33
\$\begingroup\$

Robber's thread

Your task as cops is to select three things:

  • A programming language

  • An OEIS sequence

  • A byte set

You are then to secretly write a program in that language that computes the nth term sequence using only the bytes in the set. You will reveal the three pieces of information so that robbers can try to figure out the program.

Answers will be scored by the number of bytes in the byte set with more score being good. Cracked answers have an automatic score of 0. If your answer is uncracked after one week you may reveal the intended program and mark your answer as "safe".

Answers must be capable of calculating every term in the b-files of the sequence, but are not required to calculate any terms after.

Sequences can be either 1 or 0 indexed for both cops and robbers.

Here is a Python Script that checks if your code matches the given byte set.

\$\endgroup\$
16
  • 1
    \$\begingroup\$ Lower score is better? Or higher? So in essence we're trying to find a restricted character set that makes it difficult for someone else to make a program in the chosen language, after we've already worked out a solution? \$\endgroup\$
    – BradC
    Jul 27, 2017 at 16:56
  • 1
    \$\begingroup\$ Do we need to use all the bytes in our set? I would guess yes but it should probably be specified in the challenge. \$\endgroup\$
    – Shaggy
    Jul 27, 2017 at 17:06
  • 1
    \$\begingroup\$ @Shaggy generally no, you can includes others for red herrings, but robbers can use everything \$\endgroup\$
    – Stephen
    Jul 27, 2017 at 17:08
  • 1
    \$\begingroup\$ Can robbers use the same byte twice or more? \$\endgroup\$
    – Mr. Xcoder
    Jul 27, 2017 at 17:19
  • 2
    \$\begingroup\$ @Azulflame The b-files are files that are associated with each sequence the can be accessed by replacing the A with a b and appending a .txt. For example oeis.org/b4.txt would access the b-files for that sequence. \$\endgroup\$
    – Wheat Wizard
    Jul 27, 2017 at 23:47

72 Answers 72

12
\$\begingroup\$

Haskell, A209229, (cracked)

11 characters (including newline):

s<=[ ]
how!

Outputs True/False as an indicator function for powers of 2:

1 => True
2 => True
3 => False
4 => True
5 => False
6 => False
7 => False
8 => True
9 => False
...

Inputs are positive integers.

\$\endgroup\$
4
  • \$\begingroup\$ Does 0 give False? \$\endgroup\$
    – H.PWiz
    Jul 27, 2017 at 21:33
  • \$\begingroup\$ @H.PWiz My code doesn't work for 0, your crack can do whatever you want for it. \$\endgroup\$
    – xnor
    Jul 27, 2017 at 21:35
  • \$\begingroup\$ Cracked! Very clever~ \$\endgroup\$
    – Lynn
    Jul 27, 2017 at 22:00
  • \$\begingroup\$ @Lynn Nicely done! \$\endgroup\$
    – xnor
    Jul 27, 2017 at 22:16
5
\$\begingroup\$

Python 2, A000045 (Cracked)

ml:= input(as,forge)

it contains a whitespace and a newline
Try it Online!
Intended solution

\$\endgroup\$
6
  • 1
    \$\begingroup\$ +1 I am surprised this doesn't use + at all \$\endgroup\$
    – Mr. Xcoder
    Jul 27, 2017 at 17:49
  • 2
    \$\begingroup\$ Congrats on 10k btw :) \$\endgroup\$
    – Adnan
    Jul 27, 2017 at 17:55
  • \$\begingroup\$ Cracked \$\endgroup\$
    – Adnan
    Jul 27, 2017 at 18:05
  • \$\begingroup\$ @Mr.Xcoder Don't know python that well, but would plus work? \$\endgroup\$
    – JAD
    Jul 27, 2017 at 18:10
  • \$\begingroup\$ @JarkoDubbeldam No, sum would \$\endgroup\$
    – Mr. Xcoder
    Jul 27, 2017 at 18:11
5
\$\begingroup\$

Haskell, A000045 (cracked)

I made up my mind, I think I like t more than s.

