The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this, we first formalise what it means to be a ladder.

Informally, a ladder escape (say, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, no matter how far away from the ladder escape the ladder starts. So strictly speaking, to check that a pattern is a 4th row ladder escape, we must check that the attacker can connect to the edge from an infinite set of positions. This raises the issue of how one can check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page for 2nd, 3rd, 4th, and 5th row ladders. It might be possible to resolve it for 6th row ladders but this has not yet been done, partly because such ladders are of little practical use. For 7th row ladders we run into a new difficulty – Blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which is currently unresolved. It may in theory be that there are no 7th row ladders at all.

For the purpose of our analysis, we assume that all ladders move from left to right along the red bottom edge, with Red being the attacker. Of course, the analogous analysis also applies to ladders moving in the opposite direction or along different edges.

The analysis of 2nd–4th row ladders on this page was originally contributed by the user Wccanard in 2016.

## Algebraic notation

Before we start, let us introduce some notation that will be useful.

### Open patterns

A pattern is a set of cells, each which may be empty or occupied by a stone of either color. In this article, we will only be concerned with patterns that include a red board edge. A pattern is open on the left if it comes with some cells marked "+" on its left side. No cells to the left of those "+"s may be part of the pattern. A pattern is open on the right if it comes with some cells marked "−" on its right side. No cells to the right of those "−"s may be part of the pattern. A pattern is open on both sides if it is open on the left and on the right. A pattern is closed if it is not open on either side. For example, the following four patterns are open on the right, open on both sides, open on the left, and closed, respectively:

The cells labelled "+" (if any) are called the left boundary of the pattern, the cells labelled "−" are called its right boundary, and the carrier of a pattern consists of all cells that are part of the pattern (empty or not), except the boundaries.

Suppose P is a pattern that is open on the right, Q is a pattern that is open on the left, and the right boundary of P has the same number of cells and shape as the left boundary of Q. Then we write P+Q for the pattern obtained by joining P and Q along their common boundary. More specifically, P+Q is obtained as follows: delete the right boundary from P and the left boundary from Q. Then glue the patterns P and Q together along the line where the boundaries were deleted from each. For example:

+
=

It is important to note that the carrier of P+Q consists of just the carriers of P and Q, without the boundary cells that have been deleted. The purpose of the boundary cells "+" and "−" is just to indicate where the patterns will be attached. It is possible to add more than two patterns, for example:

+
+
=

Note that addition is only well-defined if P and Q can be glued together without their carriers overlapping. We will be careful to ensure that this is always the case. However, the addition is associative, i.e., (P + Q) + R is well-defined if and only if P + (Q + R) is well-defined, and in that case, they are equal.

### The shift operation

If P is a pattern that is open on the left, we write 1 + P for the pattern obtained from P by shifting its left boundary one column to the left, and adding a column of empty cells where the boundary used to be. More generally, for any integer n ≥ 0, we write n + P for iterating this operation n times, i.e., for shifting the left boundary of P to the left by n columns and filling the additional space with empty cells. For example:

1 +
=

4 +
=

Note that empty cells are only added to the height of the boundary. For a pattern that is open on the right, we can do exactly the same thing on the other side, i.e., P + n is obtained by shifting the right boundary to the right by n columns, filling the additional space with empty cells. For example:

+ 1     =

+ 4     =

We call this the shift operation. Note that it is associative: If P and Q have matching boundaries, then (P + n) + Q = P + (n + Q).

### The reduce operation

If P is a pattern that is open on the left, we write ↑ + P for the pattern obtained from P by erasing the topmost "+" cell from its left boundary. The cell that formerly contained the "+" is no longer part of the pattern (i.e., it is not replaced by an empty cell). For example:

↑ +
=

We call this the reduce operation. The shift and reduce operations can be combined with each other and with addition of patterns. For instance, n + ↑ + m + P is the pattern obtained from P by first shifting its left boundary by m cells to the left, then reducing the size of that boundary by one cell, and then shifting it by another n cells to the left. For example:

2 + ↑ + 3 +
=

There is no issue at all with defining a 2nd row ladder. Informally, a 2nd row ladder looks like this:

Note that at each point in the ladder, Blue's move is forced. Red can choose to continue pushing the ladder as long as she wants to. We formally define a second row ladder as follows:

Definition. A second row ladder is a pattern like this:

L2:

Here, the red stone is on the second row, and we call it the ladder stone. Red's goal is to connect the ladder stone to the bottom edge. The cell immediately below and to the right of the ladder stone is empty. We denote this pattern by L2.

Before we give the formal definition of a second row ladder escape, let us consider an example. The following pattern is an example of a second row ladder escape.

Of course directly underneath the pattern is the bottom (red) edge. The cells marked "+" indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is,

Red can guarantee a connection from the ladder stone (marked 1) to the bottom edge.

Let us clarify what the hexes marked "+" in the ladder escape pattern mean. They indicate the last point where the 2nd row ladder is allowed to start. So for example, saying that the pattern above is a second row ladder escape means (among other things) that Red must win the following position:

Here, Red's ladder stone is marked "1", and the claim (easily verified) is that even with Blue to play, Red can connect the ladder stone to the bottom:

The reason that the pattern is a second row ladder escape is that this escape sequence works even if the ladder is a long way away:

Even here, Red can force a connection to the edge, even if it is Blue's move, because Blue must keep defending on the first row and Red keeps attacking on the second row,

and now we are back at the previous position with the ladder right next to the escape, where we have already seen that Red can break through to the edge. We can now give a more formal definition of a second row ladder escape.

Definition. A second row ladder escape template (or simply second row ladder escape) is given by the following data. It is a pattern P that is open on the left (see algebraic notation above), with a boundary of this shape:

subject to the following axiom: for all n ≥ 0, the position L2 + n + P is an edge template, in the sense that even if it is Blue's turn to move, Red can guarantee a connection from the ladder stone marked 1 to the edge.

Concretely, this means that any position consisting of a second row ladder L2,

followed directly to the right by an arbitrary number (zero or more) of pairs of vacant hexes on the first and second rows,

followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder or rightmost column of vacant hexes onto the hexes marked "+"), allows Red to connect the ladder stone to the edge.

Terminology and notation: If we have a left-open pattern whose boundary is of the correct shape, but we are not sure whether it satisfies the axiom of a second row ladder escape, then we refer to it as a candidate for a second row ladder escape (or simply candidate if the rest is clear from the context). A candidate is valid if it is actually a ladder escape.

We also define what it means for a ladder escape template to be minimal.

Definition. A 2nd row ladder escape template is minimal if the following two things are true. First, removing any hex from the pattern, or removing a red stone from the pattern (and replacing it with an empty hex) gives a new pattern which is not a 2nd row ladder escape template any more. And second, if the two hexes directly to the right of the two cells marked "+" are both vacant hexes in the pattern, then moving the cells marked "+" one hex to the right results in a new pattern which is not a 2nd row ladder escape.

Below, we will use analogous terminology and notations for ladders and ladder escapes on the 3rd and higher rows.

The definition of a 2nd row ladder escape allows the ladder to be an arbitrary distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However, this means that we cannot directly use the definition to check that something is a 2nd row ladder escape, because this would require checking that infinitely many patterns are edge templates. Can we find some finite criterion for checking 2nd row ladder escapes? Fortunately, as every Hex player knows, the answer is yes. We have the following theorem:

Theorem 1. Consider a candidate P for a 2nd row ladder escape. Schematically:

(Here the asterisks indicate the carrier of P, which can contain any stones at all, and can be of any shape or size, as long as it includes no cells to the left of the cells marked "+"). Then P is a valid 2nd row ladder escape if and only if L2+P is an edge template for Red.

Proof. If the ladder escape is valid, then by definition, L2+n+P is an edge template for all n ≥ 0, and in particular for n = 0. This proves the left-to-right implication.

To go the other way we actually have to play some Hex, but it's pretty trivial. We must show that L2+n+P is an edge template for all n. This is an easy induction on n. For n = 0, the claim is true by assumption. If n > 0, then Blue must play directly below Red's ladder stone (or else Red will connect to the edge immediately), and now Red can play a ladder stone at distance n−1 on the second row, which is an edge template by induction hypothesis. □

### Examples

We can use Theorem 1 to prove that all of the following patterns are minimal second row ladder escapes. Most of these templates are taken from David King's website, and there are several more there. For several of the templates, the corresponding pattern on David King's site is not minimal by our definition; for these templates, we have moved the cells marked "+" to the right to make the template minimal.

