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Peter Norvig authored and GitHub committed 2021-08-10 14:00:04 -07:00
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ipynb/RaceTrack.ipynb
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@@ -16,21 +16,20 @@
">\n",
"> For each move you have a choice of nine possible destinations, corresponding to the 3x3 grid that surrounds the point formed by adding your current velocity vector to your current position. That is, you can maintain your current velocity, or you can alter your velocity (and hence your destination) by accelerating one grid point in any direction (horizontally, vertically or diagonally). For example, if your first move was up and to the right, then your nine possible second moves are shown below, although two of them will cause you to crash into the central circle and one of them is standing still.\n",
">\n",
"> ![](https://fivethirtyeight.com/wp-content/uploads/2021/08/Screen-Shot-2021-08-04-at-10.44.11-PM.png?w=400)\n",
">\n",
"> How quickly can you navigate the track? \n",
">\n",
"> ![](https://fivethirtyeight.com/wp-content/uploads/2021/08/Screen-Shot-2021-08-04-at-10.44.11-PM.png?w=200)\n",
"\n",
"\n",
"# Defining Points and Vectors\n",
"# Defining Points, Vectors, and Paths\n",
"\n",
"We will need to represent 2-D points on a grid, as well as 2-D vectors for velocity and acceleration. I will use the following choices:\n",
"- `Point(x, y)`: a point in the 2-D plane, implemented as a complex number.\n",
"- `Vector(x, y)`: a vector in the 2-D plane, also implemented as a complex number.\n",
"- `Path`: a list of points (representing the car's positions as it moves).\n",
"- `Vector(x, y)`: a 2-D vector, also implemented as a complex number.\n",
"- `Path`: the car's history of position points as it moves, implemented as a list of Points.\n",
"- `X(point)`, `Y(point)`: the x- and y-coordinates of a point, respectively.\n",
"- `zero`: a zero vector (the car's starting velocity vector).\n",
"- `accelerations`: the nine possible acceleration vectors (to change the car's velocity), mapped to the digit number that The Riddler uses to identify them.\n"
"- `zero`, `up`, `right`: vectors that stay still, move one unit up, and move one to the right, respectively.\n",
"- `accelerations`: the nine possible acceleration vectors (to change the car's velocity)."
]
},
{
@@ -42,31 +41,25 @@
"import matplotlib.pyplot as plt\n",
"from typing import List, Tuple, Dict, Set\n",
"\n",
"Point = Vector = complex\n",
"Path = List[Point]\n",
"Point = complex\n",
"Vector = complex\n",
"Path = List[Point]\n",
"\n",
"def X(point) -> float: \"x-coordinate\"; return point.real\n",
"def Y(point) -> float: \"y-coordinate\"; return point.imag\n",
"\n",
"zero = Vector(0, 0)\n",
"zero, up, right = Vector(0, 0), Vector(0, 1), Vector(1, 0)\n",
"\n",
"accelerations = {Vector(-1, 1): 1, Vector(0, 1): 2, Vector(1, 1): 3, \n",
" Vector(-1, 0): 4, Vector(0, 0): 5, Vector(1, 0): 6, \n",
" Vector(-1, -1):7, Vector(0, -1):8, Vector(1, -1):9}"
"accelerations = [up-right, up, up+right, \n",
" -right, zero, right, \n",
" -up-right, -up, -up+right]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Defining the Track\n",
"\n",
"I want to solve the particular problem posed by The Riddler, but I also want some flexibility to race on differently-shaped race tracks. I'll define the class `Track` as a subclass of a `set` (of `Point`s), with some additional attributes:\n",
"- `track.start`: the single starting point.\n",
"- `track.finish`: the two endpoints of the finish line.\n",
"- `track.radius`: the radius of the central circle (whose center is defined as the origin).\n",
"\n",
"The points in the track are all those points on a 15x15 square except the ones that fall inside the circle's radius.\n"
"*Note:* I debated whether the elements of a `Path` should be just points, or (point, velocity) pairs, because that's what uniquely identifies the state of a car. I decided it was simpler to just use points. Given a path, the velocity can be easily computed:"
]
},
{
@@ -74,18 +67,54 @@
"execution_count": 2,
"metadata": {},
"outputs": [],
"source": [
"def velocity(path) -> Vector:\n",
" \"\"\"Velocity of the car when it is at the end of the path.\"\"\"\n",
" return zero if len(path) == 1 else path[-1] - path[-2] "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Defining the Track\n",
"\n",
"I want to solve RaceTrack games both on the particular track posed by The Riddler, and on differently-shaped race tracks. I'll define the class `Track` to be a set of points, but annotated with \n",
"attribute/value pairs (which can vary for different tracks)."
