[superprismatic]

Two lovers decide to test their love

with a daisy. The particular daisy has

13 petals, and they agree to pluck

alternately, taking either one petal or

two adjacent petals. The boy picks

one, saying, "She loves me." The girl

picks two adjacent petals, leaving

groups of 8 and 2, saying, "He loves me

not." How should the boy continue if

he wants to end up in love in spite of

any move his opponent (?) makes?

SUPERPRISMATIC CLARIFICATION: The boy

must pick the last of the petals to get

what he wishes. Also, "adjacent petals"

refers to adjacency on the original

13-petal configuration in which each

petal only has 2 adjacent petals -- one

on either side.

[Guest]

He should always pick one petal, one with the adjacent petal missing. On the last pick he may take two.

[Guest]

He should always pick one petal, one with the adjacent petal missing. On the last pick he may take two.

I think your scheme needs rethinking

Your scheme won't work if it goes:

Boy (1 from 8) = 7 left

Girl (2 from group of 2)

Boy = 1 (as in scheme - 6 left)

Girl = 1 from 6

Boy = 1 (as in scheme = 4 left)

Girl = 1 *** OOPS 3 left. Boy loses

Thinking of the strategy...

[Guest]

I think your scheme needs rethinking

Your scheme won't work if it goes:

Boy (1 from 8) = 7 left

Girl (2 from group of 2)

Boy = 1 (as in scheme - 6 left)

Girl = 1 from 6

Boy = 1 (as in scheme = 4 left)

Girl = 1 *** OOPS 3 left. Boy loses

Thinking of the strategy...

If there are three petals all adjacent (in a row), the boy could pick the middle petal, making it impossible for the girl to pick the last two. In this case, the boy would win.

[bushindo]

If there are three petals all adjacent (in a row), the boy could pick the middle petal, making it impossible for the girl to pick the last two. In this case, the boy would win.

I need my day's allowance of Walter Penney puzzle!

Lemma: if the boy can reduce the groups on the daisy to an even number of 1's, an even number of 2's, an even number of 3's, or any combination of even groups of 1 to 3, he can win. Likewise, if the girl can reduce the groups to combination of even groups of 1 to 3, she can win.

Start from the group of 8, let the boy pick the fourth petal, leaving the daisy with groups of 2, 3, and 4. You can easily see that doesn't matter what the girl pick next, the boy can reduce it to his winning combination within 1 move or 2 moves.

[Guest]

I need my day's allowance of Walter Penney puzzle!

Lemma: if the boy can reduce the groups on the daisy to an even number of 1's, an even number of 2's, an even number of 3's, or any combination of even groups of 1 to 3, he can win. Likewise, if the girl can reduce the groups to combination of even groups of 1 to 3, she can win.

Start from the group of 8, let the boy pick the fourth petal, leaving the daisy with groups of 2, 3, and 4. You can easily see that doesn't matter what the girl pick next, the boy can reduce it to his winning combination within 1 move or 2 moves.

Mmmm...I almost had it. I was at the point of even number of ones.. but couldn't get the next leap

[Guest]

If there are three petals all adjacent (in a row), the boy could pick the middle petal, making it impossible for the girl to pick the last two. In this case, the boy would win.

I meant in the context of the original proposed solution (read it carefully)...

[Guest]

His first move is to take one petal. For all other moves, he will perform the opposite move that the girl chooses. If she takes one petal, he will subsequently take two. If she takes two petals, he will take one. After he takes one for his first move, there will be 12 petals remaining. After any round, 3 more will be removed (girl takes 2, boy takes 1 or vice versa). The rounds will go from 12 to 9 to 6 to 3. It will be the girls turn when there are 3 petals. Following the same rule one more time will win the game.

[Guest]

remove one petal so that there are groups of 4 3 & 2

[Guest]

Oh, I didn't read the question completely. Sorry

[Guest]

Making the groups 2, 3, 4 does not necessarily produce a win. The girl can take 2 from the group of 4 to make groups of (2,3,2).

Start analysis from the bottom up,

The final winning move will occur with a single group of 1 or 2 petals.

(1) and (2) are winning positions.

Let the the numbers in the parenthesis represent separated groups of the designated sizes.

The positions: (1,1) and (3) are losing positions because they put the other player in the winning position no matter what.

The position (2,1) is a winning position because it can put the other player in a losing position (1,1).

The position (2,2) is a losing position because for any possible action, it will put the other player in a winning position,

since (2) and (1,2) are both winning positions.

Following this line of reasoning: these are all winning positions:

(1), (2), (1,2), (1,3), (1,1,1), (1,1,2), (4), (5), (2,3), (1,2,2), (2,2,2), (2,4), along with more

By adding the petals that could be removed in one turn, and double checking, the following configurations are shown to all be losing positions:

(1,4), (1,1,3), (1,1,1,1), (6), (2,5), (2,2,3), (3,3), (2,5).

I have not built on my list further, but this is enough to prove that putting the girl in (2,3,4) is putting the girl in a winning position (meaning if she makes all the right choices, she will win regardless of your choices).

She would simply need to take two petals from the group of 4 to put you in the losing position (2,3,2) or (2,2,3).

The results of any possible move you could make would put the girl in another winning position:

(2,3) which is a winning position since it can put the enemy in losing position (2,2)

(1,2,3) which is a winning position since it can put the enemy in losing position (1,1,3)

(2,2,2) which is a winning position since it can put the enemy in losing position (2,2)

(2,2,1) which is a winning position since it can put the enemy in losing position (2,2).

