3D Medusa

3D Medusa extends Simple Colouring from a single digit to multiple digits. By connecting conjugate pairs AND bi-value cells, we build a more powerful colour web that can find eliminations impossible to spot otherwise.

Note: This technique uses chain colouring to display candidates. Square (▢) and Diamond (◇) shapes represent the two colour partitions.

How It Works

Building the Web

Unlike Simple Colouring which only uses strong links for ONE digit:

1. Start with any strong link (conjugate pair) for any digit
2. Extend through conjugate pairs (same digit, different cells)
3. Also extend through bi-value cells (same cell, different digits)
4. Colour alternates at each step: Square (Blue) ↔ Diamond (Orange)

The Connections

Same-digit connection: If R1C1 has Square (Blue) 7, and R1C1-R1C9 is a conjugate pair for 7, then R1C9 gets Diamond (Orange) 7.

Same-cell connection: If R1C1 has Square (Blue) 7 and R1C1 is bi-value {7, 9}, then R1C1 gets Diamond (Orange) 9.

This creates a web spanning multiple digits and cells.

How to Spot 3D Medusa

Multi-Digit Focus

The key to spotting 3D Medusa is connecting bi-value cells with conjugate pairs:

Step-by-Step Scanning

1. Start with bi-value cells — These bridge different digits
2. Extend through conjugate pairs — Same digit in exactly 2 cells of a unit
3. Alternate colours — Each link flips the colour
4. Check for contradictions — Rules 1, 4, 6 prove a colour FALSE
5. Check for eliminations — Rules 2, 3, 5 remove specific candidates

Visualisation Tip

The app displays 3D Medusa webs with:

• Solid lines between conjugate pairs (same digit, different cells)
• Ovals around bi-value cell candidates (different digits, same cell)
• Square (Blue)/Diamond (Orange) shapes on candidate vertices

Example

This example demonstrates a compact 3D Medusa web with just 4 cells and 2 digits.

Pattern Analysis:

• Bi-value cells: R1C5 and R1C7 both contain {6, 8}
• Web vertices: R1C5, R1C7, R3C5, R3C7
• Digits: 6 and 8
• Rule applied: Rule 5 (Sees Both Colours)
• Elimination: R5C5~8

puzzle: S9B090203044Y074Y01050807060N050N090204050R0R02C38J6Q035U0706094M02120104460403024Z6R1F5U05090108057Q9E0402062E1I09081U0402120701020Z078303830408061I170R071F08120902
mode: guided
technique: 3D Medusa
initial:
  layers:
    hints: true
steps:
  - text: >
      3D Medusa extends Simple Colouring by connecting multiple digits through bi-value cells. Look for cells with exactly two candidates.
    hint: subtle
    technique: MED

  - text: >
      R1C5 and R1C7 are bi-value cells, both containing {6, 8}. Enable the net display to see the colour web.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R1C5, R1C7]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Start colouring: R1C5 gets Square (Blue) 6 and Diamond (Orange) 8. Since it's bi-value, the two candidates get opposite colours (shown as an oval).
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R1C5]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Extend via conjugate pairs: R1C5-R1C7 is a conjugate pair for BOTH 6 and 8. So R1C7 gets Diamond (Orange) 6 and Square (Blue) 8 — colours flip for both!
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R1C5, R1C7]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Continue extending: R3C5 links to R1C5 (column 5, digit 8) getting Square (Blue) 8. R3C7 links to R1C7 (column 7, digit 6) getting Square (Blue) 6.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R1C5, R1C7, R3C5, R3C7]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Rule 5 check: R5C5 has candidate 8 but is NOT in the web. Does it see both colours for digit 8?
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R5C5]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      R5C5 sees R1C5 (Diamond 8, same column) AND R3C5 (Square 8, same column AND box). One must be true, so R5C5 cannot be 8. Eliminate R5C5~8.
    hint: detailed
    technique: MED
    state:
      selection:
        cells: [R5C5, R1C5, R3C5]
settings:
  showCandidates: true
  showControls: true
  showDescription: true
  navigation: numbered
  coordinateFormat: rncn

