Difference between revisions of "Solutions to puzzles"

Revision as of 03:23, 14 January 2024

Piet Hein

See Solutions to Piet Hein's puzzles

Claude Berge

See Solutions to Claude Berge's puzzles

Bert Enderton

Puzzle 1

The unique winning first move is Red b4. (Blue can easily defend the bridges at e3/e4 and g2/g3. Red d3 is defeated by blue b5, red b5 is defeated by blue d3, and all other red moves are defeated by blue b4).

Notice that the red stone at a6 is already connected south via the ladder escape at e5. Also, red threatens to connect north via a double threat at b3 and d3; each of these would connect to the northern edge via Template IIIa. The two templates overlap at d1, but blue d1 is defeated via the following sequence of forcing moves:

Puzzle 2

A winning move is Red e3. This connects to the top edge via Template IIIa, and to the bottom edge via Template J5. The two templates overlap at one point d3, but if Blue plays there, Red can reply f3 to seal the connection up and down.

Puzzle 3

The unique winning first move is Red c3! (e2 is defeated by e3, d3 is defeated by e1, e3 is defeated by e2, and b4 is defeated by d3).

The following seem like horizontal's (Blue's) best tries from the above position.

Puzzle 4

Red 1. c4 is the only way to prevent Blue from connecting immediately by b5, c3 or c4. Red threatens to connect c4 to the top and bottom edges with assistance from the ladder escapes at f2 and f5.

Blue 2. c5 is met by Red 3. a6, which connects to the bottom edge with assistance from the ladder escape at f5 and threatens to connect to c4 or climb to a4 and then connect to the top edge with assistance from the ladder escape at f2.

Blue may attempt a ladder escape fork by playing in the edge template between c4 and the top edge, but 2. c3 is met by 3. b4 and 2. c2 is met by 3. b3, which maintain the connection to the top edge and block Blue's ladder escape.

Eric Demer

Puzzle 1

The unique winning move is red a3. a2 is captured to the top edge, and Red threatens b4. e4 is a ladder escape, so b3 loses to a4 and b5 loses to b4. b4 capture-dominates a4 and a5. Thus Blue plays b4.

Red responds with e2. This connects to the bottom edge by edge template IV2d, and connects to the top edge as shown.

Uniqueness:

The shaded area is a virtual connection from c3, so Red must either play in it or play e2.

The rest of this paragraph uses edge template III2a extensively, and so we do not mention it each time it's used. If Red plays a5 or b4, then Blue plays b3, winning via d3 or e1. If Red plays a4, then Blue plays c4, connecting left via a3 or a5. a3 is Red's winning move. If Red plays b3, then Blue plays c2, connecting left via a3 or b4. If Red plays c2, then Blue plays c4, winning.

Red's only remaining try is e4. Blue responds with c4. After that, consider edge template III2a to the left edge.

If Red intrudes there other than at *, then Blue responding at * captures those 6 cells. If Red intrudes at * before Red plays c2, then Blue responding at c2 still captures the template. Thus, until Red has played c2, Red has no useful way of even threatening to disconnect c3 from the left edge, so one gets

At this point c3+c4 escapes the ladder from b2.

Puzzle 2

The unique winning move is blue d2.

This connects to the left edge via edge template IV2g, so Red must play either in there or e2. Against most intrusions in that template, Blue just responds with c3. Specifically, c3 almost makes this a second order template, since even if Red starts with two moves in it, the only way Red can stop Blue from reconnecting is if those moves are in the two cells marked "*":

Specifically, Red must have defended against b4. If Red did that with b4 or a5, then Blue reconnects with b3 or c1. Thus, Red needs a4. Red must have also defended against b3. If Red did that with b3 or a3, then Blue reconnects with c1. Thus, Red also needs b2.

Since Blue c3 connects to the right edge via edge template III2b, Red must play one of the shaded cells in the following diagram.

Blue responds to e2 with d4. Whether before or after that exchange, Blue responds to each of b2,a4 with b3. At least two of b2,a3,a4 would still be empty, so that allows Blue to climb, for example like this:

Before the e2,d4 exchange, Blue responds to c3 with d4. After the e2,d4 exchange, Blue can respond to c3 with d3. In each case, Blue will have d2 and d4 captured to the right edge, so Red must defend against c1. If Red does so with b2, then b3 is an escape fork for the ladder from b5. If Red does so anywhere else, then a2 still escapes that ladder.

Thus, both before and immediately after the e2,d4 exchange, none of Red's intrusions into the edge template IV2g work, so Red must try e2 followed by d3. However, against that, Blue just plays b5, and ladders to a2.

Uniqueness:

Red's main threat is d3, bridging to edge template III2d. d2 is Blue's winning move. e2 and d3 lose to c3, since d5 escapes the ladder. d4 and c5 lose to b4, forming edge template IV2g.

If Blue plays c4, then Red plays b4. b4 will be captured to the bottom edge, so Blue must defend against c2. c3 loses to b3, since e1 escapes the ladder. The other four all lose to d3.

This leaves just b5. Red reponds with b4.

That has a connection to the bottom edge, via a5 or c4, and threatens c2. If Blue plays c4, then Red plays a5, killing b5 and thereby reverting to the "Blue plays c4, then Red plays b4" case. If Blue intrudes in the other three cells of the connection from b4 to the bottom edge, then Red plays c4, re-establishing that connection and connecting to the top edge via edge template IV2a. Thus, Blue must defend against c2. c3 loses to b3, since e1 escapes the ladder. The other four all lose to d3, since that connects via d4 or a5.

Puzzle 3

The unique winning move is blue b2.

That captures a2 and a3, and threatens e1, so Red must play in c1,c2,e1. The b2-d1 bridge is bolstered at c1 (by Red's edge), so c1 is now worse than c2.

If Red plays c2, then Blue plays d2. That captures e1 and e2, so Red must respond with c1. Blue then plays b4, which connects left via e5 or b3, and connects right via edge template IV2e.

Thus Red plays e1. Blue responds with e2.

Red must play in c1,c2,e1. As before, we discard c1. If Red plays c2, then Blue plays d2, killing e1 and thereby reverting to the "Red plays c2, then Blue plays d2" case. Thus Red plays d2. Blue responds with c4, connecting left via a5 or c3 and connecting right via e3 or d5.

Uniqueness:

I use it twice here, in already-dense prose, so I start with what I will call "the c4 ziggurat-defense":

If Blue plays in the e3 ziggurat other than at c4, then Red can respond at c4, after which Red can keep both c4 and e3 connected down inside the ziggurat: Either c4 is captured down, in which case e3 connects via one of {d3,d4,e4}, or Red can defend the bridge, in which case c4+e3 connects down via one of {b5,c5,e4}.

With that out of the way, suppose Blue plays anywhere other than the five cells marked "*" in the following diagram.

Red responds with b2, capturing b1 and c1. Blue gets one more move before it becomes Red's turn again, and Blue's two moves can't be Blue occupying both c4 and a5.

One of Blue's two moves must be in {a5,a3,b3}, and the other of Blue's two moves must defend against d2, since red d2 would connect down via edge template IV2i.

If that other move from Blue is not in the e3 ziggurat, then Red wins by laddering to e3: If Blue played a5 then Red plays b4, else Red defends the a4-b2 bridge.

If Blue played in the e3 ziggurat other than at c4, then Red uses the c4 ziggurat-defense, after which Red wins with d2 or the ladder along the bottom. Thus, one of Blue's two moves must be c4. This means Blue's other move can't be a5, so it must have been a3 or b3, to defend against red a5. In either case, Red plays b4, winning via either d2 or the other of a3,b3.

Thus, Blue must have played one of the five cells marked "*" in the following diagram.

b2 is Blue's winning move, and against the other four, Red plays d2.

If Blue has played b1 or c1, then we claim that the entire red-shaded region below is captured by Red:

To prove this, it is sufficient to show that Red has a strategy in the shaded region that keeps both a4 and d2 connected to the bottom edge. Because by the time the board is completely filled, if a4 and d2 are connected to the bottom edge, no minimal blue winning path can pass through the shaded area. Therefore all blue stones in that area will be dead.

To show that a4 and d2 are simultaneously connected down, consider any blue move in the region. If Blue plays b4, then Red responds with a5. In that case, d2 still connects down via edge template IV2i. Thus, Blue tries somewhere else in this region. Red responds at b4, and Blue gets one more move before it becomes Red's turn again.

If Blue's two moves were a5 and b5, then Red plays d4, connecting both d2 and b4 to the bottom with a trapezoid. Otherwise, Blue played at most one of a5,b5. If Blue played exactly one of those two, then Red plays the other, killing whichever of those two was Blue. In that case, Blue again gets one more move before it goes back to being Red's turn, so for showing this capture, we are left with the case where neither of Blue's was was in {a5,b5}. Since b4 is red, a5 and b5 are directly red-captured, so one of Blue's two moves must be c3, since otherwise Red plays there. The order in which Blue played Blue's two moves doesn't matter, so one can pretend Blue played c3 first. In that case, before Blue played Blue's other move, Red had a bridge to e3 and a connection from e3 due to the ladder escape at b4. Thus, Red can respond to restore the connection down from d2.

Due to this capture, the positions simplify to the following,

where the shading indicates that one of b1,c1 is blue and the other is empty.

Blue must play e1. After that, if c1 is empty, Red plays at c1 and is connected by a crescent, and if b1 is empty, Red plays at a2 and is connected by a different crescent and edge template II. This shows that both 1. b1 and 1. c1 are losing for Blue.

The remaining moves for Blue are a5 and c4. If Blue plays a5, then we claim that within the red-shaded region in the following diagram, Red can keep d2 connected down while ensuring that red b4 would connect down.

The proof is similar to our proof of capture above. Specifically, before Blue plays there, d2 has a template down and red b4 would connect down since e3 escapes the ladder. If Blue plays b4, then red b4 can't happen and d2 still connects down, since b4 isn't in the template. If Blue plays c3, then Red plays d4, capturing d2 down and guaranteeing a ladder escape for red b4. The d2-e3 bridge is bolstered at e2, so e2 is worse than d3. If Red plays in the e3 ziggurat other than at c4, then Red uses the c4 ziggurat-defense, after which d2 connects down via c3 or e2, and red b4 would also connect down. If Blue plays c4, then Red plays d4. That captures d2 down, so red b4 would connect down via b5 or c3.

Given that, Red wins as follows:

Thus​ 1. a5​ loses, so Blue's only remaining try is ​1. c4.

Red threatens

and

so Blue must play one of the cells shaded below.

If Blue plays in {d5,d3,e2}, then Red plays b4, which captures a5 and b5, so Blue must defend against b2. Blue playing in {a3,b2,b1} loses to c2, Blue playing c1 loses to b3 - threatening e1 and a2 - and Blue playing b3 loses to b2, since that connects d2 to the top via e1 or c2.

If Blue plays e1 or c2, then Red plays

and then wins with whichever of e1,c2 Blue did not play.

If Blue plays c1, then Red plays b3 and defends that bridge. That connects to the top with a2 or e1 and to the bottom with a5 or d4.

If Blue plays b3, then Red plays b2. That captures b2 and d2 to the top, so Red wins with b4 or d4.

Lastly, if Blue plays a5, then Red wins as follows.

Puzzle 4

The solution to puzzle 4 from this section has not yet been typed up.

Puzzle 5

To find Red's winning move, let's use mustplay analysis to narrow down the possibilities. Blue has several threats, which are shown with their respective carriers:

Bridge to ziggurat:

b4 and double threat:

b4 and a different double threat:

Ladder and climbing from b2:

c4 with edge template III2b and ladder:

Red must play in the intersection of these carriers, i.e., at a4, a5, or b4.

Red a4 loses to Blue c3, with double threats towards both edges:

Red a5 loses to Blue c2 as follows. Apart from attacking Blue's bridge, which Blue can simply defend, Red's moves 3 and 5 are forced, after which Blue wins by double threat:

This leaves b4 as the only possible move for Red. Indeed b4 is winning. We leave the verification of this to be done later.

Puzzle 6

The unique winning move is c4. In the follow-up, Black must play d3 to get edge template IV2b::

Puzzle 7

The unique winning move is c6:

Indeed, now Blue is connected left by double threat at d4 or b6, and right by f6 or a ladder escape fork from f4. The uniqueness of the winning move is discussed in more detail in the article on the mustplay region.

Other authors

Puzzle 1

There are two main solutions. In the first line, Blue at (*) on move 7 or 9 is also winning.

In the second line, Blue has several winning moves starting move 9, but we show just one example.

Surprisingly, these are the only winning first moves for Blue. For example, 1. h8 is a losing move, but the refutation is subtle and depends on play in the lower-left obtuse corner:

However, Red's refutation fails when Blue plays 1. g8 instead, because after Red 12, Blue can instead play 13. d8! at (*), which threatens to connect to g8 (which wouldn't have been possible had Blue started with 1. h8).

