Lantern Grid practice: resolve the shared wall, then audit the last bulb.
Try a new five-bulb challenge before opening the answers. Compare two fourth-bulb choices and predict three diagnostic counts with a playable board and editable prediction fields.
This is a try-first challenge on a third clue arrangement. The aim is to explain why a placement is forced, not to follow a list of clicks. You will solve a shared-neighbour decision, compare three unfinished or faulty patterns, and predict the game's diagnostic counts before checking them. Keep the answer panels closed until you have written a reason for each choice.
Set up the shared-walls challenge
Choose Practice this example beside the article, or Load challenge board in the game. Confirm Toggles 0 and these clues. This is a direct practice entry for the third board in the existing five-puzzle collection, different from the boards in lessons 1 and 2. It does not add a new game or expand the collection.
· · · 0 · 0 · · 1 · · · · · 1 · 2 · · 0 1 0 2 · 1
A dot is a white cell; every digit is a wall. Rows count downward and columns to the right. There are fifteen white cells. You can place and remove bulbs, use Undo, or Retry the same clues. Use paper for exclusions: the game does not store pencil marks. The three prediction fields below the challenge board accept your estimates for dark cells, mismatched wall counts and visible bulb pairs.
Challenge 1: finish the board before opening the route
Find the unique bulb pattern. For each bulb, give a local reason or a light-source argument. In particular, explain the shared neighbour R4C1: how does the 1 at R5C1 influence the 2 at R4C2? Finish with Check lighting, which alone records the win. A valid pattern need not use a prescribed order of moves.
Before revealing the solution, write your bulb coordinates and answer these questions:
- Which two bulbs are forced by the 2 at R5C3?
- Why can the 1 at R2C4 no longer use R2C3 or R3C4 after those two bulbs?
- What rules out a bulb at R3C2 once the shared neighbour is supplied?
- Which single bulb finally lights the dark top-left area?
Challenge 2: compare two fourth bulbs
Retry. Place bulbs at R4C3, R5C4 and R2C5, in that order. You should have Toggles 3 and 9 of 15 white cells lit. Compare a fourth bulb at R3C2 with a fourth bulb at R4C1. Test each choice from the same three-bulb checkpoint; Undo removes the trial bulb before the second trial.
For each trial, note the illumination total and all three diagnostic counts. Then decide whether the position can still be completed. A choice that leaves only one mismatched wall may be worse than a choice that leaves several dark cells. Support that judgment using the two walls that share R4C1; do not decide from the number of bright cells.
Challenge 3: forecast three diagnostic triples
For each pattern below, predict dark cells / mismatched wall counts / visible bulb pairs. Count a mismatched numbered wall once even when its error is more than one bulb. Count each visible pair once. Underfilled walls count as mismatches too. The totals refer to these exact patterns, not to a later repair.
| Pattern | Bulb coordinates | Your task |
|---|---|---|
| A | R4C3, R5C4, R2C5, R4C1 | Are exact wall counts enough? |
| B | R4C3, R5C4, R2C5, R3C2 | Can this brighter branch be completed? |
| C | R4C3, R5C4, R2C5, R4C1, R1C2, R3C4 | Does extra light preserve a valid finish? |
In the live challenge, enter your three estimates and choose Compare predictions. It compares your values with the current board without adding a toggle, moving a bulb or recording a win. Use Check lighting separately to test the actual pattern. Editing, checking, Undo, Retry, New puzzle or loading a board clears the prediction fields and message. Nothing is stored after refresh.
Pattern C contains the five-bulb solution plus a trial bulb. If you have already registered a win, Undo the final solution move to reopen the board before adding it back and then adding the extra bulb, or Retry and build C directly. Disabled buttons after a win are an end state, not a broken practice entry.
