Some Variations on a Theme of Irina Mel ’ nichuk Concerning the Avoidability of Patterns in Strings of Symbols

The set of all doubled patterns on n or fewer letters can be avoided on an alphabet with k letters, where k is the least even integer strictly greater than n+ 1, with the exception of n = 4. The set of all doubled patterns on 4 or fewer letters can be avoided on the 8-letter alphabet. The set of all avoidable patterns on n or fewer letters can be avoided on an alphabet with 2(n + 2) letters. Mathematics Subject Classifications: 68R15


Introduction
By a word we understand here a finite sequence of elements, usually referred to as letters, drawn from some set A, usually referred to as an alphabet.We are interested in the combinatorial properties of words.Accounts, fairly recent, of the state of the combinatorial theory of words can be found in the monographs of Allouche and Shallit (2003) [1], of Lothaire (2005) [16], of Lothaire (2002) [15], and of Lothaire (1997) [14].
In this paper we will only deal with words of positive length.For a given alphabet A, we use A + to denote the set of all words on A that have positive length.Words can be concatenated to form other words.The operation of concatenation is associative, and when A + is endowed with this operation it becomes the semigroup freely generated by A. This entails that every map that assigns to each letter in A a word from B + can be extended to a unique morphism from A + to B + .Let u and w be words.We say that u is a subword of w provided either u is an initial segment of w or u is a final segment of w or there are words x and y so that w = xuy.
Let w be a word on the alphabet B and v be a word on the alphabet A. We way that v encounters w provided there is a morphism h : B + → A + so that h(w) is a subword of v.We think of w as a pattern or template and h(w) has an instance of the pattern or template.So for v to encounter w means that an instance of the pattern w can be found among the subwords of v-that is within v itself.If v does not encounter w we say that v avoids w.We say that w is avoidable on the k-letter alphabet provided there are infinitely many words in A + that avoid w, where A is an alphabet with k letters.Similarly, we say that a set Σ of words is avoidable on the alphabet with k letters provided there are infinitely many words on the k-letter alphabet, each of which avoids every word belonging to Σ.
The notion of avoidable words was introduced independently by Bean, Ehrenfeucht, and McNulty [3] in 1979 and by Zimin [26] in 1982, and in these papers the notion of avoidability was given algorithmic characterizations.In §3 below, one of these characterizations will be more fully examined.The notion of avoidable words was inspired by work carried out by Axel Thue (see [24,25]) in the early years of the 20 th century.Thue proved that the pattern xx is avoidable on the 3-letter alphabet and that xxx is avoidable on the 2-letter alphabet.
The word xx is the simplest example of a doubled word.In general, we say a word w is doubled provided every letter that occurs in w occurs at least twice.Suppose that w is doubled.We say that w is the mesh of w provided k is the smallest natural number such that whenever x is a letter and u is a word of length greater than k in which x does not occur, then xux is not a subword of w.In Bean, Ehrenfeucht, and McNulty [3] it was proved that For any positive natural number k, the set of all doubled words of mesh k on a countably infinite alphabet is avoidable on the alphabet with 8k + 16 letters.This was the earliest global avoidability theorem for doubled words.It is easy to see by an inductive argument, that every word on an n-letter alphabet with length at least 2 n must have a subword that is doubled.Because the length of a word is an upper bound on its mesh, we also have The set of all doubled words on an alphabet with no more than n letters is avoidable on an alphabet with 8 • 2 n + 16 letters.
The method used by Zimin [26] is more efficient than that of Bean, Erhrenfeucht, and McNulty.The following is implicit in Zimin's work: The set of all doubled words on an alphabet with no more than n letters is avoidable on an alphabet with 6 • 2 n + 14 letters.
As an immediate corollary of the 1989 work of Baker, McNulty, and Taylor [2] we have The set of all doubled words on an alphabet with no more than n letters is avoidable on an alphabet with 9 • n + 20 letters, if only because this bound was proven to hold for avoidable words generally.
