# Monday's Blocking Donimoes Problem

I've designed a set of dominoes puzzles that I call Donimoes. You slide the dominoes like the cars in Nob Yoshigahara's Rush Hour puzzle, always along their long axis.

## The Blocking Puzzle's Goal

The goal is to slide all the dominoes into a rectangle, without sliding any matching numbers next to each other.

## Moves

Move a domino one space along its long axis so that none of its numbers match an adjacent number on a neighbouring domino. In this example, the lower domino can move to the right, because the three doesn't match the two, and the four doesn't match the 3. You couldn't move it another space to the right, because then the threes would be right next to each other.

## Stay Connected

All the dominoes in the puzzle have to be connected in one solid group, diagonal connections don't count. When you move a domino, it can be disconnected during the move, as long as it is connected at the start and the end of the move. Remember that it can only move one space at a time, though.

## Example Problem

Here's a small example problem:

Find a set of dominoes, set them up to match the diagram, and then slide them into a rectangular shape. If you need help, here's the solution:

You can write that solution using this notation:

01D, 12R, 12R, 12R, 01U, 60U

For each step, move the listed domino left, right, up, or down.

## Today's Problem

Now here's a more challenging problem for you to solve. Post your solution as an answer.

If you like this puzzle, watch for new problems every day this week.

• Am I right in assuming that 24D and 24U are both invalid first moves because 24D would pair the 2's and 24U would disconnect? – LeppyR64 Jul 22 '19 at 15:25
• @LeppyR64 i think so – Omega Krypton Jul 22 '19 at 15:25
• Fun puzzle. Just a word about etiquette: we're a bit cautious about self-promotion (or indeed other-promotion) around here; see this Meta post for some guidelines. Strictly, encouraging readers here to visit your website is off-limits, but I don't propose to make a fuss about it on this occasion:-). – Gareth McCaughan Jul 22 '19 at 15:46
• Wish I could upvote this twice - love the Rush Hour games, and this adds a whole other dimension to it! Man, I'm gonna lose so much time looking through more of these... – Stiv Jul 22 '19 at 15:47
• @LeppyR64 that's right. – Don Kirkby Jul 22 '19 at 16:20

Since the 'donimoes' are limited to moving along their long axis, I believe this can be done in 12 moves as follows:

32L 32L 24U 10L 02U 10L 10L 10L 02D 24D 25L 32L

Visually:

• Good answer, and I like your creative use of a spreadsheet (I assume) for the solution diagram. – Don Kirkby Jul 23 '19 at 4:29
• Indeed - Excel is a good friend! :) – Stiv Jul 23 '19 at 5:15

[EDITED to add:] As Stiv kindly points out in comments, this is a non-solution because I broke one of the rules. Or maybe, as LeppyR64 kindly points out in comments, I just failed to note that I was moving things multiple times, in which case ... it's identical to Stiv's already-posted solution. either way, I'm a twit. I'm leaving it here because I find it healthy to look like an idiot when I have been an idiot. I might delete it later.

I have a solution in

nine moves

as follows:

32 left
24 up
10 left
02 up
10 left
02 down
24 down
25 left
32 left

with this final configuration:

• But then your second move breaks the 'Stay connected' rule since 'diagonal connections don't count'... The 10/32 block ends up separated from the rest... – Stiv Jul 22 '19 at 15:41
• Yow, so it does. I thought I'd checked for that, but obviously not. Good catch. – Gareth McCaughan Jul 22 '19 at 15:42
• Got me a few times too! I have a solution in 12 but not sure if it can be done in fewer... – Stiv Jul 22 '19 at 15:43
• Are these not the same solutions except Gareth's solution compresses duplicate 1-space moves into one move? Stiv has 32L 32L where Gareth has 32 L. Stiv has 10L 10L 10L where Gareth has 10L. – LeppyR64 Jul 22 '19 at 15:44
• Ah, but maybe -- of course I can't remember now -- I actually intended my "32 left" to mean moving it left twice, in which case rather than being bad because it's invalid my solution was bad because it duplicated an existing one. (I just noticed that the numbers of moves were substantially different and didn't check what Stiv's actual moves were. My bad.) – Gareth McCaughan Jul 22 '19 at 15:48