I tried to solve this question but couldn't find a simple solution without passing all rows and find which numbers are on the same line.
Is there a simple way to find triangles?
this is my solution for finding a triangle:
How can I change it to be more "pythonic"? (or even better method for solving it)
from sympy.solvers import solve
from sympy import Symbol
from collections import Counter
vals = [8,17,19] # the triangle
dicl = [] #list of dics
for v in vals:
dic = {}
dic['val'] = v
v1 = v
done = 0
stepsb = 0
while done == 0: #going backword untill reaching the big triabgle edges
x = Symbol('x')
k = solve((x**2 + x)/2 +1 - v1, x)
k = list(filter(lambda x:x>0, k))
if k[0]%1 == 0:
done = 1
else:
v1 -= 1
stepsb += 1
dic['line'] = k[0]
dic['stepsb'] = stepsb #dist from the left edge
dic['stepsf'] = (k[0]**2 + 3*k[0] + 2)/2 - v #dist from the right edge
dicl.append(dic)
print(dic)
lines = [l['line'] for l in dicl]
mc = Counter(lines).most_common(1)[0][0] #finding the numbers on the same line
minv = min([l['val'] for l in dicl if l['line'] == mc])
maxv = max([l['val'] for l in dicl if l['line'] == mc])
stb = [l['stepsb'] for l in dicl if l['val'] == minv][0]
stf = [l['stepsf'] for l in dicl if l['val'] == maxv][0]
for k in dicl:
if k['stepsb'] == stb and k['stepsf'] == stf:
print("good")
break
A first step could be to search for a formula that translates the one-dimensional point number t to an x,y coordinate.
So, search for an n such that n*(n+1)/2 < t:
from sympy import solve, Eq
from sympy.abc import n, t
f = Eq(n * (n + 1), 2 * t)
print(solve(f, n))
This shows as positive root: (sqrt(8*t + 1) - 1)/2.
To be strict smaller, a formula that copes with small approximation errors, could be:
floor((sqrt(8*t + 1) - 1)/2 - 0.0000001
The following idea is, given a list of indices:
convert them to xy coordinates
find their center (sum and divide by the length of the list)
find the distances of each xy to the center
check that all distances are equal
To convert to an xy position, note that the height of an equilateral triangle with base 1 is sqrt(3)/2, so the distances between the y-positions should be multiplied by that factor. The x-positions need to be centered which can be achieved by subtracting n/2.
import math
def find_xy(t):
# convert the numerical position into an xy coordinate in the plane
# first find largest n such that n*(n+1)/2 < t
n = math.floor((math.sqrt(8 * t + 1) - 1) / 2 - 0.0000001)
return (n + 1) * math.sqrt(3) / 2, t - n * (n + 1) // 2 - n/2
def sq_dist(p, q):
return (p[0] - q[0]) ** 2 + (p[1] - q[1]) ** 2
def center(points):
# find the center of a list of points
l = len(points)
x = sum(p[0] for p in points)
y = sum(p[1] for p in points)
return x / l, y / l
def is_regular(tri_points):
points = [find_xy(t) for t in tri_points]
cent = center(points)
dists = [sq_dist(cent, p) for p in points]
return max(dists) - min(dists) < 0.000001
Note that this code finds geometric figures for which all the points lie on a circle. This doesn't work for the parallelogram. The actual question also has some extra criteria: all edges should follow the grid lines, and all edges need to be equal in length.
Therefore, it is useful to have 3 coordinates for each point: the row, the column and the diagonal (the 3 directions of the grid).
The length in each direction, is just the maximum minus the minimum for that direction. These lengths are called d_r, d_c and d_d in the code below.
Checking for a valid triangle, the 3 lengths need to be equal. One way to check this, is to check that the minimum of the lengths is equal to the maximum.
For a valid parallelogram, two lengths need to be equal, and the third should be the double. Checking that the maximum length is twice the minimum length should cover this. But, because this can already be reached using 3 points, we should also check that for a given direction, there are exactly 2 points at the minimum and 2 at the maximum. Summing all points and comparing twice the sum of maximum and minimum should accomplish this.
For a valid hexagon, the 3 lengths should be equal. So, the same test as for the triangle: the minimum of the lengths equal to the maximum. And also the test on the sums is needed, as 4 points can already fulfil the length conditions.