So let's use these 30 bytes (including newline) instead:

abcdeFgh|jklmnopqrtTuvwxyz
=()

Please note that the general challenge description demands that

Answers must be capable of calculating every term in the b-files of the sequence [...].

In this case, the b-file goes up to the 2000th number, which is way beyond what can be computed using Int.

\$\endgroup\$
2
  • \$\begingroup\$ Cracked! \$\endgroup\$
    – Lynn
    Jul 28, 2017 at 13:28
  • \$\begingroup\$ Cracked \$\endgroup\$
    – H.PWiz
    Jul 28, 2017 at 15:15
4
\$\begingroup\$

Octave, A000290, Cracked!

The sequence is the square numbers: 0, 1, 4, 9, 16, 25, 36, 49 ... (so that you don't have to check the link).

'()/@^_
\$\endgroup\$
2
  • \$\begingroup\$ Cracked! \$\endgroup\$ Jul 27, 2017 at 19:14
  • \$\begingroup\$ That was indeed the intended solution :) \$\endgroup\$ Jul 27, 2017 at 22:08
4
\$\begingroup\$

Haskell, A000045 (Cracked)

Everyone likes Fibonacci numbers, I like Haskell...

I have carefully selected 30 bytes for you: the lowercase letters except f, i and t, you get the uppercase letters F and T and the pipe symbol | instead, and the three symbols =(), and newline. Here they are again:

abcdeFgh|jklmnopqrsTuvwxyz
=()
\$\endgroup\$
1
  • \$\begingroup\$ Cracked \$\endgroup\$
    – xnor
    Jul 27, 2017 at 22:17
4
\$\begingroup\$

Haskell, A034262, 43 bytes, cracked

!"#%',.=?ABCDEFGHIJKLMNOPQRSTUVWXYZ[]_{|}~

Computes a(n) = n³ + n.

\$\endgroup\$
3
  • \$\begingroup\$ Can't get the b-file now. Is returning an Int good enough? \$\endgroup\$ Jul 28, 2017 at 9:28
  • 1
    \$\begingroup\$ Cracked? \$\endgroup\$ Jul 28, 2017 at 9:42
  • \$\begingroup\$ @ChristianSievers Yes, that's exactly my solution. \$\endgroup\$
    – Laikoni
    Jul 28, 2017 at 10:21
4
\$\begingroup\$

Haskell, A009056 (cracked)

Another simple one, now again with enough letters to make it look like ordinary Haskell and maybe for you to amaze me by finding a solution completely different from mine.

The sequence is Numbers >=3 and the charset consists of these 30 bytes:

{[abcdefghijklmnopqr uvwxyz]}.

The crack has some nice techniques. I was just thinking of this:

head . flip drop [ floor pi .. ]

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Cracked. \$\endgroup\$
    – nimi
    Aug 5, 2017 at 19:24
3
\$\begingroup\$

C (C99), A000005, 25 Bytes#, Cracked!

These are the bytes for a full problem, takes n as a command line argument and outputs answer to stdout.(Space is included in bytecount).

<=>,;!"()*%+acdfhimnoprt 
\$\endgroup\$
3
  • \$\begingroup\$ ok I was wondering lol. as a command line argument? \$\endgroup\$ Jul 27, 2017 at 19:25
  • \$\begingroup\$ Yes, I'll make that clearer \$\endgroup\$
    – dj0wns
    Jul 27, 2017 at 19:27
  • 1
    \$\begingroup\$ Cracked! this was fun :D \$\endgroup\$ Jul 27, 2017 at 19:40
3
\$\begingroup\$

Unary, A002275, 1 Byte

Byte set:

0

I had to at least try it :3

(I don't think it will be particularly hard seeing as every command in the language is available)

