Parse a list of signed unary numbers

Question

Unary numbers typically only represent nonnegative integers, but we can extend them to represent all integers as follows:

• A positive integer N is represented as N 1's: 5 -> 11111
• A negative integer -N is represented as a 0 followed by N 1's: -5 -> 011111
• Zero is represented as 0

We can then represent a list of these numbers unambiguously if we use 0 as the separator:

3,-2,0,1
111,011,0,1
111 0 011 0 0 0 1
11100110001

Your task: take a string representing such a list of signed unary numbers, and translate it into a list of decimal numbers.

Details

You may assume that the input is a complete list of signed unary numbers. In particular, your program will not have to handle 1) empty input or 2) input that ends with a separator.

You may assume that the magnitude of each number will not exceed 127. For languages with maximum sizes of strings or lists, you may assume that the input and output will fit in your language's data structures, but your algorithm should theoretically work for a list of any size.

Your program or function may perform I/O in any of the standard ways. Input may be a string or a list of characters, single-character strings, integers, or booleans. You may use any two characters to represent 1 and 0; if you don't use 1 and 0, please specify which characters you're using.

Output must be decimal numbers in any reasonable list format (in particular, there must be some kind of a separator between numbers). Negative numbers should be indicated with a minus sign, although if your language has a different format for negative integers I will also accept that. Zero may be represented in the output as 0 or -0.

Test cases

1 -> 1
0 -> 0 (or -0, and similarly for the other test cases)
011 -> -2
1101 -> 2,1
1100 -> 2,0
11001 -> 2,-1
110001 -> 2,0,1
11100110001 -> 3,-2,0,1
00000001 -> 0,0,0,-1
01111011111111001111111111111110111111111111111100111111111111111111111110111111111111111111111111111111111111111111 -> -4,8,-15,16,-23,42
01111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 -> -127

Answer 1

Python 2, 73 70 bytes

A function that takes a string as input and returns a string representation of a Python list. Zero can be represented both by 0 and -0 (when it comes last):

lambda s:`map(len,s.split('0'))`.replace('0, ','-').replace('--','0,')

Explanation

1. split the input string s on zeroes.
2. Take the length of each string in the resulting list (using map).

That takes us a long way. Zeroes were separators after all. And the numbers were unary, so len conveniently converts those to decimal. But now we've messed up all the non-separator uses of 0. Luckily, all non-separator uses were leading zeroes so they came after a separator-zero and gave us zero-length strings ('00'.split('0') == ['', '', '']). Those zero-length strings then also became 0 because of the len.

3. Turn the list into a string (using "reverse quotes"), so we can fix up the mess more easily.
4. replace each zero that precedes another number by a negative sign on that number instead. That fixes the use of 0 as a sign but it breaks the literal zeroes. Literal zeroes were also preceded by a separator, so they've now become pairs of extra dashes on the next number.
5. replace each -- back into a 0 element in the "list".

Answer 2

Retina, 23 21 bytes

(.)0
$1 
01
-1
1+
$.&

Try it online!

The first stage (.)0<newline>$1<space> matches any character followed by a 0. The match is replaced by the first character followed by a space. This splits the string in the individual numbers.

The second stage 01<newline>-1 replaces 0's before a block of 1's to to the - sign.

The last stage 1+<newline>$.& matches all blocks of 1's and replaces them with the length of the group.

Here is an example with the output of the individual stages.

Answer 3

Vim, 56 bytes

:s/\v(0?1*)0?/\1\r/g|%s/0/-/|%s/1*$/\=len(submatch(0))
D

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I haven't posted in vim in a while. I'm mostly using vim because V is a pain sometimes. Because the count command, which is perfect for getting the number of '1's on the line will overwrite any '0's on the line, so we can't negate it afterwards.

Explanation:

This is one byte shorter then the straightforward way:

:s/\v(0?1*)0?/\1\r/g
:%s/0/-
:%s/1*$/\=len(submatch(0))
D

due to command chaining. Since that one separates the commands, I'll use it for the explanation.

