# Find the fastest path through this Brazilian concretist poem

One day, a small jet was flying over the concretist poem "Velocidade" (Speed), by the Brazilian poet Ronaldo Azeredo.

In order to cover every vowel present in this poem, it goes in straight line from one letter to another either orthogonally or diagonally adjacent, never making a turn sharper than 45 degrees*.

Assuming the letters are like points in a square lattice, and the jet travels at constant speed, what is the fastest way this can be achieved by the pilot?

Notice that no restrictions were placed in where the jet should start, except that its position goes from one letter to another. Also, the jet is not allowed to leave the matrix of letters.

* Measured by the external angle, i.e. deviation from the original path, so that a 0 degree turn leads to the same direction, while a 180 degree turn makes the jet go in the reverse.

• ...Where does the pilot begin? Commented Sep 21, 2015 at 13:26
• Can the plane leave the square once it has entered? Commented Sep 21, 2015 at 14:34
• You'd have to end on the E at the bottom-right corner, because otherwise there's no way for you to turn out of it.
– user88
Commented Sep 21, 2015 at 15:37
• OP doesn't say it's not allowed to use vowels several times. (This would not be solvable otherwise.) Commented Sep 21, 2015 at 16:07
• Do you mean "sharper than 135 degrees"? All your answers seem to heed this constraint. Commented Sep 21, 2015 at 18:31

Here's a solution with

54 steps.

$38 \space horizontal/vertical + 16 \space diagonal => length \space 60.627$
Thanks to f'' for length optimization.

A brute-forced solution with

47 steps.

$28 \space horizontal/vertical + 19 \space diagonal => length \space 54.87$

• With three extra vowel touches in this one, do you think hypothetically there should be a solution with 51, or is that not guaranteed? Commented Sep 21, 2015 at 18:13
• @lorimer I think the extra vowels are no more different than the consonants. It's possible there is a better solution. Commented Sep 21, 2015 at 18:18
• It looks like you have a 90 degree turn at the bottommost 'O' Commented Sep 21, 2015 at 20:46
• You can save 4(sqrt2-1) by changing the triple crossing to three vertical lines.
– f''
Commented Sep 21, 2015 at 20:57
• You can also save another 2(sqrt2-1) by changing the LC/OO crossing in the lower middle to two horizontal segments.
– f''
Commented Sep 21, 2015 at 21:00

OK. Touching every vowel once, using coordinates A-J horizontal and 0-9 vertically, with the origin in the top left:

J9->J1, I0, H0->A7, A8, B9->H9
I9->I2, H1, G1->B6, B7, C8->G8
H7->H3, G2, F2->C5, C6, D7->F7
G6->G4, F3->D3, C4, C5, D6->E6, F5

That's 71 steps. It's probably not optimized perfectly, so I'll keep looking, but it's at least a starting point.

EDIT: This one has 59 but touches some vowels more than once. Counting as distance (cf. Sleafar's comment) it would be 68.11 units with 22 diagonal moves.

J9->J1, I0, H0, G1, G2, H3, I4->I8
H9->B9, A8, A7, B6, C6->E8->G8, H7
H7->H5, G4, F4->C7, C8, D9, E9, F8
F7->I4->I2, H1, G1, F2->F5, E6

This one doesn't waste as much time by avoiding U-turns, but instead grabs what it can that's nearby.

ETA3: nevermind, the edit would have missed one 'E'. i'll quit second-guessing myself now

I think I am missing something obvious, since the solution couldn't be that easy

16 diagonal + 16 vertical/horizontal steps.

• "never making a turn sharper than 45 degrees" Commented Sep 22, 2015 at 3:11
• where? I am brain-dead here, and can see only 90 degrees and 135 degrees Commented Sep 22, 2015 at 3:25
• Since this point has been brought up before, I edited my post to clarify it Commented Sep 22, 2015 at 3:41