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| 1 | + |
| 2 | +# coding: utf-8 |
| 3 | + |
| 4 | +# In[1]: |
| 5 | + |
| 6 | + |
| 7 | +#fractal is one of the interesting topics in geometry |
| 8 | +#it is usually described by a recursive function |
| 9 | +#voila,here we are! |
| 10 | +import matplotlib.pyplot as plt |
| 11 | + |
| 12 | + |
| 13 | +# In[2]: |
| 14 | + |
| 15 | + |
| 16 | +#compute euclidean distance |
| 17 | +def euclidean_distance(point1,point2): |
| 18 | + return ((point1[0]-point2[0])**2+(point1[1]-point2[1])**2)**0.5 |
| 19 | + |
| 20 | + |
| 21 | +# In[3]: |
| 22 | + |
| 23 | + |
| 24 | +#simple solution to get coefficients of the equation |
| 25 | +def get_line_params(x1,y1,x2,y2): |
| 26 | + |
| 27 | + slope=(y1-y2)/(x1-x2) |
| 28 | + intercept=y1-slope*x1 |
| 29 | + |
| 30 | + return slope,intercept |
| 31 | + |
| 32 | + |
| 33 | +# In[4]: |
| 34 | + |
| 35 | + |
| 36 | +#standard solution to quadratic equation |
| 37 | +def solve_quadratic_equation(A,B,C): |
| 38 | + x1=(-B+(B**2-4*A*C)**0.5)/(2*A) |
| 39 | + x2=(-B-(B**2-4*A*C)**0.5)/(2*A) |
| 40 | + return [x1,x2] |
| 41 | + |
| 42 | + |
| 43 | +# In[5]: |
| 44 | + |
| 45 | + |
| 46 | +#analytic geometry to compute target datapoints |
| 47 | +def get_datapoint(pivot,measure,length,direction='inner'): |
| 48 | + |
| 49 | + #for undefined slope |
| 50 | + if pivot[0]==measure[0]: |
| 51 | + y1=pivot[1]+length |
| 52 | + y2=pivot[1]-length |
| 53 | + x1=pivot[0] |
| 54 | + x2=pivot[0] |
| 55 | + |
| 56 | + #for general cases |
| 57 | + else: |
| 58 | + |
| 59 | + #get line equation |
| 60 | + slope,intercept=get_line_params(pivot[0],pivot[1], |
| 61 | + measure[0],measure[1],) |
| 62 | + |
| 63 | + #solve quadratic equation |
| 64 | + A=1 |
| 65 | + B=-2*pivot[0] |
| 66 | + C=pivot[0]**2-length**2/(slope**2+1) |
| 67 | + x1,x2=solve_quadratic_equation(A,B,C) |
| 68 | + |
| 69 | + #get y from line equation |
| 70 | + y1=slope*x1+intercept |
| 71 | + y2=slope*x2+intercept |
| 72 | + |
| 73 | + if direction=='inner': |
| 74 | + |
| 75 | + #take the one between pivot and measure points |
| 76 | + datapoint=min([(x1,y1),(x2,y2)], |
| 77 | + key=lambda x:euclidean_distance(x,measure)) |
| 78 | + else: |
| 79 | + |
| 80 | + #take the one farther away from measure points |
| 81 | + datapoint=max([(x1,y1),(x2,y2)], |
| 82 | + key=lambda x:euclidean_distance(x,measure)) |
| 83 | + |
| 84 | + return datapoint |
| 85 | + |
| 86 | + |
| 87 | +# In[6]: |
| 88 | + |
| 89 | + |
| 90 | +#recursively plot symmetric pythagorean tree at 45 degree |
| 91 | +# https://larryriddle.agnesscott.org/ifs/pythagorean/pythTree.htm |
| 92 | +def pythagorean_tree(ax,top_left,top_right,bottom_left, |
| 93 | + bottom_right,current_angle,line_len,n): |
| 94 | + |
| 95 | + #plot square |
| 96 | + ax.add_patch(plt.Rectangle(xy=bottom_left,width=line_len,height=line_len, |
| 97 | + angle=current_angle,)) |
| 98 | + |
| 99 | + if n==0: |
| 100 | + return |
| 101 | + else: |
| 102 | + |
| 103 | + #find mid point |
| 104 | + #midpoint has to satisfy two conditions |
| 105 | + #it has to be on the same line as bottom_left and bottom_right |
| 106 | + #assume this line follows y=kx+b |
| 107 | + #the midpoint is (x,kx+b) |
| 108 | + #bottom_left is (α,kα+b),bottom_right is (δ,kδ+b) |
| 109 | + #the euclidean distance between midpoint and bottom_left should be |
| 110 | + #half of the euclidean distance between bottom_left and bottom_right |
| 111 | + #(x-α)**2+(kx+b-kα-b)**2=((α-δ)**2+(kα+b-kδ-b)**2)/4 |
| 112 | + #apart from x,everything else in the equation is constant |
| 113 | + #this forms a simple quadratic solution to get two roots |