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [],
"source": [
"class Track(Set):\n",
" \"\"\"The set of legal points on the track, with any attributes.\"\"\"\n",
" def __init__(self, items, **kwds):\n",
" self.update(items)\n",
" self.__dict__.update(kwds)\n",
"\n",
"radius = 3\n",
"track = Track({Point(x, y) for x in range(-7, 8) for y in range(-7, 8) \n",
" if abs(Point(x, y)) >= radius}, \n",
" start=Point(0, -5), \n",
" finish=[Point(0, -3), Point(0, -7)], \n",
" radius=radius)"
" self.__dict__.update(kwds)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"For The Riddler's track, the track points are all those points on a 15x15 square except the ones that fall inside the circle's radius. The track also has three attributes:\n",
"- `track.start`: the single starting point.\n",
"- `track.finish`: the two endpoints of the finish line.\n",
"- `track.radius`: the radius of the central circle (whose center, without loss of generality, is the origin)."
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [],
"source": [
"side = range(-7, 8)\n",
"points = {Point(x, y) for x in side for y in side if abs(Point(x, y)) >= 3}\n",
"track = Track(points, start=Point(0, -5), finish=[Point(0, -3), Point(0, -7)], radius=3)"
]
},
{
@@ -94,12 +123,12 @@
"source": [
"# Visualization\n",
"\n",
"I think it will be helpful to be able to visualize paths on the track, so I'll define a function to do that now, even before attempting to find solutions:"
"It will be really helpful to be able to visualize paths on the track:"
]
},
{
"cell_type": "code",
"execution_count": 3,
"execution_count": 5,
"metadata": {},
"outputs": [],
"source": [
@@ -109,15 +138,16 @@
" plt.xticks([]); plt.yticks([])\n",
" ax.set_aspect('equal')\n",
" ax.add_artist(plt.Circle((0, 0), track.radius, alpha=1/3, color='k', ec='w'))\n",
" ax.plot(*XY(track), 'k,' ) # grid points\n",
" ax.plot(*XY(track), 'k,' ) # grid points\n",
" ax.plot(*XY([track.start]), 'ks', ms=10) # start point\n",
" ax.plot(*XY(track.finish), 'k:') # finish line\n",
" for path in paths: # paths\n",
" ax.plot(*XY(track.finish), 'k-') # finish line\n",
" for path in paths: # paths\n",
" ax.plot(*XY(path), 'o-') \n",
" plt.title(f'{len(paths)} paths of {set(len(path) - 1 for path in paths)} moves')\n",
" \n",
"def XY(points) -> Tuple[List[float], List[float]]: \n",
" \"\"\"A tuple of: (the x-coordinates of points, the y-coordinates of points).\"\"\"\n",
" # This converts points into the format that plt.plot expects \n",
" return [X(p) for p in points], [Y(p) for p in points]"
]
},
@@ -126,21 +156,22 @@
"metadata": {},
"source": [
"Below is my recreation of The Riddler's diagram of the first two moves. Here and throughout I'll use the convention:\n",
"- `p` is a point indicating a move\n",
"- `p` is a point indicating a move.\n",
"- `v` is the velocity of the move.\n",
"- `p1` is where the next point would be if the velocity remains constant.\n",
"- `p2` is where the next point actually is with acceleration applied to change the velocity.\n",
"\n",
"In this example, `p` is up and right from the start to `Point(1, -4)`, so the velocity is `Vector(1, 1)`."
"In this example, `p` is up and right from the start, and from there there are 9 possible values for `p2`."
