Pure maths, physics, chemistry, etc.: brain-training tasks that have nothing to do with trade [Part 2] - page 4
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Alexey, if the permutation coding does not solve the problem because one bit is missing, then this bit must be taken from the initial conditions (the first map declared by the helper).
I don't know the solution to the problem. I know how to solve it :)
In fact, the last card may give a bit to the trickster.
...
In fact, the last card may also give the magician the bit he is looking for.
This is more constructive, but the solution must be concrete. It should be clear from the solution what the helper should do and what the magician should do. What is obvious to the helper is not always obvious to the magician (the helper may genuinely think he has passed the bit on, but the magician may not realise it).
Any prior arrangements are acceptable, of course.
I know how to solve it :)
That's strong. It's like I know too, but I haven't yet found a flawless way of transmitting that bit that doesn't allow any ambiguity for the trickster. But I'm sure it's possible.
This is more constructive, but the solution must be concrete.
Alas, I can only design software :(
All initial conditions are in the problem. Specifically stipulated that there is no cheating.
Okay. Then it goes like this. Since it is the helper who decides which card to keep as the fifth card, he can pass on the last missing bit, for example, in this way:
1. If the arithmetic mean of the four declared cards is shifted up, relative to the "centre" of the deck - then the fifth card belongs to the "top" subset.
2. -------------------///-----------------------------//--------------------------------/-- вниз, --------------------------//----------------------//---------------------------"нижнему" ---//-------///------
I argue that from the set chosen by the viewers, the helper can always choose as the fifth [at least] one that matches that coding.
Waiting for a counter-example, ready to refute.
;)
Yes, I considered that option as well. Counterexample:
Consider the case where a subset of small cards has 3 cards, a subset of large cards has 2. Which card should I defer?
If the helper sets aside the small one, the mean may well be skewed towards the large one. Too bad.
If the helper sets aside a large, then the one remaining large might not be enough to make it heavier than three small ones. Also bad.
Waiting for a counter-example, ready to refute.
Yeah. While I was smoking, I constructed it myself.
for example 1, 2, 24, 47, 48.
When any card is selected, the balance either shifts to the opposite side, or the set is strictly balanced (which again leads to a bit missing).
But there's something to it, maybe you just need to change the wizard to a stochastic... :))
Are your cards numbered 1 to 52? Let's agree so there's no confusion.
It's just that mine are 0 to 51...