The king can stand on any space at the edge of a square 201 spaces wide by traveling to it as directly as possible with his 100 moves; diagonally toward it until he shares an X or Y coordinate, then orthogonally toward it the remainder of the distance. All spaces inside this square can also be reached by using the same method while wasting excess moves. If the king reaches his target space with an excess of turns remaining, he can burn them 3 at a time by moving in a triangle, and 2 at a time by moving back and forth. If the king is next to his target space with exactly 2 turns remaining, he must move onto it indirectly by first moving onto an adjacent space. This means he can reach any space within the square, for a total of 40401 spaces.
Other pieces: Rook and Queen: All squares Bishop: All squares with the colour as the one the bishop is on Pawn: 1 square (if no enemy piece) or 201 squares (if captures are included) or 400 squares (including initial double-step move) or all squares (if promoted to a queen or rook) King: 40401 squares Hopefully didn't get them wrong as I already did for a lot of times while writing this comment
My answer to the end of the video. The king can go up 100 spaces, can go down 100 space (+ his lane) So 201 lanes total We can do same with columns so there are also 201 columns. So I multiply these numbers : 201² = 40401. Thats my answer. And for n moves, the king can go to S(n)= (2n+1)² squares when n≠1 And with S(1)= 8
Answer about the king (just guessing, didn't think about it a lot): It will be able to reach all squares inside a giant square that's of side 201 (100 in each direction from the starting position), so 40.401 squares?
There is a rather remarkable recurrence due to Vincenzo Librandi that characterizes this problem... S(n) = 3*S(n-1) - 3*S(n-2) + S(n-3). Can you see why?
The king can stand on any space at the edge of a square 201 spaces wide by traveling to it as directly as possible with his 100 moves; diagonally toward it until he shares an X or Y coordinate, then orthogonally toward it the remainder of the distance.
All spaces inside this square can also be reached by using the same method while wasting excess moves.
If the king reaches his target space with an excess of turns remaining, he can burn them 3 at a time by moving in a triangle, and 2 at a time by moving back and forth.
If the king is next to his target space with exactly 2 turns remaining, he must move onto it indirectly by first moving onto an adjacent space.
This means he can reach any space within the square, for a total of 40401 spaces.
For king, it goes 1, 9, 25 ie odd squares. so its (2n-1)^2
n was defined as the number of moves, so for n=1 you get 8 (or 9 for odd squares)
Then you get (2n+1)^2 instead
Technically it goes 1,8,25...
(2n+1)^2 for any number that is not equal to 1, and when it is 1, we substitute the formula with 8
Other pieces:
Rook and Queen: All squares
Bishop: All squares with the colour as the one the bishop is on
Pawn: 1 square (if no enemy piece) or 201 squares (if captures are included) or 400 squares (including initial double-step move) or all squares (if promoted to a queen or rook)
King: 40401 squares
Hopefully didn't get them wrong as I already did for a lot of times while writing this comment
Very cool video with nice idea. I have been entertained and further educated.
Lil horsie be ridinnnnnn 🐴🐴🐴
I love the variety on your channel!
My answer to the end of the video.
The king can go up 100 spaces, can go down 100 space (+ his lane)
So 201 lanes total
We can do same with columns so there are also 201 columns.
So I multiply these numbers : 201² = 40401.
Thats my answer.
And for n moves, the king can go to S(n)= (2n+1)² squares when
n≠1
And with S(1)= 8
Answer about the king (just guessing, didn't think about it a lot):
It will be able to reach all squares inside a giant square that's of side 201 (100 in each direction from the starting position), so 40.401 squares?
It feel like the main part, which would be proofing the observed pattern is missing in the video. still great video
There is a rather remarkable recurrence due to Vincenzo Librandi that characterizes this problem... S(n) = 3*S(n-1) - 3*S(n-2) + S(n-3). Can you see why?
Nice video!
You made a mistake there in the title
Enehhehe.