Pivot partitioning by scanning

Algorithmic problem: Pivot partioning by scanning

Prerequisites: None

Type of algorithm: loop

Auxiliary data: Five index pointers, [math]\displaystyle{ m_1,m_2,i_1,i_2,i_3 \in \mathbb N. }[/math]

Abstract View

Invariant:

1. The values of [math]\displaystyle{ m_1 }[/math] and [math]\displaystyle{ m_2 }[/math] only depend on the input sequence and are, hence, constant throughout the loop. More specifically, in there are
   1. [math]\displaystyle{ m_1-1 }[/math] elements less than [math]\displaystyle{ p }[/math],
   2. [math]\displaystyle{ m_1-m_2 }[/math] elements equal to [math]\displaystyle{ p }[/math], and
   3. [math]\displaystyle{ n - m_2 }[/math]greater than [math]\displaystyle{ p }[/math].
2. Before and after each iteration, it is:
   1. [math]\displaystyle{ 1 \le i_2 \le m_1 \le m_2 \le i_3 \le i_3 \le n+1 }[/math];
   2. [math]\displaystyle{ S[j] \lt p }[/math] for [math]\displaystyle{ j \in {1,...,i_1-1} }[/math]; if [math]\displaystyle{ i_1 \lt m_1 }[/math], it is [math]\displaystyle{ S[i_1] \nless p }[/math];
   3. [math]\displaystyle{ S[j] =p }[/math] for [math]\displaystyle{ j \in {m_1,...,i_2-1} }[/math]; if [math]\displaystyle{ i_2 \lt m_2 }[/math], it is [math]\displaystyle{ S[i_2] \ne p }[/math];
   4. [math]\displaystyle{ S[j] \gt p }[/math]for [math]\displaystyle{ j \in {m_2,...,i_3-1} }[/math]; if [math]\displaystyle{ i_3 \le m_1 }[/math], it is [math]\displaystyle{ S[i_3] \ngtr p }[/math].

Variant: At least one of [math]\displaystyle{ i_1,i_2, }[/math] and [math]\displaystyle{ i_3 }[/math] is increased by at least [math]\displaystyle{ 1 }[/math] increases by [math]\displaystyle{ 1 }[/math].

Break condition: It is [math]\displaystyle{ i_1=m_1, i_2=m_2 }[/math] and [math]\displaystyle{ i_3 = n+1\lt /math == Induction basis == '''Abstract view:''' \lt math\gt \forall v,w \in V }[/math] we set

1. [math]\displaystyle{ M(v,v):=0, \forall v \in V }[/math]
2. [math]\displaystyle{ M(v,w):=l(v,w), (v,w) \in A }[/math]
3. [math]\displaystyle{ M(v,w):= +\infty }[/math], otherwise

Implementation: Obvious.

Proof: Nothing to show.

Induction step

Abstract view:

Implementation:

Correctness: Let [math]\displaystyle{ p }[/math] denote a shoortest [math]\displaystyle{ (v,w }[/math]-path subject to the constraint that all internal nodes are taken from [math]\displaystyle{ \{u_1, \ldots, u_n \} }[/math].

1. If [math]\displaystyle{ p }[/math] does not contain [math]\displaystyle{ u_i }[/math], [math]\displaystyle{ p }[/math] is even a shortest [math]\displaystyle{ (v,w) }[/math]-path such that all internal nodes are taken from [math]\displaystyle{ \{u_1, \ldots , u_{i-1} \} }[/math]. In this case, the induction hypothesis implies that the value of [math]\displaystyle{ M(v,w) }[/math] immediately before the i-th iteration equals the length of [math]\displaystyle{ p }[/math]. Clearly, the i-th iteration does not change the value of [math]\displaystyle{ M(v,w) }[/math] in this case.
2. On the other hand, suppose [math]\displaystyle{ p }[/math] does contain [math]\displaystyle{ u_i }[/math]. Due to the prefix property, the segment of [math]\displaystyle{ p }[/math] from [math]\displaystyle{ v }[/math] to [math]\displaystyle{ u_i }[/math] and from [math]\displaystyle{ u_i }[/math] to [math]\displaystyle{ w }[/math] are a shortest [math]\displaystyle{ (v,u_i) }[/math]-path and a shortest [math]\displaystyle{ (u_i,w) }[/math]-path, respectively, subject to the constraint that all internal nodes are from [math]\displaystyle{ \{u_1, \ldots , u_{i-1} \} }[/math]. The induction hypothesis implies that tehese lengths equal the values [math]\displaystyle{ M(v,u_i) }[/math] and [math]\displaystyle{ M(u_i, w) }[/math], respectively.

Complexity

Statement: [math]\displaystyle{ \mathcal{O}(n^3) }[/math]

Proof: The overall loop has [math]\displaystyle{ n }[/math] iterations. In each iteration, we update all [math]\displaystyle{ n^2 }[/math] matrix entities. Each update requires a constant number of steps.