In 1985 Irina Mel'nichuk [17] outlined a proof of the following theorem, which gives bounds sharper than any of those above.
Mel'nichuk's Global Avoidabilitiy Theorem for Doubled Patterns on n Letters.Let n be a positive natural number.The set of all doubled patterns on the n-letter alphabet is avoidable on the alphabet with 3 n+1 2 letters.
In 2015 Michael Lane [13] provides an exposition, in English, of Mel'nichuk's proof that fills in the all details not available in Mel'nichuk [17].
The first purpose of this paper is to establish a theorem that improves Mel'nichuk's theorem.Our second purpose is to offer a minor improvement to Mel'nichuk's Global Avoidability Theorem for Avoidable Words on n Letters.
Let n be a positive natural number.The set of all avoidable patterns on the n-letter alphabet is avoidable on an alphabet with 4 n+2 2 letters. 1efore taking up these tasks, it is interesting to observe that in contrast to the global or simultaneous avoidability results mentioned above, one can ask, more locally, of an individual word w, for the smallest k so that w is avoidable on the alphabet with k letters.Denote this smallest value by µ(w) and call it the avoidability index of w.In case the word w is doubled, the situation has been substantially clarified.In 1984 A. G. Dalalyan [12] proved that every doubled word is 4-avoidable and that any doubled word in which at least 6 distinct letters appear is 3-avoidable.Dalalyan's results were rediscovered by Bell and Goh [4] and the results of Bell and Goh were enhanced by Blanchet-Sadri and Woodhouse [5] in 2013.A substantial advance was obtained in 2015 by Michael Lane [13].He proved Every doubled word on n or fewer letters of length at least min(2n + 1, 12) is avoidable on the 3-letter alphabet.
All the doubled words on these small alphabets are seen to be 3-avoidable (and even 2avoidable in some cases).So Lane's result left as unsettled only certain doubled words of length 8 on 4 letters and certain doubled words on 5 letters that have length 10-this left a list of roughly 100 doubled patterns to check.In 2016 Ochem [21], working independently of Lane, but using largely similar methods, was finally able to prove that each doubled word is 3-avoidable.It appears that many doubled words are actually 2-avoidable.The problem of characterizing the doubled words that are 2-avoidable remains open.
For words w in general, not just doubled words, the situation is much less clear.Indeed, among the most vexing problems, first raised the the mid-1970's, are Problem 0. Is the function µ on words over a countably infinite alphabet that returns the avoidability index of w (or ∞ when w is unavoidable) a computable function?If it is, what is its computational complexity?
Problem 1.Does the function µ have a finite upper bound?
The only progress on these problems has been in the investigation of the avoidability index on small alphabets.The avoidability of words on alphabets of size no more than 2 has been systematically investigated by Schmidt [23], Roth [22], and Cassaigne [8,9] and is now completely understood.For alphabets of size 3 the task was begun by Cassaigne in his 1994 dissertation [9].Clark in his 2001 dissertation [11] proved that the avoidability index of each avoidable word on a three letter alphabet is no more than 4, and, in 2006, Ochem [20] completed the work started by Cassaigne by verifying the avoidability index as 2 for the avoidable words not settled by Cassaigne.So the avoidability index of any avoidable word on the three letter alphabets is either 2 or 3. Finally, Clark [10,11] has devised a word with avoidability index 5.No avoidable word of larger avoidability index is known.

Global avoidability of doubled words
The Global Avoidablitiy Theorem for Doubled Patterns on n Letters.Let n be a positive natural number.The set of all doubled words on n or fewer letters is avoidable on • the alphabet with 8 letters, if n = 4; • the alphabet with n + 2 letters, if n is even and n = 4; • the alphabet with n + 3 letters, if n is odd.
Proof.The case when n = 4 is exceptional.We handle it at the end of the proof.
