Clean up headers, added exercise.

UKF chapter's headings were all messed up in the PDF output. I added
an exercise to use the unscented transform on the 5,000 point nonlinear
monte carlo example from the beginning of the chapter.

10_Unscented_Kalman_Filters/Unscented_Kalman_Filter.ipynb

@@ -1,7 +1,7 @@
 {
  "metadata": {
   "name": "",
-  "signature": "sha256:a65be81c259be7ffd74061efd23d3345108262b7f803512ced8660b900899d6e"
+  "signature": "sha256:5dcd9e2170b6dd3b4854a622f15d1432c9897ce652b0994c6b8ab0a32b67d2a7"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -259,13 +259,13 @@
        ],
        "metadata": {},
        "output_type": "pyout",
-       "prompt_number": 1,
+       "prompt_number": 40,
        "text": [
-        "<IPython.core.display.HTML at 0x7fed97b51240>"
+        "<IPython.core.display.HTML at 0x7facfd875a90>"
        ]
       }
      ],
-     "prompt_number": 1
+     "prompt_number": 40
     },
     {
      "cell_type": "markdown",
@@ -309,13 +309,13 @@
       {
        "metadata": {},
        "output_type": "display_data",
-       "png": 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truncated
+       "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed7b6ac390>"
+        "<matplotlib.figure.Figure at 0x7facfaeb5d68>"
        ]
       }
      ],
-     "prompt_number": 2
+     "prompt_number": 41
     },
     {
      "cell_type": "markdown",
@@ -347,13 +347,13 @@
       {
        "metadata": {},
        "output_type": "display_data",
-       "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed7b68a208>"
+        "<matplotlib.figure.Figure at 0x7facfadfdeb8>"
        ]
       }
      ],
-     "prompt_number": 3
+     "prompt_number": 42
     },
     {
      "cell_type": "markdown",
@@ -396,11 +396,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed7b683d68>"
+        "<matplotlib.figure.Figure at 0x7facfade7c88>"
        ]
       }
      ],
-     "prompt_number": 4
+     "prompt_number": 43
     },
     {
      "cell_type": "markdown",
@@ -408,11 +408,22 @@
      "source": [
       "Here on the left we show an ellipse depicting the $1\\sigma$ distribution of two state variables. The arrows show how three randomly sampled points might be transformed by some arbitrary nonlinear function to a new distribution. The ellipse on the right is drawn semi-transparently to indicate that it is an *estimate* of the mean and variance of this collection of points - if we were to sample, say, a million points the shape of the points might be very far from an ellipse. \n",
       "\n",
-      "Let's look at that by running a bunch of points through a nonlinear function.\n",
-      "\n",
-      "###### Exercise: Plot 2D points\n",
-      "\n",
-      "Write a function to pass points through the state transition function:\n",
+      "Let's look at that by running a bunch of points through a nonlinear function."
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise: Plot 2D points"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "Write a function to pass randomly generated points with a Gaussian distribution through the state transition function:\n",
       "\n",
       "$$\\mathbf{Fx} = \\begin{bmatrix}1 & 1 \\\\ .1x &y\\end{bmatrix}\\begin{bmatrix}x\\\\y\\end{bmatrix}$$ \n",
       "\n",
@@ -431,14 +442,20 @@
      "language": "python",
      "metadata": {},
      "outputs": [],
-     "prompt_number": 5
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Solution"
+     ]
     },
     {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
-      "###### Solution\n",
-      "\n",
       "Here is my answer."
