What's on your mind?
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UrBaconMeCr8zy wrote: »My fridge handle broke
Well, I hope you soon get a handle on that situation.
I knew I liked you.0 -
Bacon bits0
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Butt chins.
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How to determine the best strategy for each player in the following two-player game. There
are three piles, each of which contains some number of coins. Players alternate turns,
each turn consisting of removing any (non-zero) number of coins from a single pile.
The goal is to be the person to remove the last coin(s).1 -
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UrBaconMeCr8zy wrote: »UrBaconMeCr8zy wrote: »UrBaconMeCr8zy wrote: »My fridge handle broke
Well, I hope you soon get a handle on that situation.
That, was a pun
Come on, that was great!
I read it twice becuase I wasn't sure
Well then ...
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IslandGal3 wrote: »A vacation
Take me with you!1 -
caco_ethes wrote: »Two pills
blue or red?0 -
What's on my mind.....
I'm really hungry this morning but I wants a full on breakfast.
Bacon eggs hashbrowns... with a side of pancakes.
A kind of breakfast where you so full all day till dinner.
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0
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food. I need some.0
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How to determine the best strategy for each player in the following two-player game. There
are three piles, each of which contains some number of coins. Players alternate turns,
each turn consisting of removing any (non-zero) number of coins from a single pile.
The goal is to be the person to remove the last coin(s).
Player 1:
Take all of pile 1.
Player 2 will take whatever.
Take all but 1 coin in pile 2.
Player 2 will take something out of pile 3.
Take all but 1 coin in pile 3.
Player 2 must choose the coin in pile 2 or 3, leaving the last coin for you.0 -
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IslandGal3 wrote: »A vacation
Me too.... 49 days and counting!1 -
caco_ethes wrote: »How to determine the best strategy for each player in the following two-player game. There
are three piles, each of which contains some number of coins. Players alternate turns,
each turn consisting of removing any (non-zero) number of coins from a single pile.
The goal is to be the person to remove the last coin(s).
Player 1:
Take all of pile 1.
Player 2 will take whatever.
Take all but 1 coin in pile 2.
Player 2 will take something out of pile 3.
Take all but 1 coin in pile 3.
Player 2 must choose the coin in pile 2 or 3, leaving the last coin for you.
Call a triplet an E-triplet (the “E” stands for “even”) if it has the
following property: When the three numbers are written in base 2, there is an even
number (that is, either zero or two) of 1’s in each digit’s place. Then a triplet is a
losing position (LP) if and only if it is an E-triplet.
Proof:
Let us establish the following three facts concerning E-triplets:
1. Removal of any number of coins from one pile of an E-triplet will turn the
triplet into a non-E-triplet.
2. Given a non-E-triplet, it is always possible to remove coins from one pile to
turn the triplet into an E-triplet.
3. (0,0,0) is an E-triplet.
These facts may be demonstrated as follows:
1. This is true because any two numbers in an E-triplet uniquely determine the
third.
2. We will describe how to turn any non-E-triplet into an E-triplet. Write the
three numbers of coins in base 2, and put them in a column, with the unit’s
digits aligned, as we did above. Starting from the left, look at each digit’s
column until you find a column with an odd number (that is, either one or
three) of 1’s. Let this be the nth column (counting from the right).
If there is one 1 in the nth column, label the number containing this 1 as
A. If there are three 1’s, then arbitrarily pick any of the numbers to be A.
Remove coins from A by switching the 1 in the nth column to a 0, and also by
switching any 1’s to 0’s, or 0’s to 1’s, in other columns to the right of the nth
column, in order to produce an even number or 1’s is all columns. We have
now created an E-triplet.
Note that this switching of 1’s and 0’s does indeed correspond to removing
(as opposed to adding) coins from A, because even if all the columns to the
right of the nth column involve switching 0’s to 1’s, this addition of 2n−1 − 1
coins is still less than the subtraction of the 2n−1
coins arising from the 1-to-0
switch in the nth column.
3. This is true, by definition of an E-triplet.
The first two of these facts show that if player X receives an E-triplet on a given
turn, then player Y can ensure that X receives an E-triplet on every subsequent
turn. Therefore, X must always create a non-E-triplet, by the first of the three
facts. X therefore cannot take the last coin, because he cannot create the E-triplet
(0, 0, 0). Therefore, an E-triplet is a losing position.
And @MeeseeksAndDestroy is a thespian.1 -
caco_ethes wrote: »How to determine the best strategy for each player in the following two-player game. There
are three piles, each of which contains some number of coins. Players alternate turns,
each turn consisting of removing any (non-zero) number of coins from a single pile.
The goal is to be the person to remove the last coin(s).
Player 1:
Take all of pile 1.
Player 2 will take whatever.
Take all but 1 coin in pile 2.
Player 2 will take something out of pile 3.
Take all but 1 coin in pile 3.
Player 2 must choose the coin in pile 2 or 3, leaving the last coin for you.
