# Posts Tagged ‘polyform change paths’

## Pentominoes on paths and trees

February 3rd, 2017

Here’s a path that could be taken by a chess king. All subpaths of length four describe a different pentomino:

This led from the grid of pentomino painting instructions that I posted previously. Consider a string of arrows for which the subsequences of length 4 include instructions for producing all 12 pentominoes. (This is somewhat analogous to a de Bruijn sequence.) For the case shown, the string is ←↑↖→→→→↓↓←↓←↘↗↓, although the graphic seems more illuminating than the arrow string here.

Instead of a path, we could have a tree of pentominoes:

Along with the constraints that each pentomino occurs exactly once, and no square is used more than once, I wanted to limit the number of branches per node. The root node having three branches might be considered a flaw, but this was the best I could do.

## 3×3 block Pentominoes and Hexominoes

January 21st, 2011

There are 8 pentominoes and 8 hexominoes that fit in a 3×3 cell. The combined set seems to cry out to be presented in a 4×4 grid of 3×3 blocks, with the pentominoes and hexominoes in checkered positions:

The best I could think to do was make a figure that is connected, hole-free, and has a rotationally symmetrical pattern of connections between blocks. I had hoped to make them into a geomagic square, but now I’m pessimistic about that working. And the trick from my magic 45-ominoes of making all rows and columns have the same number of cells in polyominoes won’t work here because the total number in each row of 3×3 blocks is 22, which isn’t divisible by 3.

I’ve looked before at problems involving moving a single cell at a time to cycle through a set of polyforms. Because this set has equal numbers of pieces at two consecutive sizes, it invites using adding and removing single cells, rather than moving them, as the action for taking one piece to the next in a path:

Because the X pentomino has only one possible predecessor or successor, it cannot be part of a cycle, but it is still possible to make a path through all of these pentominoes and hexominoes with the X as one of its endpoints.

## Hexiamond Minimal Covers

December 16th, 2010

I define a cover of a set of polyforms as a shape or shapes into which each polyform in the set can fit. I’ve written up an exploration of related problems on my recreational mathematics non-blog. Most of these problems use pentominoes or other polyominoes; it lately occurred to me that I had done a disservice to the hexiamonds, which are just as worthy of attention. So here’s a minimal (ten cell) cover of the twelve hexiamonds:

A proof of its minimality is simple. There are two ways that the bar and hexagon hexiamonds can be superimposed with maximal overlap:

Each of these contains nine cells, and neither is a cover of all of the hexiamonds. (The one on the left is missing the butterfly and chevron hexiamonds, the one on the right, just the butterfly. ) Therefore any cover must contain at least ten cells, so the one given above is minimal. (The nomenclature I’m using for individual hexiamonds is given on this MathWorld page.)

In fact, there are five ways to complete a cover by adding a single triangle (marked in blue below) to one of the figures above:

And these are the minimal covers produced:

The animation at the top of this post displays a cycle where a single cell is moved at each step. (I posted about this previously with the pentominoes in a minimal cover.) The hexiamonds have the handy property that their number is exactly twice their size, which leads naturally to the following problem:

Problem #16: Starting with a hexiamond with all cells labeled, find a sequence of single cell moves that cycles through all twelve hexiamonds, returning to the first hexiamond, and that moves each labeled cell exactly twice. Bonus points if the set of cells used is either a minimal cover or a 2×3 parallelogram. Bonus points as well for satisfying conditions like those of problems #10 and #11 in the post linked above. (All cells return to their starting positions, or all cells end up in a cyclic permutation of their starting positions.)

## All Pentominoes in 5

December 13th, 2010

I’ve been thinking about variations on the problem of cycling through all twelve pentominoes by moving a single cell at a time. (I wrote about this in a previous post.) Constraining the way that the squares are allowed to move led to something almost like a chess problem.

The problem:

Starting with the above position, take five turns as follows:

A turn consists of moving one white knight, then moving one black knight, according to standard chess rules.

After each turn, the squares occupied by the ten knights must form two separate pentominoes.

After the fifth turn, all twelve pentominoes must have appeared exactly once. (This includes the two that are present in the starting position.)

[I may make a separate post discussing and spoiling the puzzle later.]

## Pentomino Cover Cycles

March 17th, 2010

What’s the smallest shape into which any of the 12 pentominoes can be placed? I call this old chestnut the “minimal pentomino cover” problem, and I’ve spent a lot of time working on a number of variations on it. For the purpose of introducing and illustrating the basic problem to my dear readers, I wanted to use an animated GIF file showing all of the pentominoes in turn being placed on a minimal cover.

An aesthetically pleasing way to cycle through the pentominoes would be to move one square at a time. This is in fact possible:

A couple of variations on the problem of finding such a cycle suggest themselves:

#9: Minimize the total distance the squares move per cycle. The taxicab metric seems to be more sensible and simpler than Euclidean distances here. I made no attempt to do any minimization in the above solution, so I’m sure there is room for improvement.

#10: If you gave every square in the pentominoes a distinct color, and kept the color the same when a square moved, you could keep track of where the squares end up at the end of a cycle. During the cycle illustrated above, two pairs of squares switch places. Is there a cycle of single-square moves through the pentominoes that ends with each square in the same place it began?

Notice that the central square can never move, because the only pentomino placement without the central square is one of the P pentomino, for which the only valid square movements turn it into a U pentomino. It would need movements to two different pentominoes to be part of a cycle.

For both of the above problems, the other 9 square pentomino cover would also be a valid substrate:

Since this one has no immobile squares, another problem using it may be solvable:

#11: Find a cycle where the permutation of the squares from one cycle to the next is cyclic (in the second sense in the linked article.) That is, successive iterations of the cycle will eventually take each square in a pentomino to all of the other positions in that pentomino.

Some very good news: I’ve been invited to the 9th Gathering for Gardner conference in Atlanta later this month. The Gathering for Gardner is an invitation-only conference  held in honor of Martin Gardner, who brought recreational mathematics to a generation through his columns in Scientific American. That generation was not my generation, but it was impossible to miss his imprint on later writers, and I’ve picked up used copies of several of the collections of his columns. A large proportion of the names on the spines on my recreational mathematics bookshelf are represented among the invitees, so this will be really special for me.