Brendan's Sudoku solver done with left-to-right comprehensions

sudoku.js

Object.prototype.copy = function () {
    let o = {}
    for (let i in this)
        o[i] = this[i]
    return o
}

// Containment testing for arrays and strings that should be coherent with their __iterator__.
Array.prototype.contains = String.prototype.contains = function (e) {
    return this.indexOf(e) != -1
}

Array.prototype.repeat = String.prototype.repeat = function (n) {
    let s = this.constructor()
    for (let i = 0; i < n; i++)
        s = s.concat(this)
    return s
}

String.prototype.center = function (w) {
    let n = this.length
    if (w <= n)
        return this
    let m = Math.floor((w - n) / 2)
    return ' '.repeat(m) + this + ' '.repeat(w - n - m)
}

Array.prototype.toString = Array.prototype.toSource
Object.prototype.toString = Object.prototype.toSource

function all(seq) {
    for (let e of seq)
        if (!e)
            return false
    return true
}

function some(seq) {
    for (let e of seq)
        if (e)
            return e
    return false
}

function cross(A, B) {
    return [for (a of A) for (b of B) a+b]
}

function dict(A) {
    let d = {}
    for (let e of A)
        d[e[0]] = e[1]
    return d
}

function set(A) {
    let s = []
    for (let e of A)
        if (!s.contains(e))
            s.push(e)
    return s
}

function zip(A, B) {
    let z = []
    let n = Math.min(A.length, B.length)
    for (let i = 0; i < n; i++)
        z.push([A[i], B[i]])
    return z
}

rows = 'ABCDEFGHI'
cols = '123456789'
digits   = '123456789'
squares  = cross(rows, cols)
unitlist = [for (c of cols) cross(rows, c)]
           .concat([for (r of rows) cross(r, cols)])
           .concat([for (rs of ['ABC','DEF','GHI']) for (cs of ['123','456','789']) cross(rs, cs)])
units = dict(for (s of squares)
              [s, [for (u of unitlist) if (u.contains(s)) u]] )
peers = dict(for (s of squares)
              [s, set([for (u of units[s]) for (s2 of u) if (s2 != s) s2])])

// Given a string of 81 digits (or . or 0 or -), return a dict of {cell:values}.
function parse_grid(grid) {
    grid = [for (c of grid) if ('0.-123456789'.contains(c)) c]
    let values = dict(for (s of squares) [s, digits])

    for (let [s, d] of zip(squares, grid))
        if (digits.contains(d) && !assign(values, s, d))
            return false
    return values
}

// Eliminate all the other values (except d) from values[s] and propagate.
function assign(values, s, d) {
    if (all(for (d2 of values[3]) if (d2 != d) eliminate(values, s, d2)))
        return values
    return false
}

// Eliminate d from values[s]; propagate when values or places <= 2.
function eliminate(values, s, d) {
    if (!values[s].contains(d))
        return values // Already eliminated
    values[s] = values[s].replace(d, '')
    if (values[s].length == 0)
	return false  // Contradiction: removed last value
    if (values[s].length == 1) {
	// If there is only one value (d2) left in square, remove it from peers
        let d2 = values[s][0]
        if (!all(for (s2 of peers[s]) eliminate(values, s2, d2)))
            return false
    }
    // Now check the places where d appears in the units of s
    for (let u of units[s]) {
        let dplaces = [for (s of u) if (values[s].contains(d)) s]
	if (dplaces.length == 0)
	    return false
	if (dplaces.length == 1)
	    // d can only be in one place in unit; assign it there
            if (!assign(values, dplaces[0], d))
                return false
    }
    return values
}

// Used for debugging.
function print_board(values) {
    let width = 1 + Math.max.apply(Math, [for (s of squares) values[s].length])
    let line = '\n' + ['-'.repeat(width*3)].repeat(3).join('+')
    for (let r of rows)
        print([for (c of cols)
                values[r+c].center(width) + ('36'.contains(c) && '|' || '')]
              .join('') + ('CF'.contains(r) && line || ''))
    print()
}

easy = "..3.2.6..9..3.5..1..18.64....81.29..7.......8..67.82....26.95..8..2.3..9..5.1.3.."

print_board(parse_grid(easy))

// Using depth-first search and constraint propagation, try all possible values.
function search(values) {
    if (!values)
        return false    // Failed earlier
    if (all(for (s of squares) values[s].length == 1))
        return values   // Solved!

    // Choose the unfilled square s with the fewest possibilities
    // XXX Math.min etc. should work with generator expressions and other iterators
    // XXX Math.min etc. should work on arrays (lists or tuples in Python) as well as numbers
    let a = [for (s of squares) if (values[s].length > 1) values[s].length + s].sort()
    let s = a[0].slice(-2)

    return some(for (d of values[s]) search(assign(values.copy(), s, d)))
}

hard = '4.....8.5.3..........7......2.....6.....8.4......1.......6.3.7.5..2.....1.4......'

print_board(search(parse_grid(hard)))