Elimination Rules

3D Medusa has six rules (compared to Simple Colouring's two):

Rule 1: Twice in a Cell

If a cell has TWO candidates of the SAME colour:

• That colour is FALSE
• Eliminate ALL candidates of that colour from the web
• (Or set all candidates of the opposite colour)

Pattern Analysis:

• Web vertices: R1C5, R1C8, R2C3, R2C5, R2C7, R3C3, R4C8
• Digits: 4, 5
• Contradiction: R2C3 has Square (Blue) 4 AND Square (Blue) 5 — same colour twice in one cell!
• Result: Square (Blue) is FALSE — eliminate all Square candidates

puzzle: S9B1O4C0750CIC201160301091A2C2U2E2K08061K485I6T046V9G2Q9U2I060I030A02622Y6I0501466Y6Q6Y1I090B2O684E094Y051I012I3I2M0A0E02E6D61ID609051Q6Y6U72620201082M023FAEAN051I9M
mode: guided
technique: 3D Medusa
initial:
  layers:
    hints: true
steps:
  - text: >
      Rule 1 finds contradictions when the same colour appears twice in one cell. Look for R1C8{4,5} — a bi-value cell.
    hint: subtle
    technique: MED

  - text: >
      Enable the net display. R1C8{4,5} connects digits 4 and 5. Multiple strong links extend from here.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R1C8]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Assign colours to R1C8: Square (Blue) 4, Diamond (Orange) 5 (bi-value link shown as oval). Via row 1, R1C5 gets Square (Blue) 5.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R1C5, R1C8]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Extend digit 4: R1C8(Square 4)→R2C7(Diamond 4) via box 3. R1C8(Square 4)→R4C8(Diamond 4) via column 8.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R1C8, R2C7, R4C8]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Extend digit 5: R1C5(Square 5)→R2C5(Diamond 5) via column 5. R2C5(Diamond 5)→R2C3(Square 5) via row 2.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R1C5, R2C5, R2C3]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Continue digit 4: R2C7(Diamond 4)→R2C3(Square 4) via row 2. Now R2C3 has BOTH Square 4 AND Square 5... wait, let's check further.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R2C3, R2C7]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      R2C3 now has Square 4 (from R2C7) AND Square 5 (from R2C5). Two Squares in one cell!
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R2C3]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Rule 1 contradiction! R2C3 receives Square 4 AND Square 5 — same colour twice in one cell. Square (Blue) must be FALSE.
    hint: detailed
    technique: MED
    state:
      selection:
        cells: [R2C3]

  - text: >
      Eliminate ALL Square (Blue) candidates: R1C5~5, R1C8~4, R2C3~4, R2C3~5. Diamond (Orange) is TRUE.
    hint: detailed
    technique: MED
    state:
      selection:
        cells: [R1C5, R1C8, R2C3]
settings:
  showCandidates: true
  showControls: true
  showDescription: true
  navigation: numbered
  coordinateFormat: rncn

Rule 2: Twice in a Unit

If the SAME colour appears TWICE in a unit for the SAME digit:

• That colour causes a contradiction
• That colour is FALSE
• Eliminate all candidates of that colour; set all candidates of the opposite colour

Pattern Analysis:

• Web vertices: R1C5, R2C1, R3C1, R3C6, R5C1, R5C5, R6C6
• Digits: 3, 5, 6
• Contradiction: R5C1 and R5C5 both have Diamond (Orange) 6 — both in row 5!
• Result: Diamond (Orange) is FALSE — eliminate all Diamond candidates

puzzle: S9B091R1M141O08111107122N2I0I0S1A11060H5G505007011U040I034464050R032J0I2K061I3I09083E022V3317013K3G16093U082O0S044Y030L0509365U0L5005510X642N1L4H0I070I45064C0V154X19
mode: guided
technique: 3D Medusa
initial:
  layers:
    hints: true
steps:
  - text: >
      Rule 2 finds contradictions when the same colour appears twice in a unit for the same digit. Look for bi-value cells to start the web.
    hint: subtle
    technique: MED

