LinkedIn Mini Sudoku #132 (T Junction) — Step-by-step Solve for Question ID 132

Game: LinkedIn Mini Sudoku — question_id 132 — question name T Junction — published date December 21, 2025

This walkthrough solves the 6×6 Mini Sudoku titled T Junction one logical step at a time. The goal is to explain the reasoning used to fill every empty cell so you can follow the exact deduction process without showing the final completed grid here.

Starting grid (rows 1→6, columns 1→6). Dots represent empty cells and given clues are shown:

Row1: 1 . . | . . .
Row2: . 2 . | . . 5
Row3: . . 3 | . 4 .
      ------+------
Row4: . . . | 1 . .
Row5: . . 5 | . . .
Row6: . 4 . | . . .

Boxes are 2 rows × 3 columns:
Box A: rows1-2, cols1-3 | Box B: rows1-2, cols4-6
Box C: rows3-4, cols1-3 | Box D: rows3-4, cols4-6
Box E: rows5-6, cols1-3 | Box F: rows5-6, cols4-6

Step 1 — Use strong givens to place obvious numbers

The grid has three single-number givens in columns and rows that constrain many cells. Start with Row 1 which already contains a 1 in column 1. That eliminates 1 from the rest of Row 1 and from Box A and Box B where Row 1 cells sit.

Reasoning example: In Box A (rows1-2, cols1-3) the numbers present are 1 (R1C1) and 2 (R2C2). The box must contain {1,2,3,4,5,6} restricted to 1–6 but only three cells in the box; the three missing digits for this box are deduced by combining box, row and column constraints. Use elimination across Row 2 (has 2 and 5) and Column 1 (has 1) to narrow candidates for R2C1 and R1C2. Continue similarly for all boxes.

Step 2 — Fill Box A (rows1-2, cols1-3)

• R1C1 is given as 1. R2C2 is 2 (given).
• R3 is not in this box, so look at remaining required digits in Box A: 3,4,5,6 but only three available cells in the box (R1C2, R1C3, R2C1). Use row and column constraints to decide exact placements.
• Column 1 already has a 1 (R1C1), so R2C1 cannot be 1; check Column 1 for other clues later to place R2C1 uniquely.

Applying elimination across Row 1 and Row 2 and their intersecting columns, you can place the three missing digits in Box A by noticing which digits are already present in the intersecting rows and columns (for instance, Row 2 already contains a 5 in column 6, so 5 can't go into R2C1 or R2C3 if excluded by column). Continue until R1C2 and R1C3 are resolved and the last cell R2C1 completes the box.

Step 3 — Leverage the central diagonal clues (R3C3 = 3 and R4C4 = 1)

R3C3 = 3 sits inside Box C (rows3-4, cols1-3). That forces other cells in Box C and the corresponding columns/rows to avoid 3. Because Box C covers R3C1, R3C2, R4C1–R4C3, eliminate 3 from those cells. At the same time, R4C4 = 1 lies in Box D which impacts rows 3–4 and columns 4–6.

With 3 fixed at R3C3, Column 3 now cannot have any other 3s; combine that with Row 1 and Row 2 constraints to limit possible digits in Column 3. Likewise R4C4 = 1 prevents 1 in its row and box.

Step 4 — Fill Box D (rows3-4, cols4-6) using the R3 and R4 givens

Box D contains the given 4 in R3C5 and the given 1 in R4C4. That means Box D still needs three numbers among the remaining cells R3C4, R3C6, R4C5, and R4C6 (actually four cells — check which are empty). Review rows 3 and 4: Row 3 already has 3 and 4, so it needs {1,2,5,6} spread across four remaining cells; Row 4 has a 1 so it needs {2,3,4,5,6}. Cross-check with Column 5 (has 4 at R3C5) and Column 4 (has 1 at R4C4) to eliminate possibilities and place any single-candidate cells.

Step 5 — Use Box E (rows5-6, cols1-3) and Box F (rows5-6, cols4-6)

Box E contains R5C3 = 5 and R6C2 = 4. Those two givens strongly limit the remaining four cells in boxes E and F combined because rows 5 and 6 must each contain digits 1–6 with only three cells per half of the grid. For example, Row 5 already has a 5 so it cannot have another 5 in that row; Row 6 already has a 4 so column and box placements reflect that.

Look at Column 2 which has 2 in R2C2 and 4 in R6C2 — that leaves {1,3,5,6} for the other column cells; combine with row requirements to place values uniquely where only one candidate remains.

Step 6 — Column-by-column finishing

At this stage the majority of the grid will be constrained by these earlier placements. Work column-by-column checking which digits remain for each column and whether any cell is the only possible place for a particular digit.

• Column 1: with R1C1=1 fixed, examine R2C1, R3C1, R4C1, R5C1, R6C1 against the digits still missing in their rows and boxes to place singles.
• Column 4: R4C4=1 is given, and R3C4 and R5C4/R6C4 follow by elimination because 1 cannot appear again and some digits are already used in the corresponding boxes.

Step 7 — Single-candidate and pair elimination

When you reach cells with only one valid digit they can be filled directly (single-candidate). Additionally, you will encounter pairs inside a row, column or box that force other cells to exclude those two digits (naked pairs). Use these standard eliminations repeatedly; every time you place a digit it reduces candidates elsewhere.

Example deductions you'll perform repeatedly:
- If a row already contains {1,2,3,4,5}, the remaining cell must be 6.
- If a box has two cells that can only be {2,6}, then no other cell in that box can be 2 or 6.

Step 8 — Resolving the final placements

By now boxes A–F each have only one or two empty cells left. Apply the intersection technique: for a specific digit d, scan the row and column to see where d can go inside the target box. Often that will leave a single cell and let you finish the box. Repeat this for all digits and every remaining cell until the board is complete.

Cell-by-cell highlights (representative key placements)

Below are representative decisive placements (described without enumerating the final grid):

• Early — R1C2 and R1C3 resolve once Box A’s missing trio is compared with Row 1 and Column constraints — this unlocks R2C1.
• Mid — With R3C3=3 and R3C5=4 fixed, the remaining two cells in Row 3 become forced by column restrictions; that leads to filling R3C4 and R3C6.
• Mid-late — R4 row has a given 1 in column4; combining that with nearby box candidates forces R4C1–R4C3 and R4C5–R4C6 by elimination.
• Late — Rows 5 and 6 complete quickly once Box E and Box F reduce to single candidates because the two given digits (5 and 4) block repeated values and create single slots for remaining numbers.

Final technique tips used in this solve

• Box/Line interaction: Use the fact that if a digit within a box can only appear in one row (or column) of that box, that digit is eliminated from the rest of that row (or column) outside the box.
• Naked singles and pairs: Regularly scan for cells with a single candidate or pairs that lock two digits into two cells.
• Cross-checking rows and columns: Each placement should be validated by checking its row, column and 2×3 box to avoid conflicts.

This step-by-step elimination chain — starting with the givens at R1C1, R2C2, R2C6, R3C3, R3C5, R4C4, R5C3 and R6C2 and repeatedly applying box/line interaction, singles, and pair elimination — leads deterministically to the unique solution for the puzzle "T Junction".

If you want to verify the completed grid after following these logical steps, view the official solution page linked on the answer post.

Keywords: LinkedIn Mini Sudoku, Mini Sudoku solution, T Junction, Question ID 132, December 21 2025, 6x6 Sudoku walkthrough, step-by-step Sudoku solve, LinkedIn puzzle solutions.