Difference between revisions of "Switchback"

Revision as of 04:03, 9 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.

Red makes a switchback as follows:

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

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:

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):

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:

Red's piece 7 is connected to the bottom. Play continues:

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.

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:

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:

Once again, yielding would not have helped Blue.

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:

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:

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:

Note that Red is connected to the edge by edge template D-5b. 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:

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:

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:

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.

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).

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.

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.