public AvidWindow[] GetPossibleLaunchWindows(int courseOffset, AvidWindow targetWindow, AvidWindow crossingVector)
{
if (AvidWindow.IsNullOrZero(targetWindow))
{
return(new[] { GetOppositeWindow(crossingVector) }); //This may still result in undefined vector if both are zero.
// This is fine. If both ships aren't moving and sit in the same hex, launch window can't be determined. This situation should
// not arise in any real scenario, but we still need to handle it.
}
if (courseOffset == 0 || targetWindow.Equals(crossingVector) || AvidWindow.IsNullOrZero(crossingVector))
{
return(new [] { targetWindow });
}
var pathingResult = _avidPathfinder.GetShortestPaths(_avidModel, targetWindow, crossingVector, AvidPathingOptions.DiagonalTransitionsLimitWithPolar);
if (!pathingResult.PathExists)
{
return(new AvidWindow[] { });
}
return(pathingResult.AllShortestPaths.Select(pi => pi.PathNodes[courseOffset])
.Where(lan => lan.Window.MinDistance >= lan.NodeDistance) // This is to exclude windows, that could be reached through diagonal transition
// but can also reached by two non-diagonal transitions. So, in the final result for course offset = 2, it would look like two windows
.Select(lan => (AvidWindow)lan.Window).Distinct().ToArray());
}
public int CountWindows(AvidWindow start, AvidWindow destination)
{
if (start.Equals(destination) || AvidWindow.IsNullOrZero(start) || AvidWindow.IsNullOrZero(destination))
{
return(0);
}
var pathingResult = _avidPathfinder.GetShortestPaths(_avidModel, start, destination, AvidPathingOptions.DiagonalTransitionsLimitWithPolar);
if (!pathingResult.PathExists)
{
throw new Exception("Failed to find path from start to destination. That must be the problem with the algorithm.");
}
if (Math.Abs(pathingResult.MinimalDistance - GetDistanceFromAngle(start, destination) * 2) > 1)
{
//throw new Exception("Distance inconsistent.");
}
// Since all normal transitions on the avid are two points, we need to halve the result.
return(pathingResult.MinimalDistance / 2);
}
public int CountWindows(AvidWindow start, AvidWindow destination)
{
if (start.Equals(destination) || AvidWindow.IsNullOrZero(start) || AvidWindow.IsNullOrZero(destination))
{
return 0;
}
var pathingResult = _avidPathfinder.GetShortestPaths(_avidModel, start, destination, AvidPathingOptions.DiagonalTransitionsLimitWithPolar);
if (!pathingResult.PathExists)
{
throw new Exception("Failed to find path from start to destination. That must be the problem with the algorithm.");
}
if (Math.Abs(pathingResult.MinimalDistance - GetDistanceFromAngle(start, destination) * 2) > 1)
{
//throw new Exception("Distance inconsistent.");
}
// Since all normal transitions on the avid are two points, we need to halve the result.
return pathingResult.MinimalDistance / 2;
}
public AvidWindow[] GetPossibleLaunchWindows(int courseOffset, AvidWindow targetWindow, AvidWindow crossingVector)
{
if (AvidWindow.IsNullOrZero(targetWindow))
{
return new[] { GetOppositeWindow(crossingVector) }; //This may still result in undefined vector if both are zero.
// This is fine. If both ships aren't moving and sit in the same hex, launch window can't be determined. This situation should
// not arise in any real scenario, but we still need to handle it.
}
if (courseOffset == 0 || targetWindow.Equals(crossingVector) || AvidWindow.IsNullOrZero(crossingVector))
{
return new [] { targetWindow };
}
var pathingResult = _avidPathfinder.GetShortestPaths(_avidModel, targetWindow, crossingVector, AvidPathingOptions.DiagonalTransitionsLimitWithPolar);
if (!pathingResult.PathExists)
{
return new AvidWindow[] { };
}
return pathingResult.AllShortestPaths.Select(pi => pi.PathNodes[courseOffset])
.Where(lan => lan.Window.MinDistance >= lan.NodeDistance) // This is to exclude windows, that could be reached through diagonal transition
// but can also reached by two non-diagonal transitions. So, in the final result for course offset = 2, it would look like two windows
.Select(lan => (AvidWindow)lan.Window).Distinct().ToArray();
}
