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This idea is not mine, though I don't know where it originated. I once met it in a programming contest very long ago (1998, if I remember correctly). The task is to write a program in your favorite language that outputs 2012 and only 2012. The catch is that the program must still output 2012 after any one of its characters is modified. The modification can be either insertion, deletion or replacement. Of course, the modification will be such that the program is still syntactically valid.

Since I don't know all the programming languages, I have to ask the audience to help me and test the answers submitted.

Added: Many have commented that my definition of acceptable modifications is too vague. Here's my second attempt: The allowed modifications will leave your program syntactically valid and will not cause it to crash. There, I think that should cover all the compile-time, link-time and run-time errors. Though I'm sure that there will be some odd edge case in some language anyway, so when that comes up we'll look at it individually.

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  • \$\begingroup\$ How about runtime errors like reading a most likely invalid address? \$\endgroup\$ – aaaaaaaaaaaa Jan 2 '12 at 21:15
  • \$\begingroup\$ @PeterTaylor - There are 3 answers already, the C one with heavy revising. You can look for inspiration there. \$\endgroup\$ – Vilx- Jan 2 '12 at 21:58
  • 2
    \$\begingroup\$ Anyone figured out if this is impossible in APL or GolfScript or similarly terse? \$\endgroup\$ – Jeff Burdges Jan 3 '12 at 8:03
  • 12
    \$\begingroup\$ This has got me thinking about DNA and redundancy and the potential for cosmic rays to flip bits in my programs. Interesting stuff. \$\endgroup\$ – Jon Purdy Jan 3 '12 at 8:55
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    \$\begingroup\$ ... You need some sleep. \$\endgroup\$ – Vilx- Jan 3 '12 at 8:57

40 Answers 40

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C, 53 characters

main(){register*a="2012";(puts("2012"==a?a:"2012"));}

A bit longer than the scripting language answers, and follows the same basic principle. Relies on the fact that due to the constraints of C's syntax, the only letters that would be possible to change without making the program invalid are within the strings!

Edit: Shaved off 4 characters.

Reedit: Robustness increased, shaved off one character.

Re-redit: Longer, but more robust. Just try your & now! (Correctly this time).

Update: Shortened a bit; defeats most approaches seen so far.

Update: Changed int to void; should defeat the last possible approach to break it I can think of.

Update: I thunk of another approach; replacing the last a (may it rot) with 0; using two-letter names should deal with that problem.

Update: Last update revoked; assuming changes causing runtime errors are disallowed; a will work just fine.

Update: Backtracking some more; attempting to dereference *a will also segfault; so using void to tease a compile error out of it should not be necessary.

Update: And a final shortening; that relies on the string "2012" being placed at but one address (which is common); and that literal strings are read-only (also common).

Update: It cost me two characters, but I defeated your puny semi-colon!

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  • 2
    \$\begingroup\$ @PaulR: Now changed to register void; The register is there to prevent &; the void is there to prevent *. \$\endgroup\$ – Williham Totland Jan 2 '12 at 21:38
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    \$\begingroup\$ OK... correct me if this is invalid C, but it just worked on my VS2010. Insert a ; between puts and (. \$\endgroup\$ – Vilx- Jan 3 '12 at 10:58
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    \$\begingroup\$ +1, Congratulations, I think you've nailed it. I wrote a simple test harness that tried to compile and run every possible one-char variation of your code, and every single one of them either did not compile, crashed or printed 2012! (I only used printable ASCII characters for the insertion and substitution tests, but I doubt expanding the repertoire would help.) \$\endgroup\$ – Ilmari Karonen Jan 3 '12 at 14:06
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    \$\begingroup\$ @IlmariKaronen: Heh, that's a lot of effort; thanks for going through all that. :) \$\endgroup\$ – Williham Totland Jan 3 '12 at 14:14
  • 1
    \$\begingroup\$ @Titus: If you mean switching == for != in the ternary, that gets you nowhere; instead of outputting a it now outputs "2012". \$\endgroup\$ – Williham Totland Jan 23 '17 at 12:00
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Haskell, 19

(\xx@2012->xx)$2012

or, as a full program,

29

main=print$(\xx@2012->xx)2012


For a bit more fun:

(\l@(_:_:t:[])->t:l)['0'..'2']


To get one that can't be modified in such a way that yields merely a runtime error, we can encode the information in the lengths of lists which can't be modified using just one-character-changes, i.e.

map(head.show.length)[[(),()],[],[()],[(),()]]

To make it more modifiable (safely), we can also use the number itself as list element – just need to make it strings to prevent exchanging commas for plus':

map(head.show.length)[["2012","2012"],[],["2012"],["2012","2012"]]

As this string is just the result of the expression, we can also again substitute it with that – not a problem thanks to Haskell's lazyness

64 72

a=map(head.show.length.init)[[a,a,a],[a],[a,a],[a,a,a]]
main=putStrLn a

The init acts as "minus one, but non-negative" here.


We can also include the type system in the redundancy scheme and then write the number in a way that could be modified with one-character changes...

u :: Enum a => (a,[b])->(a,b)
u(a,[b]) = (succ a , b)
p :: (a,b)->(a,[b])
p(a,b) = (a,[b])

ι :: (Int,())           -- Integral type to make sure you can't make it 0/0
ι = (\n -> (n-n,()))0

twothousandandtwelve = map(head.show.fst) [ u.p.u.p$ι , ι , u.p$ι , u.p.u.p$ι ]

(GHCi> twothousandandtwelve ≡> "2012")

You could now change any one u to p vice versa, but that would always mess up the deepness of list stackings in the second tuple element and thereby trigger a compile-time error.

This idea could be expanded further in such a way that whole texts could be encoded compactly, easy to read and edit, and still safe from modifing single characters.


And yet another one...