So for now, we stipulate that n = 4. First, consider the case n = 1.The set of all doubled words on this alphabet is {x 2 , x 3 , x 4 , . . .}, where x is the sole letter.Axel Thue showed that this set is avoidable on the alphabet with 3 letters.This settles the case since 3 4 = 1 + 3.
So we take up the case when n > 1.Let k be the least even natural number so that The plan of our proof is to start with an alphabet A with k letters.We take We will construct infinitely many words on A, each of them avoiding every doubled pattern on the n-letter alphabet.We will do this by a method introduced by Axel Thue [25], which has now become standard.Namely, we will define a map Ψ : A + → A + so that the sequence Ψ(a 0 ), Ψ 2 (a 0 ), Ψ 3 (a 0 ), . . . is an infinite list of words of increasing length such that each word on this list avoids every doubled word on the n-letter alphabet.The map Ψ will be easy to describe, but before doing this it helps to look ahead to see what this the electronic journal of combinatorics 25(2) (2018), #P2.22 map needs to be like.So for the moment suppose that Ψ is in hand.Now consider a doubled word w that, contrary to our hopes, is encountered by Ψ +1 (a 0 ).This means that there will be a morphism h such that h(w) is a subword of Ψ +1 (a 0 ).
The situation at hand is illustrated in Figure 1.In this figure, points at the top are (some of) the letters of Ψ (a 0 ) arranged as they are in Ψ (a 0 ).The points at the bottom are the letters of w; they comprise a copy of w.Of course Ψ takes each letter at the top to a word (indicated here as a line segment) in the middle of the diagram.Together they constitute a subword of Ψ (a 0 ).Likewise, h takes each letter at the bottom to a word in the middle.The diagram is drawn in a way to suggest that all the Ψ images of letters have the same length-this will indeed be an attribute of Ψ when we finally define it.The (|)'s that appear once in the Ψ-image of each letter from A indicate specially chosen subwords of length 2 that we will call Mel'nichuk decisive representatives of the letter they come from by way of Ψ.We will rig matters so that these decisive representative never straddle a border between h-images of adjacent letters of w.As seen in the diagram, some letters of w have images that may engulf at least one Mel'nichuk decisive representative, while others do not.Now obtain w from w by deleting all the letters whose images under h engulf no decisive representative.The diagram suggests how to construct a morphism h so that h (w ) is a subword of Ψ (a 0 ).Then an appeal to induction will finish the proof.
Rather than an appeal to induction, it is convenient to prove the theorem indirectly.So, in pursuit of a contradiction, we assume the theorem fails.We pick a doubled word w, as short as possible, on no more the n letters, that is encountered by Ψ t (a 0 ) for some natural number t.For this w, we pick as small as possible so that Ψ +1 encounters w.Finally, we pick a morphism h so that h(w) is a subword of Ψ +1 (a 0 ).
The essential requirement placed on Ψ by the idea sketched above is that the image of each letter x in A should have a large number of distinct subwords of length 2 that will permit us to identity a decisive representative of x.Since at most n letters occur in w, there will only be at most n possibilities for the first letter of h(ξ), as ξ runs through the letters occurring in w.So we must have at more than n suitable choices for the decisive representatives.The main difficulty, is that the k decisive representatives must be pairwise distinct.