]
},
{
"cell_type": "code",
"execution_count": 4,
"execution_count": 6,
"metadata": {},
"outputs": [
{
"data": {
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truncated
"image/png": "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 truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -150,8 +181,9 @@
}
],
"source": [
"p = Point(1, -4)\n",
"p1 = p + Vector(1, 1)\n",
"v = up + right\n",
"p = track.start + v\n",
"p1 = p + v\n",
"p2s = [p1 + a for a in accelerations]\n",
"\n",
"plot([[track.start, p, p2] for p2 in p2s])"
@@ -163,17 +195,14 @@
"source": [
"# Circumnavigating the Track\n",
"\n",
"The goal of the race is to circumnavigate the center circle. I can test if a path goes from the fourth [quadrant](https://mathworld.wolfram.com/Quadrant.html) to the first, second, third, and back to the fourth. The predicate `is_solution(path)` does that.\n",
"What exactly does it mean to mean to \"circumnavigate the track in a counterclockwise loop\"? I'll define a circumnavigation (and thus, a *solution*) as a path goes from the plane's fourth [**quadrant**](https://mathworld.wolfram.com/Quadrant.html) to the first, second, third, and back to the fourth. (Points with an x-coordinate of exactly 0 are considered to be in the first or fourth quadrant.) \n",
"\n",
"<center><div><img src=\"https://upload.wikimedia.org/wikipedia/commons/thumb/e/e3/Cartesian-coordinate-system-with-quadrant.svg/242px-Cartesian-coordinate-system-with-quadrant.svg.png\" width=120></div></center>\n",
"\n",
"\n",
"If you wanted to generalize this to other race tracks, you could change the definition of `quadrant` (there could be more than 4 \"quadrants,\" and they could have any shape), and change the `goals` in `is_solution`."
"<center><div><img src=\"https://upload.wikimedia.org/wikipedia/commons/thumb/e/e3/Cartesian-coordinate-system-with-quadrant.svg/242px-Cartesian-coordinate-system-with-quadrant.svg.png\" width=120></div></center>\n"
]
},
{
"cell_type": "code",
"execution_count": 5,
"execution_count": 7,
"metadata": {},
"outputs": [],
"source": [
@@ -189,26 +218,31 @@
"\n",
"def quadrant(p: Point) -> int: \n",
" \"\"\"What quadrant of the 2-D plane is this point in?\"\"\"\n",
" return (1 if Y(p) >= 0 else 4) if X(p) >= 0 else (2 if Y(p) >= 0 else 3)"
" if X(p) >= 0:\n",
" return 1 if Y(p) > 0 else 4\n",
" else: \n",
" return 2 if Y(p) > 0 else 3"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"To generalize this to other race tracks, you could change the definition of `quadrant` (there could be more than 4 \"quadrants,\" and they could have any shape), and change the `goals` in `is_solution`.\n",
"\n",
"# Legal Moves\n",
"\n",
"As stated earlier, I want to solve the specific circular track, and also make the code general enough that it can handle other tracks. I'll do that by splitting up responsibility for determining legal moves:\n",
"- `all_moves(path)` is general; it returns the points on the track that can be reached by applying the 9 accelerations.\n",
"- `circular_track_moves(path)` is specific; it limits moves in the following ways:\n",
" - It won't make moves that stay in one place (they are legal, but wasteful and not needed on this track).\n",
"As stated earlier, I want to solve the specific Riddler track, and also make the code general enough that it can handle other tracks. I'll do that by splitting up responsibility for determining legal moves:\n",
"- `all_moves(path, track)` is general; it returns track points that can be reached with any acceleration.\n",
"- `riddler_track_moves(path, track)` is specific; it limits moves in the following ways:\n",
" - It won't make moves that stay in one place (they are legal, but wasteful on this track).\n",
" - It won't pass from the fourth quadrant backwards over the finish line into the third (legal, but wasteful).\n",
" - It won't let a move's line segment intersect the central circle (even if the start and end points of the move are on valid grid points outside of the circle).\n"
" - It won't let a move's line segment intersect the central circle (even if the start and end points of the move are on valid track points outside of the circle).\n"
]
},
{
"cell_type": "code",
"execution_count": 6,
"execution_count": 8,
"metadata": {},
"outputs": [],
"source": [
@@ -217,29 +251,25 @@
" p1 = path[-1] + velocity(path)\n",
" return {(p1 + a) for a in accelerations} & track\n",
" \n",
"def circular_track_moves(path, track=track) -> Set[Point]:\n",
" \"\"\"Reasonable moves on the circular track.\"\"\"\n",
"def riddler_track_moves(path, track=track) -> Set[Point]:\n",
" \"\"\"Reasonable moves on The Riddler's track.\"\"\"\n",
" p = path[-1]\n",
" return {p2 for p2 in all_moves(path, track)\n",
" if p2 != p\n",
" and not (quadrant(p) == 4 and quadrant(p2) == 3) # Don't go backwards\n",
" and not intersects_circle(p, p2, track)}\n",
"\n",
"def velocity(path) -> Vector:\n",
" \"\"\"Velocity of the car when it is at the end of the path.\"\"\"\n",
" return zero if len(path) == 1 else path[-1] - path[-2] "
" and not intersects_circle(p, p2, track)}"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The math for solving for the intersection of a line segment and a circle [looks complicated](https://stackoverflow.com/questions/6091728/line-segment-circle-intersection), so instead I just define N+1 points along the line from point *p* to point *p2* and check if any of those points is within the radius of the circle (which by definition is centered on the origin). If N is too small this could miss an intersection, but that shouldn't be a problem for our uses."
