Can you outgolf me? (Robbers section)

Question

Robbers section

The cops section can be found here.

Challenge

Your task is to outgolf the submissions of the cops in the same language and the same version (for example, Python 3.5 ≠ Python 3.4, so that is not allowed). A submission is outgolfed when the length in bytes is shorter than the original submission. You only need to golf off at least 1 byte in order to crack a submission. E.g. if the task was to perform 2 × n, and the submission was the following:

print(2*input())

You could outgolf the cop by doing the following:

print 2*input()

Or even this (since lambda's are allowed):

lambda x:2*x

Post this with the following header:

##{language name}, <s>{prev byte count}</s> {byte count}, {cop's submission + link}

For example:

Python 2, 16 12 bytes, Adnan (+ link to submission)

lambda x:2*x

Computes A005843, (offset = 0).

In that case, you have cracked the submission.

Scoring

The person with who cracked the most submissions is the winner.

Rules

• The crack submission must be in the same language as the cop submission.
• The same input should result into the same output (so a(2) = 4 should remain 4).
• For languages such as Python, you can import libraries that are standard included within the language. (So, no numpy/sympy etc.)
• Input and output are both in decimal (base 10).

---

Note

This challenge is finished. The winner of the Robbers section is feersum. The final scores for the CnR are shown below:

• feersum: 16 cracks
• Dennis: 12 cracks
• Leaky Nun: 6 cracks
• Lynn: 4 cracks
• miles: 3 cracks
• Martin Ender: 2 cracks
• Emigna: 2 cracks
• jimmy23013: 1 crack
• Sp3000: 1 crack
• randomra: 1 crack
• alephalpha: 1 crack
• nimi: 1 crack
• Destructible Watermelon: 1 crack
• Dom Hastings: 1 crack

Answer 1

Cheddar, 7 6 bytes, Downgoat

(<<)&1

This seems to work, but it's always possible that I don't understand the language correctly.

Answer 2

Jelly, 5 4 bytes, George V. Williams

RÆḊḞ

Try it here.

A hidden feature!

If I remembered correctly, ÆḊ(A) = sqrt(det(AA^T)) is n! times the n dimensional Lebesgue measure of a simplex formed by n input point and the origin in m dimensional space. When n=1 it degenerate to the Euclidean distance. Not that weird after all...

Answer 3

Hexagony, 91 33 bytes, Blue

1""?{\>{+/</+'+./_'..@'~&/!}'+=($

Unfolded:

    1 " " ?
   { \ > { +
  / < / + ' +
 . / _ ' . . @
  ' ~ & / ! }
   ' + = ( $
    . . . .

Try it online!

Still looks somewhat golfable but I figured I'd post it before FryAmTheEggman beats me to it. ;)

Explanation

Here are some colour-coded execution paths:

However, these are unnecessarily convoluted due to golfing. Here is the exact same code with a saner layout:

That's better. And finally, here is a memory diagram, where the red arrow indicates the initial position and orientation of the memory pointer (MP):

The gist is that I'm iteratively computing Fibonacci numbers on the three edges labelled f(i), f(i+1) and f(i+2) while keeping track of the iterator on the edges A, B and C. While doing so the roles of these edges are swapped out cyclically after each iteration. Let's see how this happens...

The code starts on the grey path which does some initial setup. Note that f(i) already has its correct initial value of 0.

1   Set edge f(i+1) to 1.
""  Move the MP to edge A.
?   Read input n into edge A.
)   Increment n.

Now the green path is the main loop. _ and > are just mirrors.

(     Decrement n.
<     If the result is zero or less, continue on the red path, otherwise
      perform another iteration of the main loop.
{     Move the MP to edge f(i+2).
+     Add edges f(i) and f(i+1) into this edge, computing the next Fibonacci number.
'     Move the MP to the edge opposite A.
~     Multiply by -1 to ensure that it's non-positive (the edge may have a positive
      value after a few iterations).
&     Copy the current value of n from A.
'     Move back and to the right again.
+     Copy n by adding it to zero. Since we know that the other adjacent edge
      is always zero, we no longer need to use ~&.
'+'+  Repeat the process twice, moving n all the way from A to B.
=     Reverse the orientation of the MP so that it points at f(i) which now
      becomes f(i+2) for the next iteration.

This way, the MP moves around the inner triplet of edges, computing successive Fibonacci numbers until n reaches zero. Then finally the red path is executed:

{}    Move the MP to f(i).
!     Print it.
@     Terminate the program.

---

Diagrams generated with Timwi's HexagonyColorer and EsotericIDE.