main = print N2012
data Counter = Τv |Πy |Υj |Cε |Ho |Φϑ |Ωm |Sg |Πl |Pt |Yϑ |Γσ |Km |Φz |Εα |Av |Ζρ |Ηρ |Εv |Κs |Rζ |Γϑ |Οc |Dι |Rυ |Λd |Bγ |Wt |Xε |Ωη |Ιa |Hζ |Ed |Qj |Wπ |Κw |Qu |Γο |Oι |Mσ |Ωκ |Yg |Kυ |Aj |Du |Λζ |Nζ |Θτ |Pε |Yf |Βa |Τγ |Qx |Jη |Pδ |Iq |Ωn |Fv |Kl |Ψη |Δj |Θσ |Hd |Θq |Υs |Ht |Fρ |Jh |Lζ |Hμ |Υι |Ρζ |Ρv |Dυ |Wo |Iχ |Iζ |Γy |Kr |Sσ |Iμ |Μο |Xw |Εμ |Cσ |Yξ |Aq |Jf |Hσ |Oq |Hq |Nυ |Lo |Jκ |Ρz |Οk |Θi |Θα |Αη |Gh |Lξ |Jm |Ων |Zu |Μc |Qη |Κγ |Αψ |Χζ |Hρ |Γρ |Uϑ |Rj |Χγ |Rw |Mω |Πζ |Θρ |Ωd |Υh |Nt |Tη |Qψ |Θω |Εχ |Iw |Σx |Ηn |Mτ |Xt |Yx |Φε |Hh |Wη |Mf |Ψχ |Νγ |Βξ |Aϑ |Qp |Τϑ |Φm |Uy |Gy |Cd |Bχ |Λl |Οτ |Εa |Df |Li |Aι |Yi |Νκ |Vc |Γx |Φρ |Φp |Nξ |Kf |Tw |Λξ |Φn |Λa |Oψ |Υχ |Fψ |Xω |Τq |Οσ |Σj |Θψ |Το |Νr |Ιπ |Τi |Dτ |Φf |Μn |Χm |Ηε |Wa |Αχ |Uδ |Λf |Ρu |Qk |Wα |Uρ |Τζ |Lg |Qy |Τν |Jϑ |Βδ |Mε |Μι |Πβ |Bη |Eκ |Κz |Ηh |Fδ |Σp |Εγ |Qφ |Μτ |Νχ |Ψν |Pw |Χz |Εϑ |We |Nπ |Tυ |Wg |Bh |Tρ |Ζν |Λm |Ag |Dσ |Πι
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               Cf |Επ |Fo |Οh |Tσ |Ηv |Sα |Ζq |Dk |Jπ |Ιm |Mj |Oi |Ψa |Qγ |Rn |Dξ |De |Γk |Ψm |Lα |Cl |Θο |Γq |Λc |Tx |Nm |Ki |Υο |Χr |Φs |Κi |Φλ |Vq |Αω |Ch |Tμ |Xb |Ζπ |Ym |Ζn |Eω |Ξj |Υκ |Τg |Uo |Ai |Sy |Τe |Ητ |Tτ |Λg |Bp |Δq |Χo |Pπ |Dγ |Δγ |Yπ |Ys |Ωδ |Ψσ |Sζ |Πξ |Rφ |Hj |Uf |Td |Ξk |Xψ |Οj |Cx |Φπ |Gλ |Φδ |Ej |Yψ |Ae |Φφ |Jγ |Qχ |Ξγ |Δp |Σg |Is |Eσ |Λπ |Cδ |Ιe |Cυ |Oh |Hm |Tb |Qi |Οl |Bε |Eψ |Hn |Ja |Σν |Γr |Ηu |Ζξ |Ζb |Nu |Θξ |Κd |Qο |Lq |Λw |Ηf |Kξ |Ευ |Rr |Τm |Εξ |Ψp |Χh |Ξi |Fπ |Μφ |Fu |Cξ |Aα |Pγ |Sk |Cω |Ηr |Αp |Ββ |Bx |Fp |Tζ |Pω |Λp |Lm |Jp |Bl |Φc |Vf |Τz |Εy |Λμ |Rd |Νf |Πρ |Ηx |Μψ |Γη |Bα |Συ |Iσ |Γt |Κξ |Io |Ζφ |Γl |Θf |Γλ |Υγ |Ψh |Xg |Tn |Iu |Bφ |Πχ |Λq |Χπ |Bϑ |Εm |Κφ |Λt |Ιu |Ρs |Ιβ |Ωg |Yν |Lσ |Ζι |Eι |Aτ |Φa |Pα |Θz |Ψκ |Θs |Θη |Ηl |Φζ |Bt |Ρυ |On |Ξε |Tf |Gp |Mα |Μi |Kβ |Σο |Ωξ |Νl |Iz |Fk |Dj |Bπ |Nz |Xr |Mp |Χω |Sϑ |Hu |Αμ |Js |Βn |If |Τw |Ηz |Σz |Po |Yj |Ημ |Yβ |Σm |Do
                |Ηχ |Κg |Θo |Ζh |Ψj |Ψu |Ωφ |Δμ |Γa |Bν |Ιε |Oz |Νq |Υp |Qλ |Υc |Υy |Kc |Kh |Ew |Wγ |Νβ |Ωλ |Οξ |Zι |Yr |Sυ |Γπ |Bm |Μj |Pa |Os |Χδ |Κδ |Εx |Iγ |Eη |Fλ |Tγ |Yλ |Hξ |Φq |Τξ |Ql |Δn |Zn |Ot |Sa |Φψ |Nμ |Ξr |Ξc |Φj |Gl |Oλ |Rπ |Am |Mο |Gx |Fd |Cg |Χu |Lι |Wv |Ζt |Jυ |Pσ |Σκ |Wκ |Pv |Ιg |Ωι |Δx |Φl |Eb |Δυ |Cr |Nχ |Ογ |Νφ |Gu |Ασ |Λi |Rτ |Eh |Xη |Md |Wm |Tt |Πα |Υe |Βk |Ju |Dρ |Χβ |Οs |Γi |Kι |Κe |Mm |Χf |Oκ |Vb |Γβ |Οy |Vv |Νϑ |Hl |Λα |Wξ |Om |Βφ |Ρp |Φβ |Βb |Αυ |Υδ |Χφ |Pλ |Νρ |Υλ |Ul |Kγ |Qc |Νm |Πz |Hφ |Es |Ψπ |Xm |Xξ |Tν |Eλ |Ao |Ak |Ka |Ζη |Xk |Γψ |Βπ |Fβ |Βρ |Xx |Βζ |Iτ |Pϑ |Εb |Ψγ |Τk |Gm |Yn |Xν |Νu |Hϑ |Εr |Τπ |Uw |Mh |Og |Μυ |Tj |Λν |Qm |Xn |Ην |Νi |Kη |Zv |Ιι |Ση |Yk |Dx |Aχ |Ou |Fy |Cα |Θl |Γκ |Ax |Vκ |Cn |Cλ |Ξϑ |Wε |Υl |Ψt |Ωa |Θe |Ξω |Ηo |Ll |Bζ |Kw |Αβ |Δc |Oυ |Βj |Jβ |Νε |Eϑ |Ξg |Tz |Cc |Ry |Sρ |Ψz |Yα |Pq |Υg |Jn |Vμ |Σk |Ck |Ωt |Zg |Pι |Hω |Λλ |Aμ |Wλ |Ιλ |Βc |Ξa |
               Jk |Πϑ |Ιt |Εψ |Hε |Ωϑ |Εη |Ie |Κω |Yc |Iβ |Ου |Hg |Θr |Nn |Uμ |Ζv |Ζχ |Jρ |Pο |Ng |Be |Δv |Fζ |Ρe |Qe |Cq |Κf |Θλ |Tϑ |Ξq |Me |Βq |Oα |Θc |Qr |Δt |Dm |Yu |Ru |Σh |Λr |Yy |Εε |Μχ |Mφ |Δδ |Kφ |Cγ |Ζσ |Iω |Au |Wb |Κc |Πq |Ωω |Pυ |Γn |Nγ |Cv |Βχ |Φg |Gο |Ug |Kο |Βκ |Wμ |Hτ |Hχ |Ue |Οw |Sμ |Sm |Υω |Yb |Χa |Ιi |Κν |Πu |Κψ |Uτ |Lβ |Fj |Pn |Εf |Τσ |Qε |Ψo |Λρ |Oϑ |Πν |Ts |Ηο |Μρ |Ff |Ψβ |Ne |Nκ |Bλ |Bσ |Mx |Πp |Υσ |Ιn |Αz |Fz |Ηa |Uν |Mζ |Δϑ |Yι |Ζe |Ψα |Tο |Βg |Lπ |Ζf |Αλ |Em |Θh |Gπ |Γω |Kω |Tξ |Σn |So |Im |Φυ |Ξb |Ii |Λι |Xz |Kδ |Μω |Uυ |Wf |Χb |Sλ |Lγ |Οη |Ιs |Xβ |Pκ |Bc |Ιp |Od |Αn |Va |Tω |Ζw |Ιτ |Θε |Ρi |Gι |Τh |Υx |Nτ |Δη |Εφ |Kx |Xa |Gν |Ft |Yt |Qd |Gσ |Ξυ |Εs |Nσ |Νc |Λj |Υu |Ρc |Ψξ |Δm |Qβ |Μu |Υb |Nk |Ωτ |Κr |Δd |Iλ |Πa |Ωρ |Χν |Μh |Jξ |Μμ |Fc |Iφ |Zr |Ux |Φb |Πo |Gd |Eζ |Αα |Νν |Λz |Vη |Pψ |Ωf |Lρ |Cb |Ν |Α |Χ |Ω |Zτ |Τκ |Αε |Bβ |Uι |Fi |Ui |Βx |Ωq |Βp |Λh |Uu |Ωw |Xp |Ζβ |Λτ
 | N2012 deriving(Enum); instance Show Counter where show = show . fromEnum
\$\endgroup\$
  • 1
    \$\begingroup\$ I'm afraid this compiles just fine modified, although an exception is thrown at runtime. \$\endgroup\$ – Jeff Burdges Jan 3 '12 at 19:13
  • 3
    \$\begingroup\$ @JeffBurdges: sure, I take this to be included in "and will not cause it to crash". \$\endgroup\$ – ceased to turn counterclockwis Jan 3 '12 at 21:30
  • 1
    \$\begingroup\$ For the a=map(head.show.length)[[a,a],[],[a],[a,a]] solution, insert a between []. I really like this post, though! Very clever solutions. \$\endgroup\$ – Dillon Cower Jan 4 '12 at 0:15
  • 9
    \$\begingroup\$ I've verified that all 5825 variations of your 29-character program (replacing or inserting ASCII characters 32-126) work as expected. Here's the test script I used. It can easily be tweaked to test other programs, including other languages. Warning: It took almost 1 hour to run on my laptop :) \$\endgroup\$ – hammar Jan 4 '12 at 0:30
  • 1
    \$\begingroup\$ [a] -> [] in the 64-char solution \$\endgroup\$ – John Dvorak Sep 14 '14 at 16:28
14
\$\begingroup\$