"The math for solving for the intersection of a line segment and a circle [looks complicated](https://stackoverflow.com/questions/6091728/line-segment-circle-intersection), so instead I just define N+1 points along the line from point *p* to point *p2* and check if any of those points is within the radius of the circle. I chose N to be big enough that I believe this approach won't miss any intersection points on our small track."
]
},
{
"cell_type": "code",
"execution_count": 7,
"execution_count": 9,
"metadata": {},
"outputs": [],
"source": [
@@ -259,51 +289,62 @@
"source": [
"# A Frontier of Possible Paths\n",
"\n",
"Now I want to expand paths by adding legal moves to the end until I find solution(s)–paths that circumnavigate the circle. I could keep all my paths in a `list`, but there will be an exponential number of them (up to 9<sup><i>n<i/></sup> paths of length *n* and I estimate *n* will be around 10 to 13).\n",
"Now I want to expand paths by adding legal moves to the end until I find solution(s). I could keep *all* my paths in a `list`, but there will be an exponential number of them: O(9<sup><i>n<i/></sup>) paths of length *n*. \n",
"\n",
"Instead I will keep track of a *frontier*, which I define as a mapping of `{(endpoint, velocity): path}`. The idea is that there may be many paths that arrive at the some endpoint with the same velocity in the same number of moves. I only need to keep one of them, because they can all be continued in exactly the same ways.\n",
"I am not looking for all solutions, just one fastest solution (or possibly a small sampling of them). So I am willing to discard some partial paths if I can prove that they can never lead to a faster solution than another partial path.\n",
" \n",
"*Note:* Before I added the limitation that moves can't pass backwards from the fourth quadrant to the third, I had **a bug**: sometimes the frontier would capture a `(endpoint, velocity)` pair for a path that was in the third quadrant but hadn't circumnavigated, and was replacing a path to the same point that had circumnavigated. That meant I was discarding some valid paths."
"The key idea is that there may be many partial paths that arrive at the some endpoint with the same velocity in the same number of moves. I only need to keep one of them, because they can all be continued in exactly the same ways.\n",
"\n",
"I will define a data structure that I call a `Frontier`, which is a dict of `{(endpoint, velocity): path}`. By the nature of dicts, this keeps only one path for each `(endpoint, velocity)` pair. \n",
" \n",
"Here's how I expand a frontier, adding one more move in all possible ways, and discarding duplicates:"
]
},
{
"cell_type": "code",
"execution_count": 8,
"execution_count": 10,
"metadata": {},
"outputs": [],
"source": [
"Frontier = Dict[Tuple[Point, Vector], Path]\n",
"\n",
"def expand(frontier, legal_moves=circular_track_moves, track=track) -> Frontier:\n",
"def expand(frontier, legal_moves=riddler_track_moves, track=track) -> Frontier:\n",
" \"\"\"The {(endpoint, velocity): path} frontier extended one move in all legal ways.\"\"\"\n",
" return {(p, velocity([path[-1], p])): path + [p] \n",
" for path in frontier.values() \n",
" for p in legal_moves(path, track)}"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"*Note:* Before I added the limitation that moves can't pass backwards from the fourth quadrant to the third, it was possible that two different paths would arrive at the same point in the third quadrant with the same velocity. If the path that was discarded had circumnavigated and the path that was kept had not, that was **a bug** that discarded some valid solutions."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Searching for the Fastest Solutions\n",
"\n",
"Now all I have to do to solve the problem is repeatedly expand the frontier until one or more solutions are found. The first solutions found are guaranteed to be the fastest (i.e., they are the paths with the fewest possible moves), because we are expanding paths one move at a time, and the previous time did not find any solutions. In the end, I return a list of all the solutions in the frontier, but there may be many other paths that are not returned, because the frontier only keeps one path for each `(endpoint, velocity)` pair, and there are only so many ways to cross the finish line. "