The shaded cell is not part of the template, and can be occupied by Blue.

In the below templates, the stone marked "↓" indicates a stone connected to the bottom edge, but the connection is not shown. The connection from 10 to the edge must not use any of the empty hexes in the pattern.

There is a minor issue with defining ladders on the 3rd and higher rows. We want a definition that is useful in practice and not too restrictive. For example, we surely want this to be a third row ladder:

even though there are a few blue stones on the first row. It is intuitively clear (and also provably true) that these blue stones cannot be of any help to Blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder to have no stones on the first three rows to the right of the ladder (until we reach the escape), we do not want to also guarantee that there are no stones on the first row to the left of the ladder. We formally define third row ladders as follows.

Definition. A third row ladder is a pattern like this:

L3:

The red stone is again called the ladder stone, and Red's goal is to connect the ladder stone to the bottom edge. We denote this pattern by L3.

The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red ladder stone. This is a minimal requirement, because for example if one of these cells were filled,

then in reality the game could look like this,

and Blue can block the ladder with this move.

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because Blue can defend against a third row ladder in more than one way: Blue can at some stage decide to yield to the second row. The following definition is unsurprising.

Definition. A third row ladder escape is given by the following data. It is a pattern P that is open on the left, with a boundary of the shape

To be a third row ladder escape, the pattern must satisfy the property that for all n ≥ 0, L3 + n + P is an edge template connecting the Red ladder stone to the bottom edge, with Blue to move. Or in words: for all n ≥ 0, if we insert a third row ladder at distance n into the pattern at the three hexes marked "+", the resulting position is an edge template connecting the ladder stone to the bottom edge.

Like for second row escapes, a pattern that has the required shape for a ladder escape, but it is not (yet) known to be a valid ladder escape, is called a candidate.

In pictures, for the pattern

(where the carrier is schematically indicated by stars) to be a 3rd row ladder escape, it must give rise to an edge template when we attach a 3rd row ladder at distance 0,

or at distance 1,

or at distance 6,

or at any other distance.

Just like for second row ladder escapes, we again find ourselves in the situation that trying to use the definition to check that something is a 3rd row ladder escape involves checking that infinitely many positions are edge templates. Once again, we have a theorem that allows us to replace this by a finite condition.

Theorem 2. Consider a candidate P for a 3rd row ladder escape. Assume that (a) L2+↑+P is an edge template and (b) L3+P is an edge template, each such template connecting the ladder stone to the bottom edge. Then P is a valid third row ladder escape.

Proof. First note that by Theorem 1, because L2+↑+P is an edge template, P escapes all 2nd row ladders. Now under the assumptions of the theorem, we must show that L3+n+P is an edge template for all n ≥ 0. We prove this by induction on n. For n = 0, the claim is true by assumption (b). Now suppose the claim is true for some n. To show the claim for n+1, consider the position L3+n+1+P. The first three columns of this position look like this:

(This is followed by n more columns of three empty hexes and by the pattern P). Blue has three possible moves in a triangle under stone 1, and Blue needs to play one of these or he will lose instantly. We analyze all three moves in turn.

For the first, Red pushes the ladder and will connect to the edge because by induction hypothesis, L3+n+P connects to the edge, so stone 3 connects to the edge, and so stone 1 does too.

For the second, Red just wins outright, i.e., we do not need to use the induction hypothesis.

And for the third, Red responds like this. Since stone 3 is a 2nd row ladder stone, it is connected to the edge because, as we noted above, ↑+P is a 2nd row ladder escape. Therefore stone 1 is also connected.

The induction is now complete, showing that P is a 3rd row ladder escape. □

Remark: in more concrete terms, Theorem 2 states that a pattern P is a 3rd row ladder escape if the pattern becomes an edge template (connecting the ladder stone to the edge) when we attach each of the following two patterns to its left boundary:

A:

B:

In contrast to the situation with 2nd row ladders, while Theorem 2 is sufficient to show that a position is a 3rd row ladder escape, it is not necessary. For example, consider the following third row ladder escape template P:

One can check directly that L3+2+P and L2+↑+2+P are both edge templates, so that 2+P is a 3rd row ladder escape by Theorem 2. In particular, L3+n+P is an edge template for all n ≥ 2. Moreover, one can check that L3+P and L3+1+P are also edge templates, so that P is a valid 3rd row ladder escape.

It is, however, not a valid 2nd row ladder escape for ladders at distance 0, because in the position L2+↑+P,

the ladder stone marked "1" cannot connect to the edge.

Theorem 2 is therefore not sufficient to check that a given pattern is a 3rd row ladder escape. We need to work a little harder to get a necessary and sufficient condition for 3rd row ladder escapes.

Lemma 3 (2nd to 3rd row jump). Any 3rd row ladder escape also escapes 2nd row ladders that start at distance 2 or greater. More specifically, if L3+P is an edge template, then so is L2+↑+2+P.

The lemma is perhaps easier understood in pictures: given any 3rd row ladder escape, replacing the three cells marked "+"

by the pattern

yields a 2nd row ladder escape.

Proof. Assume L3+P is an edge template. We must show that L2+↑+2+P is an edge template. It looks as follows, with P added to it:

But Blue must play 2, and Red can jump to 3. Then 3 is a 3rd row ladder stone, and is connected to the edge because L3+P is an edge template by assumption. Therefore, 1 is also connected.

We now finally get a necessary and sufficient condition for 3rd row ladder escapes in the following theorem.

Theorem 4. Consider a candidate P for a 3rd row ladder escape. Then P is a valid 3rd row ladder escape if and only if L3+P, L3+1+P, and L3+2+P are edge templates.

Proof. The left-to-right implication is trivial, since by definition, if P is valid then L3+n+P is an edge template for all n, including n = 0, 1, 2. For the opposite implication, assume that L3+P, L3+1+P, and L3+2+P are edge templates. By Lemma 3, L2+↑+2+P is an edge template. By Theorem 2 and the assumption about L3+2+P, 2+P is a 3rd row ladder escape. It follows that L3+n+P is an edge template for all n ≥ 2. Since we additionally assumed this to be the case for n = 0 and n = 1, P is a valid third row ladder escape. □

As a matter of fact, Theorem 4 is not tight. We can get the following better result. However, the proof of Theorem 4 generalizes more easily to 4th row and higher ladders, which is why it is of interest.

Theorem 5. Consider a candidate P for a 3rd row ladder escape. Then P is a valid 3rd row ladder escape if and only if L3+P and L3+1+P are edge templates.

Proof. The left-to-right implication is trivial. For the right-to-left implication, by Theorem 4, it suffices to show that L3+2+P is an edge template. Indeed, consider Blue's options in the position L3+2+P:

As usual, there are only two possible moves for Blue to avoid losing immediately. If Blue moves at 2, then Red can respond at 3, which connects to the edge because L3+1+P is an edge template by assumption.

If Blue instead moves at 2, then Red responds as follows, which connects to the edge because L3+P is an edge template by assumption.

### Examples

Here are some examples of third row ladder escapes. Again most of these are taken from David King's website. For several of the ladder escape templates, the version shown on David King's website is not minimal by our definition; in these cases, we have moved the cells marked "+" to the right to make the template minimal. All of the templates in this section have been proven to be third row ladder escapes using Theorem 5. All of them are minimal. As before, a stone marked "↓" is assumed to be connected to the bottom row, but the connection is not shown. Any shaded cells are not part of the pattern and can be occupied by Blue.

The following third row ladder escapes also escape 2nd row ladders at distance 0 or greater:

The following third row ladder escape also escapes 2nd row ladders at distance 1 or greater (but not at distance 0):

The following third row ladder escapes also escape 2nd row ladders at distance 2 or greater (but not at distance 0 or 1):

The version of this last pattern on David King's website has the cells marked "+" (he uses arrows) sloping in the other direction; the location that is shown here makes the template minimal.

