Mathematic Grid Puzzle

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Socoder -> Concept/Design -> Mathematic Grid Puzzle
Fri, 16 Mar 2018, 05:17
Jayenkai	Mathematic Grid Puzzle OK, I have a grid of levels onscreen.. (GridW, GridH) (in this case, 9x9) The spot in the middle of the grid should be level 1. Around it are 2,3,4 and 5, then higher and higher, until the four corner points are 78,79,80 and 81 I need a quick and easy Level=LevelAt(x,y) function, ensuring every point is a unique value. ...Anyone!?! -=-=- ''Load, Next List!''
Fri, 16 Mar 2018, 05:53
spinal	So something like this - http://jsfiddle.net/hitbyatruck/c4Kd6/ (stole from here - https://stackoverflow.com/questions/398299/looping-in-a-spiral) basically just draw the spiral and count along while doing it? -=-=- Check out my excellent homepage!
Fri, 16 Mar 2018, 06:01
rskgames	int Level[9][9] = { { 70,71,72,73,74,75,76,77,78 }, { 69,41,42,43,44,45,46,47,79 }, { 68,40,20,21,22,23,24,48,80 }, { 67,39,19, 7, 8, 9,25,49,81 }, { 66,38,18, 6, 1, 2,10,26,50 }, { 65,37,17, 5, 4, 3,11,27,51 }, { 64,36,16,15,14,13,12,28,52 }, { 63,35,34,33,32,31,30,29,53 }, { 62,61,60,59,58,57,56,55,54 } };
Fri, 16 Mar 2018, 06:09
Jayenkai	Yeah, I'm starting to think that resorting to a pre-gen lookup is going to be easiest, but it also means I'm stuck to that size. If I decide to bump things up to a 31x31 grid of levels, it's going to be a pain to redo it all. I was hoping for maths to solve this. I don't think it's doable. -=-=- ''Load, Next List!''
Fri, 16 Mar 2018, 06:20
rskgames	I am trying to come up with the generator.
Fri, 16 Mar 2018, 08:44
Pakz	Is this map not like the floodfill method? Probably not though.
Fri, 16 Mar 2018, 10:09
Jayenkai	Took a break and baked a cake. cakeage Back to coding, and managed this. Graphics 512,512,0 gw=15 gh=15 Repeat Cls For x=1 To gw For y=1 To gh txt$=SpiralAt(x,y,gw,gh) dx=x(512/(gw+1)) dy=y(512/(gh+1)) dx=dx-(TextWidth(txt)/2) DrawText(txt,dx,dy) Next Next Flip Until KeyDown(KEY_ESCAPE) Function SpiralAt(fx,fy,gw,gh) ' Find Center cx=((gw+1)/2) cy=((gh+1)/2) px=cx;py=cy;d=3 layer=0 count=1 Repeat found=0 If px=fx And py=fy Then found=1 If found=0 If d=3 Then py=py-1;If py<cy-layer Then layer=layer+1;d=4 If d=2 Then px=px-1;If px-1<cx-layer Then px=cx-layer;d=3 If d=1 Then py=py+1;If py+1>cy+layer Then py=cy+layer;d=2 If d=0 Then px=px+1;If px+1>cx+layer Then px=cx+layer;d=1 If d=4 Then d=0 If px<1 Or py<1 Or px>gw Or py>gh Then Return EndIf If found=0 Then count=count+1 Until found=1 ret=layer Return count End Function --v Uses the spiral path finding, and seems to run fast enough in BMax that I can use it in the menu for my game. Also, it's lacking the "Four outer corners = highest values" that I'd've liked. .. It'll do.. I suppose. Bah, humbug! -=-=- ''Load, Next List!''
Fri, 16 Mar 2018, 10:33
rockford	Mmmmmm... cake!