We can regard words on the alphabet A as right-directed paths where the vertices have been labeled with letters for A. What we require is a graceful labeling.In 1982 Bloom and Hsu [6] introduced the notion of graceful labelings of directed graphs.In 1985 Bloom and Hsu [7] observed that one-way directed paths of even length have a graceful labelings.A single example suffices to see the general case.Consider the graceful numberings of the directed path with 10 vertices displayed in Figure 2. Observe that the edge labels are constructed from the vertex labeling as follows: if x → y, then the label of the edge from x to y is the natural number r < 10 so that y − x ≡ r (mod 10).Another way to say this is that the label is y − x as computed in cyclic group Z 10 .The numberings shown in Figure 2 are graceful in the sense that all the edge labels are distinct.More is true.The labeling of the path at the bottom can be obtained from the labeling of the path at the top.Actually, we give two ways to see how these two labellings are related.In the first way, to obtain the bottom labeling simple add 7 (in Z 10 ) to the label of each vertex in the top labeling.It is evident that the labels on the edges will not change when this is done.In this way, we can obtain 10 different graceful vertex labellings with the same edge labeling.The second way to see the connection is to realize that the vertex labeling occurs around an oblong.Each of these labellings amount to numbering the vertices counterclockwise around the oblong, starting at some vertex with 0. In the numbering at the top, we started with the leftmost of the lower vertices, while in the numbering at the bottom, we started at the fourth vertex from the left among the lower vertices.With this second viewpoint in mind, it is easy to see that if 0 is on the lower level, but not the leftmost vertex, then 9 will be somewhere on the directed path preceding 0. On the other hand, if 0 is on the upper level, then 1 will precede 0 on the directed path.
The remarks above hold in general.A graceful numbering of a directed path with k vertices, where k is even, can be made by labeling the leftmost vertex with 0 and running counterclockwise around the oblong.The last vertex on the path will get the label k/2.Using this graceful labeling, we can devise others as described above.There will be k of them and they will all generate the same edge labels.Moreover, for any one of these graceful labelings, if 0 is on the lower level (but not the left end of the path) then k − 1 precedes 0 on the path, whereas, if 0 is on the upper level, then 1 precedes 0 on the path.
In the above constructions, if we replace the label i by a i , for each i < k, we get k graceful words from the k gracefully labeled left-directed paths.In case k = 10, here is what they look like: a 0 a 9 a 1 a 8 a 2 a 7 a 3 a 6 a 4 a 5 a 1 a 0 a 2 a 9 a 3 a 8 a 4 a 7 a 5 a 6 a 2 a 1 a 3 a 0 a 4 a 9 a 5 a 8 a 6 a 7 a 3 a 2 a 4 a 1 a 5 a 0 a 6 a 9 a 7 a 8 a 4 a 3 a 5 a 2 a 6 a 1 a 7 a 0 a 8 a 9 a 5 a 4 a 6 a 3 a 7 a 2 a 8 a 1 a 9 a 0 a 6 a 5 a 7 a 3 a 8 a 3 a 9 a 2 a 0 a 1 a 7 a 6 a 8 a 3 a 9 a 4 a 0 a 3 a 1 a 2 a 8 a 7 a 9 a 4 a 0 a 5 a 1 a 4 a 2 a 3 a 9 a 8 a 0 a 5 a 1 a 6 a 2 a 5 a 3 a 4 Observe that the word on each row begins with one of our k letters.We denote the word on the i th -row by a i .It is important to observe that the word a p a q , where p = q, is a subword of exactly one a i .Indeed, q − p (calculated in Z k ) is the edge-label associated with two adjacent columns in our array of words.In the left of these adjacent columns a p occurs exactly once.What this will mean is that each of the k − 1 the subwords of a i that have length 2 are available as potential decisive representatives of a i .
We define our map Ψ so that Ψ(a i ) = a i a i , for each i < k.
So Ψ(a i ) is a word of length k + 1 that begins and ends with the letter a i and in which every letter a j with j = i occurs exactly once.Also, there are k − 1 subwords of length 2 that are available to be chosen as decisive representatives of a i .Since k − 1 > n, chose one of these available words of length 2 so that its right letter is not the leftmost letter in h(ξ), for any letter ξ occurring in w.At this point, our proof would be essentially complete, except for the possibility that w is empty.To prevent this, we have show that there is some letter ξ of w so that some decisive representative is a subword of h(ξ).But since decisive representative cannot straddle the h-image of adjacent letters in w, it will be enough to prove the next lemma.

Some decisive representative is a subword of h(w).
the electronic journal of combinatorics 25(2) (2018), #P2.22 Proof.It follows easily by induction on t that no two adjacent letters in Ψ t (a 0 ) are identical.
