Langford's ProblemPlease cite this page if republishing any of my text or diagrams. See the Creative Commons License below.Please send additions, corrections, or notifications to john @ timehaven . us. Thanks. Results from 2018!Rory Molinari reports on Planar Langford pairings: P(2, 31), and musings on planarity. I will integrate his results into this page before G4G13! See [LINK] for his full report.Results from 2017Boris Dimitrov reports that he has confirmed Knuth's results for planar solutions up to 28 pairs. See Planar Solutions below.Results from 2015July 2015 — Team AssarpourLiu computedL(2,27) = 111,683,611,098,764,903,232 L(2,28) = 1,607,383,260,609,382,393,152Interestingly, before that, in April 2015, Team Krajecki computed L(2,27) and sent results via email. When Team AssarpourLiu executed their method for 27 and 28, the results for 27 was different than Team Krajecki's number, so, it was up to one or both of the parties to resolve the discrepancy. The Team AssarpourLiu paper might be downloaded from [arXiv.org]. We have a May 2017 PDF of The Team AssarpourLiu paper [HERE].
Results from 2016November. Team Krajecki writes to say they had arrived at the same number as Team AssarpourLiu, but their program didn't display it correctly. You can find the details about their algorithm and the computation time in the paper on [SuperFri].The above numbers have been added to the Table of Solutions below. As of December 9, 2016, the page is up to date. What is Langford's Problem!?Langford's problem is named for Scottish mathematician C. Dudley Langford who once observed his son playing with colored blocks. He noticed that the child had piled three pairs of colored blocks such that there was one block between the red pair, two blocks between the blue pair, and three blocks between the green pair (stacked up on the right):Langford added a yellow pair and came up with (below):
These solutions for 3 and 4 pairs are unique. Reversing the order is not significant, because all you have to do is stand on your head, or, walk around to the other side of the arrangement and view it from that side. Generalizing from colors to numbers, the above became 312132 and 41312432. Langford's 1958 submission to Mathematical Gazette gave one arrangement each for 7, 8, 11, 12, and 15 pairs of numbers. He noted that arrangements didn't seem possible for 5, 6, 9, or 10 pairs, so he called for a theoretical treatment — What values of 'n' were solvable?  
Roy O. Davies finds key to the Solvability of 'n'The proof of why 5, 6, 9, 10, etc are impossible is fairly simple, and can be found in the Roy O. Davies paper, ON LANGFORD's PROBLEM II in Mathematics Gazette, 1959, v43.In general, Davies proved that 'n' must be a multiple of four, or one less. In the remarks at the end of his paper, Davies wondered how many solutions there were for different n's, and stated (wrongly!) that for n=7, there were 25 solutions. Details. Consider in general the pair of k's in the arrangement, where k is any number from 1 to n. If the first k is in position p, then the other k must be in position p + k + 1. For example, the one's might be in position 8 and 10 in an arrangement, with position 9 separating them. Let's say that all the various values for p (the position of the first number for a given pair) are stored in an array 'P', so that P_{1} holds the position of the leading (leftmost) 1, and P_{2} holds the position of the leading 2. The positions of the leading numbers of each pair are kept in P_{1} through P_{n}. We can form the summation of numbers (P_{k}) and (P_{k}+k+1) for k=1,2,...n. But the positions of all the individual numbers must be the same numbers (1, 2, ... 2n) in some order. Therefore summation for k=1,n of (2P_{k}+k+1) can be equated with the summation of 1, 2, ... 2n, which is n(2n+1). Rearrange terms to solve for the summation of just the P_{k}'s, which has to be an integer. That's the key! That is, whatever the summation might be for a given solution, it is a whole number. The final step is left for the reader, which shows that n must be of the form 4m or 4m1 in order for things to be integral. See 'Odd/Even Parity as key to Solvability' below for a simpler explanation.  
Mathematical Games, Design Electronics, and The Daily TelegraphMartin Gardner posed n=4 in a collection of short problems in Mathematical Games, November 1967, Scientific American. He told readers that month that no solutions were possible with 5 or 6 pairs, but that there were 25 unique arrangements for n=7, without citing a reference. (There later turned out to be 26 solutions for n=7.)In December 1967, Gardner explained that Langford's Problem has been shown to be solvable when n is any multiple of four or one less. Martin said no one knew how to determine the number of distinct solutions for a given number of pairs except by exhaustive trial and error. The gauntlet was down, and the stage was set for a popular computing problem. It should be noted that at around the same time, the problem also appeared in Design Electronics and the Daily Telegraph, both British publications. So people in England and elsewhere found themselves working on Langford's Problem.  
