Stack Exchange Vote Simulator

Question

Write a program or function that takes in a string only containing the characters ^ and v (you can assume there will be no other characters). Read from left to right this string represents the sequence of mouse clicks a single user made while viewing a Stack Exchange question or answer for the first time.

Every ^ represents a click of the upvote button and every v represents a click of the downvote button. (For working examples look slightly left.)

Assume that no voting limitations are in effect so all the clicks are registered correctly.
Print or return:

• 1 or +1 if the post ends up being upvoted.
• 0 if the post ends up not being voted on. (-0 and +0 are not valid)
• -1 if the post ends up being downvoted.

Posts start with zero net votes from the user and the buttons change the net votes as follows:

Net Votes Before    Button Pressed    Net Votes After
1                   ^                 0
1                   v                 -1
0                   ^                 1
0                   v                 -1
-1                  ^                 1
-1                  v                 0

The shortest code in bytes wins.

Test cases:

[empty string] -> 0
^^ -> 0
^v -> -1
^ -> 1
v -> -1
v^ -> 1
vv -> 0
^^^ -> 1
vvv -> -1
^^^^ -> 0
vvvv -> 0
^^^^^ -> 1
vvvvv -> -1
^^^^^^ -> 0
vvvvvv -> 0
^^v -> -1
^v^ -> 1
^vv -> 0
vv^ -> 1
v^v -> -1
v^^ -> 0
^vvv^^vv^vv^v^ -> 1
^vvv^^vv^vv^v^^ -> 0
^vvv^^vv^vv^v^^^ -> 1
^vvv^^vv^vv^v^^v -> -1
^vvv^^vv^vv^v^^vv -> 0
^vvv^^vv^vv^v^^vvv -> -1
^vvvvvvvvvvvv -> 0
^^vvvvvvvvvvvv -> 0
^^^vvvvvvvvvvvv -> 0
vvv^^^^^^^^^^^^ -> 0
vv^^^^^^^^^^^^ -> 0
v^^^^^^^^^^^^ -> 0

Answer 1

Gol><> 0.3.11, 13 12 11 bytes

iEh`^=:@)+M

Try it online. Even though this will work fine in the next update, I've listed it as 0.3.11 just in case.

Explanation

i               Read char
 Eh             If EOF, halt and output top of stack as num
   `^=          Push 1 if char is ^, else 0
      :@        Dup and rotate, giving [is^ is^ votecount]
        )       Compare greater than, pushing 1 or 0 as appropriate
         +M     Add and subtract 1

Note that the first use of @ pulls a 0 from the bottom of the stack to initialise the vote count for the first iteration

To illustrate with a full table:

Votes before    Button    Is ^?    Compare <    Add     Subtract 1
     1            ^         1         0          1          0
     1            v         0         0          0         -1
     0            ^         1         1          2          1
     0            v         0         0          0         -1
    -1            ^         1         1          2          1
    -1            v         0         1          1          0

Answer 2

x86 machine code, 24 bytes

31 C0 8A 11 84 D2 75 07 C0 E0 02 C0 F8 06 C3 41 38 C2 74 EC 88 D0 EB EA

This is a function using the fastcall calling convention, which takes a string and returns an 8-bit integer.

I tested it with the following C program, which must be compiled for 32-bit mode.

#include <stdio.h>
#include <inttypes.h>

 __attribute__ ((aligned (16))) const unsigned char fun[] = {

    0x31,  //xor eax,eax
        0xC0,
    0x8A, //mov [ecx],dl
        1 | 2<<3,
    0x84, //test dl, dl
        0xC0 | 2<<3 | 2,
    0x75, // jnz
        7,
    0xC0, //shl al 2
        0xC0 | 4<<3,
        2,
    0xC0, //sar al 6
        0xC0 | 7<<3,
        6,
    0xC3, //ret
    0x41, //inc ecx
    0x38, //cmp al,dl
        0xC0 | 2,
    0x74, //je
        -20,
    0x88, //mov dl,al
        0xC0 | 2<<3,
    0xEB, //jmp
        -22,
};

