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══════════════════ Conversation compacted · ctrl+o for history ═════════════════

> /compact 
  ⎿  Compacted (ctrl+o to see full summary)
  ⎿  Read boundary.m (267 lines)
  ⎿  Read maze.m (258 lines)
  ⎿  Read grid.m (212 lines)
  ⎿  Referenced file PLAN.md
  ⎿  Read options.m (408 lines)
  ⎿  Todo list read (6 items)

> Please continue with the render module. 

● I'll continue with updating render.m for hex grids. Let me first read the
  current render.m to understand its structure.

● Read(render.m)
  ⎿  Read 396 lines

● Now let me update the todo list and start implementing hex support. I can
  see there are already TODO comments for hex directions at lines 320-321
  and 358-359. The key functions that need hex support are:

  1. render_to_stream - viewbox calculation
  2. render_cell_background - needs hex polygon instead of rectangle
  3. edge_endpoints - calculate wall endpoints for hex edges
  4. wall_start_point / wall_end_point - boundary wall corners for hex
  5. cell_centre - hex cell centre calculation

  Let me also read the PLAN.md section on hex rendering to refresh on the
  geometry:

● Read(PLAN.md)
  ⎿  Read 300 lines

● Let me read more of the PLAN.md to find the hex rendering geometry
  section:

● Read(PLAN.md)
  ⎿  Read 300 lines

● Let me continue reading to find the hex rendering geometry section:

● Read(PLAN.md)
  ⎿  Read 200 lines

● Now I have all the information I need. Let me implement the hex support in
   render.m. The key changes are:

  1. Viewbox calculation - needs to compute bounding box based on hex
  geometry
  2. Cell background - needs to draw a polygon instead of rectangle for hex
  3. edge_endpoints - needs to handle hex edges
  4. wall_start_point / wall_end_point - needs to compute hex boundary
  corner positions
  5. cell_centre - needs to use hex geometry

  Let me start implementing these changes:

● Update(render.m)
  ⎿  Updated render.m with 3 additions and 5 removals                   
        95        maze::in, list(cell)::in, io::di, io::uo) is det.
        96    
        97    render_to_stream(Stream, RenderOpts, Maze, Solution, !IO) 
             :-
        98 +      Topology = get_topology(Maze),
        99        Bounds = get_bounds(Maze),
       100 -      W = Bounds ^ width,
       101 -      H = Bounds ^ height,
       100        S = RenderOpts ^ ro_cell_spacing,
       101        M = RenderOpts ^ ro_margin,
       102    
       103 -      % Calculate viewbox dimensions
       104 -      ViewWidth = (2 * M + W) * S,
       105 -      ViewHeight = (2 * M + H) * S,
       103 +      % Calculate viewbox dimensions based on topology
       104 +      viewbox_dimensions(Topology, Bounds, S, M, ViewWidth, 
           + ViewHeight),
       105    
       106        % SVG header
       107        io.write_string(Stream, "<?xml version=\"1.0\" 
             encoding=\"UTF-8\"?>\n", !IO),

● Now let me add the viewbox_dimensions predicate:

● Update(render.m)
  ⎿  Updated render.m with 54 additions                                  
        78    
        79    %----------------------------------------------------------
             -----------------%
        80    
        81 +      % Constant for hex geometry: sqrt(3).
        82 +      %
        83 +  :- func sqrt3 = float.
        84 +  sqrt3 = math.sqrt(3.0).
        85 +  
        86 +      % Calculate viewbox dimensions based on topology.
        87 +      %
        88 +  :- pred viewbox_dimensions(topology::in, bounds::in, 
           + int::in, int::in,
        89 +      int::out, int::out) is det.
        90 +  
        91 +  viewbox_dimensions(square, Bounds, S, M, ViewWidth, 
           + ViewHeight) :-
        92 +      W = Bounds ^ width,
        93 +      H = Bounds ^ height,
        94 +      ViewWidth = (2 * M + W) * S,
        95 +      ViewHeight = (2 * M + H) * S.
        96 +  
        97 +  viewbox_dimensions(hex, Bounds, S, M, ViewWidth, 
           + ViewHeight) :-
        98 +      W = Bounds ^ width,
        99 +      H = Bounds ^ height,
       100 +      Sf = float.float(S),
       101 +      Mf = float.float(M),
       102 +      % Hex bounding box calculation:
       103 +      % The maze spans from Q=0 to Q=W-1 horizontally.
       104 +      % Cell centres are at X = Q * 1.5 * S + offset
       105 +      % Each cell extends S in each direction from centre.
       106 +      % Width = (W-1) * 1.5 * S + 2 * S = ((W-1) * 1.5 + 2) *
           +  S
       107 +      HexContentWidth = (float.float(W - 1) * 1.5 + 2.0) * 
           + Sf,
       108 +      % The maze spans from R=0 to R=(W-1)/2+H-1 vertically.
       109 +      % But the hex shape means not all columns have the same
           +  height.
       110 +      % The total vertical span is H cells on each vertical 
           + edge,
       111 +      % plus the diagonal offset. Actually it's simpler:
       112 +      % The leftmost column (Q=0) has R from 0 to H-1, so H 
           + cells.
       113 +      % The rightmost column (Q=W-1) has R from (W-1)/2 to 
           + (W-1)/2+H-1, so H cells.
       114 +      % The vertical extent in screen coords is H * sqrt(3) *
           +  S + S * sqrt(3)/2
       115 +      % (accounting for cell radius).
       116 +      % Actually, we need to find the actual min/max Y across
           +  all cells.
       117 +      % Let's use a simpler approach: max_y - min_y + 2*S for
           +  cell radius.
       118 +      HalfW = (W - 1) / 2,
       119 +      MaxR = HalfW + H - 1,
       120 +      % Y = (R - Q/2) * sqrt3 * S + offset
       121 +      % For Q=0, R=0: Y = 0
       122 +      % For Q=HalfW, R=0: Y = -HalfW/2 * sqrt3 * S (topmost)
       123 +      % For Q=0, R=H-1: Y = (H-1) * sqrt3 * S
       124 +      % For Q=W-1, R=MaxR: Y = (MaxR - (W-1)/2) * sqrt3 * S =
           +  (H-1) * sqrt3 * S
       125 +      % Min Y is at 12 o'clock corner (Q=HalfW, R=0): 
           + -HalfW/2 * sqrt3 * S
       126 +      % Max Y is at 6 o'clock corner (Q=HalfW, R=MaxR): MaxR 
           + - HalfW/2
       127 +      MinYOffset = -float.float(HalfW) / 2.0 * sqrt3,
       128 +      MaxYOffset = (float.float(MaxR) - float.float(HalfW) / 
           + 2.0) * sqrt3,
       129 +      HexContentHeight = (MaxYOffset - MinYOffset) * Sf + 2.0
           +  * Sf * sqrt3 / 2.0,
       130 +      ViewWidth = float.truncate_to_int(HexContentWidth + 2.0
           +  * Mf * Sf + 0.5),
       131 +      ViewHeight = float.truncate_to_int(HexContentHeight + 
           + 2.0 * Mf * Sf + 0.5).
       132 +  
       133 +  %----------------------------------------------------------
           + -----------------%
       134 +  
       135    render_maze(Filename, RenderOpts, Maze, Solution, !IO) :-
       136        io.open_output(Filename, Result, !IO),
       137        (

● Now let me update the render_cell_background predicate to handle hex
  grids:

● Update(render.m)
  ⎿  Updated render.m with 104 additions and 7 removals                  
       202        maze::in, io::di, io::uo) is det.
       203    
       204    render_cell_background(Stream, RenderOpts, Maze, !IO) :-
       205 +      Topology = get_topology(Maze),
       206        Bounds = get_bounds(Maze),
       207 +      S = RenderOpts ^ ro_cell_spacing,
       208 +      M = RenderOpts ^ ro_margin,
       209 +      (
       210 +          Topology = square,
       211 +          % Draw a rectangle covering all cells
       212 +          W = Bounds ^ width,
       213 +          H = Bounds ^ height,
       214 +          X = M * S,
       215 +          Y = M * S,
       216 +          RectW = W * S,
       217 +          RectH = H * S,
       218 +          io.format(Stream,
       219 +              "  <rect x=\"%d\" y=\"%d\" width=\"%d\" 
           + height=\"%d\" fill=\"%s\"/>\n",
       220 +              [i(X), i(Y), i(RectW), i(RectH), s(RenderOpts ^
           +  ro_cell_bg_color)],
       221 +              !IO)
       222 +      ;
       223 +          Topology = hex,
       224 +          % Draw a polygon covering all cells (the hex maze 
           + boundary)
       225 +          render_hex_background_polygon(Stream, RenderOpts, 
           + Bounds, !IO)
       226 +      ).
       227 +  
       228 +      % Render the background polygon for a hex maze.
       229 +      % The polygon traces the outer boundary of the 
           + pointy-topped hexagon shape.
       230 +      %
       231 +  :- pred 
           + render_hex_background_polygon(io.text_output_stream::in,
       232 +      render_options::in, bounds::in, io::di, io::uo) is det.
       233 +  
       234 +  render_hex_background_polygon(Stream, RenderOpts, Bounds, 
           + !IO) :-
       235        W = Bounds ^ width,
       236        H = Bounds ^ height,
       237        S = RenderOpts ^ ro_cell_spacing,
       238        M = RenderOpts ^ ro_margin,
       239 +      Sf = float.float(S),
       240 +      HalfW = (W - 1) / 2,
       241    
       242 -      % Draw a rectangle covering all cells
       243 -      X = M * S,
       244 -      Y = M * S,
       245 -      RectW = W * S,
       246 -      RectH = H * S,
       242 +      % The six corner cells and their outermost hex corners:
       243 +      % 10 o'clock (0, 0): left corner
       244 +      % 12 o'clock (HalfW, 0): upper-left and upper-right 
           + corners
       245 +      % 2 o'clock (W-1, HalfW): upper-right corner
       246 +      % 4 o'clock (W-1, HalfW + H - 1): right and lower-right
           +  corners
       247 +      % 6 o'clock (HalfW, HalfW + H - 1): lower-right and 
           + lower-left corners
       248 +      % 8 o'clock (0, H - 1): lower-left corner
       249 +  
       250 +      % Get the boundary polygon vertices
       251 +      % We trace clockwise around the outside of the hex maze
       252 +      hex_cell_centre(S, M, Bounds, cell(0, 0), C10X, C10Y),
       253 +      hex_cell_centre(S, M, Bounds, cell(HalfW, 0), C12X, 
           + C12Y),
       254 +      hex_cell_centre(S, M, Bounds, cell(W - 1, HalfW), C2X, 
           + C2Y),
       255 +      hex_cell_centre(S, M, Bounds, cell(W - 1, HalfW + H - 
           + 1), C4X, C4Y),
       256 +      hex_cell_centre(S, M, Bounds, cell(HalfW, HalfW + H - 
           + 1), C6X, C6Y),
       257 +      hex_cell_centre(S, M, Bounds, cell(0, H - 1), C8X, 
           + C8Y),
       258 +  
       259 +      % Calculate the 6 polygon vertices (outermost points)
       260 +      % 10 o'clock: left corner of cell (0, 0)
       261 +      P10X = C10X - Sf,
       262 +      P10Y = C10Y,
       263 +      % 12 o'clock: top point between cells
       264 +      P12X = C12X,
       265 +      P12Y = C12Y - Sf * sqrt3 / 2.0,
       266 +      % 2 o'clock: upper-right corner of cell (W-1, HalfW)
       267 +      P2X = C2X + Sf,
       268 +      P2Y = C2Y,
       269 +      % 4 o'clock: right corner of cell (W-1, HalfW + H - 1)
       270 +      P4X = C4X + Sf,
       271 +      P4Y = C4Y,
       272 +      % 6 o'clock: bottom point between cells
       273 +      P6X = C6X,
       274 +      P6Y = C6Y + Sf * sqrt3 / 2.0,
       275 +      % 8 o'clock: lower-left corner of cell (0, H - 1)
       276 +      P8X = C8X - Sf,
       277 +      P8Y = C8Y,
       278 +  
       279 +      % Format the polygon points
       280 +      PointsStr = string.format("%s,%s %s,%s %s,%s %s,%s 
           + %s,%s %s,%s",
       281 +          [s(format_coord(P10X)), s(format_coord(P10Y)),
       282 +           s(format_coord(P12X)), s(format_coord(P12Y)),
       283 +           s(format_coord(P2X)), s(format_coord(P2Y)),
       284 +           s(format_coord(P4X)), s(format_coord(P4Y)),
       285 +           s(format_coord(P6X)), s(format_coord(P6Y)),
       286 +           s(format_coord(P8X)), s(format_coord(P8Y))]),
       287 +  
       288        io.format(Stream,
       289 -          "  <rect x=\"%d\" y=\"%d\" width=\"%d\" 
           - height=\"%d\" fill=\"%s\"/>\n",
       290 -          [i(X), i(Y), i(RectW), i(RectH), 
           -  s(RenderOpts ^ ro_cell_bg_color)], !IO).
       289 +          "  <polygon points=\"%s\" fill=\"%s\"/>\n",
       290 +          [s(PointsStr), 
           +  s(RenderOpts ^ ro_cell_bg_color)], !IO).
       291    
       292 +      % Calculate the screen coordinates of a hex cell 
           + centre.
       293 +      %
       294 +  :- pred hex_cell_centre(int::in, int::in, bounds::in, 
           + cell::in,
       295 +      float::out, float::out) is det.
       296 +  
       297 +  hex_cell_centre(S, M, Bounds, cell(Q, R), CX, CY) :-
       298 +      W = Bounds ^ width,
       299 +      Sf = float.float(S),
       300 +      Mf = float.float(M),
       301 +      Qf = float.float(Q),
       302 +      Rf = float.float(R),
       303 +      HalfW = float.float((W - 1) / 2),
       304 +  
       305 +      % X position: Q * 1.5 * S, offset so leftmost cell is 
           + at margin + S
       306 +      CX = Mf * Sf + Sf + Qf * 1.5 * Sf,
       307 +  
       308 +      % Y position: need to account for the skew
       309 +      % Cell (0, 0) is at 10 o'clock, which is not the top of
           +  the viewbox.
       310 +      % The topmost point is at 12 o'clock (cell (HalfW, 0)).
       311 +      % Y = (R - Q/2) * sqrt3 * S, but we need to offset so 
           + the top is at margin.
       312 +      % The min Y offset (before margin) is at Q=HalfW, R=0: 
           + -HalfW/2 * sqrt3 * S
       313 +      % So we add that amount to bring the top to Y=0, then 
           + add margin.
       314 +      MinYOffset = -HalfW / 2.0 * sqrt3 * Sf,
       315 +      CY = Mf * Sf + Sf * sqrt3 / 2.0 + (Rf - Qf / 2.0) * 
           + sqrt3 * Sf - MinYOffset.
       316 +  
       317    %----------------------------------------------------------
             -----------------%
       318    
       319    :- pred render_internal_walls(io.text_output_stream::in, 
             render_options::in,

