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More Contact Problems
by Walter Trice - 15 August 2006
Walter Trice
The single most common strategic decision in backgammon is whether to intensify the conflict between the opposing armies of checkers, or to try to reduce contact and emphasize the racing aspects of the game plan.

I think of checker-play problems in which this factor is the most important feature as ‘contact problems.’ Last week’s column presented a rather tricky set of variations on a single such problem. This time we’ll look at a more varied set of middle-game contact problems.

Problem 1: Red to play 2-2.

Problem 1 should be very easy. Red is behind in the race, so he would like to keep White blocked, and hopefully hit a blot in the next few rolls. Fortunately the timing is very much in his favor. Red has seven checkers outside his home board to White’s four, which means that he will be able to play his rolls more comfortably than White. White’s anchor in Red’s home board has become a liability, since White cannot use his lead in the race until he evacuates both checkers safely.

 

Tactically it is quite obvious that if Red plays 13/9(2) (best play) then White will be forced to leave a blot for Red to shoot at if he rolls 6-1, 6-2, or 6-3. Furthermore most of his 5’s and 4’s will force him to break the six point, making it much harder for him to come out on top if there is an exchange of hits. The 9 point increases the blockade against White’s anchor. Even if White gets the checkers on the midpoint home safely there is a good chance that he will be squeezed off the anchor within a few rolls, which will leave him subject to a powerful attack.

Problem 2: Red to play 6-6.

Problem 2 is a type that comes up often in holding games. Red will be trailing in the race even after the double sixes, so he would like to maintain contact. But the mechanical ‘maximum-contact’ play of 8/2(3), 7/1 would just create a timing problem for Red. His next six would force a concession in the outfield, turning control of the game over to White. Neither can Red really afford to give up the midpoint while holding the anchor, since this would release the pressure on White’s 15 point. So it seems that Red simply must use two of his sixes to jump off the anchor with 20/14(2).

 

Then what? 8/2(2) would keep White doubly blocked in the outfield, but only temporarily. Red might find himself running out of sixes soon, and might be forced to leave the first blot himself. Here (and quite typically) the confrontation between two points and one point works out in favor of the single isolated point, because two checkers for one side tie up four checkers for the other.

 

So after 20/14(2), Red must continue by giving up either the 14 point or the 13. Which one? Well we have finally come to the easy part of the decision. Since Red will still be down in the race, he should keep the 14 point and finish his play with 13/7(2). Even that extra one-pip gap does make a difference.

Problem 3: Red to play 6-6.

Almost the same position as Problem 2, but now Red will be leading in the race, so he simply plays 20/8(2), making it as easy as possibly to break the remaining contact.

Problem 4: Red to play 4-1.

In Problem 4 Red is far enough behind in the race that most of his wins have to come by hitting a blot. Red’s furthest-back checker on the 17 point is an important player in the blot-hunt, since it controls some territory that White has to run through when he gets squeezed off the 18. Red would like to make his 4 point, so that if he does hit something he will be almost sure to win the game.

 

I would expect most intermediate-level opponents to play 17/13, 5/4 here. It is perfectly safe for the moment, and it gives Red a spare checker on the midpoint that he will probably be able to play home safely next time. Furthermore Red will probably make the four anyway, and there is a very good chance that he will get at least a six-shot at White when White is forced off his outfield point.

 

But all that is just not good enough. Moving 17/13, 5/4 is a big mistake. Red should go ahead and make the four with 8/4, 5/4. Most of White’s ace-hits are going to be tremendously double-edged, with multiple return-shots and blots, and lots of jeopardy later even if Red misses everything on his first roll. White’s 1-1 would be very good for him, of course. But Red does not have the kind of huge advantage that allows him the luxury of playing to dodge every possible joker.

 

Tactically there are a lot of sequences where leaving the blot back on the 17 pays off. For instance: White doesn’t hit the blot but has to play the checker on his seven point home. Red then rolls something-or-other, plays the 8-blot home, and plays another number in his home board. Now White rolls, say, a 6-2 – and Red has 25 shots at two blots instead of just 17 shots at one blot!

 

Problem 5: Red to play 4-3.

Who could resist the ‘loose hit’ with Bar/22, 6/2*? Not even I, most likely, if this problem had come up in one of my own games. With a four to two advantage in home board points, with 4 spare checkers close to the field of attack, and with White having a blot in his home board many of us would hit in a flash, going for the quick closeout and possible gammon win.

 

But in this case the obvious attacking play would be wrong. Red is up 35 pips, which is quite a lot, so he should play to de-emphasize contact and hold his racing lead with the simple Bar/21, 13/10.

 

If he then continues by successfully escaping with his last back checker White will have to get very lucky to win, with both contact and the race working against him. Red may subsequently be able to attack and close out the White blot without having to risk much at all.

 

After Bar/21, 13/10 for Red, the only active winning plan available to White is an attack against the Red blot in his home board. But the same conditions that suggest that Red might want to attack (better board, White’s blot) make an attack by White very unlikely to succeed. Only having at least one more Red checker sent back could give White much hope of winning by active play. This is why Red prefers to have any hitting take place on White’s side of the board, for the time being.


Here are the rollouts for the positions above:

 

Problem 1: Red to play 2-2.


Problem 2: Red to play 6-6.


Problem 3: Red to play 6-6.


Problem 4: Red to play 4-1.


Problem 5: Red to play 4-3.


 

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