Sisyphus, PHP

Password:

A string golf with 1,000,000 leading whitespaces and 1,000,000 trailing whitespaces

Output: golf

Reason: 1,000,000 is the default backtracking limit of PCRE (which you can get by var_dump(ini_get('pcre.backtrack_limit'));). And preg_match will return FALSE other than 0 or 1 when this limit is breaked.

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ovs, Python 2

Password:

class m:0 m.__init__ = hex.__init__ n = 49374 m.__str__ = n.__hex__ class a: __metaclass__ = m 

Output: 0xc0de

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R, Robin Ryder

Password: as.roman(c(1, 9)))

It was roman numerals!

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Lynn, Haskell

Lynn created a stack based mini-language, the task was to generate the primes up to 500 in 60 or less operation. Here is my 55-operation password:

[0,0,2,1,4,1,0,2,1,1,4,4,4,30,0,2,0,2,0,20,10,0,3,10,2,0,3,1,0,10,3,6,6,6,6,5,4,7,5,0,7,3,10,2,0,3,20,2,1,0,3,0,3,30,2] 

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The available operations are

0 push 1 1 duplicate top of stack 2 add top two values 3 subtract 4 multiply 5 integer divide top value by second value 6 push the second value without popping it 7 swap top two values c>7 while loop, runs until top of stack is 0 the loops ends at the first instruction >=c 

0,0,2,1,4,1,0,2,1,1,4,4,4 pushes the initial number 500. The remainder of the code is best explained inside out:

6,6,6,6,5,4,7,5 is a divisibility test. Given k and n as the top two values on the stack this calculates \$\lfloor {\lfloor {n \over k} \rfloor \cdot k \over n}\rfloor\$, which is only 1 if k divides n: Try it online!

1,0,10,3,div test,0,7,3,10,2,0,3 is a primality test, or a composite test since this returns truthy (non-zero) values for composite numbers:

1 -- duplicate n 0 -- push 1 - stack: [1, k=n, n] -- in the next iterations of the loop, -- the top of stack will be the inverted result -- of the divisibility test 10 10 -- while loop: -- runs until [0, d, n] is on the stack, -- where d is the largest divisor of n <n 3 -- subtract top value (always 1) from k div -- the divisibility test 0 -- push 1 7 -- swap top two values 3 -- subtract (1 - div test result) 2 -- add the top 0 to the last k 0,3 -- subtract 1 -- if the loop ended with [0, 1, n], this is now 0 -- otherwise we have a positive number 

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0,20,10,0,3,10,2,0,3,comp. test,20,2,1 generates the next prime less than n:

0 -- push 1. This means the current number is composite -- Even if it isn't, we still want to find a prime <n 20 -- while loop. This iterates until the composite tests returns 0 10 10 -- we have an positive number on the top of the stack ... 0 3 -- by subtracting 1 until it is 0, ... 2 -- and adding this to the last prime candidate ... -- we can get rid of it. 0 3 -- subtract 1 to get new prime candidate pc comp -- check if pc is composite 20 -- end of loop, top of stack is now [0, p], with p prime 2 -- add 0+p 1 -- duplicate the prime, such that we store the result, -- and can use the value to find the next prime 

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30,0,2,0,2,next prime,0,3,0,3,30,2 repeats this until the prime 2 is found:

30 30 -- while loop 0 2 -- add 1 0 2 -- add 1 np -- find the prime less than this 0 3 -- subtract 1 0 3 -- subtract 1 -- if the prime was 2, this is now 0 -- and the while loop terminates 2 -- add the 0 to the 2 to remove it 

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R, Robin Ryder

5.099829245500619335478113833945732102551318887107339446461762721i 

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Sisyphus, Jelly

Password:

[1.1071487177940904,1.1071487177940904,1.1071487177940904,1.1071487177940904,1.1071487177940904,1.1071487177940904,1.1071487177940904,0.897846510365972] 

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The code OÆTP, when taken literally, means product(math.tan(ord(c)) for c in input). But the ord function in Jelly does nothing for numbers, so we can ignore that. Now the problem is to generate that very specific number. I figured that as product is likely to have precision loss, I'd use 2's as multiplicands. The number 1.1071487177940904 is equal to arctan(2), and I use seven copies of it to reduce the problem to arctan(x) where x<2 so that I have better chance at getting that exact result. Finally I computed arctan(answer/128) and put it as the last term of the input array, and it worked perfectly.

att, Wolfram Language (Mathematica)

Password:

flag /: Head[flag] = flag 

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.

We can adapt this kind of solution to work with any Mathematica program:

a /: _[a] = flag; a 

With this argument, any function returns flag.

JavaScript by tjjfvi

The flag is: ()* (Or ()? / (){0})

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This splits between the two x's (it matches the empty string) without actually capturing anything with the group. The return value of split includes capturing groups, which results in ["x", undefined, "x"]. This gets reduced to "xundefinedx".

Sisyphus, Python3

All you need to do is keep the processor busy for more than 9 seconds.

