Langford's Problem

Serious editing in progress - April 2014. Will be adding anecdotes, puzzles, etc learned at G4G11 ASAP. Plus I have been sitting on other material.
  • Compiling biography of Langford's life. Link here soon! (Spring 2014).
  • Stanford archive has actual artifacts that people sent to MG.
  • Langford appeared on BBC radio mystery - the combination to a Safe had been forgotten, but the person knew it was a Langford sequence... via Chris Maslanka
  • Langford somehow made money with a puzzle based on LP.
  • George Miller's PuzzlePalace.com has several puzzles based on LP.
  • Puzzles by Daniel Hardisky.
  • Donald Knuth says Jean Brette made a puzzle based on Skolem's variant using width instead of planarity, and gave a copy to David Singmaster in 1992. I confirmed this with Singmaster at G4G11. (Sounds similar to the Hardisky puzzle.)
  • Art and Music by Gerhard Hotter.
  • Browser based game Duel had character moods that were a function of unique distances. Not sure if it was an exact analog. Need a reference to this - no longer extant.
  • Virtual Manipulative by Gus Miller coming summer 2014.

Please cite this page if re-publishing any of this information. This helps the network effect.
Additions & Corrections all Welcome!


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 between the blue pair, and three blocks between the green pair:

Langford added a yellow pair and came up with:

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 P1 holds the position of the leading (leftmost) 1, and P2 holds the position of the leading 2. The positions of the leading numbers of each pair are kept in P1 through Pn.

We can form the summation of numbers (Pk) and (Pk+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 (2Pk+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 Pk'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 4m-1 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 Telegraph

Martin 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=7 by 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 (46 years ago!).

Since 1968, the number of solutions for 15, 16, 19, 20, 23, and 24 have been calculated by multiple people using various techniques.

  • 15 and 16 were counted in the 1980's.
  • 19 was counted in 1999!
  • 20 was determined in 2002
  • 23 was determined in 2004 (one team)
  • 24 was determined in 2005 (one team)
  • 27 has not been determined.

L Notation

We use the notation L(2,n) to indicate the number of unique solutions for Langford's Problem with n pairs of blocks.

V(2,n) refers to Nickerson's Variant of Langford's Problem.
L(s,n) and V(s,n) denote higher-order sequences. Both concepts are defined below.

Mike Godfrey determines L(2,20) with new Algebraic Method

In 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) Computed

In April, 2004, a grid computing experiment at Université de Reims Champagne-Ardenne 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) Computed

Michaël 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:

  • Parallel Tree Search for Combinatorial Problems: A Comparative Study Between OpenMP and MPI, Studia, regular-issues, Vol. 4, No.2, (PDF)
  • Solving the Langford problem in parallel - a shorter paper specifically about solving LP in parallel, (PDF)
  • Résolution parallèle de problèmes combinatories en mémoire partagée - Christophe Jaillet's doctoral thesis from December 2005 (PDF)

L(2,27) is Unknown

I have no word on any attempt to compute L(2,27). Perhaps with a Quantum computer? :^)

Nickerson's Variant of Langford's Problem

A variant of Langford's Problem has the second number of each k-pair occur in the kth place after the first, so that there are k-1 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?]

Higher-Order Langford Sequences

Arrangements 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 9k-1, 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??]

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 Solutions

A subset of the solutions for higher orders of n are planar... 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.

  • The single solution to n=3 is planar.
  • The one solution to n=4 [41312432] is non-planar!
  • Four of the 150 solutions for n=8 are planar.
  • Only sixteen of 17,792 solutions for n=11 are planar.
This is sequence number A125762 on OEIS.org, included in the Table of Solutions below. Knuth calculated up to n=28. The values have not been independently confirmed by a second program, in case you are looking for something to do! Send your results to my contact address below.

Exercise. Adapt the classic search algorithm or invent your own Algorithm X to generate only planar solutions for a given n.

Full Notation

We 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:
  • s indicates the entanglement number (whether pairs, triplets, quadruplets, quintuplets, sextuplets, ...)
  • n indicates the Order of the problem, i.e., the number of pairs, triplets, ...
Example: L(4,24) = 3, denotes that there are three solutions with quadruplets for n=24.

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 Solutions

A 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. Will convert to Links. jm]

11 and 111 are the trivial solutions for V(2,1) and V(3,1).

n Pairs
L(2,n)
A014552
Pairs
V(2,n)
Triplets
L(3,n)
Triplets
V(3,n)
Quads
L(4,n)
Planar
P(2,n)
A125762
1-1-1--
2------
31----1
413---0
5-5----
6------
726----0
8150252*--4
9-1,32839--
10--520--
1117,792--33-16
12108,144227,968---40
13-1,520,280----
14------
1539,809,640----194
16 326,721,800700,078,384---274
17 -6,124,491,24813,440---
18 --54,947200,343--
19 256,814,891,280-249,280869,006-2,384
20 2,636,337,861,200 5,717,789,399,488 - 4,247,790 - 4,719
21 - 61,782,464,083,584 - - - -
22 ------
23 3,799,455,942,515,488 - - - - 31,856
24 46,845,158,056,515,936 ?? - - 3 62,124
25-??--????-
26--???-????-
27??-??????????426,502
28??????????????817,717
29-??-???????-

*There is no proof of impossibility, only exhaustive search (I think).