  - text: >
      Enable the net display. R2C1{3,5} is a bi-value cell that connects digits 3 and 5. Multiple strong links extend from here.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R2C1]
      graph:
        type: strongNet
        componentIndices: [0]
        invertColours: true

  - text: >
      Assign colours to R2C1: Square (Blue) 5, Diamond (Orange) 3 (bi-value link shown as oval). Both colours start in this cell.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R2C1]
      graph:
        type: strongNet
        componentIndices: [0]
        invertColours: true

  - text: >
      Extend digit 3: R2C1(Diamond 3)→R5C1(Square 3) via column 1 conjugate pair. R5C1{3,6} is bi-value, so R5C1 gets Diamond 6.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R2C1, R5C1]
      graph:
        type: strongNet
        componentIndices: [0]
        invertColours: true

  - text: >
      Extend digit 5: R2C1(Square 5)→R3C1(Diamond 5) via column 1. R3C1(Diamond 5)→R3C6(Square 5) via row 3.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R2C1, R3C1, R3C6]
      graph:
        type: strongNet
        componentIndices: [0]
        invertColours: true

  - text: >
      R3C6{5,6} is bi-value: Square 5 → Diamond 6. Extend digit 6: R3C6(Diamond 6)→R1C5(Square 6), R3C6(Diamond 6)→R6C6(Square 6).
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R1C5, R3C6, R6C6]
      graph:
        type: strongNet
        componentIndices: [0]
        invertColours: true

  - text: >
      Continue digit 6: R1C5(Square 6)→R5C5(Diamond 6) via column 5. Now check: R5C1 has Diamond 6 AND R5C5 has Diamond 6!
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R1C5, R5C1, R5C5]
      graph:
        type: strongNet
        componentIndices: [0]
        invertColours: true

  - text: >
      Rule 2 contradiction! R5C1 and R5C5 BOTH have Diamond 6 — same colour, same digit, same row. Diamond (Orange) must be FALSE.
    hint: detailed
    technique: MED
    state:
      selection:
        cells: [R5C1, R5C5]

  - text: >
      Eliminate ALL Diamond (Orange) candidates: R2C1~3, R3C1~5, R3C6~6, R5C1~6, R5C5~6. Square (Blue) is TRUE.
    hint: detailed
    technique: MED
    state:
      selection:
        cells: [R2C1, R3C1, R3C6, R5C1, R5C5]
settings:
  showCandidates: true
  showControls: true
  showDescription: true
  navigation: numbered
  coordinateFormat: rncn

Rule 3: Two Colours in a Cell

If a cell has candidates of BOTH colours:

• One candidate will be true
• Any OTHER candidate in that cell (not in the web) can be eliminated

Pattern Analysis:

• Web vertices: R3C1, R3C2, R6C1, R6C2
• Digits: 3, 7
• Rule condition: R3C2 has Diamond (Orange) 3 AND Square (Blue) 7 — both colours in one cell!
• Elimination: R3C2~8 (uncoloured candidate in dual-colour cell)

puzzle: S9B02093F1U2Q1M08031V1743575I025A09071V2U5Y6Q016A09041U02080405070601020903060P0LBCB6480504072E2G090O04051F1F08094503044J071F101U0R064C4K03440711092B052D8K7N1G030804
mode: guided
technique: 3D Medusa
initial:
  layers:
    hints: true
steps:
  - text: >
      Rule 3 finds eliminations when both colours appear in the same cell. Look for R6C1{3,7} — a bi-value cell.
    hint: subtle
    technique: MED

  - text: >
      Enable the net display. R6C1{3,7} connects digits 3 and 7. R3C1-R6C1 and R6C1-R6C2 form conjugate pairs.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R3C1, R6C1, R6C2]
      graph:
        type: strongNet
        componentIndices: [1]
        invertColours: true

  - text: >
      Start colouring: R6C1 gets Square (Blue) 3, Diamond (Orange) 7 (bi-value link shown as oval). Via column 1, R3C1 gets Diamond (Orange) 3.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R3C1, R6C1]
      graph:
        type: strongNet
        componentIndices: [1]
        invertColours: true

