Can you outgolf me? (Cops section)

Question

Cops section

The robbers section can be found here.

^{Thanks to FryAmTheEggman, Peter Taylor, Nathan Merrill, xnor, Dennis, Laikoni and Mego for their contributions.}

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Challenge

Your task is to write 2 different programs (full programs/functions/etc.) in the same language and the same version (e.g. Python 3.5 ≠ Python 3.4, so that is not allowed), and when given n (using STDIN/function arguments/etc.), compute a(n) where a is an OEIS sequence of your choice. One of those programs is shorter than the other. You only need to submit the longer program of the two. The other one needs to be saved in case of not being cracked after 7 days. Your submission is cracked when your program has been outgolfed (whether it is by 1 byte or more).

For example, if the task you chose was to perform 2 × n, this could be a valid submission (in Python 2):

Python 2, 16 bytes, score = 15 / 16 = 0.9375

print(2*input())

Computes A005843, (offset = 0).

If your submission has been cracked, then you need to state that in your header like so:

Python 2, 16 bytes, score = 15 / 16 = 0.9375, [cracked] + link

print(2*input())

Computes A005843, (offset = 0).

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Offset

This can be found on every OEIS page. For example, for A005843, the offset is 0,2. We only need to use the first one, which is 0. This means that the function is defined for all numbers ≥ 0.

In other words, the function OEIS(n) starts with n = 0. Your program needs to work for all cases given by OEIS.

More information can be found here.

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Scoring

The score you get for your submission is equal to the following formula:

Score = Length (in bytes) of secret code ÷ Length (in bytes) of public code

The example above has the score 15 ÷ 16 = 0.9375.

The submission with the lowest score wins. Only submissions that have posted their solution will be eligible for winning.

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Rules

• The task you need to do is an OEIS sequence of your choice.
• Given n, output OEIS(n). Deviation is not allowed, so you need to produce the exact same sequence (when given n, you need to output OEIS(n)).
• Submissions that are not cracked within a period of 7 days are considered safe after the solution has been posted (submissions older than 7 days that do not have their solution posted are still vulnerable in being cracked).
• In your submission, you need to post the following things: language name, byte count, full code, so no pastebin links etc. (to prevent answers like Unary), OEIS sequence, score with lengths of both programs and additionally, the encoding that is used.
• Note: the same sequence cannot be posted twice in the same language. (For example, if the sequence A005843 has been done in Pyth, you cannot use Pyth again for that same sequence.)
• Input and output are both in decimal (base 10)

Leaderboard

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script><style>table th,table td{padding: 5px;}th{text-align: left;}.score{text-align: right;}table a{display: block;}.main{float: left;margin-right: 30px;}.main h3,.main div{margin: 5px;}.message{font-style: italic;}#api_error{color: red;font-weight: bold;margin: 5px;}</style> <script>QUESTION_ID=88979;var safe_list=[];var uncracked_list=[];var n=0;var bycreation=function(x,y){return (x[0][0]<y[0][0])-(x[0][0]>y[0][0]);};var byscore=function(x,y){return (x[0][1]>y[0][1])-(x[0][1]<y[0][1]);};function u(l,o){jQuery(l[1]).empty();l[0].sort(o);for(var i=0;i<l[0].length;i++) l[0][i][1].appendTo(l[1]);if(l[0].length==0) jQuery('<tr><td colspan="3" class="message">none yet.</td></tr>').appendTo(l[1]);}function m(s){if('error_message' in s) jQuery('#api_error').text('API Error: '+s.error_message);}function g(p){jQuery.getJSON('//api.stackexchange.com/2.2/questions/' + QUESTION_ID + '/answers?page=' + p + '&pagesize=100&order=desc&sort=creation&site=codegolf&filter=!.Fjs-H6J36w0DtV5A_ZMzR7bRqt1e', function(s){m(s);s.items.map(function(a){var he = jQuery('<div/>').html(a.body).children().first();he.find('strike').text('');var h = he.text();if (!/cracked/i.test(h) && (typeof a.comments == 'undefined' || a.comments.filter(function(b){var c = jQuery('<div/>').html(b.body);return /^cracked/i.test(c.text()) || c.find('a').filter(function(){return /cracked/i.test(jQuery(this).text())}).length > 0}).length == 0)){var m = /^\s*((?:[^,;(\s]|\s+[^-,;(\s])+).*(0.\d+)/.exec(h);var e = [[n++, m ? m[2]-0 : null], jQuery('<tr/>').append( jQuery('<td/>').append( jQuery('<a/>').text(m ? m[1] : h).attr('href', a.link)), jQuery('<td class="score"/>').text(m ? m[2] : '?'), jQuery('<td/>').append( jQuery('<a/>').text(a.owner.display_name).attr('href', a.owner.link)) )];if(/safe/i.test(h)) safe_list.push(e);else uncracked_list.push(e);}});if (s.items.length == 100) g(p + 1);else{var s=[[uncracked_list, '#uncracked'], [safe_list, '#safe']];for(var i=0;i<2;i++) u(s[i],byscore);jQuery('#uncracked_by_score').bind('click',function(){u(s[0],byscore);return false});jQuery('#uncracked_by_creation').bind('click',function(){u(s[0],bycreation);return false});}}).error(function(e){m(e.responseJSON);});}g(1);</script><link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/Sites/codegolf/all.css?v=7509797c03ea"><div id="api_error"></div><div class="main"><h3>Uncracked submissions</h3><table> <tr> <th>Language</th> <th class="score">Score</th> <th>User</th> </tr> <tbody id="uncracked"></tbody></table><div>Sort by: <a href="#" id="uncracked_by_score">score</a> <a href="#" id="uncracked_by_creation">creation</a></div></div><div class="main"><h3>Safe submissions</h3><table> <tr> <th>Language</th> <th class="score">Score</th> <th>User</th> </tr> <tbody id="safe"></tbody></table></div>

Note

This challenge is finished. The final winner is feersum with his Seed answer. Congratulations! :).

