258 lines
16 KiB
Go
258 lines
16 KiB
Go
package milazzo
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/**
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```cpp
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sealed class MyVisibility : Visibility
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{
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/// <param name="blocksLight">A function that accepts the X and Y coordinates of a tile and determines whether the
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/// given tile blocks the passage of light. The function must be able to accept coordinates that are out of bounds.
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/// </param>
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/// <param name="setVisible">A function that sets a tile to be visible, given its X and Y coordinates. The function
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/// must ignore coordinates that are out of bounds.
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/// </param>
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/// <param name="getDistance">A function that takes the X and Y coordinate of a point where X >= 0,
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/// Y >= 0, and X >= Y, and returns the distance from the point to the origin (0,0).
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/// </param>
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public MyVisibility(Func<int, int, bool> blocksLight, Action<int, int> setVisible, Func<int, int, int> getDistance)
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{
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_blocksLight = blocksLight;
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GetDistance = getDistance;
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_setVisible = setVisible;
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}
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public override void Compute(LevelPoint origin, int rangeLimit)
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{
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_setVisible(origin.X, origin.Y);
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for(uint octant=0; octant<8; octant++) Compute(octant, origin, rangeLimit, 1, new Slope(1, 1), new Slope(0, 1));
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}
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struct Slope // represents the slope Y/X as a rational number
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{
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public Slope(uint y, uint x) { Y = y; X = x; }
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public bool Greater(uint y, uint x) { return Y*x > X*y; } // this > y/x
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public bool GreaterOrEqual(uint y, uint x) { return Y*x >= X*y; } // this >= y/x
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public bool Less(uint y, uint x) { return Y*x < X*y; } // this < y/x
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//public bool LessOrEqual(uint y, uint x) { return Y*x <= X*y; } // this <= y/x
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public readonly uint X, Y;
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}
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void Compute(uint octant, LevelPoint origin, int rangeLimit, uint x, Slope top, Slope bottom)
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{
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// throughout this function there are references to various parts of tiles. a tile's coordinates refer to its
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// center, and the following diagram shows the parts of the tile and the vectors from the origin that pass through
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// those parts. given a part of a tile with vector u, a vector v passes above it if v > u and below it if v < u
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// g center: y / x
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// a------b a top left: (y*2+1) / (x*2-1) i inner top left: (y*4+1) / (x*4-1)
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// | /\ | b top right: (y*2+1) / (x*2+1) j inner top right: (y*4+1) / (x*4+1)
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// |i/__\j| c bottom left: (y*2-1) / (x*2-1) k inner bottom left: (y*4-1) / (x*4-1)
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//e|/| |\|f d bottom right: (y*2-1) / (x*2+1) m inner bottom right: (y*4-1) / (x*4+1)
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// |\|__|/| e middle left: (y*2) / (x*2-1)
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// |k\ /m| f middle right: (y*2) / (x*2+1) a-d are the corners of the tile
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// | \/ | g top center: (y*2+1) / (x*2) e-h are the corners of the inner (wall) diamond
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// c------d h bottom center: (y*2-1) / (x*2) i-m are the corners of the inner square (1/2 tile width)
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// h
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for(; x <= (uint)rangeLimit; x++) // (x <= (uint)rangeLimit) == (rangeLimit < 0 || x <= rangeLimit)
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{
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// compute the Y coordinates of the top and bottom of the sector. we maintain that top > bottom
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uint topY;
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if(top.X == 1) // if top == ?/1 then it must be 1/1 because 0/1 < top <= 1/1. this is special-cased because top
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{ // starts at 1/1 and remains 1/1 as long as it doesn't hit anything, so it's a common case
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topY = x;
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}
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else // top < 1
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{
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// get the tile that the top vector enters from the left. since our coordinates refer to the center of the
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// tile, this is (x-0.5)*top+0.5, which can be computed as (x-0.5)*top+0.5 = (2(x+0.5)*top+1)/2 =
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// ((2x+1)*top+1)/2. since top == a/b, this is ((2x+1)*a+b)/2b. if it enters a tile at one of the left
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// corners, it will round up, so it'll enter from the bottom-left and never the top-left
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topY = ((x*2-1) * top.Y + top.X) / (top.X*2); // the Y coordinate of the tile entered from the left
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// now it's possible that the vector passes from the left side of the tile up into the tile above before
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// exiting from the right side of this column. so we may need to increment topY
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if(BlocksLight(x, topY, octant, origin)) // if the tile blocks light (i.e. is a wall)...
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{
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// if the tile entered from the left blocks light, whether it passes into the tile above depends on the shape
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// of the wall tile as well as the angle of the vector. if the tile has does not have a beveled top-left
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// corner, then it is blocked. the corner is beveled if the tiles above and to the left are not walls. we can
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// ignore the tile to the left because if it was a wall tile, the top vector must have entered this tile from
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// the bottom-left corner, in which case it can't possibly enter the tile above.
