Peter Joy Updated: Save Send news by mail electrónicoTu name *
Your email *
email *
we’ll Start our tour with a riddle: I will give a series of clues to describe a historical character and encourage you to discover your identity along them.
The clues are the following: popularization of mathematics, a prolific writer, responsible for a fixed section on mathematical recreation to a scientific journal for more than 25 years, philosopher of science, founder of a movement skeptic, amateur magician, promoter of vocations mathematics.
not even a high-profile internet search engine needs more data to locate Martin Gardner, one of the heroes of youth from a multitude of teenagers who love math.
Among the enormous amount of material that accumulates in his works we can find inspiration for many articles in this section, so we will limit ourselves to an example very suitable for the current situation. In the November issue of the 1967 of the section “Mathematical Games” for the magazine “Scientific American” rescues an old issue of 1958, known as the puzzle Langford.
The problem cited by Gardner was raised by british mathematician Dudley Langford in 1958, and it occurred to him seeing play to your child with some colored blocks.
According to his own words,
there Were two blocks of each color and one day I realized that my son had stacked so that there’s only one block between the pair red, two between the pair green and three between the pair blue. Then I found that a redistribution full allowed to add a couple of blocks yellow with four blocks between them.
First solution of the puzzle of Langford using colors
Working with numbers instead of colors, Langford raised the general problem:
Given a set with 2n numbers, from 1 to n, repeated twice, is to place them in a row so that between each two numbers with the same value k, there is exactly k numbers.
The own Langford discovered other solutions with a higher number of partners, published in the journal “Mathematical Gazette” in 1958. Interestingly, the sequences 312132 and 41312432 are the only solutions (except for its symmetric) to three and four pairs, respectively. There are methods of attraction to build systems that are suitable for all cases but do not allow to determine the number of solutions.
In the course of time and thanks to the calculation power of modern computers, there is a known number of solutions for all cases up to n = 30 (30 couples of numbers). In reality, this case is simple because it does not have a solution, nor has the case n = 29. However, for n = 28, is known as from 2015 there are a total of 1.607.383.260.609.382.393.152 solutions (more than a thousand trillion, although none of them has a pint of obtained “pulse”). Still has not been found a general formula that gives the number of solutions to any amount of numbers but you have succeeded in striking formulas for some values. An animated version that shows the difficulty of the problem, even for simple cases, is made by John Miller and you can appreciate on this link.
Logo of Google that “almost” fits the sequence of LangfordLogo of MacTutor with the colors following the sequence of Langford – Source: John Miller.
you Can find a beautiful demonstration elementary (not simple) that there are only solutions when the number of pairs is a multiple of four or one unit less than a multiple of four on the blog DataGenetics. A similar problem that has a solution for all values of n consists in adding the number zero in the penultimate position of the sequence, known as succession of Langford “hooked”.
In 1966, Frank Gillespie and W. Utz defined the succession of Langford generalized, using most of two equal numbers in each sequence, and wondering what would be the cases that had solution. The presentation in society of this issue appeared in 1988 and was described, how not to, by Martin Gardner in chapter 6 of the book “Time travel and other mathematical bewilderments” (translated the same year under the title “Travels in time and other perplexities math”). This is as posed by Martin Gardner, classifying it as extremely difficult:
Pulls out the 27 cards with values from ace to nine of three of the suits of a deck of cards. The problem is to place all the cards in a row on the table with the following condition:
For any value of k between 1 and 9, between the first and second cards of the same value k there must be exactly k letters and between the second and third cards of the same value k, there must also be exactly k letters.
Thus, between the first ace and second ace must be only one letter, between the second as and the third as there should be only a letter; between the first two and the second two should be two letters, between the second two and the third two should be two letters; etc
As you can see, is the problem of Langford using sets of three letters (or numbers) instead of couples. The example proposed by Gardner is the smallest that it has a solution, in fact you have only these three solutions:
1-9-1-2-1-8-2-4-6-2-7-9-4-5-8-6-3-4-7-5-3-9-6-8-3-5-7
1-8-1-9-1-5-2-6-7-2-8-5-2-9-6-4-7-5-3-8-4-6-3-9-7-4-3
1-9-1-6-1-8-2-5-7-2-6-9-2-5-8-4-7-6-3-5-4-9-3-8-7-4-3
Interestingly, it seems that there is no solution to eight cards, as they D. P. Roselle and T. C. Thomasson Jr. in 1971 by computer programs, so that does not count as a demonstration. To ten cards, there are five solutions, and there are no more solutions with other values whenever they are used, three suits of a deck of cards. If someone raises the analogous problem using the four suits of the deck, and respecting the same rules of separation between pairs of equal values, don’t strain yourself: there is no solution.
There are many other ways to maintain the safety distance between numbers, with variations of the sequences described here, as are the inheritance of Skolem , studied independently by the Swedish mathematician Thoralf Skolem , the inheritance of Nickerson , and the generalizations with triples, cuaternas, etc, of equal numbers, as already described. Will we be able to apply these mathematical skills in the new redistribution of the society and its customs? At least, some of the artists to feel inspired by these mathematical structures, such as the German artist Gerhard Hotter , which has resulted in some of their creations their particular vision of the death of Langford (you can make a virtual tour in this gallery).
Peter Joy. University of the Basque Country/Euskal Herriko Unibertsitatea. Commission of disclosure of the Royal Mathematical Society of Spain (RSME).
The ABCdario Math is a section that emerges from the collaboration with the Commission of disclosure of the RSME.
