Robbers - square times square root

Question 1

Task: Crack the scrambled code for multiplying the square root of an integer n by the square of it!

You must post a comment in the cops' thread with a link to your working source, mentioning clearly that you have Cracked it. In your answer's title, you must include the link to the original answer.

Rules:

You may only change the order of the characters in the original source.
Safe answers cannot be cracked anymore.
The other rules mentioned in the cops' thread
Please edit the answer you crack

WINNER: Emigna - 10 submissons (had some trouble counting)

Honorable mentions: Notjagan, Plannapus, TEHTMI

Question 2

JavaScript (ES7), Neil

_26_=>_26_**6.25**.5

The hard part, of course, was figuring out what to do with all the extra characters. (And also not posting this solution in the wrong thread, like I accidentally did at first. Oopsie...)

Question 3

@SethWhite: I had to use all the characters that Neil used in his scrambled code, otherwise this wouldn't have been a valid crack.

Question 4

How does this work?

Question 5

@Arjun _26_=> defines an anonymous function taking one parameter called _26_ (variables may start with an underscore but not a number). Then the remainder is just using ** as Math.pow() to raise the input to the power of 2.5 (6.25 power 0.5).

Question 6

Ah! I was thinking that _26_ was something ES7 specific. Didn't know that variables can be in this form too! (I have never seen a variable without an alphabet). That was a very clever approach by @Neil. And you were very clever too in cracking it! Have your well-deserved +1! :)

Question 7

Python, 15 bytes, Mr. Xcoder

lambda x:x**2.5

Pretty simple. Just takes x and raises it to the 2.5th power.

Question 8

OCaml, 13 bytes, shooqie

fun f->f**2.5

Try it online!

Question 9

I was posting it right now! Cannot believe it! Good job!

Question 10

Python 3, 44 bytes, Kyle Gullion

Those *s were quite misleading. Very clever!

lambda i:i**(lambda o,r:o/r)(*map(ord,'i*'))

Due to the quite limited character set I would be very surprised if there were any other valid solutions beyond trivial renaming or reordering of arguments.

Question 11

You got me, nicely done!

Question 12

C++ (gcc), 100 bytes, Mr. Xcoder

#include<math.h>
#include"iostream"
using namespace std;int main(){float n;cin>>n;cout<<pow(n,2.5);}

Try it online!

Question 13

Well done! Please edit the answer you crack, because I cannot do it now!

Question 14

An anonymous user suggested an edit to save 8 bytes by removing the using and just doing int main(){float n;std::cin>>n;std::cout<<pow(n,2.5);}

Question 15

Save another byte #including <cmath> instead of math.h.:)

Question 16

Haskell, Leo

x=exp.(2.5*).log

A pointfree function named x. Usage: x 4 -> 32.0

Question 17

Inform 7, corvus_192

Cool, an Inform7 entry. :) I just had to give this one a try.

I'm pretty sure this is the intended solution:

R is a room.
To f (n - number): say "[n * n * real square root of n]".

Note that this solution only works if compiled with the Glulx back-end, due to the use of the real square root of function.

BTW, the double quotes and square brackets are actually unnecessary; just say n * n * real square root of n would work just as well. The periods at the end of the commands could be omitted, too; or we could keep the first period and get rid of the newlines instead. Other parts of the code we could trim away include the article "a" before "room" and the spaces before the parentheses and after the colon. Fortunately, since we've got a spare pair of brackets, we can always use them to comment out all these extra characters. ;) So this is a valid solution, too:

R is room.To f(n - number):say n * n * real square root of n[
" a . "
]

To test this solution interactively, it's convenient to append something like the following test harness to the code:

Effing is an action applying to one number.
Understand "f [number]" as effing.
Carry out effing: f the number understood.

After compiling and running the program, you can type e.g. f 4. f 6. f 9. f 25 at the > prompt and receive something like the following output:

Welcome
An Interactive Fiction
Release 1 / Serial number 170404 / Inform 7 build 6L38 (I6/v6.33 lib 6/12N) SD
R
>f 4. f 6. f 9. f 25
32.0
88.18164
243.0
3125.0
>

BTW, I just noticed that Inform (or presumably, rather, Glulx) rounds the last decimal place of f 6 wrong: the correct value is much closer to 88.18163 than to 88.18164. Fortunately, I don't think this affects the correctness of the solution(s), especially since the challenge specified "any rounding mechanism of your choice". :)

Question 18

I'm sure I found an Inform 7 fiddle a while ago, but I can't find it when I search. Do you know if there is an online interpreter anywhere?