JavaScript

I believe this is runtime error proof, any single change should either result in a compile error or a single alert saying 2012.

Edit: Code would make a runtime error on something like if("alert(2012 "==r), I moved the try section to deal with it.

Edit: Nice one Vilx-, but fixable :-) Now there is a bracket mismatch for inserting that semicolon.

Edit: But then a comma could do the same thing as the semicolon, that is a host of options, I think I have fixed it, but there is an awful lot of code now.

Edit: Simplified a bit.

Edit: One more in an infinite series of bugfixes.

Edit: This kinda feels more long and complicated than bulletproof, but it should at least take care of ~eval and !eval.

var q="alert(2012 "
var p=1
try{
    if("alert(2012 "==q){
        if(eval(((p=5,q+")")||alert(2012)))){
            if(p!==5){
                alert(2012)
            }
        }
    }
    else{
        alert(2012)
    }
}
catch(e){
    alert(2012)
}
\$\endgroup\$
  • 1
    \$\begingroup\$ Because for instance q-")" returns NaN, which eval convert to "NaN" before evaling it, which simply turns it back to NaN. Strange thing to do, but technically legit, so that doesn't invoke the catch. \$\endgroup\$ – aaaaaaaaaaaa Jan 3 '12 at 8:23
  • 4
    \$\begingroup\$ Same thing as the C solution - insert a ; between eval and (. \$\endgroup\$ – Vilx- Jan 3 '12 at 11:00
  • 1
    \$\begingroup\$ For the comment readers, the semicolon vulnerability is fixed, I believe the code to be clean now. \$\endgroup\$ – aaaaaaaaaaaa Jan 3 '12 at 23:38
  • 4
    \$\begingroup\$ Not sure if this counts or not, but putting a ~ in front of the eval causes it to echo 2012 twice instead of once. Not sure if that disobeys the rules or not :P \$\endgroup\$ – mellamokb Jan 11 '12 at 0:34
  • 2
    \$\begingroup\$ Adding a ! after eval( breaks it. \$\endgroup\$ – jimmy23013 Feb 28 '15 at 1:00
13
\$\begingroup\$

Perl, 49 chars

  do{
use strict;$_=2012;;2012==$_&&0-print||die}

Based on J B's answer, but this one actually satisfies the spec. An exhaustive check indicates that every one-character deletion, insertion or replacement either leaves the output unchanged or causes the program to crash when run (as indicated by a non-zero return value and output to stderr), at least as long as insertions and replacements are restricted to printable ASCII characters.

(Without the restriction, the task is impossible in Perl: a little-known feature of the Perl parser is that it stops when it encounters a Ctrl-D or a Ctrl-Z character, so inserting either of those in front of any file turns it into a valid Perl program that does nothing.)

Edit: Shaved off one more char by replacing 1==print with 0-print.

\$\endgroup\$
  • \$\begingroup\$ Breaks with Perl 5.28 where print starts returning 'true' instead of 1 :-P \$\endgroup\$ – J B Jan 3 '12 at 21:39
  • 3
    \$\begingroup\$ @JB: Well, you can downvote it when that happens. :) (For the benefit of other readers, this is a joke. As far as I know, there are no plans to change the return value of print in any version of Perl 5, even if it isn't explicitly documented.) \$\endgroup\$ – Ilmari Karonen Jan 3 '12 at 21:47
12
\$\begingroup\$

Brainfuck

I am trying to convince myself that this is possible, and I am fairly certain I may have stretched it a bit too far. I have made a few assumptions about my environment:

  1. An infinite loop is considered a 'crash'. A similar condition could possibly be achieved by decrementing past zero or to the left of memory location zero in certain interpreters. Many interpreters are difficult to crash at runtime. I avoid the halting problem by using only the simplest, most obvious infinite loop.
  2. Unmatched square braces are considered a compile error.
  3. This will only work in an environment where the program's output is piped back to it's own input. I use that to verify that it did indeed output '2012'. This is the only way I could get around simply deleting one of the output characters.