import math
def find_row_col_diag(t):
# convert the numerical position into an row,col,diag coordinate in the plane
# first find largest n such that n*(n+1)/2 < t
n = math.floor((math.sqrt(8 * t + 1) - 1) / 2 - 0.0000001)
row, col = n + 1, t - n * (n + 1) // 2
return row, col, row - col
def check_valid_figure(tri_points):
points = [find_row_col_diag(t) for t in tri_points]
rs = [r for (r, c, d) in points]
cs = [c for (r, c, d) in points]
ds = [d for (r, c, d) in points]
sum_r = sum(rs)
min_r = min(rs)
max_r = max(rs)
d_r = max_r - min_r
sum_c = sum(cs)
min_c = min(cs)
max_c = max(cs)
d_c = max_c - min_c
sum_d = sum(ds)
min_d = min(ds)
max_d = max(ds)
d_d = max_d - min_d
if len(points) == 3:
is_ok = max(d_r, d_c, d_d) == min(d_r, d_c, d_d)
elif len(points) == 4:
is_ok = max(d_r, d_c, d_d) == 2 * min(d_r, d_c, d_d) \
and sum_r == 2 * (min_r + max_r) and sum_c == 2 * (min_c + max_c) and sum_d == 2 * (min_d + max_d)
elif len(points) == 6:
is_ok = max(d_r, d_c, d_d) == min(d_r, d_c, d_d) \
and len(set(rs)) == 3 and len(set(cs)) == 3 and len(set(ds)) == 3
else:
is_ok = False
print(" ".join([str(t) for t in tri_points]), end=" ")
if is_ok:
print("are the vertices of a",
"triangle" if len(points) == 3 else "parallelogram" if len(points) == 4 else "hexagon")
else:
print("are not the vertices of an acceptable figure")
tri_point_lists = [[1, 2, 3],
[11, 13, 22, 24],
[11, 13, 29, 31],
[11, 13, 23, 25],
[26, 11, 13, 24],
[22, 23, 30],
[4, 5, 9, 13, 12, 7]]
for lst in tri_point_lists:
check_valid_figure(lst)
The last code can be further compressed using list comprehensions:
def check_valid_figure_bis(tri_points):
points = [find_row_col_diag(t) for t in tri_points]
rs, cs, ds = [[p[i] for p in points] for i in range(3)]
sums = [sum(xs) for xs in (rs, cs, ds)]
mins = [min(xs) for xs in (rs, cs, ds)]
maxs = [max(xs) for xs in (rs, cs, ds)]
lens = [ma - mi for mi, ma in zip(mins, maxs)]
if len(points) == 3:
is_ok = max(lens) == min(lens)
elif len(points) == 4:
is_ok = max(lens) == 2 * min(lens) and all([su == 2 * (mi + ma) for su, mi, ma in zip(sums, mins, maxs)])
elif len(points) == 6:
is_ok = max(lens) == min(lens) and all([len(set(xs)) == 3 for xs in (rs, cs, ds)])
else:
is_ok = False
return is_ok
Related
I am trying to develop a random tetrahedron generator that can take a list of four coordinates and produce a tetrahedron. Currently, I am only able to plot a tetrahedron using four pre-determined points. Here is the code I have as of now:
def rand_tetrahedron_generator(bounds, min_len, max_len):
"""
bounds: List - max length in each dimension
min_len: int - minimum length of tetrahedron
max_len: int - maximum length of tetrahedron
"""
assert len(bounds) == 3
assert min_len <= max_len
max_len = min(max_len, bounds[0], bounds[1], bounds[2])
bounds = np.array(bounds)
p1 = np.random.randint(low=0, high=bounds, size=3)
p2 = np.random.randint(low=0, high=bounds - min_len, size=3)
p3 = np.random.randint(low=0, high=p1+p2, size=3)
p4 = np.random.randint(low=0, high=p1+p2, size=3)
points = np.array([p1,p2,p3,p4])
center = np.mean(points, axis=0)
x, y, z = (np.indices((60, 60, 60))-np.array([20,25,25]).reshape(-1,1,1,1))/8
mx = midpoints(x)
my = midpoints(y)
mz = midpoints(z)
conditions = []
for p1,p2,p3 in itertools.combinations(points, 3):
a, n = surface_normal_form(p1,p2,p3)
conditions.append((mx-a[0])*n[0]+(my-a[1])*n[1]+(mz-a[2])*n[2] <= 0)
simplex = conditions[0] & conditions[1] & conditions[2] & conditions[3]
return simplex
def surface_normal_form(a,b,c):
v = b-a
w = c-b
n = np.cross(v,w)
#normal needs to point out
if (center-a)#n > 0:
n *= -1
return a, n
def midpoints(x):
sl = ()
for i in range(x.ndim):
x = (x[sl + np.index_exp[:-1]] + x[sl + np.index_exp[1:]]) / 2.0
sl += np.index_exp[:]
return x
I believe that the way I generate p1, p2, p3, and p4 is incorrect because the function sometimes generates points that are unable to form a tetrahedron. I would greatly appreciate any advice on how to solve this issue. I have also attached an image of the final result I am looking for.
Compute the kth element in spiral order for an m x n 2D array A in O(1) time.
For example...
A = [[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12]]
The spiral order is 1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7.
If k = 5, the function returns 12.
The closest solution I found was to check if k exists in the perimeter of A. If not, recursively call the function without the perimeter and subtract the perimeter length from k. Time complexity increases with the size of A (more perimeters to drop). Is there a way to go about this in O(1) time?
