Removing the rope, 2 years later

I created an example of what is happening.

In the picture you see a hexagon map, simplified.
The grey fields are obstruction. Let's say each obstruction has 2 points.

A red A indicates the attacker.
The green D indicates the defender.

This is at a range of 18.
We got some very special units that have this range.
For the sake of the problem, I am using this range as example.
(Don't tell me to remove such ranges, that is not the point here)

Rope method
We use a rope from A to D to see where the projectiles go.
At such distance, it is harder to see how the projectiles go. It is clumsy and when a 3D board is used or many units are in the way. This causes troubles.

Either way, the answer here is that the rope touches 3 obstruction hexagons.

Several ways to work with this are:
- Each touched hexagon counts only half. We have 3 points now.
- A die is rolled for every possible obstruction. Whether or not per attack die. Obstruction can be 0, 2, 4 or 6.
- An average is used of what both players want. 0, 1, 2, 3, 4, 5 or 6 can occur. However, most common is 3 again.

Within the parallelogram, thick red lines
There is only one way to work with this...
An average is used of what both players want. This time the most optimal path can have 0 in a much easier way, see the violet line.
The green line indicates what the defender might choose. This goes up to 6 hexagons or 12 obstruction points. The average is clearly 6 now.
Most down time is for the defender to determine the maximum obstruction possible.

ASAP method
As described before. The zigzag path is chosen to be as close as possible to the size of the parallelogram. In this example, both sides have the same size.
So you simply have a zigzag path. The first direction is a choice now. The attacker needs to check both ways for the most optimal path. See both the thin red and blue line. And it is 1 hexagon with 2 obstruction. Thin red line.
The defender can try to get a higher score, which is only 2 hexagon or 4 obstruction. See the thin blue line.
The average is 3. And this comes closest to the rope method. But still takes a bit of time. It is also the one that feels the fairest.

The major problem with the zigzag method is:
With one side being longer than another side. You start with the longer side. The defender has the same, but from his position.
The longer side might require 2 or more hops. And there is no way to determine how many hops you need without a rope... or counting.
You see, there is some calculation to do.
For example, if the long side of the parallelogram is 10 and the short side is only 3. That means that the trajectory of the weapons need to hop 3 to 4 times on the long side. Before it can hop 1 time in the direction of the short side.
So players need to count the long and short side. Then divide the 2 numbers.
And here we get some trouble too.

Going 3 in the long side gives us now 7L and 3W.
Although we know we need to go 1 to the side.
This is when we choose to hop 3 all the time.
7L 2W. 4L 2W. 4L 1W. 1L 1W. 1L 0W
If we pick 4 to hop, this is silly as well.
Maybe we need to re-determine every time?
Getting us a 4 hops at 4L 1W. Which creates a little different bending at the end.

And of course, the defender would get a different path.

That is why the method of simply being inside the parallelogram is still a valid and faster way in a lot of occasions. Albeit not always fair. And players start doing a little bit of a chess deployment.

In other words. The defender decides on the path.
But how to do it efficiently and fair?

This picture holds 3 examples.
(It made us realize, that perhaps we should have 12 directions instead of 6)
I hope it doesn't look to chaotic.

Basics
Let us begin with explaining what is going on here.

The middle is a red hex. This is the attacker.
On the top right we have purple hexes at a distance of 6.
On the left and right, we have green hexes at a distance of 7.
On the right, we also have violet hexes at the same distance of 7.

I will be talking about 6th and 12th directions.
A hexagon has 6 sides, and going to either side is a 6th direction.
When going over the corners, you will be crossing the line, where 2 adjacent hexes touch each other. The very next hex is at a distance of 2. This is the 12th direction.

The hexagons on the right
Here I have simply drawn some obstructing hexagons.
The pruple lines as the line of sight.
The red lines when using a rope.

The way of how the red lines touch the obstruction hexagons. Would mean different weights of obstruction. However, this was too tedious. And the rope itself at greater distances would cause troubles.

Let's begin with targeting the purple hexes.
As you can see, several lines are drawn from the red hex to the purple hexes.
The different colours indicate different situations (sorry colourblind people)
The red line is simple, it goes directly to the target.
The green lines are the same situation as the first example picture.
There is no doubt anymore, but the parallellogram should be ignored here. The green line that goes right through the middle is the main line of vision. This is the benefit of using 12 directions instead of 6. But it will make players crosseyed when other situations occur.
The orange, yellow and blue lines are all parallellograms. There is a big parallellogram to spot, and a smaller one. The big one is with the 6 directions in mind. The smaller one is with 12 directions in mind.

In all these cases, every hex that is touched by the smaller parallellogram, is considered to be of influence. Even if you would draw a line with the rope. All hexes would be touched. Would it be safe to asume that every hex is only partly crossed. Thus should only partly count?
In that case, seeing as how this vision is 2 hexes wide. Could we simply add up all obstruction points and then divide by 2? Is this useful for bigger distances?

The green hexes on the left
These are at a distance of 7. And the distance is an odd number.
What happens now is that the big parallellograms have a hex that brings the FEELING that it is going OUTSIDE. I can't describe it in a better way.
The big question here is that, should these out of the box hexes even count?

The 12th direction clearly is needed here.
It is also here that I am doubting the simpler dividing.
Is the width 2 or 2.5 for the yellow?
The green and blue clearly show 0.5 now. I can't divide by 0.5.

Towards the purple hexes. We had the green going to be divided by 1 and blue divided by 2. And this is now proven wrong.
Perhaps looking at the amount of hexes inside the parallellogram instead? Where hexes barely touching the parallellogram count for 0.5.

Sure I could start counting. But this takes way to long. No doubt it would bring the game to a full stop if the rules of counting aren't clear.
While attempting over and over. I discovered that a hexagon on the parallellogram can be touched in 4 different ways:
- From corner to corner.
- From corner to a side.
- From side to side.
- Barely with a side.

If I am very precise, these 4 ways need different weight points.
And we used to use that fact in our rope method too.
Perhaps it can be used again. Considering that a hexagon inside the parallellogram counts as 100%. What would the weight be, of the other situations?

1/1 - Inside the parallellogram.
1/1 - Target location.
1/2 - From corner to corner.
1/2 - From corner to a side.
1/2 - From side to side.
1/2 - Barely with a side.

Wow, thanking myself!
The distance of 6 yields a score of 6 for each situation.
The distance of 7 yields a score of 7 for each situation...
except yellow. Here we get a score of 8.
I would love it if anyone of you tests this. See if you can manage. See how problematic this is.

Now, we know that that one extra score means that we counted to many. Here it starts where the parallellogram with 12th directions is to much too. Should the rule be that the hexagons on the outside corners of the parallellogram being discounted now?

semi conclusion

And got accros a nice idea. Combining 2 methods into one.
I could cut in both methods too!!! See the 4th example

Let me know if this was understandable.
Let me know what you think about this method, can it still be improved?

Going up, ok. But the difference in height will reduce the chance to hit in a linear fashion. 6 means you can't shoot upwards. Also, the range gets reduced by 1 for every difference in height.

(We did had a rule that you could only attack the edge of the next elevation. But that didn't work for any player)

Behind the "delta", no vision at all.
The obstruction is really for the flatter terrain.
Although, the layer of snow on top of that 5 height is an excellent joke.

Instead of bar, i should have said range.

To know if you considered all middle's. This equals the predetermined range.

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Maybe this can be rewritten to make it more understandable. Perhaps, a better example being drawn as well?