:s/                     " Substitute
                        " Search for...
   \v                   "   Enable 'magic'. This determines whether certain atoms require a backslash or not.
                        "   Without it we would have: '\(0\?1*\)0\?', which is 2 bytes longer
      0?                "   An optional 0
        1*              "   Followed by any number of '1's
     (    )             "   (call that group 1)
           0?           "   Followed by another optional 0
             /          " Replace it with...
              \1        "   Subgroup 1
                \r      "   A newline
                  /g    " Do this for every match on the current line.

Now, each signed unary number is on an individual line. Using '11100110001' as an example, at this point we'll have:

111
011
0
1

:%s/0   " Replace every 0
     /- " With a dash  

:%s/1*$/                    " Replace every run of 1's at the end of a line
        \=len(submatch(0))  " With the length of said run

Since we added newlines at the end of each match, we had an empty line before running that. After running that, we'll have a '0' (because it matched a run of 0 '1's). So we just call D to delete this line, leaving it blank

Answer 4

Haskell, 68 66 bytes

f(x:r)|(a,b)<-span(>0)r=([(0-),(1+)]!!x$sum a):[z|_:t<-[b],z<-f t]

Try it online! Takes input as a list of zeros and ones. Example usage: f [0,0,0,1,1] yields [0,-2].

Explanation:

The pattern matching in f(x:r)|(a,b)<-span(>0)r binds x to the first element of the input, a to a (potentially empty) list of following 1s, and b to the rest of the input. Given an input [0,1,1,1,0,0,1], we get x=0, a=[1,1,1] and b=[0,0,1].

The current number is then either the sum of a negated if x=0, or the sum of a plus one if x=1. This is achieved by indexing with x into a list containing a negation and increment function, and applying the resulting function to the sum of a: [(0-),(1+)]!!x$sum a.

The rest list b is either empty or contains a separating zero and the next number. The list comprehension [z|_:t<-[b],z<-f t] tries to match b on the pattern _:t, that is forget the head element and bind the rest of the list to t. If b is empty this match fails and the list comprehension evaluates to [], which is the base case for the recursion. Otherwise the function f is recursively applied to t and the list comprehension evaluates to all elements z from the result of f t.

Answer 5

Husk, 20 18 17 15 14 bytes

Γ~:?Σṁ_Πȯ₀tΣġ/

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Explanation

Γ~:?Σṁ_Πȯ₀tΣġ/  Input is a list, say x = [0,1,1,0,0,0,1,1]
            ġ   Group by
             /  division.
                This splits x right before each 0: [[0,1,1],[0],[0],[0,1,1]]
Γ               Deconstruct into head y = [0,1,1] and tail z = [[0],[0],[0,1,1]]
   ?Σṁ_Π        Apply to y:
       Π         Product: 0
   ?Σ            If that is nonzero, take sum of y,
     ṁ_          else take sum of negated elements of y: u = -2
        ȯ₀tΣ    Apply to z:
           Σ     Concatenate: [0,0,0,1,1]
          t      Drop first element: [0,0,1,1]
         ₀       Recurse: [0,2]
 ~:             Tack u to the front: [-2,0,2]

The splitting works like this. ġ/ splits its argument between each pair of elements a,b for which /a b is falsy. /a b is division with flipped arguments, so b divided by a. The relevant values in this program are these:

• /1 1 gives 1 (truthy).
• /1 0 gives 0 (falsy).
• /0 1 gives Inf (positive infinity, truthy).
• /0 0 gives Any (a special NaN-like value, falsy).

Answer 6

Perl 5, 40 + 1 (-n) = 41 bytes

print'-'x/00/.y/1//.$"for"0$_"=~/00?1*/g

Try it online!

Answer 7

Wolfram Language (Mathematica), 80 bytes

StringCases[#<>"0",x_~~Shortest@y___~~"0":>(If[x=="0",-#,#+1]&)@StringLength@y]&

Try it online!