Question 19

@Flounderer: I don't really know of any. There are certainly online (even JS-based) players for the Glulx / Z-machine bytecode produced by the Inform 7 compiler, but I'm not aware of anything that would directly take plain Inform 7 source code and compile and run it online. The compiler / IDE is pretty easy to install, though; on Ubuntu Linux, it's as easy as apt-get install gnome-inform7.

Question 20

Mathematica, Greg Martin

f[y_]:=With[{x=
 #&@@{#(#)#^(1/(1+1))&@y,#&@@@{1^(1),-1}}
},Print[#,".",IntegerString[Round@#2,10,3]]&@@QuotientRemainder[1000x,1000]]

Thanks for leaving the rounding stuff intact!

Explanation: #(#)#^(1/(1+1))&@y does the main work of multiplying y squared, aka y(y), and y's square root, y^(1/(1+1)). The #&@@@{1^(1),-1} bit is just junk to use up the other letters, and #&@@ picks out the useful bit from the junk.

Question 21

MATL, 12 bytes, Luis Mendo

10'U&+:'n/^P

Calculate 10/4 = 2.5 with 4 coming from string length. Use this as an exponent. P is a no-op here.

Question 22

NO, 41 bytes, This Guy

NOOOOOO?NOOOOOOOOOOO!nnOOOOO
NOOOOOOOO?no

Question 23

Really? Well at least I got my language known.

Question 24

@ThisGuy: You've made some fun languages :) Not terribly useful in general though ;)

Question 25

Look at the 2nd paragraph

Question 26

R, Steadybox

y<-function(p)(p^(1/2)*p^2);

seems to be an anagram of funny(p1)-tio(^*^)/pc(2)<p2;

Question 27

Python 2, 60 bytes, Anthony Pham

print (input()**(5.0/(2*5554448893999/5554448893840))-0)

Based on discarding characters through Python 2's float division (the default for / between integers).

Question 28

C, 50 bytes, Dave

double b(float \ufa2d){return pow(\ufa2d,25e-1);%>

Try it online!

This requires -lm compiler flag, but I don't know how it would be possible to solve this without it.

Question 29

correct, and almost exactly what I had (variable was named \uad2f on mine). Well done; I thought I'd left enough red-herrings in there to keep people busy a lot longer! Also the -lm flag wasn't needed for me using Clang (I'd have mentioned it!) but you're correct that strictly speaking it's required.

Question 30

R, Flounderer

This is a crack of the 33 bytes solution of @Flounderer

scan()^(floor(pi)-1/2)-sin(7*0e1)

Usage:

> scan()^(floor(pi)-1/2)-sin(7*0e1)
1: 4
2: 
Read 1 item
[1] 32
> scan()^(floor(pi)-1/2)-sin(7*0e1)
1: 6
2: 
Read 1 item
[1] 88.18163
> scan()^(floor(pi)-1/2)-sin(7*0e1)
1: 9
2: 
Read 1 item
[1] 243
> scan()^(floor(pi)-1/2)-sin(7*0e1)
1: 25
2: 
Read 1 item
[1] 3125

Question 31

Yes, this is not the intended solution which involved sin(pi), but unfortunately it does work! +1

Question 32

It was scan()^(-floor(-sin(pi)*2e17)/10)

Question 33

RProgN 2, ATaco

]2^\š*

Apparently StackExchange needs extra characters, so here you go.

Question 34

My solution was ]š2円^*, but they both work the same way.

Question 35

HODOR, 198, This Guy

Walder
Hodor?!
hodor.
Hodor, Hodor Hodor Hodor Hodor Hodor Hodor Hodor, Hodor Hodor,
hodor,
Hodor, Hodor Hodor Hodor Hodor Hodor Hodor, Hodor Hodor,
Hodor, Hodor Hodor Hodor, hodor!,
HODOR!!
HODOR!!!

Explanation:

Start
read input into accumulator
copy accumulator to storage
Do math, option 7(nth root), n=2
swap storage and accumulator
Do math, option 6(nth power), n=2
Do math, option 3(times), storage
output accumulator as a number
end

note: I had to make some modifications to the get interpreter to run on my machine (the one you have posted doesn't seem to accept lowercase h, among some other things)

Also, I don't seem to have enough rep to comment, so if someone could let @This Guy know, I would be grateful

I think this fixed the error, the code now starts with Walder instead of Wylis, which adds the extra byte

Question 36

Almost there. Had to change a mistake in my code so yours ins't quite right.

