The 6x10 RectangleLet us revisit a classic pentomino configuration: the 6x10 rectangle. Below is one solution.
An interesting way to codify the arrangement is to consider each piece and count how many other pieces it touches ("count its neighbors"):
F - 3 I - 4 L - 3 N - 4 P - 3 T - 7 U - 3 V - 2 W - 5 X - 4 Y - 6 Z - 4Now, let's sort that arrangement numerically (greater numbers to lesser) rather than alphabetically. We can construct a 12 digit number like so: 765444433332 (yes, we are losing some information here, namely which letters went with which numbers, but I'm just interested in the numerical part).
Many questions come immediately to mind; the most obvious of which is: how does that 12 digit number compare to likewise generated numbers from other 6x10 rectangle solutions?
Here's another one:
It generates the following 12 digit number: 544444443222.
So we know with certainty that the generated numbers are not all going to be the same when considering the 6x10 rentangles. But I seriously doubt each is unique.
So what are all of the associated 12 digit numbers? This is all just another way to classify those 2339 solutions to the 6x10.
What is the largest such number generated?
How about the smallest?
If anyone out there wants to do such an analysis, please let me know the results... I'll happily post them here (with credit to you).
Instead of listing the digits out to form a 12 digit number, what if we summed them up?
The first example adds up to 48, the second: 42 (we know the sums always have to be even since neighbors always come in pairs: if a pentomino touches another, then that "touching" will be counted a second time when the second pentomino is looked at).
How do the sums of all 2339 solutions compare? It seems possible that the highest 12 digit number very well may not lend itself to the highest sum... but is this the case?
Other ConfigurationsBut why restrict ourselves to just the 6x10 retangle? Let's throw these questions wide open. Let's consider all fully connected configurations (all of the pentominoes must properly touch one another).
Having 10 or 11 neighbors throws a wrench in that base-10 generator. It can easily be remedied, of course, if one is willing to accept a base-12 notation where "A" stands for "10" and "B" is substituted for "11". With this in mind, the two configurations would generate the following numbers:
Surrounding L (left): B33333333333 (sum = 44, base-10) Surrounding I (right): B33333333322 (sum = 42)
Questions #2 and #4 above are actually somewhat trivial. By stringing the 12 pieces together in a long configuration, you can generate a minimal number of 222222222211 (sum = 22); any lesser number or sum can only be generated by disconnected figures.
But questions #1 and #3 don't seem to be quite as simple.
Anyone want to give them a shot?
- Eric Harshbarger, 1 February 2005
Update (1 Feb 2005): Within a few hours of writing this article, Aad van de Wetering of Holland sent me two text files. One of them lists all 2339 generated 12 digit strings and their sums. The second prints out all of the configurations in text format (a handy thing to have for this and other 6x10 questions).
From the first file one learns that the greatest 12 digit number generated from 6x10 retangles is 864433333333 (and there is only one solution that generates that number).
It's sum is only 46, however; less than the maximum sum found therein. That maximum is 50, which is achieved two dozen times (the least sum found is 38; occurring only 3 times).
Update (early Feb 2005): while I was in Europe, Aad van de Wetering and I traded several more emails about the topics on this page. At first he believed B33333333333 == 44 was probably the greatest sum which included a 'B' (a pieces with 11 neighbors). I replied, that, no, at least a sum of 46 could be found... I had sketched an example on graph paper, but did not have the time to update these pages right then. Here answered the next day with affirmation: he had found this example, B44333333333:
Then I mentioned that I had found various configurations whose neighborly sums had reached 52 (though, obviously, no single piece in the formations touched all the other pieces). I suspected 54 might be possible, but had no answer (I mentioned that it would likely involve a configuration that was an 8x8 square less four unit squares).
Soon, he responded with an answer which was, "far from easy to find."