Reversed Engineered from Uniqueness (Cop's Thread)

Question

Given the output of the cop's program (o), the byte-count (n) and the number of unique bytes (c) used, come up with a corresponding piece of code that is n bytes long with c unique bytes which matches the cop's output o.

This is the cops thread. Post solutions that are to-be-cracked here.

The robbers thread is located here.

Cops should post solutions like this:

#[Language], `n` Bytes, `c` Unique Bytes (c*n points) [Cracked](Link-To-Crack)

    [Output]

---
    
(Optional spoiler)

Rules

You may not take any input for your program.
The program must use at least 1 byte, but cannot exceed 255 bytes.
The output itself is also limited to 255 bytes.
The program must have consistent output results when executed multiple times.
If your submission is not cracked within 7 days, you may mark it as "safe".
- When marking it safe, post the intended solution and score it as c*n.
- Only mark an answer as "safe" if you really want the +15 for an accepted answer; it's more fun to see how long you can go.

Winning

The uncracked post with the lowest c*n score, wins the cop's thread.
This will be decided after 10 safe answers, or a few weeks.

Caveats

If you feel cocky, you may tell the user the algorithm using a spoiler tag.

Uncracked Submissions:

fetch("https://api.stackexchange.com/2.2/questions/147635/answers?order=desc&sort=activity&site=codegolf&filter=!.Fjs-H6J36vlFcdkRGfButLhYEngU&key=kAc8QIHB*IqJDUFcjEF1KA((&pagesize=100").then(x=>x.json()).then(data=>{var res = data.items.filter(i=>!/^#.*cracked/im.test(i.body_markdown)).map(x=>{const matched = /^ ?#{1,3} ?(?:(?:(?:\[|<a href ?= ?".*?">)([^\]]+)(?:\]|<\/a>)(?:[\(\[][a-z0-9/:\.]+[\]\)])?)|([^, ]+)).*[^\d](\d+) ?\[?(?:.*(?:byte|block|codel)s?)(?:\](?:\(.+\))?)? ?(?:\(?(?!no[nt][ -]competing)\)?)?/gim.exec(x.body_markdown);if(!matched){return;}return {link: x.link, lang: matched[1] || matched[2], owner: x.owner}}).filter(Boolean).reverse().forEach(ans=>{var tr = document.createElement("tr");var add = (lang, link)=>{var td = document.createElement("td");var a = document.createElement("a");a.innerHTML = lang;a.href = link;td.appendChild(a);tr.appendChild(td);};add(ans.lang, ans.link);add(ans.owner.display_name, ans.owner.link);document.querySelector("tbody").appendChild(tr);});});

<html><body><h1>Uncracked Submissions</h1><table><thead><tr><th>Language</th><th>Author</th></tr></thead><tbody></tbody></table></body></html>

Are functions allowed? (Provided that they return a string, I suppose.) — Arnauld, Commented Nov 10, 2017 at 3:32
@cairdcoinheringaahing. OP has said: "The program must have consistent output results when executed multiple times." — user72349, Commented Nov 10, 2017 at 7:39
Just a suggestion for future CnRs: challenges where the user can choose an arbitrary fixed output are quite problematic for CnRs because people can basically mash their keyboard (barring syntax restrictions) and post the result, because there's no need for the cop to understand what their program is actually doing. Otherwise, this is a neat challenge idea. :) — Martin Ender, Commented Nov 10, 2017 at 8:07
@EriktheOutgolfer As I said "barring syntax restrictions". The point wasn't that you actually post a random stream of characters, but a random valid program which you don't need to understand. And especially in many esolangs, it's very easy to write a program that outputs a decent amount of random garbage without having to understand what that program does. — Martin Ender, Commented Nov 10, 2017 at 9:23
Considering that this challenge has already more than 30 answers, I suggest adding a code snippet which lists uncracked submissions like in this post. — user72349, Commented Nov 13, 2017 at 14:07

Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Japt, 5 bytes, 5 unique bytes (25 points) (Cracked by Scrooble)

