Source permutation

Question

A permutation of a set \$S = \{s_1, s_2, \dotsc, s_n\}\$ is a bijective function \$\pi: S \to S\$. For example, if \$S = \{1,2,3,4\}\$ then the function \$\pi: x \mapsto 1 + (x + 1 \mod 4)\$ is a permutation:

$$ \pi(1) = 3,\quad \pi(2) = 4,\quad \pi(3) = 1,\quad \pi(4) = 2 $$

We can also have permutations on infinite sets, let's take \$\mathbb{N}\$ as an example: The function \$\pi: x \mapsto x-1 + 2\cdot(x \mod 2)\$ is a permutation, swapping the odd and even integers in blocks of two. The first elements are as follows:

$$ 2,1,4,3,6,5,8,7,10,9,12,11,14,13,16,15,\dotsc $$

Challenge

Your task for this challenge is to write a function/program implementing any¹ permutation on the positive natural numbers. The score of your solution is the sum of codepoints after mapping them with the implemented permutation.

Example

Suppose we take the above permutation implemented with Python:

def pi(x):
    return x - 1 + 2*(x % 2)

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The character d has codepoint \$100\$, \$\texttt{pi}(100) = 99\$. If we do this for every character, we get:

$$ 99,102,101,31,111,106,39,119,42,57,9,31,31,31,31,113,102,115,118,113,109,31,119,31,46,31,50,31,44,31,49,41,39,119,31,38,31,49,42 $$

The sum of all these mapped characters is \$2463\$, this would be the score for that function.

Rules

You will implement a permutation \$\pi\$ either as a function or program

• given an natural number \$x\$, return/output \$\pi(x)\$
• for the purpose of this challenge \$\mathbb{N}\$ does not contain \$0\$
• the permutation must non-trivially permute an infinite subset of \$\mathbb{N}\$
• your function/program is not allowed to read its own source

Scoring

The score is given by the sum of all codepoints (zero bytes may not be part of the source code) under that permutation (the codepoints depend on your language², you're free to use SBCS, UTF-8 etc. as long as your language supports it).

The submission with the lowest score wins, ties are broken by earliest submission.

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1. Except for permutations which only permute a finite subset of \$\mathbb{N}\$, meaning that the set \$\{ x | \pi(x) \neq x \}\$ must be infinite.

2. If it improves your score, you can for example use a UTF-8 encoded Jelly submission instead of the usual SBCS.

Answer 1

Jelly, score  288 250 212  199

-38 thanks to Erik the Outgolfer!

C-*+

Swaps even with odd.

The score is \$67+45+44+43=199\$ - see self-scoring here.

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Answer 2

JavaScript (ES6), Score =  276  268

$=>(--$^40)+!0

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Answer 3

Perl 6, Score: 201

*-!0+^40+!0

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Port of Arnauld's answer. This benefits from xor (+^) having the same precedence as - and +, and the use of a Whatever lambda to reduce overall characters. Other than that, I couldn't find a way of representing it differently that got a better score.

Perl 6, Score 804 702

{{(++$∉ords(q[!$%()+-2?[]_doqrsx{}∉])??++$+22-$++%2-$++%2!!++$)xx$_}()[-!$+$_]}

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The first quine-y type answer here, and I think it scores pretty well.

This produces the sequence \$23,22,25,24...\$ from the question body with the range \$1,2,3,4...21\$ inserted at the indexes of the unique sorted codepoints of the code. For example, the 30th through 35th elements of the sequence are \$50, 53, 52, 1, 55, 54\$ since the 33rd codepoint is ! and that's the lowest codepoint in my code.

Answer 4

Python 2 score: 742 698 694 points

lambda a:a^96or~~96

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-44 points thanks to Ørjan Johansen; -4 points thx to xnor.

Answer 5

Retina 0.8.2, 6 bytes, score 260

T`O`RO

Try it online! Link includes self-scoring footer. Simply swaps digits 1 and 9 and 3 and 7 in the decimal representations, so that numbers that contain no digits coprime to 10 are unaffected.

Answer 6

C# (Visual C# Interactive Compiler), 22 bytes, Score 247 245

A=>A>65?A-1+A%2*2:66-A

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Simple, if less than 66, return 66 minus input, else use the formula in the question that swaps even and odd numbers.

Answer 7

TI-BASIC, 9 bytes, score 1088 1051 1000

Ans-cos(π2fPart(2⁻¹Ans

Swaps even with odd. Even maps to Ans-1 and odd maps to Ans+1.

TI-BASIC is tokenized, so this program will have the following hex-values:

Ans   -    cos(  π    2    fPart(  2   ⁻¹  Ans
72    71   C4    AC   32   BA      32  0C  72

Thus the score is: \$113+114+195+171+49+185+49+11+113=1000\$

Output test program:

For(I,1,10
I
Ans-cos(π2fPart(2⁻¹Ans
Disp Ans
Pause
End

Which outputs:

2
1
4
3
6
5
8
7
10
9

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Notes:

• TI-BASIC's token values can be found here.

• Pause is used in the output program to better see the permutation, as the calculator only has 8 lines. Press [ENTER] to view the next permutation.

Answer 8

Charcoal, 13 bytes, score 681

⁻⁺²³²ι⊗﹪⊖ι²³³

Try it online! Link is to self-scoring version with header to map over an array of byte codes. (Charcoal has a custom code page so I've manually inserted the correct byte codes in the input.) Works by reversing ranges of 233 numbers, so that 117, 350, 583 ... are unchanged. Explanation:

     ι          Value
 ⁺              Plus
  ²³²           Literal 232
⁻               Minus
         ι      Value
        ⊖       Decremented
       ﹪        Modulo
          ²³³   Literal 233
      ⊗         Doubled

Answer 9

Haskell, score 985

(\((.),(-))->(.)*200+mod(-39+(-))200+1).(\(*)->divMod((*)-1)200)

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Answer 10

05AB1E, score: 488 in 05AB1E's code page

È·<-

Swaps odd and even like the example function.

Will try to improve the score from here.

Try it online with input in the range [1, 100] or Try it online with the codepoints.

Explanation:

È     # Check if the (implicit) input is even (1 if truthy; 0 if falsey)
 ·    # Double (2 if truthy; 0 if falsey)
  <   # Decrease by 1 (1 if truthy; -1 if falsey)
   -  # Subtract it from the (implicit) input (and output implicitly)

Answer 11

Brainfuck, 47 bytes, score 2988

,[-<+<+>>]<[->[>+<[-]]+>[<->-]<<]>[-<<++>>]<<-.

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I used the permutation given in the introduction. Since this is bijection you can use it as a simple symmetric cipher similar to ROT13 or Atbash. My solution works on unbounded cells. However, by restricting yourself to 8-bit cells, you could save 2 points by replacing [-] with [+].