  - text: >
      Continue: R6C1(Diamond 7)→R6C2(Square 7) via row 6. R3C1(Diamond 3)→R3C2(Square 3) via row 3. The web has 4 cells.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R3C1, R3C2, R6C1, R6C2]
      graph:
        type: strongNet
        componentIndices: [1]
        invertColours: true

  - text: >
      Now check R3C2: It has Square 3 (from R3C1 link). Does it also have another colour? R3C2-R6C2 is a conjugate pair for digit 7!
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R3C2, R6C2]
      graph:
        type: strongNet
        componentIndices: [1]
        invertColours: true

  - text: >
      R6C2 has Square 7, so R3C2 gets Diamond 7. Now R3C2 has BOTH Square 3 AND Diamond 7 — Rule 3 applies!
    hint: detailed
    technique: MED
    state:
      selection:
        cells: [R3C2]

  - text: >
      One of Blue/Orange MUST be true. R3C2's other candidate (8) cannot be true. Eliminate R3C2~8.
    hint: detailed
    technique: MED
    state:
      selection:
        cells: [R3C2]
settings:
  showCandidates: true
  showControls: true
  showDescription: true
  navigation: numbered
  coordinateFormat: rncn

Rule 4: Two Colours Elsewhere

If an uncoloured candidate sees both a Square (Blue) and a Diamond (Orange) candidate for the same digit:

• One of them is true
• The uncoloured candidate is eliminated

Pattern Analysis:

• Web vertices: R2C9, R3C2, R6C9, R8C2, R8C9, R9C2, R9C3, R9C8
• Digits: 4, 6, 8, 9
• Rule condition: R2C1 and R3C8 each see BOTH Square (Blue) 6 and Diamond (Orange) 6
• Eliminations: R2C16, R3C86 (both see both colours)

puzzle: S9B019E7O5W050662BE031W04035X095U6B5E4Y08AA8Y2B04032R8Y022I034A0506CY02017U09054Y0402014Y03073E02015U03CY5M5M8Q03010709080S1M1O05244Y2403011809074A18B68C060718034C01
mode: guided
technique: 3D Medusa
initial:
  layers:
    hints: true
steps:
  - text: >
      Rule 4 eliminates uncoloured candidates that see both colours. This 8-cell web spans 4 digits.
    hint: subtle
    technique: MED

  - text: >
      Enable the net display. R2C9{6,8} is a bi-value cell. Multiple bi-value cells chain together in this web.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R2C9]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Assign colours to R2C9: Square (Blue) 6, Diamond (Orange) 8 (bi-value). Via column 9, R2C9(Square 6)→R6C9(Diamond 6).
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R2C9, R6C9]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Extend digit 8: R2C9(Diamond 8)→R8C9(Square 8). R8C9{4,8} is bi-value, so Square 8→Diamond 4.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R2C9, R8C9]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Continue digit 8: R8C9(Square 8)→R9C8(Diamond 8) via row 9. R8C9(Square 8)→R8C2(Diamond 8) via row 8.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R8C2, R8C9, R9C8]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      R8C2{6,8} is bi-value: Diamond 8→Square 6. Via column 2, R8C2(Square 6)→R3C2(Diamond 6). Now we have both colours for digit 6!
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R3C2, R8C2]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      The web continues: R9C2{8,9} gives Square 8→Diamond 9, and R9C2(Diamond 9)→R9C3(Square 9). But focus on digit 6.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R9C2, R9C3]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Rule 4 check: R2C1 has candidate 6 but is NOT in the web. R2C1 sees R2C9 (Square 6, same row) AND R3C2 (Diamond 6, same box). Eliminate R2C1~6.
    hint: detailed
    technique: MED
    state:
      selection:
        cells: [R2C1, R2C9, R3C2]

  - text: >
      Also: R3C8 has candidate 6 and sees R2C9 (Square 6, same box) AND R3C2 (Diamond 6, same row). Eliminate R3C8~6.
    hint: detailed
    technique: MED
    state:
      selection:
        cells: [R3C8, R2C9, R3C2]
settings:
  showCandidates: true
  showControls: true
  showDescription: true
  navigation: numbered
  coordinateFormat: rncn

Rule 5: Two Colours Unit + Cell

If an uncoloured candidate:

• Shares a CELL with one colour (say Diamond/Orange)
• AND sees the OPPOSITE colour (Square/Blue) for the same digit in a unit

Then that candidate is eliminated. Selecting it would force BOTH colours to be false (contradiction).

