Question

One of my clients wants to use a unique code for his items (long story..) and he asked me for a solution. The code will consist in 4 parts in which the first one is the zip code where the item is sent from, the second one is the supplier registration number, the third number is the year when the item is sent and the last part is a three division alphanumeric unique character.

As you can see the first three parts are static fields which will never change for the same sender in the same year. So we can say that the last part is the identifier part for that year. This part is 3-division alpahnumeric which means starting from 000 and ending with ZZZ.

The problem is that my client, for some reasonable reasons, wants this part to be not sequential. For example this is not what he wants:

06450-05-2012-000

06450-05-2012-001

06450-05-2012-002

…

06450-05-2012-ZZY

06450-05-2012-ZZZ

The last part should produced randomly like:

06450-05-2012-A17

06450-05-2012-0BF

06450-05-2012-002

…

06450-05-2012-T7W

06450-05-2012-22C

But it should also non-repetitive. So once a possible id is generated the possibility should be discarded from the selection pool.

I am looking for an effective way to do this.

1. If I only record selected possibilities and check a newly created one against them there is always a worst case possibility that it keeps producing already selected ones, especially near the end.
2. If I create all possibilities at once and record them in a table or a file it may take a while after every item creation because it will lookup for a non-selected record. By the way 26 letters + 10 digits means 46.656 possible combinations, and there is a chance that there may be a 4th divison added which means 1.679.616 possible combinations.

Is there a more effective way you can suggest? I will use C# for coding and MS SQL for databese..

Answer 1

If it doesn’t have to be random, you could maybe simply choose a fixed but “unpredictable” addend which is relatively prime to 26 + 10 == 36 == 2²·3². This means, just choose a fixed addend divisible by neither 2 nor 3.

Then keep adding this fixed number to your previous serial number every time you need a new serial number. This is to be done modulo 46656 (or 1679616) of course.

Mathematics guarantees you won’t get the same number twice (before no more “free” numbers are left).

As the addend, you could use const int addend = 26075 since it’s 5 modulo 6.