# Coin Game

• Oct 4th 2011, 08:18 AM
VonNemo19
Coin Game
Hi.

Imagine 2 coins on a table, at positions A and B. In the game there are 8 possible moves:

\$\displaystyle M_1:\$ flip the coin at A.
\$\displaystyle M_2:\$ flip the coin at B.
\$\displaystyle M_3:\$ flip both coins.
\$\displaystyle M_4:\$ switch the coins.
\$\displaystyle M_5:\$ flip the coin at A, then switch.
\$\displaystyle M_6:\$ flip the coin at B, then switch.
\$\displaystyle M_7:\$ flip both, then switch.
\$\displaystyle I:\$ do nothing.

We may consider an operation on the set \$\displaystyle \{I,M_1,...,M_7\}\$, which consists of performing any 2 moves in succession. EG: If we switch coins, then flip at A, this is the same as first flipping over the coin at B then switching.

Write the table for \$\displaystyle (G,*)\$.

Ok, so I understand the notion of group, I understand how to create operation tables, but I'm having trouble visualizing the equivalences between a pair of elements and the associated element which is the pair's product. For example,

\$\displaystyle M_1*M_2=?\$.

Can anyone help me see this?

*edit>
OK. So, what about \$\displaystyle M_5*M_2\$ ? The way that I think about this is like, if we let the coins have sides 1 and 2 at points A and B, respectively, then if side 1 is showing at both A and B, and \$\displaystyle M_5\$ is made, then 1 becomes 2 at A and then moves to B. So, we have 1 at A and 2 at B. Now, we do \$\displaystyle M_2\$ and 1 now shows at B. This is where we started. So, is \$\displaystyle M_5*M_2=I\$ or \$\displaystyle M_5*M_2=M_4\$. To me it seems as if \$\displaystyle *\$ is not uniquely defined.
• Oct 4th 2011, 10:53 AM
Opalg
Re: Coin Game
Quote:

Originally Posted by VonNemo19
Hi.

Imagine 2 coins on a table, at positions A and B. In the game there are 8 possible moves:

\$\displaystyle M_1:\$ flip the coin at A.
\$\displaystyle M_2:\$ flip the coin at B.
\$\displaystyle M_3:\$ flip both coins.
\$\displaystyle M_4:\$ switch the coins.
\$\displaystyle M_5:\$ flip the coin at A, then switch.
\$\displaystyle M_6:\$ flip the coin at B, then switch.
\$\displaystyle M_7:\$ flip both, then switch.
\$\displaystyle I:\$ do nothing.

We may consider an operation on the set \$\displaystyle \{I,M_1,...,M_7\}\$, which consists of performing any 2 moves in succession. EG: If we switch coins, then flip at A, this is the same as first flipping over the coin at B then switching.

Write the table for \$\displaystyle (G,*)\$.

Ok, so I understand the notion of group, I understand how to create operation tables, but I'm having trouble visualizing the equivalences between a pair of elements and the associated element which is the pair's product. For example,

\$\displaystyle M_1*M_2=?\$.

Can anyone help me see this?

*edit>
OK. So, what about \$\displaystyle M_5*M_2\$ ? The way that I think about this is like, if we let the coins have sides 1 and 2 at points A and B, respectively, then if side 1 is showing at both A and B, and \$\displaystyle M_5\$ is made, then 1 becomes 2 at A and then moves to B. So, we have 1 at A and 2 at B. Now, we do \$\displaystyle M_2\$ and 1 now shows at B. This is where we started. So, is \$\displaystyle M_5*M_2=I\$ or \$\displaystyle M_5*M_2=M_4\$. To me it seems as if \$\displaystyle *\$ is not uniquely defined.

The reason you think that * is not uniquely defined is that you are treating the coins as indistinguishable, which is not the case. (If it were, then \$\displaystyle M_4\$ would be the same as \$\displaystyle I.\$)

To keep track of the situation, you need to keep track of which coin is in which position, as well as which side of each coin is showing. Suppose you call the coins x and y. Use the notation \$\displaystyle x_1\$ to denote that coin x is showing side 1, and similarly \$\displaystyle x_2,\$ \$\displaystyle y_1\$ and \$\displaystyle y_2.\$

Suppose that the system starts in the state \$\displaystyle (x_1,y_1),\$ which means that coin x is at A (showing side 1) and coin y is at B (also showing side 1). After the move \$\displaystyle M_5\$ is made, the system will be in state \$\displaystyle (y_1,x_2).\$ If you then make the move \$\displaystyle M_2,\$ the state will become \$\displaystyle (y_1,x_1).\$ So both coins show side 1, but their positions have switched. Thus \$\displaystyle M_5*M_2 = M_4,\$ and the ambiguity that troubled you has been resolved.
• Oct 4th 2011, 11:53 AM
VonNemo19
Re: Coin Game
Opalg. I'm honored that I've finally reached the point where MHF's best mathematician has found my question interesting enough to respond to. Thank you.

You've helped a lot. I've decided to continue through this elementary treatment of algebra at a snail's pace so that I don't miss a thing. Therefore, I will have many more questions on the topic.

C-Ya