# Rappresentazioni in base k del co-dominio di un polinomio: è privo di contesto?

14

Nel capitolo 4 di A Second Course in Automata Theory di Jeffrey Shallit il seguente problema è elencato come aperto:

Let $$p ( n )p(n)p(n)$$ un polinomio a coefficienti razionali tali che $$p ( n ) ∈ Np(n)∈Np(n) \in \mathbb{N}$$ per tutti gli $$n ∈ Nn∈Nn \in \mathbb{N}$$ . Dimostrare o confutare che il linguaggio delle rappresentazioni base-k di tutti i numeri interi in $${ p ( n ) ∣ n ⩾ 0 }{p(n)∣n⩾0}\{p(n) \mid n \geqslant 0\}$$ è privo di contesto se e solo se il grado di $$ppp$$ è$$⩽ 1⩽1\leqslant 1$$ .

Qual è il suo stato adesso (come ad ottobre 2018)? È provato? Che dire di alcuni casi speciali?

1
Se k=1$k = 1$ (rappresentazione unaria), anche il più semplice p(n)=n2$p(n) = n^2$ non è privo di contesto (il noto linguaggio non CF L={1n2}$L = \{ 1^{n^2} \}$ )
Marzio De Biasi

@MarzioDeBiasi La cosiddetta rappresentazione unaria non è base- 1$1$ . L'unico numero intero rappresentabile nella base effettiva 1$1$ sarebbe 0$0$ .
Emil Jeřábek sostiene Monica il

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@ EmilJeřábek: Penso che in molti contesti la base 1 sia un alias di "rappresentazione unaria" .
Marzio De Biasi,

Risposte:

10

Ovviamente $$k ≥ 2k≥2k \geq 2$$ qui.

C'era una volta un manoscritto di Horváth che sosteneva di risolvere il problema, ma non era chiaro in diversi punti e per quanto ne sapevo non fu mai pubblicato.

Per quanto ne so, il problema è ancora aperto. Una direzione dell'implicazione è facile, ovviamente.

È già risolto per k=2$k= 2$ ? (Ho un'idea per dimostrarlo per k=2$k=2$ e se funziona la stessa tecnica potrebbe probabilmente essere applicata ad altre basi)
Marzio De Biasi

Sarei molto felice di ricevere qualsiasi tuo feedback sulla mia risposta.
domotorp,

Non riesco a capire la tua soluzione richiesta, scusa.
Jeffrey Shallit,

Ho pubblicato un'altra risposta con molti più dettagli; la dichiarazione completa è piuttosto complessa, quindi ho aggiunto alcuni lemmi più semplici che hanno l'idea principale, si spera che renda il tutto più credibile.
domotorp,

3

Questo è uno schizzo della dimostrazione per $$k = 2k=2k=2$$ e $$L = { [ n 2 ] 2 ∣ n ≥ 1 }L={[n2]2∣n≥1}L = \{[n^2]_2 \mid n \geq 1\}$$ ; dove $$[ n 2 ] 2[n2]2[n^2]_2$$ è la rappresentazione binaria di $$n 2n2n^2$$ . Per una migliore chiarezza posizioniamo il bit meno significativo delle stringhe binarie sulla sinistra, ad es. $$[ 4 2 ] 2 = 00001[42]2=00001[4^2]_2 = 00001$$ .

L'idea principale è quella di supporre che $$LLL$$ sia privo di contesto, quindi provare a "semplificare" intersecandolo con un semplice linguaggio regolare $$RRR$$ ; la nuova lingua $$L ∩ RL∩RL \cap R$$ è ancora privo di contesto e dovrebbe contenere comunque rappresentazioni binarie di quadrati; quindi applichiamo il lemma di pompaggio per i linguaggi CF al fine di ottenere una stringa binaria che non è una rappresentazione di un quadrato.