Answer 4

Haskell, 5 4 bytes, xnor

(^)1

Simple currying.

Answer 5

Stack Cats, 14 13 bytes, feersum

^]T{_+:}_

with the -nm flags for +4 bytes. Try it online!

Okay, that loop was nuts. I tried several approaches, such as brute forcing over a reduced alphabet and brute forcing 3x+2 or 5x+4 and trying to extend it, but I never expected the solution to actually contain a loop.

The best way to see how this works is to add a D flag for debugging (so run with -nmD) and turn debug on for the above TIO link. A {} loop remembers the top of stack at the beginning of the loop, and exits when the top of stack is that value again. The interior of the loop does some fun subtracting and cycling of the top three elements of the stack, which is how the loop gets to run for so many iterations.

Answer 6

Sesos, 14 11 bytes, Leaky Nun

Computes n². Try it here.

Hex dump:

0000000: 16c0f7 959d9b 26e83e ce3d                         ......&.>.=

From assembly:

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

Answer 7

Woefully, 776 759 bytes, Destructible Watermelon

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I tried to read the source code for this language but it was too confusing. For one thing, ip[1] is a line number while ip[0] is the column number, while the cp coordinates are used the other way around. Yet, sometimes the value of cp is assigned to ip. I gave up on trying to understand what the program is doing and found a way to encode the identical sequence of instructions using fewer bars.

Answer 8

Brachylog, 27 21 bytes, Fatalize

yrb:1a:+a:[1]c*.
:2/.

Try it online!

Answer 9

J, 17 12 bytes, miles

+/@(]!2*-)i:

Pretty much the same as the original just more golfed. :)

i: having +1 range compared to i. is useful (and weird). If you use i. here n=0 will be incorrect but luckily i: solves that.

Try it online here.

Answer 10

M, 10 6 bytes, Dennis

R×\³¡Ṫ

Given n, it computes the n^th-level factorial of n. This was a fun exercise!

The code is capable of running as Jelly so you can Try it online.

Explanation

R×\³¡Ṫ  Input: n
R       Create the range [1, 2, ..., n]
   ³¡   Repeat n times starting with that range
 ×\       Find the cumulative products
     Ṫ  Get the last value in the list
        Return implicitly

Answer 11

Snowman, 50 44 bytes, Doorknob

((}#2nMNdE0nR1`wRaC2aGaZ::nM;aF;aM:`nS;aF*))

Try it online!

Answer 12

Haskell, 15 14 bytes, xnor

until odd succ

I spent a fruitless couple of hours learning to decipher "pointless" syntax... until I found this instead.

Or for a less mellifluous 13 bytes, until odd(+1).

Answer 13

Python 2, 43 40, xsot

g=lambda n:n<2or-~sum(map(g,range(n)))/3

Answer 14

Pyke, 11 9 bytes, muddyfish

hVoeX*oe+

Try it here!

How it works

          Implicit input: n (accumulator), n (iterations)
h         Increment the number of iterations.
 V        Do the following n + 1 times.
  o         Iterator. Pushes its value (initially 0) and increments it.
   e        Perform integer division by 2.
            This pushes 0 the first time, then 1, then 2, etc.
    X       Square the result.
     *      Multiply the accumulator and the result.
      oe    As before.
        +   Add the result to the accumulator.
            This sets the accumulator to a(0) = 0 in the first iteration and
            applies the recursive formula in all subsequent ones.

Answer 15

05AB1E, 7 4, Emigna

LnOx

From the formula for the sum of squares of positive integers 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2*n+1)/6, if we multiply both sides by 2 we get Sum_{k=0..n} 2*k^2 = n(n+1)(2*n+1)/3, which is an alternative formula for this sequence. - Mike Warburton (mikewarb(AT)gmail.com), Sep 08 2007

Answer 16

Jelly, 22 21 bytes, Dennis

_²×c×Ḥc¥@÷⁸÷’{S
‘µR+ç

I spent several hours reading Jelly source code for the last one, so I might as well put this "skill" to use. I hope @Dennis will share with us his mathematical discoveries allowing a shorter formula (assuming there is something and not only weird Jelly tricks!).

Answer 17

J, 20 19 bytes, miles

[:+/2^~+/@(!|.)\@i.

This computes the product as a sum of squared Fibonacci numbers, which are calculated as a sum of binomial coefficients.

Thankfully, @miles himself posted the code to generate Fibonacci numbers in this comment.

Answer 18

Acc!!, 526 525 bytes, DLosc

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
}

I have no idea how this works, but I was able to spot a tiny improvement.