You can still submit new cops, but be aware that they are no longer competing.

Answer 1

Seed, 5861 bytes, score = 5012 / 5861 = 0.85

The sequence is primes (A000040) with offset 1. a(1) = 2, a(2) = 3, a(3) = 5 etc.

106 4339425277766562922902283581834741289660008085947971671079664775075736459902652798498038280771739790213868067702391567527146683746696872737118568202495046682058807677834082334206717794057290386357040004038910321326390033894692316122893125849512049817771469697446796247656883761642659391998672259889315862253584980121050081936190889196979721443972131545258528161479083569474217100401074866410321578452749003295370495810488337344650619973376676577461877392643228932028285261311284691649403036725905675576380944186859525020321196255472295415627414823269713084816196540461818684672201926996228242729726187404845487167114556965625764494860789841409004737497250600337038738035289643512265772877903971230007222865264200102217827010958702813633039465646713707971175729900391272165457566364779628858903697222589748797809421482136725017988969980267265196209027326008642464190920825439635011414535065156799655809935871795742526416544853103823906918352472744460644939241345215614650110978624804796257165525598653433482592675123776747497586586903140407616610040250976121531777891358439091358523224019193934111409521776865798864774150127996987606796522094617839125169013781373842026970010007574244564780540958252950607459585834584855526028427463655493110963000525209314274839412714497954647707284352161251044088451086878301225167181712809612927720502228546704347274977503482518386819117870800284276687560241308964641752876311905619184965236937789822712948719170589044519552259399272657757694404103028213338440810122219269214268424256451648966039627998513353115348057963135398345514276156595104642595820063441019481255889600472121104059631555738973905087895006671206400595057069658845297458058584470727379036742268107372233190371861824194831387484478317333784774872796689435056263039565495723444232483369405079512770383639748492508848098080619713255928884946598796741958520788406091704951276729428229224292748702301286318784744573918534142896761450194446126776354892827260482520089107240497527796383449573487121759294618654056309957794765646022274249211113876873102681817500947681708211056388348991201016699342850495527766741633390367735662514559206616070479934780700857859919517187362382258597709910134111383911258731633002354208155277838257255571878219168563173495861133946240923601273185050088148273459064040178741322137494758164795598458780786653602794809987537740537018415433767449108362051063814315399931951925462073202072886716208053889630274798247936580024665863765351912184189125660586187615847325588786048095120241198943086897428919324650596915625795076460123743259068671341944912206659194476673792489442514470540309819938731582497982088632076086088279435186513066668502875362808653657423813387124496122632219269226944975782747969308509448942429277233562654639293567532567668357917753810024961683829653277391094269518376510962710057956994339018307344554672056556222387849272880157861877494719706801624724491116189525394408237355854147113614645956561900837121715298276123085019204014577395795131906357190097536924932784935203378709529040555114884933996346694363879974847691625806412083107877442577777402405491226347699452398189866905599648314105255526411599513768016126067224570735746339691839657336828937030584950250402550603260483421505256395736457980708347396132620971927806636308105501893575073944959824958733880580825249931469481777083600987966500968473202481877213198175820182125298036242272662171321630056435823478710070315022531849275633515412140708923196338877549535352180465031450246889723670908173572778497329815806296369714467774385173078365517895215622645081749679859298824530173433952201710212962315524645807786760255396609101229899503687886977229729747349967302227815724222900649259120496955396616388023947812556426182596252076072286860171961582235043470190275528327438941205680729222092142315348205283459886659277456757338926863444370956987436702675569004062857510888080701482282900718067707825890168959050535970636214821273965900140346587802750221148933877600652180282267212515086387728695565345543441575183083490091817551421389124038251086513387106526847199935776240422217886407416027185332010280169564289345500368555274327733580514983967396271907637608170801013991375555069570288329399237332712790289521766624379537848996471168926519414464863388365890585061582441222989105844636887033599262856636618609644981203616618819656730174147506366895579518927217154437260067784133452192099436160162797896733220282837763342940047719962882720310397266700665603774047807673735452896542215047419894928360985667680051383584281780118001522220147385455276205847620842066894760474814386271419398361771509559702341442734727141312211989794380570433135781896005067541537095546614638001539678780066976441749790924521292297473522803115912791790379839635473194794843511234906415092857115568242448079933264380632375450234146479596225552359821776361923588178896354011117990551249184457345201223244319766597339520899930287542362386381372955844126876031262062731835081542890548095759704856479235361996156162229417953890962902505112862674541020677153054937034038823204321411753183982406667628845943390275194956321260584953509501973880059966268311741789559039618821364775407403947492157311255310143283125490988585303127442698159113924719563571459841025286208880511134222538431747221840824203312684036627017414295981003169360893015436564680773233890198618904647085929678054127680367983802905553144716598061593632352021737488422700265144189474970515439967472618438343180405852959047054139020095303915498443045344690691354304662161461750826840689185141612937350984288238847592910919431788170821390987459951181698659544772214696392241600642992000900364649438402093845534643663733216626212187314397293309505677932731383013397665193960914949915855970134736764497124186701371371881061763702617034928084811708964018610410971938419410791443362686750151572343348438861493025667676713

The Befunge-98 programs were tested with this interpreter.

