Bongard Problems

[TheGreatErenan]

Well, I'm not, like, the boss or anything, but in my opinion, pretty much the only rule that really matters is that the correct solution should be plainly the correct intended solution more than any other potential solution.

For instance, with this one:

Group A:

banana
color
belvedere
hindsight
humbug
pascal

Group B:

tricycle
balrog
minus
random
wellington
kangaroo

There is the trivial solution that Group A contains words that are members of the set {banana, color, belvedere, hindsight, humbug, pascal}, whereas Group B contains words that are not members of that set. But obviously if you make up a puzzle with a solution like that, then you're kind of missing the point.

So I'd say, yes, you can have multiple parameters, but I would advise caution. Here are some notes from http://www.foundalis.com/res/invalBP.html:

• A note about "if ... then ..." rules: Strictly (and logically) speaking, these rules should be allowed. It turns out, however, that such rules can be very tricky and counter-intuitive for us, mere mortal, non logico-mathematically oriented humans. For example, consider the seemingly simple rule: "If there is a triangle then it is isosceles". So what if there is no triangle in one of the boxes at left? Apparently the box would satisfy the rule (since the premise is false), so it should be placed on the left side. We could then have a Bongard problem showing just one isosceles triangle in one of the six boxes on the left, with the other five boxes containing irrelevant (non-triangular) shapes, and all boxes on the right containing at least one non-isosceles triangle (among other shapes). Such a problem would be extremely hard to solve. Besides, it would be possible for one to postulate weird "if ... then ..." rules for most of the hard problems which contain a different, elegant solution, and thus render them trivial. So, I believe, "if ... then ..." rules should be avoided, not because they are logically impossible, but because they do not comply to the spirit of BPs, which is to probe common-sense intelligence.
• For reasons similar to the ones explained above, rules consisting of long sequences of disjunctions ("... or ... or ...", etc.) should be avoided. Again, one can turn the most elegant problems to trivial ones in this way. Two clauses, however, ("... or ...") do not seem to be an extreme thing to consider.
  

[[PETA] Doodat]

In words starting with A, only the letter A can be 3 in a row (AAA), and it must end in the letter O.
In words starting with V, only the letter V can be 3 in a row (VVV), and it must end in the letter X.

How is this?

[TheGreatErenan]

(03-13-2014 10:04 AM)[PETA] Doodat Wrote:  In words starting with A, only the letter A can be 3 in a row (AAA), and it must end in the letter O.
In words starting with V, only the letter V can be 3 in a row (VVV), and it must end in the letter X.

How is this?

Well, it is true that every member of Group A satisfies those rules, and every member of Group B does not. However, there is a much simpler rule that I had intended as the solution. So here, let me add some more members that violate your rules but which satisfy mine.

Group A:

AAVAO
AAAO
VAVAX
AVAVAO
VVVAX
VAAVAVX
VVVAAAAO
AVX


Group B:

VAAAX
VVAAO
VO
AVAX
AAAVVVO
AAVAAAVX
AAAVVVVO
AVVO

[Flarp55]

If there are more As than Vs then it must end in O. Otherwise it must end in X.

[[PETA] Doodat]

Derp. Nice job.

[Flarp55]

(03-13-2014 11:31 AM)[PETA] Doodat Wrote:  Derp. Nice job.

Actually I had gotten the same answer as you to start with. I only realized it after he had posted the extra answers.

[TheGreatErenan]

Yup. That's it.

Group A: The word ends in O if and only if there are more A's than V's.
Group B: The word ends in O if and only if there are at least as many V's as A's.

[.Memories.]

More please?

[Demon]

Here's one for math types, though this might be too much

Group A
1331
1943
5337
1703
0547

Group B
6553
4697
2690
1573
6479

[TheGreatErenan]

The only thing I can think of, though I really doubt this is the correct solution, is that if you treat them as hexadecimal numbers and convert to binary, then the number of 1's is divisible by 3 in Group A but not in Group B.