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Practice Problem; PLEASE SHOW YOUR WORK!

Posted By: Matt Cohn-Geier
Date: Thursday, 11 September 2008, at 3:56 p.m.

In Response To: Practice Problem; PLEASE SHOW YOUR WORK! (Chuck Bower)

By cluster counting: two checkers on the 6 point go forward, one goes back, one on the 8 point goes forward = 62.

The rest of the checkers = 39 + 42 = 40 + 41 = 81. 81 + 62 = 143.

I am considering learning Naccel per Stick's post, so even though I got Blue's count in a couple seconds and would use cluster counting (I don't really attribute this to Kissane, but whatever) for something this easy, I might as well try it:

6 quads in Q3, plus 6 more in Q2, plus 8 more in Q1 = 20 quads. 2 on the 4 point + 2 on the 21 point = 2 quads. 3 on the 8 point = 1 quad. Total = 23 quads, 3 pips left over = 138 + 3 pips = 141 pips. Bleh, I hate Naccel.

White's position:

The checkers in the inner board are 20 (6+3+1). The mid + bar is 63. The checker on the bar is 10. That comes to 93. The other checkers are 40, plus 5, which is 138. Again, "cluster" counting this is pretty easy for me now.

6 quads in Q3, plus 6 more in Q2, plus 6 more in Q1 = 18 quads. Two on the three point = 1 quad, three on the 8 point = 1 quad, the 22 point and 3 on the mid = 1 quad, so 21 quads so far. The 10 point and the two checkers on the ace = 1 quad, so 22 quads with 4 left over on the 23 point.

That's 132+4 = 136. Damn it ;)

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