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//
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// otherwise, with a beveled top-left corner, the slope of the vector must be greater than or equal to the
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// slope of the vector to the top center of the tile (x*2, topY*2+1) in order for it to miss the wall and
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// pass into the tile above
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if(top.GreaterOrEqual(topY*2+1, x*2) && !BlocksLight(x, topY+1, octant, origin)) topY++;
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}
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else // the tile doesn't block light
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{
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// since this tile doesn't block light, there's nothing to stop it from passing into the tile above, and it
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// does so if the vector is greater than the vector for the bottom-right corner of the tile above. however,
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// there is one additional consideration. later code in this method assumes that if a tile blocks light then
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// it must be visible, so if the tile above blocks light we have to make sure the light actually impacts the
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// wall shape. now there are three cases: 1) the tile above is clear, in which case the vector must be above
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// the bottom-right corner of the tile above, 2) the tile above blocks light and does not have a beveled
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// bottom-right corner, in which case the vector must be above the bottom-right corner, and 3) the tile above
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// blocks light and does have a beveled bottom-right corner, in which case the vector must be above the
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// bottom center of the tile above (i.e. the corner of the beveled edge).
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//
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// now it's possible to merge 1 and 2 into a single check, and we get the following: if the tile above and to
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// the right is a wall, then the vector must be above the bottom-right corner. otherwise, the vector must be
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// above the bottom center. this works because if the tile above and to the right is a wall, then there are
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// two cases: 1) the tile above is also a wall, in which case we must check against the bottom-right corner,
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// or 2) the tile above is not a wall, in which case the vector passes into it if it's above the bottom-right
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// corner. so either way we use the bottom-right corner in that case. now, if the tile above and to the right
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// is not a wall, then we again have two cases: 1) the tile above is a wall with a beveled edge, in which
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// case we must check against the bottom center, or 2) the tile above is not a wall, in which case it will
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// only be visible if light passes through the inner square, and the inner square is guaranteed to be no
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// larger than a wall diamond, so if it wouldn't pass through a wall diamond then it can't be visible, so
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// there's no point in incrementing topY even if light passes through the corner of the tile above. so we
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// might as well use the bottom center for both cases.
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uint ax = x*2; // center
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if(BlocksLight(x+1, topY+1, octant, origin)) ax++; // use bottom-right if the tile above and right is a wall
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if(top.Greater(topY*2+1, ax)) topY++;
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}
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}
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uint bottomY;
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if(bottom.Y == 0) // if bottom == 0/?, then it's hitting the tile at Y=0 dead center. this is special-cased because
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{ // bottom.Y starts at zero and remains zero as long as it doesn't hit anything, so it's common
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bottomY = 0;
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}
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else // bottom > 0
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{
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bottomY = ((x*2-1) * bottom.Y + bottom.X) / (bottom.X*2); // the tile that the bottom vector enters from the left
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// code below assumes that if a tile is a wall then it's visible, so if the tile contains a wall we have to
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// ensure that the bottom vector actually hits the wall shape. it misses the wall shape if the top-left corner
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// is beveled and bottom >= (bottomY*2+1)/(x*2). finally, the top-left corner is beveled if the tiles to the
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// left and above are clear. we can assume the tile to the left is clear because otherwise the bottom vector
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// would be greater, so we only have to check above
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if(bottom.GreaterOrEqual(bottomY*2+1, x*2) && BlocksLight(x, bottomY, octant, origin) &&
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!BlocksLight(x, bottomY+1, octant, origin))
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{
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bottomY++;
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}
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}
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// go through the tiles in the column now that we know which ones could possibly be visible
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int wasOpaque = -1; // 0:false, 1:true, -1:not applicable
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for(uint y = topY; (int)y >= (int)bottomY; y--) // use a signed comparison because y can wrap around when decremented
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{
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if(rangeLimit < 0 || GetDistance((int)x, (int)y) <= rangeLimit) // skip the tile if it's out of visual range
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{
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bool isOpaque = BlocksLight(x, y, octant, origin);
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// every tile where topY > y > bottomY is guaranteed to be visible. also, the code that initializes topY and
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// bottomY guarantees that if the tile is opaque then it's visible. so we only have to do extra work for the
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// case where the tile is clear and y == topY or y == bottomY. if y == topY then we have to make sure that
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// the top vector is above the bottom-right corner of the inner square. if y == bottomY then we have to make
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// sure that the bottom vector is below the top-left corner of the inner square
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bool isVisible =
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isOpaque || ((y != topY || top.Greater(y*4-1, x*4+1)) && (y != bottomY || bottom.Less(y*4+1, x*4-1)));
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// NOTE: if you want the algorithm to be either fully or mostly symmetrical, replace the line above with the
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// following line (and uncomment the Slope.LessOrEqual method). the line ensures that a clear tile is visible
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// only if there's an unobstructed line to its center. if you want it to be fully symmetrical, also remove
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// the "isOpaque ||" part and see NOTE comments further down
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// bool isVisible = isOpaque || ((y != topY || top.GreaterOrEqual(y, x)) && (y != bottomY || bottom.LessOrEqual(y, x)));
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if(isVisible) SetVisible(x, y, octant, origin);
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// if we found a transition from clear to opaque or vice versa, adjust the top and bottom vectors
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if(x != rangeLimit) // but don't bother adjusting them if this is the last column anyway
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{
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if(isOpaque)
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{
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if(wasOpaque == 0) // if we found a transition from clear to opaque, this sector is done in this column,
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{ // so adjust the bottom vector upward and continue processing it in the next column
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// if the opaque tile has a beveled top-left corner, move the bottom vector up to the top center.
