Haystack, 84 bytes, Cracked

0\\1-c\ // ?10F17+c8F+4+cd8F+3+c6-c1+c,c2+c8+c| 0 \1++c,c| F/c++2F8 c\8F+2+cd 

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This looks (to me) rather convoluted but if you find the right subset it's a bit too easy... oh well, just to get us started :P

Hexagony, 37 bytes

H[@;(...e<l.a;./$.>;\sN;\ac.>).;;;._y 

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Just my obligatory Hexagony entry...

For convenience, here is the unfolded code:

 H [ @ ; ( . . . e < l . a ; . / $ . > ; \ s N ; \ a c . > ) . ; ; ; . _ y 

How this works:

The program starts off with H, then we move to IP #5. This IP starts in the west corner, bouncing and wrapping around while executing (in effect) ;a;y;s; (so we've printed Hays). Then s gets incremented to a t by ) and printed, then we pass through Ne...(c before getting to a;c; (still bouncing around a small section of the hexagon). The program hits the _, reflects up through \ to l which gets decremented to a k by (, which passes through another \ before being printed and the program terminates on the @.

Verbose version

H H is entered into the current memory cell [ We move to the previous instruction pointer (starting at the / and moving NE) / No-op < Mirrors the IP back to south-west / No-op (IP now wraps to the NE corner) ; Outputs H . No-op a a is entered into the current memory cell > Mirrors the IP back to north-east a a is entered into the current memory cell . No-op ; Outputs a (IP now wraps SE corner) y y is entered into the current memory cell ; Outputs y . No-op s s is entered into the current memory cell (IP now wraps SW corner) ; Outputs s ) Increments the memory cell to t \ Mirrors the IP west ; Outputs t N N is entered into the current memory cell (IP now wraps to e) e e is entered into the current memory cell ... No-ops ( Decrements the memory cell to d (IP now wraps to . on SE edge) . No-op c c is entered into the current memory cell a a is entered into the current memory cell \ Mirrors the IP north-east > Redirects the IP east ; Outputs a \ Mirrors the IP south-west c c is entered into the current memory cell ; Outputs c _ Mirrors the IP north-west .\. No-ops l l is entered into the current memory cell ( Decrements the memory cell to k (IP now wraps to . on SE edge) .\ No-ops ; Outputs k . No-op @ Terminates the program

Brain-Flak, 146 bytes (Cracked)

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

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Intended solution, 80 bytes

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

Haskell, 168 bytes (Cracked by nimi)

hays=map;hay=zipWith;stack=head;h=stack{- hay.(hays.(stackany hay$or id).stack hay <*>hays(sum$stack haystack<$>hay))-}$words "Haystack Hayst ackH aysta ckH aystac k" 

Try it online! Evaluating the identifier h returns the string "Haystack", after some deletions h yields "Needle".

JavaScript, 95 bytes (ES6), Cracked

A function returning a string.

f=(k=b=x=35)=>x--?f(k*74837258394056219&268435455):k&2?'N'+(k^124038877).toString(b):'Haystack' 

"Haystack" demo

f=(k=b=x=35)=>x--?f(k*74837258394056219&268435455):k&2?'N'+(k^124038877).toString(b):'Haystack' console.log(f())

Jelly, 41 bytes (Cracked)

“¿ọ⁽ṅ*FỊ⁼g£¡#!ʋzoɦṪ£ṢÞḲÐɠ`”m3⁾“»jVḟ“¡!pṄ» 

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Happy hunting!

Hexagony, 32 bytes. Cracked

I couldn't solve Martin's, so I'm posting my own.

];N.@cl;e@;;(\H/;ya;_.>s.;t//<._ 

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Here it is formatted:

 ] ; N . @ c l ; e @ ; ; ( \ H / ; y a ; _ . > s . ; t / / < . _ . . . . . 

My aim with this was for both solutions to use as many IPs as possible, I got 6 for Needle and only 5 for Haystack.

