Good Puzzles Are the Reason I Check Facebook
Here’s one by Sergei Luchinin, designed for 7th graders.
Puzzle. We have an 8-by-8 chessboard, but it’s not colored in the usual checkerboard pattern. Instead, all cells in odd-numbered columns are black, and all cells in even-numbered columns are white. A limping rook is placed in the lower-left corner and can only move one cell to the right or one cell up. The rook’s goal is to reach the upper-right corner.
The question is: Are there more paths that pass through more white cells, or more that pass through more black cells?
Share:
Zarunias:
You can group the paths into pairs: Every path is paired with the path you get if you rotate the whole path by 180°. In this way every path that passes through more black cells is paired with a path that passes through more white cells and vice versa. As there is no middle cell in the grid no path is paired with itself. So the numbers must be the same.
2 November 2024, 9:41 am2024 in math puzzles. – Math with Bad Drawings:
[…] The limping rook […]
19 December 2024, 1:00 pmarmadillo:
> (Long-winded) Proof using combinatorics:
At some point, in traversing left to right, you must cross 4 black and 4 white cells in alternating order chronologically:
B…W…B…W…B…W…B…W… (done)
Next we need to allocate the vertical movements into the spaces in between.
But here we only care about the relative counts of white and black allocations. There are an equal number of white and black “buckets”, so without loss of generality we can just look at the ways to allocate 7 identical objects (vertical movements) into two buckets (corresponding to black or white), and compare the number of times where black or white wins.
B ___ W ___
White wins if you allocate 0, 1, 2, or 3 objects into white buckets.
Black wins if you allocate 4, 5, 6, or 7 objects into black buckets.
Since there are four general cases of each, we can say that neither black nor white wins on average.
> Follow-up question: Is this true for any dimension of vertically striped board?
Changing the number of rows only changes the number of objects in our example above, e.g. if it’s 6 then white wins for allocations of {0, 1, 2} to black, and black wins for allocations of {4, 5, 6} — both winning in three general cases. So the result is invariant to the number of rows.
But if the number of columns is odd, this advantages black.
> Another follow-up: For the 8×8 board, what board colorings do black and white tie?
By counting the number of paths that cross a cell, you get a grid of numbers like this:
1 8 36 120 330 792 1716 3432
8 49 168 420 840 1386 1848 1716
36 168 441 840 1260 1512 1386 792
120 420 840 1225 1400 1260 840 330
330 840 1260 1400 1225 840 420 120
792 1386 1512 1260 840 441 168 36
1716 1848 1386 840 420 168 49 8
3432 1716 792 330 120 36 8 1
For black and white to tie, the sums of their cells’ path counts must be equivalent. We can see this is true for the vertical stripes pattern, it’s also true for horizontal stripes, but it’s not true for a checkerboard pattern.
There are some crazy patterns where white and black tie, for example:
B W B W B W B W
W B W B W W B W
B W B B W B W W
W W W W B W B B
B B B W B W W W
W B W B B W B B
W W B W B B W B
W B B W B W B W
I made a claude artifact where you can create your own 🙂 https://claude.site/artifacts/0154aa1a-ef99-46e2-a2a9-49c24abd88cc
I have no idea how to do this, but I wonder if there’s a nice way to count the number of unique board colorings for which black and white tie.
19 January 2025, 12:05 pm