24c24
< _+_/100^i*100^i*9
---
> _*10-(_%100^i)*9

Answer 19

Haskell, 10 bytes, xnor

gcd=<<(2^)

Usage example: map ( gcd=<<(2^) ) [1..17] -> [1,2,1,4,1,2,1,8,1,2,1,4,1,2,1,16,1].

How it works: From the oeis page we see that a(n) = gcd(2^n, n) or written in Haskell syntax: a n = gcd (2^n) n. Functions with the pattern f x = g (h x) x can be turned to point-free via the function =<<: f = g =<< h, hence gcd=<<(2^) which translates back to gcd (2^x) x.

Answer 20

Sesos, 14 9 bytes, Leaky Nun

Computes n mod 16. Try it here.

Hex:

0000000: 17f84a 750e4a 7d9d0f                              ..Ju.J}..

Assembly:

set numin
set numout
set mask
get
jmp, sub 1, fwd 1, add 16, rwd 1, jnz
fwd 1
jmp, sub 16, fwd 1, add 1, rwd 1, jnz
fwd 1
put

Answer 21

Python, 39 17 bytes, Destructible Watermelon

lambda n:n*-~n>>1

Answer 22

05AB1E, 9 4 bytes, Emigna

>n4÷

Try it online!

Computes this function instead:

Answer 23

Hexagony, 7 6 bytes, Adnan

?!/$(@

Unfolded:

 ? ! 
/ $ (
 @ .

Try it online!

Same idea, slightly different layout.

Answer 24

MATL, 11 10 bytes, Luis Mendo

YftdAwg_p*

Instead of doing -1^length(array) it converts the elements to Boolean values (which are always 1), negates them, and takes the product of the elements.

Answer 25

Jelly, 11 10, Dennis

Ḥ’_Rc’*@RP

Vectorized version of same approach.

Try it online!

Answer 26

05AB1E, 10 8 bytes, Adnan

µNÂÂQ>i¼

Try it online.

Answer 27

Jelly, 9 8 bytes, Dennis

œċr0$L€Ḅ

Sorry! I wasn't able to find your intended solution.

This relies on the fact that C(n+k-1, k) is the number of ways to choose k values from n with replacement.

Note: This is inefficient since it generates the possible sets in order to count them, so try to avoid using large values of n online.

Try it online or Verify up to n.

I later found another 8 byte version that is efficient enough to compute n = 1000. This computes the values using the binomial coefficient and avoids generating the lists.

Ḷ+c’Ṛ;1Ḅ

Try it online or Verify up to n.

Explanation

œċr0$L€Ḅ  Input: n
  r0$     Create a range [n, n-1, ..., 0]
œċ        Create all combinations with replacement for
          (n, n), (n, n-1), ..., (n, 0)
     L€   Find the length of each
       Ḅ  Convert it from binary to decimal and return

Ḷ+c’Ṛ;1Ḅ  Input: n
Ḷ         Creates the range [0, 1, ..., n-1]
 +        Add n to each in that range
   ’      Get n-1
  c       Compute the binomial coefficients between each
    Ṛ     Reverse the values
     ;1   Append 1 to it
       Ḅ  Convert it from binary to decimal and return

Answer 28

Brachylog, 11 10 bytes, Fatalize

yb:AcLrLc.

Try it online!

Explanation

Brachylog is a Prolog-derived languages, whose greatest ability is to prove things.

Here, we prove these statements:

yb:AcLrLc.
yb:AcL       Inclusive range from 1 to input, concatenated with A, gives L
     LrL     L reversed is still L
       Lc.   L concatenated is output

Answer 29

M, 9 8 bytes, Dennis​

Ḥrc’ḊḄḤ‘

Answer 30

QBasic, 30 29 bytes, DLosc

INPUT n:?(n MOD 2)*(n+.5)+n/2

Answer 31

Pip, 24 22 bytes, DLosc

Y5T#y>aY(A_My)JkyA(ya)

Answer 32

J, 12 11 bytes, miles

##.>:@I.@#:

+/ for summing an array behaves poorly with regards to vectorization, so I tried to use base 1 instead. # happened to work for a 1-byte verb that results in 1 when called on a scalar.

Answer 33

Python 3.5, 38 36 bytes, R. Kap

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

Even if you don't accept True as 1, this is good enough.

Answer 34

J, 9 7 bytes, Leaky Nun

2&o.t.]

This is a monadic verb; it computes the y^th term of the Taylor series of the cosine function (2&o.).

Example run

   f =: 2&o.t.]
   f each 0 1 2 3 4 5 6 7
┌─┬─┬──┬─┬─┬─┬──┬─┐
│1│0│_1│0│1│0│_1│0│
└─┴─┴──┴─┴─┴─┴──┴─┘

Answer 35

J, 10 9 bytes, Leaky Nun

-:*1-~3*]

Just shifted the operators around.

