# Practice with searching

## Table of Contents

## Routing from Dreese Labs to Goodale Park

In this example, our goal is to find a route between Woodruff & Tuttle and the Goodale parking lot. The graph on the right shows how various road intersections connect to each other, and the distance between intersections (in miles).

- Initial state
- Woodruff & Tuttle
- Possible actions
- Every action is the same: drive to another intersection.
- Transition model
- Shown in the graph and the table; each intersection can access the other intersections which are linked in the graph.
- Goal criteria
- Our destination is Goodale parking lot
- Path cost
- Shown in the graph and the table; distance (in miles) to the next intersection.

### Map data

This map data is encoded in Java and Python source code if you would like to use it in a program.

Marker | Start | End | Distance (mi) |
---|---|---|---|

A | High & Goodale | Goodale parking lot | 0.22 |

B | High & 5th | High & Goodale | 0.93 |

C | I-71 5th offramp | High & 5th | 1.07 |

D | I-71 11th offramp | I-71 5th offramp | 0.52 |

E | I-71 17th offramp | I-71 11th offramp | 0.47 |

F | US-23 & 17th | I-71 17th offramp | 0.75 |

G | US-23 & 15th | US-23 & 17th | 0.12 |

H | High & 15th | US-23 & 15th | 0.49 |

I | High & Woodruff | High & 15th | 0.26 |

J | Woodruff & Tuttle | High & Woodruff | 0.46 |

K | Lane & Tuttle | Woodruff & Tuttle | 0.17 |

L | SR-315 & Lane | Lane & Tuttle | 0.74 |

M | SR-315 I-670 offramp | SR-315 & Lane | 2.05 |

N | Park & Vine | SR-315 I-670 offramp | 0.99 |

O | Park & Goodale | Park & Vine | 0.15 |

P | Goodale parking lot | Park & Goodale | 0.13 |

Q | US-23 & Goodale | US-23 & 15th | 1.76 |

R | SR-315 & King | SR-315 & Lane | 1.10 |

S | High & King | SR-315 & King | 1.02 |

T | High & 11th | I-71 11th offramp | 1.15 |

US-23 & Goodale | Goodale parking lot | 0.41 | |

High & 15th | High & 11th | 0.37 | |

High & 11th | High & King | 0.31 | |

High & King | High & 5th | 0.21 |

#### Longitude / latitude information

Location | Lon (W°) | Lat (N°) |
---|---|---|

High & Goodale | -83.00286 | 39.97384 |

High & 5th | -83.00551 | 39.98710 |

I-71 5th offramp | -82.98526 | 39.98631 |

I-71 11th offramp | -82.98519 | 39.99394 |

I-71 17th offramp | -82.98443 | 40.00072 |

US-23 & 17th | -82.99865 | 40.00133 |

US-23 & 15th | -83.00118 | 39.99973 |

High & 15th | -83.00807 | 40.00007 |

High & Woodruff | -83.00888 | 40.00409 |

Woodruff & Tuttle | -83.01748 | 40.00400 |

Lane & Tuttle | -83.01683 | 40.00631 |

SR-315 & Lane | -83.03085 | 40.00646 |

SR-315 I-670 offramp | -83.02120 | 39.97749 |

Park & Vine | -83.00469 | 39.97147 |

Park & Goodale | -83.00453 | 39.97362 |

Goodale parking lot | -83.00649 | 39.97372 |

US-23 & Goodale | -82.99826 | 39.97423 |

SR-315 & King | -83.02511 | 39.99084 |

High & King | -83.00610 | 39.99019 |

High & 11th | -83.00712 | 39.99528 |

Here is some Python code for converting two long/lat coordinates into a measure of miles between them:

# from: http://www.johndcook.com/python_longitude_latitude.html def dist((long1, lat1), (long2, lat2)): # Convert latitude and longitude to # spherical coordinates in radians. degrees_to_radians = math.pi/180.0 # phi = 90 - latitude phi1 = (90.0 - lat1)*degrees_to_radians phi2 = (90.0 - lat2)*degrees_to_radians # theta = longitude theta1 = long1*degrees_to_radians theta2 = long2*degrees_to_radians # Compute spherical distance from spherical coordinates. # For two locations in spherical coordinates # (1, theta, phi) and (1, theta, phi) # cosine( arc length ) = # sin phi sin phi' cos(theta-theta') + cos phi cos phi' # distance = rho * arc length cos = (math.sin(phi1)*math.sin(phi2)*math.cos(theta1 - theta2) + math.cos(phi1)*math.cos(phi2)) arc = math.acos( cos ) # Remember to multiply arc by the radius of the earth # in your favorite set of units to get length. return (arc * 3960.0)

## Comparison of searches

### Number of checked states (time)

### Maximum number of states in memory

### Length of path (goodness of solutions)

#### Let’s introduce an anomaly

Suppose we change the distance from High & 15th to High & 11th to 37 miles (originally was 0.37 miles). This modification is a proxy for a traffic jam between these two intersections or some other similar unplanned situation. A* will account for this change because its heuristic is the composite heuristic \(f(s) = g(s) + h(s)\), which means that \(f\) accounts for both the estimated distance \(h\) between High & 15th and High & 11th (“as the crow flies”), and the cost \(g\) of traveling that distance (the traffic jam is accounted for here).

Best-first search, on the other hand (and hill-climbing for that matter) do not take into account \(g\), so they do not plan for the traffic jam; they only plan for the “as the crow flies” distance. Thus, best-first and hill-climbing will travel across this jammed route and their solution cost will suffer as a result.