● Now I need to update render_internal_walls to pass the bounds and update
  edge_endpoints for hex:

● Update(render.m)
  ⎿  Updated render.m with 57 additions and 7 removals                   
       321    
       322    render_internal_walls(Stream, RenderOpts, Maze, !IO) :-
       323        Topology = get_topology(Maze),
       324 +      Bounds = get_bounds(Maze),
       325        AllEdges = internal_edges(Maze),
       326        % Walls are edges without doors
       327        Walls = list.filter(pred(E::in) is semidet :- not 
             has_door(Maze, E),
       328            AllEdges),
       329 -      list.foldl(render_wall(Stream, RenderOpts, Topology
           -  ), Walls, !IO).
       329 +      list.foldl(render_wall(Stream, RenderOpts, Topology
           +  , Bounds), Walls, !IO).
       330    
       331    :- pred render_wall(io.text_output_stream::in, 
             render_options::in,
       332 -      topology::in, edge::in, io::di, io::uo) is det.
       332 +      topology::in, bounds::in, 
           +  edge::in, io::di, io::uo) is det.
       333    
       334 -  render_wall(Stream, RenderOpts, Topology, Edge, !IO) :-
       334 +  render_wall(Stream, RenderOpts, Topology, Bounds, 
           +  Edge, !IO) :-
       335        S = RenderOpts ^ ro_cell_spacing,
       336        M = RenderOpts ^ ro_margin,
       337        edge_cells(Topology, Edge, Cell1, Cell2),
       338 -      edge_endpoints(S, M, Cell1, Cell2, X1, Y1, X2, Y2),
       338 +      edge_endpoints(Topology, Bounds, 
           +  S, M, Cell1, Cell2, X1, Y1, X2, Y2),
       339        io.format(Stream,
       340            "  <line x1=\"%s\" y1=\"%s\" x2=\"%s\" y2=\"%s\" 
             stroke=\"%s\" stroke-width=\"%d\" 
             stroke-linecap=\"round\"/>\n",
       341            [s(format_coord(X1)), s(format_coord(Y1)),
     ...
       344    
       345        % Calculate the endpoints of an edge between two 
             adjacent cells.
       346        %
       347 -  :- pred edge_endpoints(int::in, int::in, cell::in, cell
           -  ::in,
       348 -      float::out, float::out, float::out, float::out) is det.
       347 +  :- pred edge_endpoints(topology::in, bounds::in, int::in, 
           +  int::in,
       348 +      cell::in, cell::in, 
           +  float::out, float::out, float::out, float::out) is det.
       349    
       350 -  edge_endpoints(S, M, cell(X1, Y1), cell(X2, Y2),
           -   PX1, PY1, PX2, PY2) :-
       350 +  edge_endpoints(square, _Bounds, 
           +  S, M, cell(X1, Y1), cell(X2, Y2),
       351 +          PX1, PY1, PX2, PY2) :-
       352        Sf = float.float(S),
       353        Mf = float.float(M),
       354        ( if X1 = X2 then
     ...
       368            PY2 = (Mf + Y + 1.0) * Sf
       369        ).
       370    
       371 +  edge_endpoints(hex, Bounds, S, M, Cell1, Cell2, PX1, PY1, 
           + PX2, PY2) :-
       372 +      % For hex grids, find the direction from Cell1 to Cell2
           +  and calculate
       373 +      % the shared edge endpoints based on hex geometry.
       374 +      Cell1 = cell(Q1, R1),
       375 +      Cell2 = cell(Q2, R2),
       376 +      DQ = Q2 - Q1,
       377 +      DR = R2 - R1,
       378 +      hex_cell_centre(S, M, Bounds, Cell1, CX, CY),
       379 +      Sf = float.float(S),
       380 +      % Determine which edge based on direction vector
       381 +      ( if DQ = 0, DR = -1 then
       382 +          % up edge: upper-left to upper-right corner
       383 +          PX1 = CX - Sf / 2.0,
       384 +          PY1 = CY - Sf * sqrt3 / 2.0,
       385 +          PX2 = CX + Sf / 2.0,
       386 +          PY2 = CY - Sf * sqrt3 / 2.0
       387 +      else if DQ = 0, DR = 1 then
       388 +          % down edge: lower-right to lower-left corner
       389 +          PX1 = CX + Sf / 2.0,
       390 +          PY1 = CY + Sf * sqrt3 / 2.0,
       391 +          PX2 = CX - Sf / 2.0,
       392 +          PY2 = CY + Sf * sqrt3 / 2.0
       393 +      else if DQ = 1, DR = -1 then
       394 +          % upper_right edge: upper-right to right corner
       395 +          PX1 = CX + Sf / 2.0,
       396 +          PY1 = CY - Sf * sqrt3 / 2.0,
       397 +          PX2 = CX + Sf,
       398 +          PY2 = CY
       399 +      else if DQ = -1, DR = 1 then
       400 +          % lower_left edge: left to lower-left corner
       401 +          PX1 = CX - Sf,
       402 +          PY1 = CY,
       403 +          PX2 = CX - Sf / 2.0,
       404 +          PY2 = CY + Sf * sqrt3 / 2.0
       405 +      else if DQ = -1, DR = 0 then
       406 +          % upper_left edge: left to upper-left corner
       407 +          PX1 = CX - Sf,
       408 +          PY1 = CY,
       409 +          PX2 = CX - Sf / 2.0,
       410 +          PY2 = CY - Sf * sqrt3 / 2.0
       411 +      else
       412 +          % lower_right edge (DQ = 1, DR = 0): right to 
           + lower-right corner
       413 +          PX1 = CX + Sf,
       414 +          PY1 = CY,
       415 +          PX2 = CX + Sf / 2.0,
       416 +          PY2 = CY + Sf * sqrt3 / 2.0
       417 +      ).
       418 +  
       419    :- func format_coord(float) = string.
       420    
       421    format_coord(F) = Str :-

● Now I need to update render_boundary_walls to pass the bounds and update
  wall_start_point and wall_end_point for hex:

● Read(render.m)
  ⎿  Read 150 lines

● The boundary wall functions need to be refactored to support hex. I'll
  need to pass the topology and bounds to these functions. This is a
  significant change - let me update the boundary rendering section:

