# Mr. Binary Counterman

Mr. Binary Counterman, son of Mr. Boolean Masker & Mrs. Even Oddify, follows in his parents’ footsteps and has a peculiar way of keeping track of the digits.

When given a list of booleans, he counts the 1s and 0s separately, numbering the 1s with the odds & the 0s with the evens.

For example, when he looks at 1 1 0 0 1 0 he counts: 1st odd, 2nd odd, 1st even, 2nd even, 3rd odd, 3rd even and copies it down as 1 3 2 4 5 6

Mr. Binary Counterman thinks it looks prettier to start counting odds at 1 and evens at 2. However the pattern is more symmetric if you start counting evens at 0. You may do either. So either 1 3 2 4 5 6 or 1 3 0 2 5 4 are good given the list above.

As input you may take any representation of a boolean list or binary number, the output should be the list of resulting numbers with any delimiter. (But the list elements should be separate & identifiable.)

This is , so least bytes wins.

## Test Cases

1 0 1 0 1 0
1 2 3 4 5 6

1 1 1 1
1 3 5 7

0 0 0 0
2 4 6 8

0 1 1 0 0
2 1 3 4 6

0 1 1 0 0 1 0 1 1
2 1 3 4 6 5 8 7 9

0 0 1 0 0 1 1 1
2 4 1 6 8 3 5 7

0
2

1
1

1 1 1 0 0 0
1 3 5 2 4 6

• Can we take the bits flipped? Jun 15, 2021 at 3:30
• Absolutely! Look forward to seeing what you've come up with. Jun 15, 2021 at 3:54

# JavaScript (ES6), 29 bytes

a=>a.map(v=>b[v]+=2,b=[0,-1])


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• Have to laugh at everyone else trying to make this as convoluted as they can while you & I take the most simplistic approach possible and both our solutions are still competitive! Jun 18, 2021 at 0:27

# J, 12 bytes

+2*/:<.&/:\:


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Explanation: Self plus twice the minimum of ascending and descending ranks.

Given a boolean array 1 1 0 0 1 1 1, ascending rank /:@/: and descending rank /:@\: are computed as follows:

array:       1 1 0 0 1 1 1
asc. rank:   2 3 0 1 4 5 6
desc. rank:  0 1 5 6 2 3 4
minimum:     0 1 0 1 2 3 4


# APL(Dyalog Unicode), 9 bytes SBCS

⊢+2×⍋⌊⍥⍋⍒


Try it on APLgolf!

• Wow! Fantastic insight! Jun 15, 2021 at 2:41
• Holy wow! You should def post the APL solution separately. It's one byte away from first! Jun 15, 2021 at 4:04

# Risky, 44 bytes

__0+0+_0+0+__0+0+_0+0+__0+0+_0+0+__0+0+_0+?1__0+0+_0+0+__0+0+_0!-_0!_{1+_0+0_[2_{0+__{1


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How it works:

This is a really low level explanation:

... + __0+0+_0+?                                                  ;  the input array
1                                                ;  map with the following pairs:
... + __0+0+_0!-                               ;  [0, -1]
_                             ;  map to
0!_{1+_0+0                  ;  range with same length
_                ;  map to
[              ;  absolute value
+       ;    of the sum of
2_{0         ;      twice the index in the range and
__{1  ;      the offset (0 or -1)


That's useless, though. Here's a better description of how this works:

Risky has an operator called "map pairs" which takes an array, and maps the items according to a set of rules. The rules are arrays, starting with the item to be replaced, and with (typically) one item to map to. However, if multiple are specified, they'll be used in order.

This answer generates those mappings, which look like [[0, 2, 4, 6, ...], [1, 1, 3, 5, 7, ...]]. It does this by mapping [0, -1] to [2_{0+__{1 over a range [0, x), which is essentially (x, n) => abs(2 * x + n), where x is the number in the range and n is either 0 or -1.

