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Ecoris
Promising
Supreme Hero
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posted May 16, 2009 03:21 PM |
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I don't get it. Do you mean 10/9?
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted May 16, 2009 03:34 PM |
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What part don't you get? I mean 1.111111111111... like a "1" followed by an infinite amount of "1"s after the separator.
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Ecoris
Promising
Supreme Hero
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posted May 16, 2009 03:45 PM |
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That much is obvious. I'm aksing whether you want that to equal 10 ot 10/9?
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted May 16, 2009 04:12 PM |
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Edited by dimis at 16:16, 16 May 2009.
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Cover time on a complete graph Kn
Quote:
The expected number of steps needed to visit all N vertices of a complete N-graph is (N-1) + (N-1)/2 + (N-1)/3 + ... + (N-1)/(N-2) + 1.
True. I wanted though one more step, since the harmonic numbers appear out of nowhere:
(N-1) ( 1 + 1/2 + 1/3 + ... + 1/(N-2) + 1/(N-1) )
which is equal to
(N-1) H_{N-1}
where
H_{N-1} = 1 + 1/2 + 1/3 + ... + 1/(N-2) + 1/(N-1)
is the (N-1)-th Harmonic number.
May be some interesting references are:
Harmonic Number (equation 13 in the link gives a good approximation)
Euler-Mascheroni constant
Ok, time to go. Have a nice weekend.
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted May 16, 2009 05:26 PM |
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Quote:
That much is obvious. I'm aksing whether you want that to equal 10 ot 10/9?
Oh, I want it equal to 10.
Edit > Otherwise, it wouldn't be much of a chalenge to show, would it? 
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eCoRiS
Promising
Supreme Hero
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posted May 20, 2009 02:09 PM |
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Quote:
Oh, I want it equal to 10.
My mind is definatly not suited to solve such a problem 
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted May 22, 2009 01:30 PM |
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Oh come one, you've solved much more difficult problems than this one. You just have to think a bit out of the box and consider under which circumstances the equation can be true to reach the solution. Because obviously there's something fishy as it stands.
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted May 22, 2009 07:11 PM |
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R u looking for this?
1.(1) = 1.111111111111... = 10 (mod 80/9)
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TheDeath
Responsible
Undefeatable Hero
with serious business
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posted May 22, 2009 08:15 PM |
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In binary, 1.11111111... = 10 (means 2)
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The above post is subject to SIRIOUSness.
No jokes were harmed during the making of this signature.
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted May 23, 2009 07:51 AM |
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The New York Times had a tasty puzzle last week, I'll reprint it here. (what is it exactly that makes a puzzle tasty I wonder?)
A magician calls up an audience member to the stage. "Pick 5 cards out of this standard deck, randomly if you want. Then take them to my assistant." After the audience member takes the cards to the assistant, the assistant picks one and tells the audience member to put it in his pocket. Then the assistant arranges the other 4 cards and tells the audience member to bring the 4 cards back to the magician, place them face down on the table in front of the magician, and one-by-one, pick them up and announce them to the magician. After this the magician announces the card that is in the audience members pocket, guarenteed.
How does he do it?
This is math btw, no trickery 
I'll be gone for awhile but I'm pretty sure somebody'll figure it out.
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dimis
Responsible
Supreme Hero
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posted May 23, 2009 08:13 AM |
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nice and right on time!
/back-to-work
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted May 23, 2009 02:40 PM |
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Quote:
In binary, 1.11111111... = 10 (means 2)
Exactly.
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Ecoris
Promising
Supreme Hero
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posted May 24, 2009 02:59 PM |
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Does the assistant look at the cards?
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted May 24, 2009 06:04 PM |
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That's what I understood. And then the magician tells the number only, not the suit as well.
Otherwise, I have to rethink.
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Ecoris
Promising
Supreme Hero
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posted May 24, 2009 08:30 PM |
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Edited by Ecoris at 20:31, 24 May 2009.
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Lets say the cards are numbered 1,2,3,...,52; that doesn't change the problem.
Calculate the sum S of the five chosen cards. Write S = 5Q+R, where R is 1,2,3,4 or 5. Put the Rth lowest card of the five cards in the pocket. Let {a,b,c,d} be the remaining four cards. There are at most 16 values of e such that {a,b,c,d,e} result in {a,b,c,d} by this procedure (16 might be not be the optimal bound). But there are 24 ways to order the four cards a,b,c and d. As such, to each of the (at most) 16 possible values of e we can assign a unique ordering of a,b,c and d.
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted May 25, 2009 06:28 AM |
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I had a similar idea. The four cards that are left are treated like bits.
Now, there are 4! = 24 different ways to arrange the cards according to something that we will agree. With four bits we can represent 2^4 = 16 different numbers > 13 which are all the possibilities of the hidden card; not to mention that the guy who helps the magician chooses the card which will be hidden ...
So 24 > 16 > 13 and that is where I stopped thinking since I am really busy these days.
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Ecoris
Promising
Supreme Hero
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posted May 25, 2009 09:11 AM |
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How would you treat the four remaining cards as bits?
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted May 25, 2009 09:21 AM |
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Edited by dimis at 09:27, 25 May 2009.
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I can arrange 4 cards in 24 different ways. Let's map 15 of them to the first 15 numbers represented by 4 bits, and let's use the rest 9 permutations for the last 16th number (which is btw not needed).
Example: 4 different numbers on the cards on the table. If they are in increasing order, they represent 0000. If they are small1, big, bigger, small2 = 0001, small2, big, bigger, small1 = 0010, ..., decreasing order = eg = 1111.
If they are all from the same suit the same number (from the same suit is the other extreme above), define a similar > ordering depending on the suit; i.e. diamonds > hearts > spades > clubs, or something.
And I guess so on for the rest. Not to mention that from the 5 cards, at least two are from the same suit.
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Ecoris
Promising
Supreme Hero
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posted May 25, 2009 09:32 AM |
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But what do you gain by converting the 24 permutations into 4 bit binary numbers??
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted May 25, 2009 09:32 AM |
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Plus the guy who helps the magician picks the card to leave outside.
So, if you have 2 from the same suit, and the rest are different suits (1 from each), take out one of the two cards from the same suit. Now the ordering on suits solves the problem.
If there are 3 cards from the same suit, leave outside one of the others. The ordering of suits defines in general which suit has precedence, and among the 3 cards of the same suit you have an obvious ordering since all of them are distinct.
Finally, if there are 4 cards from the same suit, the obvious ordering is enough.
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