The second pip counting method we will learn, the criss-cross count, is another easy algorithm for calculating the relative count. Like the 321 colourless count, it starts with a crude phase that gives you a count within 5 or so pips and can be further perfected using a unit adjustment refinement phase. The main benefit of criss-cross is that all its numerical operations have tiny (single digit) numbers; by contrast, the numbers in the 321 count can end up being pretty large. You also often get to skip entire steps if things line up right. On the other hand, the 321 count only adds checkers, whereas the criss-cross count might add *or subtract* checkers on the boards. Personally, I get more mixed up having to remember whether I need a minus sign or not than I have trouble adding 2-digit numbers, so I prefer 321 for that reason. That being said, many people around me find criss-cross to be way easier, and so I am presenting it below.

Consider the position below:

We are holding the 22-pt anchor in our opponent’s board, and have rolled fabulous running numbers. However, we note that our opponent has already safetied a back checker, so they might be substantially ahead of the race, making a pure racing play (such as 22/10(2)) a large blunder. Also, typical advice for a single checker back is to “attack a blot and prime an anchor,” and this roll could be a good opportunity to start a blitz attack (8/2*(2) 7/1(2)). Knowing the pip count could be very helpful, here, in ruling out the running play.

**1. **The first step is to do the “criss-cross” calculation by adding and subtracting the number of checkers as shown in the diagram:

On our board, this calculation becomes 4 + 2 – 4 – 1 = 1.

**2.** Next we add and subtract *twice* the number of checkers on the midpoint triads, as shown below:

In this position, we need to add 2 × (3 – 4) = -2 to the 1 we calculated in step 1, for a total of -1.

**3. **Next we take into account what we will call the **stray checkers**, for simplicity. These are our checkers that are on our opponent’s side of the board (*e.g.,* on the 22-pt and 13-pt in the position above) and our opponent’s checkers on our side of the board (on the 12-pt and 2-pt). These stray checkers have the longest way to go to get home, so they each add many pips. We take our stray checkers, and subtract our opponents stray checkers, multiply it by 5 and add it to our tally. Here, this step adds 5 × (5 – 4) = 5 pips to our tally, so 5 – 1 = 4 total.

**4.** The last step is to multiply the final tally by 3, producing the approximate relative pip count. 3 × 4 = 12, and a positive number indicates that we are trailing. (The exact pip count is 12, so once again our approximate method has done ok!)

In terms of what play to make, though we are trailing before the roll, the 24 pips from the roll put us up in the race by over 10 pips, making the running play the right one.

So to recap, the steps are:

- Do the “criss-cross” calculation for 4 of the triads on the board (1 – 4 in the diagram above)
- Add 2× the difference in the number of checkers on the midpoint triads (5 – 6 above)
- Add 5× the difference of the stray checkers
- Multiply the result by 3×. As before, a negative number means we are leading, a positive number means we are trailing.

#### What about checkers on the bar?

As in the colourless count, we should pretend that any checkers on the bar are in the first triad they can access. *Our checkers* would be in the 24-pt to 22-pt triad, and* our opponent’s* would be in the 3-pt to 1-pt triad. In this count, both sets of checkers need to be considered during the criss-cross step, *and also* in the stray checkers step.

##### Further reading:

- The Google doc by Max Glaezer that, as far as I know, is the first to describe the criss-cross count.
- A video that demonstrates this count in action.

Next lesson: Unit adjustments

## Leave a Reply

You must be logged in to post a comment.