\$\endgroup\$
8
  • 1
    \$\begingroup\$ Why not include all of the bytes to maximize your score? You have scored the lowest amount possible, when it would make no difference to include all of the bytes. \$\endgroup\$
    – Wheat Wizard
    Jul 27, 2017 at 19:52
  • \$\begingroup\$ I thought we were going for a low score (sorry, I'm new here). \$\endgroup\$
    – sonar235
    Jul 27, 2017 at 19:59
  • 1
    \$\begingroup\$ Cracked \$\endgroup\$
    – Blue
    Jul 27, 2017 at 20:04
  • \$\begingroup\$ @muddyfish Is input as a character allowed for brainfuck? \$\endgroup\$
    – sonar235
    Jul 27, 2017 at 20:06
  • \$\begingroup\$ @sonar235 In general we do allow input as character for brainfuck. Although since values here exceed 255, I don't think an answer that uses them can be valid. \$\endgroup\$
    – Wheat Wizard
    Jul 27, 2017 at 20:07
3
\$\begingroup\$

Perl 5, A000030 (Cracked)

Byte set:

imnprt7 $;()<>=~.

Try it online!

\$\endgroup\$
1
3
\$\begingroup\$

JavaScript (ES6), 17 bytes, A000290 (Cracked)

Again, this is the simple square sequence a(n) = n2.

Byte set:

$()=>CI`abelotv{}
\$\endgroup\$
1
  • \$\begingroup\$ Cracked! \$\endgroup\$
    – LarsW
    Jul 27, 2017 at 22:33
3
\$\begingroup\$

Hexagony, A057077, 77 bytes

Periodic sequence 1, 1, -1, -1. As a list:

a(0) = 1
a(1) = 1
a(2) = -1
a(3) = -1
a(4) = 1
a(5) = 1
a(6) = -1
   ...

Character set (edited), which includes a newline and a space character:

!%&(),0123456789;?@ABCDEFGHIJKLMOPQRSTUVWXYZ^abcdefghijklmnopqrstuvwxyz[]#. 

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ Any reason you modified the character set? \$\endgroup\$
    – Poke
    Jul 28, 2017 at 0:49
  • 1
    \$\begingroup\$ @Poke apparently the higher the byte count the better, so I just added a bunch of non-useful commands. \$\endgroup\$
    – Adnan
    Jul 28, 2017 at 0:53
3
\$\begingroup\$

Haskell, A000045 (cracked)

It's cracked, and I won't start a new version, but if you want to play more: it's possible without y, and it's possible to be efficient.


I apologize for leading you in wrong directions by giving a g. Let's do the same without!

Here are the 29 remaining bytes (including newline):

abcdeFh|jklmnopqrtTuvwxyz
=()

Again, remember that Int won't be enough to compute the 2000th Fibonacci number which is needed because it is in the b-file.

\$\endgroup\$
1
  • \$\begingroup\$ Cracked \$\endgroup\$
    – nimi
    Jul 29, 2017 at 13:01
3
\$\begingroup\$

Haskell, A000045 (cracked)

This is kindof (as announced) not a new version, but completely different. (Right?)

I still hope I can make you rediscover my nice little observation.

This time you are asked to implement the Fibonacci sequence using a charset of size 17, which (as far as I know) contains only one two unneeded chars:

eilnt=(,).[ ]_:0!

Note that there is no newline (but feel free to show a version which has them for readability) and remember that you have to be able to compute the 2000th Fibonacci number.

\$\endgroup\$
1
  • \$\begingroup\$ Cracked. \$\endgroup\$
    – nimi
    Jul 30, 2017 at 15:12
3
\$\begingroup\$

Cubix, A000027 (SAFE) 17 points

!&')-/0;@Oiru.NSQ

My solution:

!O!;i)!/u&!!r-)0'u;;!@

watch it online here

I originally did this without .NSQ but I figured I could add them safely.

