A Puzzling Fortnight - Day 7

JeeJee

Canada5652 Posts

February 18 2010 16:26 GMT

Let's jump right into it this time.

You vs me. An even numer of coins (potentially different denominations) are arranged in a line between us, going left-to-right. We are going to play the following game:
We take turns taking a coin from either end of the coin line. The game ends when there are no more coins. The winner is the one whose little pile of coins at the end is worth more than the other's. Simple.

You go first. Devise a strategy such that your pile at the end will always at least match my pile in value.

GL!

onmach

United States1241 Posts

February 18 2010 16:50 GMT

Hmm, my naive approach. If there are 2 coins, pick the largest. If there are 4 or more coins, add up the values of the coin on the left and second coin from the right, and compare that to the value of the second coin to the left and the coin on the right. If the first number is higher, then take a coin from the left side, and if the second is larger take a coin from the right side.

ieatkids5

United States4628 Posts

February 18 2010 16:55 GMT

11111111
if all the denominations are the same, then just pick whatever

any other scenario, add up every other number starting at one end, then add up every other number starting from the other end. whichever side you start on that has a greater sum, start picking from that side.

example:
43472191

4+4+2+9=19
3+7+1+1=12

start picking from the side with the 4. your opponent has to take the 3 because if he takes the 1, then you will take the 9. after he takes the 3, take the 4. he will take the 7. take the 2. he takes the 1. you take the 9.

another example:
27261192

2+2+1+9=14
7+6+1+2=16

start from the right side and take the 2. your opponent will take the 9 because you will take it if he doesnt. take the 1. opponent will take next 1 because he cannot take the 2 on the left (or you will take the 7. take the 6. opponent takes the 2. take the 7.

edit - updated with examples

onmach

United States1241 Posts

February 18 2010 19:15 GMT

Your solution is simpler, and after looking your examples, mine was clearly defective.

JeeJee

Canada5652 Posts

February 18 2010 19:48 GMT

On February 19 2010 01:55 ieatkids5 wrote:
11111111
if all the denominations are the same, then just pick whatever

any other scenario, add up every other number starting at one end, then add up every other number starting from the other end. whichever side you start on that has a greater sum, start picking from that side.

example:
43472191

4+4+2+9=19
3+7+1+1=12

start picking from the side with the 4. your opponent has to take the 3 because if he takes the 1, then you will take the 9. after he takes the 3, take the 4. he will take the 7. take the 2. he takes the 1. you take the 9.

another example:
27261192

2+2+1+9=14
7+6+1+2=16

start from the right side and take the 2. your opponent will take the 9 because you will take it if he doesnt. take the 1. opponent will take next 1 because he cannot take the 2 on the left (or you will take the 7. take the 6. opponent takes the 2. take the 7.

edit - updated with examples

i assume you re-calculate these 2 values every time it is your turn?

edit
@onmach
which example shows inadequacy of your strategy? curious..

ieatkids5

United States4628 Posts

February 20 2010 02:01 GMT

On February 19 2010 04:48 JeeJee wrote:

Show nested quote +

i assume you re-calculate these 2 values every time it is your turn?

edit
@onmach
which example shows inadequacy of your strategy? curious..

heh oops
yes

Daigomi

South Africa4316 Posts

March 03 2010 21:25 GMT

+ Show Spoiler +

You add the value of all the odd numbered coins (coin #1, #3, etc) and all the even numbered coins (coin #2, #4, etc) together. If the odd numbered coins total the highest, you start on the left. If the even numbered coins have the highest total, you start on the right. After that you just pick on the same side as your opponent does every turn.

Please or register to reply.

A Puzzling Fortnight - Day 7

Completed

Ongoing

Upcoming