Programmers, Start your Computers!Early in 1968, as a college freshman, I (John Miller) programmed Langford's Problem and found 26 (not 25!) solutions for n=7 and 150 solutions for n=8. Four others did likewise with computers. Two others solved n=7by hand. In addition, E.J.Groth cracked n=11 (17,792 solutions) and n=12 (108,144). Martin Gardner published these results in his March 1968 column. Since 1968, the number of solutions for 15, 16, 19, and 20 have been calculated by multiple people using various techniques. 23, 24, and 27 were computed by one team. 27 and 28 were computed by yet another team.
 
L NotationWe use the notation  
Mike Godfrey determines L(2,20) with new Algebraic MethodIn February 2002, Mike Godfrey, at the University of Manchester Institute of Science and Technology, UK, came up with an algebraic method of determining L(2,n) that was way faster than the classic search algorithm. Using this new method, Godfrey and his colleague Ron van Bruchem cracked L(2,20), V(2,20), and V(2,21). The fellows also verified all previous results for L(2,n) and V(2,n). (See the tables below.)Th Godfrey Method does not enumerate solutions, but rather extracts the number of solutions by repeated evaluations of a generating function.. then dividing the result by a very large number! Here is Godfrey's letter of explanation. No plans to publish this method have been set. A Zip archive of Godfrey's FORTRAN source code is available for download HERE.  
L(2,23) ComputedIn April, 2004, a grid computing experiment at Université de Reims ChampagneArdenne used Godfrey's Method on a mix of 30 Intel and Sun machines, one with 24 processors. Five people worked on this project. See Michaël Krajecki's letter of explanation. There are nearly 4 quadrillion solutions to n=23.  
L(2,24) ComputedMichaël Krajecki, Christophe Jaillet, Alain Bui obtained L(2,24) in April 2005 after 3 months' computation with 12 to 15 processors. There are nearly 47 quadrillion solutions!References:
 
L(2,27) Computed!Christophe Jaillet wrote: We had a large experiment using besteffort distributed computing, and got L(2,27) with an average amount of 181 CPUGPU machines during 2 days (1 day and 20 hours).The team solved L(2,27) in late April 2015... reporting 111,683,606,778,027,803,456 solutions! However, this turned out to be erroneous. They had calculated the correct number, but fouled up and displayed this incorrect result. See News at top 
 
L(2,28) Solved in 2015!See News at the top of this page. (We will incorporate here in 2017.)  
Variations
Variations are explored below.  
Nickerson's Variant of Langford's ProblemA variant of Langford's Problem has the second number of each kpair occur in the kth place after the first, so that there are k1 other numbers between the two k's. The singular variant solution for n=4 is 42324311. R. Nickerson proved this is possible whenever n is of the form 4m or 4m+1. See Bibliography.Note the 1's are always together in the variant. [So the variant is like adding a pair of 0's to the classic problem  no blocks between the two zeroes. Skolem variant?]  
HigherOrder Langford SequencesArrangements can be made using triplets as well, where the outer elements of each triplet are separated from the middle element in the triplet, as in this arrangement (the 9 triplet is shown in red):3 4 7 9 3 6 4 8 3 5 7 4 6 9 2 5 8 2 7 6 2 5 1 9 1 8 1 ^ . . . . . . . . . ^ . . . . . . . . . ^Nine triplets can be arranged in 3 ways. See the table below for other results. Such triplet sequences can be done for any n>8 that is of the form 9k1, 9k, 9k+1. Mathematically, we write this as n ≡ (1,0,1) mod 9, pronounced "n is congruent to 1, 0, or 1 modulo 9". [Could say n ≡ (0,1,8) mod 9.] Details. Frank Ruskey found at least one solution with triplets for n = 26,27,28, 35,36,37, 44,45,46, 53,54,55, 62,63,64 (solutions for 26,27,28 were also found by Brad Jackson at San Jose State). The existence of these sequences is proven in the Levine paper. The triplet arrangement can also be done with the Nickerson variation. I suspect that the criteria for solvability of the variant with triplets is n ≡ (0,1,2) mod 9. [Isn't this in the paper??] By a theorem of Levine, there are no perfect 4sequences unless k == 1, 0 (mod 8). (Theorem 2 in The existence of Perfect 3sequences.) Extensive search of higher orders has discovered three solutions with quadruplets for n=24. Saito & Hayaska say no known quintuplets for n≤24, and no sextuplets for n≤21.  
