In [1]:

from sys import stderr from queue import LifoQueue as stack from queue import PriorityQueue as p_queue from queue import SimpleQueue as queue # It seems like these structures can keep "garbage" fro # previous runs, so we must clean them out before using: def gc(queue): if not queue.empty(): while not queue.empty(): queue.get()

1

The following indicates the simple structure used for our graphs. The following shows a graph with nodes 0,1,2,3,4,5,6 and directed edges: $0\to 1$ with weight 1, $0\to 2$ with weight 2, etc.

2

In [37]:

ToyGraph = {0 : {1:1, 2:1}, 1 : {3:8}, 2 : {4:2}, 3 : {4:1, 6:2}, 4 : {5:2, 3:5}, 5 : {3:1, 4:2}, 6 : {}}

3

`showGraph`

We can visualize our graphs using the networkx module. We need to load a few modules and convert our adjacency list graph to a networkx graph. This is done below in the following code which may be ignored. Things are set up to indicate the DFS ('blue'), the BFS solution ('red'), and the UCS solution ('green') some basic attempt has been made to indicte when the path overlay each other.

4

In [86]:

import networkx as nx import pylab as plt import pydot as pdot from IPython.core.display import HTML, display, Image #import pygraphviz #from networkx.drawing.nx_agraph import graphviz_layout def adjToNxGraph(G, digraph=True): """ Converts one of our adjacency "list" representations for a graph into a networkx graph. """ if digraph: Gr = nx.DiGraph() else: Gr = nx.Graph() for node in G: Gr.add_node(node) if G[node]: for adj in G[node]: Gr.add_edge(node, adj) Gr[node][adj]['weight'] = G[node][adj] return Gr def showGraph(G, start, goal, paths = [], node_labels = 'default', node_pos = 'neato', gsize = (14,14), save_file=None, digraph=True): """ paths should be an array of which paths to show: paths = ['bfs', 'dfs', 'ucs'] node_labels must be one of: 'default', 'none', or a list of labels to use. save_file must be an image file name with extension, i.e., save_file='my_graph.png' """ fig, ax = plt.subplots(figsize=gsize) # Conver G into structure used in networkx Gr = adjToNxGraph(G, digraph=digraph) if node_pos is 'project_layout': # The project graphs have a particular structure. node_pos = dict(zip(Gr.nodes(),[(b, 9 - a) for a,b in Gr.nodes()])) else: node_pos = nx.nx_pydot.graphviz_layout(Gr, prog=node_pos, root=start) edge_weight=nx.get_edge_attributes(Gr,'weight') def path_edges(path): edges = list(zip(path[:-1], path[1:])) cost = sum([Gr[z[0]][z[1]]['weight'] for z in edges]) if not digraph: edges += list(zip(path[1:], path[:-1])) return edges, cost # Process Paths: if 'bfs' in paths: bpath = getPath(bdfs(G, start, goal, search ='bfs'), start, goal) bedges, bcost = path_edges(bpath) else: bpath = [] bedges = [] if 'dfs' in paths: dpath = getPath(bdfs(G, start, goal, search = 'dfs'), start, goal) dedges, dcost = path_edges(dpath) else: dpath = [] dedges = [] if 'ucs' in paths: ucost, back = ucs(G, start, goal) upath = getPath(back, start, goal) uedges, ucost = path_edges(upath) else: upath = [] uedges = [] node_col = ['orange' if node in upath else 'purple' if node in bpath and node in dpath else 'blue' if node in dpath else 'red' if node in bpath else 'lightgray' for node in Gr.nodes()] if node_labels == 'default': nodes = nx.draw_networkx_nodes(Gr, node_pos, ax = ax, node_color=node_col, node_size=400) nodes.set_edgecolor('k') nx.draw_networkx_labels(Gr, node_pos, ax = ax, font_size = 8) elif node_labels == 'none': nodes = nx.draw_networkx_nodes(Gr, node_pos, ax = ax, node_color=node_col, node_size=50) else: # labels must be a list nodes = nx.draw_networkx_nodes(Gr, node_pos, ax = ax, node_color=node_col, node_size=400) nodes.set_edgecolor('k') mapping = dict(zip(Gr.nodes, node_labels)) nx.draw_networkx_labels(Gr, node_pos, labels=mapping, ax = ax, font_size = 8) edge_col = ['purple' if edge in bedges and edge in dedges else 'blue' if edge in dedges else 'red' if edge in bedges else 'orange' if edge in uedges else 'gray' for edge in Gr.edges()] edge_width = [3 if edge in dedges or edge in bedges or edge in uedges else 1 for edge in Gr.edges()] if digraph: nx.draw_networkx_edge_labels(Gr, node_pos, ax = ax, edge_color=edge_col, label_pos=0.3, edge_labels=edge_weight) else: nx.draw_networkx_edge_labels(Gr, node_pos, ax = ax, edge_color=edge_col, edge_labels=edge_weight) nx.draw_networkx_edges(Gr, node_pos, ax = ax, edge_color=edge_col, width=edge_width, alpha=.3) if save_file: plt.savefig(save_file) plt.show() result = "DFS gives a path of length {} with cost {}<br>".format(len(dpath) - 1, dcost) if 'dfs' in paths else "" result += "BFS gives a path of length {} with cost {}. BFS always returns a minimal length path.<br>".format(len(bpath) - 1, bcost) if 'bfs' in paths else "" result += "UCS gives a path of length {} with cost {}. UCS alway returns a minimal cost path.".format(len(upath) - 1, ucost) if 'ucs' in paths else "" display(HTML(result)) # Need display in Jupyter