Since h(w) is a subword of Ψ +1 (a 0 ) we have three alternatives to consider.
Alternative: Ψ(a i ) is a subword of h(w), for some i < k.
In this case the decisive representative of a i will be a subword of h(w).
Alternative: h(w) is a subword of Ψ(a i ), for some i < k.
We reject this alternative, since Ψ(a i ) has no doubled subwords, but h(w) must be doubled since w is doubled.
Alternative: h(w) is a subword of Ψ(a i a j ), for some i, j < k with i = j.
In view of the second alternative, here we know that h(w) straddles the boundary between Ψ(a i ) and Ψ(a j ).It is harmless to suppose that i = 0.So Ψ(a 0 a j ) is To keep out of the first Alternative, we need only consider that h(w) is a subword of the following word: In this word every letter occurs exactly 2 times.Since w is doubled, h(w) is doubled.So any letter that occurs in h(w) occurs at least twice.Since h(w) straddles the middle of the displayed word, we know that both a 0 and a j occur in h(w).The other occurrence of a 0 is in a j .If a k−1 precedes a 0 in a j , then a k−1 must occur (twice) in h(w).But it follows that every letter that occurs in w, must occur exactly twice in w.This will force all k of the letters to occur in h(w).But then a j will be a subword of h(w).Hence the decisive representative of a j will be a subword of h(w).
So it remains to consider the case when a k−1 does not occur in h(w).In this case, we know that a 1 must precede a 0 in a j .This forces all the letters except a k−1 to occur in h(w).So we must have the following situation: Consider any letter a i with i different from each of 0, a j , and a k

2
. Pick a letter ξ of w so that a i occurs in h(ξ).Now a i occurs once in a 0 and once in a j .Because a 0 and a j have no subwords of length 2 in common, it must be that h(ξ) has length 1.The same also applies when i = j.So consider a 0 .Pick the letter ξ of w so that a 0 occurs in h(ξ).Now the word a 0 a j can occur only once in h(w).However, there is one situation when the the electronic journal of combinatorics 25(2) (2018), #P2.22 word ak 2 a 0 can occur twice.This happens when k 2 ≡ 3 (mod k).This means that k = 6.In turn, this means that n is either 3 or 4.But recall that n = 4.So if n is 3, it might be that h(ξ) has length 2, where ξ is the letter so that a 0 occurs in h(ξ).In this event, the length of h(ζ) is 1 for each letter ζ = ξ of w.It follows that the length of h(w) is no greater than 2(3 + 1).So we have 2(k − 1) 2(3 + 1).This means that k 5.But in this case, k = 3 + 3 = 6.So we must reject this case.It follows that h(ξ) has length 1 for each letter ξ of w.But then the length of h(w) cannot be greater than 2n.So we would have 2(k − 1) 2n.This is impossible since we have chosen k so that k − 1 > n.
In this way, the lemma is established.
With the Lemma A in hand, we see that not all the letters in w are deletable.So w is not empty.By the minimality in the choice of w we see that the length of w and the length of w must be the same.This means that no letters have been deleted.But then w = w and we see that Ψ (a 0 ) encounters w (with the help of h ).This violates the minimality in the choice of .At last, this is the contradiction needed to complete our indirect proof of the theorem, when n = 4.
For the exceptional case n = 4, observe that each doubled word on no more than 4 letters is also a word on no more than 5 letters.Now n = 5 is an unexceptional instance of the theorem and 5 + 3 = 8.So the set of all doubled words on no more than 4 letters is avoidable on an alphabet with 8 letters.
There is only one case where Mel'nichuk's bound is sharper than the bound proved here: n = 1.The bound given in the present theorem is 4, whereas Mel'nichuk's bound is 3, which is actually the bound established by Axel Thue [25].Table 1 provides  n is even and n = 4 The referee has sketched an argument that, when n = 4, the true value is 5.