L'intersezione di $$LLL$$ con parole regolari che contiene solo un piccolo numero finito di $$111$$ cifra non è promettente. Si scopre che fino a quattro $$111$$ cifre $$( R = { 0 ∗ 1 } , { 0 ∗ 10 ∗ 1 } , { 0 ∗ 10 ∗ 10 ∗ 1 } , { 0 ∗ 10 ∗ 10 ∗ 10 ∗ 1 } )(R={0∗1},{0∗10∗1},{0∗10∗10∗1},{0∗10∗10∗10∗1})(R = \{ 0^*1 \}, \{ 0^*10^*1 \}, \{ 0^*10^*10^*1 \}, \{0^*10^*10^*10^*1\} )$$ otteniamo un linguaggio CF; e con cinque $$111$$ cifre abbiamo un problema apparentemente difficile di teoria dei numeri.

L'approccio promettente è quello di intersecare $$LLL$$ con $$R = 10 +1 +0+1R=10+1+0+1R = 1\,0^+\,1^+\,0^+\,1$$ ; ciò equivale a limitare$$LLL$$ ai quadrati:

$$n 2 = 2 0 + 2 a ( ( 2 b - 1 ) + 2 b + c ) , 1 < a , b , c$$

n2=20+2a((2b1)+2b+c),1<a,b,c

(quadrati dispari informali la cui rappresentazione binaria contiene tutti $$000$$ s tranne una sequenza di $$111$$ s nel mezzo).

    n        n^2  n                  n^2
39       1521  111..1             1...11111.1
143      20449  1111...1           1....1111111..1
543     294849  11111....1         1.....111111111...1
2111    4456321  111111.....1       1......11111111111....1
8319   69205761  1111111......1     1.......1111111111111.....1
33023 1090518529  11111111.......1   1........111111111111111......1
LSB          MSB   LSB                         MSB


Con alcuni sforzi, possiamo dimostrare quanto segue:

Teorema: il numero $$2 0 + 2 a ( ( 2 b - 1 ) + 2 b + c ) ;0 < c , 3 < a < b20+2a((2b−1)+2b+c);0 è un quadrato se e solo se

$$b = 2 a - 3 , c = a - 3$$

b=2a3,c=a3

(la prova è piuttosto lunga, la pubblicherò sul mio blog)

A questo punto, possiamo facilmente dimostrare che $$L ∩ RL∩RL \cap R$$ non è privo di contesto usando il lemma di pompaggio (possiamo pompare al massimo due "segmenti" della stringa $$100..0011 ... 1100.001100..0011...1100.001100..0011...1100.001$$ ). Quindi anche $$LLL$$ non è privo di contesto.

Probabilmente la stessa tecnica può essere applicata a qualsiasi base $$kkk$$ .

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Il risultato per nella base 2 è noto da molto tempo. La sfida è quella di far funzionare una costruzione come questa per ogni polinomio che associa numeri interi a numeri interi e per ogni base. (n2)$(n^2)$
Jeffrey Shallit,

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Siamo così simili, ho anche trascorso i miei ultimi due giorni a pensare a questo problema, anche se ho adottato un approccio completamente diverso.
domotorp,

3

Penso di avere una prova. La prova segue da questo lemma.

Lemma. Per una lingua senza contesto $$LLL$$ se per infinitamente molte $$nnn$$ ci sono $$n 6n6n^6$$ parole di uguale lunghezza le cui prime $$n 2n2n^2$$ lettere sono uguali e le loro ultime $$nnn$$ lettere sono diverse (a coppie), allora c'è una $$BBB$$ tale che ci sono infinitamente molte coppie $$u , v ∈ Lu,v∈Lu,v\in L$$ di uguale lunghezza che differiscono solo nelle loro ultime lettere $$B.BB$$

Quindi se $$uuu$$ e $$vvv$$ rappresentano numeri binari, la loro differenza sarà al massimo $$2 B2B2^B$$ infinitamente spesso, il che è impossibile per i polinomi. D'altra parte, con una certa teoria dei numeri si può dimostrare che la condizione è soddisfatta per ogni polinomio con valore intero $$ppp$$ : prendere qualsiasi $$x 1 , … , x n 6x1,…,xn6x_1,\ldots,x_{n^6}$$ per cui $$f ( x i ) ≠ f ( x j )f(xi)≠f(xj)f(x_i)\ne f(x_j)$$ , e quindi aggiungere un numero sufficientemente grande $$NNN$$ a ciascuno di essi per ottenere le parole desiderate $$f (xi+N)f(xi+N)f(x_i+N)$$.

