We use the following slightly technical setup to encode the bounds we can use.Let be the closure of the set of all pairs such that there exists such that for all . Let be the interior of .
Finally, we need the following straightforward limit evaluation.
Proof.
We start by quantifying the “extra room” implicit in the strict inequality and the condition . More precisely, using 10 and the condition , which, in particular, implies , as well as the conditions , and , we deduce that there exist such that
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| | | (11) |
We choose implicitly to be sufficiently large as a function of and to satisfy the inequalities throughout the proof.Let , , . Note that by (11) we have and so . Similarly, and .Note that
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| | | | (12) |
where the last inequality holds as long as is sufficiently large as a function of and .
We will prove by induction on for fixed that if is a candidate with such that
| | | (13) |
then is -good. By (12) this implies the theorem.
If or then is trivially -good, implying, in particular the base case of our statement. Thus we move on to the induction step and assume .
We assume without loss of generality that for every , as otherwise we can replace by increasing the left side of (13) as
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and both terms on the right side of this identity increase after the replacement. Thus
| | | (14) |
for every For readers familiar with[CGMS23] we will indicate how the steps in our proof correspond to steps of their book algorithm. The observation in this paragraph corresponds to the degree regularization step of the book algorithm.
Next we use (13) to obtain a lower bound on by first upper bounding .If then contains a red or a blue as by 9 (3) and so is -good.Thus we may assume that
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| | | | (15) |
Our next goal is to show that most of the vertices of have blue degree at most as otherwise using 8 we can apply the induction hypothesis to the candidate where is a large blue book in guaranteed by8 and (3). This part of the proof corresponds to the big blue step.
Let
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with the first two being the parameters we will use in 8.Note that where is a constant depending only on and . As , by (3) we can choose large enough so that
| | | (16) |
Let . Suppose first that . Then by 8, contains a red or a blue book with and . In the first case is -good, and thus we may assume that the second case holds.As by (14), we have Thus
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where the last inequality holds by the choice of as
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Thus is -good by the induction hypothesis, implying that is -good.
Thus we may assume that . We now move on to the last part of the argument, corresponding to either the red step or the density increment step of the book algorithm, which are treated in the same way in our implementation.
Let By 7, we have
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As
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we have
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Thus there exists such that and
| | | (18) |
Let
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Note that by (14), and
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implying
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If and then
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This implies that is -good by the induction hypothesis, and so is -good.
Thus we may assume that either or , i.e.
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Symmetrically, we also have
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It follows from (19) that
| | | (20) |
The remainder of the proof is occupied with showing that (20) contradicts (11).As , , the function increases for and we can lower bound the left side of (20) by replacing by and by , implying
| | | (21) |
Note that
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and so
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where the last inequality holds whenever (and thus ) is large enough as a function of .It follows that (21) in turn implies i.e.
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contradicting (11).∎