Difference between revisions of "Switchback"

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(Removed some unnecessary moves from diagrams.)
(C5-without-C1 switchback: Corrected Red's strategy)
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<hexboard size="6x10"
 
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   contents="R h2 B h6 R a3 B a4 R 1:b3 B 2:b4 R 3:c3 B 4:c4 R 5:d3 E x:g3 a:d5 b:e5"
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   contents="R h2 B h6 R a3 B a4 R 1:b3 B 2:b4 R 3:c3 B 4:c4 R 5:d3 E x:e3 y:g3 a:d5 b:e5"
 
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At this point, if Blue continues to push the ladder, Red will play at "x" to form [[Fifth_row_edge_templates#V-2-d|edge template V-2d]] and gets the switchback. The only possible moves for Blue to prevent the switchback are "a" and "b". If Blue moves at "b", then Red gets a foldback underneath. Therefore, Blue's best option is "a". Now Red gets neither the switchback nor a foldback underneath. Red still gets a [[ladder escape fork]] for 2nd row ladders arriving from the left.
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At this point, if Blue continues to push the ladder, Red will play at "x" and then at "y" to form [[Fifth_row_edge_templates#V-2-d|edge template V-2d]] and get the switchback. The only possible moves for Blue to prevent the switchback are "a" and "b". If Blue moves at "b", then Red gets a foldback underneath. Therefore, Blue's best option is "a". Now Red gets neither the switchback nor a foldback underneath. Red still gets a [[ladder escape fork]] for 2nd row ladders arriving from the left.
  
 
== 2nd-to-6th row switchback ==
 
== 2nd-to-6th row switchback ==

Revision as of 16:06, 25 October 2020

A switchback is a situation in which a ladder moves back two or more rows and changes direction. The attacker is still in control after the switchback. Although it is not always a ladder escape, it often can be and is usually a strong play.

For example, consider the following situation. Assume the piece on d1 is in some way connected to the top.

abcdefgh1234

Red makes a switchback as follows:

abcdefgh123413587246

Now the ladder continues from right to the left on the 4th row:

abcdefgh1234131191210

Note here how Red was able to connect back to the d1 piece. This is not always possible, but even if it isn't, the switchback can be used to create a long line connected to the edge and several rows back from it, a distinct advantage.

A3 switchback

A single piece at a3 (or at the equivalent cell on the opposite site of the board) escapes 2nd row ladders. It can also be used as a 3rd-to-5th row switchback:

971358246

Note that at no point in the 3rd row ladder, Blue could have yielded, because Red's piece could have escaped the resulting 2nd row ladder outright.

For an example, see also A3 escape trick.

Additionally, even if some of the corner is occupied by the opponent, a single piece at a3 can still be used as a 2nd-to-4th row switchback (but it does not escape 2nd row ladders in this case, nor serve as a 3rd-to-5th row switchback):

7136524

A4 switchback

We have already seen in the first example above that a single Red piece at a4 (or the equivalent cell on the opposite side of the board) can be used as a 2nd-to-4th row switchback. It also works as a 3rd-to-5th row switchback, as follows:

913587246

Red's 7 is connected to the bottom, and 9 is the ladder stone in the opposite direction.

Note that if Blue had instead decided to yield the ladder to the second row at any point, the outcome for Blue would have been worse: in that case, Red can perform a 2nd-to-4th row switchback which reconnects to Red's 3rd row ladder.

1191310258746

A5 switchback

A single Red piece at a5 (or the equivalent cell on the opposite side of the board) can be used as a 2nd-to-4th row switchback and as a 3rd-to-5th row switchback. The 2nd-to-4th row switchback works as follows:

913587246

Note that Red's piece 9 is connected to the bottom edge by double threat at the two cells marked "*". Play then continues leftward along the 4th row.

The 3rd-to-5th row switchback works as follows:

913587246

Once again, yielding would not have helped Blue.

B5 switchback

A single Red piece at b5 (or the equivalent cell on the opposite side of the board) can be used as a 2nd-to-4th row switchback and as a 3rd-to-5th row switchback for the same reasons as a5. In addition, it can be used as a 4th-to-6th row switchback as follows:

51432ba

Note that Red's 3 is connected by edge template V-2a. The cell marked "*" is not needed for the switchback. If Blue tries to yield to a 3rd row ladder at any point, Red can play the 3rd-to-5th row switchback and connect. Also, playing "a" prior to move 2 does not help Blue since Red can just respond at "b".

A6 switchback

A single Red piece at a6 (or the equivalent cell on the opposite side of the board) can be used as a 2nd-to-4th row switchback as follows:

9115108631472

The red stone marked "3" is the unique winning move for Red; in particular, Red cannot push the ladder past "1". After this, Blue has several possible responses, of which only one is shown above. However, Red can complete the switchback (or connect) in all cases.