Explanation:

This sequence is just "The Positive Integers". However, Cubix has three input commands, i, which reads in a single char (pushing -1 if input is empty), A, which reads in the rest of the input as chars (pushing a -1 to the bettom of the stack), and I, which reads the next number off the input (pushing 0 if there isn't a match). So naturally, I only provided i which reads in digits as their ascii value. uh-oh. Additionally, -1 is the usual marker for end of input, in conjunction with ? so I got rid of ?, forcing me to use ! (skip next instruction if TOS is not zero) for control flow. Finally, I thought I needed & to concatenate digits for printing with O (which outputs the top of stack as a number), but I realize now that that wasn't necessary either!

Another part of the challenge was originally to not have . the no-op character, but you can use pairs of ! instead if you're careful:

    ! O
    ! ;
i ) ! / u & ! !
r - ) 0 ' u ; ;
    ! @
    . .

i) : read input, increment.

!/ : if top of stack is zero (end of input), turn left

left: ;O.@ : pop top of stack, output as number, halt.

otherwise:

u'0: push char code of 0 to top of stack

)-r: increment, subtract, and rotate

;; : pop top of stack twice

u& : concatenate digits

!! : net zero effect, now we are at i) again.

\$\endgroup\$
2
  • \$\begingroup\$ Cubix has some noop characters right? Why don't you add those to increase your score? Or is part of the challenge that there are no noops. \$\endgroup\$
    – Wheat Wizard
    Jul 27, 2017 at 20:34
  • \$\begingroup\$ @WheatWizard I decided to be merciful and add . but it can be done without it. \$\endgroup\$
    – Giuseppe
    Jul 27, 2017 at 21:23
3
\$\begingroup\$

Seed, A005408 (Odd Numbers) - Safe

Here's a slightly more challenging one. You may use any characters valid in Seed:

[0-9 ]

You shouldn't be able to brute force this one in a week unless you have a monster of a computer. Good luck! It's crackable.

Hint

This is the Befunge-98 program used in my solution: 9&2*1-.@ (the 9 can be removed, but my solution just happens to have one)