5

6

In [39]:

showGraph(ToyGraph,0,4,gsize=(8,8))

7

Here these are implemented using a basic stack (DFS) and queue (BFS). Often the goal is simply to answer "Is there a path between the start node and the goal node?", but sometimes we also wish to get hold of an actual path, for this a *back pointer* structure is introduced. When we expand the fringe at the current node we indicate for each new element of the fringe where it came from, a dictionary is used for this as well. A path can then be reconstructed backwards starting at the current element and working back to the start node which indicates the history of what nodes were expended to reach the current node. In this way a complete search tree can be reconstructed from the back pointer structure.

8

In [40]:

def bdfs(G, start, goal, search = 'dfs'): """ This is a template. Taking fringe = stack() gives DFS and fringe = queue() gives BFS. We need to add a priority function to get UCS. Usage: back_pointer = bdfs(G, start, goal, fringe = stack()) (this is dfs) back_pointer = bdfs(G, start, goal, fringe = queue()) (this is bfs) """ # There is actually a second subtle difference between stack and queue and that # has to do with when one revises the pack_pointer. Essentially, this amounts to # defining a priority function where queue prioritizes short paths, fat search trees # while dfs prioritizes long paths, skinny search trees. depth = {} if search is 'dfs': fringe = stack() weight = -1 # We are pretending all edges have weight -1 else: fringe = queue() weight = 1 # We are pretending all edges have weight 1 gc(fringe) # Make sure there is no garbage in the fringe closed = set() back_pointer = {} current = start depth[start] = 0 fringe.put(current) while True: # If the fringe becomes empty we are out of luck if fringe.empty(): print("There is no path from {} to {}".format(start, goal), file=stderr) return None # Get the next closed element of the closed set. This is complicated # by the fact that our queue has no delete so items that are already # in the closed set might still be in the queue. We must make sure not # to choose such an item. while True: current = fringe.get() if current not in closed: break if fringe.empty(): print("There is no path from {} to {}".format(start, goal), file=stderr) return None # Add current to the closed set closed.add(current) # If current is the goal we are done. if current == goal: return back_pointer # Add nodes adjacent adjacent to current to the fringe # provided they are not in the closed set. if G[current]: # Check if G[current] != {}, bool({}) = False for node in G[current]: if node not in closed: node_depth = depth[current] + weight if node not in depth or node_depth < depth[node]: back_pointer[node] = current depth[node] = node_depth fringe.put(node) def dfs(G, start, goal): return bdfs(G, start, goal, search = 'dfs') def bfs(G, start, goal): return bdfs(G, start, goal, search = 'bfs')

9

The following reconstructs the path from the back pointers.

