I'm still working on this project (:з」∠) and the pictures below haven't been updated yet (其实是当时手边没有合适的三阶魔方，只好拿四阶来做实验，然后这个readme是当时写的，整体完成后忘记换图片了，悄悄用中文写，希望不会有人看到）
After I finish this competition, I'm going to put the pure vision part in another branch and use only two pictures to solve the cube.
This is a small project for a vision and control competition, which use opencv and raspberry pi to do the vision part and add another pca9685 to control 6 servo to rotate the cube.The vision part use only two cameras to figure out the state of the cube. It can recognize three faces of a cube at a time.
Because of the limited funds and the mechanical structure, I have to use two webcams to do the cube recognition. So I have to detect 3 faces of the cube at one time.
I use 3s video to get a stable result of recognition.For every frame, I use GaussianBlur() to get rid of the noise and use Canny() to detect the edges.
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GaussianBlur image
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Canny image
Next I use findCounters() and approxPolyDP() to get the square counters.For each square counter, I use the angle of its longest diagonal to define which face it belongs to. And there is also a trickey function in it.Because the cam and the cube are settled so I can use the stable postion of the squares in the unstable frame and get the most detaild square counter data.
- Squares image
For each face, there is a key sticker, which belongs to three faces of the corner block.I use the postion data of every stickers to define where the corner bolck is.After I find the key block.I can use the data of it to map the whole cube and find the bound of each face.I use warpPerspective() to flatten mapping and get the imagess for three faces.
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key block
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cube bound
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roi image
- remapped image
After that, I write a simple color reduce funtion just for this program. I turn the image into the HSV color space and use its Hue value and the Saturation value to define the color of each sticker. These color data are stored in a sticker array and will be combined to get all the color position in a string.So that I can use the two phase algorithm to solve the cube.
- color reduction
- recognize result
I use a kociemba binary file from https://github.com/muodov/kociemba
Example：DRLUUBFBRBLURRLRUBLRDDFDLFUFUFFDBRDUBRUFLLFDDBFLUBLRBD Example output: D2 R' D' F2 B D R2 D2 R' F2 D' F2 U' B2 L2 U2 D R2 U
the cube looks like this
|************|
|*U1**U2**U3*|
|************|
|*U4**U5**U6*|
|************|
|*U7**U8**U9*|
|************|
|************|************|************|************|
|*L1**L2**L3*|*F1**F2**F3*|*R1**R2**R3*|*B1**B2**B3*|
|************|************|************|************|
|*L4**L5**L6*|*F4**F5**F6*|*R4**R5**R6*|*B4**B5**B6*|
|************|************|************|************|
|*L7**L8**L9*|*F7**F8**F9*|*R7**R8**R9*|*B7**B8**B9*|
|************|************|************|************|
|************|
|*D1**D2**D3*|
|************|
|*D4**D5**D6*|
|************|
|*D7**D8**D9*|
|************|
and the input order is:
U1, U2, U3, U4, U5, U6, U7, U8, U9, U refered to servo 1
R1, R2, R3, R4, R5, R6, R7, R8, R9, R refered to servo 2
F1, F2, F3, F4, F5, F6, F7, F8, F9, F refered to servo 3
D1, D2, D3, D4, D5, D6, D7, D8, D9, D refered to servo 4
L1, L2, L3, L4, L5, L6, L7, L8, L9, L refered to servo 5
B1, B2, B3, B4, B5, B6, B7, B8, B9. B refered to servo 6
Solution string consists of space-separated parts, each of them represents a single move:
A single letter by itself means to turn that face clockwise 90 degrees.
A letter followed by an apostrophe means to turn that face counterclockwise 90 degrees.
A letter with the number 2 after it means to turn that face 180 degrees.
e.g. R U R’ U R U2 R’ U
https://stacks.stanford.edu/file/druid:yj296hj2790/Ng_Rubiks%20Cube_Reconstruction_from_Images.pdf
https://stacks.stanford.edu/file/druid:yj296hj2790/Ng_Rubiks%20Cube_Reconstruction_from_Images.pdf