Unfortunately, if you get any stricter I fear this will be impossible. Here is my solution:

++++++++++++++++++++++++++++++++++++++++++++++++++
.--.+.+.
,--------------------------------------------------[]
,------------------------------------------------[]
,-------------------------------------------------[]
,--------------------------------------------------[]
,[]EOF = 0

Basically, you can change the output code, or the verification code, but not both. One of them is guaranteed to work. If one of them doesn't it will 'crash'.

\$\endgroup\$
  • 5
    \$\begingroup\$ Ugh, brainfuck! You just HAD to, didn't you? XD \$\endgroup\$ – Vilx- Mar 20 '12 at 16:15
  • 8
    \$\begingroup\$ would you have preferred he used whitespace instead? \$\endgroup\$ – NRGdallas Sep 28 '12 at 22:16
7
\$\begingroup\$

Python2

import sys;xx='2012';(
1/(sys.stdout.write(xx=='2012' and xx or 2012)==None))

I had to change Ray's test script slightly to test this as the stdout redirect was breaking it. Passing empty dicts to exec avoids polluting the namespace

exec(prog, {}, {})
\$\endgroup\$
  • \$\begingroup\$ Great! You make it! \$\endgroup\$ – Ray Feb 6 '13 at 6:49
5
\$\begingroup\$

Brain-Flak, 44 + 3 = 47 bytes [Non-Competing]

This uses Brain-Flak's -A flag and outputs the characters 2012 to STDOUT

((((((((()()()){}){}){}()){})[()])[()])()())

Try it online!

Alternative, 50 bytes

(((((()()()()){}){}){}){})({({}[()])}{}[()()()()])

Try it online!

Explanation

Any single character modification to either of the codes above will cause the program to error.

\$\endgroup\$
  • 3
    \$\begingroup\$ Ha! The right language for the challenge. \$\endgroup\$ – DLosc Jan 23 '17 at 4:37
4
\$\begingroup\$

Sisi, non-competing

Finally I think I found one of my languages that works. It's horrendously long, and the language is newer than the question, but it still feels like an accomplishment.

1 set xx 2012
2 set y xx=2012
3 jumpif y 55
4 set xx 2012
828 set x xx
829 set ax xx
830 set xa xx
831 set axx xx
832 set xax xx
833 set xxa xx
834 set bx xx
835 set xb xx
836 set bxx xx
837 set xbx xx
838 set xxb xx
839 set cx xx
840 set xc xx
841 set cxx xx
842 set xcx xx
843 set xxc xx
844 set dx xx
845 set xd xx
846 set dxx xx
847 set xdx xx
848 set xxd xx
849 set ex xx
850 set xe xx
851 set exx xx
852 set xex xx
853 set xxe xx
854 set fx xx
855 set xf xx
856 set fxx xx
857 set xfx xx
858 set xxf xx
859 set gx xx
860 set xg xx
861 set gxx xx
862 set xgx xx
863 set xxg xx
864 set hx xx
865 set xh xx
866 set hxx xx
867 set xhx xx
868 set xxh xx
869 set ix xx
870 set xi xx
871 set ixx xx
872 set xix xx
873 set xxi xx
874 set jx xx
875 set xj xx
876 set jxx xx
877 set xjx xx
878 set xxj xx
879 set kx xx
880 set xk xx
881 set kxx xx
882 set xkx xx
883 set xxk xx
884 set lx xx
885 set xl xx
886 set lxx xx
887 set xlx xx
888 set xxl xx
889 set mx xx
890 set xm xx
891 set mxx xx
892 set xmx xx
893 set xxm xx
894 set nx xx
895 set xn xx
896 set nxx xx
897 set xnx xx
898 set xxn xx
899 set ox xx
900 set xo xx
901 set oxx xx
902 set xox xx
903 set xxo xx
904 set px xx
905 set xp xx
906 set pxx xx
907 set xpx xx
908 set xxp xx
909 set qx xx
910 set xq xx
911 set qxx xx
912 set xqx xx
913 set xxq xx
914 set rx xx
915 set xr xx
916 set rxx xx
917 set xrx xx
918 set xxr xx
919 set sx xx
920 set xs xx
921 set sxx xx
922 set xsx xx
923 set xxs xx
924 set tx xx
925 set xt xx
926 set txx xx
927 set xtx xx
928 set xxt xx
929 set ux xx
930 set xu xx
931 set uxx xx
932 set xux xx
933 set xxu xx
934 set vx xx
935 set xv xx
936 set vxx xx
937 set xvx xx
938 set xxv xx
939 set wx xx
940 set xw xx
941 set wxx xx
942 set xwx xx
943 set xxw xx
944 set yx xx
945 set xy xx
946 set yxx xx
947 set xyx xx
948 set xxy xx
949 set zx xx
950 set xz xx
951 set zxx xx
952 set xzx xx
953 set xxz xx
954 set xxx xx
955 print xx

About Sisi

Sisi is a toy language inspired by assembly and QBasic. It is good for this challenge because its syntax is extremely limited.

  • It has only four commands: set, print, jump, and jumpif.
  • All commands have fixed arity.
  • All lines must have line numbers, which are strictly increasing.
  • Expressions are only allowed in set statements. They can only contain (at most) one operation, which must be binary. In particular: changing print xx to print -xx is a syntax error.
  • Variable names must be composed of lowercase letters.
  • Most importantly: there is no comment syntax!

The program

The core of the program is this part:

1 set xx 2012
2 set y xx=2012
3 jumpif y 55
4 set xx 2012
955 print xx

We store 2012 in xx, then test whether that was successful and store the test result in y. If y is truthy, jump to line 55. (Jumps to nonexistent line numbers simply fast-forward to the next line.)

