Pivot partitioning by scanning

Algorithmic problem: Pivot partioning

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_2-m_1 }[/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_1 \le m_1 \le i_2 \le m_2 \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], none of them is decreased.

Break condition: It is [math]\displaystyle{ i_1=m_1, i_2=m_2 }[/math] and [math]\displaystyle{ i_3 = n+1 }[/math]

Induction basis

Abstract view:

1. Set [math]\displaystyle{ m_1 }[/math] and [math]\displaystyle{ m_2 }[/math] according to the loop invariant.
2. Initialize [math]\displaystyle{ i_1,i_2 }[/math] and [math]\displaystyle{ i_3 }[/math].

Implementation:

1. Set [math]\displaystyle{ m_1:=1 }[/math] .
2. For [math]\displaystyle{ j \in {1,...,n} }[/math]: If [math]\displaystyle{ S[j]\lt p }[/math] increase [math]\displaystyle{ m_1 }[/math] by [math]\displaystyle{ 1 }[/math].
3. Set [math]\displaystyle{ m_2:m_1 }[/math]
4. For [math]\displaystyle{ j \in {1,...,n} }[/math]: If [math]\displaystyle{ S[j]=p }[/math] increase [math]\displaystyle{ m_2 }[/math] by [math]\displaystyle{ 1 }[/math].
5. [math]\displaystyle{ i_1:=1 }[/math].
6. [math]\displaystyle{ i_2:=m_1 }[/math].
7. [math]\displaystyle{ i_3:=m_2 }[/math].
8. While [math]\displaystyle{ i_1\lt m_1 }[/math] and [math]\displaystyle{ S[j]\lt p }[/math], set [math]\displaystyle{ i_1:=i_1+1 }[/math].
9. While [math]\displaystyle{ i_2\lt m_2 }[/math] and [math]\displaystyle{ S[j]=p }[/math] , set [math]\displaystyle{ i_2:=i_2+1 }[/math].
10. While [math]\displaystyle{ i_3\lt n+1 }[/math] and [math]\displaystyle{ S[j]\gt p }[/math], set [math]\displaystyle{ i_3:=i_3+1 }[/math] .

Proof: Obvious.

Induction step

Abstract view:

1. Move each of [math]\displaystyle{ i_1,i_2 }[/math], and [math]\displaystyle{ i_3 }[/math] forward as long as the invariant is not violated.
2. If the algorithm is not done yet, identify two out of the three index pointers [math]\displaystyle{ i_1,i_2 }[/math], and [math]\displaystyle{ i_3 }[/math]such that at least one of these two can be moved one step forward after a swap of the elements at these two positions.

Implementation:

1. If [math]\displaystyle{ i_1=m_1 }[/math] and [math]\displaystyle{ i_2=m_2 }[/math] and [math]\displaystyle{ i_3=n+1 }[/math], terminate the algorithm and return [math]\displaystyle{ m_1,m_2 }[/math].
2. If [math]\displaystyle{ i_1\lt m_1 }[/math] :
   1. If [math]\displaystyle{ S[i_1]=p }[/math]:
      1. Swap [math]\displaystyle{ S[i_1] }[/math] and [math]\displaystyle{ S[i_2] }[/math].
      2. Set [math]\displaystyle{ i_2:=i_2+1 }[/math] until [math]\displaystyle{ i_2=m_2 }[/math] or else [math]\displaystyle{ S[i_2] \ne p }[/math].
   2. Otherwise (that is, [math]\displaystyle{ S[i_1]\gt p }[/math]):
      1. Swap [math]\displaystyle{ S[i_1] }[/math] and [math]\displaystyle{ S[i_3] }[/math].
      2. Set [math]\displaystyle{ i_3:=i_3+1 }[/math] until [math]\displaystyle{ i_3=n+1 }[/math] or else [math]\displaystyle{ S[i_3] \ngtr p }[/math].
   3. While and , set .
3. Otherwise (that is, [math]\displaystyle{ i_1=m_1 }[/math]):
   1. Swap [math]\displaystyle{ S[i_2] }[/math] and [math]\displaystyle{ S[i_3] }[/math].
   2. Set [math]\displaystyle{ i_2:=i_2+1 }[/math] until [math]\displaystyle{ i_2=m2 }[/math] or else [math]\displaystyle{ S[i_2] \ne p }[/math] .
   3. Set [math]\displaystyle{ i_3:=i_3+1 }[/math] until [math]\displaystyle{ i_3=n+1 }[/math] or else [math]\displaystyle{ S[i_3] \ngtr p }[/math].

Correctness: Due to the various "otherwise" clauses, all possible cases are covered. Therefore, it remains to show that the invariants are maintained in each of the considered cases. As the values of and are never changed after initialization, all three parts of the first invariant are trivially maintained. Since (resp., , ) is only increased in case (resp., , ), Invariant 2.1 is maintained as well. So, we will focus on Invariants 2.2-2.4. If Step 2 applies, maintenance of Invariants 2.2-2.4 might be obvious. So consider Step 3. If Step 3 applies, we know that , but or (or both). Due to , all elements less than are correctly placed at the positions . Therefore, both cases, the case and and the case and are impossible, because there cannot be exactly one wrongly placed element. In other words, it is and in Step 3. Again due to , it is and . Steps 2.1.2 and 2.2.2 ensure and , so we have and . This observation justifies the procedure of Step 3.