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30 replies to this topic

#21 EventHorizon

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Posted 01 June 2012 - 03:30 PM

Spoiler for I said guess order doesn't matter, but that should be clarified.

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#22 EventHorizon

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Posted 02 June 2012 - 01:59 AM

Spoiler for Another way to frame the question

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#23 augustinus

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Posted 02 June 2012 - 03:21 PM

do we must answer with either red or black? can i answer with blue or some other color that will be false nevertheless, it will make the number of trials become 26 for better or worse
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#24 plainglazed

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Posted 02 June 2012 - 05:00 PM

I believe the answer to be one. She states in the puzzle that she cut the deck in two. It dose not mention that she shuffled the deck. It is well known that all new decks of card come in suits, alternating betwwen red and black. with the ace of spades being the bottom card. Since you have only two suits in each deck, you should start with red count out 13 cards and then switch to black for the last 13. If she shuffled the answer would be close to 26 to the 26th power.


hey bub, watch where you're going. i mean, welcome to BrainDen. get your point, should have probably stated she shuffled the deck. this puzzle was kinda a continuation of a previous one so one could presume the deck to have been previously used.

Spoiler for ?playing with half the deck missing


all true, save for that last bit. had figured this to be the logical starting point

this is my first Post in the forum :thumbsup:

Spoiler for first try


hello Assassin.MTS - ah yes, you've pointed out the dilemna quite well and welcome to the Den.

do we must answer with either red or black? can i answer with blue or some other color that will be false nevertheless, it will make the number of trials become 26 for better or worse


and @ augustinus - welcome to you as well. had originally intended only black and red as responses but am curious as to your method mentioned above.


sorry to have apparently abandoned this thread. have had a tough end of the week. didnt quote WoS above and am still working out Phil's extension of your reply. when first working out this puzzle, seemed to think that in the worst case, a comprehensive binary reduction would still take 26 trials. am rethinking/still thinking. and at EventHorizon, not a programmer but hope to try to follow your code to hopefully gain some insight into how that works.

maybe should have worded the question differently and asked for descriptions of methods that did better than 27 trials rather than asking for the minimum number of trials. probably still would have gotten to this same place fairly quickly - i.e. in way over my head.
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#25 EventHorizon

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Posted 03 June 2012 - 03:28 AM

I can beat bub's 1. Just ask, "wait, how many cards are there to guess for now?" And it would be 2 to the 26th power... assuming the only possible replies were "twenty-six" or "you didn't get all of them." That problem would be like a blind person trying to solve a rubik's cube, and asking "is this it?" and the other person saying "nope." (Was it the movie UHF that had that?)

all true, save for that last bit. had figured this to be the logical starting point

Agreed. I was actually writing a similar post when I noticed Time Out already did the job before my post was ready. I thought about commenting on bonanova's "factorial or exponential" comment in that post... something about needing to confiscate an EH silver star :)

and at EventHorizon, not a programmer but hope to try to follow your code to hopefully gain some insight into how that works.

I used the binary representation of an integer to hold the binary string guesses, and simply brute force went through each possibility. That's why it is too slow with 26 cards when going through each possible guess's value for each possible target binary string when looking for the best one (though I could speed that up if I get motivated to). I do use MIT's HackMEM method for counting the number of bits (to speed that part up), which is pretty cool. Just figured out how that works. So nothing else all that insightful in the code itself.
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#26 WitchOfSecrets

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Posted 04 June 2012 - 04:35 PM

What if our search idepends on the previous responses somehow? If we hear that there are X black cards, can we use that information to modify our strategy?
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#27 EventHorizon

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Posted 05 June 2012 - 02:19 AM

What if our search idepends on the previous responses somehow? If we hear that there are X black cards, can we use that information to modify our strategy?

I assume this is in reponse to my statement about order doesn't matter. I did explain that statement in post #21. The search does depend on the set of previous guesses and responses (just not on the order of them to get to that point) to decide the next sequence of cards to guess. So if we guess all red, and we get a response (which tells us the total red and total black cards)... the search should definitely be different based on the response given (though perhaps not initially in that exact scenario... I'm not sure yet).
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#28 Time Out

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Posted 05 June 2012 - 08:01 PM

Spoiler for this is for self improvement only. you guys are way ahead of me in the logician stakes:) ?.27 or less

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#29 plainglazed

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Posted 06 June 2012 - 07:13 PM

So sometimes/more often than not you would do better than 27 trials but no guarantee. This is actually more along the lines of what i was thinking of to reduce the number of trials. Can fairly simply guarantee in at most twenty one trials, the order of the twenty six cards. As for EH's 14-ish, starting to get a better grip on the ideas but as far as my high speed population counter - am still processing...
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#30 phil1882

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Posted 07 June 2012 - 07:36 PM

Spoiler for reducing the possibilities

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