Definition 3.0.1. Define a function f2 fromDn+1,k+1intoDn+1,k by the following:
1. The set Dn+1,k+1 consists of all n + 1-Dyck paths with k + 1 flaws, 0≤k≤n. Each path in Dn+1,k+1 can be factorized into AdBuC, where A is a subpath all above the x-xais, d is the first down step below the x-xais, B is a subpath all bellow the x-xais, say with k2 flaws, 0≤k2≤k, u is the first up step contact the x-axis after B and below the x-axis, say uBd with k2 + 1 flaws, and C is the remaining path with k− k2 flaws, 0≤k≤n.
Figure 3.1: f (AdBuC) = BuAdC
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The function f2 is one-to-one.
Proof. Let|B| = 2i, 0≤i≤n; |u| = 1; |A| = 2j, 0≤j≤n − i; |d| = 1; |C| = 2(n − i − j − 1).
To see Figure 3.3 when we start on (0, 0) to walk along the two path BuAdC and B′uA′dC′. The (2i + 1)st step of BuAdC is above the x-axis, but the (2i + 1)st step of B′uA′dC′ is still below the x-axis. This is a contradiction as two paths are the same.
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To see Figure 3.5 when we start on (0, 0) to walk along the path the two path BuAdC and B′uA′dC′. The (2r + 1)ststep of BuAdC is below the x-axis, but the (2r + 1)st step of B′uA′dC′ is still above the x-axis. This is a contradiction as two paths are the same.
Thus, we have proof that i = r.
∴ |B| = |B′| and B = B′.
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y = 1. This is a contradiction as two paths are the same.‧
y = 1. This is a contradiction as two paths are the same. Thus, we have proof that j = s.∴ |A| = |A′| and A = A′. Since B = B′and A = A′,
∵ BuAdC = B′uA′dC′ ⇒ C = C′
∴ AdBuC = A′dB′uC′ Therefore f2 is one-to-one.
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flaw will be restored and connected to original lift behind.Note: A up-step above the x-axis is called lift. The lifts are counted from right to left and top to bottom.
Proof. The function f2 from Dn+1,k+1into Dn+1,k. Suppose P has k + 1 flaws inDn+1,k+1, and Q = AdBuC, where A, dBu, C have 0 flaw, k + 1− j flaws, j flaws, respectively.
(i.e.A, dBu, C have n− k − i lifts, 0 lift, i lifts, respectively.) The (k + 1)th flaw is in subpath dBu.Then f2(Q) has k flaws on Dn+1,k, and f2(Q) = BuAdC, where B, uAb, C have k− j flaws, 0 flaw, j flaws, respectively.
A
Figure 3.10: the new lif t will be connected to original lif t behind
Notice that the path f2(Q), u is connected to the (n− k)thflaw in subpath A behind. u becomes the (n− k + 1)st lift.
Therefore, no matter how many times we use f2, u is conneted to the (n− k + 1)stlift in subpath A behind. u is still the (n− k + 1)st lift.
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Figure 3.11: Rising f rom; F alling to; F alling f rom; Rising to
Lemma 3.0.3. If P is a path inDn+1,k, then f2−1(P ) has k + 1 flaws.Moreover, the (n− k + 1)st lift in P will be dropped.
i.e. The (n− k)th lift u and the first down-step d falling to the x-axis on the right side of u inDn+1,k are dropped to the first up-step u rising to the x-axis and the first down-step d falling from the x-axis inDn+1,k+1.
Show by formula:
P reimage under f2
−−−−−−−−−−−−−−−→
Figure 3.12: BuAdC P reimage under f2
−−−−−−−−−−−→ AdBuC
In P , the (n − k + 1)st lift u and the first down-step d falling to the x-axis on the left side of u inDn+1,k, we can observe that there is an empty area enclosed by the u, d, the x-axis, and the horizontal line y = 1. After switching two portions Bu and Ad, another empty area is enclosed by the d is the first down-step falling from the x-axis, and u is the first up-step rising to the x-axis, and the horizontal line y = −1 in f2−1(P ). And the remaining segments which are
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The function f2 is onto.
Proof. Claim: f2 is onto. (f2:Dn+1,k+1 → Dn+1,k)
For any path P = BuAdC in Dn+1,k which 0≤k≤n. We choose the (n− k + 1)st lift u and choose the first down-step d falling to the x-axis on the right side of u. We switch the portions Bu and Ad then we can drop the (n− k + 1)stlift (i.e. the (k + 1)stflaw) and get a new path Q = AdBuC inDn+1,k+1, by Lemma 3.2、Lemma 3.3. Where Q has at most n + 1 flaws and P has at least one flaw. In fact, if P has k flaws then Q has k + 1 flaws.
So every P inDn+1,k, we can find a path Q inDn+1,k+1, such that f2(Q) = P . Therefore f2is one-to one and onto.
Hence, f2is one-to-one and onto by Theorem 3.1 and Theorem 3.2.
Note: Let f2−1 be the inverse function of f2.