10

In [41]:

def getPath(backPointers, start, goal): current = goal s = [current] while current != start: current = backPointers[current] s += [current] return list(reversed(s))

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12

In [73]:

showGraph(ToyGraph, 0, 6,paths = ['bfs','dfs'], gsize=(8,8))

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DFS gives a path of length 4 with cost 10

BFS gives a path of length 3 with cost 11. BFS always returns a minimal length path.

BFS gives a path of length 3 with cost 11. BFS always returns a minimal length path.

The following make is easy to try out our algorithms on random graphs. There is a probability $p$ chance that the i -> j in this graph, where $0\leq p \leq 1$. A weight function can be provided to put weights on edges.

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In [43]:

from random import random as rand from random import randint as randi def genRandDiGraph(n, p = .5, weights = lambda i,j: 1, digraph=True): G = {} # Initialize empty graph. for i in range(n): G.setdefault(i, {}) if digraph: for j in range(n): if rand() < p and j != i: # Simply choose whether or not to put # a directed edge j -> i G[i][j] = weights(i,j) else: for j in range(i + 1, n): # In case G[j] has not been initiated G.setdefault(j,{}) if rand() < p: # Simply choose whether or not to put # an directed edge j -> i G[i][j] = weights(i,j) G[j][i] = G[i][j] return G

15

Play around with different weight functions. If you do not assign a weght function, all weights default to 1 and you can verify that BFS and UCS return shortes lenth paths, since now shortest length and minimal cost are the same. Setting weighs to -1, i.e. `weights = lambda i,j: -1`

is interesting as UCS then wants to find a maximal "length" path. You can set `digraph=True`

or `digraph=False`

and see what the difference is.

16

In [60]:

RandomG = genRandDiGraph(15, .4, weights=lambda i,j : randi(1, 15), digraph=True) showGraph(RandomG, 4, 3, gsize=(10,10), digraph=True)

17

In [61]:

showGraph(RandomG, 4, 3, paths=['bfs','dfs'], gsize=(10,10), digraph=True)

18

DFS gives a path of length 11 with cost 80

BFS gives a path of length 2 with cost 14. BFS always returns a minimal length path.

BFS gives a path of length 2 with cost 14. BFS always returns a minimal length path.

Here we must switch from regular queue and stack to the priority queue and introduce the cost function. Often the goal is simply to get the least cost of a path, but sometimes we wish to have the path itself so we keep track of back pointers as in the BFS/DFS so we can reconstuct the path. UCS is gauranteed to produce a path of minimal cost.

$A^*$ search is much like UCS, the only difference is that an additional heuristic is used to modify the priority function, so that the cost function and the priority function are no longer the same. For example, in the "project" graphs below, an additional $P(\text{node}) = P(\text{current}) + \text{cost}(\text{current}, \text{node}) + h(\text{node})$ might be used as the priority where $h(\text{node})$ is the Manhatten distance from the node to the goal. This can prevent some unecessary exploration in irrelevant directions. We won't implement $A^*$ search here.

19

In [46]:

def ucs(G, start, goal, trace=False): """ This returns the least cost of a path from start to goal or reports the non-existence of such path. This also retuns a pack_pointer from which the search tree can be reconstructed as well as all paths explored including the one of interest. Usage: cost, back_pointer = ucs(Graph, start, goal) """ # Make sure th queue is empty. (Bug in implementation?) fringe = p_queue() gc(fringe) # If we did not care about the path, only the cost we could # omit this block. cost = {} # If all we want to do is solve the optimization back_pointer = {} # problem, neither of these are necessary. cost[start] = 0 # End back_pointer/cost block current = start fringe.put((0, start)) # Cost of start node is 0 closed = set() while True: # If the fringe becomes empty we are out of luck if fringe.empty(): print("There is no path from {} to {}".format(start, goal), file=stderr) return None # Get the next closed element of the closed set. This is complicated # by the fact that our queue has no delete so items that are already # in the closed set might still be in the queue. We must make sure not # to choose such an item. while True: current_cost, current = fringe.get() if current not in closed: # Add current to the closed set closed.add(current) if trace: print("Add {} to the closed set with cost {}".format(current,current_cost)) break if fringe.empty(): print("There is no path from {} to {}".format(start, goal), file=stderr) return None # If current is the goal we are done. if current == goal: return current_cost, back_pointer # Add nodes adjacent to current to the fringe # provided they are not in the closed set. if G[current]: # Check if G[current] != {}, bool({}) = False for node in G[current]: if node not in closed: node_cost = current_cost + G[current][node] # Note this little block could be removed if we only # cared about the final cost and not the path if node not in cost or cost[node] > node_cost: back_pointer[node] = current cost[node] = node_cost if trace: print("{current} <- {node}".format(current,node)) # End of back/cost block. fringe.put((node_cost, node)) if trace: print("Add {} to fringe with cost {}".format(node,node_cost))