As Mel'nichuk observed, the bound must be at least n + 1.When n = 1, the bound must be at least n + 2 = 3 and this bound can be achieved, as Axel Thue showed in 1906.It is conceivable that n + 2 is a sharp bound, but no proof is known that this bound can be acheived when n is odd; when n = 4 the referee contends that the bound is 4 + 1 = 5.Neither is it known that n + 1 will not suffice in every case except n = 1.

Global Avoidability of Avoidable Words
After her 1985 work on avoiding doubled patterns, Irina Mel'nichuk turned to avoiding patterns generally.It appears she has never put her methods in this direction into the electronic journal of combinatorics 25(2) (2018), #P2.22 the literature.However, in 1991 Mikhail Volkov gave a presentation of her methods at Marquette University and Pavel Goralčik brought her methods to Paris.An account of Mel'nichuk's methods can be found in Chapter 3 of Lothaire [15], where the following theorem is credited to Irina Mel'nichuk.

Mel'nichuk's Global Avoidability Theorem for Avoidable Words on n Letters.
Let n be a positive natural number.The set of all avoidable patterns on the n-letter alphabet is avoidable on an alphabet with 4 n+2 2 letters.
The bound 4 n+2 2 is equal to 2(n + 2) when n is even and to 2(n + 3) when n is odd.Here we can only make a small improvement: Mel'nichuk's bound in the even case works for the odd case as well.She submitted an abstract to the Colloquium on Universal Algebra held in Szeged in August 1989 in which she asserted that the bound n + 6 would serve, but included no proof.
The Global Avoidability Theorem for Avoidable Words on n Letters.Let n be a positive natural number.The set of all avoidable patterns on the n-letter alphabet is avoidable of an alphabet with 2(n + 2) letters.
We need the notion of reducibility in the proof of this theorem.We associate with each pattern w a bipartite graph, called the adjacency graph of w, as follows.The two parts of this graph are called the left alphabet and the right alphabet.In the adjacency graph their is an edge joining the letter x in the left alphabet with y is the right alphabet provided the length 2 word xy is a subword of w.For example, for the word a 0 b 0 a 1 b 1 a 2 b 2 a 3 b 3 a 4 b 4 has the adjacency graph displayed in Figure 3.
This word has a simple adjacency graph, mostly due to the fact the each letter appears in the order only once.A subset F of the alphabet is free for w provided whenever ξ and ζ are letters occurring in w there is no path in the adjacency graph connecting ξ in the the electronic journal of combinatorics 25(2) (2018), #P2.22 left alphabet with ζ in the right alphabet.The set {a 0 , a 1 , a 2 , a 3 , a 4 } is free for the word used in Figure 3.
We say that a word w reduces in one step to a word u if and only if u can be obtained from w be deleting all occurrence of the letters belonging to some set free from w.We way that w reduces to u if and only if u can be reached from w by a series of finitely many one-step reductions.
The following theorem was proved in 1979 by Bean, Ehrenfeucht, and McNulty [3] and independently in 1982 by Zimin [26].
The Characterization Theorem for Unavoidable Patterns.A word w is unavoidable if and only if w is reducible to the empty pattern.
Zimin [26] provided an additional equivalent condition.The proof of the Global Avoidability Theorem for Words on n Letters amounts to a proof that unavoidable words are reducible-one direction of the proof of the Characterization Theorem.
Proof of the Global Avoidability Theorem for Words on n Letters.This proof follows the plan used in the proof of the Global Avoidability Theorem for Doubled Patterns.
Case: k is odd.
In each display, the subscripts on the a's on each successive row can be obtained by adding 1 modulo k to the a in preceding row.One the other hand, each b i occurs in only one column, in each case.Every Ψ-image of a letter has length k.In each case, any word of length 2 can occur in at most one Ψ-image of a letter.It will be useful below to note here that reading across the Ψ-image of any letter one observes that the a's and b's alternate.This last property still holds when reading across Ψ t (a 0 ), for any natural number t.