Pyth, 44 bytes (Cracked)

Kr."Dn2û"2sf!/+rrK2 2r."EL8"2Tr."AhÐP­®Z"2 

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Java (OpenJDK 8), 226 217 bytes (Cracked)

First ever code golf, probably very easy but it was a fun challenge!

String d(){int h=3609000-5055+911,m=557558,s=15441301-157*10000;String d="0"+h*2+""+m*20+""+s*7,x="",y;for(int g=0;g<d.length();g+=3){y="";for(int e=0;e<3;e++)y+=d.charAt(e+g);x+=(char)Integer.parseInt(y);}return x;} 

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dc, 148 bytes (Cracked)

6 93 3 9 2 2**+*+483622 1 2 3 3*+3*+89 47*+*+3 5 2* 269 158 9**107 97*2 4*++2 3 3*+42 14 2**+*+5*+5 2148 1 6 2*+*+68262 5 280 7 2 3 3*+5 2**+*+*+*+P 

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It is rather simple, but I hope it will be at least a little fun to solve :з

JavaScript, 119 bytes (ES6), Cracked

A function returning a string. Quite long and not so hard, but hopefully fun.

_=>(+{}+['H'])[+[3]]+(+[][[]]+[])[+!!3]+(+[][[]]+['y'])[3]+(+[][[]]+['s'])[-~2]+(~![]+['t'])[2]+(+[][[]]+[])[+!!3]+'ck' 

"Haystack" demo

let f = _=>(+{}+['H'])[+[3]]+(+[][[]]+[])[+!!3]+(+[][[]]+['y'])[3]+(+[][[]]+['s'])[-~2]+(~![]+['t'])[2]+(+[][[]]+[])[+!!3]+'ck' console.log(f())

Python 2.7.2, 103 / 117 bytes, Cracked

Function Version (117 bytes):

def e(): a,b,s=20070763850923833476353301471991752,0b1010100010010011,"" while a>0: s=chr(a%b)+s a//=b print s 

Program Version (103 bytes):

a,b,s=20070763850923833476353301471991752,0b1010100010010011,"" while a>0: s=chr(a%b)+s a//=b print s 

This should print Haystack well. Tested on Python Fiddle.

Btw this is the first attempt.

Not sure if the program version is counted as a snippet, so I put both versions here.

Python 2.7.10 with Numpy 1.12.1, 208 209 bytes (cracked)

It appears that there is a Needle and a Haystack in Numpy! Here is the Haystack; see if you can find the Needle. I hope you have as much fun searching for the Needle as I had hiding it.

import numpy print "".join([dir(numpy)[int(i)][1-0] for i in numpy.poly1d([-1*1433/252e1,-3232/1920.,4026./72/2/3.,613/(6*4.)*1,-4723./1.8e2,-9763/120.,-2689/(-1+5*17.),1+138*.4*2])(numpy.arange(-12/3,13%9))]) 

It outputs as specified:

Haystack 

You can repl.it.

Java 8, 321 bytes, Cracked

v->{String h="Haystack";int x=-7;return x<0?h:new String(new java.math.BigInteger(new byte[]{(byte)((~-~-~-~-~-~-~-~-~-~1^-x++*x)+151),new Byte("2"+"1+\"0+\"".length()+(x=h.length()*4/x)+"-x-7")}).toByteArray())+(new StringBuffer("hidden".substring(++x%3^4,--x-x--).replaceFirst("dd","e"+(char)(x*211%+93))).reverse());} 

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More readable:

v->{ String h="Haystack"; int x=-7; return x<0? h : new String(new java.math.BigInteger(new byte[]{ (byte)((~-~-~-~-~-~-~-~-~-~1^-x++*x)+151), new Byte("2"+"1+\"0+\"".length()+(x=h.length()*4/x)+"-x-7") }).toByteArray()) +(new StringBuffer("hidden".substring(++x%3^4,--x-x--) .replaceFirst("dd","e"+(char)(x*211%+93)) ).reverse()); } 

Not sure if it's too long/hard.. Then again, Java in general is pretty long to begin with, so hiding the 'Needle' properly of course increases the byte-count quite a bit..
If no one cracks it, I'll add some spoiler-tips later on.

