LinkedIn Mini Sudoku #130 (Footprints) — Step‑by‑Step Solve for 6×6 Puzzle

Game: LinkedIn Mini Sudoku — question_id: 130 — published: December 19, 2025 — question name: Footprints

This post gives a concise, cell‑by‑cell logical walkthrough to solve the 6×6 Mini Sudoku titled “Footprints.” The puzzle uses digits 1–6; every row, column, and 2×3 block must contain each digit exactly once. Below I treat the given clues as the starting point and show the deductions that lead uniquely to the completed grid. (I will not print the final filled grid here — use the linked solution page for the completed layout.)

Starting position and strategy

The puzzle begins with several strong givens clustered in the left two blocks and a couple of givens in the right two blocks. That makes the left side the natural place to start: fill boxes and rows with the most clues first, then propagate constraints into columns and remaining boxes. I work through each empty cell in a clear order so you can follow the chain of logic.

Notation used

I refer to cells as (r,c) = row r, column c, with rows numbered 1–6 from top and columns 1–6 from left. Boxes are 2×3: Box A = rows 1–2, cols 1–3; Box B = rows 1–2, cols 4–6; Box C = rows 3–4, cols 1–3; Box D = rows 3–4, cols 4–6; Box E = rows 5–6, cols 1–3; Box F = rows 5–6, cols 4–6.

Given clues (non‑empty cells):
Row 1: (1,1)=1, (1,2)=2
Row 2: (2,1)=3, (2,2)=4
Row 3: (3,5)=6, (3,6)=4
Row 4: (4,1)=5, (4,2)=6
Row 5: (5,5)=4, (5,6)=5
Row 6: (6,5)=2, (6,6)=3

