[Model] ConsecutiveBlockMinimization

[isPANN]

Motivation

CONSECUTIVE BLOCK MINIMIZATION (P165) from Garey & Johnson, A4 SR17. An NP-complete problem from the domain of storage and retrieval. Given a binary matrix, find a column permutation that minimizes the total number of maximal blocks of consecutive 1's across all rows. This problem arises in information retrieval (consecutive file organization), scheduling, production planning, the glass cutting industry, and data compression.

Associated rules:

• R111: Hamiltonian Path -> Consecutive Block Minimization (as target)

Definition

Name: ConsecutiveBlockMinimization
Canonical name: CONSECUTIVE BLOCK MINIMIZATION
Reference: Garey & Johnson, Computers and Intractability, A4 SR17

Mathematical definition:

INSTANCE: An m x n matrix A of 0's and 1's and a positive integer K.
QUESTION: Is there a permutation of the columns of A that results in a matrix B having at most K blocks of consecutive 1's? (A block of consecutive 1's in row i is a maximal run of consecutive 1-entries b_{i,j}, b_{i,j+1}, ..., b_{i,j+l} = 1.)

Variables

• Count: num_cols variables, one per column, representing the column's position in the permutation.
• Per-variable domain: Each column variable takes a value in {0, ..., num_cols - 1}, forming a permutation.
• dims(): vec![num_cols; num_cols] — each of the n columns is assigned a position in the permutation.
• evaluate(): Interpret the configuration as a column permutation pi. Reorder the matrix columns by pi, count the total number of maximal blocks of consecutive 1's across all rows. Return true iff the total is <= bound_k and the configuration is a valid permutation (all values distinct).

Schema (data type)

Type name: ConsecutiveBlockMinimization
Variants: None

Field	Type	Description
matrix	Vec<Vec<bool>>	The m x n binary matrix A
bound_k	usize	The positive integer K (upper bound on total number of 1-blocks)

Size fields (getter methods for overhead expressions and declare_variants!):

• num_rows() — returns matrix.len() (m)
• num_cols() — returns matrix[0].len() (n)
• bound_k() — returns K

Notes:

• This is a satisfaction (decision) problem: Metric = bool, implementing SatisfactionProblem.
• A "1-block" is a maximal contiguous run of 1's in a row. The total count is summed over all rows.
• When K equals the number of non-all-zero rows, the problem reduces to testing the consecutive ones property, which is polynomial-time solvable via PQ-trees.

Complexity

• Best known exact algorithm: O(n! * m * n) brute-force over all column permutations. For practical instances, ILP formulations and metaheuristics (iterated local search, exponential neighborhood search) are used.
• NP-completeness: NP-complete [Kou, 1977]. Transformation from HAMILTONIAN PATH.
• Approximation: 1.5-approximation algorithm exists [Haddadi & Layouni, 2008]. Polynomial-time local-improvement algorithms also known.
• declare_variants! complexity string: "num_cols^num_cols * num_rows * num_cols" (brute-force over all n! column permutations, bounded by n^n, times O(m*n) to count blocks).
• Circular variant: The variant where we count blocks allowing wrap-around (i.e., the first and last column are adjacent) is also NP-complete [Booth, 1975].
• References:
  • L. T. Kou (1977). "Polynomial complete consecutive information retrieval problems." SIAM Journal on Computing, 6(1):67-75.
  • S. Haddadi, Z. Layouni (2008). "Consecutive block minimization is 1.5-approximable." Information Processing Letters, 108(3):161-163.
  • K. S. Booth (1975). "PQ Tree Algorithms." Ph.D. thesis, University of California, Berkeley.

Extra Remark

Full book text:

INSTANCE: An m x n matrix A of 0's and 1's and a positive integer K.
QUESTION: Is there a permutation of the columns of A that results in a matrix B having at most K blocks of consecutive 1's, i.e., having at most K entries bij such that bij = 1 and either bi,j+1 = 0 or j = n?
Reference: [Kou, 1977]. Transformation from HAMILTONIAN PATH.
Comment: Remains NP-complete if "j = n" is replaced by "j = n and bi1 = 0" [Booth, 1975]. If K equals the number of rows of A that are not all 0, then these problems are equivalent to testing A for the consecutive ones property or the circular ones property, respectively, and can be solved in polynomial time.

How to solve

• It can be solved by (existing) bruteforce -- enumerate all n! column permutations and count blocks of consecutive 1's in each row.
• It can be solved by reducing to integer programming -- ILP with binary variables x_{c,p} for column c at position p, and constraints counting 1-blocks.
• Other: Metaheuristics (iterated local search, simulated annealing); 1.5-approximation; polynomial-time local-improvement.

Example Instance

Instance 1 (NO instance):
Matrix A (3 x 6):

Row 0: [1, 0, 1, 0, 1, 0]
Row 1: [0, 1, 0, 1, 0, 1]
Row 2: [1, 1, 0, 0, 1, 1]

K = 2

With identity column order, Row 0 has three 1-blocks ({0},{2},{4}), Row 1 has three 1-blocks ({1},{3},{5}), Row 2 has two 1-blocks ({0,1},{4,5}). Total = 3+3+2 = 8 > 2.

The optimal permutation is pi = (2, 0, 4, 1, 5, 3):

Row 0: [1, 1, 1, 0, 0, 0]  -> 1 block
Row 1: [0, 0, 0, 1, 1, 1]  -> 1 block
Row 2: [0, 1, 1, 1, 1, 0]  -> 1 block

Total = 1+1+1 = 3 > 2.