      ]
     },
@@ -460,7 +477,7 @@
       "mean = (0, 0)\n",
       "p = np.array([[32, 15],[15., 40.]])\n",
       "\n",
-      "xs ,ys = multivariate_normal(mean=mean, cov=p, size=5000).T\n",
+      "xs ,ys = multivariate_normal(mean=mean, cov=p, size=50000).T\n",
       "fxs, fys = [], []\n",
       "for x,y in zip(xs,ys):\n",
       "    fx, fy = f(x,y)\n",
@@ -475,15 +492,18 @@
       "computed_mean_y = np.average(fys)\n",
       "\n",
       "plt.subplot(121)\n",
-      "plt.scatter (xs,ys, marker='.')\n",
+      "plt.scatter (xs,ys, marker='.', alpha=0.1)\n",
       "plt.axis('equal')\n",
       "\n",
       "plt.subplot(122)\n",
-      "plt.scatter(fxs, fys, marker='.')\n",
-      "plt.scatter(mean_fx[0], mean_fx[1], marker='v', s=120,c='r')\n",
-      "plt.scatter(computed_mean_x, computed_mean_y, marker='*',s=120,c='r')\n",
+      "plt.scatter(fxs, fys, marker='.', alpha=0.1)\n",
+      "plt.scatter(mean_fx[0], mean_fx[1], \n",
+      "            marker='v', s=300,c='k', label='mean')\n",
+      "plt.scatter(computed_mean_x, computed_mean_y, \n",
+      "            marker='*',s=120,c='r', label='computed')\n",
       "plt.ylim([-10,290])\n",
       "plt.xlim([-150,150])\n",
+      "plt.legend(loc='best', scatterpoints=1)\n",
       "plt.show()\n",
       "print ('Difference in mean x={:.3f}, y={:.3f}'.format(\n",
       "       computed_mean_x-mean_fx[0], computed_mean_y-mean_fx[1]))"
@@ -494,27 +514,40 @@
       {
        "metadata": {},
        "output_type": "display_data",
-       "png": 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+       "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76b94a90>"
+        "<matplotlib.figure.Figure at 0x7facf319c390>"
        ]
       },
       {
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "Difference in mean x=0.051, y=42.372\n"
+        "Difference in mean x=-0.008, y=43.321\n"
        ]
       }
      ],
-     "prompt_number": 6
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "This plot shows the strong nonlinearity that occurs with this function, and the large error that would result if we linearized the function at (0,0). We would reasonably expect an EKF to diverge rapidly with this function."
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Choosing Sigma Points (continued)"
+     ]
     },
     {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
-      "This plot shows the strong nonlinearity that occurs with this function, and the large error that would result if we linearized the function at (0,0). We would reasonably expect an EKF to diverge rapidly with this function.\n",
-      "\n",
       "In the chart above we used 5,000 points to generate this solution. While the answer is quite accurate, computing 5,000 points for every update would cause our filter to be very slow. So, what would be fewest number of sampled points that we can use, and what kinds of constraints does this problem formulation put on the points? We will assume that we have no special knowledge about the nonlinear function as we want to find a generalized algorithm that works for any function. For reasons that come clear in the next section, we will call these points *sigma points*.\n",
       "\n",
       "Let's consider the simplest possible case and see if it offers any insight. The simplest possible system is *identity* - the transformation does not alter the input. In mathematical notation this is just $f(x) = x$. It should be clear that if our algorithm does not work for the identity function then the filter will never converge. In other words, if the input is  1 (for a one dimensional system), the output must also be 1. If the output was different, such as 1.1, then when we fed 1.1 into the transform at the next time step, we'd get out yet another number, maybe 1.23. The filter would run away (diverge). \n",
@@ -542,11 +575,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed7b399c88>"
+        "<matplotlib.figure.Figure at 0x7facfb08a828>"
        ]
       }
      ],
-     "prompt_number": 7
+     "prompt_number": 46
     },
     {
      "cell_type": "markdown",
@@ -582,11 +615,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76e11710>"
+        "<matplotlib.figure.Figure at 0x7facfa8e9f28>"
        ]
       }
      ],
-     "prompt_number": 8
+     "prompt_number": 47
     },
     {
      "cell_type": "markdown",
@@ -637,11 +670,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76d06080>"
+        "<matplotlib.figure.Figure at 0x7facfbe58320>"
        ]
       }
      ],
-     "prompt_number": 9
+     "prompt_number": 48
     },
     {
      "cell_type": "markdown",
@@ -650,12 +683,18 @@
       "We can see that the sigma points lie between the first and second deviation, and that the larger $\\kappa$ spreads the points out further. Furthermore, the larger $\\kappa$ weighs the mean (center point) higher than the smaller $\\kappa$, and weighs the rest of the sigma points less. This should fit our intuition - the further a point is from the mean the less we should weight it. We don't know *how* these weights and sigma points are selected yet, but the choices look reasonable."
      ]
     },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Sigma Point Algorithm"
+     ]
+    },
     {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
-      "## Choosing Sigma Points\n",
-      "\n",
       "As I already said there are several published algorithms for choosing the sigma points. I will present the method first chosen by the inventors of the UKF. \n",
       "\n",
       "Our first sigma point is always going to be the mean of our input. We will call this $\\mathcal{X}_0$. So,\n",
@@ -708,18 +747,37 @@
       "$$"
      ]
     },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "The Unscented Filter"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "Now we can present the entire Unscented Kalman filter algorithm. Assume that there is a function $f(x)$ that performs the state transition for our filer - it predicts the next state given the current state. Also assume there is a measurement function $h(x)$  - it takes the current state and computes what measurement that state corresponds to. These are just nonlinear forms of the $\\mathbf{F}$ and $\\mathbf{H}$ matrices used by the linear Kalman filter."