Planar SolutionsA subset of the solutions for higher orders of n areplanar... lines connecting all pairs can be drawn without crossing, as in the following example provided by D. E. Knuth:
The connection must be simple and direct, not doing an end run around from top to bottom.
Knuth's values have all been independently confirmed by a second program by Boris Dimitrov, who is currently (2017) working on extending these results. We have a page for Boris's work: [HERE]. Rory Molinari also confirmed Knuth's results, and extended the known solutions to P(2,31).  
End Run Planar SolutionsRory Molinari has explored a variation of planarity, where connecting lines are allowed go around the end of the arrangement in order to avoid crossing. See the END RUN section on his page [LINK].
Exercise. Adapt the classic search algorithm or invent your own Algorithm X to generate only planar solutions for a given n.  
Full NotationWe use the notation L(s,n) or V(s,n) to indicate the number of solutions for (L)angford's Problem or the Nickerson (V)ariant where:
We also sometimes use the same notation to refer to the set of all solutions, whether known or unknown. So in a scalar context, L(s,n) is an integer. In another context it could refer to the particular problem in general (for now, anyway). Let's also use P(2,n) to refer to the number of planar solutions for n pairs. Eg: P(2,8) = 4.  
Table of SolutionsA dash () in the table means that no solutions are possible for the value of n. Question marks denote unsolved values. [OEIS.org numbers are given in the table heading.]
*There is no proof of impossibility for L(3,8), only exhaustive search (I think).  
Contributors
[Women are conspicuously absent from Contributors. Am I missing any?!]
Abbreviations used in the above table  
Constructing a Single SolutionLet's say that you wanted to see one solution for n=100, just for your own satisfaction.In one paper, Roy O. Davies gives a pattern for constructing a single solution for any valid n. Using this pattern, a computer program can instantly generate a solution for n=40,000 or other large number. See the Graphics section below for some examples. A Perl program is included here for your examination. Dave Moore submits an approach based on what he has observed, for your edification. You can see a graphical representation of his method on the Graphical Representations page below. [should just add the graphics to davemoorehtml!]
Stephen Scattergood has contributed to Dave Moore's method (in an effort to complete it), and analyzed planar solutions. [Should try to better summarize Steve's work here!]
Scattergood says: Attempting to find a FormulaWilliam Bouris, working diligently for years, claims to have an approach to formulating the number of solutions for a given N. He states that Langford Pairings follow from a formula which takes into account the amount of interference of digits in a multiplicative manner. For example here is a formula that gives the number for L(2,7):
See the PDF for the formulas for L(2,3), L2,4), L(2,7), and L(2,8): [PDF]
John Says: This is most interesting, as intuitively it feels like a formula might be able to summarize the interactions or interferences. I've asked Bill to explain where the numbers in the formulas come from. I pondered this myself once, and seemed like it might take as much or more work to find a given However, having higher and higher formulas however, might lead to deeper insights to the problem. (Such as what's going on with the ABC conjecture). But I am just speculating here. I don't know what's going on. Perhaps some kind of visual is needed? JM
 
Odd/Even Parity as key to SolvabilityAn arrangement of pairs of numbers 1,1, 2,2, 3,3, ... n,n will have 2n positions numbered 1..2n. Half the positions will be oddlynumbered, the other half evenlynumbered.An even pair will take an odd position and an even position, no matter where the pair is in the arrangement. For example, if Left 2 is in position 1, the Right 2 must be in position 4. That's encouragingly balanced parity!
However, an odd pair will take either two odd positions or two even positions. Therefore the odd pairs must occur in pairs themselves to preserve Exercise. Does this mean that half of the odd pairs must be in even positions, and half are in odd positions? What does this mean for the even pairs?  