To derive your needed computation, try starting with a larger matrix, such as 10x7. This will let you observe the overall effects for a couple of laps before the end cases bite you.
First, note the extent of each lap (one revolution about the matrix). Each direction has an ever-decreasing quantity of steps, compared to the previous lap. In the 10x7 case, you start just outside the matrix. Then you take (10-0) steps right, (7-1) steps down, (10-1) steps left, and (7-2) steps up. Thus, lap 1 has a length of 4 steps less than the perimeter.
You've now chopped off 2 units in each direction, and you're again just to the left of the remaining 8x5 matrix. You take (10-2) steps right, (7-3) steps down, ... you get the idea. Each lap is 4 units shorter than the previous, until you finally consume the shorter dimension entirely (i.e. no more elements).
Now, the problem divides into two steps:
Determine which lap contains element k
Determine where element k is within that lap.
Each of these is an O(1) computation.
Consider a 10x7 (mxn) matrix, in which we get 30 elements for the the outer ring, the next outer ring would have 22 elements, then 14 elements, and the most inner ring has the remaining elements. The number of elements per ring is given by 2 x (m - (1 + (2 x (r-1))) ) + 2 x (n - (1 + (2 x (r-1))) ), for the rth ring.
Save from the most inner ring, the difference between the number of elements of each adjacent ring is 8 elements. If we want to know the number of elements of the current ring plus all the other previous ones, we get an arithmetic series.
The sum of all the elements until a given r is given by sum = (r(a1 + ar)) / 2, being a1 = 2(m-1) + 2(n-1), ar = 2 x (m - (1 + (2 x (r-1) ))) + 2 x (n - (1 + (2 x (r-1) ))). For this effect, this sum represents the lower bound of k in a given ring, so we will consider k as the sum. If we solve the equation in relation to r, using the quadratic formula to disentangle the final polynomial, we reach r = math.floor((-(m + n) + math.sqrt((m+n)**2 - 4*k)) / (-4))
Now we can know in which ring the kth element is, and also know the number of elements leading up to this ring. That is the base of the below algorithm.
The rest is working with these two pieces of information to work out an offset and figure out where that offset falls in the ring (top, right, bottom or bottom portion)
This algorithm has O(1) time complexity.
Note that the algorithm below computes the x and y on the array for a given k, m and n, which is the core part of this problem. Getting the array item from these coordinates is trivial from here.
import math
def kth_element_spiral(k: int, m: int, n: int):
first_ring_max = 2 * (m-1) + 2 * (n - 1)
ring, ring_start_elements = None, None
if (k < first_ring_max):
ring = 0
ring_start_elements = 0
else:
ring = int(math.floor((-(m + n) + math.sqrt((m+n)**2 - 4*k)) /(-4)))
ring_start_elements = int((ring * (first_ring_max + 2 * (m-(1 + 2*(ring - 1))) + 2 * (n-(1 + 2*(ring - 1))))) / 2)
offset = k - ring_start_elements
width = (m - ring*2) - 1
height = (n - ring*2) - 1
if (offset <= width): # top
x = ring + offset
y = ring
return (x, y)
elif (offset <= width + height): # right
x = m - ring - 1
y = ring + (offset - width)
return (x, y)
elif (offset <= width + height + width): # bottom
x = width - (offset - width - height)
y = n - ring - 1
return (x, y)
else: # left
x = ring
y = height - (offset - width - height - width)
return (x, y)
# Some test cases
for i in range(10*6):
print(i, kth_element_spiral(i, 10, 6))
for i in range(7*5):
print(i, kth_element_spiral(i, 7, 5))
for i in range(9*6):
print(i, kth_element_spiral(i, 9, 6))
Gist available at https://gist.github.com/lopespm/a5e6552451227f1ea04579e7ec750c4d
Using Pedro's base code and some modifications for calculating total elements using arithmetic sum, below is my take:
I simplified the formula for calculating sum of AP:
"""
T1 = 2m + 2n - 4
Tr = 2m + 2n -8r + 4 #r here is the ring
Sr = (T1 + Tr)r/2 #sum of progression till ring r
Solving:
Sr = r(2m+2n-4+2m+2n-8r+4)/2
Sr = r(4m +4n -8r)/2
Sr = -4r^2 + (2m +2n)r #this is quadratic equation analogous to ax^2 + bx + c
Since Sr is lower bound of k, lets solve it with respect to r and k
-4r^2 + (2m +2n)r <= k
-4r^2 + (2m +2n)r - k <= 0
r = floor((-(m + n) + sqrt((m+n)^22 - 4*k)) /(-4))
"""
def kth_element_spiral(A, m, n, k):#k here starts with 1
"""
inputs: A: array, m: rows, n: columns, k: position
output: Array element returned for position k
"""
if k==0:
return(-1,-1)
first_ring_max = 2*m + 2*n - 4
