Solving dependency constraints

Question

I have a classic dependency solving problem. I thought I was headed in the right direction, but now I\'ve hit a roadblock and I am not sure how to proceed.

Background

Answer 1

I'm not expert on the problem, I'm proposing a complete solution that is not optimal, as there are many things that can be optimized ..

The algorithm is simple, it's ideally a recursive set . intersection DFS :

Algorithm

Def

Define: Name as String on format [ .* ]
Define: Version as String on format [ dd.dd.dd ]
Define: Revision as { Name, Version, Requirement }
Define: Range as { min, max }
Define: Condition as { Name, Range }
Define: Requirement as Set OR as Set
Define: Component as { Name, Range, Requirement }
Define: System as Set

Input

Input: T as System aka basis
Input: C as Set aka conditions to apply

Initialization

Init: S as Set = { S[i] as Condition | S[i] = {T[i].Name,T[i].Range} }
Init: Q as Stack = { push(q) | q[i] = C[i] }

Process

for (Condition c in C)
{
    S.find(c.Name).apply(c)
}

While (Q.size > 0)
{
    Condition q = Q.pop()

    switch (T.assert(S.find(q.Name),q))
    {
      case VALID:
        S.find(q.Name).apply(q)
        q.push(S.find(q.Name).Requirement)

      case INVALID:
        S.find(q.Name).set(INVALID)

      case IDENTICAL:
      case SKIP:
    }
}

return S aka Solution

Operations

Stack.push insert an item at the front of a stack

Stack.pop remove an item from the front of a stack

System.assert(Condition a, Condition b):
    if (a is INVALID) then return SKIP
    else if (b.Range = a.Range) then IDENTICAL
    else if (b.Range - a.Range = {}) then VALID
    else INVALID

Set.find(x) search an item based on condition x

Condition.apply(Condition b) = { this.Name, intersection(this.Range,b.Range) }