The following is a rather strange example of a third row ladder escape:

One of the reasons this example is strange is that if the ladder starts at distance 1 or greater, the pattern is no longer minimal. Since the ladder must come from the left, this template therefore does not come up in practice. Nevertheless, it satisfies our formal definition of a 3rd row ladder escape. This template also escapes 2nd row ladders at distance 2 or greater (but not at distance 0 or 1).

Definition. A fourth row ladder is a pattern like this:

L4:

Again, the red stone is called the ladder stone and Red wants to connect the ladder stone to the bottom edge. We denote this pattern by L4.

The key part of the definition is that the 6 hexes forming a triangle below the ladder stone are all vacant. Note that even filling in one of these can invalidate the ladder: even if we fill in the bottom left corner of the triangle,

then Blue has this move,

which is easily seen to stop the ladder. To establish the ladder, Red needs at a minimum those 6 vacant hexes under her ladder stone.

The definition of a 4th row ladder escape is entirely analogous to that of 2nd and 3rd row ladder escapes.

Definition. A fourth row ladder escape is given by a pattern P that is open on the left with a boundary of this shape.

Moreover, it must satisfy that for all n ≥ 0, L4 + n + P is an edge template connecting the Red ladder stone to the bottom edge, with Blue to move. In other words, adding a 4th row ladder at distance n, for any n ≥ 0, connects the red ladder stone to the bottom edge, with Blue to move.

As before, a candidate is a pattern that has the correct shape, but is not (yet) known to be a valid escape.

We have already encountered all of the relevant ideas. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual Hex, which this time is quite fun!

Theorem 6. Consider a candidate P for a 4th row ladder escape. If L2+↑+↑+P, L3+↑+P, L4+P, and L4+1+P are valid edge templates, then P is a 4th row ladder escape.

Proof. The idea of the proof is the same as for 3rd row ladders. First observe that by Theorems 1 and 2, since L2+↑+↑+P and L3+↑+P are edge templates, ↑+P escapes all 3rd row ladders and ↑+↑+P escapes all 2nd row ladders. We must prove that L4+n+P is an edge template for all n ≥ 0. We proceed by induction on n. The base cases n = 0 and n = 1 hold by assumption. For the induction step, assume the claim is true for n ≥ 1. We need to prove the claim for n+1.

Consider the position L4+n+1+P. It looks like this, with n−1 additional columns of four vacant hexes and the pattern P attached on the right:

We need to prove that the ladder stone 1 is connected to the edge.

The five moves marked 2 below all lose instantly to Red 3:

The two moves marked 2 below also lose instantly:

The move marked 2 below can be answered by Red 3, moving us to position L4+n+P, which is an edge template by the induction hypothesis.

Move 2 below leads us to a 2nd row ladder, which ↑+↑+P escapes, so 5 (and therefore 1) is connected to the edge:

Both moves marked 2 below lead us to a 3rd row ladder, which ↑+P escapes, so 3 (and therefore 1) is connected to the edge:

Move 2 below also leads to a 3rd row ladder (note Blue 4 must be in the triangle left and below from Red 3; Blue can also play out the bridge between 1 and 3 but this doesn't help):

The final choice for move 2 below also gives a 2nd row ladder.

This completes the induction, so P is a 4th row ladder escape. □

Remark: In more concrete terms, Theorem 6 states that P is a 4th row ladder escape if adding each of the following patterns to P gives an edge template.

A:
B:
C:
D:

Remark: Theorem 6 is analogous to Theorem 2. It gives a sufficient, but not a necessary condition for a candidate to be a 4th row ladder escape. Once again, the criterion in Theorem 6 can be checked in a finite amount of time. To get a theorem with a necessary and sufficient condition, we need another "jump lemma":

Lemma 7 (3rd to 4th row jump). Any 4th row ladder escape also escapes 3rd row ladders that start at distance 3 or greater. More specifically, if L4+P and L4+1+P are an edge templates, then so is L3+↑+3+P.

Proof. Consider the position L3+↑+3+P, which looks as follows, with P added to it:

There are only two possible moves for Blue that don't lose immediately. If Blue moves at 2, then Red can respond at 3, which is a 4th row ladder stone and connects to the edge because L4+1+P is an edge template by assumption.

In Blue moves instead at 2 in the following diagram, then Red can respond as shown.

Now Red's stone 9 is a 4th row ladder stone. Although the additional red stone 5 does not belong in the L4 template, this stone can only help Red. By assumpion, L4+P is an edge template, and so stone 9, and therefore stone 1, is connected to the edge. □

We then arrive at a necessary and sufficient condition for fourth row ladder escapes.

Theorem 8. Given a candiate P for a 4th row ladder escape. Then P is a valid 4rd row ladder escape if and only if L4+P, L4+1+P, L4+2+P, ..., L4+6+P are edge templates.

Proof. The proof is similar to that of Theorem 4. Again, the left-to-right implication is trivial. For the right-to-left implication, assume that L4+P, L4+1+P, L4+2+P, ..., L4+6+P are edge templates. By Lemma 7 applied to P and 2+P, we know that L3+↑+3+P and L3+↑+5+P are edge templates. By Lemma 3 applied to ↑+3+P, we know that L2+↑+2+↑+3+P is an edge template, and therefore also L2+↑+↑+5+P, which differs from L2+↑+2+↑+3+P only in that it contains two additional empty hexes. Since L2+↑+↑+5+P, L3+↑+5+P, L4+(5+P), and L4+(6+P) are edge templates, we know by Theorem 6 that 5+P is a valid 4th row ladder escape. Therefore, L4+n+P is an edge template for all n ≥ 5. Since we assumed this to be also true for n = 0, 1, 2, 3, 4, it follows that P is a valid 4th row ladder escape. □

Remark: Like Theorem 4, it is likely that Theorem 8 is not tight, in the sense that there probably exists an even simpler condition that is necessary and sufficient for 4th row ladder escapes (perhaps analogous to Theorem 5). Also, in practice, Theorem 6 is often easier to check since it involves fewer conditions.

### Examples

Here are some examples of fourth row ladder escapes. Most are taken from David King's website. In each case we have moved the column of "+"s as far as possible to the right to yield a minimal template. The validity of all of these escapes has been proved using Theorem 8.

The following fourth row ladder escape templates also escape 2nd and 3rd row ladders at distance 0 or greater:

The following fourth row ladder escape templates also escape 2nd row ladders and 3rd row ladders at distance 1 and greater.

The following fourth row ladder escape templates also escape 2nd row ladders at distance 2 and greater, as well as 3rd row ladders at distance 1 or greater. The stone marked "↓" is assumed to be connected to the bottom edge, although the connection is now shown.

The following fourth row ladder escape templates also escape 2nd row ladders at distance 1 and greater, as well as 3rd row ladders at distance 0 or greater.

The following fourth row ladder escape template also escapes 2nd row ladders at distance 2 and greater, as well as 3rd row ladders at distance 0 or greater.

Definition. A fifth row ladder is a pattern like this:

L5:

As usual, the red stone is called the ladder stone and Red's goal is to connect it to the bottom edge. We denote this pattern by L5.

Unlike in the case of 2nd, 3rd, and 4th row ladders, this time it is not sufficient for a triangle of cells below and to the right of the ladder stone to be empty. We also need three additional empty cells to the left of this triangle. This is a minimal requirement; if even one of these cells is occupied by Blue, for example like this:

then Blue can block the ladder with this move:

The main line is complex; see for example this Little Golem discussion thread. Many of the main lines of defense involve Blue playing an upside-down version of Tom's move, for example like this:

Note that Blue's 1 is connected to 5 by double threat at "*", and 7 is Tom's move upside-down, i.e., with the top line of blue stones serving as the "edge". Therefore, to establish the ladder, Red needs at minimum the specified 13 vacant hexes under the ladder stone.

The definition of a 5th row ladder escape is as expected.

Definition. A fifth row ladder escape is a pattern P that is open on the left with boundary of this shape:

It must satisfy the following axiom: for all n ≥ 0, L5 + n + P connects the red ladder stone to the bottom edge, with Blue to move. As usual, a candiate is such a pattern that satisfies everything except perhaps the axiom.