Pattern Analysis:

• Web vertices: R1C2, R4C1, R4C5, R4C6, R5C3, R5C5, R5C6, R7C1, R7C2
• Digits: 1, 4, 6
• Rule condition: R5C5 has uncoloured 4, shares cell with Diamond (Orange) 6, AND sees Square (Blue) 4 at R4C6
• Elimination: R5C5~4 (selecting it would eliminate both colours)

puzzle: S9B034ZABB70502AI9MCY0205AA03BE7U3601CY7N430406B7070502031F0903023E0R082I0505071HB6CA7V7X037N040L089E03059F062DAB1H05049F08039E9H9F037P059F069H080408047N9F02039F0506
mode: guided
technique: 3D Medusa
initial:
  layers:
    hints: true
steps:
  - text: >
      Rule 5 is subtle — it combines cell and unit relationships. Look for R4C1{1,6} — a bi-value cell that starts the web.
    hint: subtle
    technique: MED

  - text: >
      Enable the net display. R4C1{1,6} connects to multiple strong links for both digits 1 and 6.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R4C1]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Assign colours to R4C1: Square (Blue) 1, Diamond (Orange) 6. Via row 4, R4C6 gets Diamond 1. R4C6{1,4} is bi-value, so Square 4.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R4C1, R4C6]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Extend digit 6: R4C1(Diamond 6)→R4C5(Square 6), R4C1(Diamond 6)→R5C3(Square 6), R4C1(Diamond 6)→R7C1(Square 6).
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R4C1, R4C5, R5C3, R7C1]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Continue: R5C3(Square 6)→R5C5(Diamond 6). R4C5(Square 6)→R5C5(Diamond 6) confirms. Now R5C5 has Diamond 6 in the web.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R5C3, R5C5, R4C5]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      Rule 5 check: R5C5 also has candidate 4 (NOT in the web). Does this 4 see Square 4 anywhere? Yes — R4C6 has Square 4!
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R5C5, R4C6]
      graph:
        type: strongNet
        componentIndices: [0]

  - text: >
      R5C5's candidate 4 shares a CELL with Diamond 6 AND sees Square 4 at R4C6 (same box). If R5C5=4, it eliminates Diamond 6 from R5C5 AND Square 4 from R4C6 — both colours gone! Eliminate R5C5~4.
    hint: detailed
    technique: MED
    state:
      selection:
        cells: [R5C5]
settings:
  showCandidates: true
  showControls: true
  showDescription: true
  navigation: numbered
  coordinateFormat: rncn

Rule 6: Cell Emptied by Colour

If setting a colour true would empty another cell (remove all its candidates):

• That colour is FALSE
• The opposite colour is TRUE (setting all its candidates)

Pattern Analysis:

• Web vertices: R1C3, R1C5, R3C1, R3C4, R4C3, R5C4, R5C5, R5C8, R8C8, R8C9, R9C4, R9C7
• Digits: 1, 2, 5, 9
• Rule condition: R4C7{2,9} — both candidates see Square (R4C3 Square 2, R9C7 Square 9)
• Result: If Square (Blue) is true, R4C7 becomes empty — Square is FALSE, Diamond is TRUE

puzzle: S9B0F4M4J04B77Q4I0G02440D09440G0501030F4P4O0G45060O0D094I7T070L0F05087O0D7R4O4O067N7N0D074K4MBN4I0D03020GBM06BN4K094K4K040F0C0A0G0D010307B67O064K82070F4KBO0C01B84404
mode: guided
technique: 3D Medusa
initial:
  layers:
    hints: true
steps:
  - text: >
      Rule 6 finds a cell that would be emptied if a colour is true. This web spans 12 cells and 4 digits (1, 2, 5, 9).
    hint: subtle
    technique: MED