• No way! You used Risky for a serious problem. And it's not a half bad score despite having half the symbols. Congrats!! Jun 15, 2021 at 3:58
• Interesting language filled with 0s Jun 15, 2021 at 7:26

# Jelly, 8 bytes

,CÄḤ×ƊS_


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-2 bytes thanks to Bubbler

## How it works

,CÄḤ×ƊS_ - Main link. Takes a binary list B on the left
C       - Complement. Flip the bits of B
,        - Pair with B: [B, B']
Ä      -   Cumulative sum of each
Ḥ     -   Unhalve
×    -   Multiply modified B by B and modified B' by B'
S  - Columnwise sum
_ - Subtract B, elementwise

• Nice!! I was about to tackle it in Jelly the same way. I made an example-based explanation for my APL answer. I wonder if this is really the best way to do it. Jun 15, 2021 at 1:41
• @AviFS Yep, I saw your explanation and went "Oh, that's basically my approach" :P Jun 15, 2021 at 1:43
• 8 bytes Jun 15, 2021 at 2:54

# Japt-m, 8 bytes

?J±2:T±2


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• J is initially -1,
• T is initially 0, and,
• ± is the shortcut for +=.

# Jelly, 9 bytes

,CỤ€⁺«/Ḥ_


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My convoluted port of my own APL/J answer.

# Jelly, 9 bytes

CỤỤ«ỤỤ$Ḥ_  Try it online! Small modification of caird's 10-byter port. # 05AB1E, 8 6 bytes η^_O·α  Try it online! η # prefixes of the input ^ # XOR the first value of the input with the first prefix, second value of input with second prefix, ... _ # boolean negate O # sum each modified prefix · # double all integers α # absolute difference to the input  η^_ can be replaced with δQÅl (equality table; lower-triangular matrix), which is a byte longer but might be shorter in some other language. δQÅlO·α  Try it online! • Nice! Tied for first with cairdcoinheringaahing’s Jelly answer Jun 15, 2021 at 6:40 # PowerShell Core, 28 bytes $b=0,-1;$args|%{($b[$_]+=2)}  Try it online! Port of Arnauld's solution, thanks ! ### Initial implementation, 35 bytes switch($args){0{++$e*2}1{$o++*2+1}}


Takes the input as a list of 0/1's
Returns a list of integers

## Explanation

switch($args){ # For each argument passed as an integer 0{++$e*2}      # if it is 0, output an even number, starting from 2
1{$o++*2+1}} # if it is 1, output an odd number, starting from 1  # Jelly, 7 6 bytes ċṪ$ƤḤ+


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Basically a translation of an old version of ovs' 05AB1E answer.

## Explanation

ċṪ$ƤḤ+ Main monadic link Ƥ Map over prefixes$    (
ċ        Count the occurences of
Ṫ         the last item after removing it
$) Ḥ Unhalve + Add the original list  • Holy gamoly— 7 bytes! This is the winning answer out of all 16 so far. Look forward to reading the explanation, congrats! Jun 15, 2021 at 7:04 • @AviFS Added :) Jun 15, 2021 at 7:13 • Latecomer here checking if the solution I found was here before posting it - this six byter is the same, byte for byte. Jun 16, 2021 at 21:04 # 8086 machine code, 21 18 bytes 00000000: b0 01 b2 02 d1 eb 72 01 92 aa 40 40 72 01 92 e2 ......r...@@r... 00000010: f3 c3 ..  Function.  [bits 16] [cpu 8086] section .text ; nasm syntax ; INPUT: ; DI: destination byte array ; BX: bit pattern (little endian) ; CX: count of bits ; OUTPUT: ; stored to DI global mrbitctr mrbitctr: ; Count odd bits in AL mov al, 1 ; Count evens in DL mov dl, 2 .loop: ; Shift right BX one bit. This will put ; the lowest bit in CF. shr bx, 1 ; Was the bit set? If so, jump. jc .no_swap ; Even: swap .swap: ; Pull the old switcheroo to select evens xchg ax, dx ; Odd: don't swap .no_swap: ; Store to DI and increment stosb ; Add 2 to AL by incrementing AX twice ; Note: INC does not affect the carry flag inc ax inc ax ; swap back if even jc .no_swap_back .swap_back: xchg ax, dx .no_swap_back: ; Loop CX times. loop .loop .end: ret  • 3 bytes: swap twice # Dyalog APL, 24 23 bytes (⍵×¯1+2×+\⍵)+N×2×+\N←~⍵  The list of evens and the list of odds are generated separately and added elementwise with the + in the middle. Here's what it looks like with the problem's example input: Evens: ⍵ The input → 1 1 0 0 1 0 ~ Negate it → 0 0 1 1 0 1 N← Let N be the negated list → 0 0 1 1 0 1 +\ Take the running sum → 0 0 1 2 2 3 2× Multiply by two → 0 0 2 4 4 6 N× Multiply by the negated list → 0 0 2 4 0 6 Odds: ⍵ The input → 1 1 0 0 1 0 +\ Take the running sum → 1 2 2 2 3 3 2× Multiply by two → 2 4 4 4 6 6 ¯1+ Subtract 1 → 1 3 3 3 5 5 ⍵× Multiply by the list → 1 3 0 0 5 0 Together: (~⍵)×2×+\~⍵ Evens → 0 0 2 4 0 6 (⍵×¯1+2×+\⍵) Odds → 1 3 0 0 5 0 + Add elementwise → 1 3 2 4 5 6  Try it online! • Not a big deal, but it's generally considered polite practice here to wait for a week or more before answering your own questions. Jun 15, 2021 at 1:08 • Oh no, it is? Thanks for letting me know, @Jonah! Jun 15, 2021 at 1:10 # J, 27 26 25 bytes <@({.-~2*#\)/.~@/:~/:&;/:  Try it online! • /.~@/:~ Sort and group by value • ({.-~2*#\) Create 2 4 6 ... to the length of each group, and subtract the first element of each group from that (vectorized), so that the the 1 group becomes 1 3 5 ... • The grouping screws up the order though, so we have to... • /:&;/: Resort it according to the grade up of the original input, which makes it correct again. ## J, bonus: translation of AviFS's APL answer into J (28 bytes) ([:+/**-&1 0)@,:&(2*]*+/\)-.  Try it online! Just because I liked it and wanted to see how they'd compare. # Vyxal, 15 9 bytes ₌⇧⇩⇧$⇧∵d+


