A circular disk is cut into n distinct sectors, each shaped like a piece of pie and all meeting at the center point of the disk. Each sector is to be painted red, green, yellow, or blue in such a way that no two adjacent sectors are painted the same color. Let Sn be the number of ways to paint the disk using these four colors.
Consider a pie cut into j sectors with Sj different ways to color it. If you take one of the sectors, w and subdivide it into two sectors w1 and w2, then wedge new sector, x between them, we now have a pie of j - (original sector w) + (3 new sectors w1, w2 and x) = j+2 sectors. To create a proper coloring of this new pie, let w1 and w2 be colored the same color that w was. Sector x can now be colored any of the three remaining colors. Therefore there are 3Sj legal colorings for this new pie with j+2 sectors.
b) 4 pts. List step by step instructions to turn a properly colored pie of j sectors into a properly colored pie of j+1 sectors (similar to the problem just demonstrated above). How many different ways could you color that pie? You must use Sj in your expression.
This is what I got, If we divide any sector, let say sector w into two halves w1 and w2 making it a pie of j+1 sector. Let say we keep the color of w1 same as color of w, then w2 has two possible color choices. Does that make 2Sj legal colorings for this new pie?
Also, I thought if we divide a sector(let's say w) in two halves(Let's say w1 and w2) and color the first half(w1) same as the orignial sector (w), w2 has 2 color choice, but sector next to has now one more color choice. I am very confused.