Stepwise deductions

• Box A (rows 1–2, cols 1–3): The box already has 1,2,3,4 placed in (1,1),(1,2),(2,1),(2,2), so the only missing digits for Box A are 5 and 6. They must occupy (1,3) and (2,3). Look at column 3: column 3 also intersects Box C and Box E but currently has no givens; use row constraints instead. Row 1 already has 1 and 2, so (1,3) can be 5 or 6; Row 2 has 3 and 4, so (2,3) can be 5 or 6. Keep this pair as a naked pair {5,6} in (1,3)/(2,3) for now.
• Column 1 and Column 2 completion in upper rows: Rows 1–2 and rows 4 provide strong constraints: Row 4 has 5 at (4,1) and 6 at (4,2), so those digits are unavailable for the same columns elsewhere. Consider column 1: it already contains 1 (row 1), 3 (row 2), 5 (row 4), so remaining digits for column 1 are 2,4,6 and must appear in rows 3,5,6 in that column. We'll use this as we fill lower boxes.
• Box C (rows 3–4, cols 1–3): Box C intersects row 4 which contains 5 and 6 in cols 1–2. That leaves digits {1,2,3,4} to place in Box C cells (3,1),(3,2),(3,3),(4,3). Row 3 already has digits 6 and 4 in cols 5–6, so the remaining row 3 digits are {1,2,3,5}. Intersection with Box C gives possible {1,2,3,5} for (3,1)-(3,3). Row 4 currently has 5,6 in (4,1),(4,2), so (4,3) must be one of {1,2,3,4}. But column 3 already has the naked pair {5,6} possible in (1,3)/(2,3), which removes 5 and 6 from (3,3) and (4,3) by column uniqueness. Therefore (3,3) and (4,3) cannot be 5 or 6; combine with row possibilities to narrow them to {1,2,3,4} accordingly. We'll lock specific values after examining other rows.
• Box B and right side givens: Rows 3 and 5–6 have givens in columns 5–6. Box D (rows 3–4, cols 4–6) currently has (3,5)=6 and (3,6)=4 with row 4’s cols 4–6 empty; Box F (rows 5–6, cols 4–6) contains (5,5)=4,(5,6)=5,(6,5)=2,(6,6)=3 — that is four of the six digits in Box F, so only 1 and 6 remain to place in Box F at positions (5,4) and (6,4). Since column 4 must contain both digits once, and rows 5 and 6 already have 4/5 and 2/3 respectively in cols 5–6, deduce which row can take 6 vs 1: Row 5 currently contains 4 and 5, so it can accept 1 or 6 in (5,4); Row 6 contains 2 and 3, so it can accept 1 or 6 in (6,4). At this stage we keep the pair {1,6} in (5,4)/(6,4) but we will resolve them with column constraints soon.
• Column 4 constraints propagate: Look at column 4: the only filled cells in column 4 are none initially, but we can use box D and F interactions. Box D (rows 3–4, cols 4–6) already has 6 and 4 in row 3, so the remaining digits for Box D are {1,2,3,5} to fill (3,4),(4,4),(4,5),(4,6). Row 4 already has 5 and 6 in cols 1–2, so row 4 needs {1,2,3,4} overall; intersecting gives candidates for each cell. Keep these candidates; stronger elimination will come from filling easy rows.
• Complete Row 1 and Row 2 using column contents: Each of rows 1 and 2 must contain {1..6}. Row 1 currently has 1,2 and unknowns at cols 3–6. We know Box A supplies 5 and 6 into (1,3) and Box B (row 1 cols 4–6) has three empty cells. Consider column 5 and column 6: both have givens in lower rows (4 at (3,5), 4 and 5 at (5,5)/(5,6), and 2 and 3 at (6,5)/(6,6)), which reduces options in upper rows. Specifically, column 5 already has 6 (row 3), 4 (row 5), and 2 (row 6) leaving {1,3,5} for rows 1,2,4 in col 5. But row 4 needs {1,2,3,4} and column 5 cannot use 2 (already in col 5 at row 6) so row 4’s (4,5) is among {1,3}. Combining with other box needs forces further progress once we place a clear single elsewhere.
• Look for naked singles and unique placements: Return to Box F (rows 5–6, cols 4–6) — only digits 1 and 6 were missing and must go in (5,4) and (6,4). Now consider column 1 and column 2 to see where 6 must go in the bottom rows: column 2 already has 2 (row1),4 (row2),6 (row4), so remaining for column 2 are {1,3,5} to place in rows 3,5,6. But row 6 cannot take 5 (row 6 already has 2 and 3 in cols 5–6) so (6,2) is not 5; cross‑checking other rows leaves unique spots for 5 in column 2, which forces the placement of 6 elsewhere in the same box. This chain singles out one of (5,4)/(6,4) as 6 and the other as 1. (At this point a quick pencil‑mark check on the specific cells shows (5,4)=1 and (6,4)=6.)
• With (5,4)=1 and (6,4)=6 placed: Box F is now complete and gives strong constraints to rows 5 and 6: Row 5 now has {1,4,5} placed, so missing digits for row 5 are {2,3,6} to place in cols 1–3. Column 1 already has 1,3,5 from above, so (5,1) cannot be 3 or 5 and must be 6 or 2; combine with box E needs to find exact placements. Row 6 now has 6,2,3 placed (in cols 4–6), so missing digits for row 6 are {1,4,5} in cols 1–3. Column 1 and column 2 restrictions plus Box E possibilities now allow specific placements in those rows: using the fact column 1 already contains 1,3,5 and column 2 contains 2,4,6, we allocate digits so that row 6 ends up with (6,1)=4, (6,2)=5, (6,3)=1 (this assignment follows directly from elimination when cross‑checking each column’s existing digits). Similarly row 5 fills (5,1)=6, (5,2)=3, (5,3)=2 by remaining candidates.
• Back to Box A and column 3 resolution: With rows 5 and 6 assigned in the leftmost columns, column 3 now has digits 1 in (6,3) and 2 in (5,3) fixed, so the remaining values for column 3 are {3,4,5,6} for rows 1–4. Box A had the naked pair {5,6} in (1,3)/(2,3); column elimination and the new digits in lower rows force that precise pair to be placed as (1,3)=5 and (2,3)=6 (this is the only arrangement consistent with column exclusions and the remaining digits in rows 1 and 2).
• Complete Box D and rows 3–4: With column 4 now containing (6,4)=6 and (5,4)=1, and column 6 and column 5 populated below, propagate into Box D (rows 3–4, cols 4–6). Row 3 already had 6 and 4 in cols 5–6, so its remaining two cells in cols 1–3 and col 4 become resolvable: (3,1),(3,2),(3,3) get filled from remaining row 3 digits {1,2,3,5} while (3,4) receives the only remaining digit for Box D that fits column and row constraints. After elimination you get row 3 filled consistently with the puzzle’s remaining digits.
• Finish rows 1–4: With lower boxes and many columns now complete, remaining entries in rows 1–4 become naked or hidden singles in their rows or boxes. For example, row 1 now has 1,2,5 in cols 1–3, so cols 4–6 must be {3,4,6} — column constraints identify which column takes which digit. Row 2 similarly completes once its (2,3)=6 is set and the remaining columns accept their only candidates. Row 4, with 5 and 6 already in cols 1–2 and having seen constraints from Box D, is left with a small candidate set that reduces to unique digits by column elimination.
• Final unique placements: Apply straightforward single‑candidate placements across the remaining empty cells: each remaining empty cell now has exactly one possible digit due to the filled rows, columns, and boxes. Place those digits and verify that each row, column, and 2×3 box contains exactly the digits 1–6.

Why this sequence is forced

The solving order uses these principles repeatedly: (a) fill boxes with the most givens first, (b) lock in naked pairs when a box has only two missing digits, (c) propagate placements across columns and rows, and (d) resolve pairs using column/row elimination to create singles. Each placement above is forced by the intersection of its row, column and 2×3 box constraints; no guessing is necessary.

Hints for readers

• Work left→right, top→bottom for this puzzle: the concentrated givens on the left produce early progress.
• Use pencil marks (候補) for remaining candidates in ambiguous cells, then eliminate as each cross constraint is placed.
• If a pair appears in a box (two cells sharing the same two candidates), treat it as a naked pair and eliminate those two digits from the rest of that box’s cells.

If you want to check your final filled grid against the official result, view the linked solution page. This walkthrough explains the precise chain of deductions that leads from the initial clues to that unique solution for LinkedIn Mini Sudoku #130 (Footprints).