Question 37

fixed the start command

Question 38

Though that you had it but you went for a different method than me that uses the same number of bytes. Your answer would do what you wanted it to do but it isn't the same as mine. Also I had to change the name because of this so I made an edit to your posts.

Question 39

for clarification, does this still count as a crack, or do I need to match up exactly?

Question 40

I'm not exactly sure. I'm going to ask in the comments. +1 your solution is making me think really hard!

Question 41

C#, 172 bytes, raznagul

Hardest part was figuring out what to do with all the leftovers.

using System;using S=System.Console;class PMabddellorttuuv{static void Main(){S.Write(Math.Pow(double.Parse(S.ReadLine()),2.5));Func<double> o;int q=1,M=q*2,b,e;q*=(q*M);}}

Question 42

Nice! I got stuck on the leftovers as well. Didn't think to add them to the class name, doh!. Good job!

Question 43

+1 Not what I had intended, but very nice solution. ;)

Question 44

EXCEL, 26 Bytes pajonk

=SQRT(A1)*A1^2/1/ISNA(IP2)

A1 as input IP2 contain a second input with a #N/A Error in this case ISNA(IP2) belongs to 1

For an additional () we can do this

=SQRT(A1)*A1^2/ISNA(PI(1/2))

Question 45

Is second input allowed in such challenge?

Question 46

@pajonk The rules say "You can take input in any standard way" so I have assume that you make this trick with a second input

Question 47

@pajonk I have improve my post it could been that you have make a little mistake

Question 48

The standard way for taking one number is (I think) taking one input. In my opinion the second input would be unfair and against the rules. PS There's no mistake in number of brackets.

Question 49

@pajonk I can not use combinations with SIN and PI cause there are not enough brackets. =SQRT(A1)*A1^2/SIN(PI()/2) If you set the #NA Error through formating or something else I would see it as additional second input. SQRT and ISNA are the only two functions that make a little sense. But please ask the man who had developed the question

Question 50

Python 3.6, 64 bytes, Mr. Xcoder

php38af4r2aoot2srm0itpfpmm0726991i= (lambda x:x**2.5*1*1/1);

Maybe not what was intended, but works ;)

$ python3
Python 3.6.1 (default, Apr 4 2017, 09:36:47) 
[GCC 4.2.1 Compatible Apple LLVM 7.0.2 (clang-700.1.81)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> php38af4r2aoot2srm0itpfpmm0726991i= (lambda x:x**2.5*1*1/1);
>>> php38af4r2aoot2srm0itpfpmm0726991i(6)
88.18163074019441
>>> php38af4r2aoot2srm0itpfpmm0726991i(4)
32.0
>>> php38af4r2aoot2srm0itpfpmm0726991i(25)
3125.0

Not enough rep to comment on the cops' thread's answer yet, sorry... Would appreciate if someone could do it for me, thanks!

Question 51

@Mr. Xcoder , as said above, can't comment on your post in the cops' thread, sorry about that :)

Question 52

Came to post a similar solution from math import pi as pp0012223467899;f=lambda x:x**2.5*1*(1)/1 just to see I've been beaten to the punch. I've added a link to this post on the cops thread for you.

Question 53

Röda, 28 bytes, fergusq

f(n){push n^0.5,n^2|product}

Question 54

Python 2.7, Koishore Roy

s=e=x=y=input()**0.5
print'%.3f'%(y**(5.0))

Question 55

Or y=x=e=s, both work :)) Good job anyways!

Question 56

Good job. :P Now I need to work on a new code. Darn!

Question 57

R, Flounderer

This is a crack of @Flounderer's 31-byte solution:

`[.`=function(`]`)`]`^`[`(lh,9)

Ok that was a tough one. It creates a function called `[.`. The argument to the function is called `]` which is elevated to power 2.5 by using the 9th element of the built-in time-serie lh ("a regular time series giving the luteinizing hormone in blood samples at 10 mins intervals from a human female, 48 samples." that is used as example in one of R's base packages). lh[9] is here on top of it replaced by its equivalent `[`(lh, 9). De-obfuscated by substituting f for the function name and n for the argument name, the function then becomes f=function(n)n^lh[9].

Usage:

> `[.`=function(`]`)`]`^`[`(lh,9)
> `[.`(4)
[1] 32
> `[.`(6)
[1] 88.18163
> `[.`(9)
[1] 243
> `[.`(25)
[1] 3125

Question 58

Python 2, 44 bytes, Anthony Pham

print int(raw_input())**(0+000000000000.5*5)

Takes input from raw_input, converts to int and raises to power 2.5

Question 59

JavaScript, fəˈnɛtɪk

n=>n**("ggggg".length*2**(-"g".length))// ""((((((()))))))***,-...;;=====>Seeeeegggghhhhhhhhhilllnnnnnnorrrsstttttttttttu{}

Gets 5/2 through 5 times 2 to the negative first power, where 5 and 1 were received from the length of strings. Took the easy way out in a sense by commenting out the extraneous characters.

