Tag Info

24

My first attempt was done by hand. It used 28 commands: but this was not optimal. I have now done a computer search to find optimal solutions. It found 180 solutions of length No shorter solutions exist. I'll illustrate the following solution: To see how it works, start with a robot on every available square, and send the commands. The robots should all ...

14

13

I have a Here are the details: And the final UUL makes sure every path has touched the T.

12

Here's my go (click to enlarge): The pattern of reds is and the white and black patterns are so it's enough to count half the pieces only. :-)

11

Answer Attempt 1 At first I thought This can be achieved as follows, using the same technique as Daniel Mathias previously: Adjusting this a little, we can get After a lot more fiddling around with this, I realised it seems that Attempt 2 I then tried Attempt 3 Finally, the third option is The mistake was

9

Can this be improved?

8

Sorry that I ignored that comment of yours in the other post. In fact the answer is "yes": every number can be made. Proof: Note that the number $\alpha = \log_32$ is irrational. This implies that the sequence $(\{ k\alpha\})_{k \geq 0}$ is dense in the interval $[0,1)$, where $\{\cdot\}$ denotes the fractional part. Now let $n$ be an integer and consider ...

7

Assuming that the walls replace a cell of a grid like in the 9x9 version and are not located between two cells. How I got there:

6

This grid requires 18 steps: Explanation

5

Taking the basics of Jaap Scherphuis solution, I made some changes. "Once the robot reaches the target, the game will terminate immediately." If I understood this rule correctly, not all robots have to end up on the target square after the same number of moves. Once a robot reaches the target, he has completed the game and we don't have to care about him ...

5

18:

3

Intuitively, it seems like

3

Here's a quick upper bound: You can achieve this by

3

This should be optimal, but haven’t proved it...

3

I started making the numbers progressively by hand. The even numbers are easy to compute. (Twice of another lower number). Odd numbers which are divisible by 3 also make unbroken chains (3>6>12>24>48>96 and 9>18>36>72). $1$ 1>$2$ 1>2>4>8>16>5>10>$3$ 1>2>$4$ 1>2>4>8>16>$5$ 3>$6$ 4>8>16>32>64>21>$7$ 4>$8$ 7>14>28>$9$ 5>$10$ 10>20>40>...

2

This was probably not the most exciting question. Anyway there are many solutions. Here are some examples: Interestingly one cannot add a single other bishop of any color.

2

Here are some notes on my own approach: 1. Prep work I did a lot of prep work ahead of time so that the combinatorial engine could run as efficiently as possible. For example, I took each word in my word list and shifted it in a circular fashion into all possible positions around the wheel. I stored these results in a lookup array for quick referencing ...

1

I count compare 1 from light pile and 1 from heavy pile to ensure light pile is actually the light pile of marbles. take light marble from step 1 and compare with second from light pile. take both light marbles from step 2 and compare with two others from light pile. take the 4 from step 3 and compare with 4 others from light pile. take the 8 from ...

1

I thought about something here : There is another answer I thought of first, and I wanted to share. And last but not least, the ultimate move :

1

Looks like we have mazes we need to solve simultaneously: and These are the difficult cases, so all the other configurations (that are solvable in the first place) should be automatically solved if we can get all of these. So we need a move sequence that has Here's one: Which has Is it optimal? No idea. I'm going to need some more coffee if I'm to ...

Only top voted, non community-wiki answers of a minimum length are eligible