Output: 3628801

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 10, 2017 at 18:44

Shaggy

43.5k4 gold badges36 silver badges94 bronze badges

\$\begingroup\$ Cracked \$\endgroup\$
– Khuldraeseth na'Barya
Commented Nov 10, 2017 at 22:06
\$\begingroup\$ @Scrooble, not the intended solution but nicely done. Will update as cracked when I get back to a computer. I'm thinking now the 4 byte version I had might have been a bit trickier, may yet post it. \$\endgroup\$
– Shaggy
Commented Nov 10, 2017 at 22:51
\$\begingroup\$ What was the intended solution? \$\endgroup\$
– Khuldraeseth na'Barya
Commented Feb 28, 2018 at 22:07

Add a comment |

emanresu A · Accepted Answer · 2021-08-27 20:41:41Z

Vyxal, 4 bytes, 4 unique, score 16, Cracked by Aaron Miller

Output:

aaaabbabaabcbabccabcaabcdbabcdcabcddabcdaabcdebabcdecabcdedabcdeeabcdeaabcdefbabcdefcabcdefdabcdefeabcdeffabcdefaabcdefgbabcdefgcabcdefgdabcdefgeabcdefgfabcdefggabcdefgaabcdefghbabcdefghcabcdefghdabcdefgheabcdefghfabcdefghgabcdefghhabcdefghaabcdefghibabcdefghicabcdefghidabcdefghieabcdefghifabcdefghigabcdefghihabcdefghiiabcdefghiaabcdefghijbabcdefghijcabcdefghijdabcdefghijeabcdefghijfabcdefghijgabcdefghijhabcdefghijiabcdefghijjabcdefghijaabcdefghijkbabcdefghijkcabcdefghijkdabcdefghijkeabcdefghijkfabcdefghijkgabcdefghijkhabcdefghijkiabcdefghijkjabcdefghijkkabcdefghijkaabcdefghijklbabcdefghijklcabcdefghijkldabcdefghijkleabcdefghijklfabcdefghijklgabcdefghijklhabcdefghijkliabcdefghijkljabcdefghijklkabcdefghijkllabcdefghijklaabcdefghijklmbabcdefghijklmcabcdefghijklmdabcdefghijklmeabcdefghijklmfabcdefghijklmgabcdefghijklmhabcdefghijklmiabcdefghijklmjabcdefghijklmkabcdefghijklmlabcdefghijklmmabcdefghijklmaabcdefghijklmnbabcdefghijklmncabcdefghijklmndabcdefghijklmneabcdefghijklmnfabcdefghijklmngabcdefghijklmnhabcdefghijklmniabcdefghijklmnjabcdefghijklmnkabcdefghijklmnlabcdefghijklmnmabcdefghijklmnnabcdefghijklmnaabcdefghijklmnobabcdefghijklmnocabcdefghijklmnodabcdefghijklmnoeabcdefghijklmnofabcdefghijklmnogabcdefghijklmnohabcdefghijklmnoiabcdefghijklmnojabcdefghijklmnokabcdefghijklmnolabcdefghijklmnomabcdefghijklmnonabcdefghijklmnooabcdefghijklmnoaabcdefghijklmnopbabcdefghijklmnopcabcdefghijklmnopdabcdefghijklmnopeabcdefghijklmnopfabcdefghijklmnopgabcdefghijklmnophabcdefghijklmnopiabcdefghijklmnopjabcdefghijklmnopkabcdefghijklmnoplabcdefghijklmnopmabcdefghijklmnopnabcdefghijklmnopoabcdefghijklmnoppabcdefghijklmnopaabcdefghijklmnopqbabcdefghijklmnopqcabcdefghijklmnopqdabcdefghijklmnopqeabcdefghijklmnopqfabcdefghijklmnopqgabcdefghijklmnopqhabcdefghijklmnopqiabcdefghijklmnopqjabcdefghijklmnopqkabcdefghijklmnopqlabcdefghijklmnopqmabcdefghijklmnopqnabcdefghijklmnopqoabcdefghijklmnopqpabcdefghijklmnopqqabcdefghijklmnopqaabcdefghijklmnopqrbabcdefghijklmnopqrcabcdefghijklmnopqrdabcdefghijklmnopqreabcdefghijklmnopqrfabcdefghijklmnopqrgabcdefghijklmnopqrhabcdefghijklmnopqriabcdefghijklmnopqrjabcdefghijklmnopqrkabcdefghijklmnopqrlabcdefghijklmnopqrmabcdefghijklmnopqrnabcdefghijklmnopqroabcdefghijklmnopqrpabcdefghijklmnopqrqabcdefghijklmnopqrrabcdefghijklmnopqraabcdefghijklmnopqrsbabcdefghijklmnopqrscabcdefghijklmnopqrsdabcdefghijklmnopqrseabcdefghijklmnopqrsfabcdefghijklmnopqrsgabcdefghijklmnopqrshabcdefghijklmnopqrsiabcdefghijklmnopqrsjabcdefghijklmnopqrskabcdefghijklmnopqrslabcdefghijklmnopqrsmabcdefghijklmnopqrsnabcdefghijklmnopqrsoabcdefghijklmnopqrspabcdefghijklmnopqrsqabcdefghijklmnopqrsrabcdefghijklmnopqrssabcdefghijklmnopqrsaabcdefghijklmnopqrstbabcdefghijklmnopqrstcabcdefghijklmnopqrstdabcdefghijklmnopqrsteabcdefghijklmnopqrstfabcdefghijklmnopqrstgabcdefghijklmnopqrsthabcdefghijklmnopqrstiabcdefghijklmnopqrstjabcdefghijklmnopqrstkabcdefghijklmnopqrstlabcdefghijklmnopqrstmabcdefghijklmnopqrstnabcdefghijklmnopqrstoabcdefghijklmnopqrstpabcdefghijklmnopqrstqabcdefghijklmnopqrstrabcdefghijklmnopqrstsabcdefghijklmnopqrsttabcdefghijklmnopqrstaabcdefghijklmnopqrstubabcdefghijklmnopqrstucabcdefghijklmnopqrstudabcdefghijklmnopqrstueabcdefghijklmnopqrstufabcdefghijklmnopqrstugabcdefghijklmnopqrstuhabcdefghijklmnopqrstuiabcdefghijklmnopqrstujabcdefghijklmnopqrstukabcdefghijklmnopqrstulabcdefghijklmnopqrstumabcdefghijklmnopqrstunabcdefghijklmnopqrstuoabcdefghijklmnopqrstupabcdefghijklmnopqrstuqabcdefghijklmnopqrsturabcdefghijklmnopqrstusabcdefghijklmnopqrstutabcdefghijklmnopqrstuuabcdefghijklmnopqrstuaabcdefghijklmnopqrstuvbabcdefghijklmnopqrstuvcabcdefghijklmnopqrstuvdabcdefghijklmnopqrstuveabcdefghijklmnopqrstuvfabcdefghijklmnopqrstuvgabcdefghijklmnopqrstuvhabcdefghijklmnopqrstuviabcdefghijklmnopqrstuvjabcdefghijklmnopqrstuvkabcdefghijklmnopqrstuvlabcdefghijklmnopqrstuvmabcdefghijklmnopqrstuvnabcdefghijklmnopqrstuvoabcdefghijklmnopqrstuvpabcdefghijklmnopqrstuvqabcdefghijklmnopqrstuvrabcdefghijklmnopqrstuvsabcdefghijklmnopqrstuvtabcdefghijklmnopqrstuvuabcdefghijklmnopqrstuvvabcdefghijklmnopqrstuvaabcdefghijklmnopqrstuvwbabcdefghijklmnopqrstuvwcabcdefghijklmnopqrstuvwdabcdefghijklmnopqrstuvweabcdefghijklmnopqrstuvwfabcdefghijklmnopqrstuvwgabcdefghijklmnopqrstuvwhabcdefghijklmnopqrstuvwiabcdefghijklmnopqrstuvwjabcdefghijklmnopqrstuvwkabcdefghijklmnopqrstuvwlabcdefghijklmnopqrstuvwmabcdefghijklmnopqrstuvwnabcdefghijklmnopqrstuvwoabcdefghijklmnopqrstuvwpabcdefghijklmnopqrstuvwqabcdefghijklmnopqrstuvwrabcdefghijklmnopqrstuvwsabcdefghijklmnopqrstuvwtabcdefghijklmnopqrstuvwuabcdefghijklmnopqrstuvwvabcdefghijklmnopqrstuvwwabcdefghijklmnopqrstuvwaabcdefghijklmnopqrstuvwxbabcdefghijklmnopqrstuvwxcabcdefghijklmnopqrstuvwxdabcdefghijklmnopqrstuvwxeabcdefghijklmnopqrstuvwxfabcdefghijklmnopqrstuvwxgabcdefghijklmnopqrstuvwxhabcdefghijklmnopqrstuvwxiabcdefghijklmnopqrstuvwxjabcdefghijklmnopqrstuvwxkabcdefghijklmnopqrstuvwxlabcdefghijklmnopqrstuvwxmabcdefghijklmnopqrstuvwxnabcdefghijklmnopqrstuvwxoabcdefghijklmnopqrstuvwxpabcdefghijklmnopqrstuvwxqabcdefghijklmnopqrstuvwxrabcdefghijklmnopqrstuvwxsabcdefghijklmnopqrstuvwxtabcdefghijklmnopqrstuvwxuabcdefghijklmnopqrstuvwxvabcdefghijklmnopqrstuvwxwabcdefghijklmnopqrstuvwxxabcdefghijklmnopqrstuvwxaabcdefghijklmnopqrstuvwxybabcdefghijklmnopqrstuvwxycabcdefghijklmnopqrstuvwxydabcdefghijklmnopqrstuvwxyeabcdefghijklmnopqrstuvwxyfabcdefghijklmnopqrstuvwxygabcdefghijklmnopqrstuvwxyhabcdefghijklmnopqrstuvwxyiabcdefghijklmnopqrstuvwxyjabcdefghijklmnopqrstuvwxykabcdefghijklmnopqrstuvwxylabcdefghijklmnopqrstuvwxymabcdefghijklmnopqrstuvwxynabcdefghijklmnopqrstuvwxyoabcdefghijklmnopqrstuvwxypabcdefghijklmnopqrstuvwxyqabcdefghijklmnopqrstuvwxyrabcdefghijklmnopqrstuvwxysabcdefghijklmnopqrstuvwxytabcdefghijklmnopqrstuvwxyuabcdefghijklmnopqrstuvwxyvabcdefghijklmnopqrstuvwxywabcdefghijklmnopqrstuvwxyxabcdefghijklmnopqrstuvwxyyabcdefghijklmnopqrstuvwxyaabcdefghijklmnopqrstuvwxyzbabcdefghijklmnopqrstuvwxyzcabcdefghijklmnopqrstuvwxyzdabcdefghijklmnopqrstuvwxyzeabcdefghijklmnopqrstuvwxyzfabcdefghijklmnopqrstuvwxyzgabcdefghijklmnopqrstuvwxyzhabcdefghijklmnopqrstuvwxyziabcdefghijklmnopqrstuvwxyzjabcdefghijklmnopqrstuvwxyzkabcdefghijklmnopqrstuvwxyzlabcdefghijklmnopqrstuvwxyzmabcdefghijklmnopqrstuvwxyznabcdefghijklmnopqrstuvwxyzoabcdefghijklmnopqrstuvwxyzpabcdefghijklmnopqrstuvwxyzqabcdefghijklmnopqrstuvwxyzrabcdefghijklmnopqrstuvwxyzsabcdefghijklmnopqrstuvwxyztabcdefghijklmnopqrstuvwxyzuabcdefghijklmnopqrstuvwxyzvabcdefghijklmnopqrstuvwxyzwabcdefghijklmnopqrstuvwxyzxabcdefghijklmnopqrstuvwxyzyabcdefghijklmnopqrstuvwxyzzabcdefghijklmnopqrstuvwxyz