Abuses the mechanic of StringCases, since it does not check overlapping patterns. Since we search from left to right, without overlaps, we always get only the integers we need.

Explanation

#<>"0"

Append a zero at the end

StringCases

Find all of the following pattern...

x_~~Shortest@y___~~"0"

A single character (call it x), followed by the shortest possible zero-length or longer string (call it y), followed by a zero.

(If[x=="0",-#,#+1]&)@StringLength@y

Apply to matching pattern: take the length of y. If x is zero, then negate the value. Else, increment one.

This covers 00 as well, since y would be an empty string, and we would compute -0 ( == 0).

Answer 8

Brain-Flak, 94 (70?) bytes

([]){{}({(<()>)()}{}(<>)<>{<([{}]<>{})><>})
{{}<>([{}])(<>)}{}{}([])}{}<>([]){{}({}<>)<>([])}<>

Try it online!

This is actually surprisingly terse for brain-flak.

Here is a commented/readable version:

([])

{

    #Pop the Stack height
    {}

    (
        #If there isn't a leading 0, evaluate to 1...
        {
            (<()>)

            ()
        }

        #Pop the 0
        {}

        #Push a 0 onto the alternate stack
        (<>)
        <>

        #Run of '1's
        {
            #Decrement the alternate stack
            <([{}]<>{})>
            <>
        }

        #And push it here
    )

    #Was there a not leading 0?

    {
        {}

        #Invert the value on the alternate stack
        <>([{}])(<>)
    }

    #Pop 2 zeros
    {}{}


    ([])

}{}<>

#Push stack height
([])

#Reverse the stack
{

    {}

    ({}<>)

    <>([])

}<>

If the output can be in reverse, we can do this for 70 instead:

([]){{}({(<()>)()}{}(<>)<>{<([{}]<>{})><>}){{}<>([{}])(<>)}{}{}([])}<>

This tip of mine is almost perfect for this situation. But it doesn't quite work since we have to push a 0 before doing the operation (counting the '1's), and the operation happens in a loop. The shortest I could come up with utilizing this tip is:

([]){{}({(<()>)()}{}(<>)<>{<([{}]<>{})><>})
{{}<>([{}])(<>)}{}{}(<>())<>([])}{}<>{{}({}<>)<>}<>

which is also 94 bytes.

Answer 9

Jelly,  19  18 bytes

^{There must be a better way...}

®ḢN$Ḣ©?ṄEȧ
ṣ0L€ÇL¿

A full program printing each number followed by a linefeed.

Try it online!

How?

®ḢN$Ḣ©?ṄEȧ - Link 1, print first number and yield next input: list of numbers, X
           -                              e.g. [8,0,15,16,...] or [0,4,8,0,15,16,...]
      ?    - if...
    Ḣ      - condition: yield head and modify  8([0,15,16,...])   0([4,8,0,15,16,...])  
     ©     -            (copy to register)     8                  0
®          - then: recall from the register    8
   $       - else: last two links as a monad:
 Ḣ         -         yield head and modify                        4([8,0,15,16,...])
  N                  negate                                      -4
       Ṅ   - print that and yield it           8                 -4
        E  - all equal (to get 0 to be truthy) 1                  1
         ȧ - AND the (modified) input          [0,15,16,...]      [8,0,15,16,...]
           -   (ready to be the input for the next call to this link)

ṣ0L€ÇL¿ - Main link: list e.g. [0,1,0,0,0,0,1,1]
ṣ0      - split at zeros       [[],[1],[],[],[],[1,1]
  L€    - length of €ach       [0,1,0,0,0,2]
      ¿ - while...
     L  - condition: length                           1  1  1  0  ([0,1,0,0,0,2], [0,0,0,2], [0,2], [])
    Ç   - action: call the last link (1) as a monad  -1  0 -2     ( - 1            - 0        - 2)

Answer 10

Acc!!, 252 237 bytes

N
Count i while _/48 {
Count n while 48/_ {
Write 45
50+N
}
_+49/_*50
Count u while _%50/49 {
_+100-_%50+N
}
_/50-1
Count h while _/200 {
Write _/200+48
_%200+1
}
Count t while _/20+_%2 {
Write _/20+48
_%20-_%2
}
Write _/2+48
Write 9
N
}

Uses -0. Outputs numbers separated by tab characters, with a trailing tab. Try it online!

