3
$\begingroup$

enter image description here

Quarter rolling a die means triping it on any of the 4 sides adjacent to its base or face in contact with the flat surface. Using the assembled die and quarter chessboard:

  1. You can place and align the 1 x 1 die on any square of the board with its top face pips as initial count.

  2. By quarter rolling to any adjacent square once, you just add the top pips number of die on that square to previous count until all 16 square has pip numbers to be added.

How to get a maximum die roll (sum of all top face number of pips) by quarter rolling a standard die on every square of a 4 x 4 checkerboard?

Improve this question
$\endgroup$
2
  • $\begingroup$ Is it allowed to pass on a square twice? If so do you consider the first number of the last? $\endgroup$
    – Florian F
    Jul 26, 2020 at 11:16
  • $\begingroup$ @Froilan -it is only allowed to role on any square once. You can find many possible ways (mirror or symmetric) to pass all 16 square starting on - corner , side & middle square. $\endgroup$
    – TSLF
    Jul 27, 2020 at 18:19

1 Answer 1

Reset to default
4
$\begingroup$

Early result:

Starting anywhere in this grid, with the die in the appropriate orientation, follow the path for a sum total of 70 pips.
enter image description here
I started with 6-5-4-6 from various positions. This is the highest total I have achieved. Should have bonus points for being a closed Hamiltonian circuit.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Was this answer accepted because it is optimal? If so, how can you prove it? $\endgroup$
    – melfnt
    Jul 27, 2020 at 8:21
  • $\begingroup$ @melfnt An exhaustive search using computer programming proves this to be optimal. $\endgroup$
    – Daniel Mathias
    Jul 27, 2020 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.