The girl will make her appropriate move and put you in position (2,2) or position (1,1,3).

Let's examine position (2,2), possible results after your move are to place the girl in positions:

(1,2) or (2)

If you place her in (1,2)

she will place you in (1,1), you will take 1, and she will win.

If you place her in (2), she will take both petals and win.

Let's examine position (1,1,3):

The possible results you can put her in are:

(1,3)

(1,1,2)

(1,1,1)

If you place her in (1,3), she will put you in (1,1) and she will win.

If you place her in (1,1,2), she will put you in (1,1) and she will win.

If you place her in (1,1,1), she will put you in (1,1) and she will win.

Therefore if she can put you in position (2,3,2) she can win no matter what you do if she makes the right moves.

By putting her in position (2,3,4), you allow her to place you in position (2,3,2) and win.

I have not found the correct answer, but I believe I have shown that making groups of 2, 3, and 4 will not necessarily secure a win.

To solve the puzzle, we would just need to find a losing position that can be created from (2,8) in one move.

Edited August 18, 2009 by mmiguel1

[bushindo]

Making the groups 2, 3, 4 does not necessarily produce a win. The girl can take 2 from the group of 4 to make groups of (2,3,2).

Start analysis from the bottom up,

The final winning move will occur with a single group of 1 or 2 petals.

(1) and (2) are winning positions.

Let the the numbers in the parenthesis represent separated groups of the designated sizes.

The positions: (1,1) and (3) are losing positions because they put the other player in the winning position no matter what.

The position (2,1) is a winning position because it can put the other player in a losing position (1,1).

The position (2,2) is a losing position because for any possible action, it will put the other player in a winning position,

since (2) and (1,2) are both winning positions.

Following this line of reasoning: these are all winning positions:

(1), (2), (1,2), (1,3), (1,1,1), (1,1,2), (4), (5), (2,3), (1,2,2), (2,2,2), (2,4), along with more

By adding the petals that could be removed in one turn, and double checking, the following configurations are shown to all be losing positions:

(1,4), (1,1,3), (1,1,1,1), (6), (2,5), (2,2,3), (3,3), (2,5).

I have not built on my list further, but this is enough to prove that putting the girl in (2,3,4) is putting the girl in a winning position (meaning if she makes all the right choices, she will win regardless of your choices).

She would simply need to take two petals from the group of 4 to put you in the losing position (2,3,2) or (2,2,3).

The results of any possible move you could make would put the girl in another winning position:

(2,3) which is a winning position since it can put the enemy in losing position (2,2)

(1,2,3) which is a winning position since it can put the enemy in losing position (1,1,3)

(2,2,2) which is a winning position since it can put the enemy in losing position (2,2)

(2,2,1) which is a winning position since it can put the enemy in losing position (2,2).

The girl will make her appropriate move and put you in position (2,2) or position (1,1,3).

Let's examine position (2,2), possible results after your move are to place the girl in positions:

(1,2) or (2)

If you place her in (1,2)

she will place you in (1,1), you will take 1, and she will win.

If you place her in (2), she will take both petals and win.

Let's examine position (1,1,3):

The possible results you can put her in are:

(1,3)

(1,1,2)

(1,1,1)

If you place her in (1,3), she will put you in (1,1) and she will win.

If you place her in (1,1,2), she will put you in (1,1) and she will win.

If you place her in (1,1,1), she will put you in (1,1) and she will win.

Therefore if she can put you in position (2,3,2) she can win no matter what you do if she makes the right moves.

By putting her in position (2,3,4), you allow her to place you in position (2,3,2) and win.

I have not found the correct answer, but I believe I have shown that making groups of 2, 3, and 4 will not necessarily secure a win.

To solve the puzzle, we would just need to find a losing position that can be created from (2,8) in one move.

There is a mistake in your reasoning.

You correct showed that (2,2,2) is a winning position. The mistake is that you assumed that if the boy gets to the state (2,3,2), he would remove 1 from the edge of the group of 3, hence handing the winning position of (2,2,2) to the girl. However, if the boy were to remove the middle petal from the group of 3, he would hand the losing position of (2,1,1,2) to the girl.

In general, if the boy can reduce the configurations to an even number of groups of size 1, an even number of group of size 2, or an even number of groups of size 3, then he can win. He can also win if he can produce any combination of even groups of size 1 to 3 (i.e (1,1,2,2) ). Induction will show that group size doesn't matter, he can win as long as he has an even number of groups for any size.

Edited August 18, 2009 by bushindo

[Guest]

You are correct, bushindo.

Thanks!

[superprismatic]

I need my day's allowance of Walter Penney puzzle!

Lemma: if the boy can reduce the groups on the daisy to an even number of 1's, an even number of 2's, an even number of 3's, or any combination of even groups of 1 to 3, he can win. Likewise, if the girl can reduce the groups to combination of even groups of 1 to 3, she can win.

Start from the group of 8, let the boy pick the fourth petal, leaving the daisy with groups of 2, 3, and 4. You can easily see that doesn't matter what the girl pick next, the boy can reduce it to his winning combination within 1 move or 2 moves.

You found a second solution. Penney gave another.

Take 1 from the end of the 8, leaving 7 and 2. It works as well.