Solution:

62 92671758747582594731336103958852355343308794409787718910287760272065096600068486400261521203099179296478278113800406388237579729434074471528101978922110199511009255327569364221068648720732186414156697930516237153253745234146558781777104311285708042469572129997820696177040412749585193035961972308024909384538547357820271391461203855177879703963391294547499579588457829374981409596253284387318191154655397249791533591896711203680125312645807793061567274893660125978667479654664977040722935418267606762108334976561590548772755653088127344268269983549959628254712562135604114391709222466418283973346968039685907258341712475120187026707300070769277380483828579629391533415119380882514570806683534933872011332303802477012040660361613689139008855327957705058672774790021218679288003003953301651226513713984857174365383390364296326192225244927665294515693697694918935732394438095829822147927645949273829493190176397786165741955566462476231578299385726525505407052332068778469428870102672560545990553686935179657522071350801304923521681690806124866463401094200444841941834667455137491597902735287855498886460945851544063102556545691787612423033525861765804657417395955322217721677429700032333887700477665924915189639029356029794151144702621112140447347270986003871777552705154393697526621456025974679633450745341583481291685834000335168972075093212539251102818038917942913311300883294570091156661153874804268309393591292001433191276766990017144340677002050765359295580546228905861008474333888247511333039470305173620221481374758754343560048199433044290376988914313248904786418615239832295700238599693805552407166251501198275363727855984193340187485162706203747898935844148656997727555488455764358003951396850496841760348138874901474547533715922587211143833052692993182786162665394965914056238514702648647904702501871758140636318131208564891924287008550289224318358936576352473100482724524675417108540029486047223784009872784235439805791496176981701859374772960623187174667015174831665360382067784289660747175586412802848517818731070091826086320292632019033525579172665790335268736167170506003176022610987557889205903933680970434653929602313812168432779881423599218075810156457004870273456214668951969634696002866863369645150677406566613367576078149751561615160777945725724620047443832859087000460506626402089973036918592151204779260519899343451226942874643654023265001514280212345984966126290887141500898797940093805650642580450926977375576911590855135774911449619005627413806680159169643085790457809525639117624947749945044091079624534522626841372604654172723500062361904864176709974716350878399949908529715899937417421315012456868864220900338162700464737416505300734198857624165994112815507157337074226022552948626042899845891195024145834980781844015548398775284084741665926642729256313545870065439195137107807599897817556866239630270351410298105991743248934632486671734759038305157913444368204353943206369388913837519310828223093441519335111533635957953613758894822654736600526811789875376813119426924959017038654104216784121093688306563643326587639486472221258233221666454164763738631579246841130247019172136121041002571694545781948282785399495873501148416357057693713305042834246973535325571882393889489457235864027134943913383832461393499203435931881991959787045205816313165984531168137210464591653390767999403651750434079431253272021002273680565508340556736474927770564408343360602057949991306130979603289667991356237787412616305022971197507815004141825394838946364485315703051195195566893733123391158803413838803831010122718476800229685100410524315094662633390222854924884579083098055980542607258318868514903610787510921796157291630993618714015755412239872758443166948699744841899868754369627081727143351807615097715679652005998467491609044596871312950634152039807480021814406950780706131231897491212637759991818212542181136384052857191779658528790835620632765143337026858373660057972387266312097135260115742458792764792668883627539340807572869610941154184473111399152964165437112713815173281951728792354570851956468302291939952274005357250989986640723863408051924618400882866539701190471828299028566020683682444415198672952980294639217217840535225987439355834087974716313911977302809235338769491339553247328065401203243450045946392960085318343121705830317674151229536850726617093615850507955559652374337057819549481320081981520577039493601331233500403284295119207704095876958023271178964331413629547646937335760969491450824461526563643617594783473684358594189269252499897162333533284912320654686655888508024970105099967896167541978181602786701854274646885561632089896312016789257459673121974866871919820865433343707787147414982407950775979279255414469970743690769124215210050618943726165676550098723299244096267839544684847323547847832349290874282817429866612456451105673214159820212156069771415582214200701894487126822756864305461967035982308878073752362075553218935807632264803200753661147341613284071218919438723527468202903770806766095252957940538229987302177328543423522712562396242285027178395886649344

Answer 2

Jelly, 5 bytes, score 0.8 (4 / 5) [cracked!]

R²Sƽ

Computes A127721.

Try it online!

---

Here's the solution:

RÆḊḞ

• implicitly take input n
• R: list from [1, 2, ..., n]
• ÆḊ: In our case, this will actually return the square root of the sum of squares (which may be useful in future golfing)!
• Ḟ: floor the result.

The documentation for ÆḊ reads:

ÆḊ: Determinant, extended to non-square matrices.