Sat, 17 Mar 2018, 06:54
rskgames	I have uploaded my generate spiral grid cpp code and executable GenerateSpiralGrid (.zip) Below I have given usage and default output. Usage: GenerateSpiralGrid <number Of grids> <Direction> <SPIN> <Corner Special Case> Where <number Of grids> -> Any odd Number <Direction> -> can be 1-4. 1-EAST, 2-SOUTH, 3-WEST, 4-NORTH <SPIN> -> can be 1-2. 1-Clockwise, 2-Anticlockwise <Corner Special Case> -> can be 1-2. 1-Normal, 2-Corners will have max values. Grid : 5x5 Direction : North Spin : Clockwise 24 25 10 11 12 23 9 2 3 13 22 8 1 4 14 21 7 6 5 15 20 19 18 17 16
Sat, 17 Mar 2018, 08:05
TomToad	Here is a version in BlitzMax that will create a grid that you can then check against. It progresses outward and ends up with 78, 79, 80, and 81 in the corner. SuperStrict 'FillGrid will create a grid of the appropriate size and fill it with the values needed Function FillGrid:Int[,](width:Int,Height:Int) Type TCell 'To hold a single cell Field x:Int Field y:Int Function Create:TCell(x:Int, y:Int) Local Cell:TCell = New TCell Cell.x = x Cell.y = y Return Cell End Function End Type Local Grid:Int[Width,Height] Local List:TList = CreateList() Local Cell:TCell = TCell.Create(Width/2,Height/2) 'First cell will be in the middle of the grid List.AddLast(Cell) 'add it to the list Grid[Cell.x,Cell.y] = 1 'Place 1 in the first grid position Local Value:Int = 2 'Start the next value at 2 While Not List.IsEmpty() 'Iterate over the list until it has no more cells Cell = TCell(List.RemoveFirst()) 'remove the cell at the head of the list 'Check all 8 directions to see if the cell is empty and not past the edge. Add the next value ' if so and push the cell onto the list 'First the verticle/horizontal directions If Cell.x < Width-1 And grid[Cell.x+1,Cell.y] = 0 grid[Cell.x+1,Cell.y] = Value 'assign this grid the next value Value :+ 1 'increate the value by one List.AddLast(TCell.Create(Cell.x+1,Cell.y)) End If If Cell.y < Height-1 And grid[Cell.x,Cell.y+1] = 0 grid[Cell.x,Cell.y+1] = Value Value :+ 1 List.AddLast(TCell.Create(Cell.x,Cell.y+1)) End If If Cell.x > 0 And grid[Cell.x-1,Cell.y] = 0 grid[Cell.x-1,Cell.y] = Value Value :+ 1 List.AddLast(TCell.Create(Cell.x-1,Cell.y)) End If If Cell.y > 0 And grid[Cell.x,Cell.y-1] = 0 grid[Cell.x,Cell.y-1] = Value Value :+ 1 List.AddLast(TCell.Create(Cell.x,Cell.y-1)) End If 'This will check the diagonals If Cell.x < Width-1 And Cell.y < Height-1 And grid[Cell.x+1,Cell.y+1] = 0 grid[Cell.x+1,Cell.y+1] = Value Value :+ 1 List.AddLast(TCell.Create(Cell.x+1,Cell.y+1)) End If If Cell.x < Width-1 And Cell.y > 0 And grid[Cell.x+1,Cell.y-1] = 0 grid[Cell.x+1,Cell.y-1] = Value Value :+ 1 List.AddLast(TCell.Create(Cell.x+1,Cell.y-1)) End If If Cell.x > 0 And Cell.y < Height - 1 And grid[Cell.x-1,Cell.y+1] = 0 grid[Cell.x-1,Cell.y+1] = Value Value :+ 1 List.AddLast(TCell.Create(Cell.x-1,Cell.y+1)) End If If Cell.x > 0 And Cell.y > 0 And grid[Cell.x-1,Cell.y-1] = 0 grid[Cell.x-1,Cell.y-1] = Value Value :+ 1 List.AddLast(TCell.Create(Cell.x-1,Cell.y-1)) End If Wend 'Once all the cells have been filled, return the grid Return Grid End Function Local Grid:Int[,] = FillGrid(9,9) ' create the grid Local s:String For Local y:Int = 0 To 8 s = "" 'clear the string at the start of each row For Local x:Int = 0 To 8 s :+ Right(" "+Grid[x,y],2)+" " 'build the string for each item in the row Next Print s 'Print the row Next --v The output for a 9x9 grid looks like this 81 77 75 73 71 72 74 76 79 70 49 45 43 41 42 44 47 56 68 40 25 21 19 20 23 30 54 66 38 18 9 5 7 12 28 52 64 36 16 4 1 2 10 26 50 65 37 17 8 3 6 11 27 51 67 39 24 15 13 14 22 29 53 69 48 35 33 31 32 34 46 55 80 63 61 59 57 58 60 62 78 --v Edit: Reading back, it looks like you are trying to implement this in BlitzBasic/Blitz3D. I need to get ready for work now. I'll try to adapt it later.
