Loop Trace advanced practice: force an edge, then count the loops.
Try a new board without zero clues, test a missing-edge assumption, and distinguish one valid loop from a number-perfect drawing with two loops.
A zero clue gives an immediate exclusion. What do you do when there is no zero? This practice asks you to test an assumption, carry its consequences through two dots, and check the result against a cell. A second, separate diagnostic asks why a drawing can have perfect numbers and no loose ends without being a valid answer.
Both diagrams are new for this lesson. The first is a solvable challenge with one valid loop; the second is a deliberately invalid diagnostic, not another puzzle you are expected to solve. Try the questions before expanding their answers. “Advanced” describes the kind of reasoning practised here, not a measured difficulty rating.
Challenge 1: solve without a zero clue
Choose Practice this example beside the article or Load advanced board in the game. Begin with Edits 0 and no marks. Your three clue rows should be:
3 1 3 3 2 3 3 1 3
Cell coordinates refer to the three rows and columns of squares. Dot coordinates refer to the four rows and columns of dots. H1-2 is the horizontal segment joining the second and third dots of the top boundary. V1-2 drops from the second top dot to the dot below it.
Keep deductions about unused edges on paper for the checkpoints below. Crosses are optional, but every change of mark adds an edit. Our route counts only the listed line changes.
- At the top-left corner cell, can either of its two boundary edges be unused? Explain using its clue of three and the corner dot.
- After applying that reasoning at both top corners, suppose H1-2 stays unused. What happens to V1-2, V1-3 and the top-middle clue of one?
- Use the matching bottom row and the middle side cells to finish. Before closing the last connection, predict the remaining open ends and matching clues.
Reveal the corner argument and missing-edge contradiction
The top-left corner dot has only two possible incident edges, H1-1 and V1-1. In a valid finished loop they are either both present or both absent. If both were absent, the top-left cell could have at most its two other sides, not the required three. Therefore both boundary edges are lines. The same argument forces H1-3 and V1-4 at the top-right corner.
Now suppose H1-2 were unused. At dot row 1, column 2, the existing H1-1 would have to continue through V1-2. At dot row 1, column 3, H1-3 would have to continue through V1-3. Those two vertical lines border the top-middle cell, giving it at least two sides. Its clue is one: a contradiction. H1-2 must therefore be a line.
Once H1-2 is drawn, both adjoining top dots already have two lines. V1-2 and V1-3 must stay unused. The top-middle one is reached, so H2-2 is unused too. Each top corner three now forces its bottom edge: H2-1 on the left and H2-3 on the right.
You can reproduce the contradiction rather than merely accepting the argument. Retry, draw H1-1, V1-1, H1-3 and V1-4, then add V1-2 and V1-3 while leaving H1-2 unmarked. At Edits 6 the top-middle cell reads two of one. There are two line groups, four open ends and no branching dot. The problem is an overfilled cell, not a branch. Undo twice returns to the four-edit corner checkpoint. Do not keep the two temporary vertical lines in your solution.
Challenge 2: complete the board, then audit your answer
Return to Edits 0 before using the full route below, or continue from the appropriate checkpoint without repeating edges already drawn. Try to explain why each added line is necessary before opening the worked answer.
Reveal the full deductions and sixteen-line route
- H1-1 → V1-1 → H1-3 → V1-4. The two top corner threes force these four boundary edges. They make two separate bent paths; a temporary separation is allowed.
- H1-2 → H2-1 → H2-3. The contradiction in Challenge 1 forces H1-2. The two top interior verticals and H2-2 remain unused, so the two corner threes force H2-1 and H2-3.
- H4-1 → V3-1 → H4-3 → V3-4. Apply the same corner argument to the bottom-left and bottom-right threes. Each corner requires both of its boundary edges. There are now three temporary line groups.
- H4-2 → H3-1 → H3-3. If H4-2 were unused, both V3-2 and V3-3 would be required at the bottom interior dots and would overfill the bottom-middle one. Thus H4-2 is a line. V3-2, V3-3 and H3-2 stay unused. The bottom corner threes then force H3-1 and H3-3.