Theorem 9. Consider a candiate P for a fifth row ladder escape. Assume L5+P, L5+1+P, L5+2+P, L4+↑+P, L4+↑+1+P, L3+↑+↑+P, and L2+↑+↑+↑+P are all edge templates. Then P is a 5th row ladder escape.

Proof. The proof idea is the same as for 3rd and 4th row ladders, but there are a lot more cases to consider. First, note that by previous theorems, ↑+P escapes all 4th row ladders, ↑+↑+P escapes all 3rd row ladders, and ↑+↑+↑+P escapes all 2nd row ladders. We prove by induction on n that L5+n+P is an edge template for all n ≥ 0. The base cases n = 0, 1, 2 are true by assumption. For the induction step, assume the claim is true for n ≥ 2. We need to prove the claim for n+1.

Consider the position L5+n+1+P, which looks like this (followed by an additional n−2 columns of five empty hexes and the pattern P):

The eight moves marked 2 below all lose instantly to Red 3 by edge template IV-1-a:

The three moves marked 2 below also lose instantly by edge template IV-1-a:

The six moves marked 2 below give a 4th row ladder, which ↑+P escapes.

This leaves us with 11 more moves to consider. If Blue pushes the ladder by making the move marked 2 below, Red can answer 3, moving us to position L5+n+P, which is an edge template by the induction hypothesis.

Move 2 below gives a 3rd row ladder, which ↑+↑+P escapes.

If Blue plays move 2 below, Red can answer 3, forcing Blue to respond in the ziggurat below and to the left of stone 3. If Blue plays in any of the cells marked 4, Red plays 5 and gets a 4th row ladder, which ↑+P escapes. Blue could have also first intruded upon the bridge between 1 and 3, but this does not help. From now on, we tacitly ignore bridge intrusions that are not helpful to Blue.

If, on the other hand, Blue plays 4 below, then Red gets a 2nd row ladder, which ↑+↑+↑+P escapes.

If Blue plays move 2 below, Red gets a 2nd row ladder, which ↑+↑+↑+P escapes.

If Blue plays move 2 below, Red can answer 3, forcing Blue to respond in one of the hexes marked 4. Then Red can play 5 and gets a 3rd row ladder, which ↑+↑+P escapes.

Similarly, if Blue plays move 2 below, Red can answer 3, forcing Blue to respond in one of the hexes marked 4. Then Red can play 5 and gets a 3rd row ladder, which ↑+↑+P escapes.

If Blue plays move 2 below, Red can answer 3. There are only four hexes where Blue can respond without losing outright. If Blue moves in one of the three hexes marked 4, then Red gets a 3rd row ladder, which ↑+↑+P escapes.

If Blue instead moves in the hex marked 4 below, then the sequence plays out slightly differently, but Red still gets a 3rd row ladder.

If Blue plays move 2 below, Red can answer 3. Then Blue must respond in any of the hexes marked "+", or else Blue will immediately lose to edge template III1b or a ziggurat.

If Blue responds in one of the two hexes marked 4, Red gets a 3rd row ladder.

If Blue instead responds with move 4 below, Red gets a 2nd row ladder.

If Blue instead responds with move 4 below, Red still gets a 2nd row ladder.

If Blue plays move 2 below, Red can answer 3. Then Blue must respond in one of the hexes marked "+" to avoid immediately losing to edge template III1b.

If Blue responds in one of the hexes marked 4 below, Red gets a 3rd row ladder.

If Blue instead responds in one of the hexes marked 4 below, Red gets a 2nd row ladder.

If Blue instead responds at 4, Red gets a 3rd row ladder.

If Blue instead responds at 4, Red gets a 2nd row ladder.

If Blue instead responds at 4, Red gets a 2nd row ladder.

If Blue plays move 2 below, Red can answer 3. Then Blue must respond in one of the hexes marked "+" to avoid immediately losing to edge template III1b or a bridge.

If Blue responds in one of the hexes marked 4 below, Red gets a 2nd row ladder.

If Blue instead responds at 4 below, Red also gets a 2nd row ladder.

Finally, if Blue plays move 2 below, Red gets a 2nd row ladder.

This completes the induction, so P is a 5th row ladder escape. □

Remark: In more concrete terms, Theorem 9 states that P is a 5th row ladder escape if adding each of the following patterns to P gives an edge template.

A:
B:
C:
D:
E:
F:
G:

Like Theorems 2 and 5, Theorem 9 gives a sufficient, but not necessary condition for 5th row ladder escapes. We do not currently have a necessary and sufficient condition. One problem is that we have no appropriate "jump lemma" from 4th to 5th row ladders. In fact, we can prove that no such jump lemma is possible.

Lemma 10 (No jumping from 4th to 5th row). Suppose Red is the attacker in a 4th row ladder. Given enough Blue pieces on the 6th row, and enough space on the right, jumping is not an option for Red. If Red tries to jump, Blue can block the ladder, and Red will get at most a 2nd row ladder in the opposite direction.

Proof. If Red tries to jump, Blue can play as follows.

This is exactly an upside-down version of the situation in Theorem 16 below. No matter where Red plays next, Blue can prevent Red from connecting. The hexes marked "*" are not required by Blue (i.e., they could be occupied by Red). Under optimal play, Red gets at most a 2nd row ladder in the opposite direction as follows:

See the proof of Theorem 16 for a detailed discussion of all the possible moves. □

Lemma 10 is a significant obstacle to establishing a necessary and sufficient criterion for 5th row ladder escapes. We do have the following generalization of Theorem 9, which gives a weaker sufficient condition (it is perhaps also necessary, but this has not been shown):

Theorem 11. Given a candiate P for a 5th row ladder escape. If there is some n ≥ 0 such that L5+P, L5+1+P, ..., L5+n+P, L5+n+1+P, L5+n+2+P, as well as L4+↑+n+P, L4+↑+n+1+P, L3+↑+↑+n+P and L2+↑+↑+↑+n+P, are edge templates, then P is a valid 5th row ladder escape.

Proof. This follows directly from Theorem 9 applied to n+P. □

### Examples

Here are some examples of fifth row ladder escapes. The validity of these escapes has been proved using Theorem 11. These escapes are minimal.

The following fifth row ladder escapes also escape 2nd to 4th row ladders at distance 0 or greater:

The following fifth row ladder escape also escapes 2nd row ladders at distance 2 or greater, 3rd row ladders at distance 0 or greater, and 4th row ladders at distance 1 or greater:

The following fifth row ladder escape also escapes 2nd row ladders at distance 2 or greater, 3rd row ladders at distance 0 or greater, and 4th row ladders at distance 2 or greater:

## Sixth row ladders and up

Because of the 2nd-to-6th row switchback trick, 6th and higher row ladders do not exist in the usual sense. More specifically, even if we allow an arbitrary amount of empty space under the ladder stone, it is not possible for the attacker to keep pushing the ladder. Consider the following situation:

Let's assume there is an arbitrary amount of empty space in the bottom 4 rows to the left of this diagram. The stone marked "1" is connected to the top, and looks like it could be the ladder stone for a potential 6th row ladder. If such a ladder were possible, the red stones on the M-file should certainly escape it.

From Blue's point of view, Blue is the attacker in an upside-down 2nd row ladder. Blue can therefore use an upside-down version of the 2nd-to-6th row switchback trick. To do so, Blue plays at 2:

If both Red and Blue keep playing optimally, the best that Red can get is a pair of parallel 2nd and 4th row ladders in the opposite direction:

Here, stone 10 is connected to the line of blue stones along the top, so Red has no way of connecting right. Red can now push a 4th row ladder from 5, and/or a 2nd row ladder from 13. There is not enough space for Red to immediately perform Tom's move. So unless Red has a ladder escape somewhere to the left of this diagram, or unless there's enough space on the 5th row somewhere to the left of this diagram to perform Tom's move, Red fails to connect to the edge.

Note that this argument does not show that 6th row ladders are categorically impossible. It only shows that the "usual" notion of ladder does not work. It is conceivable that 6th row ladders are possible under additional assumptions. For example, there might be a notion of 6th row ladder that requires additional space on the 7th row to its right, or on the 5th row to its left. It is currently unknown whether any viable notion of 6th row ladder exists.