  - text: >
      Enable the net display. The web connects through multiple bi-value cells: R4C3{1,2}, R5C4{1,9}, R5C5{1,9}, R8C9{5,9}.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R4C3, R5C4, R5C5, R8C9]
      graph:
        type: strongNet
        componentIndices: [1]
        invertColours: true

  - text: >
      Key cells for Rule 6: R4C3 has Diamond 2, and R9C7 has Diamond 9. Both are Diamond vertices for different digits.
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R4C3, R9C7]
      graph:
        type: strongNet
        componentIndices: [1]
        invertColours: true

  - text: >
      Now examine R4C7. It contains only candidates {2, 9}. It's NOT in the web (uncoloured).
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R4C7]
      graph:
        type: strongNet
        componentIndices: [1]
        invertColours: true

  - text: >
      R4C7's candidate 2 sees R4C3 (Diamond 2, same row). R4C7's candidate 9 sees R9C7 (Diamond 9, same column). Both see Diamond!
    hint: subtle
    technique: MED
    state:
      selection:
        cells: [R4C7, R4C3, R9C7]
      graph:
        type: strongNet
        componentIndices: [1]
        invertColours: true

  - text: >
      Rule 6 contradiction: If Diamond (Orange) is true, R4C3=2 and R9C7=9. This eliminates BOTH candidates from R4C7, making it empty — impossible!
    hint: detailed
    technique: MED
    state:
      selection:
        cells: [R4C7]

  - text: >
      Diamond must be FALSE. Therefore Square (Blue) is TRUE. Set all Square candidates and eliminate all Diamond candidates.
    hint: detailed
    technique: MED
    state:
      selection:
        cells: [R1C5, R3C1, R5C4, R5C5, R8C8, R8C9, R9C4]
settings:
  showCandidates: true
  showControls: true
  showDescription: true
  navigation: numbered
  coordinateFormat: rncn

Tips

1. Start from Simple Colouring failures — If single-digit colouring doesn't help, try 3D Medusa
2. Include bi-value cells — These are crucial for connecting different digits
3. Check all rules — Rules 1 and 3 are most common
4. Large webs are powerful — The more connections, the more likely to find contradictions

Complexity

3D Medusa is one of the most complex standard techniques:

• Requires tracking colours across multiple digits
• Web can span most of the grid
• Multiple rules to check at each step

More Puzzles

• 3D Medusa ex. 1

• 3D Medusa ex. 2

• 3D Medusa ex. 3

• 3D Medusa ex. 4

• 3D Medusa ex. 5

• 3D Medusa ex. 6

• 3D Medusa ex. 7

• 3D Medusa ex. 8

• 3D Medusa ex. 9

• 3D Medusa ex. 10

• 3D Medusa ex. 11

• 3D Medusa ex. 12

• 3D Medusa ex. 13

• 3D Medusa ex. 14

• 3D Medusa ex. 15

• 3D Medusa ex. 16

• 3D Medusa ex. 17

• 3D Medusa ex. 18

• 3D Medusa ex. 19

• 3D Medusa ex. 20

• 3D Medusa ex. 21

• 3D Medusa ex. 22

• 3D Medusa ex. 23

• 3D Medusa ex. 24

• 3D Medusa ex. 25

• 3D Medusa ex. 26

• 3D Medusa Two Colours in a Cell ex. 1

• 3D Medusa Two Colours in a Cell ex. 2

• 3D Medusa Two Colour Units + Cell ex. 1

• 3D Medusa Two Colour Units + Cell ex. 2

• 3D Medusa Twice in a Unit Same Colour ex. 1

• 3D Medusa Twice in a Unit Same Colour ex. 2

• 3D Medusa Twice in a Unit Same Colour ex. 3

• 3D Medusa Twice in a Cell Same Colour ex. 1

• 3D Medusa Twice in a Cell Same Colour ex. 2

• 3D Medusa Two Colours Elsewhere ex. 1

• 3D Medusa Two Colours Elsewhere ex. 2

• 3D Medusa Two Colours Elsewhere ex. 3

Related Techniques

• Simple Colouring — Single-digit version
• Y-Wing — Uses bi-value cells in simpler pattern