Question 60

Python 3.6, c..

f=lambda x:x**2.5or'1*77*8+8/5/(((aafothipie.xml)))'

Try it online!

Stuffs all the unneeded characters into a string after the or. Since x**2.5 is nonzero and so truthy, the part after the or isn't evaluated. Any syntactically valid expression would be OK here.

Python parses 2.5 and or as separate tokens in 2.5or, though the syntax highlighter for the code doesn't recognize this.

Question 61

I think you meant x**2.5? Damn Python, I'm doing the next one without strings.

Question 62

C#, 112 bytes, Jan Ivan

using System;class P{static void Main(){var b=Math.Pow(double.Parse(Console.ReadLine()),2.5);Console.Write(b);}}

Question 63

05AB1E, 47 bytes, Okx

).2555BFHIJJKKKPQRS``„cg...ghi...lsw...x}T...Áöž«‚1n1t*

Try it online!

Question 64

Ah, I knew there would be a workaround the method I used to 'secure' the code. Well done!

Ilmari Karonen 21k5 gold badges55 silver badges101 bronze badges · Accepted Answer · 2017-04-03 23:57:46Z

9

\$\begingroup\$

JavaScript (ES7), Neil

_26_=>_26_**6.25**.5

The hard part, of course, was figuring out what to do with all the extra characters. (And also not posting this solution in the wrong thread, like I accidentally did at first. Oopsie...)

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edited Apr 13, 2017 at 12:39

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answered Apr 3, 2017 at 23:57

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Ilmari Karonen

21k5 gold badges55 silver badges101 bronze badges

\$\endgroup\$

4

1

\$\begingroup\$ @SethWhite: I had to use all the characters that Neil used in his scrambled code, otherwise this wouldn't have been a valid crack. \$\endgroup\$

Ilmari Karonen
– Ilmari Karonen

2017年04月04日 17:39:16 +00:00
Commented Apr 4, 2017 at 17:39
\$\begingroup\$ How does this work? \$\endgroup\$

Arjun
– Arjun

2017年04月05日 07:30:29 +00:00
Commented Apr 5, 2017 at 7:30
\$\begingroup\$ @Arjun _26_=> defines an anonymous function taking one parameter called _26_ (variables may start with an underscore but not a number). Then the remainder is just using ** as Math.pow() to raise the input to the power of 2.5 (6.25 power 0.5). \$\endgroup\$

Joe
– Joe

2017年04月05日 09:24:13 +00:00
Commented Apr 5, 2017 at 9:24
\$\begingroup\$ Ah! I was thinking that _26_ was something ES7 specific. Didn't know that variables can be in this form too! (I have never seen a variable without an alphabet). That was a very clever approach by @Neil. And you were very clever too in cracking it! Have your well-deserved +1! :) \$\endgroup\$

Arjun
– Arjun

2017年04月05日 10:01:13 +00:00
Commented Apr 5, 2017 at 10:01

Add a comment |

Rules:

WINNER: Emigna - 10 submissons (had some trouble counting)

49 Answers 49

JavaScript (ES7), Neil

Python, 15 bytes, Mr. Xcoder

OCaml, 13 bytes, shooqie

Python 3, 44 bytes, Kyle Gullion

C++ (gcc), 100 bytes, Mr. Xcoder

Haskell, Leo

Inform 7, corvus_192

Mathematica, Greg Martin

MATL, 12 bytes, Luis Mendo

NO, 41 bytes, This Guy

Python 2, 60 bytes, Anthony Pham

C, 50 bytes, Dave

R, Flounderer

RProgN 2, ATaco

HODOR, 198, This Guy

C#, 172 bytes, raznagul

EXCEL, 26 Bytes pajonk

Python 3.6, 64 bytes, Mr. Xcoder

Röda, 28 bytes, fergusq

Python 2.7, Koishore Roy

R, Flounderer

Python 2, 44 bytes, Anthony Pham

JavaScript, fəˈnɛtɪk

Python 3.6, c..

C#, 112 bytes, Jan Ivan

05AB1E, 47 bytes, Okx

Fireball, 8 bytes, Okx

Haskell, 64 bytes, @nimi

R, steadybox

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