For 7th row ladders the situation is even worse. As explained in open problems about edge templates, no amount of space under the ladder (even if we demand that the entire 5th row is clear) is known to guarantee a red connection if Blue just ignores the ladder and plays elsewhere. Thus, it is possible that 7th row ladders do not even exist in theory. Of course they do not occur in practice either.

## Second-to-fourth row switchbacks

### Definition of switchback

Informally, a 2nd-to-4th row switchback is a pattern that allows the attacker to turn around a 2nd row ladder into a ladder on the 4th row in the opposite direction. For example, in the following situation, suppose ladder stone marked "1" is connected to the top, with Blue to move.

Red pushes the 2nd row ladder to d3, the breaks at f3:

At this point, Blue is forced to play 8, and then a new ladder starts in the opposite direction:

In this example, the ladder reconnects to Red's original group, although in general this does not need to be the case (even if the switchback doesn't connect, Red has just created a parallel edge 4 cells from the original edge - a large advantage for Red in any case).

To formalize the concept of a 2nd-to-4th row switchback, consider a 2nd row ladder.

L2:

This is the same ladder as defined in the section of second-row ladders above; only this time, Red's goal will be slightly different. To explain Red's goal, we show a slightly larger area around L2:

This time, Red's goal will be to do at least one of the following two things: either connect the red ladder stone to the edge, or else, occupy the cell marked "a" with a red stone that is connected to the edge, without using the cells marked "b" or any cells to their left. We refer to this as the switchback condition. We also call "a" the switchback cell and "b" the gap cells. With this in mind, we now give the definition of a 2nd-to-4th row switchback template.

Definition. A 2nd-to-4th row switchback template (or simply 2-to-4 switchback) is given by the following data. It is a pattern P with a left boundary of shape

and satisfying the following axiom: L2+↑+↑+n+P satisfies the switchback condition for all n ≥ 0.

As usual, a candidate is a pattern that has the correct shape, but is not (yet) known to be a valid switchback template.

As in previous sections, we write L2+↑+↑+n+P for the pattern obtained from P by moving the four hexes marked "+" to the left by n columns (leaving 4 rows of empty space), then removing the top two cells marked "+" (they are not part of the pattern) and replacing the remaining cells marked "+" by L2. Note that the switchback cell "a" and gap cells "b" are not part of L2. They are simply three cells on the board whose position is defined relative to L2. Depending on the value of n, they may or may not end up being inside the pattern P.

### Characterization of switchbacks

The following theorem gives a finite necessary and sufficient condition for a pattern to be a switchback.

Theorem 12. Given a candidate P for a 2-to-4 switchback. Then P is a valid 2-to-4 switchback if and only if L2+↑+↑+P satisfies the switchback condition.

Proof. The left-to-right implication is trivial. For the opposite implication, we will show, by induction on n, that L2+↑+↑+n+P satisfies the switchback condition for all n ≥ 0. The base case n = 0 holds by assumption. Now suppose that the claim is true for some n. To show the claim for n+1, consider the position L2+↑+↑+n+1+P. It looks as follows, with n more empty columns and P appended on the right:

Here, the ladder stone is marked "1". Blue has no choice but to push the ladder, and Red also pushes:

At this point, the induction hypothesis guarantees that Red can either connect 3 to the edge, or else that Red can occupy and connect the switchback cell "a" while keeping "b" empty:

If 3 is connected to the edge, then so is 1, and we are done. Otherwise, "a" is connected to the edge and "b" is empty. Thus, the board looks like this, with "a" now acting as a ladder stone:

Since Red's stones on the 2nd row are already connected to the top, and 1 is connected to the bottom, Blue has no choice but to respond at 2. Then Red can play 3.

Now the switchback condition for L2+↑+↑+n+1+P is satisfied, proving the theorem. □

### Examples

Any 2nd row ladder escape template trivially also works as a switchback template (with the location of the cells marked "+" adjusted as necessary; they may need to be moved to the left if there isn't space for the two additional "+"s in the pattern). Since such a template escapes 2nd row ladders outright, there is no need for the second part of the switchback condition.

The following are examples of 2nd-to-4th row switchback templates that are not second row ladder escapes. They are minimal.

In the last two templates, the shaded hex is not part of the template, and can be occupied by Blue. The following template is useful for obtuse corners:

In the following template, the stone marked "↓" is assumed to be connected to the bottom edge, but the connection is not shown. The blue stone is not technically part of the pattern; however, if this cell were empty, the pattern would already work as a 2nd row ladder escape.

## Third-to-fifth row switchbacks

### Definition of switchback

The definition of 3rd-to-5th row switchbacks is similar to that of 2nd-to-4th row switchbacks. Consider a 3rd row ladder.

L3:

We define the locations of the switchback cell "a" and gap cells "b" relative to L3:

Again, the switchback condition states that with Blue to move, Red can either connect the ladder stone to the edge, or else Red can occupy the switchback cell and connect it to the edge, without using the gap cells or anything to their left.

Definition. A 3nd-to-5th row switchback template (or simply 3-to-5 switchback) is given by the following data. It is a pattern P, open on the left with boundary

and subject to the requirement that L3+↑+↑+n+P satisfies the switchback condition for all n ≥ 0.

### Characterization of switchbacks

The following theorem gives a finite necessary and sufficient condition for a pattern to be a switchback. It is analogous to the corresponding theorem for 3rd row ladders.

Theorem 13. Given a candidate P for a 3-to-5 switchback. Then P is a valid 3-to-5 switchback if and only if L3+↑+↑+P and L3+↑+↑+1+P satisfy the switchback condition.

Proof. The left-to-right implication is trivial. For the opposite implication, we will show, by induction on n, that L3+↑+↑+n+P satisfies the switchback condition for all n ≥ 0. The base cases n = 0 and n = 1 hold by assumption. Now suppose that the claim is true for n and n+1. To show the claim for n+2, consider the position L3+↑+↑+n+2+P. It looks as follows, with n more empty columns and P appended on the right:

The ladder stone is marked "1". As usual for 3rd row ladders, Blue must either push or yield, or else Red will connect to the edge outright. If Blue pushes, then so does Red:

By induction hypothesis, L3+↑+↑+n+1+P satisfies the switchback condition, so either Red can connect to the edge, or else Red can get a new ladder stone 1 in the switchback cell for L3+↑+↑+n+1+P, which allows Red to satisfy the switchback condition for L3+↑+↑+n+2+P with stone 3.

The other option is for Blue to yield. (We will see later that when n is large enough, yielding in this situation is actually a terrible idea for Blue, since it will allow Red to use P to connect to the edge. But this is not relevant for the current proof).

Now by the induction hypothesis, L3+↑+↑+n+P satisfies the switchback condition, so either Red can connect to the edge, or else Red can get a new ladder stone 1 in the switchback cell for L3+↑+↑+n+P. This allows Red to satisfy the switchback condition for L3+↑+↑+n+2+P with stone 5.

This finishes the proof of the theorem. □

One may ask whether every 3-to-5 switchback template also works as a 2-to-4 switchback template. This is indeed the case at sufficient distance, due to the following jumping lemma.

Lemma 14 (2nd to 3rd row switchback jump). Any 3-to-5 switchback template is also a 2-to-4 switchback template at distance 4 or greater. More specifically, if P is a 3-to-5 switchback template, then ↑+4+P is a 2-to-4 switchback template.

Proof. Suppose P is a 3-to-5 switchback template, and consider Q = ↑+4+P. By Theorem 12, we must show that L2+↑+↑+Q satisfies the switchback condition. The position L2+↑+↑+Q looks like this, with P attached on the right:

After Blue pushes the ladder at 2, Red plays 3, which is essentially Tom's move. While this move is not sufficient to connect Red to the edge, it creates enough trouble to allow Red to get the desired switchback in the presence of P.

Let us consider Blue's options. If Blue moves outside the area marked "x", Red simply pushes the ladder and connects, using 3 as a ladder escape.

If Blue moves in any of the cells marked 4, Red gets the switchback without using P:

If Blue moves at 4, Red also gets the switchback without using P:

If Blue moves at 4, Red also gets the switchback without using P. Note that 3 is connected by edge template III2b.