"Now all I have to do to solve the problem is repeatedly expand the frontier until one or more solutions are found. The first solutions found are guaranteed to be the fastest (i.e., they are the paths with the fewest possible moves), because we are expanding paths one move at a time. In the end, `search` returns a list of all the solutions in the frontier, but there may be many other paths that are not returned, because the frontier only keeps one path for each `(endpoint, velocity)` pair, and there are only so many ways to cross the finish line. "
]
},
{
"cell_type": "code",
"execution_count": 9,
"execution_count": 11,
"metadata": {},
"outputs": [],
"source": [
"def search(track=track, path=None, legal_moves=circular_track_moves) -> List[Path]:\n",
"def search(track=track, path=None, legal_moves=riddler_track_moves) -> List[Path]:\n",
" \"\"\"Find the shortest possible solution paths.\"\"\"\n",
" if path is None: path = [track.start]\n",
" frontier = {(path[-1], velocity(path)): path}\n",
" solutions = []\n",
" while not solutions:\n",
" frontier = expand(frontier, legal_moves, track)\n",
" solutions = list(filter(is_solution, frontier.values()))\n",
" solutions = [path for path in frontier.values() if is_solution(path)]\n",
" return solutions"
]
},
@@ -318,12 +359,12 @@
},
{
"cell_type": "code",
"execution_count": 10,
"execution_count": 12,
"metadata": {},
"outputs": [
{
"data": {
"image/png": 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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -342,7 +383,7 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"We see that the fastest solution is 12 moves, and there are at least 7 ways to get there (with perhaps other ways discarded).\n",
"We see that the fastest solution is 12 moves, and there are at least 7 ways to get there (with perhaps others discarded).\n",
"\n",
"# Submitting your Answer\n",
"\n",
@@ -354,12 +395,12 @@
"> 4 5 6\n",
"> 7 8 9\n",
"\n",
"I will define two functions to translate back and forth between my `Path` and The Riddler's list of `Digits`: "
"I already started the tricky work by defining the nine `accelerations` to be in this order. I finish the tricky work by defining two functions to translate back and forth between my `Path` and The Riddler's `Digits`: "
]
},
{
"cell_type": "code",
"execution_count": 11,
"execution_count": 13,
"metadata": {},
"outputs": [],
"source": [
@@ -367,7 +408,7 @@
"\n",
"def digits_from_path(path) -> Digits:\n",
" \"\"\"The digits that the path represents.\"\"\"\n",
" return [accelerations[velocity(path[:i + 1]) - velocity(path[:i])]\n",
" return [1 + accelerations.index(velocity(path[:i + 1]) - velocity(path[:i])) \n",
" for i in range(1, len(path))]\n",
"\n",
"def path_from_digits(digits, start=track.start) -> Path:\n",
@@ -375,14 +416,14 @@
" v = zero # Starting velocity\n",
" path = [start]\n",
" for d in digits:\n",
" v += next(a for a in accelerations if accelerations[a] == d)\n",
" v += accelerations[d - 1]\n",
" path.append(v + path[-1])\n",
" return path"
]
},
{
"cell_type": "code",
"execution_count": 12,
"execution_count": 14,
"metadata": {},
"outputs": [
{
@@ -397,7 +438,7 @@
" [3, 3, 1, 7, 7, 5, 7, 8, 9, 9, 3, 3]]"
]
},
"execution_count": 12,
"execution_count": 14,
"metadata": {},
"output_type": "execute_result"
}
@@ -412,17 +453,17 @@
"source": [
"# Alternative Solutions\n",
"\n",
"I have identified six different opening sequences of moves that each lead to solutions with the fastest possible time of 12 moves. I don't understand why every plot has exactly 7 solution paths; I suppose it is just because the finish line is crowded and there are only so many ways to get there."
"I have identified six different opening sequences of moves that each lead to solutions with the fastest possible time of 12 moves. I don't understand exactly why in each case there exactly 7 solution paths are reported; I suppose it is just because the finish line is crowded and there are only so many ways to get there."