Radiation hardening

  • If the assignment in line 1 is modified, then y is falsey, the jump doesn't happen, and line 4 sets xx to 2012.
  • If the assignment in line 2 or the jump condition in line 3 is modified, we don't care: xx will get set to 2012 whether we take the jump or not.
  • The jump target in line 3 can be changed to as small as 5 or as large as 955. Any possible modification gets it to the print on line 955 sooner or later. It isn't possible with one modification to jump backwards (creating a loop) or past the end of the program.
  • If the assignment in line 4 is modified, we don't care: line 1's assignment will be correct and we will jump past line 4.
  • If line 955 is modified, we may have a problem. The one unfortunate thing about Sisi is that uninitialized variables default to 0, so a modification like print ax isn't an error. The ugly but effective solution is lines 828-954, which assign 2012 to every possible variable with an edit distance of 1 from xx. This ensures that any modification to the final print xx will still print 2012.
  • If a line number is modified, either: 1) it will be out of order and be a syntax error, or 2) it won't affect the program flow. The main modification we might be worried about--changing line 4 to 94, thereby inserting it after the jump to 55--doesn't matter because all it does is assign 2012 to xx again.
\$\endgroup\$
  • \$\begingroup\$ Good one! I can't think of a way to circumvent this. "A" for effort from me, that's for sure! :) \$\endgroup\$ – Vilx- Jan 23 '17 at 10:08
3
\$\begingroup\$

Python

A little verbose (now with a fix for that pesky semicolon between print and (!):

a=[0]
(lambda x:print(
set(['2012']).intersection(set(["2012"])).pop()))(
*a)
\$\endgroup\$
  • \$\begingroup\$ @Jeff Burdges: Yeah, with this one I'm banking on runtime errors being disallowed as well. :) \$\endgroup\$ – Dillon Cower Jan 3 '12 at 7:05
  • \$\begingroup\$ I'm afraid it's still syntactically correct if the intersection yields the empty set and pop throws a runtime error. Add more 2012s on both sides. \$\endgroup\$ – Jeff Burdges Jan 3 '12 at 7:10
  • \$\begingroup\$ @DC: I would say that an exception is not the same as a runtime error, although it seems exceptions are called errors in Python. pop()ing the empty set isn't an error in and of itself, but it does raise an 'Error'. \$\endgroup\$ – Williham Totland Jan 3 '12 at 9:37
  • 1
    \$\begingroup\$ An exception becomes an error when it's not caught. \$\endgroup\$ – Ilmari Karonen Jan 7 '12 at 0:27
  • 1
    \$\begingroup\$ inserting a comma after pop outputs <built-in method pop of set object at 0xb6fe93ac> () \$\endgroup\$ – gnibbler Feb 4 '13 at 23:23
3
\$\begingroup\$

T-SQL 2012, 55

DECLARE @n CHAR(4)='2012'PRINT IIF(@n='2012',@n,'2012')
\$\endgroup\$
  • \$\begingroup\$ What SQL dialect is this? Looks like Transact-SQL (aka SQL Server), but it has invalid syntax for that. \$\endgroup\$ – Vilx- May 3 '12 at 10:22
  • \$\begingroup\$ @Vilx-: I tested it in SQL Server 2008 myself when I was trying to break it, and it worked just fine. \$\endgroup\$ – mellamokb May 4 '12 at 14:23
  • \$\begingroup\$ @mellamokb - That's odd. It's exactly what I did, and it complained that you cannot assign a default value to variable (you need a separate SET statement), and there is no IIF (only CASE). \$\endgroup\$ – Vilx- May 4 '12 at 18:15
  • 1
    \$\begingroup\$ @Vilx-: Ah, you are correct. I realize now I was actually testing on SQL Azure. With SQL Server 2008 I get those same errors as you. However, it seems to work fine in SQL Server 2012/Azure. Here's a demo with SQL 2012: sqlfiddle.com/#!6/d41d8/27 \$\endgroup\$ – mellamokb May 7 '12 at 14:10
  • \$\begingroup\$ OK, congratulations, I can't find any obvious ways of breaking this! :) That's not to say there aren't any. ;) \$\endgroup\$ – Vilx- May 7 '12 at 15:12
3
\$\begingroup\$

I've make a python script to judge all the python solutions. The script breaks the first and the third python solutions, with many different methods. And the second solution use lambda to guard itself, is unbreakable. The second python solution is in python 3. I modified it into python 2 format and then judge program broke it.

Here is the judge script.

from StringIO import StringIO
import sys

def run(prog, atexp=True):
    stdout = sys.stdout
    fakeOut = StringIO()
    try:
        sys.stdout = fakeOut
        # exec prog # running exec directly will break some solutions by mistake
        exec(prog, {}, {}) 
        return fakeOut.getvalue().rstrip() == '2012'
    except:
        return atexp
    finally:
        sys.stdout = stdout
    return True

def judge(prog):
    RED = '\x1b[31m'
    WHITE = '\x1b[37m'
    ans = True
    chars = """abcdefghijklmnopqABCDEFGHIJKLMNOP`~1234567890!@#$%^&*()_+-=[]{};:'"<>,./?"""
    # attack by replace
    if not run(prog):
        print "Are you joking...This program don't print 2012 itself."
        return False
    p = list(prog)
    for i in xrange(len(prog)):
        for c in chars:
            p[i] = c
            r = run(''.join(p))
            if not r:
                print 'Attack by replace success'
                print 'origin:\n'+prog
                print 'modified:\n'+''.join(p[:i])+RED+c+WHITE+''.join(p[i+1:])
                ans = False
        p[i] = prog[i]
    # attack by delete
    for i in xrange(len(prog)):
        p = prog[:i]+prog[i+1:]
        r = run(''.join(p))
        if not r:
            print 'Attack by delete success'
            print 'origin:\n'+prog
            print 'modified:\n'+''.join(p[:i])+RED+'{|}'+WHITE+''.join(p[i:])
            ans = False
    # attack by insert
    for i in xrange(len(prog)+1):
        p = list(prog)
        p.insert(i, ' ')
        for c in chars:
            p[i] = c
            r = run(''.join(p))
            if not r:
                print 'Attack by insert success'
                print 'origin:\n'+prog
                print 'modified:\n'+''.join(p[:i])+RED+c+WHITE+''.join(p[i+1:])
                ans = False
    if ans: 
        print 'Amazing! judgement passed'
    return ans

p1="""xx='2012'
print(xx if xx=='2012' else
'2012')
"""
p2="""import sys
a=[0]
(lambda x:sys.stdout.write(
set(['2012']).intersection(set(["2012"])).pop()))(
*a)
"""

p3="""print sorted(['2012',
'2012','2012']).__getitem__(1)
"""
p4="""print 2012
"""
judge(p3) 
\$\endgroup\$
  • \$\begingroup\$ cool! my answer sucks and blows at the same time \$\endgroup\$ – gnibbler Feb 4 '13 at 22:34
  • \$\begingroup\$ p2 always throws an exception for Python2. The output is never 2012. Your judge program won't run under Python3 as it uses print statements. \$\endgroup\$ – gnibbler Feb 4 '13 at 22:51
  • \$\begingroup\$ @gnibbler I don't noticed that p2 is in python 3. \$\endgroup\$ – Ray Feb 5 '13 at 15:10
  • \$\begingroup\$ I've worked out a new answer for Python2 that passes your script \$\endgroup\$ – gnibbler Feb 5 '13 at 23:08
3
\$\begingroup\$

Mathematica, 23 chars

Note: "/." means "replace"

2012/.Except@2012->2012
\$\endgroup\$
  • 2
    \$\begingroup\$ This might be kind of a weak answer/defeat, but change /. to // (postfix form). \$\endgroup\$ – Dillon Cower Jan 8 '12 at 19:31
  • \$\begingroup\$ @DillonCower Good catch! \$\endgroup\$ – Dr. belisarius Mar 8 '13 at 12:53
2
\$\begingroup\$

Javascript - 68, 77, 84, 80 chars

Here's a solution that should work this year :)

a="eval(alert(new Date().getFullYear()))" 
a.length==37?eval(a||a||a):alert(2012)

Here's the test fiddle.

