Encirclement Question

rreinesch

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After you overcome the pangs of sorrow for the 4-6-8 answer this:

Can the Russians encircle the 4-6-8?
 

SamB

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<sniff> Ok, I'm over that.

No.

Encirclement is very specific, and your situation does not fit the rule...
 

rreinesch

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But wouldn't some combinations satisfy case c) of encirclement? I know case c) says that fire crosses any three non-continguous hexsides, but then how does that fit with the second case c) example that crosses 2 hexsides and a hexspine, yet still satisfies encirclement?
 

Bret Hildebran

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Sure the Russians can encircle the 468. They just need 3 consecutive shots to do it in, rather than 2. In this case any 3 consecutive shots (other than Z31, Z28 & W28) will encircle the hapless German. To get encirclement w/3 shots you need one down the spine and 2 shots on either side of the opposite spine or through a face & 2 shots down the opposite adjacent faces. All sequences except the one noted above will accomplish that...
 

sdennis

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I agree with Bret, my first thought was "Is this a trick question?"
I don't have the rules handy or memorized but encirclement is possible from my general knowledge of the rule.
 

Trezza

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Piece of cake.....

Just drop the encircle counter on the unit and tell the guy to read the damn rule book.......
 

SamB

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I disagree. The unit in the illustration does not meet the requirements of Case A, B, or C for Encirclement.

Case C is three non-contiguous hex sides. There is no case C encirclement with one or more of the shots coming down the hex spine.
 

bos

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I disagree. The unit in the illustration does not meet the requirements of Case A, B, or C for Encirclement.

Case C is three non-contiguous hex sides. There is no case C encirclement with one or more of the shots coming down the hex spine.
See the example at the end of 7.7. The 4th example is Encircled via case c, but one of the three attacks is via the hexspine. I understand that this is because LOS along a hexspine is also through BOTH of the hexsides.

However, I can't see why the second "Not Encircled" example would not be encircled by that interpretation.

Case C should probably read something like: "The location does not have any two contiguous hexsides free of firers' LOS". In fact, this would cover case A as well, leaving just case B needed.
 
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Ole Boe

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See the example at the end of 7.7. The 4th example is Encircled via case c, but one of the three attacks is via the hexspine. I understand that this is because LOS along a hexspine is also through BOTH of the hexsides.
I agree about the example, but am not sure about the reason. Explaining it with LOS along a hexspine being part of both hexsides is not good enough, because it would still not fulfill C, due to the other two attacks coming from two contiguous hexsides, not two non-contigious as required.

To me it looks like there should be a case d: one hexspine and both hexsides forming the opposing hexspine.

However, I can't see why the second "Not Encircled" example would not be encircled by that interpretation.
Because by that interpretation, you cover hexsides 1, (top), 4, 6 and that is still not three non-contiguous hexsides.

Case C should probably read something like: "The location does not have any two contiguous hexsides free of firers' LOS". In fact, this would cover case A as well, leaving just case B needed.
That makes good sense. To me, this looks like a good candidate for an errata, since the text doesn't match the examples.
 

bos

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Right, I was thinking two sides could be contiguous but not the third. That allows example 4, but unfortunately also allows example 6. One way or the other, an example doesn't fit, so I agree it needs errata.

I think the examples make the intent clear, however. If there isn't > 180 degrees clear of fire in some direction, then the location is encircled.

That is, with fire coming through hexsides centered on the hexside (otherwise we'd need a ruler or protractor).
 

mgmasl

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Continuous fire from 527, 447 and 458 or 527, 447 and 628 cause encirclement.

IMO the rules doesn´t need to be clearer. The concept is so clear that to describe ALL the posible incoming fires to a hex to cause encirclement it would be everything but necesary...

Anyway, any hexside -by definition in ASLRB- includes the two vertex, and so the fire from the 527 and the 447 have to be considered as crossing the hexsides...

Miguel
 

rreinesch

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See the example at the end of 7.7. The 4th example is Encircled via case c, but one of the three attacks is via the hexspine. I understand that this is because LOS along a hexspine is also through BOTH of the hexsides.