]
},
{
"cell_type": "code",
"execution_count": 13,
"execution_count": 15,
"metadata": {},
"outputs": [
{
"data": {
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truncated
"image/png": 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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -432,7 +473,7 @@
},
{
"data": {
"image/png": 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truncated
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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -442,7 +483,7 @@
},
{
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"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -452,7 +493,7 @@
},
{
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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -462,7 +503,7 @@
},
{
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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -472,7 +513,7 @@
},
{
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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -494,12 +535,12 @@
"source": [
"# Visualizations and Tests\n",
"\n",
"To gain more confidence in the code, here are some visualizations and unit tests:"
"To gain more confidence in the code, here are some additional visualizations and unit tests:"
]
},
{
"cell_type": "code",
"execution_count": 14,
"execution_count": 16,
"metadata": {},
"outputs": [],
"source": [
@@ -512,12 +553,12 @@
},
{
"cell_type": "code",
"execution_count": 15,
"execution_count": 17,
"metadata": {},
"outputs": [
{
"data": {
"image/png": 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truncated
"image/png": 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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -537,12 +578,12 @@
},
{
"cell_type": "code",
"execution_count": 16,
"execution_count": 18,
"metadata": {},
"outputs": [
{
"data": {
"image/png": 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truncated
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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -562,12 +603,12 @@
},
{
"cell_type": "code",
"execution_count": 17,
"execution_count": 19,
"metadata": {},
"outputs": [
{
"data": {
"image/png": 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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -577,7 +618,7 @@
}
],
"source": [
"# Waypoints on a line segment that are checked to see if they are outside the circle\n",
"# Waypoints on a line segment\n",
"p, q = Point(-7, 7), Point(7, 0)\n",
"plot([waypoints(p, q)])\n",
"assert not intersects_circle(p, q)"
@@ -588,17 +629,17 @@
"metadata": {},
"source": [
"Here's a visualization for `expand(frontier)` showing the first six expansions.\n",
"Note that the growth in number of paths is much slower then O(9<sup><i>n</i></sup>). After 6 moves there are only 1,107 paths, or just 0.2% of 9<sup>6</sup> = 531,441."
"Note that the growth in number of paths is much slower then O(9<sup><i>n</i></sup>). After 6 moves there are only 1,107 paths, which is 500 times fewer than 9<sup>6</sup> = 531,441."
]
},
{
"cell_type": "code",
"execution_count": 18,
"execution_count": 20,
"metadata": {},
"outputs": [
{
"data": {
"image/png": 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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -608,7 +649,7 @@
},
{
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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -618,7 +659,7 @@
},
{
"data": {
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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -628,7 +669,7 @@
},
{
"data": {
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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -671,29 +712,17 @@
"source": [
"# A Different Track\n",
"\n",
"Although the code here is not ready to handle large, arbitrary-shaped tracks, it can handle some small variations as is, and other variations with some tweaking. Here I define `track2` to be a track with a larger central circle, and a circular ring of valid points around the circle rather than a full rectangular grid:"
"Although the code here is not ready to handle large, arbitrary-shaped tracks, it can handle some minor track variations. Here I define `track2` to be a track with a larger central circle, and a circular ring of valid points rather than a full rectangular grid:"