The key is extended to non-square matrices. The "determinant" of a non-square matrix is usually undefined, but one reasonable definition is sqrt(det(A A^T)) (which for a square matrix reduces to |det(A)|). In our case, A A^T is a 1 x 1 matrix containing the sum of squares. The square root of the determinant of that gives us exactly what we need to shave off the last byte!

Answer 3

Retina, 28 bytes, score = 0.9286... (26/28), cracked by feersum

.+
$*
^$|^((^|\3)(^.|\1))*.$

Computes A192687, (offset = 0).

Try it online! (The first line enables a linefeed-separated test suite.)

This is the difference between Hofstadter's male and female sequences. (Relevant PPCG challenge.)

This was my original code:

.+
$*
^((^.|\3)(\1)|){2,}$

This answer was a bit of a gamble anyway, because the actual solution is based on a regex I announced to be the shortest known Fibonacci-testing regex in chat a few months ago. Luckily, no one seemed to remember that. :)

Answer 4

Hexagony, 91 bytes, Score = 0.725274725 (66/91) [Cracked]

Computes A000045 (Fibonacci sequence, offset 0).

I won't be too surprised if someone manages to beat this and my more golfed version, although it should be hard.

Edit: Holy cow, @MartinEnder whipped me with a 33 byte solution.

Golfed (91):

?\]~<~.{>'"/(@{\''1<{!1>{{1}/}{'\1</={}/_\'0"/>+(}\/}(+'+'%=<>=%"=+("\/+"(+}+<>{{}=~\.....|

Formatted:

      ? \ ] ~ < ~
     . { > ' " / (
    @ { \ ' ' 1 < {
   ! 1 > { { 1 } / }
  { ' \ 1 < / = { } /
 _ \ ' 0 " / > + ( } \
  / } ( + ' + ' % = <
   > = % " = + ( " \
    / + " ( + } + <
     > { { } = ~ \
      . . . . . |

Try it online!

I'm not going to post an explanation for this, it's too horrible...

Golfed (66):

?{1}]0@._.>\>+{./'++.!.|.*'}..\}{\=++.../'"<_}\"+<./{(/\=*"=/>{=+"

Formatted:

      ? { 1 } ] 0
     @ . _ . > \ >
    + { . / ' + + .
   ! . | . * ' } . .
  \ } { \ = + + . . .
 / ' " < _ } \ " + < .
  / { ( / \ = * " = /
   > { = + " . . . .
    . . . . . . . .
     . . . . . . .
      . . . . . .

Colored:

Try it online!

Explanation:

The memory layout I used looks a little like this:

   |
   a
   |
  / \
b+a  b
/     \
      |
    input

The initialization (in black) sets a=0 and b=1. Then, the main loop:

• goes from a to the input cell - '"
• decrements the input - (
• adds b and a - {{=+
• moves it "out of the way" - "+{=*
• sets a to b - '+
• moves a "out of the way" - '+}=*
• moves b+a back to its original position - "=+
• sets b to b+a - "+
• moves a back to its original position - }+

Once the input cell reaches 0, the MP moves to a, prints, and exits.

What I could have done to save more bytes is to use &, which just sets the current cell to either its left or right neighbor. I also could have had a bit better flow control, but it's okay as it stands.

Answer 5

M, 10 bytes, score 0.6 (6 / 10) [cracked]

R‘ạḤc’*@RP

A tad cheaty since M and Jelly are quite similar, but allowed by the rules. This version is based on @LeakyNun's crack to my Jelly answer.

This calculates sequence A068943. Try it online!

Intended solution

The following code works in M/Jelly.

R¹¡PÐL

I actually had a 4-byte solution to which I added some fluff to make it seem more difficult to crack by using brute force.

R¡FP

Try it online!

This is the one I'm going to explain.

R¡FP  Main link. Argument: n

 ¡    Execute the left to the left n times, updating the return value.
R       Range; map each integer k to [1, ..., k].
      This does the following for the first values of n.
        1 → [1]
        2 → [1,2]   → [[1],[1,2]]
        3 → [1,2,3] → [[1],[1,2],[1,2,3]] → [[[1]],[[1],[1,2]],[[1],[1,2],[1,2,3]]]
  F   Flatten the resulting, nested array.
   P  Take the product of the reulting array of integers.

Answer 6

Stack Cats, 14 bytes, score = 13 / 14 = 0.929 [cracked]

That's 10 bytes of code, plus 4 for the arguments -nm.

Computes A017053. In case OEIS is down, that's a(n) = 7n + 6, starting at n = 0.

![_-_:-_-_

The full code (usable without -m argument) is ![_-_:-_-_-_-:_-_]!

The hidden solution was

!]|{_+:}_

Answer 7

Snowman, 50 bytes, score = 0.9 (45 / 50) [cracked by Lynn]

((}#NDe`nOnO|`2nMNdE0nR2aGaZ::nM;aF;aM:nS;aF,nM*))

This is a subroutine that takes a number as its argument and returns another number.

Computes A122649 (offset = 1).

Try it online!

Answer 8

Haskell, 15 bytes, score = 13/15 (0.866) (cracked)

f x=1+2*div x 2

A109613 with offset 0. Repeats each odd number twice.

1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, 13, 13, ...

Answer 9

Brachylog, 27 bytes, score = 0.666... (18/27), Cracked!