ring, ring_start_elements = None, None #ring: rings elapsed to reach position k
if (k < first_ring_max):
ring = 0
ring_start_elements = 0
else:
ring = int(math.floor((-(m + n) + math.sqrt((m+n)**2 - 4*k)) /(-4)))
#arithmetic sum to find out how many elements covered in ring, ring-1 , .. 0
ring_start_elements = int(-4*(ring**2) + 2*ring*(m+n))
offset = k - ring_start_elements
#rows left in current ring, logic: after every ring top and bottom rows removed
mod_m = (m - ring*2)
# columns left in current ring, logic: after every ring left and right rows removed
mod_n = (n - ring*2)
if (offset <= mod_n): # top row in the current ring
x = ring
y = ring + offset - 1
elif (offset <= mod_n + mod_m - 1): # rightmost column in the current ring
x = ring + offset - mod_n
y = n - 1 - ring
elif (offset <= mod_n + mod_m - 1 + mod_n - 1): # bottom row in the current ring
x = m - 1 - ring
y = ring + mod_n - 1 - (offset - (mod_n + mod_m - 1))
elif offset <= mod_n + mod_m - 1 + mod_n - 1 + mod_m-1: # leftmost column in the current ring
x = ring + mod_m - 1 - (offset - (mod_n + mod_m-1 + mod_n - 1))
y = ring
else:
return(-1, -1)#not found
return A[x][y]
# some test cases
for i in range(1,10):
print(i, kth_element_spiral([[1,2,3], [4,5,6], [7,8,9]],3, 3, i))
for i in range(1,9):
print(i, kth_element_spiral([[1,2,3,4], [5,6,7,8]],2, 4, i))
for i in range(1,5):
print(i, kth_element_spiral([[1], [2], [3], [4]], 4, 1, i))
I am now trying to calculate the poisson sphere distribution(a 3D version of the poisson disk) using python and then plug in the result to POV-RAY so that I can generate some random distributed packing rocks.
I am following these two links:
[https://github.com/CodingTrain/Rainbow-Code/blob/master/CodingChallenges/CC_33_poisson_disc/sketch.js#L13]
[https://www.cs.ubc.ca/~rbridson/docs/bridson-siggraph07-poissondisk.pdf]tl;dr
0.Create an n-dimensional grid array and cell size = r/sqrt(n) where r is the minimum distance between each sphere. All arrays are set to be default -1 which stands for 'without point'
1.Create an initial sample. (it should be placed randomly but I choose to put it in the middle). Put it in the grid array. Also, intialize an active array. Put the initial sample in the active array.
2.While the active list is not empty, pick a random index. Generate points near it and make sure the points are not overlapping with nearby points(only test with the nearby arrays). If no sample can be created near the 'random index', kick the 'random index' out. Loop the process.
And here is my code:
import math
import numpy
from random import uniform
import random
from math import floor
r = 1
k = 30
grid = []
w = r / math.sqrt(2)
active = []
width = 100
height = 100
depth = 100
cols = floor(width / w)
rows = floor(height / w)
deps = floor(depth / w)
default = numpy.array((-1,-1,-1))
for i in range(cols * rows * deps):
grid.append(default)
x = width / 2
y = height / 2
z = depth / 2
i = floor(x / w)
j = floor(y / w)
k = floor(z / w)
pos = numpy.array((x,y,z))
grid[i + cols * (j + rows * k)] = pos
active.append(pos)
while (len(active) > 0) and (len(grid[grid == -1]) > 0):
randIndex = floor(uniform(0, len(active)))
pos = active[randIndex]
found = False
for n in range(k):
m1 = uniform(-2 * r, 2 * r)
m2 = uniform(-2 * r, 2 * r)
m3 = uniform(-2 * r, 2 * r)
m = numpy.array((m1,m2,m3))
sample = numpy.add(pos, m)
col = floor(sample[0] / w)
row = floor(sample[1] / w)
dep = floor(sample[2] / w)
if (col > -1 and row > -1 and dep > -1 and col < cols and row < rows and dep < deps and numpy.all([grid[col + cols * (row + rows * dep)],default])==True):
ok = True
for i in range(-1,2):
for j in range(-1, 2):
for k in range(-1, 2):
index = (col + i) + cols * ((row + j) + rows * (dep + k))
if col + i > -1 and row + j > -1 and dep + k > -1 and col + i < cols and row + j < rows and dep + k < deps:
neighbor = grid[index]
if numpy.all([neighbor, default]) == False:
d = numpy.linalg.norm(sample - neighbor)
if (d < r):
ok = False
if ok == True:
found = True
grid[col + cols * (row + rows * dep)] = sample
active.append(sample)
if found == False:
del active[randIndex]
print(len(active))
for printout in range(len(grid)):
print("<" + str(active[printout][0]) + "," + str(active[printout][1]) + "," + str(active[printout][2]) + ">")
print(len(grid))
My code seems to run forever and do not obey my condition(distance of two spheres must be larger than 2 * radius) as shown in the visualization by POV-RAY.(picture in comment)
Therefore I tried to add a print(len(active)) in the last of the while loop.
