Hello Starship Captains!
Introduction
Playing Zendo with my family gave me a new appreciation for the skill needed as a Student and a Master alike to formulate a rule correctly. I noticed some quirks about Zendo that I like to share with you.
This post will discuss the meaning of the Zendo Winning Rule.
By playing Zendo, I noticed the ability of students to come up with rules that a Master couldn't even remember. In other words, the student's rules would be more difficult than any rule a Master would come up with, or dare to come up with. This led to considerations of simple rules that are possibly equivalent to difficult rules. And thinking about this problem, I stumbled upon a hidden aspect of the Zendo Winning Rule.
The Zendo Winning Rule
The Zendo Winning Rule governs when a Student wins. It can be stated in several ways.
Zendo Winning Rule. A Student wins if at their turn they guess the Buddha Nature and
1. the Student's guessed rule is logically equivalent to the Buddha Nature (or Master's Rule).
2. the Master can not disprove the Student's guessed rule.
Note that if 1. is the case, then 2. is the case. But not vice versa. I prefer formulation 2. over 1. for reasons I will show in this post.
Summary of Game Play
If a Student has a guessing stone at the end of their turn, they can guess the rule. Once the Student and Master agree on the formulation of the rule, it is clear that formulation 1 of the Winning Rule will never be a problem to decide, given that it is possible to compare rules formally. So, as your skills in Zendo and logic improve, it is possible to always decide whether two rules are logically equivalent or not. If they are, the Student wins. There doesn't seem a problem with this Winning Rule, but there is.
If the rules are not logically equivalent, the Master must disprove the Student's (guessed) Rule and to this end he has two options:
1. Make a koan that follows the Student's Rule but does not have the Buddha Nature. The Student would expect it to be marked white, but in fact the Master will mark it black.
2. Make a koan that does not follow the Student's Rule, but has the Buddha Nature. The Student would expect it to be marked black, but in fact the Master will mark it white.
If the Master can not disprove the Student's guessed rule, it must be equal . . . and the Student wins. (This looks like a hands on way to prove that rules are equivalent, so there is no need to be skilled in formal logic.)
However, I found certain sets of rules that are NOT logically equivalent, yet can not (under any circumstance) be disproved by the Master. So, it would seem that Winning Rule 1, although sufficient, is not necessary to win! If the Master fails to disprove the Student's guess, then by Winning Rule 2, the Student wins nevertheless.
At this point an example could come in handy. See the next section.
Attributes
We know some of the attributes of pyramids, used in formulations of the Buddha Nature: color, size, groundedness, orientation, etc.
Now, for any attribute A, consider the rule schema S(A): "All pyramids in the koan have attribute A."
Example. Consider the attribute A = red, Then S(A) gives the rule: "All pyramids in the koan are red." It is easy to decide of any koan whether or not it follows this rule.
Example. Now consider the attribute A = ungrounded. Then S(A) gives the rule: "All pyramids in the koan are ungrounded." We run into a problem finding a koan that follows this rule . . .
It is quickly realized that in any koan (played on Earth), there is at least one pyramid touching the table, and hence is grounded. At least within the confines of the game rules (pyramids are not allowed to touch anything else but pyramids and the table), there is NO koan with at least one pyramid that ever will follow the Buddha Nature. The Master would have a hard time setting up the initial koans to begin with! It is impossible for any koan to follow the just stated. The only koan that follows this rule is the empty koan (consisting of no pyramids at all). A Student guessing this odd rule will be able to win with a guess of the form "the koan contains no pyramids". The Master would not be able to make a counter example, and the Student wins.
Example. Consider the attribute A = weird. Then S(A) gives the rule: "All pyramids in the koan are weird."
In this case it IS possible to make a non-empty koan following this rule, but it requires a minimum number of pyramids. The question is how many exactly. It's larger than 4, because 4 pyramids can not be configured to even form a closed loop (see Fig. 1). In this case, an interesting question arises.
