Sum of ten and the quotient of a number x and 6
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Answer:
For tree see picture but instead of BG replace all values with RRBBB and see the notes above how to roate the colours instead of the shape roation below just do transition (0,0)(1,0) (1,1).
Then 13 colours minimum change.
This will be 3 + 5 + 4 each colour change = 12 each colour change. instead of doing the 56 each colour rotation below.
Step-by-step explanation:
make 4 bits of paper and label each side -1 and 1 and central 0
above central point vertically label 1 and -1 for transition.
Then change just top one 4x = 4 more
Do the same with 2nd from top = 8 more to include top changing 4x again.
Do the same with 2nd from bottom = 12 more
Do the same with bottom = 16 more.
4+8+12+16 =40
Then repeat the process on side for blocks and we can double up
40+40 = 80.
When drawing tree clearly state columns
Top two are red only change
Top of tree will read in order so where ever you see capital R you will know its a new column for the tree and fill in when you understand the 254 transitions of a 5 column tree. Then change B + R accordingly when you get to row 6. Here you just start changing blue back to the top after row 5
Columns go down and rows go across its a new row you will need
approx 5-6 rows each colour change.
For the ones started for you below its just transition and colour stays where it is.
R x RT 45d RT (1,0) 45 RT (1,1) 45 RT (1,0)
R x R x
B x B x
B x B x
B x B x
'x' = RED Normal as you keep 0 zero central on all 4 bricks and work with top row as explained below.
Top RED brick you have to Right side place transition on every 2nd column for each colour and thenyou should find a pattern if youve numbered your papers or brcks to just see how it works. (hanging out one side call this 'off set' or call it T for transition (1.0)
RED Central 'diagonal' call it '45 degree rotate'
E ABCD RED Normal edge transition its edge so it hangs out
E ABCD RED On edge 45 degree diagonal its edge so it hangs out
2nd only change
S ABCD 1st red normal 2nd red Right side place rotate
S ABCD 1st red right side rotate 2nd red 45 degree rotate
S ABCD 1st red right side rotate 2nd central place 45 degree rotate
S ABCD 1st red right side rotate 2nd edge Normal edge transition
S ABCD 1st red right side rotate 2nd 45 degree its edge diagonal
2nd is repeated by changing 1st 4x more in bold. see what is now missing for 1st red. and copy the 2nd changes to 1st all whilst rotaing 2nd again in change .
Copy paste the work
Then apply this to 3 + 4 + 5th brick.
As the above shows 10 and when you finish 2nd and 1st it will be 14 total
Then 3 will be 14 again
and 4th brick will be 14 again
and 5th brick will be 14 again
making the transitions and rotations where all bricks in same level placement
14 x 4 = 56 transitions/rotations
Then repeat again for 1 blue at the top change
Then repeat again for 2 blue at the top for change
Then repeat again for 2 blue in the top middle
Then repeat again for 2 blue in the bottom middle
You can label tree
RR = Red rotation
BR = Blue rotation
RRE= Red rotation edge
BRE = Blue rotation
and colour code blue and red pens.
You can label transitions
RT = red transition (1,0)
RT = red transition (1,1) for central
and same for blue etc..
The end result will be 56 x 4 = 224 diagram tree
On ce you learn this it will take just 6 minutes to write out.