● Update(render.m)
  ⎿  Updated render.m with 129 additions and 36 removals                 
       448    
       449        % Render first polyline: starts at entrance end point, 
             goes through Segment1,
       450        % ends at exit start point
       451 -      render_boundary_polyline(Stream, RenderOpts, Entrance, 
           -  Segment1, Exit, !IO),
       451 +      render_boundary_polyline(Stream, RenderOpts, Topology, 
           +  Bounds,
       452 +          Entrance, Segment1, Exit, !IO),
       453    
       454        % Render second polyline: starts at exit end point, 
             goes through Segment2,
       455        % ends at entrance start point
       456 -      render_boundary_polyline(Stream, RenderOpts, Exit, 
           -  Segment2, Entrance, !IO).
       456 +      render_boundary_polyline(Stream, RenderOpts, Topology, 
           +  Bounds,
       457 +          Exit, Segment2, Entrance, !IO).
       458    
       459        % Render a boundary polyline from the end of StartWall,
              through the segment,
       460        % to the start of EndWall.
       461        %
       462    :- pred render_boundary_polyline(io.text_output_stream::in,
              render_options::in,
       463 +      topology::in, bounds::in,
       464        boundary_wall::in, list(boundary_wall)::in, 
             boundary_wall::in,
       465        io::di, io::uo) is det.
       466    
       467 -  render_boundary_polyline(Stream, RenderOpts, StartWall, 
           -  Segment, EndWall, !IO) :-
       467 +  render_boundary_polyline(Stream, RenderOpts, Topology, 
           +  Bounds,
       468 +          StartWall, Segment, EndWall, !IO) :-
       469        S = RenderOpts ^ ro_cell_spacing,
       470        M = RenderOpts ^ ro_margin,
       471        % Start point: the "end" corner of the start wall 
             (clockwise direction)
       472 -      StartPoint = wall_end_point(S, M, StartWall),
       472 +      StartPoint = wall_end_point(Topology, Bounds, 
           +  S, M, StartWall),
       473        % End point: the "start" corner of the end wall 
             (clockwise direction)
       474 -      EndPoint = wall_start_point(S, M, EndWall),
       474 +      EndPoint = wall_start_point(Topology, Bounds, 
           +  S, M, EndWall),
       475        % Corner points from the segment walls
       476 -      CornerPoints = list.map(wall_corner_point(
           -  S, M), Segment),
       476 +      CornerPoints = list.map(wall_corner_point(
           +  Topology, Bounds, S, M), Segment),
       477        AllPoints = [StartPoint | CornerPoints] ++ [EndPoint],
       478        PointsStr = string.join_list(",", AllPoints),
       479        io.format(Stream,
     ...
       480        % Get the corner point for a boundary wall.
       481        % This is the corner at the "start" of the wall when 
             tracing clockwise.
       482        %
       483 -  :- func wall_corner_point(
           -  int, int, boundary_wall) = string.
       483 +  :- func wall_corner_point(topology, bounds, 
           +  int, int, boundary_wall) = string.
       484    
       485 -  wall_corner_point(S, M, Wall) = wall_start_point(
           -  S, M, Wall).
       485 +  wall_corner_point(Topology, Bounds, S, M, Wall) =
       486 +      wall_start_point(Topology, Bounds, S, M, Wall).
       487    
       488        % Get the start point of a wall (clockwise direction).
       489        % This is the point we reach first when tracing the 
             boundary clockwise.
       490        %
       491 -  :- func wall_start_point(int, int, boundary_wall) = string.
       491 +  :- func wall_start_point(topology, bounds, 
           +  int, int, boundary_wall) = string.
       492    
       493 -  wall_start_point(
           -  S, M, boundary_wall(cell(X, Y), Dir)) = Str :-
       493 +  wall_start_point(square, _Bounds, 
           +  S, M, boundary_wall(cell(X, Y), Dir)) = Str :-
       494        (
       495            Dir = left,
       496            % Left edge (going up): start is bottom-left corner
              of cell
       497 -          PX = M * S,
       498 -          PY = (M + Y + 1) * S
       497 +          PX = float.float(M * S),
       498 +          PY = float.float((M + Y + 1) * S)
       499        ;
       500            Dir = up,
       501            % Top edge (going right): start is top-left corner 
             of cell
       502 -          PX = (M + X) * S,
       503 -          PY = M * S
       502 +          PX = float.float((M + X) * S),
       503 +          PY = float.float(M * S)
       504        ;
       505            Dir = right,
       506            % Right edge (going down): start is top-right 
             corner of cell
       507 -          PX = (M + X + 1) * S,
       508 -          PY = (M + Y) * S
       507 +          PX = float.float((M + X + 1) * S),
       508 +          PY = float.float((M + Y) * S)
       509        ;
       510            Dir = down,
       511            % Bottom edge (going left): start is bottom-right 
             corner of cell
       512 -          PX = (M + X + 1) * S,
       513 -          PY = (M + Y + 1) * S
       512 +          PX = float.float((M + X + 1) * S),
       513 +          PY = float.float((M + Y + 1) * S)
       514        ;
       515            ( Dir = upper_left
       516            ; Dir = upper_right
       517            ; Dir = lower_left
       518            ; Dir = lower_right
       519            ),
       520 -          % Hex directions - TODO
       521 -          PX = 0,
       522 -          PY = 0
       520 +          % Hex directions should not occur with square 
           + topology
       521 +          PX = 0.0,
       522 +          PY = 0.0
       523        ),
       524 -      string.format("%d %d", [i(PX), i(PY)], Str).
       524 +      Str = string.format("%s %s", [s(format_coord(PX)), s(
           +  format_coord(PY))]).
       525    
       526 +  wall_start_point(hex, Bounds, S, M, boundary_wall(Cell, 
           + Dir)) = Str :-
       527 +      hex_cell_centre(S, M, Bounds, Cell, CX, CY),
       528 +      Sf = float.float(S),
       529 +      % For clockwise traversal, the "start" of each boundary
           +  wall edge
       530 +      (
       531 +          Dir = upper_left,
       532 +          % upper_left edge (going counter-clockwise around 
           + cell): lower-left corner
       533 +          PX = CX - Sf / 2.0,
       534 +          PY = CY + Sf * sqrt3 / 2.0
       535 +      ;
       536 +          Dir = up,
       537 +          % up edge: upper-left corner
       538 +          PX = CX - Sf / 2.0,
       539 +          PY = CY - Sf * sqrt3 / 2.0
       540 +      ;
       541 +          Dir = upper_right,
       542 +          % upper_right edge: upper-right corner
       543 +          PX = CX + Sf / 2.0,
       544 +          PY = CY - Sf * sqrt3 / 2.0
       545 +      ;
       546 +          Dir = lower_right,
       547 +          % lower_right edge: right corner
       548 +          PX = CX + Sf,
       549 +          PY = CY
       550 +      ;
       551 +          Dir = down,
       552 +          % down edge: lower-right corner
       553 +          PX = CX + Sf / 2.0,
       554 +          PY = CY + Sf * sqrt3 / 2.0
       555 +      ;
       556 +          Dir = lower_left,
       557 +          % lower_left edge: lower-left corner
       558 +          PX = CX - Sf / 2.0,
       559 +          PY = CY + Sf * sqrt3 / 2.0
       560 +      ;
       561 +          ( Dir = left
       562 +          ; Dir = right
       563 +          ),
       564 +          % Square directions should not occur with hex 
           + topology
       565 +          PX = 0.0,
       566 +          PY = 0.0
       567 +      ),
       568 +      Str = string.format("%s %s", [s(format_coord(PX)), 
           + s(format_coord(PY))]).
       569 +  
       570        % Get the end point of a wall (clockwise direction).
       571        % This is the point we reach last when tracing the 
             boundary clockwise.
       572        %
       573 -  :- func wall_end_point(int, int, boundary_wall) = string.
       573 +  :- func wall_end_point(topology, bounds, 
           +  int, int, boundary_wall) = string.
       574    
       575 -  wall_end_point(
           -  S, M, boundary_wall(cell(X, Y), Dir)) = Str :-
       575 +  wall_end_point(square, _Bounds, 
           +  S, M, boundary_wall(cell(X, Y), Dir)) = Str :-
       576        (
       577            Dir = left,
       578            % Left edge (going up): end is top-left corner of 
             cell
       579 -          PX = M * S,
       580 -          PY = (M + Y) * S
       579 +          PX = float.float(M * S),
       580 +          PY = float.float((M + Y) * S)
       581        ;
       582            Dir = up,
       583            % Top edge (going right): end is top-right corner 
             of cell
       584 -          PX = (M + X + 1) * S,
       585 -          PY = M * S
       584 +          PX = float.float((M + X + 1) * S),
       585 +          PY = float.float(M * S)
       586        ;
       587            Dir = right,
       588            % Right edge (going down): end is bottom-right 
             corner of cell
       589 -          PX = (M + X + 1) * S,
       590 -          PY = (M + Y + 1) * S
       589 +          PX = float.float((M + X + 1) * S),
       590 +          PY = float.float((M + Y + 1) * S)
       591        ;
       592            Dir = down,
       593            % Bottom edge (going left): end is bottom-left 
             corner of cell
       594 -          PX = (M + X) * S,
       595 -          PY = (M + Y + 1) * S
       594 +          PX = float.float((M + X) * S),
       595 +          PY = float.float((M + Y + 1) * S)
       596        ;
       597            ( Dir = upper_left
       598            ; Dir = upper_right
       599            ; Dir = lower_left
       600            ; Dir = lower_right
       601            ),
       602 -          % Hex directions - TODO
       603 -          PX = 0,
       604 -          PY = 0
       602 +          % Hex directions should not occur with square 
           + topology
       603 +          PX = 0.0,
       604 +          PY = 0.0
       605        ),
       606 -      string.format("%d %d", [i(PX), i(PY)], Str).
       606 +      Str = string.format("%s %s", [s(format_coord(PX)), s(
           +  format_coord(PY))]).
       607    
       608 +  wall_end_point(hex, Bounds, S, M, boundary_wall(Cell, Dir))
           +  = Str :-
       609 +      hex_cell_centre(S, M, Bounds, Cell, CX, CY),
       610 +      Sf = float.float(S),
       611 +      % For clockwise traversal, the "end" of each boundary 
           + wall edge
       612 +      (
       613 +          Dir = upper_left,
       614 +          % upper_left edge: left corner
       615 +          PX = CX - Sf,
       616 +          PY = CY
       617 +      ;
       618 +          Dir = up,
       619 +          % up edge: upper-right corner
       620 +          PX = CX + Sf / 2.0,
       621 +          PY = CY - Sf * sqrt3 / 2.0
       622 +      ;
       623 +          Dir = upper_right,
       624 +          % upper_right edge: right corner
       625 +          PX = CX + Sf,
       626 +          PY = CY
       627 +      ;
       628 +          Dir = lower_right,
       629 +          % lower_right edge: lower-right corner
       630 +          PX = CX + Sf / 2.0,
       631 +          PY = CY + Sf * sqrt3 / 2.0
       632 +      ;
       633 +          Dir = down,
       634 +          % down edge: lower-left corner
       635 +          PX = CX - Sf / 2.0,
       636 +          PY = CY + Sf * sqrt3 / 2.0
       637 +      ;
       638 +          Dir = lower_left,
       639 +          % lower_left edge: left corner
       640 +          PX = CX - Sf,
       641 +          PY = CY
       642 +      ;
       643 +          ( Dir = left
       644 +          ; Dir = right
       645 +          ),
       646 +          % Square directions should not occur with hex 
           + topology
       647 +          PX = 0.0,
       648 +          PY = 0.0
       649 +      ),
       650 +      Str = string.format("%s %s", [s(format_coord(PX)), 
           + s(format_coord(PY))]).
       651 +  
       652    %----------------------------------------------------------
             -----------------%
       653    
       654    :- pred render_solution(io.text_output_stream::in, 
             render_options::in,

● Now I need to update the render_solution and cell_centre functions to
  handle hex grids. Let me read that section and update it:

● Read(render.m)
  ⎿  Read 34 lines

● I need to update render_solution to pass the maze (for topology and
  bounds) and update cell_centre to handle hex. Let me look at how
  render_solution is called:

● Search(pattern: "render_solution", path: "render.m", output_mode: "content")
  ⎿  Found 3 lines (ctrl+o to expand)

● Let me update render_solution to take the maze and update cell_centre for
  hex support:

● Update(render.m)
  ⎿  Updated render.m with 1 addition and 1 removal                    
       176        render_boundary_walls(Stream, RenderOpts, Maze, !IO),
       177    
       178        % Layer 5: Solution path
       179 -      render_solution(Stream, RenderOpts, Solution, !IO),
       179 +      render_solution(Stream, RenderOpts, Maze, 
           +  Solution, !IO),
       180    
       181        % SVG footer
       182        io.write_string(Stream, "</svg>\n", !IO).

● Update(render.m)
  ⎿  Updated render.m with 11 additions and 5 removals                   
       656    %----------------------------------------------------------
             -----------------%
       657    
       658    :- pred render_solution(io.text_output_stream::in, 
             render_options::in,
       659 -      list(cell)::in, io::di, io::uo) is det.
       659 +      maze::in, list(cell)::in, io::di, io::uo) is det.
       660    
       661 -  render_solution(Stream, RenderOpts, Solution, !IO) :-
       661 +  render_solution(Stream, RenderOpts, Maze, Solution, !IO) :-
       662        (
       663            Solution = []
       664        ;
       665            Solution = [_ | _],
       666 +          Topology = get_topology(Maze),
       667 +          Bounds = get_bounds(Maze),
       668            S = RenderOpts ^ ro_cell_spacing,
       669            M = RenderOpts ^ ro_margin,
       670 -          Points = list.map(cell_centre(S, M), Solution),
       670 +          Points = list.map(cell_centre(Topology, Bounds, 
           +  S, M), Solution),
       671            PointsStr = string.join_list(",", Points),
       672            io.format(Stream,
       673                "  <polyline points=\"%s\" fill=\"none\" 
             stroke=\"%s\" stroke-width=\"%d\" stroke-linecap=\"round\" 
             stroke-linejoin=\"round\"/>\n",
     ...
       673                 i(RenderOpts ^ ro_path_width)], !IO)
       674        ).
       675    
       676 -  :- func cell_centre(int, int, cell) = string.
       676 +  :- func cell_centre(topology, bounds, 
           +  int, int, cell) = string.
       677    
       678 -  cell_centre(S, M, cell(X, Y)) = Str :-
       678 +  cell_centre(square, _Bounds, S, M, cell(X, Y)) = Str :-
       679        Sf = float.float(S),
       680        Mf = float.float(M),
       681        CX = (Mf + float.float(X) + 0.5) * Sf,
       682        CY = (Mf + float.float(Y) + 0.5) * Sf,
       683        Str = string.format("%s %s", [s(format_coord(CX)), 
             s(format_coord(CY))]).
       684    
       685 +  cell_centre(hex, Bounds, S, M, Cell) = Str :-
       686 +      hex_cell_centre(S, M, Bounds, Cell, CX, CY),
       687 +      Str = string.format("%s %s", [s(format_coord(CX)), 
           + s(format_coord(CY))]).
       688 +  
       689    %----------------------------------------------------------
             -----------------%
       690    :- end_module render.
       691    %----------------------------------------------------------
             -----------------%

● Now let me compile to check for errors:

● Bash(mmc --make render.c 2>&1)
  ⎿  Making Mercury/int3s/render.int3
     Making Mercury/ints/render.int
     Making Mercury/cs/render.c

● Let me now compile the whole program to make sure everything works
  together:

● Bash(mmc --make amaze 2>&1)
  ⎿  Making Mercury/os/render.o
     Making amaze

● Excellent! Let me now test with a hex maze:

● Bash(./amaze -t hex -s 5 -r 42 /tmp/hex_test.svg && cat /tmp/hex_test.svg)
  ⎿  Error: Exit code 1
     Uncaught Mercury exception:
     Software Error: map.lookup: key not found
     	Key Type: grid.cell
     	Key Value: cell(0, -1)
     	Value Type: grid.cell
     Stack dump not available in this grade.

● There's an issue with cell(0, -1) being looked up. This is happening in a
  map lookup, likely during maze generation. The cell (0, -1) is outside the
   bounds of the hex maze. Let me check the maze bounds checking - the issue
   is that cell(0, 0) is supposed to be valid but adjacency in the up
  direction from (0, 0) gives (0, -1) which should be outside the maze.

  Let me check if the problem is in how boundary.m defines the hex maze's
  entrance cell. Looking at boundary.m:

● Read(boundary.m)
  ⎿  Read 70 lines

● I see - Edge2 includes cells from Q=0 to HalfW with R=0, all with
  direction up. But wait, for Q=0, R=0, the cell's up edge would connect to
  (0, -1), which is outside the maze. Looking at the PLAN.md, for the 10-12
  o'clock edge, the boundary cells should have their top edge on the
  boundary.

  Actually, I think the issue is something different. Let me check where the
   error is occurring by looking at the key being used. The error says
  cell(0, -1) was being looked up - this seems like it's happening during
  maze generation, probably in the union-find structure.

  Let me check what value W has. For size 5, W=5, H=5. HalfW = (5-1)/2 = 2.

  For cell (0, 0), is it valid?
  - Q >= 0: yes
  - Q < W: yes (0 < 5)
  - R >= Q - (W-1)/2: 0 >= 0 - 2 = -2, yes
  - R <= Q + H - 1: 0 <= 0 + 5 - 1 = 4, yes

  So (0, 0) is valid. But (0, -1):
  - Q >= 0: yes
  - R >= Q - (W-1)/2: -1 >= 0 - 2 = -2, yes (!)
  - R <= Q + H - 1: -1 <= 0 + 4 = 4, yes

  Wait, (0, -1) passes the constraints! That's wrong. The issue is that (0,
  -1) should not be in the maze, but with the current constraints it is.

  Looking at the corner cells:
  - 10 o'clock: (0, 0)
  - 12 o'clock: (2, 0)  -- for W=5
  - 2 o'clock: (4, 2)
  - 4 o'clock: (4, 6)
  - 6 o'clock: (2, 6)
  - 8 o'clock: (0, 4)

  For (0, -1) to be outside the maze, there needs to be a constraint that's
  being violated. But wait - the top-right constraint says R >= Q - (W-1)/2.
   For (0, -1), we have -1 >= 0 - 2 = -2, which is true.

  I think I need to re-examine the constraints. Looking at the 10 o'clock
  corner cell (0, 0), the valid cells at Q=0 should be R from 0 to H-1. So
  for Q=0:
  - Min R should be 0 (at 10 o'clock)
  - Max R should be H-1 (at 8 o'clock)

  So the minimum R at Q=0 is 0, not -2. The constraint R >= Q - (W-1)/2
  gives the wrong minimum for Q=0 when (W-1)/2 > 0.

  Looking back at the plan, the constraint R >= Q - (W-1)/2 should give R >=
   0 for Q = (W-1)/2 (at 12 o'clock). But for Q=0, it gives R >= -(W-1)/2,
  which is wrong.

  The issue is that the diagonal constraint works for the top-right diagonal
   (from 10 to 2 o'clock), but for Q < (W-1)/2, we need an additional
  constraint. Let me look at this more carefully.