20

In [47]:

showGraph(ToyGraph, 0, 6, paths=['bfs','dfs','ucs'], gsize=(8,8))

21

DFS gives a path of length 4 with cost 10

BFS gives a path of length 3 with cost 11. BFS always returns a minimal length path.

UCS gives a path of length 5 with cost 8. UCS alway returns a minimal cost path.

BFS gives a path of length 3 with cost 11. BFS always returns a minimal length path.

UCS gives a path of length 5 with cost 8. UCS alway returns a minimal cost path.

In [48]:

showGraph(RandomG, 4, 3, paths=['bfs','dfs','ucs'], gsize=(10,10))

22

DFS gives a path of length 10 with cost 82

BFS gives a path of length 2 with cost 18. BFS always returns a minimal length path.

UCS gives a path of length 5 with cost 14. UCS alway returns a minimal cost path.

BFS gives a path of length 2 with cost 18. BFS always returns a minimal length path.

UCS gives a path of length 5 with cost 14. UCS alway returns a minimal cost path.

Here we generate the sort of graphs used in the class project.

23

In [49]:

from random import choices import numpy as np def genWeights(n, m, MIN, MAX, radius=[1,0,0,1], block_prob = 0): """ This generates an instance of our problem of size mxn with integer entries roughly from MIN to MAX with some Inf's is block_prob is non-zero. radius=[L,R,D,U] indicates how many grid points in each direction to include in smoothing. The default is to average a grid point and its closest N,NE,E neighbor """ rng = list(range(MIN, MAX + 1)) + [np.infty] prbs = [(1 - block_prob)/(len(rng) - 1) if i < len(rng) - 1 else block_prob for i in range(len(rng))] wts = np.random.choice(rng, (m,n), p=prbs) wts_smoothed = np.array(wts) # This makes a new copy for i in range(m): for j in range(n): tmp = wts[max(i-radius[0],0):min(i+radius[1]+1,m),max(j-radius[2],0):min(j+radius[3]+1,n)] if rand() < 0.3: wts_smoothed[i,j] = np.mean(tmp[tmp != np.infty]) else: wts_smoothed[i,j] = np.mean(tmp) return np.round(wts_smoothed)

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Generate a 10x10 instance of our problem using values between 1 and 15. I have opted to omit any 'inf' nodes in the generation and decided to add them in as a smiley face. You could make a bigger case, say 100 x 100 and design the masking array so you get a "weighted" maze.

25

In [50]:

wts = genWeights(10, 10, 1, 30, block_prob=0) # Add some nodes that are impossible to reach. Think of # these as walls in a maze. mask = np.array([ [0, 0, 0, 0, 1, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 0, 0, 1, 1, 0, 0], [1, 0, 1, 0, 0, 0, 0, 1, 0, 1], [1, 0, 0, 0, 0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 1, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0, 0, 0, 1, 0], [0, 0, 1, 1, 1, 1, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 1, 0, 0, 0, 0] ]) wts[mask == 1] = np.infty

