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 What price glory? (Posted on 2015-01-22)
Two math. wizards play in the following manner:
From a row of integers 0,1,2,…1023,1024 A erases 512 numbers of his choice, - following this B erases 256 numbers of B’s choice.
Step 3: A erases 128 numbers, etc…
So at Step 10 player B chooses one of the 3 remaining numbers and erases it to define the amount of (dollars, pounds, euros, marbles) to be paid by A i.e. the difference between the two remaining numbers.

Clearly, A chooses a strategy to minimize this amount while
his opponent wants to maximize the outcome.

Assuming both follow the best strategy (Which?),
what will be the outcome of the game?

 No Solution Yet Submitted by Ady TZIDON No Rating

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 If there isn't an even better strategy... (spoiler?) | Comment 1 of 5
It would seem at first glance that A would take the top half or the bottom half to minimize the spread of the numbers, while B would remove the middle set.

After A's first move, 0 through 512 will remain.

B will remove 65 through 128, leaving 0 - 64 and 129 - 512.

A will reduce that to 0 - 64.

After B's and A's next moves, 0 - 16 will be left. After their next two, 0 - 4. After their next, 0, 1 and 2.

B will then remove the 1. At this point there will be two numbers left: 0 and 2.  Other, similar, choices could have been made along the way, but when two numbers are left, they should differ by 2.

But hold on: There's a better strategy for B:

B would take every second remaining number. After A has reduced the set to 0 through 512, B could remove the remaining odd numbers. Then A takes away the top half; then B takes away numbers not divisible by 4, etc.

After A: 0 - 512
After B: 0 - 512 odd
After A: 0 - 256 odd
After B: 0 - 256 divisible by 4
After A: 0 - 128 div by 4
After B: 0 - 128 div by 8
After A: 0 - 64 div by 8
After B: 0 - 64 div by 16
After A: 0 - 32 div by 16

That's just 0, 16 and 32. After B removes the 16, the final difference will be 32.

 Posted by Charlie on 2015-01-22 11:13:59

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