Surprisingly, I think I discovered the bug as the length of the active list just keep increasing! (It is supposed to be the same length as the grid) I think the problem is caused by the active.append(), but I can't figure out where is the problem as the code is literally the 90% the same as the one made by Mr.Shiffman.
I don't want to free ride this but I have already checked again and again while correcting again and again for this code :(. Still, I don't know where the bug is. (why do the active[] keep appending!?)
Thank you for the precious time.
Im needing to solve a whole range of 8x8 and 9x9 matrices so thought I could build a python program to make the whole thing easier.
So far I have managed to create:
from __future__ import division
import numpy as np
def solveEqns(A,v):
def lu( A ):
#Factor A into LU by Gaussian elimination with scaled partial pivoting
n, m = np.shape( A )
if n != m:
print "Error: input matrix is not square"
return None
# Generate initial index vector
p = range( n )
# Determine the largest (in magnitude) element in each row. These
# factors are used to scale the pivot elements for comparison purposes
# when deciding which row to use as a pivot row.
s = [0] * n
for i in xrange( n ):
smax = 0.0
for j in xrange( n ):
smax = max( smax, abs( A[i][j] ) )
s[i] = smax
# Begin Gaussian elimination.
for k in xrange( n - 1 ):
# Find the remaining row with the largest scaled pivot.
rmax = 0.0
for i in xrange( k, n ):
r = abs( A[p[i][k]] / s[p[i]] )
if r > rmax:
rmax = r
j = i
# Row j has the largest scaled pivot, so "swap" that row with the
# current row (row k). The swap is not actually done by copying rows,
# but by swaping two entries in an index vector.
p[j], p[k] = ( p[k], p[j] )
# Now carry out the next elimination step as usual, except for the
# added complication of the index vector.
for i in xrange( k + 1, n ):
xmult = A[p[i],k] / A[p[k],k]
A[p[i],k] = xmult
for j in xrange( k + 1, n ):
A[p[i],j] = A[p[i],j] - xmult * A[p[k],j]
# All done, return factored matrix A and permutation vector p
return ( A, p )
def solve( A, p, b ):
#Solves Ax = b given an LU factored matrix A and permuation vector p
n, m = np.shape( A )
if n != m:
print "Error: input matrix is not square"
return None
# Forward solve
x = np.zeros( n )
for k in xrange( n - 1 ):
for i in xrange( k + 1, n ):
b[p[i]] = b[p[i]] - A[p[i],k] * b[p[k]]
# Backward solve
for i in xrange( n - 1, -1, -1 ):
sum = b[p[i]]
for j in xrange( i + 1, n ):
sum = sum - A[p[i],j] * x[j]
x[i] = sum / A[p[i],i]
# All done, return solution vector
return x
lu(A)
return solve(A,p,v)
def circuit():
A = np.array([[1,0,0,0,0,8,0,0,0],[0,1,0,0,5,0,0,0,0],[0,1,0,0,5,0,0,0,0],[0,0,0,1,-1,1,0,0,0],[0,0,1,0,0,0,1,-1,0],[0,0,1,0,0,0,1,0,-1],[0,1,0,0,-1,0,0,0,1],[1,0,0,0,0,-1,0,1,0],[1,-1,0,1,0,0,0,0,0]])
v = np.array([9,-12,-0.5,0,0,0,0,0,0])
I = solveEqns(A,v)
return I
to solve the 9x9 matrix A at the end. This is one of the easier ones i need to solve so can solve it outside of python to check if the results coming through are accurate.
Im getting a traceback error on line 26 of:
Traceback (most recent call last):
File "<ipython-input-110-6daf773db1e3>", line 1, in <module>
solveEqns(A,b)
File "C:/Users/SamMc/Documents/Python Scripts/q6u1510416 v4.py", line 65, in solveEqns
lu(A)
File "C:/Users/SamMc/Documents/Python Scripts/q6u1510416 v4.py", line 26, in lu
r = abs( A[p[i][k]] / s[p[i]] )
TypeError: 'int' object has no attribute '__getitem__'
which i cant figure out why its not pulling through a number from the matrix.
Any help would be greatly appreciated.
Thanks
Sam
you might use gauss elimination via scaled pivoting. the code is shown below.
import numpy as np
def gauss_pivot(a,b,tol=1.0e-12):
"""
x = gaussPivot(a,b,tol=1.0e-12).