+ybL:L:[1]co~c[A:B]hl-?,A*.

Computes A010551 (offset = 0).

You can try it online here.

Explanation

Since the vast majority of people don't know this language, and since I'm posting this answer mostly to get people looking into it (See: Brachylog's Wiki), I'll provide a short explanation of the code above:

+           Add 1 to the input N
ybL         L = [1, 2, ..., N+1]
:L:[1]c     Construct a list [1, 2, ..., N+1, 1, 2, ..., N+1, 1]
o           Sort the list from smallest to biggest
~c[A:B]     A concatenated to B results in that sorted list
hl-?,       The length of A is N + 1
A*.         The output is the result of the product of all elements of A

Answer 10

Haskell, 4 bytes / 5 bytes (cracked by Leaky Nun)

Let's start simple. A000012, the sequence of all 1's (offset 0). Here's a table:

0 -> 1
1 -> 1
2 -> 1
3 -> 1
4 -> 1
 ...

5 bytes:

f n=1

Answer 11

Java 7, 53 bytes, score = 0.9623 (51/53) Cracked

int f(int n){return n<1?3:n<2?0:n<3?2:f(n-2)+f(n-3);}

Computes the Perrin sequence, A001608

f(0)=3
f(1)=0
f(2)=2
f(n)=f(n-2)+f(n-3)

First code golf attempt, feel free to point out any mistakes.

Explanation after cracking:

miles shortened the first three values (0,1,2) -> (3,0,2) to

n<2?3-3*n:n<3?2

while my own solution was a slightly over-engineered

n<3?(n+4)%5*2%5

Combining the two tricks, you get

n<3?3-3*n%5

for a 47-byte

int k(int n){return n<3?3-3*n%5:f(n-2)+f(n-3);}

which looks pretty compact for Java :)

Answer 12

Cheddar, 7 bytes, score = 0.8571 (6/7), [cracked]

n->2**n

Pretty simple, just the powers of two. OEIS A000079

Try it online!

Answer 13

J, 17 bytes, score = 0.8235 (14/17) (cracked)

(+/@(!+:)&i.-)@>:

Computes A002478.

The funny thing is that this version was originally the short one to be kept secret.

Answer 14

Pyke, 11 bytes, score = 0.45 (5/11) [cracked]

hZRVoeX*oe+

Calculates OEIS A180255

Try it here!

Answer 15

Python 2, 43 bytes, score = 0.9302 (40/43), cracked

f=lambda n:n==1or-(-sum(map(f,range(n)))/3)

Computes A072493

Let's see if anybody can golf off all 3 bytes.

Answer 16

Jelly, 9 bytes, score 0.5556 (5 / 9) [cracked]

r©0+’Ac®Ḅ

This calculates sequence A119259. Try it online!

Answer 17

Brachylog, 11 bytes, score = 10 / 11 = 0.909, cracked!

ybLrb:Lrcc.

Computes A173426.

Try it online!

Explanation

ybL      L = [1, 2, …, Input]
rb       Get [Input - 1, …, 2, 1]
:Lrc     Concatenate [1, 2, …, Input] and [Input - 1, …, 2, 1]
c.       Concatenate [1, 2, …, Input, Input - 1, …, 2, 1] into a single integer

Answer 18

J, 20 bytes, score = 0.75 (15/20) (cracked)

[:*/0 1(]{:,+/)^:[~]

Computes A001654.

My solution for 15 bytes was

*&(+/@:!&i.-)>:

Answer 19

Sesos, 14 bytes, score = 0.8571 (12/14) (cracked)

0000000: 16f8be 760e1e 7c5f3b 07ddc7 ce3f                  ...v..|_;....?

Computes A000290.

Try it online!

Hints

This binary file has been generated by the following assembler:

set numin
set numout
get
jmp,sub 1,fwd 1,add 1,fwd 1,add 1,rwd 2,jnz
fwd 1
jmp
  sub 1,fwd 1
  jmp,sub 1,fwd 1,add 1,fwd 1,add 1,rwd 2,jnz
  fwd 1
  jmp,sub 1,rwd 1,add 1,fwd 1,jnz
  rwd 2
jnz
fwd 3
put

Answer 20

MATL, 11 bytes, score = 0.8181 (9/11), cracked

YftdA-1bn^*

Computes the Möbius function, or A087811 (offset 1).

Try it online!

Resources for robbers

I think this one should be easy, but anyway here is some help.

Code explanation:

Yf     % Implicit input. Push array of prime factors, with repetitions
t      % Duplicate
d      % Compute consecutive differences
A      % 1 if all those differences are nonzero, 0 otherwise
-1     % Push -1
b      % Bubble up array of prime factors to the top of the stack
n      % Number of elements
^      % -1 raised to that
*      % Multiply. Implicitly display

Language documentation.

MATL chatroom.

Answer 21

Jelly, 11 bytes, score 0.5455 (6 / 11) [cracked]

Ḷ+’cḶ
r1*ÇP

This calculates sequence A068943. Try it online!

Answer 22

MarioLANG, 87 bytes, score = 0.839 (73/87), cracked

;
)-)+(< >>
-)===" ""====
>>+([!)( >-(+(
"====#[(("== [
!-) - <!!![)<<)
#======###====:

Computes A000217, the triangular numbers. Offset 0.