Proof of the lemma. Take a large enough $$nnn$$ such that there are $$n6n6n^6$$ words of equal length, $$w1,…,wn6w1,…,wn6w_1,\ldots,w_{n^6}$$, that satisfy the conditions. For each $$wiwiw_i$$ fix a way in which it can be generated from the context-free grammar. (Warning! I'm not an expert of this field, so I might not use the proper terms.)

Say that the application of a rule $$A→BCA→BCA\to BC$$ splits two letters $$bbb$$ and $$ccc$$ of the final word, if the $$bbb$$ and $$ccc$$ are both derived from $$AAA$$, but $$bbb$$ is derived from $$BBB$$, while $$ccc$$ is derived from $$CCC$$. Each rule splits at most $$O(1)O(1)O(1)$$ letters of $$wiwiw_i$$ from each other.

In any $$wiwiw_i$$, there will be $$Ω(n)Ω(n)\Omega(n)$$ consecutive letters among the first $$n2n2n^2$$ letters that are split from each other by some consecutive rules such that no two letters among the last $$nnn$$ letters are split from each other while applying these rules. If we write these rules collectively for letter $$wiwiw_i$$ as $$Ai→B1iB2i…BniAi→B1iB2i…BniA_i\to B_i^1B_i^2\ldots B_i^n$$, then no letter from the last $$nnn$$ letters is derived from $$BjiBjiB_i^j$$ for $$j, and $$B1iB2i…Bn−1iB1iB2i…Bn−1iB_i^1B_i^2\ldots B_i^{n-1}$$ are all converted into some part of the first $$n2n2n^2$$ letters. We can apply the pumping lemma to the rule $$Ai→B1iB2i…BniAi→B1iB2i…BniA_i\to B_i^1B_i^2\ldots B_i^n$$ if $$nnn$$ is large enough.

There are only $$(n22)(n22)\binom{n^2}2$$ choices for the interval of $$Ω(n)Ω(n)\Omega(n)$$ letters, $$O(n)O(n)O(n)$$ options about what the pumping lemma gives (as it has $$O(1)O(1)O(1)$$ length), so by the pigeonhole principle there will be two words for which these are all the same. But then after pumping we can obtain an arbitrarily long common initial part for these two words, while we know that they'll differ only in their last $$nnn$$ bits.

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Note. This is a much more detailed version of my other answer, as that didn't seem to be comprehensible enough. I've tried to convert it to resemble more standard pumping lemmas, but the full proof got way to complex. I recommend to read the statement of the first two lemmas to understand the main idea, then the statement of the Corollary, and finally the end, where I prove why the Corollary implies the answer to the question.

The proof is based on a generalization of the pumping lemma. The lemma that we need is quite elaborate, so instead of stating it right away, I start with some easier generalizations, eventually building up to more complicated ones. As I've later learned, this is very similar to the so-called interchange lemma.

Twin Pumping Lemma. For every context-free language $$LLL$$ there is a $$ppp$$ such that from any $$ppp$$ words $$s1,…,sp∈Ls1,…,sp∈Ls_1,\ldots,s_p\in L$$ we can select two, $$sss$$ and $$s′s′s'$$, that can be written as $$s=uvwxys=uvwxys=uvwxy$$ and $$s′=u′v′w′x′y′s′=u′v′w′x′y′s'=u'v'w'x'y'$$ such that $$1≤|vx|≤p1≤|vx|≤p1\le |vx|\le p$$, $$1≤|v′x′|≤p1≤|v′x′|≤p1\le |v'x'|\le p$$ and every word $$ˉuˉv1…ˉvnˉwˉxn…ˉx1ˉy∈Lu¯v¯1…v¯nw¯x¯n…x¯1y¯∈L\bar u\bar v_1\ldots \bar v_n \bar w\bar x_n \ldots \bar x_1 \bar y\in L$$, where $$ˉww¯\bar w$$ can be either $$www$$ or $$w′w′w'$$, and similarly, $$ˉviv¯i\bar v_i$$ can be either $$vvv$$ or $$v′v′v'$$ and $$ˉxix¯i\bar x_i$$ can be either $$xxx$$ or $$x′x′x'$$, but only such that $$ˉvi=vv¯i=v\bar v_i=v$$ if and only if $$ˉxi=xx¯i=x\bar x_i=x$$ (thus $$ˉvi=v′v¯i=v′\bar v_i=v'$$ if and only if $$ˉxi=x′x¯i=x′\bar x_i=x'$$), and $$ˉu=uu¯=u\bar u=u$$ if and only if $$ˉy=yy¯=y\bar y=y$$ and $$ˉu=u′u¯=u′\bar u=u'$$ if and only if $$ˉy=y′y¯=y′\bar y=y'$$. Moreover, if instead of $$ppp$$, we are given $$p(n+44)p(n+44)p\binom {n+4}4$$ words of length $$nnn$$, we can additionally suppose for the selected two words that $$|u|=|u′||u|=|u′||u|=|u'|$$, $$|v|=|v′||v|=|v′||v|=|v'|$$, $$|w|=|w′||w|=|w′||w|=|w'|$$, $$|x|=|x′||x|=|x′||x|=|x'|$$ and $$|y|=|y′||y|=|y′||y|=|y'|$$.