Only slightly evil :}

\$\begingroup\$ cracked \$\endgroup\$
– Aaroneous Miller
Commented Aug 27, 2021 at 13:40 — Aaroneous Miller, Commented Aug 27, 2021 at 13:40

Arnauld · Accepted Answer · 2017-11-23 07:29:30Z

1

JavaScript (ES6), 35 bytes x 16 unique = 560 points - Safe!

Some kind of compromise between a reasonable size and an uncrackable algorithm (as it originally was meant to be ^^).

Expected output (255 bytes):

246913680357025702570358136924703692581470370360360370471582604715938260493827161505050505161728495173962851852963075208642087643221000001123457802570360483728396297532100123581594064200024727410136175571756954621413918119481621442132286383494753789149870

Intended solution

f=(x=0xff)=>--x&&(1/7/x+'')[7]+f(x)

O.innerText = f()

pre {
  white-space: pre-wrap;
  white-space: -moz-pre-wrap;
  word-wrap: break-word;
}

<pre id=O></pre>

Unique characters: 0, 1, 7, f, =, (, x, ), >, -, &, /, +, ', [, ]

edited Nov 23, 2017 at 7:29

answered Nov 11, 2017 at 9:35

Arnauld

198k20 gold badges179 silver badges649 bronze badges

\$\begingroup\$ Near match, with some tweak may be done (Mathematica) (may be spoiler). \$\endgroup\$
– user202729
Commented Nov 13, 2017 at 9:31
\$\begingroup\$ The function looks so weird... Probably not quadratic or polynomial, but I guess it's convex. The positive side grows much faster than the negative side. [Almost safe now] \$\endgroup\$
– user202729
Commented Nov 17, 2017 at 14:49

Add a comment |

Martin Ender · Accepted Answer · 2017-11-23 10:04:58Z

Alice, 7 bytes, 6 unique, 42 points, safe

-10764@

Try it online!

Solution

gZ/
@o@

Try it online!

g is the easiest way to get a decently big number onto the stack quickly, because it pops two implicit zeros from the stack and then reads the code point at that cell, which is 113 for g itself.

Z then packs 0 and 113 into a single integer, which is a good way to get an even bigger number. That number happens to be -10764.

The number gets printed in Ordinal mode, but then we run into g again in Ordinal mode. The label Alice is looking for is just an empty string. It finds that immediately in the bottom left corner and extracts the @ as a string. When we hit o again, that @ is printed as well.

Afterwards, the IP finds its way to the bottom right corner where the other @ terminates the program.

Bubbler · Accepted Answer · 2018-08-30 04:09:17Z

1

J, 8 Bytes x 6 Unique = 48 Points, Cracked

Output:

23 _0.677447j0.296961 _0.677447j_0.296961 0.125003j0.726137 0.125003j_0.726137 _0.379097j0.630438 _0.379097j_0.630438 0.518498j0.521654 0.518498j_0.521654

Hint:

It has to do with primes

edited Aug 30, 2018 at 4:09

Bubbler

78.4k5 gold badges149 silver badges469 bronze badges

answered Dec 1, 2017 at 3:20

Bolce Bussiere

1,0907 silver badges9 bronze badges

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Stephen · Accepted Answer · 2017-11-10 15:24:37Z

0

Explode, 24 total bytes, 7 unique, 168 points

Output:

\}AtAY)_)MqGq/Y,Y}AwA_)\)JqDq/Y,Y}AtA_)\)MqDq/Y,Y}AtA_)\)MqDq2Y,YzAtAb)\)JqDq2Y,YzAtAb)\)JqDq2Y,YzAtAb)\)JqDq2Y,YzAtAb)\)JqDq2Y,YzAtAb)\)JqDq2Y,YzAtAb)\)JqDq2Y,YzAtAb)\)JqDq2Y,YzAtAb3)yqobUPK>1,'zmhcVID?2% {na\WJ=83&ytobUPK>1,'zmhcVID?2% {na\WJ

Now that I've read the question correctly, you can truly see the output explode.

edited Nov 10, 2017 at 15:24

answered Nov 10, 2017 at 15:19

Stephen

14k3 gold badges52 silver badges117 bronze badges

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Bolce Bussiere · Accepted Answer · 2017-11-19 02:26:43Z

0

><>, 16 Bytes * 11 Unique = 176 Points

Output:

; ;;;;;;;;   ;
;;;
;;;;;;;;;;;;;;;;;;; ;!;";#;$;%;&

Shouldn't be too hard.

answered Nov 19, 2017 at 2:26

Bolce Bussiere

1,0907 silver badges9 bronze badges

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Brainf**k, 220 bytes, 7 unique (1540 points), safe

Edit: Sorry that I counted the bytes wrong.

output (88 bytes)

rh3NY8C]'t/FS:\7pb5D?yc,G>A@OsWU_6x*gaRo)Hz-w.ul+T&IM<P;BvjK%imn1[Xk2Ef(Z|Q{dL^`e4q0V9J=

Just do some brainf**k for fun.

