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.

  • \$\begingroup\$ How about runtime errors like reading a most likely invalid address? \$\endgroup\$ Commented Jan 2, 2012 at 21:15
  • \$\begingroup\$ @PeterTaylor - There are 3 answers already, the C one with heavy revising. You can look for inspiration there. \$\endgroup\$
    – Vilx-
    Commented Jan 2, 2012 at 21:58
  • 2
    \$\begingroup\$ Anyone figured out if this is impossible in APL or GolfScript or similarly terse? \$\endgroup\$ Commented Jan 3, 2012 at 8:03
  • 19
    \$\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
    Commented Jan 3, 2012 at 8:55
  • 19
    \$\begingroup\$ ... You need some sleep. \$\endgroup\$
    – Vilx-
    Commented Jan 3, 2012 at 8:57

21 Answers 21


C, 53 characters


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!

  • 2
    \$\begingroup\$ @PaulR: Now changed to register void; The register is there to prevent &; the void is there to prevent *. \$\endgroup\$ Commented Jan 2, 2012 at 21:38
  • 6
    \$\begingroup\$ OK... correct me if this is invalid C, but it just worked on my VS2010. Insert a ; between puts and (. \$\endgroup\$
    – Vilx-
    Commented Jan 3, 2012 at 10:58
  • 41
    \$\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\$ Commented Jan 3, 2012 at 14:06
  • 3
    \$\begingroup\$ @IlmariKaronen: Heh, that's a lot of effort; thanks for going through all that. :) \$\endgroup\$ Commented Jan 3, 2012 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\$ Commented Jan 23, 2017 at 12:00

Haskell, 19


or, as a full program,



For a bit more fun:


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.


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':


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

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
  • 1
    \$\begingroup\$ I'm afraid this compiles just fine modified, although an exception is thrown at runtime. \$\endgroup\$ Commented Jan 3, 2012 at 19:13
  • 3
    \$\begingroup\$ @JeffBurdges: sure, I take this to be included in "and will not cause it to crash". \$\endgroup\$ Commented Jan 3, 2012 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\$ Commented Jan 4, 2012 at 0:15
  • 10
    \$\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
    Commented Jan 4, 2012 at 0:30
  • 1
    \$\begingroup\$ [a] -> [] in the 64-char solution \$\endgroup\$ Commented Sep 14, 2014 at 16:28

Perl, 49 chars

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.

  • \$\begingroup\$ Breaks with Perl 5.28 where print starts returning 'true' instead of 1 :-P \$\endgroup\$
    – J B
    Commented Jan 3, 2012 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\$ Commented Jan 3, 2012 at 21:47


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
    if("alert(2012 "==q){
  • 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\$ Commented Jan 3, 2012 at 8:23
  • 4
    \$\begingroup\$ Same thing as the C solution - insert a ; between eval and (. \$\endgroup\$
    – Vilx-
    Commented Jan 3, 2012 at 11:00
  • 1
    \$\begingroup\$ For the comment readers, the semicolon vulnerability is fixed, I believe the code to be clean now. \$\endgroup\$ Commented Jan 3, 2012 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
    Commented Jan 11, 2012 at 0:34
  • 2
    \$\begingroup\$ Adding a ! after eval( breaks it. \$\endgroup\$
    – jimmy23013
    Commented Feb 28, 2015 at 1:00


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'.

  • 8
    \$\begingroup\$ would you have preferred he used whitespace instead? \$\endgroup\$
    – NRGdallas
    Commented Sep 28, 2012 at 22:16
  • \$\begingroup\$ Fail if adding . after any ]. You can claim . don't output 0 to resolve \$\endgroup\$
    – l4m2
    Commented Jun 24, 2022 at 6:47


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, {}, {})
  • \$\begingroup\$ Great! You make it! \$\endgroup\$
    – Ray
    Commented Feb 6, 2013 at 6:49

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!


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

  • 3
    \$\begingroup\$ Ha! The right language for the challenge. \$\endgroup\$
    – DLosc
    Commented Jan 23, 2017 at 4:37

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.
  • \$\begingroup\$ Good one! I can't think of a way to circumvent this. "A" for effort from me, that's for sure! :) \$\endgroup\$
    – Vilx-
    Commented Jan 23, 2017 at 10:08

T-SQL 2012, 55

DECLARE @n CHAR(4)='2012'PRINT IIF(@n='2012',@n,'2012')
  • \$\begingroup\$ What SQL dialect is this? Looks like Transact-SQL (aka SQL Server), but it has invalid syntax for that. \$\endgroup\$
    – Vilx-
    Commented May 3, 2012 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
    Commented May 4, 2012 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-
    Commented May 4, 2012 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
    Commented May 7, 2012 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-
    Commented May 7, 2012 at 15:12

Ruby 1.9 - 43 chars

b if qq!=[:p,2012]

Not tested, so break away.


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.


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.


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){


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

Try all these modifications here, to verify they all still print 2012.

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 -almost- never win anything on this SE xD).

  • \$\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\$
    – user41805
    Commented Jan 23, 2017 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\$ Commented Jan 23, 2017 at 14:34

Mathematica 24

this is a simple fix to @Dr.belisarius answer (unfortunately, I can't add comments yet)


Changing Replace (/.) to RepeatedReplace (//.) fixes the problem @Dillon found, since /// is a syntax error.