Algorithmic Solution NotesOn this page, details are given about the systematic enumeration of Langford sequences.  
Graphical RepresentationsOn this page, you'll find graphics representing various aspects of Langford sequences. For example, the following graphic shows the state of 19 as it was on Jan 30, 1997. Blue represents even numbers, yellow, odd.
 
The Bridges of Königsburg?Here's a fun graph of the path through 8 pairs of blocks [5286235743681417]. Don't traverse the black lines — follow the red arrows. Notice the cycle of 4 blocks between the two 4's, and so on. It's like taking a tour of a city, without walking over any of the bridges!
The above graph was part of my talk at G4G11. (See Bibliography)  
ForceDirected GraphsA Langford graph has 2n nodes and 3n1 edges. The graph will have 2 nodes of degree 2, and 2n2 nodes of degree 3. This is because the two end blocks have either a left or right connection (not both), plus each end block has has one edge for their twin, making two edges. Interior blocks all have a left, right, and pairing edge, for three edges.Click below to play with a forcedirected graph of a solution of L(2,11). Can you tell which dots represent the two end blocks?  
A Very Odd ObservationThis concerns the number of algorithmic operations required to get to the first solution of L(n,2). n=39 and n=47 take billions of operations, while other n's take only hundreds or less. To make things more bizarre, adjacent n's sometimes take the exact number of operations to reach their first solution even though it couldn't possibly be the same search tree.  
Art, Music, and Puzzles based on Langford's Problem
 
Hardware for Langford's ProblemFor my senior thesis at WSU (1972), I designed (but did not build) a sequential logic circuit for the search algorithm.Saito & Hayasaka at Miyagi Tech College in Japan designed and built special purpose computers to check for the existence of solutions for various higher order tuples. See reference in bibliography. In the 1960s, E. J. Groth used Langford pairings to construct circuits for integer multiplication. Need reference! Knuth mentions in his Volume 4a.  
C. D. Langford BiographyCharles Dudley Langford was born in Highgate (near London) in October 16, 1905. He contracted polio at age 12.Langford started out as an industrial chemist and was a member of Royal Chemical Society. He became a passionate Maths teacher who thought about all kinds of mathy problems, publishing 30 notes in Mathematical Gazette. He coauthored a paper on Hinged Dissections with Martyn Cundy, and corresponded with Harry Lindgren. At some point, he moved to Ayrshire, Scotland. Little did we know that Langford was dying during the time we were cranking away on his problem using computers in 1967 and 1968. He died January 11, 1969 at age 63, in Girvan. His son, Charles Andrew Langford was born in 1938 in Girvan. Langford is buried in Girvan, Ayrshire district of Scotland, in the East Doune (Girvan) Cemetery. At G4G11, Chris Maslanka told me that Langford appeared on BBC radio mystery — the combination to a safe had been forgotten, but one person knew it was a Langford sequence beginning with three particular numbers (or something like that). He also said that Langford somehow made money with a puzzle based on LP.  
BibliographyI have put the bibliography onto a separate page for those who are interested in fetching it.The proofs of possibility take advantage of the summations formed by the positions of pairs and their requisite separation in an arrangement. For it to all hold true (integral) n must be of the form 4m or 4m1. The Roy Davies paper (and others) gives patterns for constructing a single langford sequence (arrangement) for any given n, as a demonstration (construction) proof for a theorem that the solutions are not only possible but at least one exists for each feasible value of n. Martin Gardner revisits Langford's Problem in Mathematical Magic Show, published by Alfred A. Knopf, ISBN 088385449X, first and second editions. Volume 1 of The Art of Computer Programming [TAOCP] came out in 1968. Volume 4a (Chapter 7) of TAOCP, Combinatorial Searching, uses Langford's Problem to illustrate! Here's a PDF of paper that I presented at G4G11, Langford's Problem, Remixed: [PDF]. Here is a video of my 12 minute G4G11 talk on Langford's Problem: Please Honor this Creative Commons LicenseExcept where otherwise noted, content on this site is licensed under a Creative Commons Attribution 4.0 International license [CC BY 4.0]. All I ask is notification and attribution.Send additions & corrections to john @ timehaven . us. Thanks. John E. Miller, Portland, Oregon.