^{Amount of time writing the actual algorithm: 20 minutes. Amount of time debugging my decimal output code: 45 minutes. :^P}

With comments

I don't know if these comments explain the code very well--they're based on my notes to myself while I was writing it, so they assume some understanding of how Acc!! works. If anything needs more explanation, let me know and I'll try to make it clearer.

# We partition the accumulator _ as [number][flag][nextchar]
# [flag] is a 2-value slot and [nextchar] a 50-value slot
# So [nextchar] is _%50, [flag] is _/50%2, [number] is _/100
# [flag] is 1 if we're in the middle of reading a number, 0 if we're between numbers
# It is also used for outputting as decimal (see below)
# Possible input characters are 0, 1, and newline, so [nextchar] is 48, 49, or 10

# Read the first character
N
# Loop while the character we just read is 0 or 1 and not newline
Count i while _/48 {
  # What we do in the loop depends on the combination of [flag] and [nextchar]:
  # 0,48 (start of number, read 0) => write minus sign, [flag] = 1, read another char
  # _,49 (read 1) => increment [number], [flag] = 1, read another char
  # 1,48 (middle of number, read 0) => write/clear [number], status = 0, read another
  #      char
  # 1,10 (middle of number, read <cr>) => ditto; the next read will be 0 for eof, which
  #      means the acc will be less than 48 and exit the loop

  # Process leading 0, if any
  Count n while 48/_ {
    # acc is 48: i.e. [number] is 0, [flag] is 0, [nextchar] is 48 (representing a 0)
    # Output minus sign
    Write 45
    # Set [flag] to 1 (thereby exiting loop) and read [nextchar]
    50+N
  }
  # If number starts with 1, then we didn't do the previous loop and [flag] is not set
  # In this case, acc is 49, so we add (50 if acc <= 49) to set [flag]
  _+49/_*50

  # Process a run of 1's
  Count u while _%50/49 {
    # [nextchar] is 49 (representing a 1)
    # Increment [number] and read another
    _+100-_%50+N
  }

  # At this stage, we know that we're at the end of a number, so write it as decimal
  # This is "easier" (ha) because the number has at most three digits
  # We shift our partitioning to [number][flag] and set [flag] to 0
  _/50-1

  # Output hundreds digit if nonzero
  # Since [number] is _/2, the hundreds digit is _/200
  Count h while _/200 {
    Write _/200+48
    # Mod 200 leaves only tens and units; also, set [flag] to 1
    _%200+1
  }
  # Output tens digit (_/20) if nonzero OR if there was a hundreds digit
  # In the latter case, [flag] is 1
  Count t while _/20+_%2 {
    Write _/20+48
    # Mod 20 leaves only units; clear [flag] if it was set
    _%20-_%2
  }
  # Write units unconditionally
  Write _/2+48

  # Write a tab for the separator
  Write 9
  # Read another character
  N
}

Answer 11

Python 2, 96 92 bytes

lambda s:[[len(t),-len(t)+1]['1'>t]for t in s.replace('10','1 ').replace('00','0 ').split()]

Try it online!

Thx to ovs and DLosc for 2 bytes each.

Answer 12

R, 119 bytes

function(x){n=nchar
y=regmatches(x,regexec("(0?)(1*)0?([01]*)",x))[[1]]
cat((-1)^n(y[2])*n(y[3]),"")
if(y[4]>0)f(y[4])}

Try it online!

The code uses this solution from stackoverflow for a related problem (Thanks to jeales for the idea). The output is a space-separated string printed to stdout.

