There is an inverted triangle colored dots. The colors are red, blue and yellow. If the two dots above are the same color, then the third dot below matches both. If the two dots above are different colors, then the third one below is different from either.

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Above are some examples

If you make a colored row of only 3 dots, can you predict the outcome? Is there a pattern? How about 4 dots. Try the same for 5 and 6 dots and see if you can find any simplifications. Use this pattern to predict a row of 28 dots.

My Progress:

I read a paper on this online and it seems the for an even number of dots for the top row the product of the two yields the final dot. For odd ones I can carve out two triangles and get the problem. However, I need a solution based on the simplification of 3/4 dot rows.

I listed the possibilities for 3 dots:

  • If all three colors are different then the middle one is the final one (e.g. B R Y = R)
  • If two colors are involved then
    • If the odd one is in the middle, it is the color not involved (e.g. YBY = R)
    • If the odd one is the first or last it is that one (e.g YBB = Y)

However I don't understand how I can apply the above pattern to 4, 5, and then 28 dots to come up with a quicker solution.

  • $\begingroup$ What is the "outcome" you want to predict? If you start with a row of three colored dots, then the "outcome" is a triangular array of six colored dots, right? $\endgroup$ – bof Apr 29 '17 at 5:22
  • $\begingroup$ Outcome is gonna be one dot. $\endgroup$ – qaispak Apr 29 '17 at 6:13
  • 1
    $\begingroup$ You define the "outcome" to be the color of the last dot? Why didn't you say so in the first place. $\endgroup$ – bof Apr 29 '17 at 6:17
  • $\begingroup$ Ah, I just noticed! $28=1+3^3$. $\endgroup$ – Lord Shark the Unknown Apr 29 '17 at 12:30

Think of the colours as integers modulo $3$. So you can think of the rows as vectors over $\Bbb F_3$. If one row is then $(a_1,\ldots,a_n)$ then the next row is $(-a_1-a_2,-a_2-a_3,\ldots,-a_{n-1}-a_n)$. The next one will be $(a_1+2a_2+a_3,a_2+2a_3+a_4,\ldots)$. Starting with top row $(a_1,\ldots,a_n)$ I reckon the colour in last row will be $$(-1)^{n-1}\sum_{k=1}^n\binom{n-1}{k-1}a_k$$ modulo $3$.

  • $\begingroup$ I need to understand in terms of pattern/breaking down into groups of three. $\endgroup$ – qaispak Apr 29 '17 at 5:18
  • $\begingroup$ That is precisely what @LordSharktheUnknown has given you. Each color is represented by an element of $\mathbb{F}_3$, a $3$ element field. You can choose how the colors are assigned yourself. $\endgroup$ – infinitylord Apr 29 '17 at 6:29

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