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Predict Step"
+     ]
+    },
     {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
-      "## The Unscented Filter\n",
-      "\n",
-      "Now we can present the entire Unscented Kalman filter algorithm. Assume that there is a function $f(x)$ that performs the state transition for our filer - it predicts the next state given the current state. Also assume there is a measurement function $h(x)$  - it takes the current state and computes what measurement that state corresponds to. These are just nonlinear forms of the $\\mathbf{F}$ and $\\mathbf{H}$ matrices used by the linear Kalman filter.\n",
-      "\n",
-      "### Predict Step\n",
       "For the predict step, we will generate the weights and sigma points as specified above.\n",
-      "$$\\mathcal{X} = sigma\\_function(\\mu, \\Sigma) \\\\\n",
-      "W = weight\\_function(n, \\kappa)$$\n",
+      "$$\\begin{aligned}\n",
+      "\\mathcal{X} &= sigma\\_function(\\mu, \\Sigma) \\\\\n",
+      "W &= weight\\_function(n, \\kappa)\\end{aligned}$$\n",
       "\n",
       "We pass each sigma point through the function f.\n",
       "\n",
@@ -740,11 +798,18 @@
       "\\end{aligned}$$"
      ]
     },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Update Step"
+     ]
+    },
     {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
-      "### Update Step\n",
       "Now we can perform the update step of the filter. Recall that Kalman filters perform the update state in measurement space. So, the first thing we do is convert the sigma points into measurements using the $h(x)$ function.\n",
       "\n",
       "\n",
@@ -867,7 +932,7 @@
      "language": "python",
      "metadata": {},
      "outputs": [],
-     "prompt_number": 10
+     "prompt_number": 49
     },
     {
      "cell_type": "markdown",
@@ -912,7 +977,7 @@
      "language": "python",
      "metadata": {},
      "outputs": [],
-     "prompt_number": 11
+     "prompt_number": 50
     },
     {
      "cell_type": "markdown",
@@ -939,7 +1004,7 @@
     },
     {
      "cell_type": "heading",
-     "level": 3,
+     "level": 2,
      "metadata": {},
      "source": [
       "Implement the Filter"
@@ -1016,13 +1081,13 @@
       {
        "metadata": {},
        "output_type": "display_data",
-       "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76e0eeb8>"
+        "<matplotlib.figure.Figure at 0x7facf185c2e8>"
        ]
       }
      ],
-     "prompt_number": 12
+     "prompt_number": 51
     },
     {
      "cell_type": "markdown",
@@ -1048,13 +1113,13 @@
       {
        "metadata": {},
        "output_type": "display_data",
-       "png": 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truncated
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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76e31898>"
+        "<matplotlib.figure.Figure at 0x7facfad775f8>"
        ]
       }
      ],
-     "prompt_number": 13
+     "prompt_number": 52
     },
     {
      "cell_type": "markdown",
@@ -1097,11 +1162,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76e146d8>"
+        "<matplotlib.figure.Figure at 0x7facfafed518>"
        ]
       }
      ],
-     "prompt_number": 14
+     "prompt_number": 53
     },
     {
      "cell_type": "markdown",
@@ -1220,11 +1285,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76d4eb00>"
+        "<matplotlib.figure.Figure at 0x7facfb02c6d8>"
        ]
       }
      ],
-     "prompt_number": 15
+     "prompt_number": 54
     },
     {
      "cell_type": "markdown",
@@ -1252,11 +1317,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed7b3d36a0>"
+        "<matplotlib.figure.Figure at 0x7facf310b320>"
        ]
       }
      ],
-     "prompt_number": 16
+     "prompt_number": 55
     },
     {
      "cell_type": "markdown",
@@ -1289,7 +1354,7 @@
      "language": "python",
      "metadata": {},
      "outputs": [],
-     "prompt_number": 17
+     "prompt_number": 56
     },
     {
      "cell_type": "heading",
@@ -1352,11 +1417,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76b0de80>"
+        "<matplotlib.figure.Figure at 0x7facfb000a58>"
        ]
       }
      ],
-     "prompt_number": 18
+     "prompt_number": 57
     },
     {
      "cell_type": "heading",
@@ -1403,11 +1468,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76cbc400>"
+        "<matplotlib.figure.Figure at 0x7facf183ec88>"
        ]
       }
      ],