Puzzle 2

Note that Red is connected upwards by Template J5. The only way to prevent Red from connecting downwards is to play in the cell marked (*).

Perhaps surprisingly, (*) is a winning move, while any intrusion into the template on the first move is losing. If Red responds as expected at 2.c5, Blue's response is to intrude Red's upward connection at 3.b4. This leaves Blue in a strong position.

If Red instead plays 2.g4, Blue can respond with 3.d5, again leaving Blue in a strong position.

Other possible moves by Red also have a strong response, for example 2.c6 3.c4, or 2.d4 3.d5, or 2.d5 3.c4, or 2.e4 3.e5.

Puzzle 3

To simplify the analysis, it helps to note that because c9 connects to the left and a3 connects to the top, the left and top areas of the board are completely settled. Moreover, i2 captures j1 and j2 and kills h3, which in turns captures i3 and j3. Additionally, f6 and g5 are captured by Red. Therefore, the situation of the puzzle is equivalent to the following:

Therefore, whatever Red can do, Red must do in the lower right corner. Here is how Red can win:

Puzzle 4

To simplify the analysis, it helps to note that Blue's g2 group is already connected to the right edge (see below), and Red's c6 group is already connected to the bottom edge. Moreover, Blue does not have any prospects for escaping a 2nd row ladder approaching the lower left from under c6. For these reasons, the position in the puzzle is equivalent to the following:

Therefore, everything Blue can do must be done in the upper left corner. The winning move is c4:

Blue 1. c4 threatens to connect left by b5 or c3 and right by d5, d4 or e3. If Red replies in the region a2-c2-c4-a6, other than at c3, to threaten c4's connection left, then Blue 3. c3 connects to the left edge, and to the right via e2 or d5. The only reply outside the region a2-c2-c4-a6 that prevents the group of 4 blue pieces connecting immediately to the left edge by d5, d4 or e3 is Red 2. d4. However, Red 2. d4 and 2. c3 can be answered by 3. e2.

To see why Blue's g2 group is connected right, Blue's main threats are 1. g5, connecting via Edge template IV1d and by the ladder escape fork 1. i1 2. j1 3. i3 4. i2 5. g4. Red can only meet both these threats by playing in overlap of these templates, in the marked cells.

However these replies can be met by:

2. i3 3. i2 4. h2 5. g4 6. h3 7. h5

2. j3 3. i2 4. h2 5. g4 6. h3 7. h4 8. i3 9. i5

2. g4 3. i1 4. j1 5. i2 6. j2 7. i3 8. j3 9. i5 10. i4 11. h5 12. h4 13. f6

2. h4 then either 3. g4 threatening 5. i2 and 5. g6 - or simply 3. i2 4. h2 5. g4 6. h3 7. g6 - forming a crescent and connecting with the assistance of the second row ladder escape at i2

2. j2 3. h2 4. i2 (capture-dominates the alternatives i1 & j1) 5. h5 - a double ziggurat connection

[Note that although we used Edge template IV1d from g5 in the above explanation to minimise the overlap and the resulting case analysis, we don't need the whole template or the method of connection used for that. If however we use the smaller and simpler Edge template IV1a from g5, then the overlap with the ladder escape fork template is larger and we need to consider the additional replies h3, i2 and j1 (although j1 is capture-dominated by i2, so any winning variations against i2 also work against j1). Although 3. g5, connecting via Edge template IV1d, works against all these replies, there are simpler and quicker responses, e.g.: 2. h3 3. i4; 2. i2 3. h3 4. i3 5. h5 or 4.j3 5. i5; 2. j1 3. i3. So the analysis above only demonstrates that Blue is connected, rather than showing the smallest connection template or the quickest or simplest method of connection.]

Puzzle 5

Red's mustplay region consists of these 8 cells, because if Red plays anywhere else, Blue plays at d4 and wins immediately.

The unique winning move is Red c4. Red then connects for example like this:

Another move that looks plausible is Red d4, but it fails to this:

Puzzle 6

The unique winning move is b7. In the following line, Red's moves below (1, 3, 5, and 7) are the only moves that preserve the win. Note that moves 1 and 3 are minimaxing moves, while move 7 is a foiling move that blocks the ladder escape created from Blue 6.

See Also

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