VAR A$=@2012
GOTO A$@2012

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.

  • \$\begingroup\$ OK... I don't know anything about SmileBASIC, so this is purely a guess: ?-A$VAR C - could this output -2012? \$\endgroup\$
    – Vilx-
    Commented Mar 4, 2018 at 9:58
  • \$\begingroup\$ That won't work because A$ is a string. \$\endgroup\$
    – 12Me21
    Commented Mar 4, 2018 at 15:39
  • \$\begingroup\$ Wait, I have an idea! What about: ?B$VAR C? :) \$\endgroup\$
    – Vilx-
    Commented Mar 4, 2018 at 15:58
  • \$\begingroup\$ That will trigger an error because "B$" hasn't been declared. \$\endgroup\$
    – 12Me21
    Commented Mar 4, 2018 at 16:12

Sinclair ZX81/Timex TS1000/1500 BASIC, 54 tokenized BASIC bytes used

 1 LET A$="2012"
 2 PRINT "2012" AND A$="2012";"2012" AND A$<>"2012"

Regardless of the value in A$ 2012 is still shown on the screen. I have used string literals because it's faster for writing to the screen (in BASIC at least) and also saves BASIC bytes. So A$ can contain any string value possible on the ZX81 tokenized "one-touch" entry system.


Rust, 105 bytes

fn main(){use std::io::{Write,stdout};stdout().write_all({let a=b"2012";if a==b"2012"{a}else{b"2012"}});}

Try it online

Most of the length comes from writing directly to stdout vs using the print! macro. Unfortunately, the format string in print!("{}", x); can be easily modified and is checked at compile time so you can't pass in the value as a string directly.

  • \$\begingroup\$ I don't know any Rust, so this looks solid. One question - the place where it goes {a} - can't something be done with that? Like {!a} or {~a} or something? I'm just shooting blind here, but perhaps there's some operator that can be used on a string? \$\endgroup\$
    – Vilx-
    Commented Aug 1, 2020 at 0:11
  • \$\begingroup\$ @Vilx- Neither of those operators are valid in that location. a's type is &[u8] (which is a slice) and write_all is expecting a &[u8] so it's not really possible with a single character change to modify a in that location without causing a compiler error, as far as I can tell at least. \$\endgroup\$
    – TehPers
    Commented Aug 2, 2020 at 3:39

Tcl, 55 chars.

if {$a=={puts 2012}} [set a {puts 2012}] {
puts 2012

Check if the code is unmodified, then execute it, or print 2012.
I consider accessing a undefined variable as syntax error.

## Tcl, 138 Characters
set a {set a {$a};if {\$a=={$a}} {puts 2012}};if {[info ex a]&&$a=={set a {$a};if {\$a=={$a}} {puts 2012}}} [subst -noc $a] {
puts 2012

Ok, this is a quine variant: either the code is unmodified, then execute it, or simply print 2012

The last line is a little bit special: It is a comment, but the } closes the brace.


Lenguage, 176848577745260300319721504 bytes

Not really fun though.

No modification/edit a byte: ----------[+>+++++<]>.--.+.+.+

Add a byte: ----------[+>+++++<]>.--.+.+.-

Remove a byte: ----------[+>+++++<]>.--.+.+<]

JavaScript, 48 bytes


Try it online!

  • \$\begingroup\$ This language almost feels like cheating here. :D \$\endgroup\$
    – Vilx-
    Commented Jun 25, 2022 at 17:58

Haskell, 65

import Data.List

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"=~
  • \$\begingroup\$ You Perl solution can be broken by replacing the first $& with something like $0. \$\endgroup\$
    – Howard
    Commented Jan 3, 2012 at 8:29
  • \$\begingroup\$ And the Haskell one breaks if you replace the comma with a plus sign. Valiant attempt though. \$\endgroup\$
    – Jeff Burka
    Commented Jan 3, 2012 at 9:33
  • \$\begingroup\$ I think nub$filter(\x->x==2012||x==2012)(2012:[2012]) might be safe for the Haskell solution. \$\endgroup\$
    – Jeff Burka
    Commented Jan 3, 2012 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\$ Commented Jan 3, 2012 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\$ Commented Jan 6, 2012 at 12:26

C# .NET, 83 bytes

class P{static void Main(){var q="2012";System.Console.Write(q=="2012"?q:"2012");}}

Try Online

  • \$\begingroup\$ Aaand it fails. Should output 2012, not 2014. XD But, yes, the approach seems to be valid. I cannot think of a way to break it. \$\endgroup\$
    – Vilx-
    Commented Aug 20, 2019 at 18:30
  • \$\begingroup\$ I feel stupid now \$\endgroup\$ Commented Aug 20, 2019 at 19:49
  • \$\begingroup\$ Aaand this version fails because I can do "2012"=="2012"?"2014":"2012". You need the var. :) \$\endgroup\$
    – Vilx-
    Commented Aug 20, 2019 at 20:21
  • \$\begingroup\$ Well I should probably stop trying to make this shorter then xD \$\endgroup\$ Commented Aug 20, 2019 at 21:55

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