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ID with the below structure
XXXXXXX-1234-5678-9012-345-C
where
XXX XXXX - series, can be used to split ID's between the groups based on the way how it was generated. Or we can just use it to increase number of combinations and decrease the collision probability
initially we can should generate all the ids within the same series.
Options for the series:
- all all the latin letters - 8000 810,000 combinations
- numbers and only some letters that looks the same in Cyrillic and Latin alphabet and can be entered via the phone keypad: A, E, H, K, M, P, T, X - 512 combinations104,976 combinations
- numbers - 1000 10,000 combinations
It is proposed to go with option 2 with the possibility to extend to option 1 if needed in future
1234-5678-9012-345 - randomly generated numbers
Number of combinations - 1100,000,000,000,000
Options for numbers:
...
Total number of possible combinations excluding series will be:
51251,200,000,000,000,000
Probability of collisions: 0.1% 01% probability in the range of around 5.5 millions.
Probability of collisions including series: 0.0001% probability in the range of around 3 24 millions.
Solution
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2 - TO GO!
The same as solution1, but
XXXX-12E4-5A8-P01X-T3 -52A8-P01-3
XXXX-12E4-52A8-P01 - randomly generated numbers and letters A, E, H, K, M, P, T, X.
...
In this case we might not check if ID is unique.
But this This number is shorter but not so convenient for human to work with.
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from math import log1p, sqrt def birthday(probability_exponent, bits): probability = 10. ** probability_exponent outputs = 2. ** bits return sqrt(2. * outputs * -log1p(-probability)) print birthday(-6, 68) |