Sat, 17 Mar 2018, 08:28
Jayenkai	Actually, the final implementation is being done in the new C++ Framework. I tend to test things in Blitz/Max because.. well, it's just nice and quick, isn't it!! For now, the Spiral method will do. But I've been thinking and planning all night, trying to come up with a more substantial way to do it, and I think I might have the idea somewhat working, involving layers and a little bit of trig!! But I'm not coding it, today. Instead I'm focusing on getting the finishing touches added to the game.. SFX, Music, etc. It's a bit too quiet, at the minute!! -=-=- ''Load, Next List!''
Sun, 18 Mar 2018, 08:05
shroom_monk	The solution that makes most sense will depend on how many times per frame you want to look these values up, and how much you care about efficiency. Pre-computing the results once, storing them in an array, and looking them up as needed at runtime is probably the most efficient solution. However, I was intrigued by the puzzle of computing the value of a single position on the fly, so here is a C++ solution that produces a quick spiral with some relatively concise code (though I'm sure it could easily be shrunk down): #include <cstdio> // printf #include <cassert> // assert #include <cstdlib> // abs int GetSpiralGridValue(int x, int y, int size) { // Grid must be square, and its width odd, for a spiral pattern to make sense assert(size % 2 == 1); // Where is the centre of the grid, and what is our offset from it? int const centre = (size - 1) / 2; int const offsetX = x - centre; int const offsetY = y - centre; // The largest absolute offset value determines which 'layer' of the spiral we're on int const absOffsetX = abs(offsetX); int const absOffsetY = abs(offsetY); int const layer = (absOffsetX > absOffsetY) ? absOffsetX : absOffsetY; // Where does this layer start (i.e. what is its 'due east' position)? int const layerStart = (layer == 0) ? 1 : (2 * layer - 1) * (2 * layer - 1) + 1; // The result now depends on where we are if (offsetX == 0 && offsetY == 0) { // The very centre return 1; } else if (offsetX == layer) { // The east side if (offsetY >= 0) { // Start of this layer return layerStart + offsetY; } else { // End of this layer return layerStart + 8 * layer + offsetY; } } else if (offsetY == layer) { // The south side return layerStart + 2 * layer - offsetX; } else if (-offsetX == layer) { // The west side return layerStart + 4 * layer - offsetY; } else if (-offsetY == layer) { // The north side return layerStart + 6 * layer + offsetX; } else { // ??? assert(false); return 0; } } int main() { int const gridSize = 9; for (int y = 0; y < gridSize; ++y) { for (int x = 0; x < gridSize; ++x) { printf("%3d ", GetSpiralGridValue(x, y, gridSize)); } printf("\n"); } return 0; } --v 70 71 72 73 74 75 76 77 78 69 41 42 43 44 45 46 47 79 68 40 20 21 22 23 24 48 80 67 39 19 7 8 9 25 49 81 66 38 18 6 1 2 10 26 50 65 37 17 5 4 3 11 27 51 64 36 16 15 14 13 12 28 52 63 35 34 33 32 31 30 29 53 62 61 60 59 58 57 56 55 54 --v -=-=- A mushroom a day keeps the doctor away... Keep It Simple, Shroom!