]
},
{
"cell_type": "code",
"execution_count": 19,
"execution_count": 21,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"[[6, 6, 3, 1, 1, 2, 4, 7, 4, 7, 7, 8, 9, 9, 9, 9, 5, 3, 6, 3, 3],\n",
" [6, 6, 3, 1, 1, 2, 4, 7, 4, 7, 7, 8, 9, 9, 9, 9, 5, 3, 3, 6, 6],\n",
" [6, 6, 3, 1, 1, 2, 4, 7, 4, 7, 7, 8, 9, 9, 9, 9, 5, 3, 3, 6, 3]]"
]
},
"execution_count": 19,
"metadata": {},
"output_type": "execute_result"
},
{
"data": {
"image/png": 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truncated
"image/png": 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truncated
"text/plain": [
"<Figure size 432x288 with 1 Axes>"
]
@@ -704,16 +733,40 @@
],
"source": [
"radius2 = 10\n",
"track2 = Track({Point(x, y) for x in range(-14, 15) for y in range(-14, 15) \n",
" if radius2 <= abs(Point(x, y)) < 13}, \n",
" start=Point(0, -12), \n",
" finish=[Point(0, -radius2), Point(0, -12)], \n",
" radius=radius2)\n",
"side2 = range(-14, 15)\n",
"start2 = Point(0, -radius2 - 2)\n",
"track2 = Track({Point(x, y) for x in side2 for y in side2 \n",
" if radius2 <= abs(Point(x, y)) <= radius2 + 3}, \n",
" radius=radius2, start=start2, \n",
" finish=[Point(0, -radius2), Point(0, -radius2 - 3)])\n",
"\n",
"solutions2 = search(track2)\n",
" \n",
"plot(solutions2, track=track2)\n",
"\n",
"plot(solutions2, track=track2)"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"[[6, 6, 3, 1, 1, 2, 4, 7, 4, 7, 7, 8, 9, 9, 9, 9, 5, 3, 3, 9, 3],\n",
" [6, 6, 3, 1, 1, 2, 4, 7, 4, 7, 7, 8, 9, 9, 9, 9, 5, 3, 6, 3, 3],\n",
" [6, 6, 3, 1, 1, 2, 4, 7, 4, 7, 7, 8, 9, 9, 9, 9, 5, 3, 3, 6, 9],\n",
" [6, 6, 3, 1, 1, 2, 4, 7, 4, 7, 7, 8, 9, 9, 9, 9, 5, 3, 3, 6, 6],\n",
" [6, 6, 3, 1, 1, 2, 4, 7, 4, 7, 7, 8, 9, 9, 9, 9, 5, 3, 3, 6, 3],\n",
" [6, 3, 3, 2, 1, 7, 4, 7, 1, 7, 8, 8, 7, 9, 9, 9, 9, 3, 3, 3, 3]]"
]
},
"execution_count": 22,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"[digits_from_path(path) for path in solutions2]"
]
},
@@ -721,12 +774,12 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"# Additional tests:"
"# Additional tests"
]
},
{
"cell_type": "code",
"execution_count": 20,
"execution_count": 23,
"metadata": {},
"outputs": [
{
@@ -735,38 +788,39 @@
"'tests pass'"
]
},
"execution_count": 20,
"execution_count": 23,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"path1 = [track.start, Point(1, -4)] # First move\n",
"start = track.start\n",
"path1 = [start, Point(1, -4)] # First move\n",
"\n",
"assert track.start == Point(0, -5) == 0-5j\n",
"assert X(track.start) == 0 and Y(track.start) == -5 \n",
"assert velocity([track.start]) == 0\n",
"assert start == Point(0, -5) == 0-5j\n",
"assert X(start) == 0 and Y(start) == -5 \n",
"assert velocity([start]) == 0\n",
"assert velocity(path1) == Vector(1, 1)\n",
"assert XY(path1) == ([0, 1], [-5, -4])\n",
"\n",
"assert quadrant(track.start) == 4\n",
"assert quadrant(Point(1, 2)) == 1\n",
"assert quadrant(Point(1, -4)) == 4 == quadrant(start)\n",
"assert quadrant(Point(1, 2)) == 1\n",
"assert quadrant(Point(-5, 4)) == 2\n",
"assert quadrant(Point(-1, -1)) == 3\n",
"assert quadrant(Point(-1, -1))== 3\n",
"\n",
"assert waypoints(0, 10, 10) == [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]\n",
"assert waypoints(0, 100, 10) == [0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100]\n",
"assert waypoints(zero, Point(10, 20), 10) == [\n",
" 0j,1+2j,2+4j,3+6j,4+8j,5+10j,6+12j,7+14j,8+16j,9+18j,10+20j]\n",
"\n",
"# There are 9 moves from the start point\n",
"assert all_moves([track.start]) == {\n",
"assert all_moves([start]) == {\n",
" -1-4j, -1-5j, -1-6j, \n",
" 0-4j, 0-5j, 0-6j, \n",
" 1-4j, 1-5j, 1-6j}\n",
"\n",
"# But 3 moves are backwards and 1 stays in the same place; those are disallowed\n",
"assert circular_track_moves([track.start]) == {0-6j, 1-6j, 1-5j, 0-4j, 1-4j}\n",
"assert expand({(track.start, zero): [track.start]}) == {\n",
"assert riddler_track_moves([start]) == {0-6j, 1-6j, 1-5j, 0-4j, 1-4j}\n",
"assert expand({(start, zero): [start]}) == {\n",
" (0-6j, 0-1j): [0-5j, 0-6j],\n",
" (1-6j, 1-1j): [0-5j, 1-6j],\n",
" (1-5j, 1+0j): [0-5j, 1-5j],\n",