Professor: Welcome back.
Class: Welcome back to you, too, Professor.
Rachel: So what are you going to talk about today?
Professor: Well, the last two sessions had to do with queens winning queens.
Ryan: Were you planning to present more of that?
Professor: Not exactly.
Lucian: OK, what were you thinking of showing?
Professor: Situations where two rooks win the queen.
Ryan: There are plenty of those.
Zephyr: Especially since two rooks are worth about a pawn more than a queen.
Lucian: In relative value.
Thomas: Gosh, ten pawns to nine.
Rachel: But the queen’s a powerhouse.
Hale: And what about if the rooks are in disarray?
Ryan: The queen might pick off one of them.
Professor: What do you say we skip the superficial stuff and see our first problem?
Zephyr: OK, Professor, whatever you say.
Question 1: How should White save the f5-rook?
Ryan: That’s easy.
Hale: It’s just another skewer setup.
Rachel: True, but it’s still nice.
Hale: “Nice” is in the eyes of she who gives the skewer.
Professor: Nice or not, him or her, let’s move to our next setup.
Question 2: How can White free up the g3-rook?
Rachel: So what if the g3-rook is pinned?
Lucian: Rachel is right. That pin isn’t going to last.
Lucian was also right. The class had no trouble in finding the correct answer. They even joked about it. As always, the group was ready for more, examples and jokes.
Question 3: How can the rooks snare the queen?
Hale: The problem is simple.
Thomas: Yeah, those rooks are animals.
Ryan: Animals? Oh, you mean “pigs.”
Zephyr: That’s what some people call them.
Lucian: If they’re on the seventh rank.
Rachel: I wonder what they call two rooks on the seventh.
Lucian: I call it a sty.
Professor: Shall we get out of the mud and try the next problem?
Rachel: Is this our last problem?
Professor: No, it’s merely the antepenultimate one.
Ryan: Oh goody, after this one, there are two more problems to go.
Question 4: How does White win the queen?
Ryan: This is a bit harder.
Lucian: Not for me.
Not surprisingly, Lucian was correct. After a few back and forth wisecracks, the class was clamoring for more.
Professor: I think it’s time for our penultimate problem.
Question 5: How does White win the queen?
It took but a few minutes and the class had the answer. The students were beaming with glow.
Ryan: Didn’t we have a problem like this one before?
Lucian: Of course. It was during the session on staircases.
Hale: Can we see another problem?
Zephyr: The final one, no doubt?
Professor: Yes. Here it is -- the ultimate one -- at least for today.
Question 6: How do the rooks get the better of the queen?
Like the time before, and the time before that, the class easily solved the problem. And they had a few parting comments.
Rachel: So the seventh rank is triumphant once again.
Hale: The doubled rook action doesn’t hurt either.
Ryan: And notice: we didn’t fall for that stalemate shot.
Professor: No you didn’t. Well, I guess that’s that.
Lucian: No more problems for today?
Professor: Not for this class.
Zephyr: You have another class?
Ryan: Somewhere else?
Rachel: On the same day?
Hale: With different students?
Lucian: Are we stronger than they are?
Thomas: Are we your favorite students?
Professor: Yes, yes, yes, yes, yes, and yes, in that order.
Answers below -- Try to solve NM Pandolfini's puzzles first!
Answer 1: White saves the f5-rook by sacrificing it! Here the queen is trapped by 1. Ra5!. If 1…Qxa5 (driving the queen onto a bad square), then 2. Rh6+ Kd5 (or either of Black’s other two legal moves) 3. Rh5+ skewers king and queen.
A number of similar ideas for this way of winning the queen were first worked out by German composer Josef Kling (1811 – 1876).
Answer 2: In this familiar problem composed by the Italian Aristide Dall’ Ava, the g3-rook gets unpinned, thanks to a deflection.
After 1. Ra8 + Ke7 (or 1…Kd7), 2. Ra7!, White’s own pin drives the queen off the b8-h2 diagonal). After the essentially forced 2…Qxa7, White has a winning skewer, 3. Rg7+.
Answer 3: This group of queen-traps was first explored by French composer Henri Rinck (1870-1952).
The winning idea begins with the confining 1. Rf7. The main line continues 1…Qe8 (moving the queen to d8, c8, or b8 doesn’t help) 2. Rh7+ Kg8 3. Rag7+ Kf8 4. Rh8+, where the sac of one rook allows the other rook to win the queen, again by a skewer.
Answer 4: In this 1942 composition by the Czech Ladislav Prokeš (1884 – 1966), Black’s queen is soon lost.
The main winning variation is 1. Rgc2 Qe1 (1…Qe3 actually loses the queen sooner) 2. Rc8+ Kg7 3. Rb7+ Kf6 4. Rc6+ Kf5 (note that 4…Ke5 runs into the skewer 5. Re7+) 5. Rb5+ Kf4 6. Rc4+ Kf3 7. Rb3+ Kf2 8. Rc2+ Kf1 9. Rb1, pinning and winning the queen.
Also note that 1…Qd1 fares no better, as in the exemplary line 2. Rb8+ Kg7 3. Rc7+ Kf6 4. Rb6+ Ke5 5. Rc5+ Ke4 6. Rb4+ Ke3 7. Rc3+ Ke2 8. Rb2+ Ke1 9. Rb1, again pinning and winning the queen.
Answer 5: Here we have another version of a Rinck problem we’ve seen before (as in the lesson on staircase checking sequences).
The key winning line here is 1. Rg7+ Ka6 (1…Kb8 loses to 2. Rh8+) 2. Kd7+ Kb7 3. Ke6+ Kc6 4. Kf7+ Kd7 5. Kg6+ Ke6 6. Kh7+ Kf5 7. Rh5+, another winning skewer check.
Answer 6: In this final problem, White begins his entrapment of the queen by 1. Rg7!. The rook shuts down the queen’s use of the a1-h8 diagonal, while being immune from capture (1…Qxg7? allows 2. Rc8 mate).
Moreover, after 1. Rg7, White threatens 2. Ra7 + Kb8 3. Rb6+, followed by Ra8+.
Black can do no better than 1…Qb8 2. Rgc7 (threatening Rc7-c8) Qb7+ (the queen is safe because 3. Rxb7 is stalemate) 3. Ka5, when 3…Qb6+ fails to the simple 4. Kxb6.
Positions where a lone queen has to face off against two rooks, with no other friendly or enemy units on the board, are rare, but they do occur.
When other forces are involved, especially passed pawns, different factors (such as king safety, open lines, and promotion threats) come into play.
Generally, the queen tends to seek a centralized post, so that it can attack and defend in all directions. Meanwhile, the rooks need to coordinate as a team, so that they can confine, control, and threaten mate.