RE3 (PSX) Water Control Room Puzzle crack tool

/*
 * One of the final puzzles in Resident Evil 3 (PSX). This puzzle is obnoxious.
 * Use this tool to crack the puzzle if you've been beating your
 * head against a wall and lost track of what the initial puzzle state was
 *
 * To use: (rudimentary javascript knowledge recommended)
 * 1. paste this script into a javascript console but don't hit enter
 * 2. go to the top of the pasted script and make sure the four arrays match the game
 * 3. hit enter
 *
 * The gauges are in 8 groups of 2, and shift 2 units, as such there are 8^3=512
 * combinations possible printarray() prints the array as it appears in-game,
 * i.e. in 8 groups of 2
 *
 * When bruteforce() is run, it will tell you how many times you need to shift each
 * channel left to solve the puzzle.
 * Shifts in this crack are always left because why over-complicate things.
 * And since this is mod 8, you can subtract from 8 to get the equivalent right
 * shift count.
 * In the state shown below, the output is 0, 6, 2, which also means 0 2R 2L
 *
 * -Braden
 */

let want = [2,1,1,1,3,0,2,0,2,1,3,1,1,0,2,1];
let A    = [1,0,1,1,1,0,0,0,1,1,1,0,0,0,1,1];
let B    = [1,0,1,0,1,0,1,1,0,0,1,0,0,0,0,0];
let C    = [1,0,0,0,1,1,0,0,1,0,1,0,0,0,1,0];

function shift(array){
    return array.slice(2).concat(array.slice(0,2));
}

function printarray(arr){
    let outbuf = "";

    for(let i = 0; i < arr.length; i += 2)
        outbuf += arr[i] + "" + arr[i+1] + " ";

    console.log(outbuf);
}

function test(a,b,c){
    let result = a.map((x, i) => x + b[i] + c[i]);

    printarray(want);
    printarray(result);

    for(let i = 0; i < result.length; ++i)
        if(result[i] != want[i])
            return false;

    return true;
}

function nshift(arr, n){
    let out = arr;

    for(let i = 0; i < n; ++i)
        out = shift(out);

    return out;
}

function ntest(a,b,c){
    return test(nshift(A, a), nshift(B, b), nshift(C, c));
}

function bruteforce(){
    /* cubic loop, 512 iterations max */
    for(let a = 0; a < 8; ++a)
    for(let b = 0; b < 8; ++b)
    for(let c = 0; c < 8; ++c)
        if(ntest(a,b,c))
            return console.log(`${a} left, ${b} left, ${c} left`);
}

bruteforce();