Answer 36

Java, 53 51 bytes, JollyJoker

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

Combines the first two if statements into 3-3*n for n less than 2.

Answer 37

Retina, 28 27 bytes, Martin Ender

.+
$*
((^.?|\3)(^|\1)){99}$

Where it says {99} I wanted to use +, but that mysteriously causes the match to fail.

The regex matches strings which have a length that is a sum of a prefix of the Fibonacci numbers. The main group matches lengths starting with 1, 2, 3, ..., which skips one of the initial 1s, so I tried to improve it by getting it to match length 1 twice at the beginning, so that it was not necessary to have a special case.

Answer 38

05AB1E, 13 12 bytes, Adnan

µ•vÉ•DNìÙ{Q½

Try it online

Answer 39

Python 3, 77 40 bytes, Joe

g=lambda n,c=1:+(n==0 or n>0<g(n-c,c+1))

Answer 40

MarioLANG, 87 86 bytes, Business Cat

Don't know how I did it, but here it is I think

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

Answer 41

CoffeeScript, 83 50 bytes, DerpfacePython

s=(n)->0|r<0||Array(0|(n/9)+2).join ((n-1)%9+1)+''

Tested via http://coffeescript.org/.

Answer 42

Python, 69 32 bytes, HiggsBot

f=lambda n:0**n or(4*n-2)*f(n-1)

Test it on Ideone.

Two bytes saved by @xnor. Thanks! Two more could be saved by returning True instead of 1.

Answer 43

Actually, 28 11 bytes, Mego

19,`;τ(+`nX

Try it online!

How it works

This uses the recursive definition from A048696: a(0) = 1, a(1) = 9, a(n) = 2a(n-1) + a(n-2).

19           Push 1 and 9 on the stack.
  ,          Read n from STDIN and push it on the stack.
   `    `n   Do the following n times:
    ;          Push a copy of the topmost integer.
     τ         Multiply it by 2.
      (        Rotate the bottom-most integer on top of the stack.
       +       Add the two topmost integers.
             Each iteration turns the stack [a(k-2) a(k-1)] into [a(k-1) a(k)].
             After n iterations, we are left with [a(n) a(n+1)].
          X  Discard a(n+1) from the stack.

Answer 44

05AB1E, 7 6 bytes, Adnan

!¹L!/O

or

!žr<*ï

Note that both of these return 1 for 0, but so did the cop program.

Answer 45

Acc!!, 522 514 bytes, DLosc

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+2-j {
_+(_%10-_/100^j%10)*(100^j-1)
}
}
_/100
Count j while _/100^j {
Write _/100^j%10+48
}

Not so tiny improvement this time, but I still have no idea what I'm doing.

28,29c28,29
< Count j while i+1-j {
< _+(_%10-_/100^(j+1)%10)*(100^(j+1)-1)
---
> Count j while i+2-j {
> _+(_%10-_/100^j%10)*(100^j-1)

Since (100^j-1) is zero when j = 0, we can loop over the range [0, ..., i + 2) instead of looping over [0, ..., i + 1) and incrementing j. This saves eight bytes.

Answer 46

Python, 46 45 bytes, Sp3000

f=lambda n,k=1:n and-~f(n-(k+(k&-k)&k>0),k+1)

Test it on Ideone.

Answer 47

Excel, 16 12 bytes, Anastasiya-Romanova 秀

=FACT(n)*2^n

I don't actually have Excel, but from what I read online, this should work.

Answer 48

M, 18 17 bytes, Dennis

Tragically, Dennis has missed a trivial modification this time. Only the first line is different.

Ḥrc
‘Ḥc0r$×Ç:‘+\S

Answer 49

Acc!!, 512 511 bytes, DLosc

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 {
_+3^(3*k+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
}

This is the improvement.

15c15
< _+27^k*3^(i%2)*78
---
> _+3^(3*k+i%2)*78

Answer 50

J, 18 13 bytes, miles

(2&-%-.-*:)t.

I checked the shortest most popular J answer for Fibonacci numbers: a generating function. Unsurprisingly, the same approach is also good for Lucas numbers.

Answer 51

JavaScript(ES6), 77 76, user81655

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

Just a 1-byte peephole golf.

Answer 52

Acc!!, 523 521 bytes, DLosc

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+i%2*6
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
}

I have no idea how this works, but I was able to spot a tiny improvement.

13c13
< _+3^(1+i%2)
---
> _+3+i%2*6