Contributors

PROBLEMDATEPERSONCOMPUTERTIMELANGWhere
L(2,3-4)?C. Dudley LangfordHand???
L(2,7)-11959Roy O. DaviesHand, missed 1???
L(2,7-8)May-67Dave MooreTRW-1305m,40mFORTRANRolls Royce
L(2,7-8)Nov-67Glen F. Stahly????
L(2,7-8)Nov-67John MillerIBM 1130?FORTRANGonzaga
L(2,7-8)Nov-67Malcolm Holtje
L(2,7-8)Nov-67Robert Smith
L(2,7)Nov-67Thomas Starbird
L(2,7-12)Nov-67E. J. GrothSDS 930<dFORTRANMotorola
L(2,11-12)1968?John MillerIBM 1130?AsmGonzaga
L(2,15)Sep-80John MillerVAX 11/780?PascalL&C
L(2,15)Feb-87Frederick GrothCommodore 6415.5 dAsmHome
L(2,16)Feb-87Frederick GrothCommodore 64122.4 dAsmHome
L(2,15)Jul-89Andrew BurkeCogent XTM?COGI
L(2,16)Jul-89Andrew BurkeCogent XTM120hCOGI
L(2,16)May-94John MillerDec Alpha??L&C
L(2,19)May-99 Rick Groth Team Mac/Pentium2 moCDistributed
L(2,19)Jul-99John MillerDEC Alpha2.5 years!CL&C,...
L(2,19)Mar-02Ron van Bruchem Pentium~6HGodfrey's?
L(2,20)Feb-02 Godfrey/van Bruchem AMD/Pentium1 WeekFORTRANUMIST/home
L(2,23)Apr-04 Krajëcki Team Sun/Intel4 daysJava/CONFIITREIMS
L(2,24)Apr-05 Krajëcki Team 12-15 processors3 monthsJava/CONFIIT?REIMS?
V 2
V(2,4-5)Feb-69John MillerIBM 1130??L&C
V(2,8-9)Feb-69John MillerIBM 1130??Gonzaga
V(2,12-13)Mar-89John MillerVAX??L&C
V(2,16)Feb-99John BoyerIntel??U.Victoria
V(2,17)Feb-99John BoyerIntel??U.Victoria
V(2,20)Mar-02 Mike Godfrey Pentium III65.5 HFORTRANUMIST/home
V(2,21)Mar-02 Godfrey/van Bruchem AMD/Pentium< 1 weekFORTRANUMIST/home
L 3
L(3,9-10)Mar-89John MillerVAX??L&C
L(3,17)Aug-89John MillerVAX??L&C
L(3,17)Apr-97Frank Ruskey???U.Victoria
L(3,18)Apr-97Frank Ruskey???U.Victoria
L(3,19)May-97Frank Ruskey???U.Victoria
V 3
V(3,9-11)Mar-89John MillerVAX??L&C
V(3,18)1999Boyer/WatsonIntel??U.Victoria
V(3,19)Feb-99John BoyerIntel??U.Victoria
V(3,20)Mar-99John BoyerIntel??U.Victoria
L 4
L(4,24)~1979Saito & Hayaskacustom18.5m?Miyagi Tech
L(4,24)2004 Richard Noble Intel2 yQuickBASICRetired
PLANAR SOLUTIONS
P(2,n)2007? Donald Knuth ???Stanford

[Women are conspicuously absent from Contributors. Am I missing any?!]

Abbreviations used in the above table
SDS = Scientific Data Systems.
L&C = Lewis & Clark College, Portland, Oregon.
U.Victoria = University of Victoria, British Columbia, Canada.
OGI = Oregon Graduate Institute, Portland, Oregon.
UMIST = University of Manchester Institute of Science and Technology in the UK.
REIMS = LERI Resycom - Université de Reims Champagne-Ardenne Département de Mathématiques et Informatique.

Constructing a Single Solution

Let'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 dave-moore-html!]

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: One learns a lot by going after this problem... Not necessarily solving much about the problem, but a lot about whatever methods you tried to use.

Odd/Even Parity as key to Solvability

An 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 oddly-numbered, the other half evenly-numbered.

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 parity if there is any hope of covering all the positions in a full arrangement. So, there must be an even number of odd pairs. This happens only when n is 3, 4, 7, 8, 11, 12, 15, 16, ..., 4m-1, 4m.

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 Notes

On this page, details are given about the systematic enumeration of Langford sequences.

Graphical Representations

On 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.

A Very Odd Observation

This 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.

Bibliography

I 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 4m-1.

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 0-88385-449-X, first and second editions.

Charles Dudley Langford Biography

Work in Progress, including obituary from Math Gazette. Will be posted Spring or Summer 2014. Not complete enough yet to edit into final form. May have to travel to Scotland! :^)

Hardware for Langford's Problem

For my senior thesis at WSU, 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.]


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[Need to use a Creative Commons license for this.]
Till then, ©2014, John E. Miller, Portland, Oregon.
Contact: miller@lclark.edu
Created: 1995?
Updated: 08-Apr-2006
Updated: as results came in!
Updated: 14-Feb-2014