Project Euler Forum

Can you explain how F(1,3) = 4.

I way I am reading the problem, I get F(1,3) = 1
The frog starts in Square 1 (leftmost). The frog has to visit Square 3 (rightmost), thus leaving Square 2 as the only possible unvisited square.
Leaving only 1 journey, Square 1 to Square 3, followed by Square 3 to Square 1.

Where have I misunderstood the problem

pointywalnut wrote:Can you explain how F(1,3) = 4.

I way I am reading the problem, I get F(1,3) = 1
The frog starts in Square 1 (leftmost). The frog has to visit Square 3 (rightmost), thus leaving Square 2 as the only possible unvisited square.
Leaving only 1 journey, Square 1 to Square 3, followed by Square 3 to Square 1.

Where have I misunderstood the problem

"... so that at most one square remains unvisited."
Notice "at most", so trips with no unvisited squares are also counted.

Martin

Thanks, missed that, now I can attempt to solve it.

I am still confused on how F(1,3) = 4. The way I am reading the problem, it would be F(1,3) = 3 being this sequence:

1 square at a time = covers every square and returns back home
2 squares at a time = Square 1 to Square 3, back to Square 1.
3 squares at a time = Square 1 to Square 3 since it can't go outside the limit, then back to Square 1 since it can't go outside the limit.

Is the 3 squares scenario wrong? Am I missing something else? Thanks

The frog is able to change the length of his hop, so for instance he can hop 2 squares in a single hop on the way over, then take 2 smaller hops back.

Ah I get it. The way I was reading it, whatever it chooses for going to the right, it would be the same to the left. And the 3 squares scenario was wrong as well. Figured out F(1,3) = 4 now. Thank you

Hi,
please can someone check these results:
F(1,10)=13557
F(2,6)=28419
F(3,6)=4825827
Thanks

All of these results are correct.

jpatou wrote:Hi,
please can someone check these results:
F(1,10)=13557
F(2,6)=28419
F(3,6)=4825827
Thanks