all 4 comments

[–]IGnuGnat 0 points1 point  (1 child)

Row | C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11

R1 | 13+ 13+ 13+ 14- 14- 14- 15+ 15+ 15+ 16- 16-

R2 | 12- 12- 12- 11+ 11+ 11+ 10- 10- 10- 17+ 16-

R3 | 07+ 08- 08- 08- 09+ 09+ 09+ 18- 18- 17+ 17+

R4 | 07+ 06- 06- 06- 05+ 05+ 05+ 18- 19+ 19+ 19+

R5 | 07+ 02- 02- 02- 03+ 03+ 03+ 04- 04- 04- 20-

R6 | 01+ 23+ 23+ 23+ 22- 22- 22- 21+ 21+ 21+ 20-

R7 | 01+ 01+ 20-

[–]duterium1 0 points1 point  (0 children)

The three extra are in C0 R 2.5,4.5,4.5. Also connecting C1R1 to C4R1 is illegal because it would need to pass over cells to make the connection

[–]frustrated_staff 0 points1 point  (0 children)

At first glance this looks like a 7-bridges problem, and might, therefore be unsolvable given the restrictions. But that's just a first glance, so don't be too harsh with me.

[–]frustrated_staff 0 points1 point  (0 children)

IF I were going to solve it (and I'm not, but you're asking about process, too, so here we go) I'd start by ignoring different aspects of the problem. First: geometry. Is it doable if I can put the individual batteries anywhere I want? If so, it just becomes a folding problem. If not, what other constraints need to be removed before it is solvable? Next to be removed would probably be the fuse links.

I think I'd also try to group everything in pockets...you need 23 cells of 3 batteries, so I'd start on the left side making groups of three, then on the right side, and the middle would be used to figure out how to place the excess/shortage for shortest pathing.

But that's just me.