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Let me just go into the XOOPSCRC or QCP checksum for you breifly. The checksum is based in one of the oldest fractal systems on the planet called the i-ching. It has a complexity wrapper around it and is bitwise so it has 2 planes.

Firstly to work out the base of the checksum, in a single seed there is 62/63 characters or nodes, with either 1 or two wild cards, which are replaced by digitals 0 - 9 when they occur. So saying there is two wildcards or aa/AA in a base for the enumeration you would do the following maths to work out the base of the algorithm.

Base of algorithm variant 1 = 62 x 2 + 10 = base 134

Base of algorithm variant 2 = 63 x 2 + 10 = base 136

Why 64 well the i-ching is 64bit having 64 different binary combinations in it, consisting of 6 binary components or elements per character being parsed. To work out if you have any collisions you multiple the number of characters your outputing by the number of nodes per character which is 6. But remember there is a nuclear process in each node which has a combination of 4 so the number of combinations in the i-ching is 6 * 10 ^ 2. Why the power to function of 2 well the nuclear process in this formula is broken down and also has a XOR cipher function in it as well. Now because of this ciper function in the base of it, you also have to put this to the power of 2 to power of 2 due to the bit wise so the actual base of the algorithm is:

Base of algorithm variant 1 = (62 x 2 + 10) ^ 2 ^ 2 = base 322417936

Base of algorithm variant 2 = (63 x 2 + 10) ^ 2 ^ 2 = base 342102016

This means no collisions occur in the formula as the square root of the base divided by 10 divided by 64 divided by 2

No collisions at *square root((62 x 2 + 10) ^ 2 ^ 2) / 10 / 64 = 28.05625 *or with variant 2 *square root((63 x 2 + 10) ^ 2 ^ 2) / 10 / 64 = 28.9* these variables are floored coming to 28 in total for no collision point, anything less than 28 has more collisions which you can work out in a percentile so for example of 100% of data types 12/28*100 = 42.857142857142857142857142857143 meaning that of 100% of data there is a chance of odds of 42.9% that a reoccurring fingerprint will occur.

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