This statement can be proved essentially the same way as the pumping lemma, we just need to pick some $$sss$$ and $$s′s′s'$$ for which the same rule is pumped. This can be done if $$ppp$$ is large enough since there are only a constant number of rules. In fact, we don't even need that the same rule is pumped, but only that the non-terminal symbol is the same in the pumped rule. For the moreover part, notice that for a word of length $$nnn$$ there are only $$(n+44)(n+44)\binom {n+4}4$$ options it can be broken into five subwords, thus the statement follows from the pigeonhole principle.

Next we give another way of generalizing the pumping lemma (and later we'll combine the two).

Nested Pumping Lemma. For every context-free language $$LLL$$ there is a $$ppp$$ such that for any $$kkk$$ any word $$s∈Ls∈Ls\in L$$ can be written as $$s=uv1…vkwxk…x1ys=uv1…vkwxk…x1ys=uv_1\ldots v_kwx_k\ldots x_1y$$ such that $$∀i 1≤|vixi|≤p∀i 1≤|vixi|≤p\forall i~1\le |v_ix_i|\le p$$ and for every sequence $$(ij)mj=1(ij)mj=1(i_j)_{j=1}^m$$ the word $$uvi1…vimwxim…xi1y∈Luvi1…vimwxim…xi1y∈Luv_{i_1}\ldots v_{i_m}wx_{i_m}\ldots x_{i_1}y\in L$$.

Note that the indices $$ijiji_j$$ can be arbitrary from $$111$$ to $$kkk$$, the same index can occur multiple times. The proof of the Nested Pumping Lemma is essentially the same as the original pumping lemma's, we just need to use that we obtain the same non-terminal symbol from itself $$kkk$$ times - this is true if we do $$p(k−1)+1p(k−1)+1p(k-1)+1$$ steps (instead of the $$ppp$$ from the original pumping lemma). We can also strengthen Ogden's lemma in a similar way.

Nested Ogden's Lemma. For every context-free language $$LLL$$ there is a $$ppp$$ such that for any $$kkk$$ marking any at least $$pkpkp^k$$ positions in any word $$s∈Ls∈Ls\in L$$, it can be written as $$s=uv1…vkwxk…x1ys=uv1…vkwxk…x1ys=uv_1\ldots v_kwx_k\ldots x_1y$$ such that $$∀i 1≤∀i 1≤\forall i~1\le$$ '# of marks in $$vixivixiv_ix_i$$'$$≤pk≤pk\le p^k$$ and for every sequence $$(ij)mj=1(ij)mj=1(i_j)_{j=1}^m$$ the word $$uvi1…vimwxim…xi1y∈Luvi1…vimwxim…xi1y∈Luv_{i_1}\ldots v_{i_m}wx_{i_m}\ldots x_{i_1}y\in L$$.

Unfortunately, in our application $$pkpkp^k$$ would be too large, so we need to weaken the conclusion to allow non-nested $$viviv_i$$-$$xixix_i$$ pairs. Luckily, using Dilworth, the structure stays simple.