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// otherwise, move it up to the top left. the corner is beveled if the tiles above and to the left are
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// clear. we can assume the tile to the left is clear because otherwise the vector would be higher, so
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// we only have to check the tile above
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uint nx = x*2, ny = y*2+1; // top center by default
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// NOTE: if you're using full symmetry and want more expansive walls (recommended), comment out the next line
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if(BlocksLight(x, y+1, octant, origin)) nx--; // top left if the corner is not beveled
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if(top.Greater(ny, nx)) // we have to maintain the invariant that top > bottom, so the new sector
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{ // created by adjusting the bottom is only valid if that's the case
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// if we're at the bottom of the column, then just adjust the current sector rather than recursing
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// since there's no chance that this sector can be split in two by a later transition back to clear
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if(y == bottomY) { bottom = new Slope(ny, nx); break; } // don't recurse unless necessary
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else Compute(octant, origin, rangeLimit, x+1, top, new Slope(ny, nx));
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}
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else // the new bottom is greater than or equal to the top, so the new sector is empty and we'll ignore
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{ // it. if we're at the bottom of the column, we'd normally adjust the current sector rather than
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if(y == bottomY) return; // recursing, so that invalidates the current sector and we're done
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}
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}
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wasOpaque = 1;
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}
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else
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{
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if(wasOpaque > 0) // if we found a transition from opaque to clear, adjust the top vector downwards
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{
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// if the opaque tile has a beveled bottom-right corner, move the top vector down to the bottom center.
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// otherwise, move it down to the bottom right. the corner is beveled if the tiles below and to the right
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// are clear. we know the tile below is clear because that's the current tile, so just check to the right
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uint nx = x*2, ny = y*2+1; // the bottom of the opaque tile (oy*2-1) equals the top of this tile (y*2+1)
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// NOTE: if you're using full symmetry and want more expansive walls (recommended), comment out the next line
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if(BlocksLight(x+1, y+1, octant, origin)) nx++; // check the right of the opaque tile (y+1), not this one
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// we have to maintain the invariant that top > bottom. if not, the sector is empty and we're done
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if(bottom.GreaterOrEqual(ny, nx)) return;
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top = new Slope(ny, nx);
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}
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wasOpaque = 0;
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}
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}
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}
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}
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// if the column didn't end in a clear tile, then there's no reason to continue processing the current sector
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// because that means either 1) wasOpaque == -1, implying that the sector is empty or at its range limit, or 2)
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// wasOpaque == 1, implying that we found a transition from clear to opaque and we recursed and we never found
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// a transition back to clear, so there's nothing else for us to do that the recursive method hasn't already. (if
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// we didn't recurse (because y == bottomY), it would have executed a break, leaving wasOpaque equal to 0.)
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if(wasOpaque != 0) break;
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}
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}
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// NOTE: the code duplication between BlocksLight and SetVisible is for performance. don't refactor the octant
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// translation out unless you don't mind an 18% drop in speed
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bool BlocksLight(uint x, uint y, uint octant, LevelPoint origin)
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{
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uint nx = origin.X, ny = origin.Y;
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switch(octant)
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{
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case 0: nx += x; ny -= y; break;
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case 1: nx += y; ny -= x; break;
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case 2: nx -= y; ny -= x; break;
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case 3: nx -= x; ny -= y; break;
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case 4: nx -= x; ny += y; break;
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case 5: nx -= y; ny += x; break;
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case 6: nx += y; ny += x; break;
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case 7: nx += x; ny += y; break;
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}
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return _blocksLight((int)nx, (int)ny);
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}
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void SetVisible(uint x, uint y, uint octant, LevelPoint origin)
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{
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uint nx = origin.X, ny = origin.Y;
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switch(octant)
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{
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case 0: nx += x; ny -= y; break;
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case 1: nx += y; ny -= x; break;
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case 2: nx -= y; ny -= x; break;
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case 3: nx -= x; ny -= y; break;
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case 4: nx -= x; ny += y; break;
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case 5: nx -= y; ny += x; break;
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case 6: nx += y; ny += x; break;
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case 7: nx += x; ny += y; break;
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}
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_setVisible((int)nx, (int)ny);
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}
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readonly Func<int, int, bool> _blocksLight;
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readonly Func<int, int, int> GetDistance;
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readonly Action<int, int> _setVisible;
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}
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```
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*/
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