If Blue moves at 4, Red also gets the switchback without using P. Note that 3 is connected to the edge by double threat.

This leaves only one option for Blue. If Blue moves at 4, then Red responds as follows:

By hypothesis, since P is a 3-to-5 switchback template, Red can either connect 3 to the edge, or else get a connected red stone at "a", with "b" empty:

In either case, 7 is connected to the edge, so Red has the desired switchback. □

Corollary 15. In a 3rd row ladder at distance 5 or greater to a 3-to-5 switchback, Blue cannot yield.

Proof. If Blue yields, then Red can switch back the resulting 2nd row ladder to the 4th row by the previous lemma. This will reconnect to Red's original 3rd row ladder, and therefore connect Red to the edge. In a diagram:

Note that Blue must play 6 for the same reason as in the lemma. Since Red will either connect 5 or "a" to the edge, 7 is also connected. Rather than just giving Red a switchback, 7 is actually connected to 1 by a crescent. □

Here is another interesting fact about 3-to-5 switchbacks. Given enough space, the defender of a 3rd row ladder cannot yield without giving the attacker a switchback.

Theorem 16. Given enough space to the right of a 3rd row ladder and two empty rows above it, if the defender tries to yield, the attacker can achieve a 3-to-5 switchback without requiring any addtional stones.

Proof. Let 1 be the ladder stone of a 3rd row ladder, and assume there is at least as much space as indicated in the following diagram.

If Blue yields at 2, then Red can play as follows:

If Blue's next move is outside of the area marked "x", then Red connects to the edge outright, as follows:

Therefore, Blue must move in one of the hexes marked "x" above. This leaves nine possible moves for Blue.

If Blue moves at one of the hexes marked 6 below, then Red connects by edge template IV2b:

If Blue moves at 6, then Red pushes the second row ladder twice and connects by Tom's move:

If Blue moves at 6, then Red gets 2nd and 4th row parallel ladders, which connect by Tom's move:

If Blue moves at 6, then Red connects by a crescent and Tom's move:

If Blue moves at 6, then Red connects by crescent and ziggurat:

If Blue moves at 6, then Red connects by shopping cart and Tom's move:

If Blue moves at 6, the situation is almost identical:

Note that in all cases so far, Red connected outright, i.e., didn't need a switchback. The final remaining possibility is for Blue to move at 6 in the following diagram. Then Red gets the switchback. Note that 7 is connected to the edge by edge template IV2b.

This finishes the proof of the theorem. □

### Examples

Every 3rd row ladder escape template is also a 3-to-5 switchback template (possibly with the location of the column of "+"s adjusted), but it need not be minimal. Here are some examples of 3-to-5 switchback templates that are not 3rd row ladder escapes.

The shaded cell is not part of the template and can be occupied by Blue.

## Second and fourth row parallel ladders

Parallel ladders, especially on the 2nd and 4th rows, are quite common in Hex. For example, consider this situation, with Blue to move and the Red stone connected to the top:

Play may proceed as follows:

Now Red has a choice: she can either continue pushing the 4th row ladder from 2, or the 2nd row ladder from 6. However, having parallel ladders puts Red in a stronger position than having a 2nd row ladder or a 4th row ladder alone. As we will see, there exist ladder escape templates than can escape a parallel ladder, but can neither escape a 2nd row ladder nor a 4th row ladder on its own.

Note. Unlike with single-row ladders, in the case of a parallel ladder, Red actually has a choice whether to push the 2nd row ladder or the 4th row ladder. For this reason, our formal definition of a parallel ladder follows a slightly different approach than that we took for single-row ladders above. Whereas above, we always assumed that Blue was next to move (and the ladder stone was already in a pushing position), here, we will assume that Red is next to move. This affects the definition of the ladder pattern, in that the ladder stones do not yet have empty space below them.

Definition. A parallel ladder on the 2nd and 4th rows, or 2-4 parallel ladder for short, is a pattern like this:

L24:

The red stones on the second and fourth rows are called the ladder stones, and Red's goal is to connect at least one of these stones to the bottom edge, with Red to move first. (We can assume that both ladder stones are already connected to the top). We denote this pattern by L24. There is also a variant of L24 that looks like this:

L24a:

L24 and L24a are equivalent, and for simplicity we will only use L24.

Definition. An escape template for parallel ladders on the 2nd and 4th rows, or 2-4 parallel ladder escape for short, is a pattern P with a left boundary of shape

satisfying the following axiom: adding a 2-4 parallel ladder at distance n, for any n ≥ 0, allows Red to connect, in the sense that it guarantees the connection of at least one of the red ladder stones to the bottom edge, with Red to move.

As before, a candidate is a pattern that has the correct shape, but is not (yet) known to be a valid escape.

Fortunately, 2-4 parallel ladders are easy to analyze; they are almost as simple as 2nd row ladders. The reason is that, just as for 2nd row ladders, the defender has no choice; he must always push, because as we will see, yielding is not an option. We get a simple and clean theorem with a necessary and sufficient condition for 2-4 parallel ladder escapes.

Theorem 17. Consider a candidate P for a 2-4 parallel ladder escape. Then P is a valid 2-4 parallel ladder escape if and only if L24+P, L24+1+P, and L24+2+P allow Red to connect (with Red to move).

Proof. If the ladder escape is valid, then by definition, L24+n+P allows Red to connect for all n, including n = 0, 1, 2. So the left-to-right implication is trivial. To prove the right-to-left implication, assume L24+P, L24+1+P, and L24+2+P allow Red to connect. We prove by induction that L24+n+P allows Red to connect for all n ≥ 0. The cases n = 0, 1, 2 are true by assumption. Now suppose the claim is true for some n ≥ 2. We must show the claim for n+1. To do so, consider the position L24+n+1+P. The first six columns of this position look like this:

This is followed by n−2 more columns of four empty hexes and by the pattern P. Red starts by pushing the 4th row ladder.

Now consider Blue's possible moves. If Blue moves in any of the hexes marked 2 below (or elsewhere on the board), Red wins outright (i.e., without using the induction hypothesis).

If Blue moves in any of the hexes marked 2 below, Red also wins outright:

If Blue moves at 2 below, Red also wins outright:

This means that the only possible move that is not immediately losing for Blue is to push the 4th row ladder. In this case, Red can respond as follows:

This position allows Red to connect by induction hypothesis, finishing the proof. □

It is clear that every 2nd row ladder escape and every 4th row ladder escape is also an escape for 2nd-and-4th row parallel ladders, since Red can decide to push only the 2nd row ladder, or only the 4th row ladder. In addition, 2nd-to-4th row switchback templates also work as 2-4 parallel ladder escapes. This is intuitively clear, as Red can simply push the 2nd row ladder and switch it back to the 4th row, where it will connect with the 4th row of the parallel ladder. The following theorem proves this more formally, using the definitions.

Theorem 18. Every 2nd-to-4th row switchback template is also a 2-4 parallel ladder escape.

Proof. Assume P is a 2nd-to-4th row switchback template. To show that P is a 2-4 parallel ladder escape, we must show that L24+n+P allows Red to connect with Red to move, for all n ≥ 0. Consider the position L24+n+P, which looks as follows, with an additional n blank columns and P on the right:

Red plays as follows. At this point, since n+P is a 2nd-to-4th row switchback template, Red can either connect 3 to the edge, or get a connected stone at "a" with "b" empty.

This allows Red to connect at least one of the ladder stones, as required. □

### Examples

As mentioned above, every 2nd row ladder escape, every 4th row ladder escape, and every 2nd-to-4th row switchback template works as a 2-4 parallel ladder escape. But there are some examples of 2-4 parallel ladder escapes that are none of the above. The most famous of these is Tom's move, which states that a sufficient amount of empty space is enough for a 2-4 parallel ladder to connect to the edge. Specifically, the following is a 2-4 parallel ladder escape template:

Here are some other examples of 2-4 parallel ladder escapes that are neither 2nd nor 4th row ladder escapes nor 2nd-to-4th row switchbacks. They can be shown to be valid by Theorem 17, and are minimal. Unlike Tom's move, these ladder escapes don't require space on the 5th row.

