Strategy 2 Dots and Boxes
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Knox (Computer) at 20051103
The following assumes that you have read or already the stuff included in the post Basic Strategy (with good diagrams). This is the last post of this nature I intend to make.
Chains of Length Two
We defined a long chain as one that contains three or more boxes. The reason is that no matter what edge your opponent plays in such a chain, you can always take all but two of the boxes followed by a doubledealing move that declines the last two boxes. This is not true for chains containing just two boxes because your opponent might play the middle edge leaving us without a doubledealing move that ends our turn in the same chain. Such a move is called a hardhearted handout.
** * * * * ***       *** *** *** Hardhearted Handouts
If you have won the longchain fight (the fight for control) and you are forced to play in a chain of two boxes, then you should always play a hardhearted handout giving your opponent no choice. If you instead play a halfhearted handout (see the following diagram), your opponent might reply with a doubledealing move taking control from you – the doubledealing move leaves a doublecross move which switches the desired parity to the opposite of that specified by the longchain rule.
*** ** * ***     *** *** *** Halfhearted Handouts
Sacrificing for Control
Beginning players generally don't sacrifice boxes when there are still free moves around. Hence, they are initially surprised to find out that most games between the top players involve such sacrifices. One common reason for a sacrifice is to get the desired long chain parity. Such a sacrifice could create a chain, destroy a threatened chain, merge two chains into one, or prevent two chains from merging. Rather than try to give a comprehensive list of examples, I'll give examples of what seems to be the two most common types of control sacrifices.
Sacrifices to Create a Chain
Consider the following two positions.
**** * * * ***   * **** * * * * *     * * *** * ****    * * * * * **** *   *** ** * ****
In both cases the player to move wants an even number of long chains and there is presently only one. The only way to force a second chain is by a sacrifice that converts the threatened quad into a long chain.
**** * * * ***   * **** * * * * *     * * *** * ****     * * * * * **** *   *** ** * ****
After the free box is taken on the left, there is a second long chain of length three. On the right, there is a chain of length four after the free boxes are taken.
Sacrifices to Destroy Chains
**** *  ** * * *** * *** * *      ** * * **** * * * * *       *** * * * * * ** ***      * *** *** * * ** * *
In all three positions above, the player to move wants an odd number of long chains. In each position, there is currently one long chain but the opponent is threatening to create a second long chain. These threats should be stopped by the following sacrifice moves.
**** *  ** * * *** * *** * *       ** * * **** * * * * *        *** * * * * * ** ***       * *** *** * * ** * *
Preemptive sacrifices
Consider the following position.
* * * **  **** * O  ** ** * O * **** ** ***
The player to move has two boxes and wants an even number of long chains but there is no way to change that fact that there are three long chains. Therefore, the player to move has lost control. If he plays the free move, he loses 7 to 9. Instead he should play in the unfinished chain!
* * * **   **** * O  ** ** * O * **** ** ***
Why? He is going to be forced to be the first to play in a long chain. So instead of waiting for the chain to grow, he plays in it now effectively shortening the chain. In this instance, the preemptive sacrifice turns the 7 to 9 loss into an 8 to 8 tie.
Preemptive sacrifices are common in the games among top players. If you can force control, the top players simply will not let you grow your chains very long. A preemptive sacrifice is often the most effective or only effective way to do that. Before I continue, a very quick digression.
There are three types of loony moves.
A halfhearted handout.
A move which offers a long chain.
A move which offers a cycle.
Fact: If your opponent has just made a loony move, then you can obtain at least 1/2 of the remaining boxes.
The preemptive sacrifice is a loony move and therefore it can't win win under best play unless you will be ahead at the point where your opponent makes his doubledeal decision. Consequently, preemptive sacrifices are generally more effective when played after your opponent has sacrificed some boxes to win the long chain fight. Occasionally, a preemptive sacrifice is best met (after taking the offered chain) by a return preemptive sacrifice.
