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

Pip, 24 22 bytes, DLosc

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

Answer 2

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 3

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 4

J, 10 9 bytes, Leaky Nun

-:*1-~3*]

Just shifted the operators around.

Answer 5

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 6

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 7

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 8

05AB1E, 13 12 bytes, Adnan

µ•vÉ•DNìÙ{Q½

Try it online

Answer 9

Python 3, 77 40 bytes, Joe

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

Answer 10

MarioLANG, 87 86 bytes, Business Cat

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

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

Answer 11

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 12

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 13

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 14

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 15

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 16

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 17

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 18

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 19

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 20

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 21

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 22

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