[TheMagicMan]

(Original post by Llewellyn)
Before I potentially waste a lot of time doing it all wrong, is the answer to i)

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16/3 (i.e. this is the maximum value for 4 quantities)?

I think perhaps I have been somehow unclear in the game rules...how did you get that?

[Llewellyn]

(Original post by TheMagicMan)
I think perhaps I have been somehow unclear in the game rules...how did you get that?

It's probably my interpretation of the rules that is the problem, I'll get back to this later though.

[Blutooth]

Is the answer to ii, 2?

[Slumpy]

(Original post by TheMagicMan)
Here's a nice question:

i)Take 4 quantities , .

We play a game according to these rules: Player A adds a total of 1 to the quantities, apportioning it however he likes. Player B then reduces any two consecutive quantities, or the first and the last, to 0. They then alternate 'turns'. Assume Player B plays optimally. Player A wins if he makes any one of the quantities greater than a certain value . Player B's goal is to stop Player A from winning. The game continues indefinitely. What is the maximum value of such that Player A can win?

ii) Using the same rules, what is the maximum value of if there are 5 quantities?

iii) And if there are quantities?

I have a very ugly and lengthy solution for iii) so bonus points for elegance on that one

After a little bit of work, I'm guessing at

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i-2
Anything up to 2 can be won(1/3,0,1/3,1/3, then goes to 0,0,1/3,0 goes to 2/3,0,2/3,0 then repeatedly split the 1 you add between the two places so they're equal). Once above x-1, add 1 to the number left. I'm not sure if you can reach 2 however, pretty sure you could make some argument about any number above 1 should be sent to 0 by B, and you can't get more than 1 there at once. Close?

[TheMagicMan]

(Original post by Slumpy)
After a little bit of work, I'm guessing at

Spoiler:

Show

i-2
Anything up to 2 can be won(1/3,0,1/3,1/3, then goes to 0,0,1/3,0 goes to 2/3,0,2/3,0 then repeatedly split the 1 you add between the two places so they're equal). Once above x-1, add 1 to the number left. I'm not sure if you can reach 2 however, pretty sure you could make some argument about any number above 1 should be sent to 0 by B, and you can't get more than 1 there at once. Close?

Spoiler:

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This is correct. Anything beneath 2 can be reached, but 2 itself cannot.

[Slumpy]

(Original post by TheMagicMan)

Spoiler:

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This is correct. Anything beneath 2 can be reached, but 2 itself cannot.

Boom! Part ii will be attempted tomorrow...

[Blutooth]

(Original post by Slumpy)
After a little bit of work, I'm guessing at

Spoiler:

Show

i-2
Anything up to 2 can be won(1/3,0,1/3,1/3, then goes to 0,0,1/3,0 goes to 2/3,0,2/3,0 then repeatedly split the 1 you add between the two places so they're equal). Once above x-1, add 1 to the number left. I'm not sure if you can reach 2 however, pretty sure you could make some argument about any number above 1 should be sent to 0 by B, and you can't get more than 1 there at once. Close?

I have a simple proof that the value must be 2 or below, only a couple of line for ii.

[Slumpy]

(Original post by Blutooth)
I have a simple proof that the value must be 2 or below, only a couple of line for ii.

Post it up? I get the feeling you can argue in a similar way to what I did before, but it's not hugely elegant.

[gff]

(Original post by TheMagicMan)
..

Have to finish couple of things these days, so I'm half missing.

Looking at that, considering the optimality of B and the number of quantities, I reckon it is fairly easy to build a geometric progression?
You can easily build a symmetry so that B doesn't have a choice, and then every consecutive turn A does not have a choice.

So, if you start with , next you have ... hence, general term and the limit of that is 1.
Therefore, taking into account the final turn, all values below 2 can be obtained.

For part two, I'm getting some strange series. Would you reckon you get a lower value for than 2?

If so, I can try to find something about the third part, but I have to finish few pages of a coursework before that tonight. And to EAT - always forget that.

[TheMagicMan]

(Original post by gff)
Have to finish couple of things these days, so I'm half missing.

Looking at that, considering the optimality of B and the number of quantities, I reckon it is fairly easy to build a geometric progression?
You can easily build a symmetry so that B doesn't have a choice, and then every consecutive turn A does not have a choice.

So, if you start with , next you have ... hence, general term and the limit of that is 1.
Therefore, taking into account the final turn, all values below 2 can be obtained.

For part two, I'm getting some strange series. Would you reckon you get a lower value for than 2?

If so, I can try to find something about the third part, but I have to finish few pages of a coursework before that tonight. And to EAT - always forget that.

On the 5 quantity one the maximum value is at least what it is on the 4 quantity one as you can play the same strategy ignoring one quantity if you want

[Blutooth]

(Original post by TheMagicMan)
On the 5 quantity one the maximum value is at least what it is on the 4 quantity one as you can play the same strategy ignoring one quantity if you want

Part ii) is proving rather tricky to prove. Though I can show for part 2 that the x must be less than 2.25. Still can't home in on the 2.


for an even number n, i think the highest value you can get for x is 1+ 1/2 +1/3 +1/4+....+1/(n/2) +1/2
or if the number is odd 1+1/2+1/3+...+1/((n-1)/2) +1/2

Can't prove this is the highest though, though looking at the method I can sort of see why this number should be the highest, ie at each step you are focusing on filling fewerpots and at the same time minimising the amount of water thrown out by the thrower until you have one pot remaining.

[King-Panther]

Hello, what is meant differentiate with respect to x, does that mean calculate when dy/dx = 0?

(Original post by raheem94)
...

[gff]

(Original post by King-Panther)
Hello, what is meant differentiate with respect to x, does that mean calculate when dy/dx = 0?

If your is given by , then it simply asks you to find . You would equate to zero in order to find stationary points -- i.e. local minima or maxima.

[raheem94]

(Original post by King-Panther)
Hello, what is meant differentiate with respect to x, does that mean calculate when dy/dx = 0?

If you are said to differentiate the above equation with respect to 'x', then you need to find dy/dx.



You still look to be confused about that turning points concept.

When you are asked to find the turning points, then you need to first find the dy/dx (i.e. differentiate the 'y' expression with respect to 'x') and then make dy/dx equal to zero.

[King-Panther]

(Original post by raheem94)


If you are said to differentiate the above equation with respect to 'x', then you need to find dy/dx.



You still look to be confused about that turning points concept.

When you are asked to find the turning points, then you need to first find the dy/dx (i.e. differentiate the 'y' expression with respect to 'x') and then make dy/dx equal to zero.

(Original post by gff)
..

O.k, so I differentiate to find x?

The equation is y=x^1/2(8+x)

[raheem94]

(Original post by King-Panther)
O.k, so I differentiate to find x?

Can you tell me which question are you talking about?

[King-Panther]

(Original post by raheem94)
Can you tell me which question are you talking about?

The equation is y=x^1/2(8+x)

[raheem94]

(Original post by King-Panther)
The equation is y=x^1/2(8+x)

This is the equation, what does the question asks you to find? What is the complete question?

[King-Panther]

(Original post by raheem94)
This is the equation, what does the question asks you to find? What is the complete question?

differentiate with respect to x

i got, dy/dx=(4x^-1/2)+(3/2x^1/2)

[raheem94]

(Original post by King-Panther)
differentiate with respect to x

i got, dy/dx=(4x^-1/2)+(3/2x^1/2)

Yes, you did it correctly and got the correct answer .