Reveal the forced deductions and full five-bulb route
The 2 at R5C3 has exactly two adjacent white cells: R4C3 and R5C4. Both require bulbs. Their wall neighbours block any line between them. R4C3 illuminates R2C3 up column 3; R5C4 illuminates R3C4 up column 4. Neither of those lit cells can hold a bulb without seeing an existing bulb. The 1 at R2C4 therefore needs R2C5, its only remaining available white neighbour.
Now examine the bottom-left 1 at R5C1. Its only adjacent white cell is R4C1, so place that bulb. The 2 at R4C2 now has exactly its two required bulbs, at R4C1 and R4C3. Its remaining white neighbour R3C2 must stay bulb-free. A satisfied 2 is an exclusion rule as well as a count.
R1C1 is still dark. Its uninterrupted top-row source set is R1C1, R1C2 and R1C3; the wall below R1C1 stops its column immediately. The zero at R2C1 excludes R1C1, and the zero at R1C4 excludes R1C3. R1C2 is the only possible source. It also lights R2C2 and R3C2 down column 2, stopping at the 2 wall in row 4.
Retry and follow this route to compare exact checkpoints. Check lighting between rows if useful; it adds no toggle. All five checkpoints have 0 visible bulb pairs.
- R4C3: Toggles 1; 5 white cells lit; 6 wall counts mismatched.
- R5C4: Toggles 2; 7 cells lit; 4 mismatches.
- R2C5: Toggles 3; 9 cells lit; 2 mismatches.
- R4C1: Toggles 4; 11 cells lit; all wall counts exact.
- R1C2: Toggles 5; all 15 cells lit; all wall counts exact.
Use Check lighting after the fifth toggle. The checked finish locks all fifteen white-cell buttons and moves focus to Undo last toggle. Undo removes R1C2, returning to Toggles 4 and four dark cells, with all wall counts still exact. Retry starts again at zero toggles on the same clues.
Reveal the three prediction triples, final map and self-check
| Pattern | Lit cells | Dark / walls / pairs | Result |
|---|---|---|---|
| A | 11 of 15 | 4 / 0 / 0 | Incomplete: exact wall numbers leave a dark region. |
| B | 13 of 15 | 2 / 1 / 0 | Impossible without removing the trial at R3C2. |
| C | 15 of 15 | 0 / 2 / 1 | Invalid despite full illumination. |
In A the dark cells are R1C1, R1C2, R2C2 and R3C2. In C the extra bulb at R3C4 sees R5C4 along column 4. It also overfills the 1 walls at R2C4 and R3C5, producing two mismatched walls. The zero at R4C5 stays exact: it is diagonal from R3C4, so that bulb does not count beside it. Removing just R3C4 restores the valid five-bulb pattern, but you must Check lighting again to register the win.
+ B + 0 + 0 + + 1 B + + + + 1 B 2 B + 0 1 0 2 B 1
All fifteen white cells receive light, all ten numbered walls are exact and none of the five bulbs sees another. Independent enumeration finds exactly one full bulb pattern for these clues. This is a unique-pattern statement, not a claim that only one order of clicks works.
Why is a bulb at R1C1 not a substitute for R1C2? It would violate the zero immediately below at R2C1. From pattern A it lights only R1C1 and R1C2 newly, leaving R2C2 and R3C2 dark; the wall below blocks its downward beam. Its diagnostic triple is 2 / 1 / 0. Tracing the exact rays explains both the forbidden placement and its incomplete coverage.
Review your reasoning, not just the final display
Keep one reason for each forced bulb, explain why B cannot be repaired by adding bulbs, and distinguish an illuminated empty cell from a bulb. If your predictions differ, rebuild the listed pattern after Retry before changing your arithmetic. An old extra bulb changes the counts even when the clue layout is correct.
This challenge combines several deductions on one small fixed board. It is not a measured difficulty rating, a guarantee about larger puzzles, or a promise of improved ability. The prediction comparison is a local learning aid, not a solver or a stored score. Load beginner board and Load strategy board select the earlier lessons; Load challenge board restores this one. New puzzle continues through the same five prepared rounds.