Find Drivers License Number Nj
Driver's License numbers in New Jersey aren't random. They follow, where A is the first letter of the person's last name, ffff is some mapping of the remaining letters of the last name to a four digit numeric, lll is a mapping of the full first name to a three digit numeric and ii is a code representing the middle initial (according to the below table: 6 7 8 - 1 a j 2 b k s 3 c l t 4 d m u 5 e n v 6 f o w 7 g p x 8 h q y 9 i r z Where the number corresponding to the initial is 10.column number + row number. Mm corresponds to the month born, and yy to the year born. E is the eye color (a value 1-8 corresponding to BRO, BLU, GRY, GRN, BLK, etc.)The only thing I don't understand is how the names are mapped to the integer values.
I only have 5 examples for the last name mappings: (ignoring the first letter because it doesn't play into the mapping aab - 0001ackson - 0062eals - 2024eimel - 2278ounds - 6810For first names, I only have four: Alexander - 019Richard - 655John - 407Matthew - 529Does anyone have any ideas how the implementation is done, or even a general mapping function that will hash a max 25 length string to a four digit or three digit number while maintaining lexicographical order ( 0000ackson - 0035eals - 1547emiel - 1722ounds - 5695Get a list of the top 10,000 most common surnames. Order by the second letter, and then check the index. This produces the following mappings: aab - 0005ackson - 0128eals - 2813emiel - 3235ounds - 7588Each letter subdivides the 10,000. The first number (according to 1-26) cuts it into one of 26 pieces. The second cuts the piece into one of 26, and so on and so forth. This produces the following mappings: aab - 0000ackson - 0028eals - 1536emiel - 1648ounds - 5656Convert each of the first four letters to 1-26. Concatenate all of them, multiply the resulting number by 10,000, and divide by 26262626.
Find Drivers License Number Texas
This produces the following mappings: aab - 0003ackson - 0392eals - 1908emiel - 1953ounds - 5792Do the above with 0-25, divide by 25252525. This produces the following mappings: aab - 0000ackson - 0008eals - 1584emiel - 1631ounds - 5623Additional SamplesWhile I believe all of the above samples are correct, I tried to track down more authentic sample data points. Ones that I can guarantee are below:Last Names avis - 0921eals - 2024olff - 6247orello - 6581First Names Alexander - 019Andrew - 042Gabriel - 270Lena - 456. In case you're still trying to figure this out, I've made some progress. With assistance from u/jccool5000 on reddit , who has a collection of over 900 samples mostly from Ontario.
AFAIK, Ontario and NJ share the same encoding - Quebec, not so sure. I did some data manipulation to figure this out.Starting with the numbers of the last name, 1st of 4 digits corresponds to the 2nd letter of the last name, as the 1st is already coded directly to the first letter of the license number. 0 = A1 = B C D2 = E3 = F G H4 = I J K5 = L M N6 = O7 = P Q R8 = S T9 = U V W X Y ZThe remaining three numerical digits codes the second letter of the last name as well, from 000-999. However, each second-digit has its own 000-999 range.
That is to say:. Hypothetical last name XA is X0001. Hypothetical last name XAZZZ is X0999, or something close to 999. Hypothetical last name XB is X1001. Hypothetical last name XDZZZ is X1999, or something close to 999. Hypothetical last name XE is X2001. Hypothetical last name XEZZZ is X2999, or something close to 999.You can refer to the above table to see when the 999 will reset back to 000.
This is just the pattern I've found so far. I don't know how the numbers are distributed to the names.First name code is a lot simpler, but at the same time, it's also not evenly distributed. The difference with first name code is it only goes from 000 (Aaron) to probably 799 (796 for Zoe).
What I mean by not evenly distributed is names that start with A range from 000 to 071, which 071 has some names that start with BA. Meanwhile, names that begin in Y are confined to a small range of no less than 785 to no more than 792.
I don't see this above, but male or female is coded in as well. In the last five digits, the first 2 are month of birth. Males are 01-12. Females 50 is added. So the run from 51 (january) to 62 (december)Also, my name is Alexandra, which is also 019 as is your example of alexander.The absence of a middle name is reflected as 00i know a friend with middle name alexandra has 61 = (ii)another, is Serafina middle name 82 = (ii)another, is Dorothy middle name 64 = (ii)I would suggest collecting more name samples to compare.
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