  For a hex maze with W=5, H=5:
  - The leftmost column Q=0 should have cells from R=0 to R=4 (H-1)
  - The middle columns should go lower

  Actually I think we need a different constraint. Let me think about this
  again:

  For the hex maze shape:
  - At Q=0: R goes from 0 to H-1
  - At Q=1: R goes from 0 to H (for odd width mazes...)

  Wait, I think I misunderstand the shape. Let me re-read the corner
  positions:
  - 10 o'clock: (0, 0)
  - 12 o'clock: ((W-1)/2, 0)  = (2, 0) for W=5
  - 2 o'clock: (W-1, (W-1)/2) = (4, 2)
  - 4 o'clock: (W-1, (W-1)/2 + H - 1) = (4, 6)
  - 6 o'clock: ((W-1)/2, (W-1)/2 + H - 1) = (2, 6)
  - 8 o'clock: (0, H - 1) = (0, 4)

  So at Q=0, R ranges from 0 (at 10 o'clock) to H-1=4 (at 8 o'clock).
  At Q=2 (middle), R ranges from 0 (at 12 o'clock) to 6 (at 6 o'clock).
  At Q=4 (rightmost), R ranges from 2 (at 2 o'clock) to 6 (at 4 o'clock).

  The top boundary goes from (0,0) to (2,0) to (4,2).
  - For Q from 0 to 2: min R = 0
  - For Q from 2 to 4: min R = Q - 2

  The current constraint R >= Q - (W-1)/2 gives:
  - Q=0: R >= -2 (should be R >= 0)
  - Q=2: R >= 0 (correct)
  - Q=4: R >= 2 (correct)

  So the constraint is correct for Q >= (W-1)/2, but wrong for Q < (W-1)/2.
  We need:
  - For Q < (W-1)/2: R >= 0
  - For Q >= (W-1)/2: R >= Q - (W-1)/2

  Or we could write it as: R >= max(0, Q - (W-1)/2)

  Let me update the in_maze_hex predicate in maze.m:

● Read(maze.m)
  ⎿  Read 20 lines

● Update(maze.m)
  ⎿  Updated maze.m with 7 additions and 4 removals                      
       151    in_maze_hex(Bounds, cell(Q, R)) :-
       152        W = Bounds ^ width,
       153        H = Bounds ^ height,
       154 +      HalfW = (W - 1) / 2,
       155        % Q in [0, W)
       156        Q >= 0, Q < W,
       157 -      % Top-right diagonal constraint: R >= Q - (W-1)/2
       158 -      R >= Q - (W - 1) / 2,
       159 -      % Bottom-left diagonal constraint: R <= Q + H - 1
       160 -      R =< Q + H - 1.
       157 +      % Top constraint: for Q < HalfW, R >= 0; for Q >= 
           + HalfW, R >= Q - HalfW
       158 +      % Combined as: R >= max(0, Q - HalfW)
       159 +      R >= int.max(0, Q - HalfW),
       160 +      % Bottom constraint: for Q <= HalfW, R <= Q + H - 1; 
           + for Q > HalfW, R <= HalfW + H - 1
       161 +      % Combined as: R <= min(Q + H - 1, HalfW + H - 1)
       162 +      R =< int.min(Q + H - 1, HalfW + H - 1).
       163    
       164    %----------------------------------------------------------
             -----------------%
       165    

● Now I need to update cells_hex as well with the same constraint:

● Read(maze.m)
  ⎿  Read 25 lines

● Update(maze.m)
  ⎿  Updated maze.m with 6 additions and 5 removals               
       211    cells_hex(Bounds) = Cells :-
       212        W = Bounds ^ width,
       213        H = Bounds ^ height,
       214 -      % R ranges from 0 to (W-1)/2 + H - 1
       215 -      MaxR = (W - 1) / 2 + H - 1,
       214 +      HalfW = (W - 1) / 2,
       215 +      % R ranges from 0 to HalfW + H - 1
       216 +      MaxR = HalfW + H - 1,
       217        solutions(
       218            ( pred(Cell::out) is nondet :-
       219                int.nondet_int_in_range(0, W - 1, Q),
       220                int.nondet_int_in_range(0, MaxR, R),
       221 -              % Apply diagonal constraints
       222 -              R >= Q - (W - 1) / 2,
       223 -              R =< Q + H - 1,
       221 +              % Apply constraints matching in_maze_hex
       222 +              R >= int.max(0, Q - HalfW),
       223 +              R =< int.min(Q + H - 1, HalfW + H - 1),
       224                Cell = cell(Q, R)
       225            ),
       226            Cells).

● Now let me rebuild and test:

● Bash(mmc --make amaze 2>&1)
  ⎿  Making Mercury/int3s/maze.int3
     Making Mercury/ints/maze.int
     Making Mercury/cs/maze.c
     … +2 lines (ctrl+o to expand)

● Bash(./amaze -t hex -s 5 -r 42 /tmp/hex_test.svg && cat /tmp/hex_test.svg)
  ⎿  <?xml version="1.0" encoding="UTF-8"?>                                
     <svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 120 161">
       <polygon points="20,46.0 60,20 100,46.0 100,115.3 60,141.2 20,115.3"
     … +80 lines (ctrl+o to expand)

● The maze is generated. However, I notice there's an issue with one of the
  boundary polyline points - 95 158.6 seems too far down. Let me check the
  rendering more carefully. But first, let me test the solution path:

● Bash(./amaze -t hex -s 5 -r 42 -p /tmp/hex_test_path.svg && cat
      /tmp/hex_test_path.svg)
  ⎿  <?xml version="1.0" encoding="UTF-8"?>                                
     <svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 120 161">
       <polygon points="20,46.0 60,20 100,46.0 100,115.3 60,141.2 20,115.3"
     … +83 lines (ctrl+o to expand)

● I see the issue - there's a strange point 95 158.6 in the boundary
  polyline, which is far outside the viewbox. Let me trace through the
  boundary wall rendering. The issue is in the bottom boundary (6 to 8
  o'clock).

  Looking at wall_start_point for hex with lower_left direction, I have:
  Dir = lower_left,
  % lower_left edge: lower-left corner
  PX = CX - Sf / 2.0,
  PY = CY + Sf * sqrt3 / 2.0

  But when tracing clockwise, the start of a lower_left edge should be the
  lower-left corner of the cell (which is where we arrive at when tracing
  clockwise around the cell). Let me think about this more carefully.

  For the lower_left direction on the boundary (6-8 o'clock edge), we're
  tracing from 6 o'clock to 8 o'clock. The boundary cells on this edge have
  their lower_left edge on the boundary. When tracing clockwise around the
  maze:
  - At 6 o'clock we're at the bottom, about to go up-left
  - The cells' lower_left edges point outward from the maze

  For a cell on this boundary, when we trace clockwise:
  - We arrive at the cell coming from the cell to the right (at the
  lower-right corner of current cell)
  - We leave heading to the cell above-left (at the left corner of current
  cell)

  So for the lower_left direction:
  - Start (where we arrive clockwise): lower-right corner
  - End (where we leave clockwise): left corner

  But I have:
  - Start: lower-left corner (wrong!)
  - End: left corner (correct)

  Let me fix this:

● Update(render.m)
  ⎿  Updated render.m with 10 additions and 8 removals                   
       530    wall_start_point(hex, Bounds, S, M, boundary_wall(Cell, 
             Dir)) = Str :-
       531        hex_cell_centre(S, M, Bounds, Cell, CX, CY),
       532        Sf = float.float(S),
       533 -      % For clockwise traversal, the "start" of each
           -   boundary wall edge
       533 +      % For clockwise traversal around the maze boundary
           +  , the "start" of each
       534 +      % boundary wall edge is where we arrive when tracing 
           + clockwise.
       535 +      % This depends on which edge of the cell is the 
           + boundary edge.
       536        (
       537            Dir = upper_left,
       538 -          % upper_left edge (going counter-clockwise around 
           - cell): lower-left corner
       538 +          % upper_left edge on boundary: arriving from below 
           + (lower-left corner)
       539            PX = CX - Sf / 2.0,
       540            PY = CY + Sf * sqrt3 / 2.0
       541        ;
       542            Dir = up,
       543 -          % up edge: upper-left corner
       543 +          % up edge on boundary: arriving from left (
           +  upper-left corner)
       544            PX = CX - Sf / 2.0,
       545            PY = CY - Sf * sqrt3 / 2.0
       546        ;
       547            Dir = upper_right,
       548 -          % upper_right edge: upper-right corner
       548 +          % upper_right edge on boundary: 
           +  arriving from above (upper-right corner)
       549            PX = CX + Sf / 2.0,
       550            PY = CY - Sf * sqrt3 / 2.0
       551        ;
       552            Dir = lower_right,
       553 -          % lower_right edge: right corner
       553 +          % lower_right edge on boundary: 
           +  arriving from above (right corner)
       554            PX = CX + Sf,
       555            PY = CY
       556        ;
       557            Dir = down,
       558 -          % down edge: lower-right corner
       558 +          % down edge on boundary: arriving from right (
           +  lower-right corner)
       559            PX = CX + Sf / 2.0,
       560            PY = CY + Sf * sqrt3 / 2.0
       561        ;
       562            Dir = lower_left,
       563 -          % lower_left edge: lower-left corner
       564 -          PX = CX - Sf / 2.0,
       563 +          % lower_left edge on boundary: 
           +  arriving from below-right (lower-right corner)
       564 +          PX = CX + Sf / 2.0,
       565            PY = CY + Sf * sqrt3 / 2.0
       566        ;
       567            ( Dir = left

● Wait, I think I need to think about this more systematically. Let me trace
   the hex boundary for a 5x5 hex maze:

  Looking at the boundary traversal order in boundary.m:
  - Edge 1: upper_left edges at Q=0, going from entrance up to 10 o'clock
  - Edge 2: up edges from 10 to 12 o'clock
  - Edge 3: upper_right edges from 12 to 2 o'clock
  - Edge 4: lower_right edges from 2 to 4 o'clock
  - Edge 5: down edges from 4 to 6 o'clock
  - Edge 6: lower_left edges from 6 to 8 o'clock
  - Edge 7: upper_left edges from 8 o'clock back to entrance

  For lower_left edges (edge 6), we're going from 6 o'clock towards 8
  o'clock. That means we're moving counter-clockwise around the cell's
  boundary (since the cell's lower_left edge points outward/down-left).

  Wait, I think the issue is that "lower_left" is a direction from the cell,
   pointing outward. The boundary edge goes from lower-left corner to left
  corner of the cell. When we trace clockwise around the maze boundary:
  - We arrive at the lower-left corner from the previous cell's boundary
  - We leave at the left corner towards the next cell

  So:
  - Start = lower-left corner
  - End = left corner

  But in my wall_start_point, I had lower-right corner which is wrong. Let
  me reconsider.

  Actually, wait. The lower_left direction is pointing from inside the cell
  to outside. The edge that the direction points to goes from:
  - left corner to lower-left corner (the edge between the cell and its
  lower-left neighbour)

  When the boundary has this edge on it, and we're tracing clockwise around
  the maze (not the cell), we need to figure out which direction we traverse
   this edge.

  At the 6-8 o'clock boundary, we're going from bottom-right to top-left of
  the maze. The cells here have their lower_left edges on the boundary. When
   tracing clockwise around the maze:
  - We come from the lower-right (from a cell that had its down edge on the
  boundary)
  - We proceed to the upper-left (to a cell that will have its upper_left
  edge on the boundary, or lower_left if still on this edge segment)

  For each cell with lower_left edge on boundary:
  - We arrive at its lower-right corner (from the cell below-right's down
  edge end)
  - We traverse to its left corner
  - We leave towards the next cell

  Actually I think my lower_left start was wrong. The cells on the 6-8 edge
  are decreasing in Q. So for cell at (Q, Q+H-1):
  - We arrive from cell (Q+1, Q+H) which is to the lower-right
  - We enter at the lower-right corner of (Q, Q+H-1)
  - We exit at the left corner of (Q, Q+H-1)

  So start = lower-right corner of cell. But lower-right corner is at (CX +
  S/2, CY + S*sqrt3/2).

  Wait but I have both down and lower_left giving start = lower-right
  corner. That's actually correct because:
  - For down direction: the cell's down edge goes from lower-right to
  lower-left
  - For lower_left direction: the cell's lower_left edge goes from left to
  lower-left

  Hmm, that's not right either. Let me look at the hex edge definitions from
   PLAN.md:

  From PLAN.md:
  - up edge: upper-left corner to upper-right corner
  - down edge: lower-right corner to lower-left corner
  - upper_right edge: upper-right corner to right corner
  - lower_left edge: left corner to lower-left corner
  - upper_left edge: left corner to upper-left corner
  - lower_right edge: right corner to lower-right corner

  So for lower_left edge: left corner to lower-left corner.

  When tracing clockwise around the maze with this edge on the boundary (6-8
   o'clock edge):
  - Coming from the cell to the lower-right (whose down or lower_right edge
  was on boundary)
  - The arrival point is the lower-left corner of the current cell (which
  matches the end of the previous cell's down edge: lower-left corner)
  - No wait, that's not right either.

  Let me think about cells on the 6-8 boundary. These are cells (Q, Q+H-1)
  for Q from HalfW down to 0.

  At Q=HalfW (6 o'clock), the cell is ((W-1)/2, (W-1)/2 + H - 1). The
  previous cell in clockwise order was on the 4-6 edge at Q=HalfW+1. That
  cell's down edge ended at its lower-left corner. The current cell's (6
  o'clock cell) lower_left edge should start from... actually we need to see
   where the previous edge ended.

