Klein 1
Ethan Klein
Math 7-Prof. Winkler
4/16/2013
Langford Sequences: The Most Trouble You Can Have Stacking Blocks
One day, a Scottish mathematician by the name of C. Dudley Langford observed his son
playing with blocks, two in each of three colors, and saw him doing something peculiar.
(Langford 1958). Langford’s son stacked the blocks in a column with one block between the red
pair, two blocks between the blue pair, and three blocks between the yellow pair, with all blocks
in the same column. Langford then found that he could rearrange the blocks in such a way that
he could add a green pair with 4 blocks in the middle. In his restatement of the puzzle, Martin
Gardner asks us to try and find this unique answer (ignoring reversal), and to perhaps find a
simple way to determine if such an arrangement exists for a given number of pairs n (Gardner).
To make the problem’s solutions easier to show and discuss, both Gardner and Langford
represent the arrangement as a series of digits (called a Langford Sequence) with one digit
between the two 1s, two digits between the two 2s, and so on. For example, the solution for n=3
is 312132.
One of the best things about Langford’s problem is that solving it feels more like playing.
Shifting blocks or cards around, as opposed to simply guessing different numbers, engages one
in the puzzle and makes the underlying patterns apparent. Playing around for a little while, one
easily finds that the Langford Sequence for n=4 is 41312432. With that first part of the puzzle
accomplished, the next step is find how we can tell whether or not a Langford sequence exists for
a number n. Unlike simply finding a Langford sequence for n provided one exists, showing
whether or not one does exist (e.g., finding an “existence condition”) is not easy. Langford
1
Klein 2
himself had to appeal to others for a theoretical treatment of the problem but himself managed to
find sequences for n up to15 without the aid of a computer.
A good way to begin looking for an existence condition for Langford sequences, as with
many puzzles, is to try and find solutions for other values of n. Clearly n=1 and n=2 have no
solution, and, as shown above, there are solutions for n=3 and n=4. Intriguingly, we can find
solutions for n=7 and n=8 without too much difficulty, but finding a Langford sequence for n=5
or 6 seems impossible. Langford, in his original paper, lists solutions for n=3,4 7,8,11,12 and 15,
but not for any of the numbers in between. Since Langford, in what seems to be a fairly extensive
search, was unable to find any such sequences for those missing values of n, it seems (though it
is not at all certain) that a Langford sequence does not exist for those numbers. It seems likely,
then, that at least one Langford Sequence exists when n is equal to 4a or 4a-1 for some integer a.
But how can we verify this? The solution I give, adapted from Roy O. Davies (Davies 1959), is
remarkably elegant, yet far from obvious, as are all the best proofs. The hidden elegance of the
solution is a testament to the quality of the entire problem.
The observation which sets us on the right track is that the only thing that determines
whether or not an arrangement of our n pairs of digits is a Langford series is where we place the
pairs. We can thus think of the problem as selecting which positions to pair together, rather than
where we place pairs of digits. If a digit r in our series is first placed at a position ar, its
counterpart is then located at position ar+r+1. With a total of 2n positions in the sequence, the
sum of these indices is the sum of 2ar+r+1 for each ar ranging from 1 to n. Obviously, the sum of
the indices is both equal and integral no matter what order we add them in, so the sum of the
numbers from 1 to 2n, equal to n(2n+1), must equal the result from the method above. If this is
not the case for a particular n, then this reveals a paradox showing that no Langford sequence
2
Klein 3
exists for that n. In fact, the sum of the indices we find from comparing the two sums above
is
(3𝑛2 −𝑛)
4
, which is only an integer when n is of the form 4m or 4m-1 for some m. Thus, we have
shown by contradiction that a Langford Sequence cannot exist when n is equal to 4a+1 or 4a+2
for some integer a, leaving open only n equal to 4a or 4a-1. Now all that remains is to show that
a Langford sequence exists for all of these allowed n, for which Davies provides a simple
construction. Looking at the problem in a new light by examining the indices rather than what
they mark reveals a hidden structure of Langford sequences from which the solution springs.
Once the solution is found it seems almost obvious. The simplicity and clarity of the idea, once
presented, are yet another asset of the problem.
With that settled, we can now move on to the most wonderful thing about Langford’s
problem. It invites any who would investigate it to come up with their own spin on the problem
and see what happens. What sequences form when we include one empty space? What happens
when we consider sequences of triplets, quartets, or more? When the intervals between pairs
change? The possibilities are truly endless. For instance, through a very similar argument to that
given above establishing an existence condition for Langford sequences one can show that when
a Langford sequence does not exist for some n¸ there is a sequence of pairs arranged like a
Langford sequence but with a space separating the last number and the rest of the sequence, even
if the lowest number in the sequence is higher than one (Priday 1959). For example,1*1, 121*2,
and 8642752468357*3 are hooked sequences. It seems utterly trivial at first, but two hooked
sequences which share no common elements snap together to form a new Langford sequence.
That’s just one of many, many fruitful paths to investigate. I urge the reader to play around for an
extended period of time with some variation of Langford’s problem. It may be frustrating, but
it’s worth the mental strain.
3
Klein 4
Langford’s problem is a stellar example of what a good mathematical puzzle is. It’s easy
enough to understand the problem that a small child can attempt to solve it. In fact, it’s so simple
that a small child inspired it. However, trying to solve the general case is at first glance
something of a daunting proposition. The solution, though, is simple and in a way rather
beautiful. Best of all, the problem is so readily extended that it can always provide a new
challenge. That’s why this standout puzzle is one of Gardner’s best.
4
Klein 5
Works Cited
Langford, C. Dudley. "Problem." Mathematical Gazette 42.341 (1958): 228. JSTOR. Web. 16
Apr. 2013. <http://www.jstor.org/stable/3610395>.
Priday, C.J., and Roy O. Davies. "On Langford's Problem." Mathematical Gazette 43.346 (1959): 25055. JSTOR. Web. 4/16 Apr. 2013. <http://www.jstor.org/stable/3610650>.
5