Call a triplet an E-triplet (the “E” stands for “even”) if it has the
following property: When the three numbers are written in base 2, there is an even
number (that is, either zero or two) of 1’s in each digit’s place. Then a triplet is a
losing position (LP) if and only if it is an E-triplet.
Proof:
Let us establish the following three facts concerning E-triplets:
1. Removal of any number of coins from one pile of an E-triplet will turn the
triplet into a non-E-triplet.
2. Given a non-E-triplet, it is always possible to remove coins from one pile to
turn the triplet into an E-triplet.
3. (0,0,0) is an E-triplet.
These facts may be demonstrated as follows:
1. This is true because any two numbers in an E-triplet uniquely determine the
third.
2. We will describe how to turn any non-E-triplet into an E-triplet. Write the
three numbers of coins in base 2, and put them in a column, with the unit’s
digits aligned, as we did above. Starting from the left, look at each digit’s
column until you find a column with an odd number (that is, either one or
three) of 1’s. Let this be the nth column (counting from the right).
If there is one 1 in the nth column, label the number containing this 1 as
A. If there are three 1’s, then arbitrarily pick any of the numbers to be A.
Remove coins from A by switching the 1 in the nth column to a 0, and also by
switching any 1’s to 0’s, or 0’s to 1’s, in other columns to the right of the nth
column, in order to produce an even number or 1’s is all columns. We have
now created an E-triplet.
Note that this switching of 1’s and 0’s does indeed correspond to removing
(as opposed to adding) coins from A, because even if all the columns to the
right of the nth column involve switching 0’s to 1’s, this addition of 2n−1 − 1
coins is still less than the subtraction of the 2n−1
coins arising from the 1-to-0
switch in the nth column.
3. This is true, by definition of an E-triplet.
The first two of these facts show that if player X receives an E-triplet on a given
turn, then player Y can ensure that X receives an E-triplet on every subsequent
turn. Therefore, X must always create a non-E-triplet, by the first of the three
facts. X therefore cannot take the last coin, because he cannot create the E-triplet
(0, 0, 0). Therefore, an E-triplet is a losing position.
And @MeeseeksAndDestroy is a thespian.
That she is, my friend. A real hot one too.1 -
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caco_ethes wrote: »How to determine the best strategy for each player in the following two-player game. There
are three piles, each of which contains some number of coins. Players alternate turns,
each turn consisting of removing any (non-zero) number of coins from a single pile.
The goal is to be the person to remove the last coin(s).
Player 1:
Take all of pile 1.
Player 2 will take whatever.
Take all but 1 coin in pile 2.
Player 2 will take something out of pile 3.
Take all but 1 coin in pile 3.
Player 2 must choose the coin in pile 2 or 3, leaving the last coin for you.
Call a triplet an E-triplet (the “E” stands for “even”) if it has the
following property: When the three numbers are written in base 2, there is an even
number (that is, either zero or two) of 1’s in each digit’s place. Then a triplet is a
losing position (LP) if and only if it is an E-triplet.
Proof:
Let us establish the following three facts concerning E-triplets:
1. Removal of any number of coins from one pile of an E-triplet will turn the
triplet into a non-E-triplet.
2. Given a non-E-triplet, it is always possible to remove coins from one pile to
turn the triplet into an E-triplet.
3. (0,0,0) is an E-triplet.
These facts may be demonstrated as follows:
1. This is true because any two numbers in an E-triplet uniquely determine the
third.
2. We will describe how to turn any non-E-triplet into an E-triplet. Write the
three numbers of coins in base 2, and put them in a column, with the unit’s
digits aligned, as we did above. Starting from the left, look at each digit’s
column until you find a column with an odd number (that is, either one or
three) of 1’s. Let this be the nth column (counting from the right).
If there is one 1 in the nth column, label the number containing this 1 as
A. If there are three 1’s, then arbitrarily pick any of the numbers to be A.
Remove coins from A by switching the 1 in the nth column to a 0, and also by
switching any 1’s to 0’s, or 0’s to 1’s, in other columns to the right of the nth
column, in order to produce an even number or 1’s is all columns. We have
now created an E-triplet.
Note that this switching of 1’s and 0’s does indeed correspond to removing
(as opposed to adding) coins from A, because even if all the columns to the
right of the nth column involve switching 0’s to 1’s, this addition of 2n−1 − 1
coins is still less than the subtraction of the 2n−1
coins arising from the 1-to-0
switch in the nth column.
3. This is true, by definition of an E-triplet.
The first two of these facts show that if player X receives an E-triplet on a given
turn, then player Y can ensure that X receives an E-triplet on every subsequent
turn. Therefore, X must always create a non-E-triplet, by the first of the three
facts. X therefore cannot take the last coin, because he cannot create the E-triplet
(0, 0, 0). Therefore, an E-triplet is a losing position.
And @MeeseeksAndDestroy is a thespian.
Put that in a video and I might watch it.
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That is a lot!1
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CoffeeAndContour wrote: »
Hmm... lemme chew on that for a minute. Maybe. How can I make insomnia something people want to talk about...
The Can't Sleep Peeps.
*Curtesy*2
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