A Pedersen commitment works like the above but with an additional property: commitments can be added, and the sum of a set of commitments is the same as a commitment to the sum of the data (with a blinding key set as the sum of the blinding keys):
C(BF1, data1) + C(BF2, data2) == C(BF1 + BF2, data1 + data2) C(BF1, data1) - C(BF1, data1) == 0
There's a mistake in that line, which confused me at first, but then I saw the same introduction of CT in your old webpage, where you have it corrected:
http://web.archive.org/web/20150628230410/https://people.xiph.org/~greg/confidential_values.txt.
It's:
C(BF1, data1) + C(BF2, data2) == C(BF1 + BF2, data1 + data2)
C(BF1, data1) - C(BF1, data1) == 0
That change of line is important, otherwise it seems like you're multiplying commitments. I don't know if Pederson's commitment multiplication makes any sense, it certainly does not in here. Only addition.
In my understanding, and please correct me if I'm wrong, there's a direct connection between blinding factor (r) and data (a). Taking the example by gmaxwell:
If data_n = {1,1,2} and BF_n = {5,10,15} then:
C(BF1, data1) + C(BF2, data2) - C(BF3, data3) == 0
It is clear that, just as data1 + data2 = data3, the same equation applies with BF_n. Therefore, if I'm spending input_1 and input_2 to create input_3, does the recipient need to know BF1 and BF2, so that they can construct BF3? Is BF3 known by both the sender and the receiver?