DBMS for algorithmic problems

Examples

Longest Common Subsequence

dp(name = lcs) {
    input = {sa: [u8], sb: [u8]}
    node {
        key = {a: uint, b: uint}
        payload = || sa[a] == sb[b] ? 1 : 0
        next = || {
            (a == sa.len || b == sb.len) && yield node.end() && return

            if sa[a] == sb[b] {
                yield link(sa[a])
                yield node({a: a + 1, b: b + 1})
            } else {
                yield node({a: a + 1, b: b})
                yield node({a: a, b: b + 1})
            }
        }

        add = |a, b| max(a, b)
        zero = max.neutral

        mul = |a, b| a + b
        one = add.neutral
    }

    node (name = end) {
        add = |a, b| max(a, b)
        zero = max.neutral
    }

    begin node({a: 0, b: 0})

    output node.end()
}