- V2-2 → V2-3. At dot row 2, column 1, V1-1 and H2-1 already meet, so V2-1 is unused. The middle-left three has H2-1 and H3-1 and therefore needs V2-2. The matching argument on the right excludes V2-4 and forces V2-3. These two interior verticals supply the centre two and join the upper and lower paths.
The order above records sixteen line changes, not a claim that sixteen is the fewest possible interface actions under every marking strategy. At Edits 15, after V2-2, there is one group with two open ends and seven matching clues. V2-3 joins those ends. At Edits 16 all nine clues match and one closed group remains, with no open ends or branching dots. The game locks the edges and focuses Undo. Undo returns to the fifteen-edit checkpoint; replacing V2-3 wins again.
| Edits | Matching clues | Groups | Open ends |
|---|---|---|---|
| 4 | 0 of 9 | 2 | 4 |
| 7 | 3 of 9 | 1 | 2 |
| 11 | 3 of 9 | 3 | 6 |
| 14 | 6 of 9 | 2 | 4 |
| 15 | 7 of 9 | 1 | 2 |
| 16 | 9 of 9 | 1 | 0 |
This is the complete edge ledger. A zero means unused; you may leave it unmarked.
H1: 1 1 1 H2: 1 0 1 H3: 1 0 1 H4: 1 1 1 V1: 1 0 0 1 V2: 0 1 1 0 V3: 1 0 0 1
Diagnostic: perfect numbers, wrong number of loops
Choose Load two-loop test. A persistent notice identifies this as a diagnostic with no valid single-loop solution. It is intentionally outside the solvable collection. Begin with no marks and these different clues:
1 0 1 3 2 3 3 2 3
Draw the two groups below. Each listed edge begins unmarked, so one activation makes it a line. Predict whether Check loop will report a win, how many clues will match, and how many open ends will remain.
- H2-1 → V2-2 → V3-2 → H4-1 → V3-1 → V2-1. Draw the left rectangle, spanning the bottom two cells in column 1.
- H2-3 → V2-4 → V3-4 → H4-3 → V3-3 → V2-3. Draw the right rectangle, spanning the bottom two cells in column 3. Leave every other edge unused.
Reveal the two-loop check and why adding a bridge fails
At Edits 12 all nine clues match. There are twelve lines, two line groups, zero open ends and zero branching dots. This is not a win: you can trace either rectangle and return to its start without reaching the other one. The rule requires one loop, not merely a collection of closed paths.
The complete ledger is:
H1: 0 0 0 H2: 1 0 1 H3: 0 0 0 H4: 1 0 1 V1: 0 0 0 0 V2: 1 1 1 1 V3: 1 1 1 1
For a further check, add H3-2 between the two rectangles. The drawing becomes one connected group, but both endpoints already had two lines. At Edits 13 there are two branching dots and only seven matching clues: the two middle-column cells each have three bordering lines instead of two. Connecting the groups by adding a segment does not turn two closed loops into one nonbranching loop. Undo once restores the twelve-edit diagnostic.
Do not search for a winning arrangement of these diagnostic clues. They were checked against every simple loop on this small grid and have none. The exercise is to explain a rejected drawing. Load advanced board to return to the separate, uniquely solvable challenge. Retry this board clears the currently selected diagram; it does not switch between the two examples.
Your self-check before leaving the lesson
Can you explain why a corner three forces both boundary edges, why assuming a boundary edge is absent can overfill a neighbouring one, and why two closed groups cannot be repaired just by adding a connecting segment? State the failed condition each time. A low edit count, a green cell, or an allowed click is not a proof of a correct answer.
The first board has been checked for exactly one valid single-loop answer. The second is an explicit diagnostic with none. These statements concern the printed clues only. Symmetry shortened the explanation on the first board; an asymmetric puzzle may need other deductions. A contradiction from one assumption proves that assumption wrong only when every intervening step follows from the rules.