Dilworth Ogden's Lemma. For every context-free language $$LLL$$ there is a $$ppp$$ such that for any $$k,ℓk,ℓk,\ell$$ marking any at least $$pkℓpkℓpk\ell$$ positions in any word $$s∈Ls∈Ls\in L$$, it can be written either as

case (i): $$s=uv1…vkwxk…x1ys=uv_1\ldots v_kwx_k\ldots x_1y$$, or as

case (ii): $$s=uv1w1x1…vℓwℓxℓys=uv_1w_1x_1\ldots v_\ell w_\ell x_\ell y$$,

such that $$∀i 1≤\forall i~1\le$$ '# of marks in $$vixiv_ix_i$$'$$≤pkℓ\le pk\ell$$ and for every sequence $$(ij)mj=1(i_j)_{j=1}^m$$,

in case (i) the word $$uvi1…vimwxim…xi1y∈Luv_{i_1}\ldots v_{i_m}wx_{i_m}\ldots x_{i_1}y\in L$$, and

in case (ii) the word $$uvi11w1xi11…viℓℓwℓxiℓℓy∈Luv_1^{i_1}w_1x_1^{i_1}\ldots v_\ell^{i_\ell}w_\ell x_\ell^{i_\ell}y\in L$$.

Proof: Take the derivation tree generating $$ss$$. Call a non-terminal recurring if it appears again under itself in the derivation tree. By expanding the rule set, we can suppose that all non-terminal symbols are recurring in the derivation tree. (This is to be understood that we might have eliminated their recurrence; this doesn't matter, the point is that they can be pumped.) There are at least $$pkℓpk\ell$$ leaves that correspond to a marked position. We look at the nodes where two marked letters split. There are at least $$Ω(pkℓ)\Omega(pk\ell)$$ such nodes. By the pigeonhole principle, at least $$Ω(pkℓ)\Omega(pk\ell)$$ correspond to the same non-terminal. Using Dilworth, $$Ω(√pk)\Omega(\sqrt{p}k)$$ of them are in a chain or $$Ω(√pℓ)\Omega(\sqrt{p}\ell)$$ are in an antichain, giving cases (i) and (ii), respectively, if $$pp$$ is large enough.

Now we are ready to state a big combination lemma.

Super Lemma. For every context-free language $$LL$$ there is a $$pp$$ such that for any $$k,ℓk,\ell$$ marking the same at least $$pkℓpk\ell$$ positions in $$N≥max(pn2k+2,pn3ℓ+1)N\ge \max(pn^{2k+2},pn^{3\ell+1})$$ words $$s1,…,sN∈Ls_1,\ldots,s_N\in L$$, each of length $$nn$$, there are two words, $$ss$$ and $$s′s'$$, that can be written as $$s=uv1…vkwxk…x1ys=uv_1\ldots v_kwx_k\ldots x_1y$$ and $$s′=u′v′1…v′kw′x′k…x′1y′s'=u'v_1'\ldots v_k'w'x_k'\ldots x_1'y'$$ OR as $$s=uv1w1x1…vℓwℓxℓys=uv_1w_1x_1\ldots v_\ell w_\ell x_\ell y$$ and $$s′=u′v′1w′1x′1…v′ℓw′ℓx′ℓy′s'=u'v_1'w_1'x_1'\ldots v_\ell'w_\ell'x_\ell'y'$$ such that the respective lengths of the subwords are all the same, i.e., $$|u|=|u′||u|=|u'|$$, $$|vi|=|v′i||v_i|=|v_i'|$$, etc., and $$∀i\forall i$$ $$vixiv_ix_i$$ contains a mark, and for every sequence $$(ij)mj=1(i_j)_{j=1}^m$$ the word $$ˉuˉvi1…ˉvimˉwˉxim…ˉxi1ˉy∈L\bar u\bar v_{i_1}\ldots \bar v_{i_m}\bar w\bar x_{i_m}\ldots \bar x_{i_1} \bar y\in L$$ OR $$ˉuˉvi11ˉw1ˉxi11…ˉviℓℓˉwℓˉxiℓℓˉy∈L\bar u\bar v_1^{i_1}\bar w_1\bar x_1^{i_1}\ldots \bar v_\ell^{i_\ell}\bar w_\ell \bar x_\ell^{i_\ell}\bar y \in L$$, respectively, where $$ˉz\bar z$$ stands for $$zz$$ or $$z′z'$$, i.e., we can freely mix the intermediate subwords from $$ss$$ and $$s′s'$$, but only such that $$ˉu=u\bar u=u$$ if and only if $$ˉy=y\bar y=y$$ etc.