  Previous cell: (HalfW+1, HalfW+1+H-1) = (HalfW+1, HalfW+H) with direction
  down.
  The down edge goes from lower-right to lower-left, so end point is
  lower-left corner.

  Current cell: (HalfW, HalfW+H-1) with direction lower_left.
  The lower_left edge goes from left to lower-left.

  These don't connect! There's a vertex in between: at 6 o'clock, we need to
   go from the previous cell's lower-left corner to the current cell's left
  corner (or lower-left corner).

  Actually I think the issue is that the corner cells have multiple boundary
   edges. At 6 o'clock, cell (HalfW, HalfW+H-1) has two boundary edges: down
   (pointing to 6 o'clock) and lower_left.

  Looking back at boundary.m Edge 5 and Edge 6:
  - Edge 5: cells (Q, Q+H-1) for Q from W-1 down to HalfW+1, direction down
  - Edge 6: cells (Q, Q+H-1) for Q from HalfW down to 0, direction
  lower_left

  So at Q=HalfW, the cell only appears in Edge 6, not Edge 5. The Edge 5
  ends at Q=HalfW+1.

  Let me trace:
  - Edge 5 ends at cell (HalfW+1, HalfW+H), direction down. End point =
  lower-left corner of this cell.
  - Edge 6 starts at cell (HalfW, HalfW+H-1), direction lower_left. Start
  point should connect to the previous end point.

  Cell (HalfW+1, HalfW+H) center vs cell (HalfW, HalfW+H-1) center:
  - DQ = HalfW - (HalfW+1) = -1
  - DR = (HalfW+H-1) - (HalfW+H) = -1

  So cell (HalfW, HalfW+H-1) is upper-left of cell (HalfW+1, HalfW+H).

  The lower-left corner of (HalfW+1, HalfW+H) should be adjacent to... hmm,
  which corner of (HalfW, HalfW+H-1)?

  For cell C1 = (HalfW+1, HalfW+H):
  - lower-left corner at (CX1 - S/2, CY1 + S*sqrt3/2)

  For cell C2 = (HalfW, HalfW+H-1) which is upper-left of C1:
  - Using adjacency: upper_left means DQ=-1, DR=0
  - But here we have DQ=-1, DR=-1, which is... upper_right of C2 equals C1?
  No...
  - Let me calculate: C1's upper_left neighbour is at (HalfW, HalfW+H), not
  (HalfW, HalfW+H-1)

  So (HalfW, HalfW+H-1) is NOT directly adjacent to (HalfW+1, HalfW+H).
  These cells share a corner but not an edge!

  This means the boundary has a vertex at 6 o'clock corner. The traversal
  goes:
  - Cell (HalfW+1, HalfW+H) down edge ends at its lower-left corner
  - Then we turn the corner (vertex point at 6 o'clock)
  - Then cell (HalfW, HalfW+H-1) lower_left edge starts at its...
  lower-right corner? or lower-left corner?

  Actually, at the 6 o'clock vertex, both cells' edges meet at that point:
  - Cell (HalfW+1, HalfW+H)'s lower-left corner
  - Cell (HalfW, HalfW+H-1)'s lower-right corner

  These should be the same point! Let me verify:
  - C1 center: (M*S + S + (HalfW+1)1.5S, ...)
  - C2 center: (MS + S + HalfW1.5*S, ...)
  - DX = 1.5*S between them

  C1 lower-left: (CX1 - S/2, CY1 + Ssqrt3/2)
  C2 lower-right: (CX2 + S/2, CY2 + Ssqrt3/2)

  CX1 - S/2 = CX2 + 1.5*S - S/2 = CX2 + S
  CX2 + S/2 ≠ CX2 + S

  So they're NOT the same point. There's a gap!

  Wait, that can't be right. Let me think about the Y coordinates too.

  For C1 = (HalfW+1, HalfW+H), R1 = HalfW + H
  For C2 = (HalfW, HalfW+H-1), R2 = HalfW + H - 1

  CY = (Rf - Qf/2) * sqrt3 * S + offset

  CY1 = ((HalfW+H) - (HalfW+1)/2) * sqrt3 * S + offset
  CY2 = ((HalfW+H-1) - HalfW/2) * sqrt3 * S + offset

  CY1 - CY2 = (1 - 0.5) * sqrt3 * S = 0.5 * sqrt3 * S

  So C1 is below C2 by 0.5sqrt3S.

  C1 lower-left corner: (CX1 - S/2, CY1 + Ssqrt3/2)
  C2 lower-right corner: (CX2 + S/2, CY2 + Ssqrt3/2)

  X: CX1 - S/2 vs CX2 + S/2
     CX1 = CX2 + 1.5S
     CX1 - S/2 = CX2 + 1.5S - 0.5*S = CX2 + S
     CX2 + S/2 ≠ CX2 + S, difference = S/2

  Y: CY1 + Ssqrt3/2 vs CY2 + Ssqrt3/2
     CY1 = CY2 + 0.5sqrt3S
     CY1 + Ssqrt3/2 = CY2 + sqrt3S
     CY2 + Ssqrt3/2 ≠ CY2 + sqrt3S, difference = sqrt3*S/2

  So these corners are indeed different points! The boundary has a "corner"
  at 6 o'clock that is not captured by the edge endpoints.

  I think the boundary wall functions need to account for the maze corner
  vertices. But wait, looking at how the boundary polylines work:

  CornerPoints = list.map(wall_corner_point(Topology, Bounds, S, M),
  Segment),
  AllPoints = [StartPoint | CornerPoints] ++ [EndPoint],

  The wall_corner_point is wall_start_point. This should add the arrival
  point for each wall in the segment. If the walls transition properly
  corner-to-corner, this should work. But at the 6 o'clock corner, we have a
   vertex that isn't the start/end of any single wall.

  Looking at the second polyline more carefully. The segment2 goes from exit
   to entrance. The exit is on the right edge (lower_right direction). Let
  me trace:

  Exit wall end → segment2 walls → entrance wall start

  The second polyline starts at wall_end_point(Exit) = end of exit wall.
  Then adds wall_start_point for each wall in segment2.
  Then ends at wall_start_point(Entrance).

  If Segment2 includes the 4-6 and 6-8 edges, then at the 6 o'clock
  transition, we have:
  - Previous wall (from Edge5) has direction down
  - Current wall (from Edge6) has direction lower_left

  When we add wall_start_point(current_wall), it gives the start of the
  lower_left edge.

  For the down direction end point, I have: lower-left corner of cell.
  For the lower_left direction start point, I had: lower-right corner of
  cell (after my fix). But these are different cells!

  The wall_corner_point function gives the arrival corner at each wall. So
  after the last down wall, we add start of next wall which is the first
  lower_left wall's start corner.

  The issue is that at the maze corners (6 o'clock, etc.), we need an extra
  vertex point because the boundary makes a sharp turn. But the current
  logic doesn't add that corner point explicitly - it relies on the
  start/end points matching up.

  Actually, looking at the hex background polygon code, I explicitly add the
   6 corner points. For the boundary walls, I should do the same - but the
  boundary wall rendering tries to be generic.

  Let me think about this differently. The issue is:

  For hex mazes, when the boundary direction changes (e.g., from down to
  lower_left), there's a vertex at the maze corner that's not the same as
  either:
  - The end of the previous boundary wall
  - The start of the next boundary wall

  This is because the cells at maze corners have TWO boundary edges, but
  only one is included in each edge segment of the boundary.

  Looking at boundary.m more carefully:
  - Edge 5 ends with cell (HalfW+1, HalfW+H) direction down
  - Edge 6 starts with cell (HalfW, HalfW+H-1) direction lower_left

  But cell (HalfW, HalfW+H-1) actually has TWO boundary edges: down AND
  lower_left. The down edge of this cell is at the exact 6 o'clock position!
   So Edge 5 should include this cell's down edge before transitioning to
  Edge 6's lower_left edges.

  Actually, let me re-check. Cell (HalfW, HalfW+H-1):
  - Its down neighbour would be at (HalfW, HalfW+H)
  - Is (HalfW, HalfW+H) in the maze? Let me check: Q=HalfW, R=HalfW+H
    - R >= max(0, Q-HalfW) = max(0, 0) = 0: HalfW+H >= 0, yes
    - R <= min(Q+H-1, HalfW+H-1) = min(HalfW+H-1, HalfW+H-1) = HalfW+H-1
    - HalfW+H <= HalfW+H-1? NO!

  So cell (HalfW, HalfW+H) is NOT in the maze. That means cell (HalfW,
  HalfW+H-1)'s down edge IS on the boundary!

  So the boundary.m code is incorrect - Edge 5 should include cell (HalfW,
  HalfW+H-1) with direction down, and Edge 6 should start from the cell
  below that (if any).

  Let me check the boundary.m Edge 5 range: reverse((HalfW + 1) .. (W - 1)).
   This goes from W-1 down to HalfW+1, so it misses HalfW.

  This is the bug! Edge 5 should go from W-1 down to HalfW (inclusive), not
  stopping at HalfW+1.

  Wait, but that would conflict with Edge 6 which starts at HalfW. Let me
  think about what edges cell (HalfW, HalfW+H-1) actually has on the
  boundary:
  - down: neighbour at (HalfW, HalfW+H) - NOT in maze, so YES this is
  boundary
  - lower_left: neighbour at (HalfW-1, HalfW+H) - is this in maze?
    - Q=HalfW-1, R=HalfW+H
    - R >= max(0, Q-HalfW) = max(0, -1) = 0, HalfW+H >= 0, yes
    - R <= min(Q+H-1, HalfW+H-1) = min(HalfW+H-2, HalfW+H-1) = HalfW+H-2
    - HalfW+H <= HalfW+H-2? NO!

  So both down AND lower_left edges of cell (HalfW, HalfW+H-1) are on the
  boundary! This cell is at the 6 o'clock corner.

  The boundary traversal should include BOTH edges of this corner cell:
  - First the down edge (coming from 4 o'clock direction)
  - Then the lower_left edge (going towards 8 o'clock direction)

  So in boundary.m:
  - Edge 5 should include Q=HalfW (so range should be HalfW .. W-1,
  reversed)
  - Edge 6 should also include Q=HalfW (it already does: 0 .. HalfW,
  reversed)

  Both include the corner cell, which is correct! One contributes the down
  edge, the other the lower_left edge.

  Let me fix boundary.m:

● Update(boundary.m)
  ⎿  Updated boundary.m with 2 additions and 2 removals                
       171            HalfW .. (HalfW + H - 1)),
       172    
       173        % Edge 5: Bottom-right edge from 4 to 6 o'clock 
             (direction down)
       174 -      % Q decreases from W-1 to HalfW+1
       174 +      % Q decreases from W-1 to HalfW
           +   (inclusive, to include 6 o'clock corner)
       175        % R = Q + H - 1 (on the bottom-left diagonal)
       176        % These cells have their `down` edge on the boundary.
       177        Edge5 = list.map(
       178            (func(Q) = boundary_wall(cell(Q, Q + H - 1), 
             down)),
       179 -          reverse((HalfW + 1) .. (W - 1))),
       179 +          reverse(HalfW .. (W - 1))),
       180    
       181        % Edge 6: Bottom edge from 6 to 8 o'clock (direction 
             lower_left)
       182        % Q decreases from HalfW to 0

● Now I need to think about the other corners too. Let me check each corner:

  12 o'clock corner: cell (HalfW, 0)
  - Edge 2 uses cells Q=0..HalfW with direction up
  - Edge 3 uses cells Q=HalfW+1..W-1 with direction upper_right
  - Does cell (HalfW, 0) have both up and upper_right edges on boundary?
    - up neighbour: (HalfW, -1) - not in maze (R < 0), so YES
    - upper_right neighbour: (HalfW+1, -1) - is this in maze?
        - Q=HalfW+1, R=-1
      - R >= max(0, 1) = 1, but -1 < 1, so NO not in maze
    - So upper_right is also on boundary!

  Edge 2 includes this cell with up, Edge 3 should also include it with
  upper_right.
  Edge 3 currently: (HalfW + 1) .. (W - 1) - does NOT include HalfW!

  Let me fix Edge 3:

● Update(boundary.m)
  ⎿  Updated boundary.m with 2 additions and 2 removals               
       156            0 .. HalfW),
       157    
       158        % Edge 3: Top-right edge from 12 to 2 o'clock 
             (direction upper_right)
       159 -      % Q increases from HalfW+1 to W-1
       159 +      % Q increases from HalfW to W-1
           +   (inclusive, to include 12 o'clock corner)
       160        % R = Q - HalfW (on the top-right diagonal)
       161        % These cells have their `upper_right` edge on the 
             boundary.
       162        Edge3 = list.map(
       163            (func(Q) = boundary_wall(cell(Q, Q - HalfW), 
             upper_right)),
       164 -          (HalfW + 1) .. (W - 1)),
       164 +          HalfW .. (W - 1)),
       165    
       166        % Edge 4: Right edge from 2 to 4 o'clock (direction 
             lower_right)
       167        % Q = W-1, R increases from HalfW to HalfW + H - 1

● Now let me check the other corners:

  10 o'clock corner: cell (0, 0)
  - Edge 1 uses cells Q=0, R=0..StartR with direction upper_left (reversed,
  so StartR down to 0)
  - Edge 2 uses cells Q=0..HalfW with direction up
  - Does cell (0, 0) have both edges on boundary?
    - upper_left neighbour: (-1, 0) - Q < 0, not in maze, YES
    - up neighbour: (0, -1) - R < 0, not in maze, YES
    - So both are on boundary!

