A7.7 Encircled?

[Hemaelstrom]

Hi all,
Just wondering which of the following hexes would qualify as encircled? My question is really about how each hexspine contributes towards determination of case c.), given that each is intrinsic to two hexsides. Can you choose in each case which hexside to count in order to come up with three non-contiguous (in which case they would all be encircled) or is there something extra in the rules that I've overlooked?
Thanks for your insights/votes!

 

[Arch71]

Only D would qualify for encirclement.

 

[aneil1234]

agree D

Or the fire comes into the hex from directly opposite hexsides obviously (Doesn't need to be a straight line - just coming in over those two hexsides)\
But I'm sure you knew that already

 

[Philippe D.]

Agreed - only D qualifies for Encirclement. Even though the rules are not clear on this - by a strict reading, even D does not qualify.

The rule, as written, works fine whenever LOS enters the target hex through hexsides and not vertices: either through 2 exactly opposite hexsides, or through 3 different hexsides that are not contiguous. Where it's not exactly clear is what happens if LOS enters through, say, hexsides 1 and 2, and through the vertex between hexsides 4 and 5: it doesn't qualify for condition (c) three non-contiguous hexsides, unless you interpret it as meaning, whenever one of the LOSes enters through a vertex, you should check that if you "count" the vertex as being part of either hexside, in both case you qualify for encirclement. That way, 1,2 and 4-5 qualifies, because both 1,2,4 and 1,2,5 qualify (in each case you have two opposite hexsides); 1,2,3-4 would not qualify, because thoug 1,2,4 qualifies, 1,2,3 does not (and keep the provision about exactly two opposite vertices, otherwise that wouldn't qualify in my version).

Another way of phrasing what is probably the intent would be this: count each LOS entering through a hexside as entering through the middle of the hexside; then, to qualify for encirclement, you should check the following condition:

For any straight line going through the target hex's center without being exactly aligned with one of the LOS entry points, there are LOS entry points on both sides of the line

(But this isn't a good wording, first because it's probably confusing to most people, and second, because it asks for a condition to be satisfied by an infinite number of lines, something that is tricky to check for a non-mathematician)

But the idea is this: there should be fire coming from both sides, no matter how you look.

 

[von Marwitz]

View attachment 3176

Agreed - only D qualifies for Encirclement. Even though the rules are not clear on this - by a strict reading, even D does not qualify.

D does qualify even by a strict reading.

As per the Index:
Hexside (one of the six lines which combine to form a hex; each hexside also contains two vertices): [Indirect Fire TEM: C1.52] [Inherent Terrain: B.6] [MF cost: A4.131] [Railroad: B32.1] [Road hexside: B3.13] [TEM: B1.16]

This means that if LOS enters the hex through a vertex, it does at the same time enter through a hexside. That said, if it enters through 3 vertices as shown in D, it will always enter through a combination of hexsides that fulfil the requirements of Encirclement (as per case b of the examples given in the ASLRB in A7.7).

von Marwitz

 

[Binchois]

Agreed...D

As for the "3 vertices" thing, take Examples B and C. If the upper arrow was angled to be slightly inside the NW vertex - so nudged clockwise towards the center of the example - then the unit would be encircled with the 6:00 firer (without need of the third, western-most firer.). In Example A, if both the upper and lower arrow were nudged slightly closer to center (clockwise and counter-clockwise, respectively), then it too would be encircled.

 

[Philippe D.]

D does qualify even by a strict reading.

As per the Index:
Hexside (one of the six lines which combine to form a hex; each hexside also contains two vertices): [Indirect Fire TEM: C1.52] [Inherent Terrain: B.6] [MF cost: A4.131] [Railroad: B32.1] [Road hexside: B3.13] [TEM: B1.16]

This means that if LOS enters the hex through a vertex, it does at the same time enter through a hexside. That said, if it enters through 3 vertices as shown in D, it will always enter through a combination of hexsides that fulfil the requirements of Encirclement (as per case b of the examples given in the ASLRB in A7.7).

von Marwitz

The problem with this reading is, if two LOSes enter through hexside 1 and vertex 3-4, then you have LOS through 3 non-consecutive hexsides, but this does not qualify for encirclement (see last example in A7.7).

 

[von Marwitz]

The problem with this reading is, if two LOSes enter through hexside 1 and vertex 3-4, then you have LOS through 3 non-consecutive hexsides, but this does not qualify for encirclement (see last example in A7.7).

I do not think that matters because one vertex can be part of two hexsides.
Thus in case of example D in the OP, these hexsides can always be interpreted to be those that enable encirclement.

von Marwitz

 

[Philippe D.]

Well, some consistency is needed. If a vertex counts as being part of both hexsides for one encirclement, it should count also for another. That's why I tried to to include the case in my description: a vertex should count as being in the least favorable hexside, not the most favorable (to the firer) - unless you have exactly opposite vertices.

The thing is, I'm pretty sure we agree on what constitutes encirclement and what does not; what we're disagreeing on is how it should be described...