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`showProjectMatrix`

27

In [163]:

import matplotlib.colors as colors # Stolen code: # https://stackoverflow.com/questions/18926031/how-to-extract-a-subset-of-a-colormap-as-a-new-colormap-in-matplotlib def truncate_colormap(cmap, minval=0.0, maxval=1.0, n=100): new_cmap = colors.LinearSegmentedColormap.from_list( 'trunc({n},{a:.2f},{b:.2f})'.format(n=cmap.name, a=minval, b=maxval), cmap(np.linspace(minval, maxval, n))) return new_cmap cmap = plt.get_cmap('Greys') new_cmap = truncate_colormap(cmap, 0.0, 0.8) # End Stolen def showProjectMatrix(wts, save_file=None): fig, ax = plt.subplots() ax.axis('off') wts_ = np.array(wts) M = np.max(wts[wts != np.infty]) wts_[wts == np.infty] = max(int(M*(1.3)),M+5) ax.matshow(wts_, cmap=new_cmap) for (i, j), z in np.ndenumerate(wts): ax.text(j, i, '{:0.0f}'.format(z), ha='center', va='center') if save_file: plt.savefig(save_file) plt.show()

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Let's look at one representation of our problem.

29

In [164]:

showProjectMatrix(wts)

30

`wtsToGraph`

There are many ways to convert the matrix above into a graph with one node for each finite element. The nodes could for example be points on a map labeled with elevation, the edges could be labeled with the absolute value of the difference of the two elevations. (For simplicity, keep all edge weights non-negative.) A minimal cost path would correspond to a walk from one location (start) to the another (goal) that has the minimal total change in elevation, that is, minimizes the "up's and down's". This is what we shall use.

31

In [79]:

def wtsToGraph(wts): G = {} m, n = wts.shape for i in range(m): for j in range(n): if wts[i, j] == np.infty: continue node = (i, j) G[node] = {} if wts[i, j] != np.infty: if i > 0 and wts[i - 1, j] != np.infty: G[node][(i - 1, j)] = abs(wts[i - 1, j] - wts[i, j]) if j > 0 and wts[i, j - 1] != np.infty: G[node][(i, j - 1)] = abs(wts[i, j - 1] - wts[i, j]) if i < m - 1 and wts[i + 1, j] != np.infty: G[node][(i + 1, j)] = abs(wts[i + 1, j] - wts[i, j]) if j < n - 1 and wts[i, j + 1] != np.infty: G[node][(i, j + 1)] = abs(wts[i, j + 1] - wts[i, j]) return G

32

33

In [80]:

G = wtsToGraph(wts) showGraph(G, (0,0), (9,9), paths=['bfs','dfs','ucs'], node_labels='none', gsize=(14,14), node_pos='project_layout', digraph=False)

34

DFS gives a path of length 20 with cost 89.0

BFS gives a path of length 18 with cost 57.0. BFS always returns a minimal length path.

UCS gives a path of length 20 with cost 51.0. UCS alway returns a minimal cost path.

BFS gives a path of length 18 with cost 57.0. BFS always returns a minimal length path.

UCS gives a path of length 20 with cost 51.0. UCS alway returns a minimal cost path.

Design your own mask array. We'll use some topo data of Mt Ord in AZ for this.

35

In [166]:

Image("https://ketchers.github.io/Python-Play/MtOrdTop.jpg",height=400, width=400)

36

In [161]:

# Design your own array grid = np.array([ [6600,6760,6660,6700], [6600,6770,6650,6640], [6800,6920,6790,6480], [7040,7080,6800,6500] ],dtype=float) # Generate a random project (just uncomment and run) #project = genWeights(4, 4, 1, 30, block_prob=0) mask = np.array([ [0,1,0,0], [0,0,0,0], [0,0,1,0], [0,0,0,0] ]) #grid[mask == 1] = np.infty

37

Now display your array and graph.

38

In [165]:

# Display project as grid showProjectMatrix(grid, save_file="proj_mat.png") # Convert array into corresponding graph G = wtsToGraph(grid) # Change labels to A-Z alphabet = map(chr,range(65,91)) # Quick way to get alphabet # Display graph showGraph(G, (0,3), (3,0), paths=['ucs'], node_labels=alphabet, gsize=(6,6), digraph=False, node_pos='project_layout', save_file="proj_graph.png") Image("https://ketchers.github.io/Python-Play/MtOrdTop.jpg",height=350, width=350)

39

UCS gives a path of length 6 with cost 500.0. UCS alway returns a minimal cost path.

In [ ]:

40