  Edge 1 ends at (0, 0) with upper_left. Edge 2 starts at (0, 0) with up.
  Both are included!

  2 o'clock corner: cell (W-1, HalfW)
  - Edge 3 uses cells with upper_right, ending at Q=W-1
  - Edge 4 uses cells with lower_right, starting at R=HalfW
  - Does cell (W-1, HalfW) have both edges on boundary?
    - upper_right neighbour: (W, HalfW-1) - Q >= W, not in maze, YES
    - lower_right neighbour: (W, HalfW) - Q >= W, not in maze, YES
    - Both on boundary!

  Edge 3 ends at (W-1, HalfW) with upper_right. Edge 4 starts at (W-1,
  HalfW) with lower_right. Both should be included!

  Edge 3: Q from HalfW to W-1, so includes W-1. Cell is (W-1, (W-1)-HalfW) =
   (W-1, HalfW). Correct!
  Edge 4: R from HalfW to HalfW+H-1, starting at R=HalfW. Cell is (W-1,
  HalfW). Correct!

  4 o'clock corner: cell (W-1, HalfW+H-1)
  - Edge 4 uses cells with lower_right, ending at R=HalfW+H-1
  - Edge 5 uses cells with down, starting at Q=W-1
  - Does cell (W-1, HalfW+H-1) have both edges on boundary?
    - lower_right neighbour: (W, HalfW+H-1) - Q >= W, not in maze, YES
    - down neighbour: (W-1, HalfW+H) - is this in maze?
        - R >= max(0, W-1-HalfW) = max(0, HalfW) = HalfW, and HalfW+H >=
  HalfW, yes
      - R <= min(W-1+H-1, HalfW+H-1) = min(W+H-2, HalfW+H-1) = HalfW+H-1
      - HalfW+H <= HalfW+H-1? NO!
    - So down is also on boundary!

  Edge 4 ends at (W-1, HalfW+H-1). Edge 5 should start at (W-1, HalfW+H-1).
  Edge 5: Q from W-1 down to HalfW. At Q=W-1, cell is (W-1, W-1+H-1) = (W-1,
   W+H-2).

  Wait, that's wrong! For W=5, H=5: (W-1, W+H-2) = (4, 7). But the 4 o'clock
   corner is at (W-1, HalfW+H-1) = (4, 2+4) = (4, 6).

  The issue is Edge 5 formula uses R = Q + H - 1. At Q=W-1=4, R = 4+5-1 = 8,
   not 6!

  Let me recalculate. The bottom-left diagonal has constraint R <= Q + H -
  1. At Q=W-1:
  - The cell on this diagonal at Q=W-1 has R = W-1+H-1 = W+H-2

  But the 4 o'clock corner has R = HalfW + H - 1, not W+H-2!

  For W=5: HalfW = 2. 4 o'clock cell: (4, 2+5-1) = (4, 6).
  The diagonal R = Q + H - 1 at Q=4: R = 4+5-1 = 8.

  These are different! The cell (4, 8) doesn't exist because:
  - R <= min(Q+H-1, HalfW+H-1) = min(8, 6) = 6

  So the valid R values at Q=4 are from HalfW=2 to 6, not up to 8.

  The 4 o'clock corner is cell (W-1, HalfW+H-1) = (4, 6), which is on Edge 4
   (lower_right at R=HalfW+H-1=6).

  The bottom-right diagonal (4-6 o'clock edge) has cells where R is at the
  maximum valid value for each Q. For Q >= HalfW, this maximum is HalfW + H
  - 1 (constant). For Q < HalfW, the maximum is Q + H - 1.

  So Edge 5 (direction down) should be:
  - For Q from W-1 down to HalfW, cells are at (Q, HalfW+H-1) not (Q,
  Q+H-1)!

  This is a bug in boundary.m! Let me fix it:

● Update(boundary.m)
  ⎿  Updated boundary.m with 2 additions and 2 removals                  
       172    
       173        % Edge 5: Bottom-right edge from 4 to 6 o'clock 
             (direction down)
       174        % Q decreases from W-1 to HalfW (inclusive, to include 
             6 o'clock corner)
       175 -      % R = Q + H - 1 (on the bottom-left diagonal)
       175 +      % R = HalfW + H - 1 (constant, the bottom row for Q >= 
           + HalfW)
       176        % These cells have their `down` edge on the boundary.
       177        Edge5 = list.map(
       178 -          (func(Q) = boundary_wall(cell(Q, Q
           -   + H - 1), down)),
       178 +          (func(Q) = boundary_wall(cell(Q, HalfW
           +   + H - 1), down)),
       179            reverse(HalfW .. (W - 1))),
       180    
       181        % Edge 6: Bottom edge from 6 to 8 o'clock (direction 
             lower_left)

● Wait, I also need to fix Edge 6! The comment says R = Q + H - 1 but for Q
  < HalfW, the bottom boundary is actually at R = Q + H - 1 which is the
  bottom-left diagonal constraint. Let me verify:

  8 o'clock corner: cell (0, H-1)
  - Edge 6 should end at (0, H-1) with direction lower_left
  - Edge 7 should start at (0, H-1) with direction upper_left

  For cell (0, H-1):
  - lower_left neighbour: (-1, H) - Q < 0, not in maze, YES boundary
  - upper_left neighbour: (-1, H-1) - Q < 0, not in maze, YES boundary

  Both edges on boundary, correct!

  Edge 6: cells (Q, Q+H-1) for Q from HalfW down to 0
  - At Q=0: (0, 0+5-1) = (0, 4) = (0, H-1). Correct for 8 o'clock!
  - At Q=HalfW=2: (2, 2+5-1) = (2, 6). Is this the 6 o'clock cell?

  6 o'clock corner is ((W-1)/2, (W-1)/2 + H - 1) = (2, 2+5-1) = (2, 6). Yes!

  So Edge 6 formula R = Q + H - 1 is correct for Q from 0 to HalfW.

  But Edge 5 now uses R = HalfW + H - 1 = 2+5-1 = 6 for all Q from HalfW to
  W-1. At Q=HalfW=2, this gives cell (2, 6), which is also the 6 o'clock
  corner. Good!

  At Q=W-1=4, Edge 5 gives cell (4, 6). Is this the 4 o'clock corner?
  4 o'clock corner is (W-1, (W-1)/2 + H - 1) = (4, 2+5-1) = (4, 6). Yes!

  Now let me rebuild and test:

● Bash(mmc --make amaze 2>&1)
  ⎿  Making Mercury/int3s/boundary.int3
     Making Mercury/int3s/render.int3
     Making Mercury/ints/boundary.int
     … +6 lines (ctrl+o to expand)

● Bash(./amaze -t hex -s 5 -r 42 /tmp/hex_test.svg && cat /tmp/hex_test.svg)
  ⎿  <?xml version="1.0" encoding="UTF-8"?>                                
     <svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 120 161">
       <polygon points="20,46.0 60,20 100,46.0 100,115.3 60,141.2 20,115.3"
     … +81 lines (ctrl+o to expand)

● Now the boundary points look more reasonable. The second polyline now has
  points that stay within the viewbox (max Y is 141.2 which is within 161).
  Let me test with the path and view the SVG:

● Bash(./amaze -t hex -s 5 -r 42 -p /tmp/hex_test_path.svg)
  ⎿  (No content)

● Read(/tmp/hex_test_path.svg)
  ⎿  Read 44 lines

● The SVG is generating. Let me also test that the square maze still works:

● Bash(./amaze -t square -s 5 -r 42 -p /tmp/square_test.svg && echo "Square maze
       generated successfully")
  ⎿  Square maze generated successfully

● Let me also test different algorithms with hex:

● Bash(for algo in dfs kruskal prim aldous-broder wilson hunt-and-kill; do
          echo "Testing $algo with hex..."…)
  ⎿  Testing dfs with hex...       
       OK
     Testing kruskal with hex...
     … +15 lines (ctrl+o to expand)

● There's an issue with aldous-broder and wilson - they're reporting
  unreachable cells. This suggests there's still a problem with the hex maze
   generation or the cell enumeration. The error says 25 or 26 visited out
  of 29, which means some cells aren't being connected.

  Let me check the cell count. For W=5, H=5, HalfW=2:
  - Q=0: R from 0 to 4, so 5 cells
  - Q=1: R from 0 to 5, so 6 cells
  - Q=2: R from 0 to 6, so 7 cells
  - Q=3: R from 1 to 6, so 6 cells
  - Q=4: R from 2 to 6, so 5 cells

  Total: 5+6+7+6+5 = 29 cells. That matches!

  The issue might be in how adjacency is being calculated, or the neighbours
   function in maze.m. Let me check if there's a connectivity issue:

● Bash(./amaze -t hex -s 5 -r 42 -a aldous-broder -d /tmp/hex_ab.svg 2>&1 | head
       -50)
  ⎿  Uncaught Mercury exception:            
     "unreachable cells: visited 25 of 29"
     Stack dump not available in this grade.

● The dump doesn't help since it fails before outputting. Let me look at the
   aldous-broder algorithm to understand what might be going wrong:

● Read(generate.aldous_broder.m)
  ⎿  Read 134 lines

● I see the problem! The count_cells function is incorrect for hex mazes:

  count_cells(Maze) = Count :-
      Bounds = get_bounds(Maze),
      Count = Bounds ^ width * Bounds ^ height.

  For hex mazes, the total cell count is NOT width * height. It's actually
  the number of cells in the pointy-topped hexagon shape. For W=5, H=5, it's
   29 cells, not 5*5=25.

  This is why aldous-broder terminates early - it thinks there are only 25
  cells when there are actually 29, so it stops walking before visiting all
  cells.

  Let me fix the count_cells function to use the actual cell count:

● Update(generate.aldous_broder.m)
  ⎿  Updated generate.aldous_broder.m with 1 addition and 3 removals
       61        %
       62    :- func count_cells(maze) = int.
       63    
       64 -  count_cells(Maze) = Count :-
       65 -      Bounds = get_bounds(Maze),
       66 -      Count = Bounds ^ width * Bounds ^ height.
       64 +  count_cells(Maze) = list.length(cells(Maze)).
       65    
       66        % Random walk until all cells are visited.
       67        % Current is the current cell.

● Now let me check if wilson.m has the same issue:

● Search(pattern: "count_cells|width.*height", path: "generate.wilson.m",
        output_mode: "content")
  ⎿  Found 4 lines (ctrl+o to expand)

● Yes, wilson.m has the same issue. Let me fix it:

● Read(generate.wilson.m)
  ⎿  Read 15 lines

● Update(generate.wilson.m)
  ⎿  Updated generate.wilson.m with 1 addition and 3 removals            
       64        %
       65    :- func count_cells(maze) = int.
       66    
       67 -  count_cells(Maze) = Count :-
       68 -      Bounds = get_bounds(Maze),
       69 -      Count = Bounds ^ width * Bounds ^ height.
       67 +  count_cells(Maze) = list.length(cells(Maze)).
       68    
       69        % Main Wilson's loop - keep adding paths until all cells
             are in the maze.
       70        %

● Now rebuild and test:

● Bash(mmc --make amaze 2>&1)
  ⎿  Making Mercury/int3s/generate.aldous_broder.int3
     Making Mercury/int3s/generate.wilson.int3
     Making Mercury/ints/generate.aldous_broder.int
     … +6 lines (ctrl+o to expand)

● Bash(for algo in dfs kruskal prim aldous-broder wilson hunt-and-kill; do
          echo "Testing $algo with hex..."…)
  ⎿  Testing dfs with hex...      
       OK
     Testing kruskal with hex...
     … +9 lines (ctrl+o to expand)

● All topology-agnostic algorithms now work with hex. Now let me add error
  handling for binary-tree and sidewinder when used with hex:

● Read(generate.m)
  ⎿  Read 84 lines

● The error handling should happen in the options module during option
  validation. Looking at the PLAN.md, Binary Tree and Sidewinder should
  report errors when used with hex topology. Let me update the options
  module to check this:

● Read(options.m)
  ⎿  Read 30 lines

● I need to add a check that validates algorithm compatibility with
  topology. Let me find a good place to add this check:

● Update(options.m)
  ⎿  Updated options.m with 15 additions and 1 removal                  
       257                OutputFile = no
       258            ),
       259    
       260 +          % Check algorithm/topology compatibility
       261            ( if
       262 +              Topology = hex,
       263 +              (Algorithm = binary_tree ; Algorithm = 
           + sidewinder)
       264 +          then
       265 +              io.format(Stderr,
       266 +                  "Error: algorithm '%s' is not compatible 
           + with hex topology\n",
       267 +                  [s(algorithm_name(Algorithm))], !IO),
       268 +              CompatibilityOk = no
       269 +          else
       270 +              CompatibilityOk = yes
       271 +          ),
       272 +  
       273 +          ( if
       274                TopologyOk = yes,
       275                SizeOk = yes,
       276 -              AlgorithmOk = yes
       276 +              AlgorithmOk = yes,
       277 +              CompatibilityOk = yes
       278            then
       279                Options = options(Topology, Width, Height, 
             EntranceOffset,
       280                    ExitOffset, Seed, Algorithm, CellSpacing, 
             Margin, ShowPath,

● Now I need to add the algorithm_name function:

● Update(options.m)
  ⎿  Updated options.m with 11 additions                      
       296    
       297    parse_algorithm("dfs", dfs).
       298    parse_algorithm("binary-tree", binary_tree).
       299 +  
       300 +  :- func algorithm_name(algorithm) = string.
       301 +  
       302 +  algorithm_name(dfs) = "dfs".
       303 +  algorithm_name(binary_tree) = "binary-tree".
       304 +  algorithm_name(sidewinder) = "sidewinder".
       305 +  algorithm_name(hunt_and_kill) = "hunt-and-kill".
       306 +  algorithm_name(aldous_broder) = "aldous-broder".
       307 +  algorithm_name(wilson) = "wilson".
       308 +  algorithm_name(kruskal) = "kruskal".
       309 +  algorithm_name(prim) = "prim".
       310    parse_algorithm("sidewinder", sidewinder).
       311    parse_algorithm("hunt-and-kill", hunt_and_kill).
       312    parse_algorithm("aldous-broder", aldous_broder).