 

[Hemaelstrom]

a vertex should count as being in the least favorable hexside, not the most favorable (to the firer) - unless you have exactly opposite vertices

Thanks for all your ideas. This from Philippe is what i was thinking is needed to clarify the rule. It gives only D as encircled and is consistent with other rules like target facing selection.

 

[jrv]

A vertex is part of both hexsides. For encirclement you can count it as part of either, at the choice of the player attempting to do the encircling. The three cases "do the right thing."

In the four examples given, none has two shots through opposite hexspines (case a in A7.7). None has two shots through hexsides with three vertices both left and right (case b in A7.7). Any shot coming in exactly on a vertex loses that vertex (it isn't either to the left or to the right of the attack); for case b you can ignore attacks that come in on a vertex because any attack through a vertex leaves at most five other, eligible vertices, which can't possibly be made into three left and three right. That leaves only case c, and here you (the firing player) may freely choose the hexside for any shots that go through a vertex.

In EX D (start with the valid encircling shot), the hexsides to the left OR to the right of each vertex may be chosen; those three hexsides are non-contiguous. "Non-contiguous" here means that *none* of the hexsides touches another. If one does not touch the others, but the other two do, that is not case c encirclement.

EX A: The top and the bottom hexsides may be chosen for the outer-most attacks, but the attack from the left will touch one of those two no matter which of the two possible hexsides is chosen.

EX B: the bottom and the bottom-left attacks have no ambiguity about which hexside is affected, and those two touch, so they can't meet case c.

EX C: the bottom hexside is given, but that means the attack on the bottom-left also chooses the bottom (i.e. there are not three hexsides) or it chooses the bottom-left hexside, in which case bottom-left and bottom are contiguous.

All of the others are valid encircling attacks too, because they involve three attacks where the attacker can specify non-contiguous hexsides. Even though the configurations of the outermost attacks are the same as some two-attack combinations that fail to encircle, because there are three attacks they encircle.

There is no need to add "most favorable" or "least favorable" to the attacker. In fact, if you add "least favorable" then case D is not encirclement, because the top-right attack could choose either hexside, but the bottom-left attack being part of the bottom hexside and the right attack being part of the bottom-right hexside makes those two attacks on contiguous hexsides, which is least favorable to the attacker, and cases a & b don't apply either.

JR

 

[Hemaelstrom]

Now i see that my confusion comes from my reading 'non-contiguous' as 'not all contiguous' rather than 'none contiguous' or 'no two touching'. Finally i think i can now say why a hex is encircled or not, without resorting to making a choice of hexsides, which is nice. Thanks JR.

 

[Philippe D.]

A vertex is part of both hexsides. For encirclement you can count it as part of either, at the choice of the player attempting to do the encircling. The three cases "do the right thing."

In the four examples given, none has two shots through opposite hexspines (case a in A7.7). None has two shots through hexsides with three vertices both left and right (case b in A7.7). Any shot coming in exactly on a vertex loses that vertex (it isn't either to the left or to the right of the attack); for case b you can ignore attacks that come in on a vertex because any attack through a vertex leaves at most five other, eligible vertices, which can't possibly be made into three left and three right. That leaves only case c, and here you (the firing player) may freely choose the hexside for any shots that go through a vertex.

Are you saying that three shots, through hexside 1 (north, say), hexside 2 (north-east), and the vertex between hexsides 3 and 4 (4 being south), would cause encirclement? If the vertex is considered to belong both to hexsides 3 and 4, then you have hexsides 1 and 4, and that causes encirclement.

The examples in the rulebook (A7.7) include one where shots come in through hexside 4 and vertex 6-1, and this is labeled as not causing encirclement. By your reasoning, it should cause encirclement (and I repeat again - I think it should not).

Or are you saying "non-contiguous" means "three hexsides, none of which is adjacent to any other"? One of the "encirclement" examples of A7.7 is labeled (c), and it has shots through hexsides 3 and 4 (adjacent), and vertex 6-1. There is no way this can be interpreted as three hexsides, none of which is adjacent to any other.

 

[Hemaelstrom]

One of the "encirclement" examples of A7.7 is labeled (c), and it has shots through hexsides 3 and 4 (adjacent), and vertex 6-1. There is no way this can be interpreted as three hexsides, none of which is adjacent to any other.

Aha, now i remember what prompted me to post this! Back to least favourable choice then

 

[jrv]

I have to take my answer back, and I have revised it. Attacks where the outermost angle is the same may fail to encircle when there are only two attacks but succeed when there are three attacks. The last A7.7 EX would be completed by a third attack through the lower-left hexside.

JR

 

[clubby]

I've read that A7.7 several times as well as looking at the examples and the definition for the words "non-contiguous" and I came the conclusion that it meant that all three are not touching. It's the only thing that properly explains the last encircled A7.7EX. None of them touching each other works as well, but not for that EX.

 

[MajorDomo]

A good way to determine encirclement when a vertex or vertices are involved is "choose the hexside most favorable to the defender for each vertex"'

Easy to see afterwards. D in the above cases.

Rich