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Mathematics


Harut

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Assuming the table is all clear, Azat, sounds to me like you would want to go first. Put your bottle at the center of the table ... then where ever your oponent puts a bottle, just put yours at the opposide side of the center in the exact same spot (in the next turn).

 

This way, your oponent is guaranteed to run out of space before you do.

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I hear ketchup bottles work just as well :lol2: But now I see your ulterior motives for posing the question!!!

 

 

Here's another one ... I thought it was pretty cool:

 

So, there are n Armenians in Glendale, each having a unique piece of gossip.

They all want to know everything.

Their only mode of communication is the phone.

When two people make a call to each other, ALL their gossip knowledge gets exchanged ... so they will both know all that each other knew before calling.

 

Given that these are highly efficient Armenians, what is the minimum number of calls required for ALL of them to learn ALL the gossip? They can only do a normal one-on-one call. There is no conference calling or text messaging or anything like that.

Edited by Sip
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I hear ketchup bottles work just as well :lol2: But now I see your ulterior motives for posing the question!!!

Here's another one ... I thought it was pretty cool:

 

So, there are n Armenians in Glendale, each having a unique piece of gossip.

They all want to know everything.

Their only mode of communication is the phone.

When two people make a call to each other, ALL their gossip knowledge gets exchanged ... so they will both know all that each other knew before calling.

 

Given that these are highly efficient Armenians, what is the minimum number of calls required for ALL of them to learn ALL the gossip? They can only do a normal one-on-one call. There is no conference calling or text messaging or anything like that.

 

I'm too old for this sort of things now. :) I'll quickly say n-1, but I know that this answer is wrong, too simplistic... and fried the brain cells that will make me think about such things without a headache.

 

I have emailed the answer to Azat, Vava and Harut so that they can take the credit for.

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It's definitely not n-1

 

That would leave too many people without ALL the gossip. Besides for the simple case of 4 glendale armenians, it would have to be a minimum of 4 calls. Taking armenians A,B,C & D:

 

A <--> B

C <--> D

then A <--> C & B <-->D and voila - they all know.

 

I think 4 is a critical number here, cause above 4 you will need a greater number of calls than the "n" number of Armenians.

 

This is the type of discrete mathematics that keep comp-sci geeks busy (and happy), right sip? :P

 

Without attempting a classical mathematics proof, i've tried to determine the formula through a series:

 

when n=x, (y)=calls are required:

 

1, (0)

2, (1)

3, (3)

4, (4) as shown above

5, (6)

6, (8)

7, (10)

8, (12)

9, (14)

10, (16)

 

OK, this is getting long. :) But I see a pattern - and it seems that the original 4 is a constant. I would say that the solution would be something like 2n-4; but it only works for values above 4. So it's probably not right - alas I have no proof, and i never liked discrete math. :o

 

Go ahead sip - give us the solution!

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  • 7 months later...

Another one from cartalk. Please dont search the web for the answer. thats cheating...

 

I'm going to give you four numbers: 2,3,4, and 5. And they're not Roman numerals, they're the regular old Arabic numerals with which we're all familiar.

 

You also get a plus sign, and an equal sign. Now, make an equation with some stuff on one side of the equal sign, and some stuff on the other.

 

At first blush, you might say, "This is easy!" 5+2=3+4...but, you're using two plus signs, and you're only allowed to use one.

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