Pure maths, physics, chemistry, etc.: brain-training tasks that have nothing to do with trade [Part 2] - page 7

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There you go, I thought I'd made up my mind, but it turns out I haven't. Didn't think of the magician again :)

Man, he's dumb, the magician...

 

Yes. Loaded...

You can use four cards to show a number from 1 to 24, and you have to show half of the deck somehow (let's say top/bottom).

The helper has options:

vvvvv

nvvvv

nnvvv

nnnvv

Nnnnnv

nnnnnnnn

It can be set off as follows:

vvvvv in

nwvv c

nnvv in - 1. open upper ones are more or equal to lower ones, so the upper one is postponed

and nnnv - 3. and here upper ones are more or equal to lower ones, but the lower one is postponed... it doesn't work.

nnnnv - 2. the lower ones are smaller in the deposited ones, it means the lower one has been deposited.

nnnnnn

.......................

Maybe by the position of the edge cards of the helper. If the top one is closer to the top edge than the bottom one, it means it is deferred from the top of the deck... But it's just a thought, you have to think about it carefully.

 

let's say there is a deck in which each card is assigned a number, so we have numbers from 1 to 52

The helper tells us four cards that can also be arranged in a row according to their numbers. (let's say it was cards 5,4,9,13)

Replace it with the numbers 1,2,3,4. That is, the smallest card - 4 - assign 1, 5 - 2, 9 - 3, 13 - 4.

The helper has the right to give out any sequence by saying these 4 cards.

building a tree

Look at the bindings 1,2 in what order they stand, if 1-2 then discard the top 26 cards, if 2-1 then discard the bottom 26 cards.

Further we look at the position of 3 in relation to the sequence 1-2, it can be 3-1-2, 1-3-2, 1-2-3. it means we can divide the rest of the 26 cards into 3 more sets.

Next, look at the position of 4 in relation to these three. it could be 4-3-1-2,3-4-1-2,3-1-4-2,3-1-2-4, 4-1-3-2, 1-4-3-2, 1-3-4-2, 1-3-2-4, 4-1-2-3, 1-4-2-3, 1-2-4-3, 1-2-3-4, that is still possible to divide into subsets. this is enough to divide even 9 cards into separate cards for this series. still remain.

 
I think the main thing is which card out of the five to choose to guess, so that it reduces the number of choices by half. Then 24 codable values are enough. Maybe the fifth card should have the minimum distance of 0...51. But if you take a distance in this system, then if for example cards 0,1,2,3,51 will fall out and putting aside card 3 will reduce the number of trials to 48 values, while we need up to 24. Therefore the distance must be taken in a "looped" system. That is, as on a clock. Then the distance between the zero and the 51st card = 1. In that case the worst arrangement would be like 0,1,2,3,X. But even in this case, depending on the chosen card 0 or 3, the number of choices is halved to 24 choices
 
has anyone considered a matrix to solve this trick? )
 

6*8=48

vertically it needs 2 bits and horizontally it needs 2 bits, just need a quick calculation (ZS: I have 4 instead of 2) ))

1 2 3 4 5 6 7 8

1

2

3

4

5

6

ZS: got it stupid )

 

about the cards... well, the suit is relatively simple... among 5 cards there will be 2 of at least one suit... the assistant starts to tell the cards from there...

the only thing left is to encrypt with 3 cards the value of the remaining card....

if a card is the highest - then let's call it 1standard high of 2 - and let the combination of the remaining 3 cards show 1-6 variations of the needed value :-)

like this :-) seems to describe everything correctly...

 
Aleksander:

about the cards... well, the suit is relatively simple... among 5 cards there will be 2 of at least one suit... the assistant starts to tell the cards from there...

the only thing left is to encrypt with 3 cards the value of the remaining card....

if a card is the highest - then we roughly call it the 1st - one-suit highest of the 2 - and the combination of the remaining 3 cards will indicate 1-6 variations of the needed value :-)

like this :-) seems to describe it correctly...


The suit is clear, but the rest can be exemplified by

we have 6 . 7 . 8 . 9 .10 all of the same suit

 
Mischek2:


The suit is clear, but the rest can be exemplified by

we have 6 . 7 . 8 . 9 . 10 all in the same suit.


Googled it http://www.141600.ru/blog/739/entry-107932-фокус-с-картами/
 
Avals:

Googled it http://www.141600.ru/blog/739/entry-107932-фокус-с-картами/


don't want to look yet

the suit is the first thing ready to be answered, the rest are cuckoo

Leha seems to be on holiday and will give us all a shit about it ))

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