While the following two patterns aren't switchbacks at distance 0 or 1, they do work as 2nd-to-4th row switchbacks at distance 2 or greater.

## Third and fifth row parallel ladders

Parallel ladders on the 3rd and 5th rows are less common than those on the 2nd and 4th rows, but they can occur. Pushing such ladders is less straightforward, as the defender has more options. Basically, as we will show, if the defender refuses to push, then the attacker can at least get a 2nd row ladder. Moreover, a 2nd-to-4th row switchback template is in that case sufficient for the attacker to connect.

Definition. A parallel ladder on the 3nd and 5th rows, or 3-5 parallel ladder for short, is a pattern like this:

L35:

The red stones on the third and fifth rows are again called the ladder stones, and Red's goal is to connect at least one of these stones to the bottom edge, with Red to move first. We denote this pattern by L35. Just like for 2-4 parallel ladders, there is an equivalent pattern for L35 that looks like this:

L35a:

Definition. An escape template for parallel ladders on the 3rd and 5th rows, or 3-5 parallel ladder escape for short, is a pattern P with a left boundary of shape

satisfying the following axiom: adding a 3-5 parallel ladder at distance n, for any n ≥ 0, allows Red to connect, in the sense that it guarantees the connection of at least one of the red ladder stones to the bottom edge, with Red to move.

As always, a candidate is a pattern that has the correct shape, but is not (yet) known to be a valid escape.

3-5 parallel ladder escapes are not quite as easy to characterize as those for 2-4 parallel ladders, because the defender has more options. We get the following theorem, which only contains a sufficient condition for a pattern to be a 3-5 parallel ladder escape.

Theorem 19. Consider a candidate P for a 3-5 parallel ladder escape. If L35+P, L35+1+P, ..., L35+3+P allow Red to connect (with Red to move), and if ↑+P is a 2nd-to-4th row switchback template, then P is a valid 3-5 parallel ladder escape.

Proof. We prove by induction that L35+n+P allows Red to connect for all n ≥ 0. The cases n = 0, ..., 3 are true by assumption. Now suppose the claim is true for 0, ..., n, where n ≥ 3. We must show the claim for n+1. To do so, consider the position L35+(n+1)+P. The position looks like this, followed by n−3 additional empty columns and P:

Red starts by pushing the 5th row ladder.

Now consider Blue's possible moves. If Blue moves anywhere except the hexes marked "x", then Red wins outright by bridging from 1 to edge template IV-1a.

If Blue moves in any of the hexes marked 2 below, Red moves at 3 and connects by ziggurat and double threat.

This leaves 10 more moves to consider.

If Blue moves at 2, Red pushes the 3rd row ladder.

Now Blue must either push at "x" or yield at "y" (or else Red will connect immediately). If Blue pushes at "x", then Red has a 3-5 parallel ladder at distance n, which connects by induction hypothesis.

If Blue instead yields at "y", then Red can push the 2nd row ladder and use the switchback to either connect 7 to the edge or get a connected stone at "a". Note that "a" is connected to either 1 or 7 by double threat, so Red connects. (As a matter of fact, Red can do better in this case and get a 2-4 parallel ladder, but it is not needed for the current proof).

If Blue moves at 2, then Red plays as follows and connects by edge template III2e:

If Blue moves at 2, then Red gets a 2nd row ladder that connects via the 2-to-4 switchback at "a". The moves 4 and 5 can also be played in the opposite order without changing the result.

If Blue moves at 2, then Red gets a 2nd row ladder that connects via the 2-to-4 switchback at "a". (In fact, Red can get a 2-4 parallel ladder, but it is not needed in this proof).

If Blue moves at 2, Red connects.

If Blue moves at 2, Red connects

If Blue moves at 2, Red connects by edge template IV2d.

If Blue moves at 2, Red can respond at 3.

Now Blue must respond at "x" or "y", or else Red will connect immediately. If Blue plays at "x", Red gets a 2nd row ladder which connects via switchback at "a". Note that 5 is connected to at least one ladder stone by double threat.

If Blue instead plays at "y", Red also gets a 2nd row ladder which connects via switchback at "a".

If Blue moves at 2, Red connects.

Finally, if Blue moves at 2, Red also connects.

This finishes the proof. □

### Examples

Like 2-4 parallel ladders, 3-5 parallel ladders have the property that they can connect to the edge outright if given enough space. There is an analog of Tom's move for 3-5 parallel ladders. The following diagram shows the amount of space required. If Red moves in the cell marked "x", Red can guarantee to connect at least one of the ladder stones marked "1" to the edge. The cell marked "x" is essentially the unique winning move (the only other winning option for Red is to push the 3rd row ladder one more hex before playing "x").

We note that this particular pattern is not technically a 3-5 parallel ladder escape. Without additional empty space on the 6th row, it only escapes 3-5 parallel ladders at distance 0 (as shown) and at distance 1. If the ladder starts further away, Blue has the option of yielding to a 2nd row ladder for which Red would need a 2-to-4 switchback template to connect.

Every 3rd row ladder escape, every 5th row ladder escape, and every 3-to-5 switchback template is also a 3-5 parallel ladder escape. Examples of 3-5 parallel ladder escapes that aren't one of the above are relatively rare, but they do exist. The following are some examples. They have been proved correct using Theorem 19, and they are minimal.

The following template escapes 3-5 parallel ladders, but it is not a 3rd-to-5th row switchback, nor does it escape 3rd, 4th, or 5th row ladders.

The following template escapes 3-5 parallel ladders, but it is not a 3rd-to-5th row switchback, nor does it escape 2nd, 3rd, 4th, or 5th row ladders.

## Second and fourth row terraced ladders

Sometimes it can happen that a ladder forms on top of another ladder, with the two rows of attacking stones not yet connected to the edge nor to each other. We call this a terraced ladder. The following is an example of a terraced ladder on the 2nd and 4th rows, with Blue to move:

Although terraced ladders look superficially similar to parallel ladders, they should not be confused. There are two important differences: (1) in a parallel ladder, the two rows of attacking stones are connected to each other, whereas in a terraced ladder, they are not, and (2) in a parallel ladder, the upper ladder is "ahead" of the lower one, whereas in a terraced ladders, the upper ladder is at the same level or behind the lower ladder.

In fact, as we noted above, from the attacker's point of view, having 2nd and 4th row parallel ladders is stronger than having only a 2nd row ladder or only a 4th row ladder. For terraced ladders, the opposite is true: a 2nd and 4th row terraced ladder is weaker than having only a 2nd row ladder or only a 4th row ladder. Nevertheless, despite being relatively weak, terraced ladders can be pushed, and there is a notion of terraced ladder escape at arbitrary distance.

Before we develop the theory of terraced ladders, it is worth noting that terraced ladders from Red's point of view are parallel ladders from Blue's point of view, and vice versa. This can be seen by putting a row of blue stones on top, giving Blue an "edge":

Indeed, from Red's point of view, Red has terraced ladders trying to connect to the bottom edge, whereas from Blue's point of view, Blue has parallel ladders trying to connect to the top edge. The fact that parallel ladders are better for Blue than individual ladders explains why terraced ladders are worse for Red than individual ladders.

In a terraced ladder, it is the defender, not the attacker, who decides whether to push the 2nd or 4th row ladder. Since the 4th row ladder can lag behind the 2nd row ladder by an arbitrary distance, there isn't just a single ladder template, but a family of them.

Definition. A 2nd and 4th row terraced ladder is any one of the following patterns:

T(0):

T(1):

T(2):

T(3):

and so on. In general, for k ≥ 1, the pattern T(k) looks like

followed by k−1 columns of
followed by

In these patterns, we refer to Red's stone on the 4th row as the top ladder stone, and to Red's rightmost stone on the 2nd row as the bottom ladder stone. Red's goal is to connect the top ladder stone to the bottom edge, assuming it is Blue's turn first. We can assume that the top ladder stone is already connected to the top edge, but we do not assume that the top and bottom ladder stones are connected to each other.

Definition. A ladder escape template for 2nd and 4th row terraced ladders, or 2-4 terraced ladder escape for short, is a pattern P with left boundary shaped like this,

This pattern P must satisfy the following axiom: for all k ≥ 0 and all n ≥ 0, T(k)+n+P guarantees a connection of the top ladder stone to the bottom edge, with Blue to move.