int main()
{
    __fastcall int8_t (*votesimulator)(char*) = fun;
    char* s[] = {
        "",
        "^^",
        "^v",
        "^",
        "v",
        "v^",
        "vv",
        "^^^",
        "vvv",
        "^^^^",
        "vvvv",
        "^^^^^",
        "vvvvv",
        "^^^^^^",
        "vvvvvv",
        "^^v",
        "^v^",
        "^vv",
        "vv^",
        "v^v",
        "v^^",
        "^vvv^^vv^vv^v^",
        "^vvv^^vv^vv^v^^",
        "^vvv^^vv^vv^v^^^",
        "^vvv^^vv^vv^v^^v",
        "^vvv^^vv^vv^v^^vv",
        "^vvv^^vv^vv^v^^vvv",
        "^vvvvvvvvvvvv",
        "^^vvvvvvvvvvvv",
        "^^^vvvvvvvvvvvv",
        "vvv^^^^^^^^^^^^",
        "vv^^^^^^^^^^^^",
        "v^^^^^^^^^^^^",
    };

    for(int i = 0; i < sizeof(s)/sizeof(*s); i++)
        printf("%d\n", votesimulator(s[i]));

    printf("\n%d\n", sizeof(fun));
    for(int i = 0; i < sizeof(fun); i++)
        printf("%02X ", fun[i]);
    return 0;
}

Answer 3

JavaScript (ES7), 47 46 44 43 37 36 bytes

^{Crossed out 44 is still regular 44 :(}

s=>[for(x of s)s=x<"v"?s!=1:!~s-1]|s

Keeps a running total in s. Uses for of loop to iterate over each character in the string and updates s based on current character and previous value.

Edits: Golfed ~s&&-1 to !~s-1. This expression has to equal 0 if s equals -1 and -1 otherwise. Saved 6 bytes thanks to @nderscore.

How the expression works:

 ~s    // Bitwise inverse. ~s==0 only if s==-1
!      // Logical negate. Casts to boolean. Equivalent to s==-1
   -1  // Subtract. Casts to number so true-1 is 1-1 and false-1 is 0-1

Answer 4

CJam, 18 14 bytes

Updated version with significant improvements contributed by Dennis:

0'jqf{-g_@=!*}

Try it online

Explanation:

0     Start value for running total.
'j    Push character between '^ and 'v for use in loop.
q     Get input.
f{    Apply block with argument to all input characters.
  -     Subtract character from 'j. This will give -12 for '^, 12 for 'v.
  g     Signum, to get 1 for '^, -1 for 'v, which is our increment value.
  _     Copy increment value.
  @     Bring running total to top.
  =     Compare. This will give 1 for the -1/-1 and 1/1 combinations where the new
        running total is 0. Otherwise, the new running total is the increment value.
  !     Negate to get 0 for the -1/-1 and 1/1 cases.
  *     Multiply result with increment value, to get new running total.
}     End block applied to input characters.

Answer 5

Befunge 93 - 55 bytes

vj#p01:>#<:1+|
>~:10g-|v:g25<
^p01"j"<1^   <
./*34-g0<@

52 characters and 3 new lines.

Tested on this interpreter.

The j is equidistant from ^ and v in ascii so it's used to make arithmetic conversions in the end, rather than space consuming conditionals.

Answer 6

brainfuck, 146 bytes

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

This program takes each byte of input and compares it against the last. If they're the same, it throws the input away and stores "0" as the "previous input", otherwise it saves it normally.

If the final result is v, it prints -. If the final result was non-zero, 1 is added to an empty cell. Finally, 48 is added to that cell and it is printed.

Answer 7

Javascript ES6, 91 48 chars

s=>~~{'^':1,v:-1}[s.replace(/(.)\1/g).slice(-1)]

Explanation: undefined ends by d.

Test:

` -> 0
^^ -> 0
^v -> -1
^ -> 1
v -> -1
v^ -> 1
vv -> 0
^^^ -> 1
vvv -> -1
^^^^ -> 0
vvvv -> 0
^^^^^ -> 1
vvvvv -> -1
^^^^^^ -> 0
vvvvvv -> 0
^^v -> -1
^v^ -> 1
^vv -> 0
vv^ -> 1
v^v -> -1
v^^ -> 0
^vvv^^vv^vv^v^ -> 1
^vvv^^vv^vv^v^^ -> 0
^vvv^^vv^vv^v^^^ -> 1
^vvv^^vv^vv^v^^v -> -1
^vvv^^vv^vv^v^^vv -> 0
^vvv^^vv^vv^v^^vvv -> -1
^vvvvvvvvvvvv -> 0
^^vvvvvvvvvvvv -> 0
^^^vvvvvvvvvvvv -> 0
vvv^^^^^^^^^^^^ -> 0
vv^^^^^^^^^^^^ -> 0
v^^^^^^^^^^^^ -> 0`
.split("\n").map(s => s.split(" -> "))
.every(([s,key]) => (s=>~~{'^':1,v:-1}[s.replace(/(.)\1/g).slice(-1)])(s)==key)

Answer history:

s=>({'':0,'^':1,v:-1}[s.replace(/^(.)\1(\1\1)*(?=.?$)|.*(.)(((?!\3).)\5)+/,"").substr(-1)])
s=>~~{'^':1,v:-1}[s.replace(/^(.)\1(\1\1)*(?=.?$)|.*(.)(((?!\3).)\5)+/,"").substr(-1)]
s=>~~{'^':1,v:-1}[s.replace(/^.*(.)(((?!\1).)\3)+|(.)\4(\4\4)*/,"").substr(-1)]
s=>~~{'^':1,v:-1}[s.replace(/^.*(.)(((?!\1).)\3)+|(.)\4(\4\4)*/,"").slice(-1)]
s=>~~{'^':1,v:-1}[s.replace(/.*(.)(((?!\1).)\3)+|(.)\4(\4\4)*/,"").slice(-1)]
s=>~~{'^':1,v:-1}[s.replace(/.*(.)(((?!\1).)\3)+|((.)\5)*/,"").slice(-1)]
s=>~~{'^':1,v:-1}[s.replace(/((.)\2)+/g,"!").slice(-1)]
s=>~~{'^':1,v:-1}[s.replace(/(.)\1/g,"!").slice(-1)]
s=>~~{'^':1,v:-1}[s.replace(/(.)\1/g,0).slice(-1)]
s=>~~{'^':1,v:-1}[s.replace(/(.)\1/g).slice(-1)]

Answer 8

Python 2, 49

lambda s:reduce(lambda x,c:cmp(cmp('u',c),x),s,0)

Iterates through with the update function

lambda x,c:cmp(cmp('u',c),x)

that takes the current vote count x and the new character c and outputs the new vote count.

The idea is to use Python 2's cmp function, which compares its two args and gives -1, 0, 1 for <, ==, > respectively. The inner one cmp('u',c) gives -1 for v and 1 for ^; any character between them suffices for 'u'. The outer one then compares that to x, which gives cmp(1,x) for ^ and cmp(-1,x) for v, which have the right values.

Direct iteration was 3 chars longer (52), though would be one char short (48) if taking an input() with quotes was allowed.

x=0
for c in raw_input():x=cmp(cmp('u',c),x)
print x

The best recursive function I found was one char longer (50)

f=lambda s:len(s)and cmp(cmp('u',s[-1]),f(s[:-1]))

Answer 9

Prolog, 159 152 bytes

Code:

v(1,^,0).
v(1,v,-1).
v(0,^,1).
v(0,v,-1).
v(-1,^,1).
v(-1,v,0).
r(X,[H|T]):-T=[],v(X,H,Z),write(Z);v(X,H,Z),r(Z,T).
p(S):-atom_chars(S,L),r(0,L).

Test it yourself:
Online Interpreter here

Example

>p("^vvv^^vv^vv^v^^vvv").
-1

>p("^vvv^^vv^vv^v^")
1

Edit: Saved 7 bytes by unifying r-clauses with OR.

Answer 10

CJam, 16 bytes

0re`W=(2%*c'a--g

This will crash after printing 0, if applicable. The error can be suppressed with the Java interpreter. If you try this online, ignore everything but the last line of output.

How it works

0                e# Push a 0 on the stack.
 r               e# Read a whitespace-separated token from STDIN.
  e`             e# Perform run-length encoding.
    W=           e# Select the last [repetitions character] pair.
                 e# This will fail for the empty string, so the
                 e# interpreter will print the stack's only element (0).
      (          e# Shift out the number of repetitions.
       2%        e# Compute its parity.
         *       e# Create a string, repeating the character 1 or 0 times.
          c      e# Cast to character.
                 e# This will fail for a zero-length string, so the
                 e# interpreter will print the stack's only element (0).
           'a-   e# Subtract the character 'a' from '^' or 'v'.
              -  e# Subtract the difference (integer) from 0.
               g e# Apply the sign function.

Answer 11

JavaScript (ES6), 64 59 58 52 bytes

f=v=>(t=/\^*$|v*$/.exec(v)[0]).length*(t<'v'?1:-1)%2

This is based on the observation that only the last stretch of the repetition (of either ^ or v) affects the result.

Thanks to Neil for golfing off 6 bytes.

Answer 12

Haskell, 40 bytes

1%'^'=0
_%'^'=1
1%_=-1
_%_=0
v=foldl(%)0

Answer 13

Python 2, 177 159 72 bytes

Still kinda new to this code golf thing.

def v(s): 
 c=0 
 for i in s:c=((0,1)[c<1],(0,-1)[c>-1])[i=="^"] 
 return c

EDIT: Fixed the incorrect behavior.
EDIT 2: Thanks @MorganThrapp for shaving off lots of bytes.

Answer 14

Scala, 75 bytes

def d(s:String)=s./:(0){case(1,94)|(-1,'v')=>0;case(_,94)=> 1;case _=> -1}

Test for implemented function.

  object Util {
        def d(s: String) = s./:(0) { 
    case (1, '^') | (-1, 'v') => 0
    case (_, '^') => 1
    case (_, _) => -1
  }      
      def main(s: Array[String]): Unit = {
        println("1 == " + d("^vvv^^vv^vv^v^^^"))
        println("1 == " + d("^vvv^^vv^vv^v^"))
        println("-1 == " + d("^vvv^^vv^vv^v^^vvv"))
        println("0 == " + d("^^^vvvvvvvvvvvv"))
        println("0 == " + d("vvv^^^^^^^^^^^^"))
      }
    }

Answer 15

APL, 17

(⊣×≠)/⌽0,2-'^ '⍳⍞

For interpreters without fork notation (like GNU APL), it would be {⍺×⍺≠⍵}/⌽0,2-'^ '⍳⍞ (19). This is probably the most boring of possible solutions because it works directly from the definition of the problem.

Answer 16

Ruby, 41 35 bytes

Regex. Only the last button pressed is interesting, so check the run-length of that. Then compare it to "a" (or any letter between ^ and v) to get 1 or -1.

->s{s[/(.?)\1*$/].size%2*(?a<=>$1)}

Answer 17

C# 6, 18 + 80 = 98 bytes

Requires:

using System.Linq;

Actual function:

int S(string v)=>v.Split(new[]{"^^","vv"},0).Last().Length<1?0:v.Last()<95?1:-1;

How it works: the code first removes everything before the last ^^ or vv. That content is not relevant because clicking the same button twice will always cancel your vote. It does this by splitting on ^^ and vv and taking the last item. If this item is an empty string (.Length<1), then the function returns 0 because all voting has been cancelled. If the string it not empty, then it just looks at the last char of the original string: it will override all previous votes. If the char code is smaller than 95, then it will be 94, ^, so it returns 1, otherwise -1.

Answer 18

Python 2.7, 79 75 88

s=input()
print (0,(1,-1)[s[-1]=='v'])[len(s[s.rfind(('v^','^v')[s[-1]=='v'])+1:])%2!=0]

Answer 19

05AB1E, 14 12 11 bytes

Îvy'^QDŠ‹+<

Port of @Sp3000's Gol><> answer.

NOTE: @Grimy already posted a shorter 8 bytes alternative for 05AB1E, so make sure to upvote him!

Try it online or verify all test cases.

Explanation:

Î            # Push 0 (later mentioned as `r`) and the input-string
 v           # Loop over the characters of the input:
  y'^Q      '#  Does the current character equal "^"?
             #  (results in 1 for truthy; 0 for falsey - later mentioned as `b`)
      D      #  Duplicate this result `b`
       Š     #  Triple swap (`r`,`b`,`b`) to (`b`,`r`,`b`)
        ‹    #  Check if the boolean `b` is smaller than the result-integer `r`
             #  (again results in 1 for truthy; 0 for falsey)
         +   #  Add them together
          <  #  Decrease this by 1
             # (Implicitly output the result-integer `r` after the loop)

Answer 20

05AB1E, 8 bytes

㤮öÓÆ.±

Try it online!

Alternative solutions with the same length: u㤮öÓÆ(, 㤮ögÓÆ(.

Answer 21

Minkolang 0.11, 28 22 bytes

0$I2&N."j"o-34*:dr=,*!

Try it here.

Explanation

0                         Push a 0 (running total)
 $I                       Push length of input
   2&N.                   Output as integer and stop if this is 0
       "j"                Push 106
          o               Take character from input (94 for ^, 118 for v)
                          <<so ^ becomes +12 and v becomes -12>>
           -              Subtract
            34*:          Divide by 12
                d         Duplicate top of stack
                 r        Reverse stack
                  =,      Push 0 if equal, 1 otherwise
                    *     Multiply
                          <<this handles two of the same vote in a row>>
                     !    Unconditional trampoline (jumps the 0 at the beginning)

Note that there is no N. at the end. That's because I let it wrap around to the beginning. When the input is empty, the final tally is output as integer and the program stops.

Answer 22

Pyth, 13 bytes

ut+Jq\^H>JGzZ

Answer 23

Mathematica, 60 bytes

Mod[#,2]Sign@#&@Tr@Last@Split@StringCases[#,{"^"->1,_->-1}]&

Answer 24

Shape Script, 26 bytes

"^"$"0>1@-"~"v"$"0<1-"~0@!

How it woks:

"^"$     split input on '^'
"
  0>         Check if the number is more than 0 (1 if true, 0 if false).
  1@-        subtract the answer from one.
"~       Join it back together, with this string in place of '^'
"v"$     Split on 'v'
"        
  0<         Check if 0 is more than the number (1 if true, 0 if false).
  1-         subtract one from the results
"~       Join it back together, with this string in place of 'v'
0@       add a zero to the stack and place it under the string just built. 
!        run the string as code

Answer 25

Perl 5, 41 bytes

40 bytes, plus 1 for -p

/(.)\1*$/;$_=((length$&)%2)*($1=~v?-1:1)

/(.)\1*$/; compares the input string to the regex /(.)\1*$/, i.e. sees whether it ends with a single character repeated some number ≥1 of times.

If so, $& is the whole repetition string and $1 is the character; otherwise (i.e. the input string is empty), those two variables are the empty string.

$1=~v?-1:1 compares $1 to the regex v and returns −1 if it matches and 1 otherwise.

And multiply that ±1 by (length$&)%2, the length of $& modulo 2.

Answer 26

C# 6, 18 + 97 95 = 115 113 bytes, no string methods, excessive LINQ

int v(string s)=>(int)s.Reverse().TakeWhile((c,i)=>i<1||c==s[s.Length-i])?.Sum(x=>x<95?1:-1)%2;

Truly deserves to be preceded by

using System.Linq;

Got the idea of using x<95?1:-1 instead of x=='^'?1:-1 from ProgramFOX's answer

Coincidences:

• The tweak I stole makes use of comparing to 95 – the byte count excluding the using statement, using said tweak
• The sum of the digits of the total byte count equals the number of digits of the total byte count written as roman numeral

Answer 27

C: 67 66 Bytes

golfed:

void f(char *v){int i=0,c,s=0;for(;v[i]!=0;i++){v[i]>94?s--:s++;}}

ungolfed:

void f (char *v)
{
    int i = 0, c, s = 0;

    for (;v[i]!=0;i++)
    {
        v[i] > 94 ? s-- : s++;
    }
}

Answer 28

Go, 179 bytes

An extremely naive solution.

package main
import(."fmt"."strings")
func main(){a:=""
i:=0
Scanln(&a)
b:=Split(a,"")
for _,e:=range b{switch i{case 1:i--
case 0:if e=="^"{i++}else{i--}
case-1:i++}}
Println(i)}

Ungolfed:

package main

import (
    ."fmt"
    ."strings"
)

func main() {
    a := ""
    i := 0
    Scanln(&a)
    b := Split(a, "")
    for _, e := range b {
        switch i {
        case 1:
            i--
        case 0:
            if e == "^" {
                i++
            } else {
                i--
            }
        case -1:
            i++
        }
    }
    Println(i)
}

