Combinatorics- Calculating Grundy Value
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I am stuck in this problem and cannot really understand my textbook.
Two players take turns to play the following game. A basket contains 5 apples, 6 oranges, and 9 pears. At each turn the players are allowed to take 1, 2 or 3 fruits of same kind. The winner takes the last fruit.
(a) Find the value of the Grundy function at the initial position.
(b) What is a first winning move?
What I have been trying,
Binary Grundy Value
A: | | | | | 5 0101 1
O: | | | | | | 6 0110 2
P: | | | | | | | | | 9 + 1001 + 1
1010 = 10 2
I calculated Grundy Value as 2(which I used XOR). So is 2 correct for problem (a)?
For problem(b), since Grundy Value is 2, first winning move is take 2 from any fruits. (?)
combinatorics discrete-mathematics
asked Dec 7 '18 at 4:20
jaykodeveloperjaykodeveloper
1258
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add a comment |
$begingroup$
I am stuck in this problem and cannot really understand my textbook.
Two players take turns to play the following game. A basket contains 5 apples, 6 oranges, and 9 pears. At each turn the players are allowed to take 1, 2 or 3 fruits of same kind. The winner takes the last fruit.
(a) Find the value of the Grundy function at the initial position.
(b) What is a first winning move?
What I have been trying,
Binary Grundy Value
A: | | | | | 5 0101 1
O: | | | | | | 6 0110 2
P: | | | | | | | | | 9 + 1001 + 1
1010 = 10 2
I calculated Grundy Value as 2(which I used XOR). So is 2 correct for problem (a)?
For problem(b), since Grundy Value is 2, first winning move is take 2 from any fruits. (?)
combinatorics discrete-mathematics
asked Dec 7 '18 at 4:20
jaykodeveloperjaykodeveloper
1258
$endgroup$
$begingroup$
Looks right. You're using the fact that the nim-value of each component (fruit type) is the number of that type of fruit modulo 4. And in general a winning move is one that leaves your opponent with nim-value 0... which in this case does mean taking 2 of any kind of fruit.
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– mjqxxxx
Dec 7 '18 at 4:28
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@mjqxxxx I didn't think that it is modulo 4. I counted one by one. But now it kinda makes sense why it uses modulo 4, because the players can take up to 3 fruits.
$endgroup$
– jaykodeveloper
Dec 7 '18 at 4:42
add a comment |
$begingroup$
I am stuck in this problem and cannot really understand my textbook.
Two players take turns to play the following game. A basket contains 5 apples, 6 oranges, and 9 pears. At each turn the players are allowed to take 1, 2 or 3 fruits of same kind. The winner takes the last fruit.
(a) Find the value of the Grundy function at the initial position.
(b) What is a first winning move?
What I have been trying,
Binary Grundy Value
A: | | | | | 5 0101 1
O: | | | | | | 6 0110 2
P: | | | | | | | | | 9 + 1001 + 1
1010 = 10 2
I calculated Grundy Value as 2(which I used XOR). So is 2 correct for problem (a)?
For problem(b), since Grundy Value is 2, first winning move is take 2 from any fruits. (?)
combinatorics discrete-mathematics
asked Dec 7 '18 at 4:20
jaykodeveloperjaykodeveloper
1258
$endgroup$
I am stuck in this problem and cannot really understand my textbook.
Two players take turns to play the following game. A basket contains 5 apples, 6 oranges, and 9 pears. At each turn the players are allowed to take 1, 2 or 3 fruits of same kind. The winner takes the last fruit.
(a) Find the value of the Grundy function at the initial position.
(b) What is a first winning move?
What I have been trying,
Binary Grundy Value
A: | | | | | 5 0101 1
O: | | | | | | 6 0110 2
P: | | | | | | | | | 9 + 1001 + 1
1010 = 10 2
I calculated Grundy Value as 2(which I used XOR). So is 2 correct for problem (a)?
For problem(b), since Grundy Value is 2, first winning move is take 2 from any fruits. (?)
combinatorics discrete-mathematics
combinatorics discrete-mathematics
asked Dec 7 '18 at 4:20
jaykodeveloperjaykodeveloper
1258
asked Dec 7 '18 at 4:20
jaykodeveloperjaykodeveloper
1258
asked Dec 7 '18 at 4:20
jaykodeveloperjaykodeveloper
1258
asked Dec 7 '18 at 4:20
jaykodeveloperjaykodeveloper
1258
asked Dec 7 '18 at 4:20
jaykodeveloperjaykodeveloper
1258
1258
$begingroup$
Looks right. You're using the fact that the nim-value of each component (fruit type) is the number of that type of fruit modulo 4. And in general a winning move is one that leaves your opponent with nim-value 0... which in this case does mean taking 2 of any kind of fruit.
$endgroup$
– mjqxxxx
Dec 7 '18 at 4:28
$begingroup$
@mjqxxxx I didn't think that it is modulo 4. I counted one by one. But now it kinda makes sense why it uses modulo 4, because the players can take up to 3 fruits.
$endgroup$
– jaykodeveloper
Dec 7 '18 at 4:42
add a comment |
$begingroup$
Looks right. You're using the fact that the nim-value of each component (fruit type) is the number of that type of fruit modulo 4. And in general a winning move is one that leaves your opponent with nim-value 0... which in this case does mean taking 2 of any kind of fruit.
$endgroup$
– mjqxxxx
Dec 7 '18 at 4:28
$begingroup$
@mjqxxxx I didn't think that it is modulo 4. I counted one by one. But now it kinda makes sense why it uses modulo 4, because the players can take up to 3 fruits.
$endgroup$
– jaykodeveloper
Dec 7 '18 at 4:42
$begingroup$
Looks right. You're using the fact that the nim-value of each component (fruit type) is the number of that type of fruit modulo 4. And in general a winning move is one that leaves your opponent with nim-value 0... which in this case does mean taking 2 of any kind of fruit.
$endgroup$
– mjqxxxx
Dec 7 '18 at 4:28
$begingroup$
Looks right. You're using the fact that the nim-value of each component (fruit type) is the number of that type of fruit modulo 4. And in general a winning move is one that leaves your opponent with nim-value 0... which in this case does mean taking 2 of any kind of fruit.
$endgroup$
– mjqxxxx
Dec 7 '18 at 4:28
$begingroup$
@mjqxxxx I didn't think that it is modulo 4. I counted one by one. But now it kinda makes sense why it uses modulo 4, because the players can take up to 3 fruits.
$endgroup$
– jaykodeveloper
Dec 7 '18 at 4:42
$begingroup$
@mjqxxxx I didn't think that it is modulo 4. I counted one by one. But now it kinda makes sense why it uses modulo 4, because the players can take up to 3 fruits.
$endgroup$
– jaykodeveloper
Dec 7 '18 at 4:42
add a comment |
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$begingroup$
Looks right. You're using the fact that the nim-value of each component (fruit type) is the number of that type of fruit modulo 4. And in general a winning move is one that leaves your opponent with nim-value 0... which in this case does mean taking 2 of any kind of fruit.
$endgroup$
– mjqxxxx
Dec 7 '18 at 4:28
$begingroup$
@mjqxxxx I didn't think that it is modulo 4. I counted one by one. But now it kinda makes sense why it uses modulo 4, because the players can take up to 3 fruits.
$endgroup$
– jaykodeveloper
Dec 7 '18 at 4:42