● I see that the parse_algorithm clauses got mixed up with the
  algorithm_name function. Let me fix this:

● Read(options.m)
  ⎿  Read 30 lines

● The clauses are interleaved incorrectly. Let me fix this by reordering:

● Update(options.m)
  ⎿  Updated options.m with 6 additions and 6 removals              
       296    
       297    parse_algorithm("dfs", dfs).
       298    parse_algorithm("binary-tree", binary_tree).
       299 +  parse_algorithm("sidewinder", sidewinder).
       300 +  parse_algorithm("hunt-and-kill", hunt_and_kill).
       301 +  parse_algorithm("aldous-broder", aldous_broder).
       302 +  parse_algorithm("wilson", wilson).
       303 +  parse_algorithm("kruskal", kruskal).
       304 +  parse_algorithm("prim", prim).
       305    
       306    :- func algorithm_name(algorithm) = string.
       307    
     ...
       307    algorithm_name(wilson) = "wilson".
       308    algorithm_name(kruskal) = "kruskal".
       309    algorithm_name(prim) = "prim".
       310 -  parse_algorithm("sidewinder", sidewinder).
       311 -  parse_algorithm("hunt-and-kill", hunt_and_kill).
       312 -  parse_algorithm("aldous-broder", aldous_broder).
       313 -  parse_algorithm("wilson", wilson).
       314 -  parse_algorithm("kruskal", kruskal).
       315 -  parse_algorithm("prim", prim).
       310    
       311    :- pred parse_size(string::in, int::out, int::out) is 
             semidet.
       312    

● Now let me compile and test:

● Bash(mmc --make amaze 2>&1)
  ⎿  Making Mercury/int3s/options.int3
     Making Mercury/ints/options.int
     Making Mercury/cs/options.c
     … +2 lines (ctrl+o to expand)

● Bash(./amaze -t hex -a binary-tree -s 5 /tmp/test.svg; echo "Exit code: $?")
  ⎿  Exit code: 0                                                      
     Error: algorithm 'binary-tree' is not compatible with hex topology

● The error message appears, but the exit code is 0 and the error appears
  after the exit code. Let me check what's happening - the program should
  exit with a non-zero code. Let me look at the main module:

● Bash(./amaze -t hex -a sidewinder -s 5 /tmp/test.svg 2>&1; echo "Exit code:
      $?")
  ⎿  Error: algorithm 'sidewinder' is not compatible with hex topology
     Exit code: 0

● The error message is being output but the exit code is 0. This is because
  when MaybeOptions = no, the main module probably just exits without
  generating a maze. Let me check:

● Read(amaze.m)
  ⎿  Read 40 lines

● I see - when MaybeOptions = no, the program just exits without setting a
  non-zero exit code. The error message is printed to stderr, but the
  program exits successfully. Let me update the main module to set an exit
  code:

● Update(amaze.m)
  ⎿  Updated amaze.m with 2 additions and 1 removal                  
       49            MaybeOptions = yes(Options),
       50            run(Options, !IO)
       51        ;
       52 -          MaybeOptions = no
       52 +          MaybeOptions = no,
       53            % Error message already printed by parse_options
       54 +          io.set_exit_status(1, !IO)
       55        ).
       56    
       57    :- pred run(options::in, io::di, io::uo) is det.

● Bash(mmc --make amaze 2>&1 && ./amaze -t hex -a binary-tree -s 5 /tmp/test.svg
       2>&1; echo "Exit code: $?")
  ⎿  Making Mercury/int3s/amaze.int3
     Making Mercury/ints/amaze.int
     Making Mercury/cs/amaze.c
     … +4 lines (ctrl+o to expand)

● Now the error handling is working correctly. Let me also verify that
  binary-tree and sidewinder still work with square topology:

● Bash(./amaze -t square -a binary-tree -s 5 -r 42 /tmp/sq_bt.svg && echo
      "binary-tree with square: OK"
      ./amaze -t square -a sidewinder -s 5 -r 42 /tmp/sq_sw.svg && ec…)
  ⎿  binary-tree with square: OK
     sidewinder with square: OK

● Now let me update the todo list and run a comprehensive test:

● Let me run a final comprehensive test with different sizes:

● Bash(echo "=== Testing various hex sizes ==="
      for size in 3 5 7 10; do…)
  ⎿  === Testing various hex sizes ===
     Size 3: OK
     Size 5: OK
     … +15 lines (ctrl+o to expand)

● All tests pass. Let me provide a summary of the hex grid implementation:

  Summary

  The hex grid support has been implemented in iteration 6. Here are the key
   changes made:

  Files Modified:

  1. grid.m - Updated hex adjacency so R increases in the down direction
  (previously completed)
  2. maze.m -
    - Fixed in_maze_hex to use correct bounds constraints: R >= max(0, Q - 
  HalfW) and R <= min(Q + H - 1, HalfW + H - 1)
    - Fixed cells_hex to use the same constraints
  3. boundary.m -
    - Fixed Edge 3 (12-2 o'clock) to include the 12 o'clock corner cell
    - Fixed Edge 5 (4-6 o'clock) to use R = HalfW + H - 1 (constant) instead
   of R = Q + H - 1
  4. render.m - Major updates:
    - Added viewbox_dimensions for topology-aware viewbox calculation
    - Added hex_cell_centre to compute screen coordinates for hex cells
    - Added hex background polygon rendering
    - Updated edge_endpoints for hex internal walls
    - Updated wall_start_point and wall_end_point for hex boundary walls
    - Updated cell_centre for hex solution path rendering
  5. generate.aldous_broder.m and generate.wilson.m -
    - Fixed count_cells to use list.length(cells(Maze)) instead of width * 
  height
  6. options.m -
    - Added algorithm/topology compatibility check to reject binary-tree and
   sidewinder with hex topology
    - Added algorithm_name function for error messages
  7. amaze.m -
    - Set exit status to 1 when options parsing fails

  Key Hex Geometry:

  - Flat-topped hexagonal cells with pointy-topped hexagon maze shape
  - Axial coordinates (Q, R) where Q increases in upper_right direction, R
  increases down
  - Six directions: up, down, upper_left, upper_right, lower_left,
  lower_right
  - Bounds constraints create a pointy-topped hexagon shape with corners at
  clock positions (10, 12, 2, 4, 6, 8 o'clock)
  - Cell spacing S is the distance between adjacent cell centres, equal to
  the hex corner-to-centre distance

> /context 
  ⎿  
      Context Usage
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     ⛁ ⛁ ⛁ ⛁ ⛁ ⛁ ⛁ ⛁ ⛁ ⛁   ⛁ System prompt: 2.8k tokens (1.4%)
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     ⛁ ⛁ ⛁ ⛁ ⛁ ⛁ ⛁ ⛁ ⛁ ⛁   ⛶ Free space: 25k (12.6%)
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     ⛝ ⛝ ⛝ ⛝ ⛝ ⛝ ⛝ ⛝ ⛝ ⛝ 

     SlashCommand Tool · 0 commands
     └ Total: 864 tokens

     

> Nice work! There is a bug with how we are rendering internal walls, 
however, so it appears as if the solution crosses through walls. This may be
 because of an inconsistency or some other error in our handling of 
coordinates. Perhaps we are conflating edges and boundary walls at some 
point? Can you think what the problem might be? 

● Looking at the rendering code, I think I see the issue. The problem is
  likely in edge_endpoints for hex grids. When we render internal walls, we
  use edge_cells to get the two cells adjacent to an edge, then compute the
  wall endpoints. But the edge endpoints calculation might be using the
  wrong cell's perspective.

  Let me trace through the logic:

  1. In render_wall, we call edge_cells(Topology, Edge, Cell1, Cell2) which
  returns the two cells on either side of the edge
  2. Then we call edge_endpoints(Topology, Bounds, S, M, Cell1, Cell2, ...)
  3. In edge_endpoints for hex, we compute the direction from Cell1 to Cell2
   and draw the wall from Cell1's perspective

  The issue is that edge_cells returns the cells in a specific order (the
  canonical cell first), but my edge_endpoints calculates direction as
  (Q2-Q1, R2-R1) and then draws from Cell1's centre. This should be
  correct...

  Wait, I think I see it now! Looking at the hex edge endpoint calculations:

  ( if DQ = 0, DR = -1 then
      % up edge: upper-left to upper-right corner
      PX1 = CX - Sf / 2.0,
      PY1 = CY - Sf * sqrt3 / 2.0,
      PX2 = CX + Sf / 2.0,
      PY2 = CY - Sf * sqrt3 / 2.0

  This draws the up edge of Cell1 when Cell2 is above Cell1 (DR = -1). But
  this is wrong! If Cell2 is above Cell1 (Cell2's R is less), then the edge
  between them should be drawn at the TOP of Cell1, which is correct. But
  wait...

  Actually, let me check edge_cells. In grid.m:

  edge_cells(Topology, edge(Cell, Dir), Cell, AdjCell) :-
      AdjCell = adjacent(Topology, Cell, Dir).

  So edge_cells returns the canonical cell first, then computes the adjacent
   cell in the canonical direction. For hex, canonical directions are up,
  upper_right, lower_right.

  So if the edge is edge(cell(2,3), up), then:
  - Cell1 = cell(2,3)
  - Cell2 = adjacent(hex, cell(2,3), up) = cell(2, 2)

  In edge_endpoints, DQ = 0, DR = 2-3 = -1, so we draw the up edge
  correctly.

  Hmm, the logic seems right. Let me look at whether the issue might be in
  how we're computing cell centres. Let me check hex_cell_centre:

● Read(render.m)
  ⎿  Read 30 lines

● I notice the Y formula has: CY = Mf * Sf + Sf * sqrt3 / 2.0 + (Rf - Qf / 
  2.0) * sqrt3 * Sf - MinYOffset

  Let me simplify this. The term - MinYOffset is - (-HalfW / 2.0 * sqrt3 * 
  Sf) = + HalfW / 2.0 * sqrt3 * Sf.

  So: CY = Mf * Sf + Sf * sqrt3 / 2.0 + (Rf - Qf / 2.0) * sqrt3 * Sf + HalfW
   / 2.0 * sqrt3 * Sf
      CY = Mf * Sf + Sf * sqrt3 / 2.0 + (Rf - Qf / 2.0 + HalfW / 2.0) * 
  sqrt3 * Sf

  Now I notice the solution path uses cell_centre which calls
  hex_cell_centre. Let me check if the internal wall drawing is actually
  using the same centre calculation... Yes, it is - edge_endpoints calls
  hex_cell_centre.

  Let me look at a specific example. Take cell (0, 2) for a 5x5 hex maze.
  HalfW = 2.

  Cell centre:
  - CX = MS + S + 01.5*S = (M+1)*S
  - CY = MS + Ssqrt3/2 + (2 - 0 + 1)sqrt3S = MS + Ssqrt3/2 + 3sqrt3S = MS + 
  3.5sqrt3*S

  Now, cell (0, 2)'s down neighbour is (0, 3).

  For the edge between (0,2) and (0,3), the wall should be at the bottom of
  cell (0,2). In edge_endpoints with Cell1=(0,2), Cell2=(0,3):
  - DQ = 0, DR = 1
  - This matches the down case
  - We draw: lower-right to lower-left corner of Cell1
  - PX1 = CX + S/2, PY1 = CY + S*sqrt3/2
  - PX2 = CX - S/2, PY2 = CY + S*sqrt3/2

  That looks correct for a flat-topped hex.

  Wait... I just realized something. In edge_cells, the returned cells are
  based on the canonical direction. But the internal edges list uses
  get_edge(Topology, Cell, Dir) which normalizes to canonical. So when we
  get an edge and then call edge_cells, we might get the cells in a
  different order than expected.

  Actually, let me check internal_edges:

● Read(maze.m)
  ⎿  Read 20 lines

● I see - internal_edges only uses canonical directions (up, upper_right,
  lower_right for hex). And get_edge with a canonical direction just returns
   edge(Cell, Dir).

  So for hex:
  - When we have Cell and Dir in canonical directions, edge_cells returns
  Cell and the adjacent cell
  - This is consistent

  Let me think about this differently. The solution path goes through cell
  centres. The internal walls are edges. The bug is that the solution
  appears to cross walls.

  One possibility: the solution path is correct, but the walls are in the
  wrong positions. Or vice versa.

  Let me check if maybe the issue is that we're drawing walls for edges that
   DO have doors (the opposite of what we want). Let me look at
  render_internal_walls:

● Read(render.m)
  ⎿  Read 30 lines

● The logic looks correct - we filter edges to get those WITHOUT doors, then
   draw walls for those.

  Wait, let me check has_door and add_door:

● Read(maze.m)
  ⎿  Read 15 lines

● Those look fine. The edge is stored in a set, and we check membership.

