Unit adjustments

A unit adjustment is an optional final step that can fine-tune the relative count from methods based on the half-crossover count (such as the colourless count or the criss-cross count). Usually, a relative pip count within 5 pips is enough to help you decide what play to make, and this step can be entirely skipped… but sometimes you need that extra bit of precision.

Let’s look at the position from the 321 colourless pip count lesson:

Recall that in that lesson, we counted that we were ahead by 21 pips (and I revealed that the exact count was −18). How can we correct this 3-pip discrepancy?

The key here is in understanding what the first stage count actually represents, and what the unit adjustment needs to do to complete it. In the first stage, we are not distinguishing between the checkers to the left, center, or right of each “triad”. Said another way, this count implicitly assumes that all of the checkers contained in each triad are in the center point of their respective triads, like this¹:

Therefore, the unit adjustment only needs to “shift” the checkers that aren’t in the center one pip towards where they belong, left or right. As it turns out, the checkers at the points in the centers of their triads (the 2-pt, 5-pt, 8-pt, etc) are counted perfectly in the first stage! This means we can ignore them completely in the second stage. That would make the board look something like this:

What’s left is to adjust our stage 1 count (for this position, -21, meaning we are leading the race by 21 pips) by only one pip for each checker remaining on this board. If the checker is closer to being borne off than the center point (for example, on the 1-pt, 4-pt, 7-pt, etc), that helps our race, and so we subtract 1; if the checker is further away (such as on the 3-pt, 6-pt, etc) it hurts our race, so we add 1.

Let’s see how this works by way of example:

• Before we start stage 2, we are leading by 21 pips.
• The 2 checkers on the 3-pt are further from the bear-off tray than the center point for their triad (the 2-pt), which hurts our race, so we lose 2 pips and are only leading by only 19 pips.
• The 2 checkers on the 4-pt are closer to the bear-off tray than their center point (5-pt) so they help our race, so we are leading by 21 pips again.
• On the 6-pt are 4 checkers lagging behind the 5-pt, which cuts our lead to 17 pips.
• On the 12-pt are 2 checkers that are trailing, cutting our lead further to 15 pips.
• On the 13-pt are 4 checkers ahead of the 14-pt, increasing our lead to 19 pips.
• On the 18-pt, −2 for 17 pips.
• On the 19-pt, +3 for 20 pips.
• On the 21-pt, −2 for 18 pips., which is the exact count we are looking for!

Shortcuts

Evidently, this stage adds many steps to the count. Although the individual calculations are simple (+1, +2, −2, etc), there are so many that it might not feel worth it to adopt this method. Thankfully, one can use symmetry to devise shortcuts to eliminate most of the calculations in many positions.

1. Any checkers that have a matching pair on the opponent’s side of the board don’t need to be counted. For example, in the position above, the checkers on the 4-pt cancel the checkers on the 21-pt, the checkers on the 12-pt cancel out 2 of the checkers on the 13-pt, etc. After applying this rule, only 7 checkers need to be counted:

This rule dramatically reduces the number of calculations! But there’s more that can be done to simplify the count.

2. If there are any checkers that make opposite contributions (i.e., leading vs. trailing) within the same triad (or even neighbouring triads), they can also be canceled out in the same manner. For example, the checkers on the 13-pt add to our race lead and cancel out the checkers on the 18-pt. Canceling these 4 checkers out leaves only 3 to count:

Each of these 3 checkers is trailing by one pip with respect to their respective center points (the 2-pt and the 5-pt). Therefore, to do the unit adjustment for this position, we need to remove 3 pips from our original race lead.

These steps might at first seem cumbersome to visualize, but it quickly becomes automatic with practice. What’s nice is that unit adjustments don’t need to be performed every single time you need a count, so you can decide how much work you are willing to put in to get an accurate count based on your situation.

What if there are checkers on the bar?

As in the previous steps, we need to give any checkers on the bar special treatment. Those checkers are also presumed to be in the center of their respective triads. Our checkers on the bar would be on the “25-pt,” so they are each 2 pips away from their center triad, so they eat away 2 pips from our race lead; our opponents checkers on the bar are on the “0-pt,” so they are 2 pips closer to bear-off than the 2-pt, and they increase our race lead by 2 pips each.

1. Note how the 5-pt has 8 total checkers on it, consisting of 6 blue checkers and 2 white ones. Since the counts are colourless, it turns out not to matter who the checker belongs
   to, only it’s distribution. ↩︎

Next lesson: Mental shift method