Suppose the minimum required pyramids is N. Then the rule is equivalent to "the koan has at least N pyramids and all koans are ungrounded." Now the problem is this. How would the Master know the correct value of N? This could be a difficult--even unsolved--geometrical question. So, if the minimum is in reality 5, but the Master can only make koans of at least 6 pyramids, the Student will win with the rule "The koan contains at least 6 pyramids and all pyramids are ungrounded." The possibility of the Student winning because of lack of koan building skills on the Master's side, is therefore real.
We are now faced with the fact that the rules the Students can come up with in order to win, are not necessarily logically equivalent to the Buddha Nature, NOR do they even have to be correct from a theoretical standpoint!
Too Few Pyramids
A rule schema like S(A) also poses another problem. With a given set of pyramids, there is sometimes only a finite number (although a very VERY large number) of koans that can be built that follow the Buddha Nature. Consider the Buddha Nature: "All pyramids in the koan are red." In a standard Zendo set, there are exactly 15 red pyramids. Therefore, the rule is equivalent to "The koan exists of no more than 15 pyramids, which are all red." Obviously, this rule is NOT logically equivalent to the Buddha Nature. A student guessing this rule will win, because the Master simply doesn't have more than 15 pyramids, which he would need to disprove the rule.
Conclusion
We see that the first winning rule isn't sufficient to allow the game to end, because there are rules the Master can think of for which he can not (is not, under any circumstance, able to) disprove any Student's guess which is not logically equivalent to the Buddha Nature. And in that case, the Student should win . . . if only for the fact the rule is too difficult, even impossible, to guess.
Therefore, the Zendo Winning Rule: "The Student wins if the Master can not disprove the Student's guessed rule" is the preferred and reasonable rule which will allow Students to win, although sometimes with rules that are NOT equivalent to the Buddha Nature, NOR always correct from a theoretical standpoint!
Isn't that the true Zen nature of the Buddha Nature?
This, I'd like to add, will only add to anybody's fascination for Zendo.
Epilogue
Can you find a koan which contains exactly and only 5 weird pyramids? The best I could come up with was 7 before I wrote this post, but today I reduced it to 6. (See Fig. 2). It is clear you can find koans that have only weird pyramids of any number bigger than 6, by increasing the circle. (See Fig. 3).
See also my next post on solving some of these problems, and perhaps making Zendo more interesting, if only for advanced Zendo players.
Tags:
Formulation 1 of the Winning Rule could be repaired by changing "logically equivalent" to "equivalent for all constructable koans". Depending on how we define the logical space of the game, there may be logically possible koans that are not actually constructable.
Yes, I agree. "Equivalent" would then mean: "uses the same color of marking stone". Note that your formulation "equivalent for all constructible koans" is now equivalent to formulation 2. This is seen as follows.
If the Student's Guess G and the Buddha Nature B correspond for all constructible koans k, i.e., G(k) = B(k) := the color of the marking stone, then the Master can not disprove G, since he can only make constructible koans.
If the Master can not disprove G, then for any koan k with G(k) <> B(k), koan k is not constructible (otherwise koan k would be a constructible counterexample), and therefore, G and B coincide for all constructible koans.
Thank you P.D.M.
Hi P.D.M.,
considering your correct observation, later I remembered that the whole point of formulation 2 was to include the factor of the (limited) Master's ability/knowledge. It took me a while to crunch this mind boggler. Here's my fix.
When I read your response, I assumed that the set C of constructible koans is known, or at least known to the Master. But this is not so. The set C(p), the set of constructible koans by person p is a subset (or a strict subset) of C. With p = M := the Master, C(M) stands for the constructible koans the Master can come up with. As we have practical time limits when we play Zendo, we have to work with the skill level of the Master at that time.
So, if the master can't disprove the Student's guessed rule, that does not mean that there isn't such constructible koan in C, just that the Master can't find it and therefore is not in C(M).
To illustrate my point, here are two versions of my previous proof. Note that both are valid proofs!
Version 1 - C
This version is closest to your proposal, but seems unusable for the fact that C is unknown.
This proves the equivalence of [the Student's Guess G and the Buddha Nature B correspond for all constructible koans k in C] and [the Master, even one with perfect knowledge, can not disprove G].
