which has at least one flaws.
Define a function f2fromX2intoY2by the following:
1. Starting from the bottom left,(0, 0), follow the path until it first travels below the diagonal y =x.
2. Continue to follow the path until it touches the diagonaly = x again. Denote by ⇀n , the first such segment that touches the diagonaly =x, in fact,⇀n must be an north segment.
3. Swap the portion of the path before⇀n with portion after⇀n .
i.e. f2(TG⇀n Q) = Q⇀n TG,
whereT is an(i, i)totally bad path,0≤i ≤n+1, G is a(j+1, j)good path,0≤ j≤n−i, the north segment⇀n is the first north touching diagonal y = x, and Q is an(n−i−j, n−i−j) path.
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To show f2(TG⇀n Q) = Q⇀n TG by graph, we have:
Figure 3.1: TG⃗nQ
Fix⇀n and switch TG with Q, we have:
Figure 3.2: Q⃗nTG
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Theorem 3.2. LetP =TG⇀n Q is an(n+1, n+1)path. Define f2 : X2 −→Y2
by f2(TG⇀n Q) = Q⇀n TG, where T is an(i, i)totally bad path,0≤i ≤n+1, G is a(j+1, j) good path,0 ≤ j ≤ n−i, the north segment ⇀n is the first north which touches the diagonal y =x, and Q is an(n−i−j, n−i−j)path. The function f2is one-to-one and onto.
Proof. Claim: f2is one-to-one.
Let P = TG⇀n Q, P′ = T′G′⇀n Q′, whereG′ is a an (l+1, l) good path, T′ is a (k, k) totally bad path, andQ′ is an (n−k−l, n−k−l)path.
If f2(TG⇀n Q) = f2(T′G′⇀n Q′) ⇒Q⇀n TG= Q′⇀n T′G′. Claim: G =G′.
Case1: l > j
Figure 3.3: Q⃗nTG Figure 3.4: Q′⃗nT′G′
When we start on(n+1, n+1), trace back the path.
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Case2: l < j
Figure 3.5: Q′⃗nT′G′ Figure 3.6: Q⃗nTG
When we start on(n+1, n+1), trace back the path.
Ww let two paths both trace back to the point(n−l, n−l+1), in the next step, the path in Figure 3.5 is above the diagonal y=x, but the path in Figure 3.6 is still below diagonal y= x.
This is a contradiction,as two paths are the same.
Thus, we have proved thatl = j.
∴ G= G′. Claim: T =T′. Case1: i>k
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Figure 3.7: Q⃗nTG Figure 3.8: Q′⃗nT′G′
We both start on(n−j, n−j+1).
Sincei > k, when we let the path in Figure 3.8 trace to(n−j−k, n−j−k+1), the next segment is⇀n , but the path in Figure 3.7 is not⇀n .
This is a contradiction,as two paths are the same.
Case2: i<k
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This is a contradiction, as two paths are the same.
Thus, we have proved thati =k.
i.e. For any path in Y2, which has at least one flaw, we choose the last north leaving the diagonal y = x, denoted by ˆn, then we switch the portions before ˆn and after ˆn. We can get a new path with at mostn flaws which is in X2.
To show by graph:
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Figure 3.11: R ˆnS preimage under f2
−−−−−−−−−→S ˆnR
To show by formula:
Q =R ˆnS preimage under f2
−−−−−−−−−→ S ˆnR= P,
whereQ has at least one flaw, P has at most n flaws.
In fact, ifQ has k flaws, P has k−1 flaws.
So, for every pathQ in Y2, we can find a path P in X2such that f2(P) =Q.
Therefore, f2is one-to-one and onto.
Lemma 3.3. The first north⇀n touching the diagonal y= x in X2is above the diagonaly =x
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Lemma 3.4. The last north ˆn leaving from diagonal y = x in Y2is the first north ⇀n touching the diagonaly= x in X2.
i.e. If⇀n is the first north touching the diagonal y= x in X2, then⇀n is the last north leaving the diagonaly=x in Y2.
Figure 3.12: Lemma3.7 2nd part
Proof. Since ⇀n is the first north touching the diagonal y = x, we can observe that there is a empty area enclosed by the first east that leaves the diagonaly =x, denoted by⇀e , the diagonal y =x,⇀n , and the diagonal y =x−1.
After swapping two portions, another empty area is enclosed by⇀n , the diagonal y =x,⇀e , and the diagonaly=x+1, so that there is no north segment can touch the diagonal y=x between
⇀n and⇀e .
And the remain segments which are behind⇀e are at most touching the diagonal y = x but not be above the diagonaly = x. Therefore, the last north ˆn leaving from diagonal y = x in Y; is the first north⇀n touching the diagonal y= x in Y2.
Definition 3.5. The setA2consists of all paths which first segment is east, and the first touching the diagonaly= x north is marked.
The setB2consists of all paths which are totally bad path.
Defineg2fromA2intoB2byg2(P) = f(k)(P), whereP has k flaws, and f(k) =|f2◦ f2{z◦...◦ f}2 k
.
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with the first segment ofP in A2. And this part will not be separated afterward.Proof. First, we prove the first part.
Since using f2 will swap the portion berfore and after⇀n , and the first segment of path is east, after using f2, ⇀n connects with the east segment.
Thus, we have proved.
Then we prove the second part.
Notice that after using the first f2, the part of ⇀n and the east segment is above the diagonal y =x.
There is another first touching the diagonaly=x north, and the part is in a(j, j)path after that north, in the next step, we use f2 again, so this part will be swapped to before that north and since it is(j, j)path, the part is still above the diagonaly =x.
Therefore, no matter how many times you use f2,⇀n connects with the first segment of P in A2 and they are not be separated afterward.
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Thus,g2is onto. Therefore, g2is one-to-one and onto.Definition 3.9. The setC2consists of all(n, n)paths which are replaced the marked north and the next east segment inB2with a dot, and all paths inB2are(n+1, n+1)path.
Leth2be the function fromB2intoC2.
i.e. P =R⇀n⇀e S is an(n+1, n+1)path, whereR and S are all totally bad paths.
h2(P) =h2(R⇀n⇀e S) = R•S
Theorem 3.10. h2is one-to-one and onto.
Proof. It is clearly obvious thath2is one-to-one and onto.
GivenQ=R•S∈ C2. We can change•into⇀n⇀e . Thus,R•S⇒ R⇀n⇀e S∈ B2.
Therefore,h2is one-to-one and onto.
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Chapter 4 Summary
In this thesis, we prove the Catalan identity in a combinatorial way. We split the paths into two portions according to the first segment. Then we construct the functions inA1−→g1 B1
h1
−→C1
which the first segment is north.
Figure 4.1: A1 −→g1 B1−→h1 C1
And the other functions in A2 −→g2 B2 h2
−→ C2which the first segment is east.
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Figure 4.2: A2 −→g2 B2 h2
−→ C2
In chapter 3, we can also use reflection along the diagonal y = x to prove the paths with east segment, since the paths in chapter 3 is reflection along the diagonaly = x to the paths in chapter 2. But it will be less clear. In this thesis, we can obverse more details and easier to understand.