As always, a candidate is a pattern that has the correct shape, but is not (yet) known to be a valid escape. If P is such a candidate, schematically of the form

then we write ∗+P to denote the pattern obtained from P by replacing the top two cells marked "+" by empty cells, and adding two new cells marked "+" just to their left. The resulting template is then of the shape required for 4th row ladder escapes.

We will reduce the problem of establishing a terraced ladder escape to finitely many cases. This is done by two lemmas. Lemma 20 states that we only need to consider finitely many values of n (the distance from the bottom ladder stone to the escape). Lemma 21 states that we only need to consider finitely many values of k (the distance from the top ladder stone to the bottom ladder stone). .

Lemma 20. Suppose P is a candidate for a 2-4 terraced ladder escape. Assume T(k)+P, T(k)+1+P, and T(k)+2+P are edge templates for all k ≥ 0 (i.e., they guarantee a connection of the top ladder stone to the edge, with Blue to move). Moreover, assume that ∗+P escapes 2nd, 3rd, and 4th row ladders (or more precisely, ∗+P escapes 4th row ladders, ↑+∗+P (which has 3 cells marked "+") escapes 3rd row ladders, and ↑+↑+∗+P escapes 2nd row ladders). Then P is a valid 2-4 terraced ladder escape.

Proof. We need to show that T(k)+n+P is an edge template for the top ladder stone, for all k,n ≥ 0. We prove this by nested induction, with the outer induction being on n, and the inner induction on k. The base cases n = 0, 1, 2 are true by assumption. Now consider some n ≥ 3, and suppose the claim is true up to n−1. We need to show the claim for n. Consider the position T(k)+n+P, which looks like this:

followed by n−3 additional empty columns and P. Here, our diagram illustrates the case k = 4, but the following arguments are valid for all k ≥ 0. The first observation is that if Blue's next move is outside of the area marked "x", Red connects to the edge immediately by a long crescent and edge template III2b:

Note that this works for all k ≥ 0, although for k = 0 and k = 1, the connection is simpler and does not require a long crescent. Therefore, Blue must move in the area marked "x".

We consider each of Blue's options in turn. If Blue moves just below the top ladder stone, then Red responds by pushing the 4th row ladder. In case k > 0, this leads to the position T(k−1)+n+P, and he claim holds by the inner induction hypothesis:

In case k = 0, the situation is worse for Blue: in this case, Red gets a bona fide 4th row ladder, which ∗+P escapes by assumption:

If Blue moves anywhere else on the 3rd or 4th row, then Red connects the two ladders and gets a second row ladder, which ↑+↑+∗+P escapes by assumption. This works for all k ≥ 0, although for illustration, we show only the case k = 4:

This leaves only 5 possible Blue moves to consider.

If Blue moves at 1, Red gets a 3rd row ladder, which ↑+∗+P escapes. Note that 2 is connected to the top ladder stone by a long crescent (for k ≥ 2) or directly (for k = 0, 1).

If Blue moves at 1, Red gets a 3rd row ladder, which ↑+∗+P escapes by assumption:

Note that there are several alternatives to Blue's move 3, but they all result in a 3rd row ladder for Red.

If Blue moves at 1, Red simply pushes the 2nd row ladder, and we are now in position T(k+1)+n−1+P, which is a template by the outer induction hypothesis.

If Blue moves at 1, Red gets a 2nd row ladder, which ↑+↑+∗+P escapes by assumption. Note again that 2 is connected to the top ladder stone.

If Blue moves at 1, Red gets a 2nd row ladder.

This finishes the proof of the lemma. □

Having reduced the distance n to finitely many cases, we would now like to reduce the parameter k to finitely many cases as well. The following lemma does this.

Lemma 21. Suppose P is a candidate for a 2-4 terraced ladder escape. Assume T(0)+P, T(1)+P, and T(2)+P are edge templates (i.e., they guarantee a connection of the top ladder stone to the edge, with Blue to move). Then T(k)+P is an edge template for all k ≥ 0.

Proof. The first step in the proof is to show that the following two interior patterns are equivalent. By "interior pattern", we mean that the bottom row of red stones does not have to be a board edge.

B(1):
B(2):

If Red plays first in the region, a red move at x captures the entire region, so x is the only move that Red needs to consider, and its outcome is the same in B(1) and B(2).

If Blue moves first in the region, all of the interior moves (i.e., in unmarked cells) are star-decomposition dominated by x. Therefore, Blue only needs to consider the moves x, y, and z. One can show that each of these three moves (x, y, and z) in region B(2) is equivalent to the corresponding move in region B(1). For example, after Blue moves at x, z dominates all of the interior moves and whoever plays there captures the interior, regardless of whether the region is B(1) or B(2).

A consequence of the fact that regions B(1) and B(2) are equivalent is that all "longer" versions of these regions are also equivalent to B(1), B(2), and each other, i.e.,

B(3):
B(4):

and so on. This is easily proved by induction, because each longer region is obtained from the previous one by replacing a subregion of the form B(1) by B(2), which we already showed to be equivalent.

Next, consider this pattern:

B(0):

We claim that B(1) is at least as good as B(0) for Red, in the sense that anything Red can achieve with B(0), Red can also achieve with B(1). (In fact, B(1) is strictly better for Red than B(0), but that fact is not required for this proof). If Red moves first in the region B(0), the move at x again captures the whole region, and therefore achieves everything Red might hope to achieve in the region. In this case, B(0) and B(1) are equivalent. If Blue moves first, the situation is slightly more complicated. We must show that B(0) is at least as good for Blue as B(1). If Blue plays at x in B(1), then Blue has the corresponding option to move at x in B(0), which works for the same reason as in the proof of the equivalence of B(1) and B(2) above. If Blue plays at z in B(1), Red can respond by pushing the ladder, which creates a position that is literally B(0). If Blue plays at y in B(1), Red can respond at x, and a case distinction shows that no matter how the remaining 3 cells are filled, filling them in the same way in B(0) gives an equivalent position.

Finally, let C(0), C(1), C(2), ... be the same patterns as B(0), B(1), B(2), ..., except with the blue stones removed from the carrier. I.e.:

C(0):
C(1):
C(2):

etc. Note that for all k ≥ 0, C(k) is at least as good for Red as B(k). Because if the neighboring cells we removed from the templates are in fact occupied by Blue, then C(k) is the same as B(k); otherwise, if they are empty or Red, it can only help Red.

In particular, since each C(k) is at least as good for Red as B(k), and each B(k) is at least as good as B(0) = C(0), it follows that if Red wins any position containing C(0), then Red also wins the corresponding position containing C(k).

The final step in the proof is now easy. Simply observe that each T(k+2) is obtained from T(2) by replacing a subpattern of the form C(0) by C(k). Therefore, in any context P where T(2)+P is winning for Red, T(k+2)+P is also winning for Red. Combining this with the remaining two base cases T(0)+P and T(1)+P, we get the lemma. □

By combining the previous two lemmas, we obtain a sufficient condition for the validity of terraced ladder escapes.

Theorem 22. Consider a candidate P for a 2-4 terraced ladder escape. Assume T(k)+n+P is an edge template for n=1,2,3 and k=1,2,3 (nine possibilities). Moreover, assume that ∗+P escapes 2nd, 3rd, and 4th row ladders (or more precisely, ∗+P escapes 4th row ladders, ↑+∗+P (which has 3 cells marked "+") escapes 3rd row ladders, and ↑+↑+∗+P escapes 2nd row ladders). Then P is a valid 2-4 terraced ladder escape.

Proof. By Lemma 21, T(k)+n+P is an edge template for n=1,2,3 and all k ≥ 0. Therefore, the hypothesis of Lemma 20 is satisfied, and thus P is valid. □

### Non-examples

Since terraced ladders are weaker than 4th row ladders, any terraced ladder escape is also a 4th row ladder escape. The question then becomes: which 4th row ladder escapes are not terraced ladder escapes? Most, but not all, of the examples of 4th row ladder escapes given above also escape terraced ladders.

The following patterns escape 4th row ladders but do not escape terraced ladders: