BRAIN TEASER HELP
you have a random sequence of number from 1-10, 2 players and u being the first player ! Each player can only choose a number close to the one that is already chosen or the last number . The winner is the one that has a larger total ( in this case above 28 since the total of the numbers is 55 ) How can u ensure u always win??
example
4 6 3 7 10 2 9 8 5 1 - You have this starting sequence
I am player one and I choose 4
player 2 can choose either 6 or 1
Assuming he choose 6 I can either choose 3 or 1 and so on...
I chose 3 and player 2 chose 7
I chose 10 and player 2 chose 2
I chose 9 and player 2 chose 8
I chose 5 and player 2 chose 1
my total :4+3+10+9+5= 31
Player 2 : 6+7+2+8+1= 24
therefore I won.
How can we generate a winning sequence every time even though the numbers r changing?
Considering that you chose the start-point and you can only go one direction, it's pretty easy to see the two paths that stem from each number choice. Essentially you are just choosing set 1 or 2 (regardless of which way they go you will end up with every other number starting from the one you choose in the beginning), so you just total each number set and ensure that you pick the winning one.
Example: if a different number set is 5 7 4 3 10 1 8 2 6 9, you either have choice one of (5, 4, 10, 8, 6) or two of (7, 3, 1, 2, 9). You just choose the set that will win based on whichever will score higher. You could mock up a pretty basic formula for this easily.
I think you're excluding the "or the last number" stipulation in kkii33's post.
So in the original example if the set is {4 6 3 7 10 2 9 8 5 1} and Player 1 chooses 4, then Player 2 has the option of either 6 ("close to the one that is already chosen") or 1 ("the last number").
Thus a possible sequence of choices could look like this:
P1: 4 P2: 1 P1: 5 P2: 6
In each case the player is essentially picking the outermost remaining un-chosen number from the left or right of the set.
This drastically increases the potential combinations beyond simply the sum of every other number starting with the left or right.
Whether there exists a formula for always picking the highest sum is beyond me, however I would be inclined to say not practically. Since each player's choice depends on the other player's previous choice, you would have to draw an elaborate decision tree I would think.
Not sure how you would express it mathematically, but the logic is you simply look at the remaining numbers in the set and pick the higher of your two options in that round.
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