Try it online!

Short Solution:

Since the crack took only 1 byte off, I'll share my solution in 73 bytes, which uses a different algorithm altogether:

;   +)-<
-   (=="
+)-<(
(=="+
> [!>)[!(
"==#===#[
!    -  <))
#=========:

Try it online!

The first program lays out all the numbers from n to 1 along the tape, then adds up all numbers until it comes across a 0-value cell. It does this by copying each cell into the two adjacent cells, decrementing the right copy, and repeating the process on it until it reaches 0. During this process, the tape looks like this (for n=5):

0 0 5 0 0 0 0
0 5 0 5 0 0 0
0 5 0 4 0 0 0
0 5 4 0 4 0 0
0 5 4 0 3 0 0
0 5 4 3 0 3 0
0 5 4 3 0 2 0
0 5 4 3 2 0 2

... and so on. Then it moves left, summing the cells until it reaches a cell with 0.

The second program only works with three cells of the tape. Until the first cell reaches 0, it does the following:

• Move the value of the first cell to the second cell.
• Decrement the second cell to 0, adding its own value to the first and third cells.
• Decrement the first cell.

After the first cell reaches 0, the third cell will contain n + (n-1) + (n-2) + ... + 2 + 1.

Answer 23

M, 18 bytes, score 0.3889 (7 / 18) [cracked]

r0+c
‘Ḥc0r$×Ç:‘+\S

This approach is based on @Sp3000's golf of my Jelly answer.

The code calculates sequence A116881 (row sums of CM(1,2)). Try it online!

Answer 24

Haskell, 28 bytes, score = 0.3571 (10/28), cracked

f n|odd n=1|1>0=2*f(div n 2)

A006519, the highest power of 2 dividing n, starting at n=1.

1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, 1, ...

While you get cracking credit for any shorter code, I consider the real puzzle to be getting down to 10 bytes.

Answer 25

05AB1E, 9 bytes, score = 0.6667 (6/9), Cracked

ÐnŠ·sÉO4/

Computes: A087811

Try it online

Answer 26

Hexagony, 7 bytes, score = 0.857 (6/7), cracked

Not designed to be the winning submission, but definitely a brain cracker. It actually isn't that hard if you think about it :p. Code:

\!?__@(

Or a more readable version:

 \ !
? _ _
 @ (

Computes A052246.

Try it online!.

Answer 27

J, 10 bytes, score = 0.9 (9/10) (cracked by Dennis)

*2%~3<:@*]

Computes A000326.

This one should be easy.

Online interpreter.

Quickref.

Answer 28

J, 9 bytes, score = 0.6667 (6/9) (cracked by Dennis)

9 o.0j1^]

Computes A056594.

Answer 29

05AB1E, 7 bytes, score = 0.571 (4/7), Cracked

0s·Ì3c;

Computes: A006331

Try it online

After golfing 5 unique versions of this sequence before finding this public version I hope I haven't left any obvious improvement out there so it'll at least be a challenge for the robbers.

Answer 30

M, 9 bytes, score 0.6667 (6 / 9) [cracked]

r©0+’Ac®Ḅ

A tad cheaty since M and Jelly are quite similar, but allowed by the rules. @miles' crack to my Jelly answer doesn't work in M; it lacks the œċ atom.

This calculates sequence A119259. Try it online!

Answer 31

QBasic (QB64), 30 bytes, score = 28 / 30 = 0.9333 (cracked)

INPUT n:p=n MOD 2:?p*n+(n+p)/2

Computes A014682, a version of the Collatz function (offset 0).

This works on the QB64 emulator with autoformatting turned off. It should work on actual QBasic as well, but I don't have a copy to test it.

---

My 28-byte version used a bitwise operator and integer division:

INPUT n:?(1AND n)*n+(n+.5)\2

Answer 32

Acc!!, 526 bytes, score = 0.5627 (296/526) (cracked)

N
Count x while _%60-46 {
(_+_%60*5-288)*10+N
}
_/60
Count i while _/27^i {
_+27^i*(_/27^i*26-18)
}
_*3+93
Count i while _/27^i/27%3 {
_-i%2*2+1
Count j while _/3^(3*j+2-i%2)%3 {
_+3^(1+i%2)
Count k while _/3^(3*k+1+i%2)%3-1 {
_+3^(3*k+1+i%2)*26
}
}
}
Count i while _/27^i/3 {
_-_/27^i/3%27*27^i*3+_/3^(3*i+1+_%3)%3*3
}
_/3
Count i while _/100^i {
_+_/100^i*100^i*9
}
Count i while _/100^i/10 {
_+_/100^i/10%10
Count j while i+1-j {
_+(_%10-_/100^(j+1)%10)*(100^(j+1)-1)
}
}
_/100
Count j while _/100^j {
Write _/100^j%10+48
}

Computes A000204, Lucas numbers with offset 1.

A couple of notes. Due to Acc!!'s input limitations, the program takes input as a decimal number terminated with a period (.); for example:

> python acc.py Program.txt
5.
11

This program has exponential time complexity, so I wouldn't recommend running it with input greater than about 14 (takes 60 seconds on my laptop). However, given enough time, the algorithm will return the correct result for any positive integer n. Thus: a crack attempt that places any size limits on the input value is invalid.

Have fun!