Proof sketch of Super Lemma: Apply the Dilworth Ogden's Lemma for each $$sis_i$$. There are $$(n+2k+22k+2)\binom {n+2k+2}{2k+2}$$ and $$(n+3ℓ+13ℓ+1)\binom {n+3\ell+1}{3\ell+1}$$ possible options, respectively, where the boundaries between the subwords of $$sis_i$$ can be. There are a constant number of non-terminals in the language, thus by the pigeonhole principle, if $$pp$$ is large enough, the same non-terminal is pumped in the $$kk$$/$$ℓ\ell$$ rules for at least two words, $$ss$$ and $$s′s'$$, that also have the same subword boundaries.

Unfortunately, the number $$NN$$ that comes from this lemma is too large for our application. We can, however, decrease it by demanding fewer coincidences among the subwords. Now we state the lemma that we'll use.

Special Lemma. For every context-free language $$LL$$ there is a $$pp$$ such that for any $$kk$$ marking the same at least $$pkpk$$ positions in $$N=pkn2N=pkn^2$$ words $$s1,…,sN∈Ls_1,\ldots,s_N\in L$$, each of length $$nn$$, there are two words, $$ss$$ and $$s′s'$$, that can be written either as $$s=uv1…vkwxk…x1ys=uv_1\ldots v_kwx_k\ldots x_1y$$ and $$s′=u′v′1…v′kw′x′k…x′1y′s'=u'v_1'\ldots v_k'w'x_k'\ldots x_1'y'$$ such that either

case (i): $$∃i such that $$|xi|=|x′i|=0|x_i|=|x_i'|=0$$, $$|uv1…vi−1|=|u′v′1…v′i−1||uv_1\ldots v_{i-1}|=|u'v_1'\ldots v_{i-1}'|$$, and $$|vi|=|v′i||v_i|=|v_i'|$$ (i.e., the two latter conditions mean that the position of $$viv_i$$ is the same as the position of $$v′iv_i'$$), OR

case (ii): $$∀i $$|xi|≥1|x_i|\ge 1$$, $$|x′i|≥1|x_i'|\ge 1$$, $$|uv1…vk−1|=|u′v′1…v′k−1||uv_1\ldots v_{k-1}|=|u'v_1'\ldots v_{k-1}'|$$ and $$|vkwxk|=|v′kw′x′k||v_kwx_k|=|v_k'w'x_k'|$$ (i.e., these two conditions mean that the position of $$vkwxkv_kwx_k$$ is the same as the position of $$v′kw′x′kv_k'w'x_k'$$),

and (for both cases) $$∀i\forall i$$ $$vixiv_ix_i$$ contains a mark, and for every sequence $$(ij)mj=1(i_j)_{j=1}^m$$ the word $$ˉuˉvi1…ˉvimˉwˉxim…ˉxi1ˉy∈L\bar u\bar v_{i_1}\ldots \bar v_{i_m}\bar w\bar x_{i_m}\ldots \bar x_{i_1}\bar y\in L$$, where $$ˉz\bar z$$ stands for $$zz$$ or $$z′z'$$, i.e., we can freely mix the intermediate subwords from $$ss$$ and $$s′s'$$, but only such that $$ˉu=u\bar u=u$$ if and only if $$ˉy=y\bar y=y$$ etc., OR

case (iii): $$ss$$ and $$s′s'$$ can be written as $$s=uv1w1x1v2w2x2ys=uv_1w_1x_1v_2w_2x_2y$$ and $$s′=u′v′1w′1x′1v′2w′2x′2y′s'=u'v_1'w_1'x_1'v_2'w_2'x_2'y'$$ such that $$|u|=|u′||u|=|u'|$$ and $$|v1w1x1|=|v′1w′1x′1||v_1w_1x_1|=|v_1'w_1'x_1'|$$, and $$∀i\forall i$$ $$vixiv_ix_i$$ contains a mark, and $$uvh1w1xh1v2w2x2y∈Luv_1^hw_1x_1^hv_2w_2x_2y\in L$$ and $$u′vh1w1xh1v′2w′2x′2y′∈Lu'v_1^hw_1x_1^hv_2'w_2'x_2'y'\in L$$.