Version 2 - C(M)
This version seems more practical and is closer to the spirit of my original proposal.
This proves the equivalence of [the Student's Guess G and the Buddha Nature B correspond for all constructible koans k in C(M)] and [the Master, assumed without perfect knowledge, can not disprove G].
Note that in this case, there might be a k' in C \ C(M) [where C \ C(M) is the set of those constructible koans not known by the Master] for which G(k') <> B(k'). But since k' is not in C(M), it is not "constructible" by the master.
This is what I meant with my remark that the Student can win with an incorrect rule from a theoretic perspective. Only if the Master were to have perfect knowledge, can the Student win with a rule that is equivalent on C, not just on C(M).
Observation
During Zendo, the interesting situation can thus arise that when the Master admits he can not find a counter example to the Student's guessed rule, that Student wins. However, when the Master subsequently reveals his (not equivalent) rule, a second student might find a koan k' not in C(M) that disproves the Student's guessed rule. It can be argued that this second student was put at a disadvantage, as he might have guessed a "more correct" version of the rule later in the game, had it continued, if only C(M) included k'.
This situation would qualify as a Master's mistake. Can we say that the second student wins, rather than the one that won at first? Do we decide on two winners? Does the Master win for declaring the second student winner? Remember, there are no losers in this game!
Zen, zen, zen . . .
Zendo is the most stimulating game I've played in years! I'm so looking forward to playing it again!
The subject of the OP came up in the final round of the Zendo tournament at Origins a number of years ago. It was the one with myself, Pace Reagan and Dan Isaac all vying for the top prize with the inscrutable Eric Zuckerman as master. (I want to say it was 2002, but I could be off by a year or two.) One of Pace's guesses was something along the lines of, "AKHTBN iff, when you ignore the blue pieces, the number of non-upright larges is equal to the number of non-large uprights." Of course, requiring the count of non-upright larges to be equal to the count of non-large uprights is equivalent to just "larges equals uprights." e.g. Let's say you have a koan that contains a large upright, a large flat and a small upright (noe of them blue). Pace would say 1=1, so the koan should be marked white, while Eric would say 2=2 so it should be marked white.
So yes...
1. The student's guess and the master's rule would mark every koan the same
2. The master couldn't provide a counterexample.
Ahh... memories.
BTW, here's my contribution to the "all weird" discussion, courtesy the Wayback Machine.
http://web.archive.org/web/20050308110146im_/http://home.att.net:80...
The two in the middle column are pretty unstable - don't bump the table. :)
Thx Ryan.
I'm delighted that this really happened and grateful you took the time to share this with us, evidencing that my thoughts weren't that silly after all. The "of course" in your example took me some minutes to see through. But then it became clear that:
If L is the set of larges, and U is the set of uprights (both not blue), |.| denotes the number of elements of a set, & denotes intersection, + denotes union, then
U + L = (U & not(L)) + (L & not(U)) + (U & L).
A Venn-diagram shows this even clearer, but I did not take the effort to draw it. Of course, it follows that
|U + L| = |U & not(L)| + |L & not(U)| + |U & L|
The statement that both mentioned rules are equivalent boils down to saying:
Lemma. |U & not(L)| = |L & not(U)| eq. U = L.
Proof. This is true, because
U = (U & not(L)) + (U & L) = L = L & not(U)) + (U & L) eq. U & not(L) = L & not(U)
by subtracting the equal set U&L on both sides. Hence, also
|U| = |L| eq. |U & not(L)| = |L & not(U)|.
EOP.
This in turn shows that we may not expect that this is clear to all people from the outset. That's why the language of set theory comes in handy here, to settle the question once and for all. And also that in such cases, we must allow for the student to win, if the Master can't find a counterexample. In most cases, it is probably true that the rules ARE (logically) identical.
I am also impressed with the 5 different koans you showed with just 2 pyramids of various sizes, and all satisfying the rule "all weird". Wow.
In any case, I'm looking forward to learning more of this kind of "confusing" rules that turn out to be equivalent to seemingly other (and perhaps simpler) rules.