With the amount of space shown here, a6 is not sufficient to switch back a 3rd row ladder. However, with some additional space, this can be done; see B6 switchback and C6 switchback below.

B6 switchback

A single Red piece at b6 (or the equivalent cell on the opposite side of the board) can be used as a 2nd-to-4th row switchback for the same reason as a6, and can also be used as a 3rd-to-6th row switchback. It Blue does not yield, the 3rd-to-6th row switchback works as follows:

7136524

If Blue yields, the situation is more complicated (depending on the exact moment when Blue yields), but Red still gets the switchback.

C6 switchback

A single Red piece at c6 (or the equivalent cell on the opposite side of the board) can be used as a 2nd-to-4th row switchback for the same reason as a6 and as a 3rd-to-6th row switchback for the same reason as b6; moreover, it can also be used as a 3rd-to-5th row switchback. If Blue does not yield, the 3rd-to-5th row switchback can be played as follows:

5761432

Note that Red is connected to the edge by edge template V-2f. The hexes marked "*" are not needed for the switchback to work. There are other possible responses by Blue that are not shown here, but in any case Red can get at least the switchback.

If Blue tries to yield to the 2nd row, Red still gets the switchback, and can sometimes connect outright. If Blue yields early enough, Red has enough room to play the 2nd-to-4th row switchback (see A6 switchback above) and connect. If Blue yields closer to the switchback stone, Red's play is less obvious and depends on the exact moment when Blue yields.

C5-without-C1 switchback

A single Red piece at c5 (or the equivalent cell on the opposite side of the board) can be used as a 2nd-to-4th row switchback and, with enough space, as a 3rd-to-5th row switchback, even if the opponent occupies c1.

The 2nd-to-4th row switchback is analogous to the A5 switchback and can be played as follows:

5761432

Note that 7 is connected to the edge by edge template IV2e. The cells marked "*" are not required for the switchback.

The 3rd-to-5th row switchback is harder to pull off and requires some space on the 6th row. It can be played as follows:

312

Within the area shown, 3 is the unique winning move (i.e., the unique move that permits Red to complete the switchback). Note that 3 is already connected to the edge by edge template V-2d. The cells marked "*" are not required for the switchback. After move 3, there are a number of possibilities depending on Blue's response. A typical sequence is the following:

56711313141081229

At this point, Red is connected to the edge by edge template IV2e. If Blue instead decides to yield the 3rd row ladder to the 2nd row, Red has enough room to play the 2nd-to-4th row switchback and connect.

The C5-without-C1 situation can also almost be used for a 4th-to-6th row switchback. While Blue can defend against this, the defense is not obvious and there is very little room for error. Therefore, this is a useful situation for the defender to be aware of. Consider the following position, with Blue to move:

132abc

Note that if Blue had tried to yield at any time prior to move 3, Red could have used the 3rd-to-5th row switchback to connect. After move 3, Blue has only three options, marked "a", "b", and "c". Option "a" is pushing the ladder. Options "b" and "c" prevent Red from getting the switchback, but Red will get a foldback underneath instead, i.e., a 2nd row ladder going right-to-left (this also serves as a ladder escape fork for any Red 2nd row ladders arriving from the left). So "a" is Blue's best option.

135xy24ab

At this point, if Blue continues to push the ladder, Red will play at "x" and then at "y" to form edge template V-2d and get the switchback. The only possible moves for Blue to prevent the switchback are "a" and "b". If Blue moves at "b", then Red gets a foldback underneath. Therefore, Blue's best option is "a". Now Red gets neither the switchback nor a foldback underneath. Red still gets a ladder escape fork for 2nd row ladders arriving from the left.

2nd-to-6th row switchback

Perhaps surprisingly, giving enough space, it is possible for the attacker in a 2nd row ladder to force a switchback to a cascading ladder on the 4th and 6th rows, without the help of any additional pieces. Red can initiate this maneuver by playing the piece marked 5 in the following diagram. The cells marked "*" are not required for this trick (i.e., they may be occupied by Blue).

51324

This position is complicated to analyze, with many possible lines of play, but it can be shown that the following sequence is optimal for both Red and Blue. In other words, Blue cannot prevent Red from getting the switchback, and Red cannot do better (within the amount of space shown here) than getting a cascading ladder on the 4th and 6th row.

1715161397145812101361124

Also note that, while the resulting right-to-left cascading ladder is on the 4th and 6th rows, it is effectively a 2nd-and-4th row cascading ladder since Red already has a solid edge of pieces on the 2nd row.