Update 1: Fixed issue where eval(+a) broke it (thanks eBusiness).

Update 2: Fixed for the '|' case (thanks again eBusiness).

Update 3: Fixed for the '>' case (thanks again eBusiness). And broke for the '1' case :(

\$\endgroup\$
  • \$\begingroup\$ You didn't guard your eval. eval(+a) and a lot of other modifications run perfectly fine but doesn't do a whole lot. \$\endgroup\$ – aaaaaaaaaaaa Jan 3 '12 at 23:33
  • \$\begingroup\$ @eBusiness: Thanks for pointing out the eval(+a) case, that should now be fixed. Also, I've been unable to find one of the modification cases where it runs fine but doesn't output 2012, so will you please give me an example? \$\endgroup\$ – Briguy37 Jan 4 '12 at 15:06
  • \$\begingroup\$ eval(0) and eval(-a) for instance does the same thing. \$\endgroup\$ – aaaaaaaaaaaa Jan 4 '12 at 15:40
  • \$\begingroup\$ Post edit it can at least be broken by replacing || with |. \$\endgroup\$ – aaaaaaaaaaaa Jan 4 '12 at 15:45
  • 5
    \$\begingroup\$ Breaks if you change the first alert to aler=. \$\endgroup\$ – jimmy23013 Sep 14 '14 at 23:25
2
\$\begingroup\$

Ruby 1.9 - 43 chars

qq=[:p,2012]
b if qq!=[:p,2012]
send(
*qq)

Not tested, so break away.

\$\endgroup\$
2
\$\begingroup\$

Excel, 14 characters (cheating slightly):

{=2012} (in a 2x1 array with one cell hidden)

Any valid change to the array will affect the contents of both cells, and attempting to change just one cell triggers an error message.

Of course, this breaks down if you take the view that it's really only one formula, as opposed to 2 formulas that are constrained to be identical.

\$\endgroup\$
2
\$\begingroup\$

Java 7

class M{
  static String c(){
    String a = "2012",
           b = "2012";
    return a.equals(b)           // 1
            ? a                  // 2
            : a.equals("2012")   // 3
               ? a               // 4
               : b;              // 5
  }

  public static void main(String[]a){
    System.out.print(c());
  }
}

Explanation:

  1. If a's value is modified it will take the following path: 1 → 3 → 5 (and it will return b's unchanged value of 2012).
  2. If b's value is modified it will take the following path: 1 → 3 → 4 (and it will return a's unchanged value of 2012)
  3. If a.equals(b) on @1 is modified to a.equals(a) or b.equals(b) it will take the following path: 1 → 2 (and it will return a's unchanged value of 2012)
  4. If a on @2 is changed to b it will take the following path: 1 → 2 (and it will return b's unchanged value of 2012)
  5. If either a or b is changed to the opposite on the lines 3, 4 or 5 it will take the following path: 1 → 2 (and it will return a's unchanged value of 2012)
  6. If the String on @3 is modified it will take the following path: 1 -> 2 (and it will return a's unchanged value of 2012)
  7. Changing M or a (in main(String[]a)) is valid without any changes to the functionality of the code.
  8. Any other change will result in a Compile-error (excluding some of the enters/whitespaces, which can be removed/added).

If you can find any way to break it following OP's rules I'd love to know, so I can think of something to fix it.
Unlike most similar questions where a single char is modified, this question allows the basic structure of the programming language in question to be intact, so Java can finally compete in one enter one (let's face it, Java will never win anything on this SE xD).

\$\endgroup\$
  • \$\begingroup\$ What is a negative sign is inserted in path 4 before the a? Will the program still output 2012, or will it output -2012? \$\endgroup\$ – Cows quack Jan 23 '17 at 14:30
  • \$\begingroup\$ @KritixiLithos It's a String, so you can't insert a - before the a or b. Here is a screenshot of the Compile error. \$\endgroup\$ – Kevin Cruijssen Jan 23 '17 at 14:34
1
\$\begingroup\$

Ruby 1.9

puts (
rx=2012
)==2012?rx:2012.send(
:abs)

I build this short but simple program which you are invited to break according to above rules. I couldn't even find a position where the output is changed (without breaking the code) if a # is inserted in order to comment out the rest of the line.

\$\endgroup\$
  • \$\begingroup\$ puts (r = 2012) == 2012 ? -r : 2012.send(:abs) prints "-2012" \$\endgroup\$ – Joanis Jan 2 '12 at 20:31
  • \$\begingroup\$ @M.Joanis But you need two additional characters: a space character and a minus. A minus alone will not work. \$\endgroup\$ – Howard Jan 2 '12 at 20:40
  • \$\begingroup\$ Ah, you're right! Compilation error. I know nothing about Ruby, I should be more careful. Sorry! \$\endgroup\$ – Joanis Jan 2 '12 at 20:47
  • 1
    \$\begingroup\$ According to irb, adding a minus before the rx results in a valid program that outputs -2012 \$\endgroup\$ – Konrad Rudolph Jan 4 '12 at 14:14
  • 2
    \$\begingroup\$ Changing puts to putc causes this to output a ? in Ruby 1.9. \$\endgroup\$ – histocrat Dec 12 '12 at 18:39
1
\$\begingroup\$

Scala, 95

(or 54 if skip non-meaningful parts)

object Main{def main(ar:Array[String]){print(((s:String)=>if(s=="2012")s else"2012")("2012"))}}
\$\endgroup\$
  • \$\begingroup\$ I'm afraid I don't know enough about Scala to find any flaws. :( Can other people, please, take a look? \$\endgroup\$ – Vilx- Sep 26 '12 at 9:07
  • \$\begingroup\$ It's not hard at this level. Most of Scala hardness is due to complexity of its type system (which I still cannot understand fully :D ) \$\endgroup\$ – Sarge Borsch Sep 26 '12 at 13:44
  • \$\begingroup\$ I don't know Scala, but (...),("2012") seemed to break it. \$\endgroup\$ – jimmy23013 Feb 28 '15 at 1:45
1
\$\begingroup\$

C#

Surely its not this simple, is it? ...

class program
{
    static void Main()
    {
        var result = "2012";

        if (result == "2012")
            Console.WriteLine(result);
        else
            Console.WriteLine("2012");
    }
}

Ok what I miss?

