• Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 1726
  • Last Modified:

Cities within a distance of a ZIP Code

The question is simple but I can't find the solution which should be answered in C++ or Visual Basic:

 After I get

a. A ZIP Code  and
b. A distance (miles)

How do I find all the cities that are lied within the provided distance from that ZIP Code?

* I already had
- The ZIP Code Lat/Long database
- City name corresponded to that ZIP Code



0
vtb2000
Asked:
vtb2000
  • 4
  • 4
  • 2
  • +3
2 Solutions
 
fridomCommented:
Well a you have a center your ZIP code you then calculat a circle around and and see what other codes lies within that distance. I would pick a book about mathematics and check the chapter about trigonometry.

Regards
Friedrich
0
 
sunnycoderCommented:
Hi vtb2000,

Friedrich has given you a way to go .. here is another ...

In questions of this kind, you can increase the response time of your program considerably by doing some preprocessing. Since you get the input at the beginning of the program, you can dynamically allocate a 2D array of dimension nXn where n is the number of cities ...

An entry [n][m] would indicate the distance from n to m ... the matrix will be symmetrical about its diagonal ... so you can compress it a bit if n is large ...

At  run time all you need to do is get n and m,  perform a  lookup and display the result

Since this sounds like a homework question, I am not providing you any code ... give it a shot ... if you get stuck, we are here to help

sunnycoder
0
 
ozoCommented:
You can estimate the distance as
arccos(sin(lat1)*sin(lat2)+cos(lat1)*cos(lat2)*cos(lon1-lon2))*69miles/degree
0
Technology Partners: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
vtb2000Author Commented:
Thanks for all the answer so far:  I just would like to add something:

1. I can find the distance between one ZIP to another (but still thanks to Ozo).
2. However, there is over 50K of ZIP Code for US, or even worse, 780K of Postal code for Canada.

So if I calculation from one ZIP to 50K of other ZIP, find out the distance between each of the pair, sort them, and get the one that lies within the range, it would be ... tough. Let's say if make a Local Match Maker program, and about 100 people do the search at the same time then my server would run extremely slow.

So, what I am looking for IS ACTUAL CLEVER CODES that attacks this problem from another angle but so far which will reduce search time.  As for me, I am stuck.  

Please help!
0
 
ozoCommented:
For a 50 mile radius, you could quickly eliminate anything furthir than 50/69 degrees away
0
 
vtb2000Author Commented:
Ozo:

Sounds like you got the perfect solution.  However, could you clarify it a little bit please.  Thanks!
0
 
ozoCommented:
If the difference in latitude or longitude of the two cites is greater than about 50/69, you can skip that city with no furthur computation
0
 
vtb2000Author Commented:
Last questions to Ozo:

1. Has your fomular been tested yet?
2. What if the distance is 100, 200, 500, or greater?  Any ratio or fomular for that elimination check?

Thanks,
0
 
ozoCommented:
ignoring the oblateness of the Earth, one degree of latitude is slightly more than 69 miles, so check in the range ±100/69, 200/69, 500/69 for latitude
±100/(69*cos(latitude)) for longitude
0
 
MortimerCatCommented:
Adding to previous suggestions:

Reduce the amount of processing by indexing your database by Latitude. You then only need process cities that lie within a band.

Pseudo Code:

Seek LATITUDE-10miles
Do





0
 
MortimerCatCommented:
Sorry, fingers slipped and I posted before I finished.

Pseudo Code

Seek Zip (LATITUDE-10miles)
Do
   Calculate Distance
   MoveNext Zip
While Zip<LATITUDE+10


Taking this further, create extra indexed fields Latitude & Longitude rounded to the nearest 5 degrees (possibly). This effectively divides the country up into a grid system. Then you would only need to process the cities that fall in the surrounding squares.


0
 
vtb2000Author Commented:
Final comments:

From a mathematical view points (I do need time to integraded your comments in). both Ozo and MortimerCat would best help to solve the problem.  So ... I love to split the points (90 for Ozo and 35 for MortimerCat).  Could it happen by EE system?

Please advise
0
 
matticusCommented:
Quote: "For a 50 mile radius, you could quickly eliminate anything furthir than 50/69 degrees away"

This reduced my query time for smaller radii, but I noticed an increase in time when the radius reached 1000+. Has anyone else experimented with this technique?
0

Featured Post

Vote for the Most Valuable Expert

It’s time to recognize experts that go above and beyond with helpful solutions and engagement on site. Choose from the top experts in the Hall of Fame or on the right rail of your favorite topic page. Look for the blue “Nominate” button on their profile to vote.

  • 4
  • 4
  • 2
  • +3
Tackle projects and never again get stuck behind a technical roadblock.
Join Now