The proof only differs from the Super Lemma's that there are $$k(n2)k\binom n2$$ possible options for a word in case (i), which leaves $$(n2)\binom n2$$ options for case (ii), while in case (iii) there are $$(n2)\binom n2$$ options.

Corollary. If for every $$pp$$ there are $$tt$$ and $$nn$$ with $$n≥p(t+1)+tn\ge p(t+1)+t$$ such that there are $$N=n3N=n^3$$ words of length $$nn$$ in a context-free language $$LL$$ whose first $$p(t+1)p(t+1)$$ letters are the same for each word, and their last $$tt$$ letters are different for each pair or words (i.e., the words look like $$si=sbegismidisendis_i=s_i^{beg}s_i^{mid}s_i^{end}$$ such that $$|sbegi|=p(t+1)|s_i^{beg}|=p(t+1)$$, $$|smidi|=n−p(t+1)−t|s_i^{mid}|=n-p(t+1)-t$$, $$|sendi|=t|s_i^{end}|=t$$, and $$∀i≠j sbegi=sbegj\forall i\ne j~s_i^{beg}= s_j^{beg}$$ and $$sendi≠sendjs_i^{end}\ne s_j^{end}$$), then there is a $$BB$$ such that there are infinitely many pairs of words $$ah≠bh∈La_h\ne b_h\in L$$ of equal length that differ only in their last $$BB$$ letters.

Proof: Take $$k=t+1k=t+1$$ and apply the Special Lemma for our $$NN$$ words using $$N=n3≥p(t+1)n2N=n^3\ge p(t+1)n^2$$, marking the first $$p(t+1)p(t+1)$$ letters (that are the same in every word) to obtain $$s=uv1…vt+1wxt+1…x1ys=uv_1\ldots v_{t+1}wx_{t+1}\ldots x_1y$$ and $$s′=u′v′1…v′t+1w′x′t+1…x′1y′s'=u'v_1'\ldots v_{t+1}'w'x_{t+1}'\ldots x_1'y'$$ OR $$s=uv1w1x1v2w2x2ys=uv_1w_1x_1v_2w_2x_2y$$ and $$s′=u′v′1w′1x′1v′2w′2x′2y′s'=u'v_1'w_1'x_1'v_2'w_2'x_2'y'$$.

If we are in case (i) of the Special Lemma, i.e., there is an $$ii$$ such that $$|xi|=|x′i|=0|x_i|=|x_i'|=0$$, $$|uv1…vi−1|=|u′v′1…v′i−1||uv_1\ldots v_{i-1}|=|u'v_1'\ldots v_{i-1}'|$$, and $$|vi|=|v′i||v_i|=|v_i'|$$, then $$uv1…vi−1=u′v′1…v′i−1uv_1\ldots v_{i-1}=u'v_1'\ldots v_{i-1}'$$ and $$vi=v′iv_i=v_i'$$ also hold, as $$vi+1wxi+1v_{i+1}wx_{i+1}$$ needs to contain a marked letter, thus the subwords preceding $$vi+1v_{i+1}$$ consist of only marked letters, and these are the same in $$ss$$ and $$s′s'$$. We can take the words $$ah=uv1…vhi…vtwxt…x1ya_h=uv_1\ldots v_i^h\ldots v_twx_t\ldots x_1y$$ and $$bh=uv1…vhiv′i+1…v′tw′x′t…x′1y′b_h=uv_1\ldots v_i^hv_{i+1}'\ldots v_t'w'x_t'\ldots x_1'y'$$ to obtain the desired pairs; since these words end the same way as $$ss$$ and $$s′s'$$, $$ah≠bha_h\ne b_h$$ and they differ only in their last bounded many letters.