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Me when APL port

## Explained

₌⇧⇩⇧$⇧∵d+ ₌⇧⇩ # grade up input, grade down input ⇧$⇧    # grade each of those up
∵d  # 2 * the minimum of those two lists
+ # added to the input


# Python 2, 43 bytes

b=[2,1]
for e in input():print b[e];b[e]+=2


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-5 bytes and fix thanks to @xnor

The program is now reusable

f(h:t)=h:f[x+mod(x-h-1)2*2|x<-t]
f e=e


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The recursion happens on “the tail of the list, but with all elements x that have the same parity as the head incremented by 2.” Like so:

  f [1,1,0,0,1,0]
= 1 : f [3,0,0,3,0]
= 1 : 3 : f [0,0,5,0]
= 1 : 3 : 0 : f [2,5,2]
= 1 : 3 : 0 : 2 : f [5,4]
= 1 : 3 : 0 : 2 : 5 : f [4]
= 1 : 3 : 0 : 2 : 5 : 4 : f []
= [1,3,0,2,5,4]


# Jelly, 15 bytes

ṢŒg2ḷ$\€ÄFị@ỤỤ$


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ṢŒg2ḷ$\€ÄFị@ỤỤ$  Main Link; take a list of 0s and 1s
Ṣ                Sort the list
Œg              Group runs of equal elements
€         For each group
$\ Cumulatively reduce by 2ḷ x => 2 (that is, all but the first element become 2) Ä Cumulative sum, vectorizing to depth 1 F Flatten ị@ Index into (reverse order) ỤỤ$  The input graded up twice


Grading up twice returns the permutation to index into another list to get the same ordering or something like that. I think that's how it works.

• Jun 15, 2021 at 1:31
• @cairdcoinheringaahing :( Jun 15, 2021 at 1:38

# K (ngn/k), 15 bytes

{x+2*(<<x)&<>x}


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A K port of @Bubbler's J and APL solutions - don't forget to upvote it!

# Ruby-p, 29 bytes

Takes input as space separated digits (or any other non-digit separator).

gsub(/\d/){($*2+?1).count$&}


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# Stax, 9 8 bytes

╜♪N·{☼►◄


Run and debug it

Inspired by Arnauld's idea.

0-indexed, takes the bits flipped.

## Explanation

AEsF{Q2+}&
AEs        swap the input with  [1,0]
F       foreach i:
{   }&  modify the element at i in 2,1
Q      print without popping


# Wolfram Language (Mathematica), 29 bytes

(a=0@-1;a[[#~Mod~2]]+=2&/@#)&


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Boring answer. Mathematica's += etc. operators have different precedence than assignment = etc. operators. This gives them higher precedence than &, so unlike = expressions they don't need to be parenthesized on the left side of &. (//=, introduced in 12.2, is slightly different from both aforementioned groups).