Intended solution:

>+[>[-]<[>+>+<<-]>>[<<+>>-]---<[>>>+<<<-]>>>[<<[<+>>+<-]>[<+>-]>-]<<<+]<[>>[-]<[-]<[>+>+<<-]>>[<<+>>-]++++++[-<------>]<->>>++++++[-<+++++[-<+++>]>]<<--[>>+<<<[->>>[-]<+<<]>>>[->+<]<[-<<+>>]<-<->]>>>[<<<<<.>>>>>-]<<<<<<]
I didn't golf this too much. This is a sequence generated by linear congruential method, but just I output bytes between 37 and 124.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 11, 2017 at 15:14

Colera Su

2,42113 silver badges29 bronze badges

\$\begingroup\$ Safe? (char limit) \$\endgroup\$
– user202729
Commented Nov 20, 2017 at 12:55
\$\begingroup\$ @user202729 Thanks for reminding me. I almost forgot this. \$\endgroup\$
– Colera Su
Commented Nov 20, 2017 at 15:52

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Python 2, 98 bytes, 30 unique, 2940 points

Output (249 bytes):

-K:&qLcVoJk`N8QS@JH5c`1]T92i_bS.bv<zZK3vI&(kH!6=NrQGTL&VE:j|g..UQC)yk>3IoNhpjc'h0.x'L!aajp;Ip}+BQ/I64Q`5ZBRwOfDtt|O,6b>>UvvTt#8"z0CzEHtiVJ{[email protected]$RLbso]?gR=ZN2;f5vTh|[email protected]|b;F=lKyFZpx(mBPrqhiw1:`LD>AI{.S;POHfTd[5d;T=`162Hc]=.m7?T_1O<Kty

Edit: I think this is going to be too hard so I'll give a hint

Edit: Slightly more informative hint

seed(9)

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 15, 2017 at 4:47

dylnan

5,3742 gold badges16 silver badges33 bronze badges

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Jo. · Accepted Answer · 2017-11-29 07:51:11Z

0

PHP, 16 Bytes, 10 Unique Bytes (160 points) -- Safe!

0.42516833158764

Intended Solution (Try it online!)

<?=M_PI/M_E/M_E;

Simply divides π by e twice.

Edit

As with ThePirateBay's answer, I marked this as safe before I saw the OP's edit about seeing how long you can go without being cracked. Sorry to any who wanted to attempt a crack.

edited Nov 29, 2017 at 7:51

answered Nov 11, 2017 at 8:32

Jo.

1,0345 silver badges10 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Python 2, 114 Bytes, 27 Unique (3078)

両寷動娯爅潟嗁琓偵桋斥獻曕檫垉哣抹妗惱畃

Function calls not counted, lambda expression counted.

Hint 1

Pseudo-random number generator with Unicode output. Beware of the offsets ;)

Hint 2

2 generators of the same kind but with different parameters are used alternatively, in the form of a_n+1 = f₁(f₂(a_n))

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 4, 2017 at 7:01

Shieru Asakoto

6,25816 silver badges40 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Jelly, 13 bytes, 5 unique, 65 points

Output is the following 21 bytes.

