Typ 3 Grammar to test if zipped representation of 2 binary numbers a and b is b= a+1

complexTyp3Grammar.js

let log = n => console.log(n)


function binary(n) {
    if (n == 0) return '0'
    if (n == 1) return '1'
    let last = String(n & 1 )
    return binary(n >> 1) + last
}

binary(7)

// zips to strings of equal length, appending 0s when neccesary
function zip(a, b) {
    if (a.length > b.length) return zip(a, '0' + b)
    if (a.length < b.length) return zip('0' + a, b)

    let aa = [...a]
    let ab = [...b]

    let res = ""
    for (let i =0; i < aa.length; i++) {
        res += aa[i] + ab[i]
    }

    return res
}

zip("hello", "world")

function create(n) {
    return zip(binary(n), binary(n+1))
}

function verify(str) {
    let a = [...str]

    let i=0
    // (a==b)*
    for (; i < a.length; i += 2) {
        if (a[i] != a[i+1]) break
    }

    // (01)
    if (!(a[i] == '0' && a[i+1] == '1')) return false
    i += 2


    // (10)*
    for (; i < a.length; i += 2) {
        if (a[i] != '1' || a[i+1] != '0' ) break
    }


    // (a==b==0)*
    for (; i < a.length; i += 2) {
        if (a[i] != '0' || a[i+1] != '0') return false
    }

    return true
}

for (let i =0; i< 100; i++)
    log(verify(create(i)))