\$\endgroup\$
  • 1
    \$\begingroup\$ What happens if you change the 8th line to "Console.WriteLine(-result);"? \$\endgroup\$ – Glenn Randers-Pehrson Sep 15 '14 at 15:25
  • \$\begingroup\$ No C# compiler here, but shouldn't changing line 7 to if (result += "2012") make it output 20122012? \$\endgroup\$ – Philipp Sep 15 '14 at 15:26
  • \$\begingroup\$ hmmmm ... i hadn't thought of that ... back to the drawing board \$\endgroup\$ – War Sep 15 '14 at 15:28
  • \$\begingroup\$ @Philipp - Nop, that wouldn't compile. You need a boolean as a value for an if statement, not a string. \$\endgroup\$ – Vilx- Sep 15 '14 at 22:51
  • 6
    \$\begingroup\$ Adding a ; after else makes it outputs two times. \$\endgroup\$ – jimmy23013 Feb 28 '15 at 2:29
1
\$\begingroup\$

PHP, 30

<?=$i=2012;$i==2012?$i:2012;?>
\$\endgroup\$
  • \$\begingroup\$ Welcome to PPCG! Nice first post! Can you add an explanation also? \$\endgroup\$ – Rɪᴋᴇʀ Jan 22 '17 at 16:02
  • 2
    \$\begingroup\$ And here's how I'd defeat it: <?=$i=2012;$i==2012?-$i:2012;?> \$\endgroup\$ – Vilx- Jan 22 '17 at 16:11
  • \$\begingroup\$ @Vilx- Nice, I hadn't thought about that. \$\endgroup\$ – jsa Jan 28 '17 at 15:25
  • \$\begingroup\$ A lame defeat for any one-line PHP solution: put the whole code in a comment <?#$i=2012;$i==2012?$i:2012;?> Solution against that: insert a newline after $i=2012. \$\endgroup\$ – Titus Mar 5 '18 at 0:01
1
\$\begingroup\$

Taxi, 396 bytes

2012 is waiting at Starchild Numerology.2012 is waiting at Starchild Numerology.Go to Starchild Numerology: w 1 r 3 l 2 l 3 l 2 r.Pickup a passenger going to Equal's Corner.Pickup a passenger going to Equal's Corner.Go to Equal's Corner: w 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery: n 3 r 1 r 1 r.Pickup a passenger going to Post Office.Go to Post Office: n 1 l 1 r.

Formatted for humans, that's:

2012 is waiting at Starchild Numerology.
2012 is waiting at Starchild Numerology.
Go to Starchild Numerology: w 1 r 3 l 2 l 3 l 2 r.
Pickup a passenger going to Equal's Corner.
Pickup a passenger going to Equal's Corner.
Go to Equal's Corner: w 1 l.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery: n 3 r 1 r 1 r.
Pickup a passenger going to Post Office.
Go to Post Office: n 1 l 1 r.

From reading other answers, the solution seems to be to pick a fragile language, set the value, check the value, print the value.

\$\endgroup\$
1
\$\begingroup\$

SmileBASIC

OPTION STRICT
VAR A$=@2012
GOTO A$@2012
IF SHIFT(A$)THEN VAR B
?A$VAR C
B=C

First, we enable strict mode. This forces you to declare all variables before using them, and prevents things like changing ?A$ to ?B$.

Next, a string variable called "A$" is created, and the value is set to "@2012". To make sure the value of A$ hasn't been messed with, the program will try to jump to a label, and the only label in the program is @2012.

Now, A$ is definitely "@2012", but before printing it, the @ needs to be removed. SHIFT() removes the first item in a string or array (just like pop(), but from the other end). To discard the value returned by SHIFT, it is passed to an IF/THEN block, which does nothing. Just in case someone tries to comment out that line, we declare a variable "B", which is used later. If the VAR is commented out, the last line will throw an Undefined variable error.

Now, A$ is printed, and there's another VAR to prevent comments. The final line just makes sure the variables B and C have both been declared.

\$\endgroup\$
  • \$\begingroup\$ OK... I don't know anything about SmileBASIC, so this is purely a guess: ?-A$VAR C - could this output -2012? \$\endgroup\$ – Vilx- Mar 4 '18 at 9:58
  • \$\begingroup\$ That won't work because A$ is a string. \$\endgroup\$ – 12Me21 Mar 4 '18 at 15:39
  • \$\begingroup\$ Wait, I have an idea! What about: ?B$VAR C? :) \$\endgroup\$ – Vilx- Mar 4 '18 at 15:58
  • \$\begingroup\$ That will trigger an error because "B$" hasn't been declared. \$\endgroup\$ – 12Me21 Mar 4 '18 at 16:12
0
\$\begingroup\$

Haskell, 65

import Data.List
nub$filter(\x->x==2012||x==2012)([2012]++[2012])

Perl, 84

use strict;use English;$ARG=
2032;s/.*/2012/unless$ARG eq 2012;$SUBSEP=2012;print;

Failed approach :

use strict;use English;"2012 2012"=~
/2012|2012|2012/;print$MATCH||$MATCH;
\$\endgroup\$
  • \$\begingroup\$ You Perl solution can be broken by replacing the first $& with something like $0. \$\endgroup\$ – Howard Jan 3 '12 at 8:29
  • \$\begingroup\$ And the Haskell one breaks if you replace the comma with a plus sign. Valiant attempt though. \$\endgroup\$ – Jeff Burka Jan 3 '12 at 9:33
  • \$\begingroup\$ I think nub$filter(\x->x==2012||x==2012)(2012:[2012]) might be safe for the Haskell solution. \$\endgroup\$ – Jeff Burka Jan 3 '12 at 9:35
  • \$\begingroup\$ I forgot the use English in perl, that'll fix it. I think several fixes work for Haskell, including switching to strings. \$\endgroup\$ – Jeff Burdges Jan 3 '12 at 13:12
  • 1
    \$\begingroup\$ The Perl solution is still pretty fragile. Some ways to break it (found while testing my own answer) include replacing print with *rint, prin:, p:int or #rint, prepending = to the first line, or even replacing s/.*/ with s/./ or y/.*/ (which would be fixed by initializing $ARG to 2012 to begin with). Also, your $SUBSEP trick still doesn't protect against inserting a * before the final semicolon, but a much easier solution is to just remove that unnecessary semicolon. \$\endgroup\$ – Ilmari Karonen Jan 6 '12 at 12:26
0
\$\begingroup\$

PHP, 43 48

 $a='2012';echo($a==date('Y')?abs($a):date('Y'));
\$\endgroup\$
  • \$\begingroup\$ Inserting a - between the ? and the $a results in -2012 being printed. \$\endgroup\$ – Gareth Jan 9 '12 at 15:05
  • \$\begingroup\$ Cute attempt with the date(), but the program has to work beyond 2012 as well. ;) \$\endgroup\$ – Vilx- Jan 10 '12 at 8:51
  • \$\begingroup\$ @Gareth - hmm :) \$\endgroup\$ – The Coder Jan 11 '12 at 20:24
  • 1
    \$\begingroup\$ @Vilx-"The task is to write a program in your favorite language that outputs 2012" where does it say it has to work beyond 2012? :) \$\endgroup\$ – The Coder Jan 11 '12 at 20:24
  • 1
    \$\begingroup\$ @TheCoder Ok, now I'm going to insert my - between the ? and the abs. ;-) \$\endgroup\$ – Gareth Jan 11 '12 at 20:46
0
\$\begingroup\$

Ruby (57 36)

EDIT And another one in 36 chars.

p((x,y='2012','2012';x=='2012'?x:y)) 

I guess using strings is pretty failsafe because that minus thing doesn't work any more.


