By Stephen Meskin
The puzzle column in this magazine almost eight years ago: Vol 29 No 3 (2017, May–June), was about tiling 4×4 squares with tetrominoes. At the end of that column, I challenged readers to try larger squares and/or different polyominoes. Taking up my own challenge, this column is about tiling 5×5 squares with pentominoes. According to Wikipedia, “a pentomino (or 5-omino) is a polyomino of order 5; that is a polygon in the plane made up of 5 equal-sized squares connected edge to edge.” We do not consider rotations and reflections to be different, so we have only 12 different pentominoes, which we label FILNPTUVWXYZ in approximate accordance with their shapes. See the pentomino Wikipedia article for diagrams of the (non-Conway) labeling scheme we are using and a general discussion of tiling.
To tile a 5×5 square requires 5 pentominoes. For our purposes, the pentominoes in a tiling can be different, some different and some the same, and in one case, all the same. As with the pentominoes themselves, we do not consider rotations and reflections of tilings to be different. We split the tilings up into two classes.
- Problem 1: (Class A) Find at least 10 different tilings of a 5×5 square, without using an I-tile. Also, given a pair of adjacent L-tiles (“LL”), if one rotates the second L-tile 180 degrees, lifts it up and moves it left a bit, then the resulting two L-tiles take up the same space as two adjacent I-tiles (“II”), so we exclude that resulting configuration of L-tiles from Class A tilings. Similarly, we exclude a pair of adjacent P-tiles adjusted to take up the same space as two adjacent I-tiles. On the other hand, you should include each of the remaining 11 pentominoes (other than an I-tile) in at least one of your 10 class A tilings.
- Problem 2: (Class B) Find at least 10 different tilings of a 5×5 square, each using at least one I-tile, or a pair of L-tiles or a pair of P-tiles configured to take up the same space as a two adjacent I-tiles (“II”). (This is a bit easier than Problem 1, since, after you have found one such tiling, a permutation of the columns (or rows) is likely to produce another.)
- Problem 3: Find additional tilings from either class.
- ANSWERS: Diagrams are not required. Even if some diagrams are provided, every answer should list the pentominoes (using the non-Conway labeling) in the tiling in order starting with the one touching the lower left corner and going around the edge of the square in a counterclockwise direction. If there is one tile that does not touch the edge, its label should be placed at the end preceded by a forward slash (/).
The diagram shows an example of a Class B tiling due to a pair of L-tiles.
Solutions may be emailed to puzzles@actuary.org.
In order to make the solver list, your solutions must be received by Feb. 1,
2025.
Solutions to Previous Puzzles—The Game of Chicken
- What is the probability of a five-card flush? To make a flush you need five cards of the same suit, which is (12/51)*(11/50)*(10/49)*(9/48) = 0.2%
- Given a flush on the board, what is the probability that neither player improves upon the flush? The probability that no one gets dealt a flush card is (39/47)*(38/46)*(37/45)*(36/44)=46.1%. However, even if a flush card gets dealt, it might not play if the five community cards are all higher. The probability that exactly one of the four hole cards is a flush card is 4*(8/47)*(39/46)*(38/45)*(37/44)=41.0%. There is a 5/6 chance the flush card plays. So (1/6)*(.410)=6.8%. We can continue this logic if there are 2, 3 or 4 flush cards, plus make an account if someone improves the hand to a straight flush. Eventually, this leads us to the answer of 53.5%.
- What is the probability of a five-card straight? There are 52 choose five, or 2,598,960 potential five-card hands. There are 10 potential straights (A-5, 2-6, 3-7…. 10-A). There are (45 – 4)=10,200 potential combinations of each straight (we need to subtract four to take out the straight-flush possibilities). So the probability of a straight is just 10,200/2,598,960 = 0.4%.
- What is the probability that no one improves upon the straight? For nine of the 10 straights, exactly four cards will make a higher straight (you can’t improve upon an ace-high straight). This probability of both players standing pat is then (43/47)*(42/46)*(41/45)*(40/44)=69.2% for the nine straights and 100% for the ace-high straight. A weighted average of this leads to 72.3%. However, one can still potentially improve by making a flush. While the probability of a five-card flush is zero, the probability of a four-card flush on the board is (12/48)*(11/44)*(10/40)*(27/36)*4=4.7%. Note we can NOT have a paired board, so this probability is slightly higher than five random cards forming a four-card flush. Given a four-card flush, the probability that neither player has a flush card (in this case any flush card will suffice) is (38/47)*(37/46)*(36/45)*(35/44)=41.4%. We need to perform the same logic for a three-card flush, where the probability of this occurring is 37.8% and the probability of neither player having both hole cards able to create a flush is 91.8%. Overall, the probability that no one gets a flush = 94.2%. As both events are independent, the final probability of no improvement is 0.942*0.723=68.0%.
Solvers: Rui Gio, Daniel Wade, Don Onnen, Clive Keating, Al Spooner, Jason Shaw