HCG, AhhsNeg3 Aarrghh

The optimized string compressor won't help you.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Feb 28, 2018 at 23:13

dylnan

5,3742 gold badges16 silver badges33 bronze badges

Add a comment |

l4m2 · Accepted Answer · 2018-03-02 14:50:49Z

0

x86 Assemble, 22 Bytes * 17 types = 374 points, safe for 3 months

00000000h: 80 8A 5E FA 59 DF 69 24 F9 73 DA 8C 50 A6 F1 60
00000010h: 93 E3 3E 87 7D 08 F6 14 9E 03 EE 96 4E 48 99 61
00000020h: 2E 7A 62 71 96 39 69 27 CF 8E F7 AE 1A 37 71 8F
00000030h: B0 52 6A 85 F3 04 47 D4 A5 F1 8E 64 D8 4E A6 C1
00000040h: 0E 1E 35 EB C6 E3 99 28 67 79 80 58 52 F5 BF E7
00000050h: 28 87 1C E7 96 5E CA 8E 63 DB 3C 59 73 11 E1 9C
00000060h: E1 1D 6A 77 6B CD 6B 41 64 EA D6 02 0B 3F 89 60
00000070h: B4 DF 90 C4 AD AA 35 D5 14 6F 47 DC 3A 31 48 B5
00000080h: 50 87 5A 4C 5D FC 07 70 4B 6A 2A 6D 87 32 7F CA
00000090h: AB EB 75 4B 7C 17 1E 7D 01 49 C3 99 AF 98 8A 59
000000a0h: 6A B1 A1 2C B4 08 5D 42 9F AA E0 04 10 38 DC 87
000000b0h: E8 CB 15 BE CE 52 5D 02 4C 3C C4 C3 FF

Answer:

00000000h: DB 06 01 01 B7 CD DD 17 D9 FE B4 02 8A 57 02 CD
00000010h: 21 FE C2 75 F1 C2

edited Mar 2, 2018 at 14:50

answered Dec 1, 2017 at 6:48

l4m2

28.3k2 gold badges25 silver badges103 bronze badges

\$\begingroup\$ Hint: 1) output uses MOV AH, xx; MOV D?, xx; INT 21 which costs at least 6 bytes; 2) return may be RET, INT 20, INT 21, but at least 1 byte required; 3) x87 used \$\endgroup\$
– l4m2
Commented Dec 2, 2017 at 5:44
1

\$\begingroup\$ I suppose the header should say "machine code" instead of "assembly"? \$\endgroup\$
– Weijun Zhou
Commented Mar 2, 2018 at 18:01
\$\begingroup\$ maybe, though from the hint comment that can be inferred.... \$\endgroup\$
– l4m2
Commented Mar 3, 2018 at 1:50

Add a comment |

Esolanging Fruit · Accepted Answer · 2018-03-03 03:03:50Z

0

CJam, 5 bytes × 5 unique = 25 points, safe

1406563064942.4553

J_#mq
This code computes sqrt(19¹⁹).
It could have been JJ_#mq for 20 points, but I didn't realize that until after I posted it.

edited Mar 3, 2018 at 3:03

answered Feb 18, 2018 at 21:55

Esolanging Fruit

15.5k4 gold badges49 silver badges92 bronze badges

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Esolanging Fruit · Accepted Answer · 2018-03-03 03:05:18Z

0

CJam, 13 bytes * 10 unique = 130 points

819457164140982707422640583089002841689

answered Mar 3, 2018 at 3:05

Esolanging Fruit

15.5k4 gold badges49 silver badges92 bronze badges

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Esolanging Fruit · Accepted Answer · 2018-03-03 03:11:50Z

brainfuck, 25 bytes * 7 unique = 175 points

 .KoÌEÚ/îXÉAÀFÓg¤Mý´r7Ö°yh^[_j|µÜ
?{¾Y±vãWÒTÝm¢Gó¦`!é¸kO:,%%,:Ok¸é!`¦óG¢mÝTÒWãv±Y¾{?
Üµ|j_[^hy°Ö7r´ýM¤gÓFÀAÉXî/ÚEÌoK.