| 114 | + #one root would be between bottom_left and bottom_right which yields midpoint |
| 115 | + #and the other would be farther away from bottom_right |
| 116 | + #this function solves the equation via (-B+(B**2-4*A*C)**0.5)/(2*A) |
| 117 | + #alternatively,you can use scipy.optimize.root |
| 118 | + #the caveat is it does not offer both roots |
| 119 | + #a wrong initial guess could take you to the wrong root |
| 120 | + bottom_mid=get_datapoint(bottom_left,bottom_right,line_len/2) |
| 121 | + top_mid=get_datapoint(top_left,top_right,line_len/2) |
| 122 | + |
| 123 | + #compute the top point of a triangle |
| 124 | + #the computation is similar to midpoint |
| 125 | + #the euclidean distance between triangle_top and top_mid should be |
| 126 | + #half of the distance between top_mid and bottom_mid |
| 127 | + triangle_top=get_datapoint(top_mid,bottom_mid, |
| 128 | + line_len/2,direction='outer') |
| 129 | + |
| 130 | + #get top left for right square |
| 131 | + #the computation is similar to midpoint |
| 132 | + #the euclidean distance between triangle_top and rightsq_topleft |
| 133 | + #should be the same as the distance between triangle_top and top_left |
| 134 | + rightsq_topleft=get_datapoint(triangle_top,top_left, |
| 135 | + line_len/(2**0.5),direction='outer') |
| 136 | + |
| 137 | + #get midpoint of the diagonal between rightsq_topleft and top_right |
| 138 | + #the computation is similar to midpoint |
| 139 | + #the euclidean distance between rightsq_diag_mid and rightsq_topleft |
| 140 | + #should be half of the distance between rightsq_topleft and top_right |
| 141 | + rightsq_diag_mid=get_datapoint(top_right,rightsq_topleft,line_len/2) |
| 142 | + rightsq_topright=get_datapoint(rightsq_diag_mid,triangle_top, |
| 143 | + line_len/2,direction='outer') |
| 144 | + |
| 145 | + #get top left and right for left square similar to right square |
| 146 | + leftsq_topleft=get_datapoint(triangle_top,rightsq_topright, |
| 147 | + line_len,direction='outer') |
| 148 | + leftsq_topright=get_datapoint(triangle_top,top_right, |
| 149 | + line_len/(2**0.5),direction='outer') |
| 150 | + |
| 151 | + #recursive do the same for left square |
| 152 | + pythagorean_tree(ax,leftsq_topleft,leftsq_topright, |
| 153 | + top_left,triangle_top,current_angle+45, |
| 154 | + line_len/(2**0.5),n-1) |
| 155 | + |
| 156 | + #recursive do the same for right square |
| 157 | + pythagorean_tree(ax,rightsq_topleft,rightsq_topright, |
| 158 | + triangle_top,top_right,current_angle-45, |
| 159 | + line_len/(2**0.5),n-1) |
| 160 | + |
| 161 | + |
| 162 | +# In[7]: |
| 163 | + |
| 164 | + |
| 165 | +#initialize |
| 166 | +top_left=(0,0) |
| 167 | +top_right=(1,0) |
| 168 | +bottom_left=(0,-1) |
| 169 | +bottom_right=(1,-1) |
| 170 | +n=5 |
| 171 | +current_angle=0 |
| 172 | +line_len=euclidean_distance(top_left,top_right) |
| 173 | + |
| 174 | + |
| 175 | +# In[8]: |
| 176 | + |
| 177 | + |
| 178 | +#viz |
| 179 | +ax=plt.figure(figsize=(10,8)).add_subplot(111) |
| 180 | +pythagorean_tree(ax,top_left,top_right,bottom_left, |
| 181 | + bottom_right,current_angle,line_len,n) |
| 182 | + |
| 183 | +#limit figure dimension for better viz |
| 184 | +plt.ylim(min([bottom_left[1],bottom_right[1]]), |
| 185 | + max([top_left[1],top_right[1]])+euclidean_distance( |
| 186 | + top_left,top_right)*(n-2) |
| 187 | + ) |
| 188 | +plt.xlim(min([bottom_left[0],top_left[0]])-euclidean_distance( |
| 189 | + top_left,top_right)*(n-1)/2, |
| 190 | + max([bottom_right[0],top_right[0]])+euclidean_distance( |
| 191 | + top_left,top_right)*(n-1)/2, |
| 192 | + ) |
| 193 | + |
| 194 | +plt.axis('off') |
| 195 | +plt.show() |
| 196 | + |
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