EDIT next try (36 chars) [never mind, adding a - after p results in -2012]

p [2012,2012].find(2012).first||2012


This one will only work this year, but also for similar contests in the future :) (15.chars)

p Time.now.year

Note that this substitution is also working:

p Time.new.year

57 chars. If you count the newlines, too it is 76 chars, though.

p( 
#
#
#
#
#
#
#
#
#
#
__LINE__.to_s.prepend(
#
#
#
#
#
#
#
__LINE__.to_s))

\$\endgroup\$
  • \$\begingroup\$ First two solutions - add a # at the start of the line. Last solution - enter a newline before p. (Note - I don't know Ruby myself, so perhaps I'm wrong). \$\endgroup\$ – Vilx- Apr 25 '12 at 13:02
  • \$\begingroup\$ damn :) you're right \$\endgroup\$ – Patrick Oscity Apr 25 '12 at 13:30
  • \$\begingroup\$ Prints "2012" for me. The quotes aren't conform - are they? \$\endgroup\$ – user unknown May 10 '12 at 3:37
  • \$\begingroup\$ Yeah you are right. You could use $><< or puts instead. Whatever, i'm not gonna try this any more, i gave up :) \$\endgroup\$ – Patrick Oscity May 10 '12 at 7:02
0
\$\begingroup\$

Q, 23

{$[2012~m:2012;m;2012]}

.

q){$[2012~m:2012;m;2012]}`
2012
q){$[1012~m:2012;m;2012]}`
2012
q){$[2012~m:1012;m;2012]}`
2012
q){$[2012~m:2012;m;1012]}`
2012
q){$[2012~\m:2012;m;2012]}`
2012
q){$[2012~/m:2012;m;2012]}`
2012
q){$[2012~'m:2012;m;2012]}`
2012

etc

\$\endgroup\$
  • \$\begingroup\$ And adding a negative (-)? -- {$[-2012~m:2012;m;2012]} -- {$[2012~m:-2012;m;2012]} -- {$[2012~m:2012;m;-2012]} \$\endgroup\$ – Gaffi May 22 '12 at 21:07
  • \$\begingroup\$ it still outputs 2012 q){$[-2012~m:2012;m;2012]}` 2012-- q){$[2012~m:-2012;m;2012]}` 2012-- q){$[2012~m:2012;m;-2012]}` 2012 \$\endgroup\$ – tmartin May 23 '12 at 9:18
  • \$\begingroup\$ This can be defeated by switching out the ;m; for something like ;1; or ;x;. \$\endgroup\$ – streetster Jun 23 '17 at 14:48
0
\$\begingroup\$

Hmm.. let me try:

class a {
  int i=2012, j=2012;
  public static void main(String[] args){if (i==j)
    {System.out.println(i);}else{System.out.println(i);}
  }
}

128 chars without the whitespace.

class a{int i=2012, j=2012;public static void main(String[] args){if(i==j){System.out.println(i);}else{System.out.println(i);}}}
\$\endgroup\$
  • 1
    \$\begingroup\$ Broken by inserting "-", i.e. System.out.println(-i) (for the first println). Also broken by e.g. replacing the first println(i) with e.g. println(1). Also potentially broken by changing i == j to i += j (or -=, etc.). \$\endgroup\$ – TLW Jan 7 at 2:09
0
\$\begingroup\$

PHP

I hope it's unbreakable :)

$a="2012";ob_start();function c(){ob_clean();return"2012";}function ee($s){return crc32($s)+printf($s);}
print(ee($a)==1367825560?die()
:c());
\$\endgroup\$
  • \$\begingroup\$ I add one character to the end of this line: echo(printf("%s",$a)==4)?die(3 and I get 20123. \$\endgroup\$ – Vilx- Sep 27 '12 at 10:59
  • \$\begingroup\$ Hmm… PHP 5.4.7 works correctly with die(3. \$\endgroup\$ – Stanislav Yaglo Sep 27 '12 at 11:00
  • \$\begingroup\$ You're right. My bad. Continuing the search... \$\endgroup\$ – Vilx- Sep 27 '12 at 11:06
  • \$\begingroup\$ PHP 5.3.3 — correct as well. die() outputs only strings, it doesn't output numbers. \$\endgroup\$ – Stanislav Yaglo Sep 27 '12 at 11:06
  • \$\begingroup\$ Wait, DUH, found it! Just change ."2012" to ."FAIL". :) \$\endgroup\$ – Vilx- Sep 27 '12 at 11:08
0
\$\begingroup\$

Groovy: 26

x=2012;print x^2012?2012:x
\$\endgroup\$
  • \$\begingroup\$ I don't know Groovy much, but would this work? x=2012;print x^2012?2012:-x \$\endgroup\$ – Vilx- Sep 27 '12 at 14:48
  • \$\begingroup\$ what's the - for? \$\endgroup\$ – Armand Sep 27 '12 at 15:39
  • \$\begingroup\$ @Alison The - is to try and modify your code with one character so that it no longer works. \$\endgroup\$ – Gareth Oct 30 '12 at 9:11
  • 1
    \$\begingroup\$ x=2012;print x^=2012?2012:x (changing ^ to ^=) changes the output to 0. \$\endgroup\$ – histocrat Dec 12 '12 at 19:23
0
\$\begingroup\$

Lua

print(- -(function(ab,ba)assert(ab==ba)return- -ab end)(2012,2012))
\$\endgroup\$
  • 1
    \$\begingroup\$ What if you delete one of those - signs? \$\endgroup\$ – Vilx- Jan 15 '13 at 15:30
  • \$\begingroup\$ Oh, and if you change one of those 2012 to something else, the assert will raise an error. Hmm... well, I guess you deserve creativity points for abusing the rules. But seriously, that's not how it was meant. \$\endgroup\$ – Vilx- Jan 15 '13 at 15:35

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