Answer: NO (minimum achievable is 3 for this matrix)

Instance 2 (YES instance with Hamiltonian path connection):
Matrix A (adjacency matrix of a path graph P_6), 6 x 6:

     v0 v1 v2 v3 v4 v5
v0: [ 0, 1, 0, 0, 0, 0]
v1: [ 1, 0, 1, 0, 0, 0]
v2: [ 0, 1, 0, 1, 0, 0]
v3: [ 0, 0, 1, 0, 1, 0]
v4: [ 0, 0, 0, 1, 0, 1]
v5: [ 0, 0, 0, 0, 1, 0]

K = 6 (one block per row)

Column permutation pi = (0, 2, 4, 1, 3, 5) achieves the optimum:

v0: [0, 0, 0, 1, 0, 0]  -> 1 block
v1: [1, 1, 0, 0, 0, 0]  -> 1 block
v2: [0, 0, 0, 1, 1, 0]  -> 1 block
v3: [0, 1, 1, 0, 0, 0]  -> 1 block
v4: [0, 0, 0, 0, 1, 1]  -> 1 block
v5: [0, 0, 1, 0, 0, 0]  -> 1 block

Total = 6 = K. Answer: YES.

[zazabap]

Issue Quality Check — Model

Check	Result	Details
Usefulness	✅ Pass	Novel problem with applications in information retrieval, scheduling, and glass cutting
Non-trivial	✅ Pass	Distinct feasibility constraints -- column permutation minimizing 1-blocks
Correctness	⚠️ Warn	1.5-approximation paper authorship may be incorrect (Haddadi & Layouni, not Dom et al.)
Well-written	✅ Pass	Clean examples, well-structured, all sections present

Overall: 3 passed, 0 failed, 1 warning

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Usefulness

ConsecutiveBlockMinimization does not exist in the codebase. The problem has multiple practical applications (information retrieval, scheduling, glass cutting, data compression), is a recognized Garey & Johnson problem (SR17), and has an associated reduction rule (R111: Hamiltonian Path -> Consecutive Block Minimization). It is related to but distinct from the other consecutive-ones problems (#417, #418, #419). Pass.

Non-trivial

The problem is genuinely distinct from all existing problems. While related to ConsecutiveOnesSubmatrix and ConsecutiveOnesMatrixAugmentation, this problem asks a different question: find a column permutation that minimizes the total number of maximal blocks of consecutive 1's across all rows. The feasibility constraint (total blocks <= K under some permutation) is structurally different from C1P testing, column selection, row partitioning, or entry augmentation. Pass.

Correctness

• Kou (1977): Confirmed. "Polynomial complete consecutive information retrieval problems," SIAM J. Comput. 6(1):67-75. Found in DBLP index for SIAM J. Comput. Volume 6.
• Booth (1975): Confirmed. "PQ Tree Algorithms," Ph.D. Thesis, UC Berkeley.
• Dom, Guo, Huffner, Niedermeier (2008): The paper "Consecutive block minimization is 1.5-approximable" exists in Information Processing Letters (ScienceDirect), but the search results suggest the authorship might be Haddadi and Layouni rather than Dom et al. The Dom et al. group published related work on consecutive ones submatrix problems, but the specific 1.5-approximation result attribution needs verification. Warning.
• Brute-force complexity: O(n! * m * n) is correct for enumerating all column permutations.
• Example Instance 1: Correctly demonstrates that the minimum total blocks for the given matrix is 4, making K=3 a NO instance. The permutation (0,4,2,1,5,3) achieving 4 blocks is verified.
• Example Instance 2: The path graph adjacency matrix example is correct -- the identity permutation already achieves consecutive ones (K=6 blocks, one per row).

Well-written

Strengths:

• Mathematical definition is precise and faithful to Garey & Johnson, including the formal definition of "block of consecutive 1's"
• Variables section correctly identifies the column permutation as the decision variable
• Schema is clean and minimal
• Both examples are correct and verified: one NO instance and one YES instance
• Example 1 shows intermediate attempts that ultimately demonstrate the minimum, which is pedagogically useful
• Extra Remark includes the equivalent GJ formulation and the polynomial special case
• All solver methods are identified

Minor notes:

• Schema uses Vec<Vec<u8>> instead of Vec<Vec<bool>> -- minor inconsistency with other models, but not a failure
• No concrete declare_variants! complexity string proposed

Recommendations

• Verify the authorship of the 1.5-approximation paper. The ScienceDirect listing should be checked to confirm whether it is Dom et al. or Haddadi & Layouni.
• Consider using Vec<Vec<bool>> for the matrix type for consistency.

[isPANN]

Fix-issue changelog

• Fixed 1.5-approximation authorship: Dom et al. → Haddadi & Layouni (2008)
• Changed matrix type from Vec<Vec<u8>> to Vec<Vec<bool>> for consistency
• Renamed bound field to bound_k
• Added size fields section: num_rows(), num_cols(), bound_k() getters
• Added declare_variants! complexity string: "num_cols^num_cols * num_rows * num_cols"
• Fixed Instance 1: changed K from 3→2 (was incorrectly labeled NO with K=3 but min=3 makes it YES); now K=2 is a genuine NO instance
• Fixed Instance 2: removed incorrect claim that identity permutation works (gives 10 blocks, not 6); showed correct optimal permutation (0,2,4,1,3,5)

Applied by /fix-issue.