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Subject: Session Report rss

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Malachi Brown
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This is a cautionary tale in addition to a session report. Hopefully I will be able to dissect what went wrong with our session and get some feedback.

It all started at around 1am, when Doug, William, and I got back to my place after playing some games at a friend's house. We wanted to play something, and we picked Zendo. William stated that he had come up with a rule that he wanted to try, so we let him be the master.

The example koans were a little more complicated than I would generally make them. The wrong one had three pyramids, but the correct one had five of them.

Doug and I started with some variations on the correct example koan. I quickly realized that three colors were required and that position was unimportant. I believe Doug had arrived at similar conclusions.

Over the course of the next two hours, we had two or three Mondos, and I believe Doug made a guess, causing William to produce a counter example that just served to confuse us more.

At some point, Doug and I decided to share information in an attempt to figure out something. We both agreed that it probably had to do with number of pips by color, with a minimum of three colors, but we were stumped beyond that, and we couldn't articulate a guess that would force William to build a koan. Remember, it was 3am by that point.

After 11 koans each, Doug and I gave up and William revealed the rule. His rule was, to paraphrase, "A koan has the Buddha nature if it has at least three colors and the pip count of each color is a consecutive number in the Fibonacci sequence. i.e. a consecutive three digit subset of 1, 1, 2, 3, 5, 8, 13, 21".

We never had a correct koan with more than three colors, nor did we have a correct koan with a single color pip count of eight.

I believe the biggest stumbling block for us was the fact that it was a sequence, not simply a formula. I had noticed that the pip counts of two colors had to add up to the pip count of the third color, but an incorrect koan that had counts of 2, 4, and 6 gave me a lot of trouble. The fact that there could be two colors with a pip count of 1 was also troubling, because all other examples had different pip counts for each color.

In retrospect, I think a better, but still very difficult rule, would have been to require the pip count of the entire koan to be one of the numbers in the sequence. It would have been much easier to map out part of the problem domain with brute force and make some guesses that forced ever larger koans out of William, eventually either exhausting the possible numbers and winning by brute force or figuring out the basis for the sequence and actually guessing the rule. I think having to not only figure out the pattern, but also replicate a subsection of the pattern in each koan was a bit too much.

Anyway, I think the moral is to not let people named William be the master when playing Zendo. Or maybe just not picking a really hard rule at 1am. Of course, the master doesn't usually think the rule will be hard.
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J
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Re:Session Report
Malachi wrote:
I believe the biggest stumbling block for us was the fact that it was a sequence, not simply a formula


That was a hard one to figure out at 1am+. I guess you could consider
it a difference formula such as:

x(n+1) = x(n) + x(n-1)

So the value at the next step is equal to the value at the current
step plus the value of the previous step.
 
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Chris Johnson
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Re:Session Report
Malachi (#460386),

Heh.

Somewhere (perhaps in the designer's notes on his website), someone told the story of a game of Drunken Master where the rule was that the pip count had to be equal to a fibonacci number. They had problems with that rule, and your friend made it much, much harder.

Making super-hard rules is easy; making appropriately hard rules is more difficult. Hopefully your friend learned something.

Chris

 
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Malachi Brown
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Re:Session Report
jmilum wrote:

That was a hard one to figure out at 1am+. I guess you could consider
it a difference formula such as:

x(n+1) = x(n) + x(n-1)

So the value at the next step is equal to the value at the current
step plus the value of the previous step.


Yes, but to find an arbitrary nth value, you have to know two prior valid consecutive values.

As I mentioned, I thought it had to do with adding like colored pip counts together to get the count for the third, but there were examples that did that but didn't work. I had trouble making the connection between the different partial sequences available to me (1,1,2; 1,2,3; 2,3,5). I thought about "even, odd, odd" and primes, but there were counter examples for both of those ideas, etc.


fnord23 wrote:

Somewhere (perhaps in the designer's notes on his website), someone told the story of a game of Drunken Master where the rule was that the pip count had to be equal to a fibonacci number. They had problems with that rule, and your friend made it much, much harder.

Making super-hard rules is easy; making appropriately hard rules is more difficult. Hopefully your friend learned something.


I'm pretty sure I read those notes at some point. I've read just about everything Zendo related I could find. I know it's easy to make hard rules and hard to make rules that are right for the current group, but it's with these two guys that it's generally safe to break out the really tricky rules.

Two of us are programmers, and the third is an engineer. We've done some tricky stuff in the past using primes or the number of letters in the name of the color and it's generally possible to solve it.

When we get tripped up is when we fail to "get" the particular angle that the rule is based on. For example, the next day I ran a rule that the product of the red pip count total times the green pip count total had to be > 5. I lost, but only because neither of them made the leap that the product was important. One of them was stuck on sums and the other declared fairly early on that he didn't think it had to do with the sum or product of the pips, so he limited himself in that case.


Even with these issues, though, I still think it's a great game to play. I just wish I could play it more often.
 
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J
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Re:Session Report
Malachi wrote:
Yes, but to find an arbitrary nth value, you have to know two prior valid consecutive values.


Good point. How about this one:

F(n) = [(1 + sqrt(5))^n - (1 - sqrt(5))^n] / (2^n*sqrt(5))

That's Binet's Fibonacci Number Formula. I didn't know there was a formula for this. Strange what browsing a board game site will teach you

I wish I could convince my wife that I NEED just one more game so I could get Zendo.......
 
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Malachi Brown
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Re:Session Report
jmilum wrote:
How about this one:

F(n) = [(1 + sqrt(5))^n - (1 - sqrt(5))^n] / (2^n*sqrt(5))

That's Binet's Fibonacci Number Formula.

I wish I could convince my wife that I NEED just one more game so I could get Zendo.......

Yeah, umm... I don't think I would ever figure that out, no matter what time of day it is.

Sometimes I browse the The On-Line Encyclopedia of Integer Sequences and think about possible Zendo rules, but most of those aren't a good idea, and can usually be reduced to a set of numbers that can be determined by brute force.

If you can't swing Zendo, you can still try it with any set of objects that have a defined set of attributes. Some reduced set of LEGO blocks, for example, could be used to play a Zendo game. Just try to keep the defined set of attributes limited.


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