Judge's notes for Bright Bracelet

Each output can be determined by a fairly short logical argument:

Dataset 1:
E must be an end of piece 4.  The best mathematically would be CCCE,
but only one piece has C's on opposite sides, so next
best is CCGE as in the example.

Dataset 2:
There are only the connections AB, AC, BD, CD, which make a circuit,
A+B+D+C = 10

Dataset 3:
Impossible:  Piece 5 must connect with E, and there are no other E's

Dataset 4:
Impossible.  No connections at all.  (Colors must match, and it was explicitly
said that the same brightness does not mean the same color.)

Dataset 5:
The minimum-sum opposite pairs for each piece are
1 BC or AD
2 BD
3 CD
4 BD
5 BD
6 CD
7 CD
8 CD
9 CD
10 CD
11 CD

Using the BC for piece 1, the counts of each color are: 10 D, 8 C, 4 B's. There
are the necessary even numbers of each connection, and they can be connected
in one cycle.  The piece number is shown between the connecting colors:
D2B1C3D4B5D6C7D8C9D10C11 and back to the D in piece 2.
This makes 10/2 = 5 D connections, 8/2=4 C's, 4/2 = 2 B's,
5*251 + 4*252 + 2*253 = 2769

Dataset 6:
Same as Dataset 6, except all brightness is one greater:
  2769 + 11 = 2780
Together Datasets 5 and 6 took more than 10 minutes on a 2.4 GHz Intel processor
with an efficently coded and optimized permutation algorithm in C++.  For
each dataset more than 20.6 billion passes are made through part of the code.

Even with a really fast computer out there, the permutation approach is
not going to work for datasets 5 and 6 together.

Dataset 7:
Again the minimal-sum opposite pairs can be listed.  This time one element
of each pair, except in the first piece, is fixed, and is preceeded by an
underscore as a reminder.  Where another pair is used in the final solution,
it is shown in parentheses.
1 AE
2 _AB
3 _BC
4 _CD (solution uses CF)
5 _DE (solution uses DG)
6 _EE (solution uses EH)
7 A_F
8 A_G
9 A_H
10 A_A
11 A_D

Again counting. there are 8A 2B 2C 3D 4E 1F 1G 1H.
The F, G, H cannot be changed, and another must be created from elements that
are not fixed.  The closest colors to change are the odd D and two E's
which can be changed to F, G, H.  A cycle can be made:
A10A2B3C4F7A8G5D11A9H6E1 and back to the A at the other end.

This leave 4 A connections and 1 each for B through H,
 4*10+20+30+40+50+60+70+80 = 390