Any regex that requires lots of backtracking will gum up the parser. The only real difficulty is not slowing it up so much that Tio runs over its 60-second limit.

'b(.*.*.*)*z|baaaaaaay' 

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JavaScript (SpiderMonkey), tsh

Password hexdump:

61 61 61 00 

(3 a followed by a null byte).

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Eric Duminil, Ruby

My password is

14127792144400463565475544498208881214759697720904563865426051592050217695592754443713601541725640031x00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000123456789 

According to the Ruby documentation, to_i discards everything after the first integer it finds.

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Alternative solution:

9164214512877268290754278122624834497733309914632715416260853069873976599113800182718102190123456789 

I factored the semiprime with cado-nfs.

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JavaScript (V8) by PkmnQ

({toString(){return this.i--;},i:43}) 

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JavaScript variables like: ({toString(){return this.i.shift();},i:[1,2,3,4,5]}) would be useful in many cops-and-robbers answers.

({valueOf:()=>43,toString:()=>42}) 

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It is strange: While, let p=({valueOf:()=>43,toString:()=>42}): ''+p results "43" but `${p}` results "42"...

ZippyMagician, Arn

Password: J0e_Biden!

I'm sure this isn't the password that ZippyMagician was thinking of, but it works at least. The flag is equal to 2⁹⁶, and the uncompressed code starts with :*:*, which raises something to the 4th power. I'm not sure what's going on in the middle, but the last few bytes of code (:i0^:i"n) calculate the value of a^b, where a and b are the indexes of the characters 0 and n in some transformed version of the input. So if the second character is 0, it's just a matter of tweaking the input until the n is in the right place.

Python 3.7, ovs's answer

@print @int.__invert__ @len @ascii @ascii @ascii @ascii @ascii @ascii @ascii @ascii @ascii @ascii @str.lstrip @min @ascii class a: pass 

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I used a brute force program to find the solution, although in the end it ends up being quite nice: Try it online!

Initially I found one with __sizeof__, but it doesn't work on TIO (being implementation-specific).

I had to made quite a few tweaks for it to fork (disable open and id, as the former will read from stdin with list(open(1)) or something similar)

It's also possible to get import inspect and quite a few other modules, but I didn't consider that possibility.

In retrospect, repr would work as well, but ascii comes before repr in my generator program.

JavaScript (Node.js) by emanresu A

[Math.PI, 2n] 

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JSON.stringify throws a TypeError if it finds a BigInt, thus executing catch(e){return v[0]===Math.PI}, and then it is simple to make that true.

JavaScript by tjjfvi

document.all.toString = ()=>"length", document.all 

document.all is an object that is unique in several ways, including being falsy.

in uses toString, so the first part can be made true by changing toString.

HyperNeutrino, Python 3

Password: import sys;sys.exit()

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Output: nothing

I'm not sure if erroring would be OK. It would not output anything to STDOUT, only STDERR.

HyperNeutrino, Python 3

Password: raise SystemExit

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Output: nothing

ThisIsAQuestion, Python 2.7

True=False The Flag 

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Output is The Flag. Sadly reassigning True doesn't work in Python 3 anymore.

Perl 5, cracks Nahuel Fouilleul

Password: "^^@@^^:@@@^"^".,).*||,!'|"

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SE - stop firing the good guys, ><>

One password is

5 8a*3+o ab*1+o ab*1-o aa*1+o ab*6+o aa*4+o aa*5+o ab*o aa*3+o 48*o ab*5+o ab*1-o aa*1+o aa*8+o aa*8+o ab*5+o 48*o aa*o aa*1+o aa*8+o aa*5+o aa*1-o aa*5+o ab*1+o ab*7+o ab*5+o 95*1+o 95*1+o 95*1+o; 

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The program was i10p. i reads one character of input 1 and 0 push 1 and 0, and p changes the value at x=1, y=0 to the inputted character.

Whit the first input the program is modified to i50p, which allows to execute arbitrary commands from input.
0-9a-f push their hexadecimal value to the stack, *+- work as expected, o outputs a value as a character and ; terminates the program.

SunnyMoon, !@#$%^&*()_+

1728 

I read the code, and it seemed to parse decimal input, divide it by 48 (not halting if it's not divisible) and print the corresponding character 3 times.

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Kevin Cruijssen, 05AB1E

The password is

2 ,*xžIž?¶ 

The .E builtin seems to push the Python code back to stack when it failed. The first input is just any number for the loop. The second input is a reversed 05AB1E program to produce the correct output:

¶ # push newline character ? # print without trailing newline žI # push 2**31 žx # push 2**6 = 64 * # multiply these numbers => 2**37 = 137438953472 , # print this with trailing newline # since there was explicit output, implicit is now disabled 

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After some more experimenting, I found a cleaner password:

3 print() 3 

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.E executed on "print()" returns None, reversed is enoN. This is then executed as 05AB1E code, which seems to return the right result. I'm not sure how though, there might some features of the legacy version used here that I don't know of. Beacuse Python was used to print the newline, the value is still implicitly outputted.

