The placement of the 7-bit streak (7 consecutive 1s or 0s) starts between b0 and b4.
If the streak is exactly 7 bits, then any bits before and after the streak must be complementary to the bits in the streak, that is, if the streak is all zeroes, then it is encompassed by 1s. With the exception of a streak starting at b0 or b4, the encompassing bits will be 1s.
The table below shows in column 'n' how many different (discrete) 11-bit lengths there are for the various positions of the streak.
| Streak start |
b0 |
b1 |
b2 |
b3 |
b7 |
b8 |
b9 |
b10 |
n |
| b0 |
0 |
0 |
0 |
0 |
1 |
0/1 |
0/1 |
0/1 |
8 |
| b1 |
1 |
0 |
0 |
0 |
0 |
1 |
0/1 |
0/1 |
4 |
| b2 |
0/1 |
1 |
0 |
0 |
0 |
0 |
1 |
0/1 |
4 |
| b3 |
0/1 |
0/1 |
1 |
0 |
0 |
0 |
0 |
1 |
4 |
| b4 |
0/1 |
0/1 |
0/1 |
1 |
0 |
0 |
0 |
0 |
8 |
For an all 0s streak there are 28 discrete 11-bit lengths, so, there must also be 28 discrete 11-bit lengths for an all 1s streak, making 56 in all. The explanation is that, with the exception of streak positions b0 and b4, where two of the remaining 4 bits (11-7=4) can be any configuration, while two bits are constrained to be 1 or 0, depending on the nature of the streak, the configurations are 00, 01, 10, 11 (4 configurations) spread between the segments outside the streak. For streak positions b0 and b4, only one bit is constrained, leaving any configuration of 3 bits: 000 001 010 011 100 101 110 111 (8 configurations) positioned either before or after the streak.