Answer 13

sed, 227 bytes

s/^.o*/<&>/
:a
s/>z\(.o*\)/><\1>/
ta
:b
s/z>/0>/g
y/z/-/
s/oooooooooo/x/g
s/ooooooooo/m9/g
s/oooooooo/m8/g
s/ooooooo/m7/g
s/oooooo/m6/g
s/ooooo/m5/g
s/oooo/m4/g
s/ooo/m3/g
s/oo/m2/g
s/o/m1/g
s/x\([^xm]\)/x0\1/g
s/m//g
y/x/o/
tb

Try it online!

Usage

0 and 1 for input are replaced with z and o respectively. Input is given from stdin, on every line.

With comments

# first integer
s/^.o*/<&>/
# find each item
:a
s/>z\(.o*\)/><\1>/
ta
# below: to decimal
:b
# to zero
s/z>/0>/g
# minus sign
y/z/-/
# to decimal others
# m letter indicates that converting a digit is done
s/oooooooooo/x/g
s/ooooooooo/m9/g
s/oooooooo/m8/g
s/ooooooo/m7/g
s/oooooo/m6/g
s/ooooo/m5/g
s/oooo/m4/g
s/ooo/m3/g
s/oo/m2/g
s/o/m1/g
# not converted yet? then to 0
s/x\([^xm]\)/x0\1/g
s/m//g
y/x/o/
# repeat until every unary is converted
tb

Answer 14

JavaScript (ES6), 60 bytes

Returns a space-separated list of integers.

s=>(0+s).replace(/00?1*/g,s=>(l=s.length,+s[1]?l-1:2-l)+' ')

Test cases

let f =

s=>(0+s).replace(/00?1*/g,s=>(l=s.length,+s[1]?l-1:2-l)+' ')

console.log(f("1")) // 1
console.log(f("0")) // 0
console.log(f("011")) // -2
console.log(f("1101")) // 2,1
console.log(f("1100")) // 2,0
console.log(f("11001")) // 2,-1
console.log(f("110001")) // 2,0,1
console.log(f("11100110001")) // 3,-2,0,1
console.log(f("00000001")) // 0,0,0,-1
console.log(f("01111011111111001111111111111110111111111111111100111111111111111111111110111111111111111111111111111111111111111111")) // -4,8,-15,16,-23,42
console.log(f("01111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111")) // -127

Answer 15

QBasic, 88 86 bytes

1u$=INPUT$(1)
z=u$<"1
IF n*z THEN?(1-2*s)*(n-s):s=0:n=0ELSE s=s-z:n=n+1
IF"!"<u$GOTO 1

This was fun. Multiple revisions starting from a 107-byte version resulted in one of the most obfuscated bits of QBasic I think I've ever written. (Edit: Oddly enough, I was able to golf 2 bytes by making the code clearer.)

Note: this program reads user input one character at a time without echoing it to the screen (a result of using INPUT$(1) instead of the usual INPUT statement). So as you're typing, you won't see the 1's and 0's, but the decimal numbers will appear as they are calculated. Make sure to hit Enter at the end of the input to see the last number and end the program.

Ungolfed version

sign = 0
num = 0
DO
  digit$ = INPUT$(1)
  isZero = (digit$ < "1")
  IF num > 0 AND isZero THEN
    PRINT (1 - 2 * sign) * (num - sign)
    sign = 0
    num = 0
  ELSE
    IF isZero THEN sign = 1
    num = num + 1
  END IF
LOOP WHILE "!" < digit$

Explanation

(AKA "What?? That still makes no sense!")

The base strategy is to run a loop that grabs one character from INPUT$(1) each time through, does stuff with it, and keeps looping as long as the character has an ASCII value greater than that of ! (i.e., wasn't a newline).

We keep track of numbers-in-progress using two variables. num is the number of characters in the current signed unary number (including any leading zero). sign is 1 if the number had a leading zero, 0 if not. Both of these need to be initialized to 0, which is great for the golfed version because numeric variables in QBasic are automatically initialized to 0.

Whenever we read a character, the first thing is to determine whether it's 1 or 0. We'll use this result twice, so we store it in isZero. Technically, that name is misleading, since the value will also be truthy if the character is a newline. Note that truthy in QBasic is -1 and falsey is 0.