Again, I consider your contribution very valuable.
I found another rule that is problematic in such a way as to give rise to rules that are logically NOT equivalent, but the Master can not give a counter example. Consider the following rules.
1a. "All pyramids in the koan touch one another."
1b. "All pyramids in the koan touch all other pyramids in the koan."
1c. "For all pyramids p and q in the koan, if p is not q, then p and q touch each other."
(These are representatives of the same intended rule. We discuss their difference below.)
2a. "The koan does not contain more than 3 pyramids AND rule 1c."
2b. "The koan does not contain more than n pyramids AND rule 1c." (n > 3).
--
Rule 1a. may be considered vague if it comes to the meaning of "touching one another" in the case of 1 pyramid. It is suggested that touching involves at least 2 pyramids. But if we exclude koans of 1 pyramid, the rule lacks a certain elegance.
So, to include a 1-pyramid koan to always obey this rule, rule 1b. is an attempt to correct the problem of rule 1a. But rule 1b. is vague in that it doesn't specify what is exactly the "all other pyramids", since we already talk about "all pyramids" doing something, so what are "all other pyramids"? The grammar seems problematic here.
So, my final attempt, using quantifiers such as "for all" and "there is (at least one)" and a conditional statement, seems to solve the awkward nature of rules 1a. and 1b. Note that if a koan has only one pyramid, then the condition in the if statement is not true (because q = p), and therefore, logically, the statement "if p is not q, then . . . " is true.
Note. These considerations show again how important it is to formulate your rules very precisely as to prevent confusion later on, when the game draws to a close and counter examples for guesses hinge on the details in the formulation. Some Zendo cards use subtle suggestions for "at least one", but don't spell it out. If you can, make sure to "translate" the rules into a correct but precise formulation.
--
Now suppose that the Master chose rule 1b. and that the student guesses that it is rule 2a. The Master can then give a counter example. (Exercise. Please, post your counter examples.)
But for what n > 3 (rule 2b.) would it be impossible for the Master to give a counter example? This question may not be so easy to answer. The intricate geometry of the pyramids is making it difficult to oversee all possible ways they can touch.
For the minimal n (and for all n greater than that) for which it is impossible to find a counter example, rule 2b. is equal to rule 1c, even if they are NOT logically equivalent.
So, again, I've shown that a student can win with a rule that is not logically equivalent to the Master's rule.
I've also shown that there are potentially an INFINITY of rules that are all logically different from each other, yet a Master can not give a counterexample to any of them.
To clarify: I don't think this a flaw of Zendo, but rather an interesting observation about the Zen nature of the Buddha nature. The above example again justifies my choice for the second Zendo Winning Rule in favor of the first.
Note 1. The empty koan is also included in the possible koans that satisfy rules 1c., 2a., and 2b.
Note 2. If the Master knows the minimal n (say it's n*) that makes rule 2b. and rule 1c. equal, he can take the rule: "The koan does not contain more than (n* - 1) pyramids AND rule 1c." This prevents rule 1c. to be correct, and students have to find the particular n*. My next project will be to find a rule that has a fairly large n*. It's because I am curious where this is leading, not because I am in need of difficult rules . . .
Note 3. Derivative rules that use these considerations are: "Any pyramid touches only and all pyramids of its color." Consequently, a very difficult rule can be formulated that uses an analogy of rule 2b. for this and in addition different n for different colors. If your mind is now in "overload" mode, just forget about this one :-)
Note 4. If I play Zendo with my friends, I am making sure that everybody wins. My last game involved 3 other players, and all of them won. We played 5 games in 2 hours. The first game took 1 hour to get started; the learning curve was steep. The second game took 1/2 hour, and it seemed everybody got more interested. With another half our to go, we started a third game. It was solved in 13 minutes, so there was room for a fourth game and that took 8 minutes. Record times! The last game took also 8 minutes and if I didn't have another appointment, all of them would have wanted another game! Most notably, the player who at first thought it was impossible to guess the correct rule (we used the Zendo rule cards), won 3 times and was totally mind blown!
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