Answer 29

C++, 90 87 bytes

int f(char* c){int r=1,T[2][3]={{2,2,1},{1,0,0}};while(*c)r=T[*c++==94][r];return r-1;}

Ungolfed:

int f( char* c )
{
     // Lookup table created from Golfing description. Need to shift values up by 1
     // because will be using them to index table later
     int lookup_table[ 2 ][ 3 ] = {{2, 2, 1}, {1, 0, 0}};
     int result = 1;                 // empty string result: 1
     while ( *c )                    // check if end of string
     {
          int  row    = *c - '^' ? 1 : 0; // if char is '^' search in second row else in first
          auto column = result;           // column to look for is current result
          result      = lookup_table[ row ][ column ];
          c++;
      }
      return result - 1;                  // make expected result
 }

hehe finally got to write c++ in C++ ;-)

Edit:

1. Exchanged tenary row calculation to simple boolean expression.

Answer 30

Braingolf, 17 bytes

VR{.#veM|}lvlMR-s

Try it online!

Thought I could get it shorter by subtracting 9 from the charcodes of each character, then using J to change vowels to 1 and non-vowels to 0, but it turned out to be the same length

Answer 31

CJam, 27 24 bytes

'^_]r+e`W=(2%\(i99-g@*W*

Try it Online.

All I took from Dennis' answer is g (sign function) .

Answer 32

Ruby, 43

->s{a=0
s.bytes{|i|b=9-i/11;a=a!=b ?b:0}
a}

9-i/11 evaluates to 1 or -1 when given the ascii codes of ^ (94) or v (118)

In test program:

f=->s{a=0
s.bytes{|i|b=9-i/11;a=a!=b ?b:0}
a}

g=gets.chomp
puts f[g]

Answer 33

C (107 bytes)

#include<stdio.h>
int c,t;main(){while((c=getchar())-'\n')(c=='^')?++t:((c=='v')?--t:0);printf("%d\n",t);}

Answer 34

Japt, 17 bytes

¬®c nL gÃrÈ-Y©Y}0

Try it online!

Unpacked & How it works

Uq mZ{Zc nL g} rXY{X-Y&&Y}0

Uq     Convert the input string into list of chars
mZ{    Map...
Zc       Get charcode
nL       100 - charcode
g        sign(100 - charcode)
}
rXY{   Reduce...
X-Y&&Y   If X == Y, return 0; Otherwise, return Y
}0     ... default being 0

Having an array of -1s and 1s at hand yields a nice reducing function.

Answer 35

PHP 5, 47 bytes

Run on the command line, using php -nR 'code here'. The option -R counts as one byte.

echo(strrpos($s=x.$argn,"^")-strrpos($s,v))%2;

Answer 36

Julia 0.6, 38 bytes

s->foldl((x,c)->cmp(cmp('u',c),x),0,s)

Try it online!

Julia version of xnor's answer. (Uses foldl instead of reduce to save a byte.)

Answer 37

Jelly, 9 bytes

ŒgṪO³_^/Ṡ

Try it online!

Assumes the program this function is in doesn't take command-line arguments.

For a 10-byte version that doesn't assume so:

ŒgṪOȷ2_^/Ṡ

Try it online!

Answer 38

Perl 5 -pF, 32 bytes

$\=int(/v/?-($\>=0):$\<1)for@F}{

Try it online!

Answer 39

Turing Machine Code, 115 bytes

init:r
accept:e
r,^
^,_,>
r,v
v,_,>
r,_
e,0,-
^,^
r,_,>
^,v
r,v,-
^,_
e,1,-
v,^
r,^,-
v,v
r,_,>
v,_
w,-,>
w,_
e,1,-

Try it here

Ungolfed:

//-------CONFIGURATION
init: startread
accept: end

//startread
startread,^
read^,_,>

startread,v
readv,_,>

startread,_
end,0,-

//read^
read^,^
startread,_,>

read^,v
startread,v,-

read^,_
end,1,-

//readv
readv,^
startread,^,-

readv,v
startread,_,>

readv,_
write-,-,>

write-,_
end,1,-

The approach I used was to compare every two adjacent characters and either remove them if they canceled, or to remove just the first if the second one were to overwrite the first. An extra state is needed so that -1 can be written, because this simulator does not support multi-symbol cells.