Solves [a]{x} = {b} by Gauss elimination with
scaled row pivoting
"""
a = np.copy(a)
b = np.copy(b)
n = len(b)
assert (np.all(np.shape(a) ==(n,n))) # check if a is a square matrix
# Set up scale factors
s = np.zeros(n)
for i in range(n):
s[i] = max(np.abs(a[i,:])) # find the max of each row
for k in range(0, n-1): #pivot row
# Row interchange, if needed
p = np.argmax(np.abs(a[k:n,k])/s[k:n]) # find which row has max item for each col k, and scale by s
if abs(a[p,k]) < tol:
raise Exception("Matrix is singular")
if p != k: # swap rows if current row does not contain max item with the one contains max item within same col
a[[k,p+k],:] = a[[p+k, k],:]
b[k],b[p+k] = b[p+k],b[k]
s[k],s[p+k] = s[p+k],s[k]
# Elimination phase of matrix a
for i in range(k+1,n):
if a[i,k] != 0.0: # skip if a(i,k) is already zero
lam = a [i,k]/a[k,k]
a[i,k:n] = a[i,k:n] - lam*a[k,k:n]
b[i] = b[i] - lam*b[k]
if abs(a[n-1,n-1]) < tol:
raise Exception("Matrix is singular")
# Back substitution phase, solution is substituted by b
x = np.zeros_like(b)
x[n-1] = b[n-1]/a[n-1,n-1]
for k in range(n-2,-1,-1):
x[k] = (b[k] - np.dot(a[k,k+1:n],x[k+1:n]))/a[k,k]
return x
a = np.random.randn(100,100)*10
b = np.random.randn(100)*10
x = gauss_pivot(a,b)
if np.allclose(np.dot(a,x), b) == True:
print("x is the correct solution")
If you want the code to perform faster you might probably replace x by b, so upon function return b contains the solution.
you might also slightly modify elimination phase so elements of matrix a below diagonal are not zeroed, since there are irrelevant during back substitution phase. Therefore, the code becomes as shown below:
import numpy as np
def gauss_pivot(a,b,tol=1.0e-12):
"""
x = gaussPivot(a,b,tol=1.0e-12).
Solves [a]{x} = {b} by Gauss elimination with
scaled row pivoting
"""
a = np.copy(a)
b = np.copy(b)
n = len(b)
assert (np.all(np.shape(a) ==(n,n))) # check if a is a square matrix
# Set up scale factors
s = np.zeros(n)
for i in range(n):
s[i] = max(np.abs(a[i,:])) # find the max of each row
for k in range(0, n-1): #pivot row
# Row interchange, if needed
p = np.argmax(np.abs(a[k:n,k])/s[k:n]) # find which row has max item for each col k, and scale by s
if abs(a[p,k]) < tol:
raise Exception("Matrix is singular")
if p != k: # swap rows if current row does not contain max item with the one contains max item within same col
a[[k,p+k],:] = a[[p+k, k],:]
b[k],b[p+k] = b[p+k],b[k]
s[k],s[p+k] = s[p+k],s[k]
# Elimination phase of matrix a
for i in range(k+1,n):
if a[i,k] != 0.0: # skip if a(i,k) is already zero
lam = a [i,k]/a[k,k]
a[i,k+1:n] = a[i,k+1:n] - lam*a[k,k+1:n]
b[i] = b[i] - lam*b[k]
if abs(a[n-1,n-1]) < tol:
raise Exception("Matrix is singular")
# Back substitution phase, solution is substituted by b
b[n-1] = b[n-1]/a[n-1,n-1]
for k in range(n-2,-1,-1):
b[k] = (b[k] - np.dot(a[k,k+1:n],b[k+1:n]))/a[k,k]
return b
To use LU decomposition instead which is more ideal for b containing more than one column, the LU code is shown below
import numpy as np
def lu_decomp(a,tol=1.0e-9):
a = np.copy(a)
n = len(a)
assert (np.all(np.shape(a) ==(n,n))) # check if a is a square matrix
seq = np.arange(n, dtype=int)
s = np.zeros((n))
for i in range(n):
s[i] = max(abs(a[i,:]))
for k in range(0,n-1):
p = np.argmax(np.abs(a[k:n,k])/s[k:n])
if abs(a[p,k]) < tol:
raise Exception("Matrix is singular")
if p != k:
a[[k,p+k],:] = a[[p+k, k],:]
s[k],s[p+k] = s[p+k],s[k]
seq[k], seq[p+k] = seq[p+k],seq[k]
# Elimination
for i in range(k+1,n):
if a[i,k] != 0.0:
lam = a[i,k]/a[k,k]
a[i,k+1:n] = a[i,k+1:n] - lam*a[k,k+1:n]
a[i,k] = lam
return a,seq
def lu_solve(a,b,seq):
n = len(a)
x = b.copy()
for i in range(n):
x[i] = b[seq[i]]
# Solution
for k in range(1,n):
x[k] = x[k] - np.dot(a[k,0:k],x[0:k])
x[n-1] = x[n-1]/a[n-1,n-1]
for k in range(n-2,-1,-1):
x[k] = (x[k] - np.dot(a[k,k+1:n],x[k+1:n]))/a[k,k]
return x
a2 = np.random.randn(500,500)*100
b2 = np.random.randn(500,20)*100
a_decomposed, seq = lu_decomp(a2)
x2 = np.zeros_like(b2)
for col in range(b2.shape[1]):
x2[:,col] = lu_solve(a_decomposed, b2[:, col], seq)
if np.allclose(np.dot(a2,x2), b2) == True:
print("x2 is the correct solution")
Both methods gives the the output,
Gauss Elimination
x is the correct solution
LU method
x2 is the correct solution
I recommend you use scipy linalg package, from scipy.linalg import solve, lu_factor, lu_solve.
