Celtic and African Art

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Celtic and African Art

• King David High School, Manchester

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What we did

• We drew some complex Celtic knots. Sam and Gary
  drew a really big one it took them ages!
• We tried to work out how many different complex
  knots are possible on different grids.

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Some of our complex knots

• We found these quite difficult
• We really liked the Josephine knots!
• Here are a few of our knots

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How many different Complex knots are there on a
given grid?

• This proved to be a really knotty problem
• We decided to simplify the problem by looking at
  just one simple case an n by 2 grid.
• We know that a 1 by 2 grid doesnt work if we use
  Chris method so we started with 2by2

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2 by 2 knots

• You will have seen that there are 3 possible
  knots A, B and C

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3by2 knots

• There are 9 possible combinations of letters
• AA, AB, AC,
• BA, BB, BC,
• CA, CB, CC
• But the ones that are the same letters in reverse
  will just give the same knot upside down!

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3 by 2 knots

• so AB and BA are really the same.
• This means that actually there are 3x3-36
  different knots
• AA, AB, AC,
• , BB, BC,
• , , CC

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3 by 2 knots

• We realised that we now didnt need to draw all
  the different knots we just needed to look at
  the possible arrangements of the letters.
• We then needed to delete all the letter sequences
  that are reverses of ones already recorded except
  for palindromic ones. These are symmetrical
  sequences that read the same backwards.

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n by 2 grids

• For example AAB and BAA are really the same
  but back to front so if you have AAB, you
  ignore BAA.
• ABA is palindromic ie it reads the same in
  reverse but wont come up twice in any list

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3 by 2 grids

• We made a list of letters 3 at a time and crossed
  out repetitions. We got
• 3x3x3 27 possible arrangements and
• 3x3 9 repeats
• so the number of different shapes is
• 27-918

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4 by 2 grids

• This was much trickier as the number of possible
  arrangements is 3x the number for the 3 by 2s
  81 altogether!
• There are 36 repetitions
• Final answer 81-36 45

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N by 2 grids

• To cut a long story short, arrangements of 5 and
  6 letters was very messy.
• Our teacher used Excel to print out all the
  possibilities for 6 letters and it took 3 pages
  of printout!
• Here is what we found

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Grid No. of Letters No. of arrangements Number of repeats Number of different knots
2x2 1 3 0 3
3x2 2 32 9 3 6
4x2 3 33 27 32 9 18
5x2 4 34 81 33 32 36 45
6x2 5 35 243 3433108 135
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Conclusion

• We got stuck because counting became too
  complicated.
• We know that the result must involve powers of 3
• We think the rule will be different for odd
  numbers of letters than for even ones because
  there are different ways of getting repeats with
  odd lengths and even ones.

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THE END!