Definition 3.0.4. The set X2consists of all paths in n + 1-Dyck paths with k + 1 flaws.Each path in X2 can be factorized into A−→
d B−→u C. The set Y2 consists of all n + 1-Dyck paths which are good paths.
Define a function g2 from X2into Y2 by the following:
1. g2(Q) = f2k+1(Q) where Q in X2, and f2k+1 = f|2◦ f2{z◦ · · · ◦ f}2 k+1 times
2. The first down-step falling from x-axis denote by−→
d and the first up-step rising to x-axis denote by −→u .
Lemma 3.0.5. Suppose that Q in X2. In f2(Q), the first down-step−→
d falling from the x-axis of Q connects with the first up-step −→u rising to the x-axis of Q to be −→u−→
d is above the x-axis.
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Theorem 3.0.3. The function g2:X2→Y2is one-to-one.
Proof. Suppose that g2(P ) = g2(Q), where P has k flaws and Q has h flaws in X2.
For the same reason, the following can be obtained f2(f2k−1(P )) = f2(f2h−1(Q))⇒ f2k−1(P ) = f2h−1(Q) Since f2 is one-to-one, use this method k + 1 times f2(P ) = f2(f2h−k(Q))⇒ P = f2h−k(Q)
For the same reason, the following can be obtained f2(f2k−1(P )) = f2(f2h−1(Q))⇒ f2k−1(P ) = f2h−1(Q) Since f2 is one-to-one, use this method h + 1 times f2(f2k−h(P )) = f2(Q)⇒ f2k−h(P ) = Q Therefore, g2 is one-to-one.
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Theorem 3.0.4. g2:X2→Y2 is onto, where Y2 is good path of n + 1-Dyck and g2(Q) = f2k+1(Q).
Proof. Give P ∈ Y2, P has 0 flaw, and −→u−→
d is the (n− k + 1)st lift in P .
Definition:The preimage of the function g2= f2−(k+1) = f|2−1◦ f2−1{z◦ · · · ◦ f2−1}
k+1 times
Since f2−1(P ) is the preimage of P under f2and has 1 flaw.
Using this way for k times, we get the path f2−k(P ) which has k flaws.
In f2−k(P ), −→u−→
d is still above the x-axis and is the (n− k + 1)st lift.We use this way again, we get the path f2−(k+1)(p) which has k + 1 flaws and−→
d B−→u is under the x-axis, as f2 is onto and by Lemma 3.5.
So we have f2−(k+1)(P ) = f2−(k+1)(f2k+1(Q)) = Q, where Q has k + 1 flaws,Q∈ X2.
Let Q = f2−(k+1)(P ) ⇒ g2(Q) = f2k+1(Q)
= f2k+1((f2−(k+1))(P ))
= P Thus g2is onto.
Hence, g2 is one-to-one and onto by Theorem 3.3 and Theorem 3.4.
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Definition 3.0.6. The set Z2 contains the good path for all n-Dyck paths which replaces
−
→u−→
d in Y2with a dot mark, and all paths in Y2 are n + 1-Dyck paths which are good path. Let h2be the function from Y2 into Z2.
i.e. P = R−→u−→
d S is (2n + 2, 0) path, where R, −→u−→
d , and S are all good paths. h2(P ) = h2(R−→u−→
d S) = R• S
Theorem 3.0.5. h2 is one-to-one and onto.
Proof. It’s clearly that h2 is one-to-one.
Given P′ = R• S ∈ Z2
We can change• into −→u−→
d . Thus R• S ⇒ R−→u−→
d S ∈ Y2. Therefore, h2is one-to-one and onto.
R ⃗u d⃗
S R S
x-axis h2
Figure 3.13: h2(R⃗u ⃗dS) = R• S
We have completed the combinatorial proof of (n + 2)Cn+1 = (4n + 2)Cn.
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Chapter 4 Summary
• In this thesis, we prove the Catalan identity in combinatorial way.
In Chapter 2, we give a bijective proof between ‵‵P aths Start with U p − step′′ and
‵‵Dotted T otally Bad P aths′′. Then we construct the functions in X1 −→ Yg1 1 h1
−→ Z1
that Paths Start with Up-step.
In Chapter 3, we give a bijective proof between‵‵P aths Start with Down− step′′ and
‵‵Dotted Good P aths′′.Then we construct the functions in X2 −→ Yg2 2 h2
−→ Z2that Paths Start with Down-step.
Proving the Catalan identity through the Dyck paths can reveal the following advantages:
1.The subpath C does not change during the process of switching of the portions Ad and Bu.
2.Since the subpath B of P in x1 is empty, a new flaw generated after
switching of the portions Ad and Bu must be followed by the original subpath C.
Since the subpath A of Q in x2 is empty, a new lift generated after
switching of the portions Bu and Ad must be followed by the original subpath C.
3.In the process of computing the preimage of a function g1(g2), the flaws (lifts) recovery mode follows the‵‵Last− in − F irst − out′′or‵‵F irst− in − Last − out′′.
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Appendix A
examples of Catalan identity
A.1 (n + 2)C
n+1= (4n + 2)C
n• n = 1
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• n = 2
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• n = 3
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