Ruby, 185 bytes, cracked by cab404

x='yGwztsPXhxDkBKlCYdFjQnpUROfoHvqmTgbaJSLcEiZrIAuMVNW' s="n=x.to_i 36;x.bytjs.jach_cons(3){|a,b,c|n+=n*b%c;n*=a^b};puts n%8675309==1388649 ?'Njjdlj':'Haystack'" eval s.tr ?j,s.size.chr 

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I'll try to come up with something sneaky, but for now here's a try at "simple but obnoxious."

Brain-Flak, 188 bytes (Cracked)

I only just saw Funky Computer Man's answer As I posted this.

It is somewhat obfuscated.

((((((((((()()()){}){}()){}){}()){}()))<({}[(()()()()){}])(([[]]({})<>)<>)>((()()())){}{})[()]))<[[]()]>((()){}){}((){}[][(<>){}<>])(<>){}(({}<>()[()])[(((()()()){}<[()]>)<(()){}>{}){}()]) 

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Intended solution, 96 bytes:

((((((((((()()()){}){}()){}){}()){}()))<({}[(()()()()){}])(([[]]({})<>)<>)>((()()())){}{})[()]))<[[]()]>((()){}){}((){}[][(<>){}<>])(<>){}(({}<>()[()])[(((()()()){}<[()]>)<(()){}>{}){}()])((((((((()()()){}){}){}()){}){}()<>)<>)(()()()){}())<>(((({}[()]<>)()))[(((()()()){}())()){}{}]) (((((((( (()()()){}){} ){}){}()){}())) ()() <>) (()()()) {} () ) < >((() ((){} <>) ) ) (({} )[(((()() ){} () ) () {} ){}()]) 

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T-SQL, 757 characters CRACKED

Apologies for deleting my previous answer--I edited it too many times as I obfuscated, and didn't want to give anything away. :)

In any case, obfuscating in SQL is a bit difficult, unless you want to do crazy stuff like this, and I wasn't that invested.

Also, I unabashedly do not apologize for naming my variables after Dota.

SQL Fiddle

create table a(l int,c int) insert into a values (1,10),(2,1),(3,8),(4,0) go ;CREATE FUNCTION b(@ varchar(max)) returns varchar(max) as begin return 'char('+@+'),'''','end go ;CREATE FUNCTION h(@ varchar(max),@a varchar(max), @b varchar(max), @c varchar(max), @d varchar(max), @e varchar(max), @f varchar(max), @g varchar(max), @h varchar(max)) returns varchar(max) as begin return replace(replace(replace(replace(@,@a,@b),@c,@d),@e,@f),@g,@h) end declare @x varchar(max),@ int=1,@y varchar(99)='' ,@D varchar(4)='Ha',@O varchar(4)='ys' ,@T varchar(3)='ta',@A varchar(4)='ck' WHILE @<=4 BEGIN set @y+=(SELECT dbo.b(c+100)from a where l=@)+' ' set @+=1 END SELECT @x='select left(dbo.h('''+@D+@O+@T+@A+''','+ left(@y,len(@y)-1) +'),char(56))' execute(@x) 

If this is the easiest answer in this thread, you're probably right. :P It's hard to trick SQL.

Ly, 40 bytes, cracked

(78)"e"&p"Ha"s"yst"l"ck"&o(100)"l"l'&'o; 

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Oh boy, another Ly CNR submission. These haven't worked very well historically (possibly due to me and not the language), but we'll see how this fares and today is no exception.