-     "prompt_number": 19
+     "prompt_number": 58
     },
     {
      "cell_type": "markdown",
@@ -1421,7 +1486,7 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Problem: Sensor Position Effects"
+      "Exercise: Sensor Position Effects"
      ]
     },
     {
@@ -1440,7 +1505,7 @@
      "language": "python",
      "metadata": {},
      "outputs": [],
-     "prompt_number": 20
+     "prompt_number": 59
     },
     {
      "cell_type": "heading",
@@ -1478,11 +1543,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76cde0f0>"
+        "<matplotlib.figure.Figure at 0x7facf1831898>"
        ]
       }
      ],
-     "prompt_number": 21
+     "prompt_number": 60
     },
     {
      "cell_type": "markdown",
@@ -1493,7 +1558,7 @@
     },
     {
      "cell_type": "heading",
-     "level": 3,
+     "level": 2,
      "metadata": {},
      "source": [
       "Exercise: Compute Position Errors"
@@ -1515,7 +1580,7 @@
      "language": "python",
      "metadata": {},
      "outputs": [],
-     "prompt_number": 22
+     "prompt_number": 61
     },
     {
      "cell_type": "heading",
@@ -1557,7 +1622,7 @@
        ]
       }
      ],
-     "prompt_number": 23
+     "prompt_number": 62
     },
     {
      "cell_type": "heading",
@@ -1599,11 +1664,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76d9e160>"
+        "<matplotlib.figure.Figure at 0x7facf18e7be0>"
        ]
       }
      ],
-     "prompt_number": 24
+     "prompt_number": 63
     },
     {
      "cell_type": "heading",
@@ -1670,11 +1735,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed7b09c748>"
+        "<matplotlib.figure.Figure at 0x7facf1898f98>"
        ]
       }
      ],
-     "prompt_number": 25
+     "prompt_number": 64
     },
     {
      "cell_type": "markdown",
@@ -1770,11 +1835,11 @@
        "output_type": "display_data",
        "png": 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        "text": [
-        "<matplotlib.figure.Figure at 0x7fed7b0ed358>"
+        "<matplotlib.figure.Figure at 0x7facfb083ef0>"
        ]
       }
      ],
-     "prompt_number": 26
+     "prompt_number": 65
     },
     {
      "cell_type": "markdown",
@@ -1805,11 +1870,145 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed7b0f3748>"
+        "<matplotlib.figure.Figure at 0x7facfad5f2b0>"
        ]
       }
      ],
-     "prompt_number": 27
+     "prompt_number": 66
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise: Visualize Sigma Points"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "Early in the chapter you were asked to pass points through a nonlinear function and to plot the mean. Use the UKF to generate sigma points and pass them through the nonlinear function, and plot the sigma points and their mean against the 5,000 points.\n",
+      "\n",
+      "Pass the points through the state transition function:\n",
+      "\n",
+      "$$\\mathbf{Fx} = \\begin{bmatrix}1 & 1 \\\\ .1x &y\\end{bmatrix}\\begin{bmatrix}x\\\\y\\end{bmatrix}$$ \n",
+      "\n",
+      "for the mean $$\\mu = \\begin{bmatrix}0\\\\0\\end{bmatrix}$$ and the covariance\n",
+      "$$\\Sigma=\\begin{bmatrix}32&15\\\\15&40\\end{bmatrix}$$"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Your solution here"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 67
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Solution"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from filterpy.kalman import unscented_transform\n",
+      "\n",
+      "def f(x,y):\n",
+      "    return np.array([x+y, .1*x**2 +y*y])\n",
+      "\n",
+      "mean = (0, 0)\n",
+      "p = np.array([[32., 15],\n",
+      "              [15., 40.]])\n",
+      "\n",
+      "\n",
+      "### compute mean via UKF\n",
+      "kappa = 0.\n",
+      "w = UKF.weights(2, kappa)\n",
+      "sigmas = UKF.sigma_points(mean, p, kappa)\n",
+      "sigmas_f = np.zeros((5,2))\n",
+      "for i in range(5):\n",
+      "        sigmas_f[i] = f(sigmas[i,0], sigmas[i,1])\n",
+      "  \n",
+      "\n",
+      "ukf_mean, _ = unscented_transform(sigmas_f, w, w, np.zeros((2,2)))\n",
+      "\n",
+      "# generate 500000 points to compute accurate mean\n",
+      "np.random.seed(73291)\n",
+      "xs ,ys = multivariate_normal(mean=mean, cov=p, size=10000).T\n",
+      "fxs, fys = [], []\n",
+      "for x,y in zip(xs,ys):\n",
+      "    fx, fy = f(x,y)\n",
+      "    \n",
+      "    fxs.append(fx)\n",
+      "    fys.append(fy)\n",
+      "\n",