In case k is odd, Ψ(a i ) begins and ends with a's with some indices and b 0 is the second letter of the image, whereas Ψ(b i ) begins and ends with a b's with some indices and the beginning is b (k+1)/2 .In case k is even, Ψ(a i ) always begins with some a and ends with some b and the second letter is always b 0 , whereas Ψ(b i ) begins with an a and ends with a b and its second letter is always b k/2 .
It is convenient to let b * be b (k+1)/2 in the odd case and to let it be b k/2 is the even case.
The even case differs in no important way from the method described in Lothaire [15].We will prove that the set comprised of Ψ t (a 0 ) as t runs through the natural numbers avoids every pattern on the alphabet with n letters that is avoidable.This is the same as proving that every pattern on the alphabet with n letters that is encountered by some Ψ t (a 0 ) is unavoidable-or, what is the same, is reducible to the empty word.
We prove the theorem indirectly.So, in pursuit of a contradiction, we assume the theorem fails.We pick a word w on no more than n letters that cannot be reduced to the empty word, with w as short as possible, that is encountered by Ψ t (a 0 ) for some natural number t.For this w, we pick as small as possible so that Ψ +1 encounters w. (Notice that Ψ 0 (a 0 ) = a 0 and a 0 reduces to the empty word.)Finally, we pick a morphism h so that h(w) is a subword of Ψ +1 (a 0 ).
With h in hand, we pick a system of decisive representatives of the letters in A ∪ B. These decisive representative can be of only two kinds: a i b j and b i a j .Just has in the earlier proof, we have a pattern w and a morphism h so that w is obtained by deleting all occurrences of some letters from w, and so that h (w ) is a subword of Ψ (a 0 ).
In the earlier proof the difficulty was that w might have been empty, but here the difficulty is that the set of letters deleted from w to obtain w might not be free for w.So we need the following lemma.
Lemma B. The pattern w reduces to w .Proof.Zimin [26] proved, as Lemma 8 in his paper, a statement that, in the context at hand, reads If there is a pattern v and a morphism f so that • ξ is a letter of w not in w if and only if no letters of w occur in f (ξ), then w is reducible to w .
Our task, then, is to provide an appropriate pattern v and an appropriate morphism f .the electronic journal of combinatorics 25(2) (2018), #P2.22 Consider any letter ξ that occurs in w.To obtain f (ξ) we modify h(ξ).In h(ξ) we might find any number of decisive representatives.If some decisive representative is a subword of h(ξ), consider the leftmost one: xy.To get f (ξ) our first step is to replace xy by xξy, if x is a b i , and by xyξ if y is a b j .Now say there are r decisive representatives remaining in h(ξ).We introduce new letters ξ i for each i < r and insert them from left to right within or after each of remaining decisive representatives in h(ξ), just as we did with ξ itself.
Of course, if ξ was a letter deleted from w, then no decisive representative is a subword of h(ξ) and in this case f (ξ) = h(ξ).We take v = f (w).At this point, the last two stipulations in Zimin's Lemma 8 are fulfilled.
It remains to show that v reduces to w .Observe that in v • each occurrence on an a i is followed, if at all, by a b j , • each occurrence of a b i is followed, if at all, by either an a j or by some ξ, perhaps with a subscript.
• each occurrence of a ξ (even those with subscripts) is followed, if at all, by an a j .
