colring strips of paper black and white

• May 24th 2008, 05:11 AM
kurt16
colring strips of paper black and white
• May 24th 2008, 08:27 AM
Soroban
Hello, Kurt!

Here's some help . . .

Quote:

Make a strip of squares.
Color some squares black. leaving the other white.
Code:

      * - - * - - * - - * - - * - - *       |    |    |:::::|    |:::::|       |    |    |:::::|    |:::::|       * - - * - - * - - * - - * - - *

For strips of various lengths, consider the possible colorings.
Try to predict the total number of possible colorings?
For strips of a particular length, how do you know if you have all the possible colorings?
Are any coloring reflections of any others? If so, how will you count them?

Suppose our strip has $n$ squares.

For each square there are two choices: color it black or leave it white.
Hence, we can make: . $2^n$ possible decisions.

Therefore, there are: . $\boxed{2^n\text{ possible strips}}$

Most strips have "reflections": . $\blacksquare\:\blacksquare\:\blacksquare\:\square\ :\square\;\text{ and } \;\square\;\square\;\blacksquare\;\blacksquare\;\b lacksquare$

Counting the reflections turned out to be quite tricky.
I had to baby-talk my way through it.

I counted the strips that had no reflections.
. . These are "palindromic" strips, read the same forward and backwards.

Consider a strip with an even number of squares, say, $n = 8$

We have: . $\_\:\_\:\_\;\_\:|\:\_\:\_\:\_\:\_$

We have 2 choices for the first square: black or white.
. . But that determines the color of the 8th square.

We have 2 choices for the second square.
. . But that determines the color of the 7th square.

. . . . . and so on.

Hence, we have $2^4$ palindromic strips.

In general, for even $n$, there are $2^{\frac{n}{2}}$ palindromic strips.

And the other $2^n - 2^{\frac{n}{2}}$ strips have reflections.

Consider a strip with an odd number of squares, say, $n = 9$

We have: . $\_\:\_\:\_\;\_\:|\:\_\:|\:\_\:\_\:\_\:\_$

We have two choices for the first square.
. . But that determines the color of the 9th square.

We have two choices for the second square.
. . But that determines the color of the 8th square.

. . . and so on.

And we have two choices for the center square.

In general for odd $n$, there are: . $2^{\frac{n+1}{2}}$ palidromic strips.

The other $2^n - 2^{\frac{n+1}{2}}$ strips have reflections.

I need a nap . . .
.
• May 24th 2008, 04:44 PM
kurt16
still need more help