Answer 33

Python, 46 bytes, score = 0.97826 (45 / 46) [cracked]

f=lambda n,k=1:n and-~f(n-("01"in bin(k)),k+1)

Calculates A101082 (offset = 1), which are:

Numbers n such that binary representation contains bit strings "10" and "01" (possibly overlapping).

---

@Dennis cracked the submission as intended, which was to replace "01"in bin(k) with some bit twiddling, namely (k&-k)+k&k>0, to check that k's binary representation isn't of the form 111...111000...000.

Answer 34

Acc!!, 512 bytes, score = 0.5625 (288 ÷ 512) (cracked)

Fourth try. Let's see how fast Dennis cracks this one. (Edit: 8 minutes, that's a record!)

N
Count x while _%60-46 {
(_+_%60*5-288)*10+N
}
_/30
Count i while _/27^i {
_+(_/27^i*26-18)*27^i
}
_*3+93
Count i while _/27^i/27%3 {
_-i%2*2+1
Count j while _/3^(3*j+2-i%2)%3 {
_+3+i%2*6
Count k while _/3^(3*k+1+i%2)%3-1 {
_+27^k*3^(i%2)*78
}
}
}
Count i while _/27^i/3 {
_-_/27^i/3%27*27^i*3+_/3^(3*i+1+_%3)%3*3
}
_/3
Count i while _/100^i {
_*10-_%100^i*9
}
Count i while _/100^i/10 {
_+_/100^i/10%10
Count j while i+2-j {
_+(_%10-_/100^j%10)*(100^j-1)
}
}
_/100
Count j while _/100^j {
Write _/100^j%10+48
}

Computes A002878, Lucas(2*n+1), offset = 0.

Due to Acc!!'s input limitations, the program takes input as a decimal number terminated with a period (.); for example:

> python acc.py Program.txt
3.
29

This program has exponential time complexity, so I wouldn't recommend running it with input greater than about 7. However, given enough time, the algorithm will return the correct result for any positive integer n. Thus: a crack attempt that places any size limits on the input value is invalid.

Answer 35

Woefully, (noncompeting, but you still get points for cracking it), 520/776= ~0.67 Cracked by Feersum.

Computes A000042 (would compute repunits, but that one has 0 as first member)

Offset 1, zero indexed (0 outputs 1, 1 outputs 11, 2 outputs 111, etc.)

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Try it online!

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I like to think that my shorter solution was a pretty nice one.

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What I did was remove the push one command from the top, replace it at the bottom and at the end of the first command, which was free because I just had to replace some pipes, which then allowed me to shorten the push nine commands some more, because it didn't need to move that much anymore

Answer 36

J, 12 bytes, score = 0.9167 (11/12) (cracked)

(#++/)@I.@#:

Computes A230877.

Answer 37

J, 18 bytes, score = 0.7222 (13/18) (cracked)

0{]1&(}.],+/)2 1"_

Computes the Lucas numbers, A000032.

Answer 38

Sesos, 14 bytes, score = 0.6429 (9/14) (cracked)

0000000: aed40b b47bc2 8e01be 8e9ddb b107                  ....{.........

Computes A130909.

Try it online!

Answer 39

Python 3.5, 38 bytes, Score = 0.86842 (33/38) (Cracked):

G=lambda n:n<1and 1or(2*n-1)**2*G(n-1)

Computes A001818.

Try It Online! (Ideone)

Answer 40

05AB1E, 10 bytes, score = 5 ÷ 10 = 0.5, cracked

This one isn't too hard actually :p. Code:

[NNÂʽ¾¹Q#

Computes the N^th non-palindromic number, which is A029742.

Try it online!

Answer 41

JavaScript (ES6), 77 bytes, score = 0.9481 (73 / 77) [Cracked]

n=>eval("for(i=0;n;a||n--)[...s=a=++i+''].map(d=>a-=Math.pow(d,s.length));i")

Computes A005188, (offset = 1).

---

Didn't end up being super interesting but I thought I'd try and submit something.

Note: This is in ES6 so there is no ** operator. I just thought this would make it more cross-browser compatible and easier to test.

Answer 42

05AB1E, 13 bytes, score = 0.769 (10 ÷ 13), cracked

This one isn't too hard either. Code:

µN•vÉ•vyK}g_½

Calculates A001742 (offset = 1).

Try it online!.

Answer 43

Python (3.4.3), 77 bytes, score = 0.7272 (56/77) cracked

import math;lambda x: 1 if x==0 else (0 if ((math.sqrt(1+8*x)-1)/2)%1 else 1)

Computes A010054, offset = 0

Answer 44

Coffeescript, 83 bytes, score = 79/83 = 0.952 [cracked!]

r=(n)->return if r<0 then 0 else Array(parseInt(n/9)+2).join ((n-1)%9+1).toString()

Computes A010785.

(Not so) Surprisingly, @DomHastings managed to shave 33 bytes off of my solution.

Wow, I suck at code-golf, don't I? Well, here's my other code:

r=(n)->return if r<0 then 0 else ((n-1)%9+1).toString().repeat(parseInt(n/9)+2)

Here's a less messy version of the code:

r=(n)->
  if r<0
   return 0 
  else 
   ((n-1)%9+1).toString().repeat(parseInt(n/9)+2)
   # ((n-1)%9+1).toString() is the repeating digit (as a string).
   # parseInt(n/9)+2 is the times the digit is repeated.