However, I can't see why the second "Not Encircled" example would not be encircled by that interpretation.
The emaphsis on hexspines and hexsides as it relates to cases a) and b) of A7.7 deals specifically with how encirclement can be achieved from 2 fires. In case a) fires directly down opposite hexspines, which can only be possible from 2 fires and case b) which deals with how 2 fires that cross a hexside can cause encirclement, namely that there must exist exactly three vertices in both a clockwise and counterclockwise direction from each of the 2 fires. In the two conditions cited as not being encircled, this is the case that it specficially fails on. Case c) empahsizes hexsides as it deals with how to determine if encirclement occurs from 3 or more fires which enter the hex - again through any three non-contigous hexsides. Since a vertex is a part of a hexside by definition in the Index, then any fire that goes through a vertex is by definition eligible to be considered as going to either of the hexsides which encompass it for this determination.
 

SamB

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The text of the rule book doesn't match the examples (as others have pointed out).

But if you just read the text of the rule book, then the unit in question cannot be encircled by the locations pictured.

And, if you look at the examples, case C doesn't match the question either.

So neither the examples nor the text would allow the unit in question to be encircled.

You may want to encircle him. You may have good sounding reality arguments that he "should" be encircled. But the rule book doesn't allow it.

JMO, but I still think that by the rules, he can't be encircled. (Of course, by the text of the rules the last case C example isn't encircled either!) :crosseye:
 

SlyFrog

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Yes, it does need to be cleaned up. I noticed it the last time I played. My opponent came up with a reading of the rule that actually explained the second "C" example under the wording, but I can not recall it right now. Nonetheless, we both agreed that it needed to be cleaned up.

EDIT: Ah, now (I think) I remember the explanation. It is that not all three attacks have to be from noncontiguous hexsides, just that one of the attacks must be noncontiguous to the other two.
 
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mgmasl

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Yes, it does need to be cleaned up. I noticed it the last time I played. My opponent came up with a reading of the rule that actually explained the second "C" example under the wording, but I can not recall it right now. Nonetheless, we both agreed that it needed to be cleaned up.

EDIT: Ah, now (I think) I remember the explanation. It is that not all three attacks have to be from noncontiguous hexsides, just that one of the attacks must be noncontiguous to the other two.
"..three non-contiguos hexsides..." IMO, the non-contiguos is applied to the 3 hexsides,.. Looking at the second diagram for c) case in the EX at 7.7 is clear because there are two contiguos and another one non-connected to any of this two contiguos hexsides.. Looking at this same EX it looks clear that the vertice is included in the hexside in this case...

in Index, "hexside: (one of the six lines which combine to form a hex; each hexside also contains two vertices)"

Miguel
 

geezer

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EDIT: Ah, now (I think) I remember the explanation. It is that not all three attacks have to be from noncontiguous hexsides, just that one of the attacks must be noncontiguous to the other two.

This interpretion seems like the good one and im surprised we have not thought of that earlier. There is still a source of misunderstanding though, because according to this the 2nd example of "not encircled" could be considered as encircled because technically, a shot through a hexspine goes through two hexsides. Maybe a little errata stating that for case C to be valid, one shot cannot count for more then one hexside would be in order here. (something better can probably be thought of)
 

James Taylor

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I'm not adverse to RB clarification if someone wants to lobby for that, but I don't think its necessary to arrive at the correct interpretation here... and that is the one put forth by Bret and others.

Encirclement is absolutely possible.

"3 non-contiguous" means 3 hexsides that are not all connected. 2 can be connected, but not all 3. Use the worse possible interpretation for the attacker when the shot comes down the hexspine.

Never seen anyone question that interpretation.

JT
 

SamB

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Encirclement is absolutely possible.

"3 non-contiguous" means 3 hexsides that are not all connected. 2 can be connected, but not all 3. Use the worse possible interpretation for the attacker when the shot comes down the hexspine.

Never seen anyone question that interpretation.
Well, I'm questioning it right now. "non-contiguous" means roughly "non-adjacent". Some of the examples labeled "encircled" have two of the three shots coming through adjacent hexsides.

I do think this should be clarified. :rolleyes:
 

SlyFrog

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I'm not adverse to RB clarification if someone wants to lobby for that, but I don't think its necessary to arrive at the correct interpretation here... and that is the one put forth by Bret and others.

Encirclement is absolutely possible.

"3 non-contiguous" means 3 hexsides that are not all connected. 2 can be connected, but not all 3. Use the worse possible interpretation for the attacker when the shot comes down the hexspine.

Never seen anyone question that interpretation.

JT
I think the question is whether it is "all 3," to be fair (though as I mentioned above, I agree with your interpretation, I just do not think it is written as clearly as it should be).
 
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