Python 3.8 pre-release, pxeger

("unittest.mock",("mock","sentinel","pxeger","name"),()) 

Probably not the intended solution.

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Ruby, Dingus' answer

system gets;exit echo '"""' 

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r3mainer, C

I'm really sorry, but this answer on Stack Overflow has code that directly undoes elementary xorshift operations... It made running xorshift in reverse much easier (I only had to copy-paste some code and write 6 lines of Python).

07RtUrVE 

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Conor O'Brien, Javascript

^^^Z_][_\\\\\]_\\]]]]]]]]]]]]]]]]]]]]]_\\\]_\\\\\\\]_\\\\\\\\\\\\\\\\\\\\\\\\]b_\\\\\\\\\]aa 

The code seems to generate a random maze (the seed is based on our input, but I don't use that) without loops. We must solve it (in at most 999 steps) with a program in a stack-based language with these 8 commands:

0, rotate stack: S.push(S.shift()) 1, pop twice, compare and push difference: [a,b]=S.splice(-2);F=a==b;S.push(a-b) 2, increment register: M++ 3, push register: S.push(M) 4, pop and discard: S.pop(); 5, reset register: M=0 6, pop to register: M=S.pop(); 7, set answer to ToS and halt: R=S.pop();O=[] 8, pop pop() numbers, and if the last equality comparison was true then insert the numbers into the program: n=S.pop();n=S.splice(-n);if(F)O=n.concat(O) 

The program receives its input as the lengths of the lines of sight in 4 directions, and must output one or more commands in base 4 (digits 0 and 3 are errors, 1 is "rotate left" and 2 is "move forward").

My algorithm checks if the line of sight to the left is zero; if it's zero, then it rotates right (by returning 21), otherwise it rotates left and moves forward (by returning 9). I might have confused left and right completely, though.

Our "program" is obtained from our input by concatenating all char codes in it (as decimal integers). However, nines are discarded. So I used characters with codes from 90 to 98 to access the commands.

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Dingus, Ruby and Dingus, Ruby

One password for both Ruby challenges by Dingus is:

G=->*x{i=0;i+=1 while x[0][i]!~x[1];x[0][i]};Q=method G[methods,/^[o-q]ri/];C=method G[q=Q[],/[o-r]ut[b-d]/];D=method G[q,/^de/];P=G[q,/^[o-r]\z/];D[P]{|x|};C[34];C[34];C[34];C[10] 

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• Gets access to putc by listing all the available (private) methods and filters them with regexen.
• Also redefines p to not do anything.
• a lambda is used instead of def
• method :method_name with [] is used instead of ()
• . cannot be used, so only the methods for Object and Kernel are available
• Since putc is available, any string could be written.

FORTRAN like code:

GREP = lambda do |l,r| i=0 while l[i]!~r do i+=1 end l[i] end PRIVATE_METHODs=method GREP[methods,/^[o-q]ri/] PUTC=method GREP[PRIVATE_METHODs[],/[o-r]ut[b-d]/] DEFINE_METHOD=method GREP[PRIVATE_METHODs[],/^de/] P=GREP[PRIVATE_METHODs[],/^[o-r]\z/] DEFINE_METHOD[P]{|x|} PUTC[34] PUTC[34] PUTC[34] PUTC[10] 

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R, Robin Ryder

function(x,y,z){if(length(ls(1))>1|length(ls())!=3)return("S");LETTERS[lengths(lapply(y,intToUtf8(x),z))*lengths(lapply(y,intToUtf8(x+32),z))]} 

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The password is pretty insecure: 94,(numeric vector of length 1),(numeric vector of length 6). In particular, 94,1,1:6 was what I used.

The first thing I did was note that the lengths have to multiply to 18, so we need to generate two lists such that the lengths are equal to 1,18,2,9,or 3,6.

Next, noting the intToUtf8(x) and intToUtf8(x+32) I found all pairs of functions that satisfy that condition with this script. I safely ruled out nrow and ncol since they would generate the same values, and the lengths must be distinct.

After that it was a matter of just trying things out; lapply always returns a list with length equal to the length of its first argument, so y had to be of length 1. Luckily, R recycles, so lapply(1,"^",1:6)==list(1^(1:6)), which has lengths equal to 6.

Finally, ~, the formula builder is very odd, x ~ y is a formula with length 3, with three elements, '~'(), x(), and y(), so lapply(1,"~",1:6 is the same as 1 ~ 1:6 which is also length 3.

Python 3, breaks pxeger's challenge

__builtins__ SyntaxError __build_class__ 

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The first line loads the dictionary of Python's builtins.

Then, we assign __build_class__ to SyntaxError. Why? Because __build_class__ is called like so

__build_class__(<class_body>, "class_name") 

While SyntaxError is called like so

SyntaxError("msg", (filename, lineno, offset, line)) 

When Python tries to call SyntaxError as if it were __build_class__, it tries to index the class name as a tuple.