Solution

8 47142938850356537668025719950342285215695875470457212766920093958965599014291682211474803930206233718243166021324700185443274842742744297183042397910454208210130742260519105651032450664856693714718759816232968955289663444878636053035645231246428917992493766505872029989544065894418166685284738859999912471659057565000392419689361271972485444731284404987238803115210428572945108908191097214978382112218242627719651728712577306652405452922782307249485947045215982784790396555726144526602778473387279897334709142853151393812930171163300904436548870603006549005515709074682843356844201220668045057407146751793470423217099467145255174198241606814578351260769359571053755888106818197239116973006365593563235361647743411341624822052103816968153274122434280200888824954875622811325064255818154979564925710534165572852442761249176778416688044630942040966271963723430245979221181930857841829694362184653939393940015797332978459794253176110314873994228261888801228999293570329618551223457182420746927212801550646743152754821640064626761542582557138452651970009253770914346130172884305622027370793496993281847017017643506435562229916984107083951938286577012273222191422054315198157936674247934699496471202544270325061352014830137178245082445717253260177560449757186762445707057028987371278573629077370632470496186218574320801798046510846708620502139560277546345198686675095078255875594169064796673074708822106659920187882062247609587560174781170641367430722951002242213604709887062481149928551745163110045572994991844223216663621203042075294195007458339984527333125093390189721042315604498435269143549420166732177200370228527273606218617171975362431824163269672003982537382982066136613799403024924018145511099557720492305303748099327810811511080314262364010281851651151072957475365629128068033597559560186625877942054704386175359499573139930378099420149452745731809033737756051947913924265484582800618244473333957173960222243311738522875022546610298627492222587971756897328087719407454153248557203886421828643453889090192355970705084245312184441674098515659253482621260617211786550204852895652236768886852209506535523414991099331857674826373947830587028494510697603296607361093480842935154672353288419699354739650168309017848485131553416956405911683526896232046773861961911767319373432460217755874481607587604361758089936007730253450733375831228127106295259261723611771334468553746160739548375950046831923765023329346333968732796413192682936767133122325481273354810124729664400173367781325488656859581438769940474229394692089519981810909719628263357284973442177568041416363386891516725592952892168077523560584005586276794967492051823290615767599202657060820223928678900774601616908031321346819422162123048834532926372862962159255934240435694566497798544870186550219886342379298214007368081326725550763589917206162393892085506551547475259270513853987294911388226039365971184089828739349642347312302559286882065147953715607221387657413593069535573044067517274676745306396611760657091792151803798859781616126637075577936361782593546481811651450365118155866449850474140044293772144065341051900055416408240857348697564252386403719942197789892382627153382011984996644288495699209129097948405810551134169739499539470610790009272281731894550593600643079188663110695127017324336488487580799309995227054576681630676222848231145106058050452439356753552872060820230589152143268436210090733908507724084889788244157692417246691477400856716677564609725979550100138132944851304473466485128295568194188600539248624248078558656162635444219199062786468487219220160009464328883337821175254405764395407405891483810757405446047244460754827113527540703326714751461178900155717130399854358953609995319006890674085682111514072508440632941090209366005504352890092326935445829213294943731517698438648298921337375443947066950275955144209037675013663600062205168551851984361951824731229113379464426979717688372371011461890139087487634586087688796134318950140718824105959727161482389914414206768064990410615468858520426399188835467970981786227122743162945627167772066100574532803061925537235491026486409948728571706557098859331941260622260924660292578136091523126589085799981416326395824628987154802653126685882440760385315869346960183809644486238810912663304360284449610976505715001267334297285036791711464142665122000857666018757370925847113798258537503747803972255591351740843663253694303946089997282812556281486820325652814785261116697261899511762129550421005941055852897451832731304151488058522478260009347144486599715629242208891126238083949281490804191584238425634093683587199278186505945727829025071885767559828670803412582690901456978557379916793144695491189633486065077294485660840922713748873840986104486221528464294334436081663106336911265802650605150347413103936140565054608396116572669757269475369570465915381045895991937087068526458273755454602799814909213983801968791431574508976448235055061998715636460946550584682626461010298101802277643570201189324102499951196290880892383380284543173390448406975616650185808619832614403133944687275834716343817926764699295672501869876060896683204343897481630037607159476461359111190545646447421653872016775582115425356868533678655969328210255235050133718364831289991244684695035034122861927276046255405376531096051541299607470934463981741370397268760811035606321455156217990078359217247117349544774085111287345436916077577032709684005131011429476229617901273007027774182864475737502587896225475248267937497254066190335088823904767397814233350286976811901982274477445433872253983823904938249655089770642137858608313524715533520654523880832056453080193644871440738737277389718589793074725139142291918837706550037934799585495183374639955887618135803388399608755212147742199481865453122900714456703147150994585431640652462593333773031385396586933380738103697887063571042512186708015948688088011290197524699274772775288900864690592106393483764109837848793374117655194139018455509931621247697015323332300969105814058088442693320033876473960017819576425062784644138417943454576404265382986995583045527928832
\$\endgroup\$
2
  • \$\begingroup\$ How did you generate this? \$\endgroup\$ Aug 6, 2017 at 23:58
  • 1
    \$\begingroup\$ @2EZ4RTZ Painfully, and with Python's source as a reference. It can probably be golfed a lot though. \$\endgroup\$
    – TehPers
    Aug 7, 2017 at 4:31
2
\$\begingroup\$

Python 2, A000045 (Cracked)

l:= input(a,forge)[]

it contains a whitespace and a newline
Try it Online!
Intended solution

\$\endgroup\$
1
2
\$\begingroup\$

R, A000142, (Cracked)

Byte set:

-()*,`=cfinotu

Intended solution:

f=function(n,c=n==n)'if'(n,f(n-(n==n),c*n),c*(n==n))

\$\endgroup\$
1
2
\$\begingroup\$

cQuents, A000027, Cracked


 !"#%&'()*+,-./0123456789:;?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~

That's right! You get all of ASCII! Wait... there's no $... what's $ again in cQuents? Oh, yeah, the index builtin. Well, tough luck :/

Intended Solution:

#|A:A

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ Cracked? \$\endgroup\$
    – Adnan
    Jul 27, 2017 at 18:56
  • \$\begingroup\$ @Adnan yup that works, I guess I shouldn't have opened it up as much as I did :P I couldn't think of any different ways to do it, good job. Added intended solution. \$\endgroup\$
    – Stephen
    Jul 27, 2017 at 18:57
2
\$\begingroup\$

CJam, A000042, cracked by Lynn

Byte set:

{})_%si
\$\endgroup\$
1
  • 1
    \$\begingroup\$ Cracked! Cute~ \$\endgroup\$
    – Lynn
    Jul 27, 2017 at 21:04
2
\$\begingroup\$

Ruby, A000004, 5 bytes, Cracked

Just messing around to start off. Should be easy for anyone familiar with Ruby. It's a low score, but whatever.

/np.$

Cracked, unintended exploit

Cracked (intended answer using n flag)

\$\endgroup\$
3
  • \$\begingroup\$ Cracked? \$\endgroup\$
    – Adnan
    Jul 27, 2017 at 20:42
  • \$\begingroup\$ @Adnan Eh, I'll accept it. Should've required the n flag (which is the reason it's in the byte sequence, and would make $. equal 1) \$\endgroup\$
    – Value Ink
    Jul 27, 2017 at 20:45
  • \$\begingroup\$ Think I have your intended solution \$\endgroup\$
    – histocrat
    Jul 27, 2017 at 21:59
2
\$\begingroup\$

Python 3, A007504 (Cracked)

This byteset, including newline:

bfuwo)nm1h[=(t+;0a
sig%pr, le:]

My code does not provide infinite output, but can compute the entire b-list.

\$\endgroup\$
6
  • 1
    \$\begingroup\$ Welcome to PPCG! \$\endgroup\$
    – Adalynn
    Jul 28, 2017 at 0:03
  • 1
    \$\begingroup\$ Cracked \$\endgroup\$ Jul 28, 2017 at 1:15
  • 1
    \$\begingroup\$ Scrap that I used a . :( \$\endgroup\$ Jul 28, 2017 at 1:21
  • 1
    \$\begingroup\$ I've fixed the crack. \$\endgroup\$ Jul 28, 2017 at 1:32
  • 1
    \$\begingroup\$ ...and migrated to Python 3. \$\endgroup\$ Jul 28, 2017 at 1:39
2
\$\begingroup\$

R, A000290, (cracked)

Byte set:

()%cfinotu
\$\endgroup\$
7
  • \$\begingroup\$ no c, but everything else for function? That's horrifying. I know how to do this except for reading in input... \$\endgroup\$
    – Giuseppe
    Jul 27, 2017 at 19:52
  • \$\begingroup\$ @Giuseppe woops, the a shouldve been a c. sorry. \$\endgroup\$
    – JAD
    Jul 27, 2017 at 19:53
  • 1
    \$\begingroup\$ This is what I get for doing this shit manually... \$\endgroup\$
    – JAD
    Jul 27, 2017 at 19:53
  • \$\begingroup\$ No worries; this is tough enough! \$\endgroup\$
    – Giuseppe
    Jul 27, 2017 at 19:54
  • \$\begingroup\$ cracked \$\endgroup\$
    – Giuseppe
    Jul 27, 2017 at 19:56
2
\$\begingroup\$

Snowman, 212 bytes, A000042

#$%&*01:;=?@CEFGHIJLMOQUVXYZbcdefghijlmnopqsuvxyz~

... plus space, 0x00-0x1f inclusive, and 0x7f-0xff inclusive.