It follows that in the adjacency graph of v, the only edges from vertices in the set A of the left alphabet have their other vertices in the set B of the right alphabet.Also, the only edges from vertices in the set B of the right alphabet have their other vertices in the set A of the left alphabet.Therefore, the set A is free for v.So delete this free set to obtain the word v .Now v is made up of b's with certain subscripts and the various ξ's that we inserted as well as their subscripted versions.The unsubscripted ξ's are exactly the letters of w and they occur in v exactly as they occur in w .To obtain w , as we desire, all we have to do is delete the b's and the subscripted versions of the ξ's from v .There are three points that make this possible.First, the b's occur in order, as their indices cycle modulo k.Second, between any occurrence of a ξ (perhaps with subscripts) and the next one (perhaps a different one) to the right, either b 0 or b * must occur and if one of these occurs then the other does not.Third, the subscripted ξ's occur in the order of their subscripts.This means that in the adjacency graph of v we have edges of the following kinds: edges from b i in the left alphabet to b i+1 in the right alphabet, where the +1 in the subscripts is computed modulo k; we also have some edges between b's in the left alphabet and ξ's in the right, as well as edges from ξ's in the be left alphabet to b's in the right.This means that each singleton {b i } is free for v .Let us delete b 1 , resulting in v .The adjacency graph of v has the properties just mentioned (unless b i = b * ) so again all the singletons of the remaining b's are free.So deleting the singletons one after another we can reduce v to v * , where the only b's remaining are b 0 and b * .In v * these now alternate with the ξ's.So in the adjacency graph of v * there are only edges from the b's in the left alphabet to certain ξ's in the right alphabet and edges from certain ξ's in the left alphabet to b's in the right alphabet.This means {b 0 , b * } is free from v * .Deleting this free set results in v • .It remains to delete from v • the subscripted versions of the ξ's.But just as the with the b i 's above, each singleton is a free set and we can delete these one at a time, since the freeness of the remaining singletons will persist from step to step. .Putting the reductions together, we see that v reduces to w .According to Zimin, w must also reduce to w , proving our lemma.
With the Lemma B in hand, we see that w reduces to w .By the minimality in the choice of w we see that the length of w and the length of w must be the same.This means that no letters have been deleted.But then w = w and we see that Ψ (a 0 ) encounters w (with the help of h ).This violates the minimality in the choice of .At last, this is the contradiction needed to complete our indirect proof of the theorem.
What was actually accomplished in this proof was that every word w of n or few letters that is encountered by any Ψ T (a 0 ) is reducible to the empty word.Every unavoidable w must be encountered in this way.So every unavoidable word reduces to the empty word.So this is also a proof of one direction of the Characterization Theorem.
The bound 2(n + 2) given in this theorem might not be the best.That coefficient 2 at the front can be attributed to the way in which the reductions were constructed in the proof of Lemma B. At least the most transparent attempts to use graceful words lead to adjacency graphs that have large portions that are complete bipartite graphs.Such graphs simply do not provide enough free sets to support the needed reductions.But it is conceivable that some other part of graph theory will provide the means to improve this bound.

Concluding Comment
Loosely speaking the first proof in this paper was obtained by combining Mel'nichuk's method with the graceful numbering of left-directed paths with an even number of vertices.The second proof amounts to a very small alteration in her method.So the imprint of Irina Mel'nichuk's thinking is heavy in the paper before you.

Note Added in Proof
Since this article was accepted two things happened that deserve mention.
In connection with the Global Avoidability Theorem for Doubled Patterns on n Letters, Pascal Ochem has kindly shared with me his method to handle the exceptional case n = 4.The upshot of his method is the best possile bound in this case: n + 1.His method relies on an application of the Perron-Frobenius Theorem and Moulin Ollagnier's 1992 article [19] that the set of words on the 5-letter alphabet that have repetition index 5  4 is infinite.This is an adaptation of ideas in Ochem's 2016 article [21].
In connection with the Global Avoidability Theorem for Avoidable Words on n Letters, after several years I was finally able to make contact with Irina Mel'nichuk.She very kindly shared with me a manuscript that she had prepared in 1996 but had never brought to final publication.In that mansucript, Mel'nichuk gave exactly the same theorem.While our two proofs share something in common, they differ in important ways-indeed, Mel'nichuk actually proves a smaller bound in the case n is even.Her article is now available on the arXiv as [18].

Figure 2 :
Figure 2: Two Graceful Labellings of Left-directed Path on 10 Vertices

2
a comparison of the bounds.The present bound Mel'nichuk's bound: 3 n+1 Parity