This contains unprintables which will probably be mangled by SE, so here's the hexdump:

00000000: 0918 2e4b 6fc2 9ac3 8c05 45c2 8cc3 9a2f  ...Ko.....E..../
00000010: c28b c3ae 58c3 8941 c380 46c3 9367 02c2  ....X..A..F..g..
00000020: a44d c3bd c2b4 7237 03c3 96c2 b0c2 9179  .M....r7.......y
00000030: 685e 5b5f 6a7c c295 c2b5 c39c 0a3f 7bc2  h^[_j|.......?{.
00000040: be08 59c2 b110 76c3 a357 c392 54c3 9d6d  ..Y...v..W..T..m
00000050: 04c2 a247 c3b3 c2a6 6021 c3a9 c2b8 c28e  ...G....`!......
00000060: 6b4f 3a2c 2525 2c3a 4f6b c28e c2b8 c3a9  kO:,%%,:Ok......
00000070: 2160 c2a6 c3b3 47c2 a204 6dc3 9d54 c392  !`....G...m..T..
00000080: 57c3 a376 10c2 b159 08c2 be7b 3f0a c39c  W..v...Y...{?...
00000090: c2b5 c295 7c6a 5f5b 5e68 79c2 91c2 b0c3  ....|j_[^hy.....
000000a0: 9603 3772 c2b4 c3bd 4dc2 a402 67c3 9346  ..7r....M...g..F
000000b0: c380 41c3 8958 c3ae c28b 2fc3 9ac2 8c45  ..A..X..../....E
000000c0: 05c3 8cc2 9a6f 4b2e 1809 01              .....oK....

dylnan · Accepted Answer · 2018-05-18 17:45:26Z

0

Jelly, 4 bytes, 4 unique, 16 points.

[1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0]

answered May 18, 2018 at 17:45

dylnan

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Reversed Engineered from Uniqueness (Cop's Thread)

This is the cops thread. Post solutions that are to-be-cracked here.

The robbers thread is located here.

Rules

Winning

Caveats

47 Answers 47

Japt, 5 bytes, 5 unique bytes (25 points) (Cracked by Scrooble)

Vyxal, 4 bytes, 4 unique, score 16, Cracked by Aaron Miller

JavaScript (ES6), 35 bytes x 16 unique = 560 points - Safe!

Intended solution

Alice, 7 bytes, 6 unique, 42 points, safe

Solution

J, 8 Bytes x 6 Unique = 48 Points, Cracked

Explode, 24 total bytes, 7 unique, 168 points

><>, 16 Bytes * 11 Unique = 176 Points

Brainf**k, 220 bytes, 7 unique (1540 points), safe

Python 2, 98 bytes, 30 unique, 2940 points

PHP, 16 Bytes, 10 Unique Bytes (160 points) -- Safe!

Edit

Python 2, 114 Bytes, 27 Unique (3078)

Jelly, 13 bytes, 5 unique, 65 points

x86 Assemble, 22 Bytes * 17 types = 374 points, safe for 3 months

CJam, 5 bytes × 5 unique = 25 points, safe

CJam, 13 bytes * 10 unique = 130 points

brainfuck, 25 bytes * 7 unique = 175 points

Jelly, 4 bytes, 4 unique, 16 points.

Your Answer

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cops-and-robbers
or ask your own question.

Linked

Hot Network Questions

Reversed Engineered from Uniqueness (Cop's Thread)

This is the cops thread. Post solutions that are to-be-cracked here.

The robbers thread is located here.

Rules

Winning

Caveats

47 Answers 47

Japt, 5 bytes, 5 unique bytes (25 points) (Cracked by Scrooble)

Vyxal, 4 bytes, 4 unique, score 16, Cracked by Aaron Miller

JavaScript (ES6), 35 bytes x 16 unique = 560 points - Safe!

Intended solution

Alice, 7 bytes, 6 unique, 42 points, safe

Solution

J, 8 Bytes x 6 Unique = 48 Points, Cracked

Explode, 24 total bytes, 7 unique, 168 points

><>, 16 Bytes * 11 Unique = 176 Points

Brainf**k, 220 bytes, 7 unique (1540 points), safe

Python 2, 98 bytes, 30 unique, 2940 points

PHP, 16 Bytes, 10 Unique Bytes (160 points) -- Safe!

Edit

Python 2, 114 Bytes, 27 Unique (3078)

Jelly, 13 bytes, 5 unique, 65 points

x86 Assemble, 22 Bytes * 17 types = 374 points, safe for 3 months

CJam, 5 bytes × 5 unique = 25 points, safe

CJam, 13 bytes * 10 unique = 130 points

brainfuck, 25 bytes * 7 unique = 175 points

Jelly, 4 bytes, 4 unique, 16 points.

Your Answer

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