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2
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cQuents, A000217, Cracked

Byteset:

$:=1;
\-

Note that this uses a feature that I haven't pushed the documentation for yet, so I'll push that tonight, if you wait until then. (Or you can slop through my source code... have fun). Documentation pushed. Would be a byte less if a recent bugfix was on TIO.

Intended solution:

=1-1:--\1$ ;$

Try It Online!

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1
2
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JavaScript (ES6), 13 10 bytes, A000045, Cracked

This should be easy.

This is the Fibonacci sequence: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.

Byte set:

$()-:<=>?[]_~

Edit:

It can be even done with the following 10 bytes:

$()-:=>?_~
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3
  • \$\begingroup\$ Cracked! \$\endgroup\$
    – user41805
    Jul 28, 2017 at 10:55
  • 1
    \$\begingroup\$ This isn't [code-golf]: "Answers will be scored by the number of bytes in the byte set with score being good." More bytes are better. \$\endgroup\$
    – LarsW
    Jul 28, 2017 at 12:55
  • \$\begingroup\$ @LarsW on the other hand, more bytes also makes it easier to crack :) \$\endgroup\$
    – JAD
    Jul 28, 2017 at 13:48
2
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R, A105311, (Cracked)

Byte set:

'%(),:=acdegilnopstx
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1
2
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Python3, A008615, Cracked

Bytemap (with newline):

n)ir=-
(u0*pt
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1
  • \$\begingroup\$ Cracked \$\endgroup\$
    – xnor
    Jul 28, 2017 at 6:45
2
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Befunge, A000142, 29 Bytes, (Cracked)

Byte Set: @.$_ ^*:\v>-1&

If you can't tell, that Byte set includes a space.

This should be moderately easy to solve.

Edit: Forgot the "A" before the OEIS

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1
  • \$\begingroup\$ Cracked \$\endgroup\$
    – KSmarts
    Jul 28, 2017 at 16:06
2
\$\begingroup\$

R, A105311, (cracked)

'%(),:=acdeginpstx

Let's try this without the l or o.

Since this has been cracked, the intended solution:

cat(diag(diag((a=scan()))%x%diag((a==a):a)),sep=''). diag is an interesting function, that can be used in three different ways. When presented with a single integer (diag(n)), it creates a NxN matrix with 1 on the diagonal. When presented with a vector (diag(1:n)), it creates an NxN matrix with the vector on the diagonal. When presented with a matrix (diag(diag(n))), it returns the diagonal as a vector. %x% computed the Kronecker product of two matrices, where each element in matrix 1 is multiplied with each element in matrix 2 separately. Doing this with a length n identity matrix and a 1:n diagonal matrix, creates a length n^2 diagonal matrix with 1:n repeated n times. diag extracts that again, and cat prints.

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7
  • 1
    \$\begingroup\$ this'll be even more fun :) \$\endgroup\$
    – Giuseppe
    Jul 28, 2017 at 13:17
  • \$\begingroup\$ does this include a newline? \$\endgroup\$
    – Giuseppe
    Jul 28, 2017 at 13:57
  • \$\begingroup\$ @Giuseppe No newlines \$\endgroup\$
    – JAD
    Jul 28, 2017 at 13:59
  • 1
    \$\begingroup\$ @Giuseppe and I just double double checked, this byteset is correct \$\endgroup\$
    – JAD
    Jul 28, 2017 at 13:59
  • \$\begingroup\$ [cracked! ](codegolf.stackexchange.com/a/136457/67312) -- I (incorrectly) used %o% before I realized that %x% existed. I guess I could have done 'i'=='i' for 1 but either way works, so. \$\endgroup\$
    – Giuseppe
    Jul 28, 2017 at 17:04

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