  Let me look at how edges are created in the generation algorithms. Let me
  check generate.kruskal.m:

● Search(pattern: "add_door|get_edge", path: "generate.kruskal.m", output_mode:
        "content")
  ⎿  Found 1 line (ctrl+o to expand)

● Read(generate.kruskal.m)
  ⎿  Read 25 lines

● Kruskal uses internal_edges(Maze) to get edges and then adds doors to
  them. The same edges should be in the same representation.

  Wait - I just realized something important! Let me check internal_edges
  again and compare with how Kruskal gets its edges:

● Read(generate.kruskal.m)
  ⎿  Read 20 lines

● Both use internal_edges(!.Maze). Good, so the edges added as doors should
  be from the same set as the edges we check for walls.

  Let me think about this differently. Maybe the issue is in the solution
  path itself. Let me check check.m:

● Search(pattern: "neighbours", path: "check.m", output_mode: "content")
  ⎿  Found 2 lines (ctrl+o to expand)

● Read(maze.m)
  ⎿  Read 19 lines

● I see it now! In neighbours, we use get_edge(Topology, Cell, Dir) to check
   if there's a door. But in internal_edges, edges are created only with
  canonical directions.

  The issue is that get_edge normalizes the edge to canonical form. So if we
   check get_edge(hex, cell(0,3), up), it returns edge(cell(0,3), up). But
  if we check get_edge(hex, cell(0,2), down), it should return the same
  edge... let me verify:

● Read(grid.m)
  ⎿  Read 25 lines

● So for hex:
  - is_canonical(up) - yes
  - is_canonical(upper_right) - yes
  - is_canonical(lower_right) - yes
  - is_canonical(down) - NO!

  So get_edge(hex, cell(0,2), down):
  - down is not canonical
  - AdjCell = adjacent(hex, cell(0,2), down) = cell(0,3)
  - CanonDir = opposite(down) = up
  - Returns edge(cell(0,3), up)

  And get_edge(hex, cell(0,3), up):
  - up is canonical
  - Returns edge(cell(0,3), up)

  Good, both return the same edge! So the edge representations should be
  consistent.

  Let me look at the edge visualization more carefully. Maybe I can add some
   debug output. Actually, let me generate a small hex maze and examine the
  SVG manually:

● Bash(./amaze -t hex -s 3 -r 42 -p /tmp/hex3.svg && cat /tmp/hex3.svg)
  ⎿ <?xml version="1.0" encoding="UTF-8"?>                                 
    <svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 90 109">
      <polygon points="20,37.3 45,20 70,37.3 70,72.0 45,89.3 20,72.0" fill=
    … +27 lines (ctrl+o to expand)


● Let me analyze this. The solution path is:
  30 54.6 -> 45 46.0 -> 45 63.3 -> 60 54.6

  These are 4 cells. Let me identify them by their coordinates...

  For a 3x3 hex maze, W=3, H=3, HalfW=1.

  The cells are:
  - Q=0: R from 0 to 2 (3 cells)
  - Q=1: R from 0 to 3 (4 cells)
  - Q=2: R from 1 to 3 (3 cells)

  Total: 10 cells.

  Cell centres are at:
  - CX = MS + S + Q1.5S = 20 + 10 + Q15 = 30 + Q*15
  - CY = MS + Ssqrt3/2 + (R - Q/2 + HalfW/2)sqrt3S
     = 20 + 8.66 + (R - Q/2 + 0.5)*17.32
     = 28.66 + (R - Q/2 + 0.5)*17.32

  For cell (0, 1): CX = 30, CY = 28.66 + (1 - 0 + 0.5)*17.32 = 28.66 + 25.98
   = 54.64 ≈ 54.6

  So the solution path starts at cell (0, 1) at (30, 54.6).

  Next point is (45, 46.0). CX = 45 means Q = (45-30)/15 = 1.
  CY = 46.0, so (R - 0.5 + 0.5)*17.32 = 46.0 - 28.66 = 17.34, so R = 1.
  Cell (1, 1).

  Next point is (45, 63.3). Same Q=1. CY = 63.3 means R = (63.3 -
  28.66)/17.32 + 0 = 2.
  Cell (1, 2).

  Final point is (60, 54.6). CX = 60 means Q = (60-30)/15 = 2.
  CY = 54.6, so (R - 1 + 0.5)*17.32 = 54.6 - 28.66 = 25.94, R - 0.5 = 1.5, R
   = 2.
  Cell (2, 2).

  So the path is: (0,1) -> (1,1) -> (1,2) -> (2,2)

  Let me verify these cells are adjacent:
  - (0,1) to (1,1): DQ=1, DR=0. This is lower_right direction. ✓
  - (1,1) to (1,2): DQ=0, DR=1. This is down direction. ✓
  - (1,2) to (2,2): DQ=1, DR=0. This is lower_right direction. ✓

  Now let me check if walls exist between consecutive path cells. The path
  goes through these edges:
  1. Edge between (0,1) and (1,1) - direction lower_right from (0,1), or
  upper_left from (1,1)
  2. Edge between (1,1) and (1,2) - direction down from (1,1), or up from
  (1,2)
  3. Edge between (1,2) and (2,2) - direction lower_right from (1,2), or
  upper_left from (2,2)

  These edges should NOT have walls (they should have doors).

  Now let me look at the internal wall lines in the SVG:
  1. 35 46.0 -> 40 54.6
  2. 35 63.3 -> 40 72.0
  3. 40 72.0 -> 35 80.6
  4. 40 37.3 -> 50 37.3
  5. 50 54.6 -> 55 63.3
  6. 40 72.0 -> 50 72.0
  7. 55 46.0 -> 65 46.0
  8. 55 63.3 -> 65 63.3

  Now, does any wall cross the solution path?

  Path segment 1: (30, 54.6) to (45, 46.0) - this goes from bottom-left to
  upper-right
  Path segment 2: (45, 46.0) to (45, 63.3) - this goes straight down at X=45
  Path segment 3: (45, 63.3) to (60, 54.6) - this goes from bottom-left to
  upper-right

  Wall 5: 50 54.6 -> 55 63.3 - this is around the area of path segment 3...

  Let me check: Path segment 3 goes from (45, 63.3) to (60, 54.6).
  Wall 5 goes from (50, 54.6) to (55, 63.3).

  These lines cross! The path at X=50 would be at Y = 63.3 -
  (63.3-54.6)(50-45)/(60-45) = 63.3 - 8.75/15 = 63.3 - 2.9 = 60.4
  The wall at X=50 is at Y = 54.6.

  Actually wait, they don't cross at the same point. At X=50:
  - Path Y = 60.4
  - Wall goes from (50, 54.6) to (55, 63.3), so wall at X=50 starts at
  Y=54.6

  These are different Y values (60.4 vs 54.6), so they don't actually
  intersect at X=50.

  Let me check X=52.5 (midpoint of wall):
  - Wall Y = (54.6 + 63.3)/2 = 58.95
  - Path at X=52.5: Y = 63.3 - 8.7*(52.5-45)/15 = 63.3 - 4.35 = 58.95

  They DO intersect at (52.5, 58.95)!

  So wall 5 crosses path segment 3. But these should both be in the same
  cells! Let me figure out which cells.

  Wall 5 (50 54.6 -> 55 63.3) - what cells is this between?

  Looking at edge_endpoints, this pattern (diagonal line going down-right)
  suggests upper_right or lower_left direction.

  For upper_right edge from cell C1 to cell C2 where C2 is at DQ=1, DR=-1:
  - PX1 = CX1 + S/2 = CX1 + 5
  - PY1 = CY1 - S*sqrt3/2 = CY1 - 8.66
  - PX2 = CX1 + S = CX1 + 10
  - PY2 = CY1

  So if PX1 = 50 and PX2 = 55... CX1 + 5 = 50, so CX1 = 45. That's cell (1,
  ?).
  And if PY1 = 54.6, CY1 - 8.66 = 54.6, so CY1 = 63.26 ≈ 63.3.

  Cell (1, R) with CY = 63.3. Using our formula: CY = 28.66 + (R - 0.5 +
  0.5)17.32 = 28.66 + R17.32
  R*17.32 = 34.64, R = 2.

  So this is the upper_right edge of cell (1, 2)! The edge goes to cell (2,
  1).

  But wait, path segment 3 goes from (1,2) to (2,2), which uses the
  lower_right edge of (1,2), not upper_right.

  So wall 5 is between (1,2) and (2,1), NOT between (1,2) and (2,2). These
  are different edges!

  The issue is that the path LOOKS like it crosses the wall visually because
   the wall line and path line intersect geometrically, but they're
  representing different things:
  - The wall is the edge between cells (1,2) and (2,1)
  - The path segment connects centres of cells (1,2) and (2,2)

  These lines intersect in 2D space, but there's no actual crossing of a
  wall on the path.

  Hmm, but this seems like a visual bug - the solution path shouldn't
  visually cross any walls even if logically it doesn't. Let me
  reconsider...

  Actually wait. A hex maze has hexagonal cells. The solution path goes
  through cell centres, and it should stay INSIDE the cells. If it crosses a
   wall visually, that's a rendering issue - either the wall is in the wrong
   place or the cell centres are wrong.

  Let me verify cell centre for (1,2):
  CX = 30 + 1*15 = 45 ✓
  CY = 28.66 + (2 - 0.5 + 0.5)*17.32 = 28.66 + 34.64 = 63.3 ✓

  Cell centre for (2,2):
  CX = 30 + 2*15 = 60 ✓
  CY = 28.66 + (2 - 1 + 0.5)*17.32 = 28.66 + 25.98 = 54.64 ≈ 54.6 ✓

  These match the path coordinates.

  Now let me check the wall endpoints for the edge between (1,2) and (2,1):
  Cell (1,2) has upper_right edge going to (2,1).
  Using edge_endpoints with Cell1=(1,2), Cell2=(2,1):
  - DQ = 1, DR = -1, which is upper_right
  - CX = 45, CY = 63.3, S = 10
  - PX1 = 45 + 5 = 50
  - PY1 = 63.3 - 8.66 = 54.64 ≈ 54.6
  - PX2 = 45 + 10 = 55
  - PY2 = 63.3

  This gives wall from (50, 54.6) to (55, 63.3). This matches wall 5! ✓

  So the wall IS in the correct position for the edge between (1,2) and
  (2,1).

  The problem is that when we draw a STRAIGHT LINE from (1,2)'s centre to
  (2,2)'s centre, that line happens to cross through the territory that
  should belong to the edge between (1,2) and (2,1).

  In a proper hex maze visualization, the solution path should follow the
  hex edges, not straight lines between cell centres. OR, we need to ensure
  that straight lines between adjacent hex cell centres don't cross other
  edges.

  Actually, for adjacent hex cells, the straight line between centres SHOULD
   stay within the two cells. Let me think about this geometrically...

  For cell (1,2) at (45, 63.3) going to cell (2,2) at (60, 54.6):
  - These cells are adjacent via lower_right (DQ=1, DR=0)
  - The shared edge is the lower_right edge of cell (1,2)

  The lower_right edge of cell (1,2) goes from:
  - right corner: (45 + 10, 63.3) = (55, 63.3)
  - lower-right corner: (45 + 5, 63.3 + 8.66) = (50, 72.0)

  So the shared edge between (1,2) and (2,2) is from (55, 63.3) to (50,
  72.0).

  The line from (45, 63.3) to (60, 54.6) should cross this edge... let me
  check.

  At X=55, the path Y = 63.3 - 8.7*(55-45)/15 = 63.3 - 5.8 = 57.5
  The edge at X=55 starts at Y=63.3.

  At X=50, the path Y = 63.3 - 8.7*(50-45)/15 = 63.3 - 2.9 = 60.4
  The edge at X=50 ends at Y=72.0.

  Hmm, so the path at (55, 57.5) doesn't actually cross the edge which is at
   (55, 63.3). The path goes BELOW the edge at X=55.

  But the path DOES cross the wall between (1,2) and (2,1) which goes from
  (50, 54.6) to (55, 63.3).

  OH! I think I see the problem now. The wall between (1,2) and (2,1) is
  being drawn, but cells (1,2) and (2,1) should be separated by a wall only
  if there's no door there. Let me verify...

  Is (2,1) even a valid cell? For W=3, H=3, HalfW=1:
  - Q=2, R=1
  - R >= max(0, Q-HalfW) = max(0, 1) = 1, and 1 >= 1 ✓
  - R <= min(Q+H-1, HalfW+H-1) = min(4, 3) = 3, and 1 <= 3 ✓

  Yes, (2,1) is valid.

  So the edge between (1,2) and (2,1) is a legitimate internal edge, and if
  it doesn't have a door, a wall should be drawn there. The maze generator
  decided not to put a door there.

  The visual issue is that the solution path from (1,2) to (2,2) passes
  geometrically close to (or through) the wall between (1,2) and (2,1). This
   is a fundamental issue with drawing straight lines between hex cell
  centres - they can visually cross walls to other adjacent cells.

  To fix this properly, we'd need to draw the solution path differently -
  perhaps routing through cell edge midpoints or using curved paths. But
  that's a significant change.

  Alternatively, this might just be an acceptable visual artifact of the hex
   grid representation. The path IS logically correct - it goes through
  cells that have doors between them.

  Let me confirm by checking: is there a door between (1,2) and (2,2)?

  The path goes (0,1) -> (1,1) -> (1,2) -> (2,2). If the maze is valid,
  there must be doors along this path. The issue is purely visual.

  Would you like me to:
  1. Accept this as a known visual limitation of straight-line path
  rendering for hex grids
  2. Implement a different path rendering approach (e.g., through edge
  midpoints)
  3. Investigate if there's an actual error in door/wall calculation

> /context 
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