They perform way faster for large matrix size. you can use the same code above but annotate them with numba jit so for large matrix the performance is way better.
from numba import jit
#jit
def gauss_pivot(a, b):
...
...
acknowledgement: codes inspired from the book numerical methods in science and engineering with Python by Prof. Jaan Kiusalaas
https://www.amazon.co.uk/Numerical-Methods-Engineering-Python-3/dp/1107033853/ref=sr_1_1?ie=UTF8&qid=1517845946&sr=8-1&keywords=numerical+method+in+science+and+engineering+with+python
I want to generate all directions from a point in a 3D grid, but I can't quite get my head around the next bit. For the record it's all stored in a single list, so I need some maths to calculate where the next point will be.
I only really need 3 calculations to calculate any of the 26 or so different directions (up, up left, up left forwards, up left backwards, up right, up right forwards, etc), so I decided to work with X, Y, Z, then split them into up/down left/right etc, to then get the correct number to add or subtract. Generating this list to get the maths working however, seems to be the hard bit.
direction_combinations = 'X Y Z XY XZ YZ XYZ'.split()
direction_group = {}
direction_group['X'] = 'LR'
direction_group['Y'] = 'UD'
direction_group['Z'] = 'FB'
So basically, using the below code, this is the kind of stuff I'd like it to do, but obviously not have it hard coded. I could do it in a hacky way, but I imagine there's something really simple I'm missing here.
#Earlier part of the code to get this bit working
#I've also calculated the edges but it's not needed until after I've got this bit working
grid_size = 4
direction_maths = {}
direction_maths['U'] = pow(grid_size, 2)
direction_maths['R'] = 1
direction_maths['F'] = grid_size
direction_maths['D'] = -direction_maths['U']
direction_maths['L'] = -direction_maths['R']
direction_maths['B'] = -direction_maths['F']
#Bit to get working
starting_point = 25
current_direction = 'Y'
possible_directions = [direction_group[i] for i in list(current_direction)]
for y in list(possible_directions[0]):
print starting_point + direction_maths[y]
# 41 and 9 are adjacent on the Y axis
current_direction = 'XYZ'
possible_directions = [direction_group[i] for i in list(current_direction)]
for x in list(possible_directions[0]):
for y in list(possible_directions[1]):
for z in list(possible_directions[2]):
print starting_point + direction_maths[x] + direction_maths[y] + direction_maths[z]
# 44, 36, 12, 4, 46, 38, 14 and 6 are all adjacent on the corner diagonals
Here's a general idea of how the grid looks with the list indexes (using 4x4x4 as an example):
________________
/ 0 / 1 / 2 / 3 /
/___/___/___/___/
/ 4 / 5 / 6 / 7 /
/___/___/___/___/
/ 8 / 9 /10 /11 /
/___/___/___/___/
/12 /13 /14 /15 /
/___/___/___/___/
________________
/16 /17 /18 /19 /
/___/___/___/___/
/20 /21 /22 /23 /
/___/___/___/___/
/24 /25 /26 /27 /
/___/___/___/___/
/28 /29 /30 /31 /
/___/___/___/___/
________________
/32 /33 /34 /35 /
/___/___/___/___/
/36 /37 /38 /39 /
/___/___/___/___/
/40 /41 /42 /43 /
/___/___/___/___/
/44 /45 /46 /47 /
/___/___/___/___/
________________
/48 /49 /50 /51 /
/___/___/___/___/
/52 /53 /54 /55 /
/___/___/___/___/
/56 /57 /58 /59 /
/___/___/___/___/
/60 /61 /62 /63 /
/___/___/___/___/
Edit: Using the answers mixed with what I posted originally (wanted to avoid converting to and from 3D points if possible), this is what I ended up with to count the number of complete rows :)
def build_directions():
direction_group = {}
direction_group['X'] = 'LR'
direction_group['Y'] = 'UD'
direction_group['Z'] = 'FB'
direction_group[' '] = ' '
#Come up with all possible directions
all_directions = set()
for x in [' ', 'X']:
for y in [' ', 'Y']:
for z in [' ', 'Z']:
x_directions = list(direction_group[x])
y_directions = list(direction_group[y])
z_directions = list(direction_group[z])
for i in x_directions:
for j in y_directions:
for k in z_directions:
all_directions.add((i+j+k).replace(' ', ''))
#Narrow list down to remove any opposite directions
some_directions = all_directions
opposite_direction = all_directions.copy()
for i in all_directions:
if i in opposite_direction:
new_direction = ''
for j in list(i):
for k in direction_group.values():
if j in k:
new_direction += k.replace(j, '')
opposite_direction.remove(new_direction)
return opposite_direction
class CheckGrid(object):
def __init__(self, grid_data):
self.grid_data = grid_data