Even when the preemptive sacrifice must lose against perfect play, it can still be your best move in the sense of minimizing your losing margin against best play and maximizing the chances that your opponent will make a gamecosting mistake, for example your opponent's doubledeal decision could be extremely difficult (the decision requires a global analysis of the entire board).
Dependent Chains
First, I need a some definitions. Two boxes are neighbors if the edge dividing them hasn't been played. For ease of description, pretend there is an “outside” box that at the start of the game is a neighbor of every box on the outer boundary of the board. A chain is independent if both ends of the chain neighbor the outside box, otherwise it is dependent. The box at which three or four dependent chains meet is called a joint. Note that all dependent chains either run between a joint and the outside box or between two joints. For this section, consider a dependent cycle to be a dependent chain where both ends are neighbors of the same joint.
Counting the number of long chains is trivial when all the chains are independent but it is no longer obvious when the chains are dependent because play on one chain can cause other chains to merge together at joints. When using the long chain rule, the following procedure always gives the correct answer for dependent chains.
Mentally eliminate dependent short chains in any order until only long chains remain. Make sure you perform any merge operations caused by the elimination of a chain. (Note that in practice, you should open up 1chains before 2chains but this has no effect on the final count obtained by this procedure.)
The count is the number of dependent long chains minus the number of joints.
To test your understanding, find the correct count in the following positions (the joints contain an “o”).
***** *** **     * * * * * * * ** * * ** ***              * * * * * * * * ** * * * * * ** o    o   o   o ** ** * * * ** * ** * ** * *        *** ** * **** *** *** (A) (B) (C) (D)
Answers: (A) 1, (B) 2, © 1, (D) 1
A final point is that you should never open up a dependent cycle in which the joint node has exactly three neighbors. Such a play is always worse than opening the attached chain.
The Exchange of Short Chains
Losing the long chain fight means that you will be the first to be forced to play in a long chain or cycle which also means that you took the last short chain. Working backwards from the last short chain taken to the first, it's not hard to see that the loser of the long chain fight is guaranteed of getting at least 1/2 of the boxes taken during the exchange of short chains. In fact the number of boxes gained can be completely characterized in terms of the parity of 1chains and 2chains as shown in the following table.
2chain parity1chain
parityevenoddeven02odd11
When you've lost the long chain fight, you can try to set up the desired parity of short chains to gain a box or two. Note that an odd 2chain parity dominates an even 2chain parity hence the loser of the long chain fight should try to get an odd number of 2chains.
References
Elwyn Berlekamp, “The Dots and Boxes Game: Sophisticated Child's Play.” A.K. Peters Ltd., Natick MA, 2000.
Elwyn Berlekamp, John Conway, and Richard Guy, “Winning Ways: For Your Mathematical Plays, Volume 3.” A.K. Peters Ltd., Natick MA, 2003

graff at 20051124
Your section, sacrifices to create a chain is entirely incorrect.
First, because the board is 4x4 boxes and not 5x5, the player to move wants the number of long chains to be odd. In both cases the second player can take the boxes and then the free move, forcing our first player to give the chains.
Secondly, in the second position, no sacrafice is needed to create the long chain. Simply play in the line above where you played the “sacrafice” and you will have a chain of length six.
I haven't had time to read the rest of the post, but I'll give you my feedback on it when I have read it.

KnoxB (Computer) at 20051124
Here's a correct version.
Sacrifices to Create a Chain
Consider the following two positions.
+++ + + + + +++  + ++++ + + ++ +    + + +++ + ++++    + + + + + ++++ +   +++ ++ + ++++
In both cases the player to move wants an even number of long chains and there is presently only one. The only way to force a second chain is by a sacrifice that converts the threatened loop into a long chain.
+++ + + + + +++  + ++++ + + ++ +    + + +++ + ++++     + + + + + ++++ +   +++ ++ + ++++
After the free box is taken on the left, there is a second long chain of length three. On the right, there is a chain of length four after the free boxes are taken.