I will add it to the explanation.

8086 machine code (MS-DOS .COM), 71 bytes, score = 0

Slight variation on previous answers on the same theme.

In binary form:

00000000 : EB 28 28 E8 00 00 5A 83 C2 09 B4 09 CD 21 C3 48 : .((...Z......!.H
00000010 : 65 6C 6C 6F 2C 20 57 6F 72 6C 64 21 24 90 90 90 : ello, World!$... 00000020 : 90 90 90 90 90 90 90 90 90 90 EB D7 D7 E8 00 00 : ................ 00000030 : 5A 83 C2 09 B4 09 CD 21 C3 48 65 6C 6C 6F 2C 20 : Z......!.Hello, 00000040 : 57 6F 72 6C 64 21 24 : World!$


cpu 8086
org 0x100

%macro CODEPART 0
call near .nextline
.nextline:
pop dx
mov ah, 0x09
int 0x21
ret

.msg        db "Hello, World!\$"
%endmacro

jump1:
jmp jump2
db 0x28

part1:
CODEPART

filler:
times 13    db 0x90

jump2:
jmp part1
db 0xd7

part2:
CODEPART


Rundown

The active part is duplicated so that there is always one untouched by radiation. We select the healthy version by way of jumps. Each jump is a short jump, and so is only two bytes long, where the second byte is the displacement (i.e. distance to jump, with sign determining direction).

We can divide the code into four parts which could be irradiated: jump 1, code 1, jump 2, and code 2. The idea is to make sure a clean code part is always used. If one of the code parts is irradiated, the other must be chosen, but if one of the jumps is irradiated, both code parts will be clean, so it will not matter which one is chosen.

The reason for having two jump parts is to detect irradiation in the first part by jumping over it. If the first code part is irradiated, it means we will arrive one byte off the mark. If we make sure that such a botched landing selects code 2, and a proper landing selects code 1, we're golden.

For both jumps, we duplicate the displacement byte, making each jump part 3 bytes long. This ensures that irradiation in one of the two last bytes will still make the jump valid. Irradiation in the first byte will stop the jump from happening at all, since the last two bytes will form a completely different instruction.

Take the first jump:

EB 28 28        jmp +0x28 / db 0x28


If either of the 0x28 bytes are removed, it will still jump to the same place. If the 0xEB byte is removed, we will instead end up with

28 28           sub [bx + si], ch


which is a benign instruction on MS-DOS (other flavours might disagree), and then we fall through to code 1, which must be clean, since the damage was in jump 1.

If the jump is taken, we land at the second jump:

EB D7 D7        jmp -0x29 / db 0xd7


If this byte sequence is intact, and we land right on the mark, that means that code 1 was clean, and this instruction jumps back to that part. The duplicated displacement byte guarantees this, even if it is one of these displacement bytes that were damaged. If we either land one byte off (because of a damaged code 1 or jump 1) or the 0xEB byte is the damaged one, the two remaining bytes will also here be benign:

D7 D7           xlatb / xlatb


Whichever the case, if we end up executing those two instructions, we know that either jump 1, code 1, or jump 2 were irradiated, which makes a fall-through to code 2 safe.

Testing

The following program takes the COM filename as an argument, and creates all variants of it, along with a file called tester.bat. When run in DOS, this BAT file will run each of the created binaries, redirecting the outputs into one text file per program.

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char **argv)
{
FILE *fin, *fout, *fbat;
int fsize;
char *data;

if (!(fin = fopen(argv[1], "rb")))
{
fprintf(stderr, "Could not open input file \"%s\".\n", argv[1]);
exit(1);
}

if (!(fbat = fopen("tester.bat", "w")))
{
fprintf(stderr, "Could not create BAT test file.\n");
exit(2);
}

fseek(fin, 0L, SEEK_END);
fsize = ftell(fin);
fseek(fin, 0L, SEEK_SET);

if (!(data = malloc(fsize)))
{
fprintf(stderr, "Could not allocate memory.\n");
exit(3);
}

fprintf(fbat, "@echo off\n");

for (int i = 0; i < fsize; i++)
{
char fname[512];

sprintf(fname, "%02d.com", i);
fprintf(fbat, "%s > %02d.txt\n", fname, i);

fout = fopen(fname, "wb");

fwrite(data, 1, i, fout);
fwrite(data + i + 1, 1, fsize - i - 1, fout);

fclose(fout);
}

free(data);
fclose(fin);
fclose(fbat);
}


DOSBox, where I tested this, does not have the fc command, so another BAT file was created to run under Windows. It compares each text file with the first one to make sure they are all identical:

@echo off

for %%x in (*.txt) do (
fc %%x 00.txt > nul
if ERRORLEVEL 1 echo Failed: %%x
)


Ocular inspection of one of the text files is still needed.

Gol><>, score 0, 38 bytes

<<H"Hello, world!"/
H"Hello, world!" <


The language is released after the challenge.

Klein 0X1, score 0, 38 bytes

Works in both 011 (real projective plane) and 001 (Klein bottle)

//
@"!dlrow ,olleH"<
@"!dlrow ,olleH"<


Try it online!

This reflects the ip over the top of the program. The geometry causes the ip to reenter the program on the opposite end of the program. This makes it a lot shorter than the traditional method which requires a full length line to reach the right side of the program.

1X1 or 20X, score 0, 39 bytes

Works in 101, 111, 200, and 201

..

@"!dlrow ,olleH"<
@"!dlrow ,olleH"<


Try it online!

Due to the longstanding bug that spaces are trimmed off the front of Klein programs two .s are needed to make this work. If it were not for this bug the 37 byte program:



@"!dlrow ,olleH"<
@"!dlrow ,olleH"<


world work in all of these geometries.

The idea here is that we choose a geometry that naturally reaches the location we want and clear its way. Particularly we want geometries that connect the east edge to either the north or south edges with the proper twisting so that the ip leaving the east edge enters on the right side of the program. There are 4 of these.

1X1, 20X, 210, or 101, 43 bytes

Works in 101, 111, 200, 201, 210, and 101

..

@"!dlrow ,olleH"<
@"!dlrow ,olleH"<


Try it online!

A modified version of the above where we move the @s out of the path so that 210 and 101 work. Once again .. is required due to a bug and could be removed for a 2 byte save if the bug were fixed.

0X0, 41 bytes

Works in both 000 (torus) and 010 (Klein bottle)

<<@"!dlrow ,olleH"//
<<@"!dlrow ,olleH"//


Try it online!

XXX, score 0, 54 bytes

Works in every geometry

.               \\
@"!dlrow ,olleH"<
@"!dlrow ,olleH"<


Try it online!