• 0@-1 is clever! Jun 17, 2021 at 9:23

# C (clang), 52 bytes

-1 thanks to @AZTECCO, by using clang instead of gcc.

f(a,l)int*a;{for(int b[]={0,-1};l--;)*a++=b[*a]+=2;}


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# C (gcc), 53 bytes

f(a,l)int*a;{for(int b[]={0,-1};l--;a++)*a=b[*a]+=2;}


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• Nice trick! You can save by switching to clang and incrementing a in assignment f(*a,l){for(int b[]={0,-1};l--;)*a++=b[*a]+=2;} Jun 15, 2021 at 20:42

# Raku, 21 bytes

*>>.&{(%.{$_}+=2)-$_}


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Maps each element to the index into an anonymous hash, incrementing that value by two (initially zero), and finally subtracting the element itself to distinguish between odd and even.This could also be extended to values beyond 0 and 1 simply by changing the 2 to another number.

# Julia, 33 30 bytes

f(x,y=[0,-1])=x.|>i->y[i+1]+=2


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# AWK, 42 bytes

d[1]=1{for(;a++<NF;d[b]+=2)$a=+d[b=$a%2]}1


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So this one is one test with a codeblock, and a naked 1 to print all the commandline arguments. It replaces each commandline argument with the appropriate even/odd counter in the codeblock.

The "test" for the codeblock is always truthy and it just used to initialize the "odd" counter to 1.

d[1]=1{                                  }


The code block runs through each commandline argument,

       for(;a++<NF;       )


Then sets that argument to the current value of the even/odd counter with:

                           $a=+d[b=$a%2]


And at the end of the loop, increments the current counter by 2 in preparation for the next match.

                   d[b]+=2


Once that's done, it just need to print out all the arguments.

                                         1


# Zsh, 23 bytes

for x
echo $[x+a$x++*2]


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Starts even numbers at 0.

Explanation:

• for x: for each $x in the input, • $[]: arithmetic expansion
• ++: increment and return original value
• a$x: the variable named a0 or a1 (which correspond to the number of 0s and 1s seen so far) • x+*2: double and add x to get the correct value • echo : print (can't use <<< because the mutation wouldn't work in the subshell it creates) # Desmos, 64 bytes a(l)=\sum_{n=1}^{[1...length(l)]}l[n] b=1-l f(l)=2ba(b)+2la(l)-l  Just implements the strategy shown in AviFS's Dyalog APL answer. Try It On Desmos! Try It On Desmos! - Prettified ## Explanation: a(l)=\sum_{n=1}^{[1...length(l)]}l[n]: Function that takes in a list $$\l\$$ and returns the running total of $$\l\$$. b=1-l: Variable that stores the inputted list, but with each bit flipped. f(l)=2ba(b)+2la(l)-l: Function that takes in a list of bits $$\l\$$ and outputs the correct answer, based on the strategy mentioned above. # R, 42 bytes function(x,y=seq(x)*2){x[x]=y-1;x[!x]=y;x}  Try it online! -8 bytes thanks to @Dominic Takes input as booleans: TRUE(1) and FALSE(0). Straightforward approach, but takes advantage of truncating the replacement to the length of items being replaced. Different approach: ### R, 47 bytes function(x,n=0:-1)for(i in x)show(n[i]<-n[i]+2)  Try it online! Takes input incremented by 1: 2 (for 1) and 1 (for 0). • 42 bytes... Jun 15, 2021 at 8:08 • @DominicvanEssen Thanks, nice one! Didn't think of leveraging the truncating of replacement vector! Jun 15, 2021 at 8:21 # Retina 0.8.2, 43 bytes (.)(?<=((\1)|.)*(1)?)$#3$*2$4¶
2
11
1+
$.&  Try it online! Link includes test cases. Takes input as a string of bits. 1-indexed. Explanation: (.)(?<=((\1)|.)*(1)?)  For each bit, count the number of preceding identical bits. If the current bit is 1 then count it separately otherwise include the current bit in the count. $#3$*2$4¶


Record a 2 for each duplicate plus one extra for a 1 bit.

2
11


Convert to unary.

1+
$.&  Convert the sum back to decimal. # Charcoal, 12 bytes ＩＥθ⁺Ｉι⊗№…θκι  Try it online! Link is to verbose version of code. Takes input as a string of bits. 0-indexed. Explanation:  θ Input string Ｅ Map over characters ι Current character Ｉ Cast to integer ⁺ Plus ⊗ Doubled № Count of ι Current character in θ Input string … Truncated to length κ Current index Ｉ Cast to string Implicitly print on separate lines  # Perl 5 (-ap), 27 bytes $_+=$x[$_]++*2for@F;\$_="@F"
`

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