Solution:

(78)"e"sl(100)"l"l&o;, remove XXXXXX XXXXX XXXXXX X X with seven leading spaces

Java, 345 bytes, Cracked

import java.util.*;interface Main{static void main(String[]args){Stack<Hay>s=new Stack();s.add(new Needle());for(int i=0;i<1000;i++)s.add(new Hay());System.out.println(s.get(s.indexOf(new Hay())+1).a);}}class Needle extends Hay{{a="Needle";}}class Hay{String a="Haystack";public boolean equals(Object o){return getClass().equals(o.getClass());}} 

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Really long and probably easy to crack, but at least it's got a Stack<Hay>!

Vyxal o, 51 bytes

‛Ha:₍₴Cv∑∑Ṙ⌊√‹½⇩⇩²C:₴«•∞yλ;ß4ġ∆Ė_□∴9«øc:3Ẏ$4ȯ∇ppĖ5e 

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Have fun!

So, here's an explanation of the code:

‛Ha # Push 'ha' :₍ # Duplicate and parallel apply... ₴ # Flatprint C # And charcode, wrapped into a list v∑∑Ṙ # Get sum, concatenate, reverse (`961aH`) ⌊√‹½ # Intify, square root, decrement and halve (15) ⇩⇩²C # Subtract 4, convert to char, which happens to be y :₴ # Dupe and flatprint. «...« # Push a string, specially chosen, see below øc # Compress :3Ẏ$4ȯ∇pp # Insert ToS (`y`) at char 3 Ė # Evaluate 5e # Every fifth character 

So, I realised I could use the compressed string ‛₍∞ as the needle program. I could easily sneak in a ‛ and ₍, but ∞ was proving to be a problem. Eventually, I managed to construct a compressed string that contained ∞ and y, and had the property that if you took every fifth character, you got 'stack'.

So yeah, this was tricky.

Vyxal, 44 bytes

17SǏ⌊0»¹∇ḃF2»*»∷J∇Y.ḊṖ¼ḭ¦]uy»S4ẇṫ½Y⌊$oḢJCṅĖǐ 

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Rattle, 55 bytes

Haystack&Needl|Ip0=2+(1*2)!Ps[bg0*3b^](`+5);$b|1,%(?3b) 

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This isn't terribly difficult but it's not trivial either (note that "Needle" is missing the last "e" near the beginning of the code).

Hint: the Needle program will not have a trailing newline, while the Haystack program does.

TI-BASIC, 119 bytes (Safe)

Output stored in Ans.

"TIBASIC→Str1 length(Ans→X Str1 For(I,X,2(7-4not(X-1 Ans+sub("ACDEHKLNSTY",inString("ABCDEFGHIJK",sub("KCBADAEDJDHIGF",1+remainder(11I-1,14),1)),1 End sub(Ans,1+X,length(Ans)-X 

Solution

Remove characters from Str1 to give it a length of 1.

"C→Str1 length(Ans→X Str1 For(I,X,2(7-4not(X-1 Ans+sub("ACDEHKLNSTY",inString("ABCDEFGHIJK",sub("KCBADAEDJDHIGF",1+remainder(11I-1,14),1)),1 End sub(Ans,1+X,length(Ans)-X 

PHP, 44 bytes, Cracked

fairly easy to crack; but I like it.

for($i=~1;$c=H_aNyesetdalcek[$i+=2];)echo$c; 

Run with -nr or try it online.

Aceto, 154 bytes (Safe)

 27\'dNU QJi9MLJ€{{x(}]J{'!o∑€xiDQxsJ(]sicpicp1.2sJJicp90I.2+D/edxi-'>xd80J0IJicx'NIx5sJsJidcpIcpL7sssJicpei7+ UUdJicpLI7sJicpx'p\p9*coJcU'p+\p 

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'N'ed'd80J0IJic'eUpppppp Try it online!

Explanation: <space>*2 - Two spaces for the hilbert curve to work right 'N pushes 'N' onto the stack 'e pushes 'e' d duplicates it 'd pushes d 80 pushes 8, 0 J concats top two values 0I pushes 0, pops, increments, pushes back on, net effect: pushes 1 J concats to '108' i converts to integer c pops and pushes ascii equiv on stack 'e pushes 'e' (again) U reverses the stack and the p's print out the stack