17
It is a 10x10 map and you need to go east 9 times and south 9 times. There are total ${18\choose9}$ = 48620 paths which are impractical to be done with pen and paper in a reasonable amount of time.
But if you observe
This can be done pretty quickly with just 10x10=100 calculations.
The answer is
Proof of work
Method
10
The letters can be replaced as follows:
And the remaining letters spell
(Arnie's owner gave me some difficulty, because
10
This works, although I'm not sure if it's unique:
How I found it
We have two 6s (red and black) with six cards between them that must be all the same colour (let's say red). Key fact:
Now, to begin with,
After that,
Also,
6
@Mohit Jain's solution is nice and provably correct. It also shows you the best way to reach any square from the top left corner.
I happened to solve the puzzle in a way that's pretty much identical, but in reverse.
Both methods are equally valid, but somehow I felt that finding the optimum path from any square to the bottom right might be a more ...
6
My solution:
5
I'm going to say Grandpa was:
because
5
16 moves is the minimum possible. (I checked by computer, but solved it first without one.)
Explanation:
4
Start by considering the leftmost three cells.
4
You can get a bit more excitement from the letters if you assign each number to a range, albeit an irregular one.
So 1=A, 2=B, 3=C,D,E, 4=F,G, 5=H, 6=I,J, 7=K, 8=L, 9=M,N,O,P,Q, and 0=R,S,T,U,V,W,X,Y,Z.
So for example PUZZLING STACK EXCHANGE becomes 90008694 00137 30351943.
To decode, reverse the process. With some of the letters fixed, guessing the ...
4
A brute force solution: We start with the two sixes with six empty places in between. We must fill those six places with exactly one of each number 1 to 5, as well as 7, while following the rules of spacing.
Now, filling in the 3:
The 4:
When placing the 5, we can easily reject any solution that would leave more than 1 empty place in each suit, since we ...
3
At the risk of flogging a dead horse, I wanted to solve it using Dijkstra's algorithm, since an answer implied that it required a computer to do so.
First, we change it to a minimize problem. We know that all solutions use 19 steps, so we simply change the values to $10-x$ and then minimize the result:
Initialize the top left to a value of 1. Everywhere ...
3
I started by considering the corners.
3
I think it's
which starts it. Then black can do one of the following
First
Second
and Finally
The title refers to
2
2
If any corner is white, there aren't enough white cells to avoid 2x2 blues in at least one other corner. Likewise for the cells adjacent to the corners. The last cell in each 2x2 corner block must then be white.
So we have
BB-BB
BwwwB
Bw-wB
BB-BB
This uses up all 5 white cells, so the remaining "." cells must be blue.
Solution:
1
Local maximum solution:
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