Do you know the play of three-dimensional four-in-a-row? Try google Somehow you understand As you can see, it is a four-in-a-row arrangement in a 4x4x4 space. I have expanded this to 4 dimensions with ease, so I would like to write about it.
I made something like this.
I created it on an iPad using Pythonista 3 that runs Python on iOS.
――Play with 4D 4th --Maintaining the board (not reset even when closed) --Saving and loading game records --CPU match
For the time being, I will mention it on GitHub, but since the environment is limited, I will not discuss the specific UI implementation.
Since the number of combinations is huge when trying to aggregate seriously, I decided to cover all the sequences by the following procedure.
--x axis → $ (1, 0, 0, 0) $ --Diagonal of y-z plane → $ (0, \ pm1, 1, 0) $ --x-z-w Diagonal of cube → $ (\ pm1, 0, \ pm1, 1) $ --Diagonal hypercube → $ (\ pm1, \ pm1, \ pm1, 1) $
It's like (all multiple issues are optional).
In one-dimensional scanning, the axis of $ (1, 0, 0, 0) $ should be shifted, so
for i in range(4): _aggregate([[1, 0, 0, 0][i:]+[1, 0, 0, 0][:i]])
If so, it's OK.
_Aggregate is a function that appears after 2)
Scanning in two or more dimensions would make readability sad if a unified implementation was used, so I gave up and wrote all the patterns.
I used ʻitertools.product
which returns the direct product to make the compound number optional. For example, for the x-z-w cube, you can represent four diagonals by usingproduct ([-1, 1], , [-1, 1], )`.
If the scanning direction is +1 for an axis, the coordinates of the starting point must be 0 in order to proceed from 0 → 1 → 2 → 3. On the contrary, if it is -1, it is necessary to start from 3 in order to proceed from 3 → 2 → 1 → 0. If it is 0, then 0, 1, 2, 3 can all be possible. I will write this obediently.
def start(val): if val == 1: return  elif val == -1: return  else: return range(4)
Scans in the specified direction from all possible starting points. The starting point is to enumerate all combinations out of multiple for each of the four axes, so ʻitertools.product` comes into play again. Specifically, it looks like this
def _aggregate(vec): for X, Y, Z, W in vec: if (X, Y, Z, W) == (0, 0, 0, 0): continue for x, y, z, w in product(start(X), start(Y), start(Z), start(W)): s = sum( self.get_cell(x+X*i, y+Y*i, z+Z*i, w+W*i).player for i in range(4) ) if s == 4: self.black += 1 elif s == -4: self.white += 1
The state of each cell is +1 for ●, -1 for ○, and 0 if nothing is placed, so if you add them all together, you can see that they are complete.
The CPU is currently very simple:
In this game, if you make two 3s at the same time, you win, so there are processes like 3 and 4. (If there are multiple places with the same priority, select them randomly) It is strong for a simple algorithm.
If you have Pythonista, please play with it. I would be grateful if anyone could tell me that they had ported it to another environment!