Answer 45

Pip, 24 bytes, score = 22 / 24 = 0.9166 (cracked)

Y53T++i>ay.:k.A(yi)A(ya)

Computes A109648 (offset = 0). Try it online!

---

My 22-byte version:

T++v>alPB A(lJk|5v)l@a

Answer 46

Actually, 28 bytes, Score = 0.39285714 (11/28) (cracked)

2√;;4*;u@D))u(*,;)@ⁿ)1-*ⁿ@-½

Computes A048696 (offset = 0). Actually uses CP437 for its encoding.

Try it online!

Dennis's solution is actually almost identical to mine:

19,`;τ(+`nX (Dennis)
19(`;)τ+`nX (Mine)

A Levenshtein distance of 3 is pretty impressive!

Try it online!

Answer 47

Jelly, 22 bytes, score 0.2727 (6 / 22) [cracked]

‘ðḤ;+;c2\×P:+’×$¥µÐ€‘S

This calculates sequence A116881. Try it online!

Answer 48

05AB1E, 7 bytes, score = 0.857 (6 / 7) [cracked]

This one shouldn't be too hard if you think about it :p. Code:

L!zO¹!*

Try it online!

Computes A002627.

Answer 49

Acc!!, 523 bytes, score = 0.55067 (288/523) (cracked)

Second try, slightly different sequence.

N
Count x while _%60-46 {
(_+_%60*5-288)*10+N
}
_/60
Count i while _/27^i {
_+27^i*(_/27^i*26-18)
}
_*3+93
Count i while _/27^i/27%3 {
_-i%2*2+1
Count j while _/3^(3*j+2-i%2)%3 {
_+3^(1+i%2)
Count k while _/3^(3*k+1+i%2)%3-1 {
_+3^(3*k+1+i%2)*26
}
}
}
Count i while _/27^i/3 {
_-_/27^i/3%27*27^i*3+_/3^(3*i+1+_%3)%3*3
}
_/3
Count i while _/100^i {
_*10-_%100^i*9
}
Count i while _/100^i/10 {
_+_/100^i/10%10
Count j while i+1-j {
_+(_%10-_/100^(j+1)%10)*(100^(j+1)-1)
}
}
_/100
Count j while _/100^j {
Write _/100^j%10+48
}

Computes A000032, Lucas numbers with offset 0.

Due to Acc!!'s input limitations, the program takes input as a decimal number terminated with a period (.); for example:

> python acc.py Program.txt
5.
11

This program has exponential time complexity, so I wouldn't recommend running it with input greater than about 14 (takes 60 seconds on my laptop). However, given enough time, the algorithm will return the correct result for any positive integer n. Thus: a crack attempt that places any size limits on the input value is invalid.

Answer 50

Acc!!, 522 bytes, score = 0.5555 (290 ÷ 522) (cracked)

Third try. Thanks to Dennis for the golfing "help" with the first two. ;^)

N
Count x while _%60-46 {
(_+_%60*5-288)*10+N
}
_/60*2
Count i while _/27^i {
_+(_/27^i*26-18)*27^i
}
_*3+93
Count i while _/27^i/27%3 {
_-i%2*2+1
Count j while _/3^(3*j+2-i%2)%3 {
_+3+i%2*6
Count k while _/3^(3*k+1+i%2)%3-1 {
_+27^k*3^(i%2)*78
}
}
}
Count i while _/27^i/3 {
_-_/27^i/3%27*27^i*3+_/3^(3*i+1+_%3)%3*3
}
_/3
Count i while _/100^i {
_*10-_%100^i*9
}
Count i while _/100^i/10 {
_+_/100^i/10%10
Count j while i+1-j {
_+(_%10-_/100^(j+1)%10)*(100^(j+1)-1)
}
}
_/100
Count j while _/100^j {
Write _/100^j%10+48
}

Computes A005248, Lucas(2*n), offset = 0.

Due to Acc!!'s input limitations, the program takes input as a decimal number terminated with a period (.); for example:

> python acc.py Program.txt
3.
18

This program has exponential time complexity, so I wouldn't recommend running it with input greater than about 7. However, given enough time, the algorithm will return the correct result for any positive integer n. Thus: a crack attempt that places any size limits on the input value is invalid.

Answer 51

Excel, 16 bytes, score = 0.75, [cracked by Dennis♦]

=FACTDOUBLE(2*n)

Computes A000165: double factorial of even numbers, (offset = 0).

Answer 52

05AB1E - 1 byte - A000027

Code: I

05AB1E - 0 bytes - A000027

Code:

Explanation:

I means that it pushes the input to the stack.

means that nothing is changed to the stack, but the input was already added to the stack by default, so it prints out the input.

Answer 53

Python, 39 bytes, 26/39=2/3=0.666 cont. points, cracked

s=lambda x:x+s(x-1)if x==1else x+s(x-1)

won't last long, but lets see just how long. Just doing the robbers a favour I guess (also because I don't know eso golf langs ;_;)

umm, I accidentally bad code but someone cracked my lower score so...

computes A000217

Answer 54

Python, 69 bytes, score = 0.97 (67÷69), cracked

f=lambda n:reduce(lambda x,y:x*y,range(1,n+1),1)
lambda n:f(n*2)/f(n)

Computes A001813