[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 611
You are missing trading opportunities:
- Free trading apps
- Over 8,000 signals for copying
- Economic news for exploring financial markets
Registration
Log in
You agree to website policy and terms of use
If you do not have an account, please register
Right, A>A. This is the Reshetian transitivity.
I will tell you how I approached the solution.
First I decided to lower the dimensionality of the problem. Let all the numbers on the faces of the cube have a pair. That is, the cube is described by a triplet of numbers.
I got an unexpected surprise. For any pair of such cubes one of the cubes always gets an advantage due to odd number of combinations.
Started to look further. Stumbled upon the "non-transitivity" of the pairs of dice. That's when the Megamouse loses to the client. And then, after they swap dice, keeps on losing. Non-transitivity is caused by the rules clause: "in case of a tie, Megamogg loses." We solve the problem cardinally: no equality. Define that the sets of numbers on faces of the cube must not overlap.
Dimensionality of the number of possible combinations falls below the plinth.A few attempts, and, we get a solution:
A( 2, 2, 5), B( 1, 4, 4), C( 3, 3, 3). This is the minimum solution. You can get many more solutions by elementary shifts (there is also a number 6).
And there's also a limit of 15 words in the whole judgement.
blue italics is mine
The question should be in the form of either "(A and !B) or (B and !C) or (C and !A)".
or in the form " (A or B or C) and (!A or !B or !C)".
I will post the checks later.
Can you elaborate on this point?
We exclude multiplicities of the number of combinations. The cube is now described by a triplet of numbers. The multiplicity factor of the combinations is 4. There were 36, now there are 9. That leaves 36. Only the original 9.
The oddness is explained.
What is the advantage? That the odd can only be divided into unequal parts?
Is there no transitivity anymore?
Incidentally, it can be useful for brevity (in a problem about a dumb, armless and stupid guard who can't understand a judgement in more than 15 words). Really, the example is from a problem about a TV set.
The judgement "(A AND X) XOR (~A AND ~X)" can be simplified:
Since ~A = A XOR 1, thenA*X XOR (A XOR 1)(XOR 1) = A*X XOR (A*X XOR A*1 XOR X*1 XOR 1) =
= (A*X XOR A*X) XOR (A XOR X) XOR 1 = (0 XOR 1) XOR (A XOR X) = ~(A XOR X)
A = You're a liar
X = You have a telly
True with TV: ~(FALSE XOR TRUE) = ~TRUE = FALSE -> will say FALSE.
True Teller without TV: ~(FALSE XOR FALSE) ~FALSE = TRUE -> will say TRUE.
Liar with TV: ~(TRUE XOR TRUE) = ~FALSE = TRUE -> will say FALSE.
Liar without TV: ~(TRUE XOR FALSE) = ~TRUE = FALSE -> says TRUE.
Is there no transitivity anymore?
I am aware of that. Is there a transitivity or not?
Yes, the result is paradoxical, but it's there. The "<" sign means "worse", but it's different every time and has a different meaning.
(2,2,5) <_1 (3,3,3)
(3,3,3) <_2 (1,4,4)
(1,4,4) <_3 (2,2,5)
Mislaid, thank you!
P.S. Justification removed. Anyone can do it, knowing that the probability of any facet falling out is 1/6.
I would solve this problem differently.
1. It is necessary to eliminate inequality of conditions for the players - there should not be equal results. Proceeding from the principle of maximum freedom of choice, this means that as a reference model the variant of numbering the faces of cubes is taken: 1 and 6, 2 and 5, 3 and 4.
2. There is a middle - 3 and 4, it is necessary to make the preceding variant something worse, and the following one - something better. This "something" may be the same - the probability in the first trial, for example.
3. Now it is necessary to "flip and glue" (Mebius), in other words, a completely different criterion is needed. Megamouse plays all the time (see problem conditions). Shall I continue? :)
I would solve this problem differently.
1. It is necessary to eliminate inequality of conditions for the players - there should not be equal results. Based on the principle of maximum freedom of choice, this means that as a reference model the variant of numbering the faces of cubes is taken: 1 and 6, 2 and 5, 3 and 4.
2. There is a middle - 2 and 5, it is necessary to make the preceding variant something worse, and the following one - something better. This "something" can be one and the same - the probability in the first trial, for example.
It has already been said about the first point, everything is clear here. There is no need for the Megamook to artificially lower its chances.
With the second one, it's not all clear - especially if we increase the number of degrees of freedom from 3 to 6.