How to use: Click any cell to toggle: ✗ (Eliminated) → ✓ (Confirmed Match) → Empty. Placing a ✓ automatically eliminates other items in the same row/column.

Deduction principles

The three golden rules of logic grid solving

01

Direct elimination (✗)

When a clue says "Lord Blackwood was not in the Library", immediately place an ✗ at the intersection of Blackwood and Library.

02

The unique match (✓)

When three out of four cells in a row are marked ✗, the remaining cell must be ✓. Confirming a ✓ eliminates all other cells in that row and column.

03

Transitive linkage

If Lady Cynthia is in the Study (✓), and the Study contained the Poison (✓), Lady Cynthia must have held the Poison.

Boolean Matrix Algebra & Worked Example

How does a 4×4 logic grid elimination matrix compute transitive deductions?

A k-category, N×N logic grid puzzle consists of k(k − 1)/2 pairwise N×N permutation submatrices (for k = 3 categories of N = 4 items: 3 submatrices of 16 cells = 48 cells total), where each submatrix M^(A,B) is a bijection with exactly one ✓ (1) and N − 1 cross-outs ✗ (0) in every row and column.

Plain-Text Boolean Elimination & Transitive Closure Equations

For categories Suspects S = {s_1..s_4}, Rooms R = {r_1..r_4}, and Weapons W = {w_1..w_4}, let M(x, y) ∈ {0 (✗), 1 (✓), ?} denote the truth value linking item x and item y:

  • Bijection / Orthogonal Cross-Elimination: M(a_i, b_j) = 1 ⇒ ∀ m ≠ j: M(a_i, b_m) = 0 and ∀ n ≠ i: M(a_n, b_j) = 0
  • Last Remaining Candidate (Forced Match): (∀ m ≠ j: M(a_i, b_m) = 0) ⇒ M(a_i, b_j) = 1
  • Positive Transitive Closure: M(s_i, r_j) = 1 ∧ M(r_j, w_k) = 1 ⇒ M(s_i, w_k) = 1
  • Negative Transitive Propagation: M(s_i, r_j) = 1 ∧ M(r_j, w_k) = 0 ⇒ M(s_i, w_k) = 0

Worked 4×4 Case Example: Suspects, Rooms & Weapons

Using the trainer’s 4 Suspects (Lord Blackwood, Lady Cynthia, Professor Vance, Dr. Sterling), 4 Rooms (Library, Study, Dining Room, Conservatory), and 4 Weapons (Dagger, Poison, Revolver, Rope), consider these 5 clues:

  1. Clue 1: “Lady Cynthia was in the Study.” → Set M(Lady Cynthia, Study) = ✓; auto-eliminate Library, Dining Room, and Conservatory for Lady Cynthia (3 ✗), and eliminate Study for Blackwood, Vance, and Sterling (3 ✗).
  2. Clue 2: “The Poison was found in the Study.” → Set M(Study, Poison) = ✓; by positive transitivity (Cynthia = Study ∧ Study = Poison), the solver automatically marks M(Lady Cynthia, Poison) = ✓.
  3. Clue 3: “Lord Blackwood was neither in the Library nor the Dining Room, and did not have the Rope.” → Since Study is already taken by Cynthia, Blackwood is eliminated from Library, Study, and Dining Room, forcing M(Lord Blackwood, Conservatory) = ✓.
  4. Clue 4: “Professor Vance was in the Library with the Dagger.” → Confirms M(Professor Vance, Library) = ✓ and M(Professor Vance, Dagger) = ✓, which by transitivity sets M(Library, Dagger) = ✓ and leaves M(Dr. Sterling, Dining Room) = ✓ as the last remaining room.
  5. Clue 5: Since Poison (Cynthia) and Dagger (Vance) are assigned and Blackwood cannot have the Rope (Clue 3), M(Lord Blackwood, Revolver) = ✓ and M(Dr. Sterling, Rope) = ✓ complete all 48 cells.

Solved 4×4 Logic Grid Truth Table & Submatrix Breakdown

Suspect (Category 1) Library Study Dining Room Conservatory Matched Room Matched Weapon Key Deduction Rule Used
Lord Blackwood ✗ (0) ✗ (0) ✗ (0) ✓ (1) Conservatory Revolver 3 room eliminations → forced 4th room + weapon exclusion
Lady Cynthia ✗ (0) ✓ (1) ✗ (0) ✗ (0) Study Poison Direct match + positive transitive link (Study = Poison)
Professor Vance ✓ (1) ✗ (0) ✗ (0) ✗ (0) Library Dagger Direct suspect-room and suspect-weapon conjunction
Dr. Sterling ✗ (0) ✗ (0) ✓ (1) ✗ (0) Dining Room Rope Last remaining column/row bijection in both subgrids