self.grid_size = calculate_grid_size(self.grid_data)
self.grid_size_squared = pow(grid_size, 2)
self.grid_size_cubed = len(grid_data)
self.direction_edges = {}
self.direction_edges['U'] = range(self.grid_size_squared)
self.direction_edges['D'] = range(self.grid_size_squared*(self.grid_size-1), self.grid_size_squared*self.grid_size)
self.direction_edges['R'] = [i*self.grid_size+self.grid_size-1 for i in range(self.grid_size_squared)]
self.direction_edges['L'] = [i*self.grid_size for i in range(self.grid_size_squared)]
self.direction_edges['F'] = [i*self.grid_size_squared+j+self.grid_size_squared-self.grid_size for i in range(self.grid_size) for j in range(self.grid_size)]
self.direction_edges['B'] = [i*self.grid_size_squared+j for i in range(self.grid_size) for j in range(self.grid_size)]
self.direction_edges[' '] = []
self.direction_maths = {}
self.direction_maths['D'] = pow(self.grid_size, 2)
self.direction_maths['R'] = 1
self.direction_maths['F'] = self.grid_size
self.direction_maths['U'] = -self.direction_maths['D']
self.direction_maths['L'] = -self.direction_maths['R']
self.direction_maths['B'] = -self.direction_maths['F']
self.direction_maths[' '] = 0
def points(self):
total_points = defaultdict(int)
opposite_directions = build_directions()
all_matches = set()
#Loop through each point
for starting_point in range(len(self.grid_data)):
current_player = self.grid_data[starting_point]
if current_player:
for i in opposite_directions:
#Get a list of directions and calculate movement amount
possible_directions = [list(i)]
possible_directions += [[j.replace(i, '') for i in possible_directions[0] for j in direction_group.values() if i in j]]
direction_movement = sum(self.direction_maths[j] for j in possible_directions[0])
#Build list of invalid directions
invalid_directions = [[self.direction_edges[j] for j in possible_directions[k]] for k in (0, 1)]
invalid_directions = [[item for sublist in j for item in sublist] for j in invalid_directions]
num_matches = 1
list_match = [starting_point]
#Use two loops for the opposite directions
for j in (0, 1):
current_point = starting_point
while current_point not in invalid_directions[j]:
current_point += direction_movement*int('-'[:j]+'1')
if self.grid_data[current_point] == current_player:
num_matches += 1
list_match.append(current_point)
else:
break
#Add a point if enough matches
if num_matches == self.grid_size:
list_match = tuple(sorted(list_match))
if list_match not in all_matches:
all_matches.add(list_match)
total_points[current_player] += 1
return total_points
Here's basically the same thing that #AnnoSielder did, but makes use of itertools to reduce the amount of code.
from itertools import product
# Get a list of all 26 possible ways to move from a given coordinate in a 3 coordinate system.
base_deltas = filter(lambda point: not all(axis ==0 for axis in point), list(product([-1, 0, 1], repeat=3)))
# Define your max axis length or your grid size
grid_size = 4
# Simple function that applys the deltas to the given coordinate and returns you the list.
def apply_deltas(deltas, coordinate):
return [
(coordinate[0]+x, coordinate[1]+y, coordinate[2]+z)
for x, y, z in deltas
]
# This will determine whether the point is out of bounds for the given grid
is_out_of_bounds = lambda point: all(0 <= axis < grid_size for axis in point)
# Define your point, in this case it's block #27 in your example
coordinate = [3, 2, 1]
# Apply the deltas, then filter using the is_out_of_bounds lambda
directions = filter(is_out_of_bounds, apply_deltas(base_deltas, coordinate))
# directions is now the list of 17 coordinates that you could move to.
Don't make thinks unnecessary complicated. Do not describe a point in 3 dimensions with 1 number - 3 coordinates means 3 numbers.
Should be something like this:
numb = 37
cube_size = 4
# convert to x - y - z
start = [0, 0, 0]
start[2] = numb / cube_size ** 2
numb = numb % cube_size ** 2
start[1] = numb / cube_size
start[0] = numb % cube_size
for x in [-1, 0, 1]:
current_x = start[0] + x
for y in [-1, 0, 1]:
current_y = start[1] + y
for z in [-1, 0, 1]:
current_z = start[2] + z
#reconvert
convert = current_x + current_y * cube_size + current_z * cube_size ** 2
print("x: " + str(current_x) + " y: " + str(current_y) + " z: " + str(current_z) + " => " + str(convert))
Simply generate your x/y/z-coordinate, then run all possibilities of add -1/0/1 to these coordinates and re-convert to your number in the grid.