Pure maths, physics, chemistry, etc.: brain-training tasks that have nothing to do with trade [Part 2] - page 6
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Well OK, how do you handle my example? Which card will you lay down and how will it fall in the range of 0 to 1 1 to 6?
P.S. The magician only knows 4 cards and you only look at them. Obviously it won't work, since without information about the fifth, you're only coding 24 combinations anyway. Or don't you?
OK, how do you handle my example? Which card will you defer and how will it fall into the range of 0 to 1 1 to 6?
Which example? The four aces and the high king?
Lehko - the middle card is deferred, coded with the remaining number 1.
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In other words, the strategy is: if the cards are chosen tightly (in a narrow range from high to low) the middle card is chosen.
If the cards are very loose (the outer range is small) then any edge card is chosen.
There may be some variants with an attempt to "knock down the coding", but they all seem to be rejected.
// Mostly refuted by having either upper or lower card selected from outer range.
Waiting for a correct counter-example with an actual collision. I don't see it myself.
Maybe the clue is in the words? You could say diamonds seven, diamonds seven, diamonds seven and pathetically "diamonds seven". And that's two full bits.
Yes, he forbids it. What can I do - he's a moderator...
Well, then - bideby-bideby-boo
https://www.youtube.com/watch?v=Q-gWnGrIHig
Lehko - the middle card is deferred, coded with the remaining number 1.
How do you encode an ace with a one-iron? OK, 4 aces and a king is too easy as there are only 4 aces in a deck of 52 :)
Three aces (crosses, diamonds, hearts) and two kings (spades, diamonds). The middle one is, let's say, ace of crosses. That leaves the king of spades, the king of diamonds, the ace of diamonds, the ace of hearts.
How will the magician know it's not the ace of spades? What if it's the king of clubs?
The king of diamonds is 46, the ace of clubs is 49. So 3 is the difference, that's the code. Right? That is, the difference between the 2nd and the deferred, with the 4 cards sorted in order.
P.S. I'm starting to get it, I'm going to make a counterexample.
Roger: Maybe the clue is in the words? You could say diamonds seven, diamonds seven, diamonds seven, and pathetically "diamonds seven". And that's two full bits.
No, the words are interpreted the same way, no cheating, I wrote about it already.
Zero bit.
Counter-example: Cards 0 to 26 (27 cards), external range 27 to 51 (25 cards).
The cards are 0,1,2,25,26. From 1 to 25 - 25 cards > 24, the outer is also wider than 24.
Cards 2 to 24 will have to be coded, i.e. 23 values (codes are 0 to 22). That's where I bummed out. Bae.
And another thing: the outer range is empty (cards are 0,1,49,50,51). Which card should I set aside? Card 0 - and pass the trickster 0?
Most importantly, if you defer the smallest, the problem is that the magician cannot determine the inside range correctly, because he does not know which one the helper has deferred - the biggest one out of 5 or the smallest one. What to do?
The problem is exactly as I said: you have to remember to take the viewpoint of both, not just one, every time.
And here's a counterexample: outer is 0 to 5, cards are 6, 30, 31, 42, 43, again outer is 44 to 51.
The helper, seeing that the left outer range is the shortest, sets aside 6 and codes 6. The conjurer sees cards 30,31,42,43 and concludes that the inner range (30 to 43, i.e. 14 values) is obviously shorter than the outer range. And calculates 31+6=37. Bummer.
P.S. Don't tell me the helper will put off card 43.
Here, though, one way or another, the cards presented must be excluded from the numbering. That leaves 48 cards. That's a good number. And something can be done with it.
I have an idea working in this direction. There is always a pair of numbers, the distance between them does not exceed 12. We know how to transmit information with three cards about the codes from 1 to 6. The missing bit to get 12 should be given by the helper by choosing from these two cards. All that remains is to figure out how.