Now, if we're in the middle of reading a number (num > 0) and we hit a zero or end of input (isZero), we need to calculate which number we've finished reading.

• sign stores 0 for positive, 1 for negative. To get 1 for positive and -1 for negative, we need 1-2*sign.
• num stores the correct magnitude for positives but one more than the magnitude for negatives (since it includes the sign marker). So we can use num-sign for the magnitude.

Multiply these together and print; then reset sign and num to 0 in preparation for reading the next number.

Otherwise (if we haven't hit a zero, or if we've hit a zero at the beginning of a number), we update sign and num as follows:

• sign becomes 1 if we're looking at a leading zero; otherwise, if we're looking at a one, it stays at whatever it already was. The golfed code is s=s-z, which amounts to the same thing:
  • If this is a leading zero, z is -1. Since s is guaranteed to be 0 (because this is the start of a new number), s-z will be 1.
  • If this is a one, z is 0. Then s-z stays at whatever value s had previously.
• num is incremented.

That's it!

Answer 16

Jelly, 15 bytes

Ż¬<\œp$ḊLCṛ¡ḢƊ€

Try it online!

Although it's something of an improvement over the previous solution, this feels clumsy. (Though not as clumsy now that it doesn't use k.)

Input as a list of zeroes and ones.

Ż                  Prepend a leading zero to the input.
 ¬                 Flip 0 <-> 1.
    œp$            Split that around the corresponding locations of truthy elements of
  <\               it cumulatively reduced by greater than.
       Ḋ           Remove the leading empty slice,
             Ɗ€    then for each remaining slice:
        L          take the length,
         C         and subtract it from 1
           ¡       a number of times equal to
          ṛ Ḣ      the first element.

Answer 17

Acc!!, 177 bytes

N
Count i while _%5 {
Count j while _%7 {
Write 45
0
}
Count j while 0^(_%7) {
_+N
}
Count j while 3-j {
Count k while 0^k*(_/4900+j/2) {
Write 48+_/4900%10
}
_*10
}
Write 9
N
}

Try it online!

Uses -0 to represent zeros and tabs as seperators.

Commented

N                                 # Read char

Count i while _%5 {               # while _ % 5 != 0:
                                  #   ( _ here is the last read char; notice that EOF
                                  #     and newline are divisible by 5 while '0' and '1'
                                  #     are not )

                                  #   Handle potential leading '0'
Count j while _%7 {               #   if _ % 7 != 0:
                                  #     ( _ is same as above; notice that _ can only be
                                  #       '0' or '1' and additionally that only '0' is
                                  #       not divisible by 7 )
Write 45                          #     output '-'
0                                 #     _ = 0
}

                                  #   Read '1's (also consumes 1 char after stream of '1's)
Count j while 0^(_%7) {           #   while _ % 7 == 0:
                                  #     ( _ here represents sum of char values since last
                                  #       '1'; notice that it is a multiple of 7 when we
                                  #       read only '1's and would only longer be a multiple
                                  #       of 7 when we read a '0' or newline )
_+N                               #     _ = _ + N  # here N is a newly read char 
}

                                  #   Output number in decimal
Count j while 3-j {               #   for j in [0, 1, 2]:
Count k while 0^k*(_/4900+j/2) {  #     if _ / 4900 + j / 2 != 0:
                                  #       ( _ / 4900 represents the 100s place of the read
                                  #         number)
Write 48+_/4900%10                #       output _ / 4900 % 10 as a digit
}                                 #
_*10                              #     _ = _ * 10  (shifts number to prepare for next iteration)
}

Write 9                           #   Output tab as a seperator

N                                 #   Read char (preparation for the loop)
}

Answer 18

Lua, 58 bytes

(...):gsub("(0?)(1*)0?",function(s,n)print(#n-2*#s*#n)end)

Try it online!

Full program, takes input from the command line and prints the numbers to stdout separated by newlines.