If we are in case (ii) of the Special Lemma, i.e., $$∀i $$|xi|≥1|x_i|\ge 1$$, $$|x′i|≥1|x_i'|\ge 1$$, $$|uv1…vt|=|u′v′1…v′t||uv_1\ldots v_t|=|u'v_1'\ldots v_t'|$$ and $$|vt+1wxt+1|=|v′t+1w′x′t+1||v_{t+1}wx_{t+1}|=|v_{t+1}'w'x_{t+1}'|$$, then $$uv1…vt=u′v′1…v′tuv_1\ldots v_t=u'v_1'\ldots v_t'$$ also holds, similarly as in the previous case. Now we can take $$ah=uv1…vtvht+1wxht+1xt…x1ya_h=uv_1\ldots v_tv_{t+1}^{h}wx_{t+1}^{h}x_t\ldots x_1y$$ and $$bh=uv1…vtvht+1wxht+1x′t…x′1y′b_h=uv_1\ldots v_tv_{t+1}^{h}wx_{t+1}^{h}x_t'\ldots x_1'y'$$; since $$|xt…x1y|=|x′t…x′1y′|≥t|x_t\ldots x_1y|=|x_t'\ldots x_1'y'|\ge t$$, these words certainly end differently and can differ only in their last bounded many letters. (Note that this is the only place where we really need that we can pump one word with a subword of the other one.)

If we are in case (iii) of the Special Lemma, i.e., $$s=uv1w1x1v2w2x2ys=uv_1w_1x_1v_2w_2x_2y$$ and $$s′=u′v′1w′1x′1v′2w′2x′2y′s'=u'v_1'w_1'x_1'v_2'w_2'x_2'y'$$ such that $$|u|=|u′||u|=|u'|$$ and $$|v1w1x1|=|v′1w′1x′1||v_1w_1x_1|=|v_1'w_1'x_1'|$$, then $$u=u′u=u'$$ and $$v1w1x1=v′1w′1x′1v_1w_1x_1=v_1'w_1'x_1'$$ also hold, similarly as in the previous cases. Now we can take $$ah=uvh1w1xh1v2w2x2y∈La_h=uv_1^hw_1x_1^hv_2w_2x_2y\in L$$ and $$bh=uvh1w1xh1v′2w′2x′2y′∈Lb_h=uv_1^hw_1x_1^hv_2'w_2'x_2'y'\in L$$; since $$v2v_2$$ contains a marked letter, $$|v2w2x2y|≥t|v_2w_2x_2y|\ge t$$, thus these words certainly end differently and can differ only in their last bounded many letters.

This finishes the proof of the Corollary. Now let's see how to prove the original question from the Corollary.

Final proof. First we show that the condition of the Corollary is satisfied for every integer valued polynomial $$ff$$. Set $$t=p−1t=p-1$$ and $$n=Cpn=C^p$$ for some large enough $$C=C(f)C=C(f)$$. The plan is to take some numbers $$x1,…,x2Nx_1,\ldots,x_{2N}$$ (where $$N=2n3N=2n^3$$) for which $$f(xi)≠f(xj)f(x_i)\ne f(x_j)$$, and then add some sufficiently large number $$zz$$ to each of them to obtain the desired words $$si=f(xi+z)s_i=f(x_i+z)$$. If the degree of $$ff$$ is $$dd$$, then at most $$dd$$ numbers can take the same value, thus we can select $$x1,…,x2Nx_1,\ldots,x_{2N}$$ from the first $$2dN2dN$$ numbers, which means that they have $$log(dn)\log(dn)$$ digits. In this case $$f(xi)=O((dN)d)f(x_i)=O((dN)^d)$$, thus each $$f(xi)f(x_i)$$ will have at most $$dlogN+O(1)=O(logn)d\log N + O(1)=O(\log n)$$ digits. If we pick $$zz$$ to be an $$n/dn/d$$ digit number, then $$f(z)f(z)$$ will have $$nn$$ digits, and for each $$f(xi+z)f(x_i+z)$$ only the last $$O(logn)O(\log n)$$ digits can differ. The $$f(xi+z)f(x_i+z)$$ will all have $$nn$$ or $$n+1n+1$$ digits, thus at least half, i.e., $$NN$$ of them have the same length; these will be the $$sis_i$$.

From the conclusion of the Corollary we obtain infinitely many pairs of numbers $$aha_h$$ and $$bhb_h$$, such that $$|ah−bh|≤2B|a_h-b_h|\le 2^B$$, which is clearly impossible for polynomials.

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