+      "computed_mean_x = np.average(fxs)\n",
+      "computed_mean_y = np.average(fys)\n",
+      "\n",
+      "\n",
+      "plt.subplot(121)\n",
+      "plt.scatter (xs[0:10000], ys[0:10000], marker='.', alpha=0.1)\n",
+      "plt.scatter(sigmas[:,0], sigmas[:,1], c='r',s=50)\n",
+      "plt.axis('equal')\n",
+      "\n",
+      "plt.subplot(122)\n",
+      "plt.scatter(fxs[0:10000], fys[0:10000], marker='.', alpha=0.1)\n",
+      "plt.scatter(computed_mean_x, computed_mean_y, \n",
+      "            marker='v',s=300,c='k', label='computed')\n",
+      "plt.scatter(ukf_mean[0], ukf_mean[1], \n",
+      "            marker='o', s=50, c='r', label='ukf')\n",
+      "plt.ylim([-10,290])\n",
+      "plt.xlim([-150,150])\n",
+      "plt.legend(loc='best', scatterpoints=1)\n",
+      "plt.show()\n",
+      "print ('Difference in mean x={:.3f}, y={:.3f}'.format(\n",
+      "       computed_mean_x-ukf_mean[0], computed_mean_y-ukf_mean[1]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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truncated
+       "text": [
+        "<matplotlib.figure.Figure at 0x7facf3148c18>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Difference in mean x=0.083, y=-0.215\n"
+       ]
+      }
+     ],
+     "prompt_number": 68
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Discussion"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "The first exercise had an error of roughly 43 in the y axis, whereas with 10,000,000 points simulated the UKF has an error of around 0.03. It is very likely that a larger number is displayed above, but that is because I set the number of points to 10,000 so the cell will run quickly for you. I find this result very astonishing - using only 5 points the UKF generates a better estimate than using 5000 points with a naive algorithm. "
+     ]
     },
     {
      "cell_type": "heading",
@@ -1898,7 +2097,7 @@
      "language": "python",
      "metadata": {},
      "outputs": [],
-     "prompt_number": 28
+     "prompt_number": 69
     },
     {
      "cell_type": "heading",
@@ -1992,13 +2191,13 @@
       {
        "metadata": {},
        "output_type": "pyout",
-       "prompt_number": 29,
+       "prompt_number": 70,
        "text": [
         "array([410, 520, 630])"
        ]
       }
      ],
-     "prompt_number": 29
+     "prompt_number": 70
     },
     {
      "cell_type": "markdown",
@@ -2083,7 +2282,7 @@
      "language": "python",
      "metadata": {},
      "outputs": [],
-     "prompt_number": 30
+     "prompt_number": 71
     },
     {
      "cell_type": "heading",
@@ -2129,7 +2328,7 @@
      "language": "python",
      "metadata": {},
      "outputs": [],
-     "prompt_number": 31
+     "prompt_number": 72
     },
     {
      "cell_type": "heading",
@@ -2204,7 +2403,7 @@
      "language": "python",
      "metadata": {},
      "outputs": [],
-     "prompt_number": 32
+     "prompt_number": 73
     },
     {
      "cell_type": "heading",
@@ -2519,7 +2718,7 @@
      "language": "python",
      "metadata": {},
      "outputs": [],
-     "prompt_number": 33
+     "prompt_number": 74
     },
     {
      "cell_type": "heading",
@@ -2555,11 +2754,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76c99470>"
+        "<matplotlib.figure.Figure at 0x7facfaefaf60>"
        ]
       }
      ],
-     "prompt_number": 34
+     "prompt_number": 75
     },
     {
      "cell_type": "markdown",
@@ -2673,11 +2872,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76ba1668>"
+        "<matplotlib.figure.Figure at 0x7facf310bd68>"
        ]
       }
      ],
-     "prompt_number": 35
+     "prompt_number": 76
     },
     {
      "cell_type": "markdown",
@@ -2783,11 +2982,11 @@
        "output_type": "display_data",
        "png": 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        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76d58358>"
+        "<matplotlib.figure.Figure at 0x7facf313e0b8>"
        ]
       }
      ],
-     "prompt_number": 36
+     "prompt_number": 77
     },
     {
      "cell_type": "heading",
@@ -2847,11 +3046,11 @@
        "output_type": "display_data",
        "png": 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truncated
        "text": [
-        "<matplotlib.figure.Figure at 0x7fed76bc7588>"
+        "<matplotlib.figure.Figure at 0x7facfbe58828>"
        ]
       }
      ],
-     "prompt_number": 37
+     "prompt_number": 78
     },
     {
      "cell_type": "markdown",