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Deedlit11 once gave following lower bound: SCG(1)>forest4(29), where forest(n) (in here I’ll use F(n) for this function) is length of the longest sequence of cubic forests such that the ith forest has at most n + i vertices. I’m going to show that this bound is very weak by using certain graph element, namely loop on vertex with edge attached to it. I’ll define several functions which will help understanding size of SCG(1). For consistency, I’ll write Gn as graph allowed to have n vertices.

First of all, I wanted to give some elements special names, to improve clarity. Vertex with no edges connected will be called dot. Dot with around it will be called simply loop. Two dots with edge between them will be called stick. Two loops with edge between them will be called handle.

LPF(n)

LPF stands for “looped paths and forests”. Define looped path as path graph with loops at both end vertices. It’s clear that LPF(n)>F(n). Consider LPF(1). Start with a handle. In G3 one of loops in handle is deleted and loop is added. Now we can delete this loop and proceed with short sequence of cubic forests. We can easily check that F(1)=5. So at G8 we are left with handle without one loop. G9 will be 9 loops. As before, we delete 1 loop, proceed with F(1) forests and we are left with 8 loops. But now, when we delete one, we can start with forest with F(1) vertices, and we can make F(F(1)) forests. Continuing with all loops we have \(LPF(1) \geq F^9(1)\).

Now, for LPF(2) we start with handle with vertex in the middle if its edge. We call such handle with n vertices n-handle. Next, we replace it with 2 handles. We reduce first handle using method from previous paragraph. Second handle can be reduced to \(F^9(1)\) loops. Reducing loops we get \(F^{F^9(1)}(1)\) forests, so \(LPF(2) \geq F^{F^9(1)}(1)\). Similarly, LPF(3) will have one exponent more.

Define fast-growing hierarchy starting with function F(n) - \(F_\alpha (n)\) (where \(F_0(n)=F(n)\)). Our method gives bound of roughly \(F_2(n)\) (note that \(LPF(2)>F_2(2)\)). It shouldn’t be surprising that we can do better. When working on LPF(3) at some point we get 1-handle. If done right, this is about \(G LPF(1)\). Now, next graph will have (up to division by a constant) LPF(1) handles. So stack of exponents is at least LPF(1) high. So \(LPF(3)>F_3(3)\). When resolving LPF(4) we can get roughly \(F_4(4)\). So we can conclude that \(LPF(n)>F_n(n)=F_\omega(n)\) (I know it isn’t really true, because for n reaching LPF(1) argument fails. I did a bit more analysis  and this bound really seems to be true).

LF(n)

LF stands for “looped forests”. I define looped tree as tree allowed to have loop attached to leaf. So looped path is looped tree. Thus it’s obvious that \(LF(n) \geq LPF(n)\). Let’s have a look at LF(4). We start with star graph \(K_{1,3}\) with loops attached to every leaf. For G5 we replace one loop with stick. For G6 we remove that stick and we add disconnected handle. We proceed to G LPF(1). Our next graph will be star graph with one loop and one edge with LPF(1) sticks attached to it. We delete one stick and continue with LPF(1) graphs. We delete another stick and we can make LPF(LPF(1)) graphs. Repeat and end with LF(4)>\(LPF^{LPF(1)}(1)\). We can prove that LF doubly diagonalizes through LPF (at least!).

CLF(n)

This is going to be last function I define. CLF stands for “circles and looped forests”. Circle is closed cycle with no outgoing edges and no diagonals. n-circle is circle with n vertices. Loop isn’t considered circle, but 2 vertices with pair of edges is circle, just like triangle, square… As for CLF(1) all I can think of is 2-circle and then LF(2) graphs. Sequence for CLF(2) will start with triangle and we replace it with 2 2-circles. First of them we reduce as before to reach single 2-circle as G LF(2). Now we use looped forests to reach LF(LF(2)). When computing CLF(3) we start with square, then triangle and 2-circle, LF(2) steps later 2 2-circles, LF(LF(2)) steps later single 2-circle and we finish with \(LF^3(2)\). So \(CLF(n)\geq LF^n(2)\).

We’ll need number CLF(10) later, so I’ll try to explain it better. When we reach G LF(2) we have 9-circle to work with. But instead of using 8-circle and 2-circle we can have multiple 8-circles. We can have almost (up to division by a constant) LF(2) circles. Take a look at the last circle. We make 7-circle and 2-circle, then replace 2-circle with LF(2) forests. Now we replace 7-circle with LF(2) 6-circles, etc. At some step, we’ll have a lot of 8-, 6-, 4- and 2-circles. We’ll have approximately LF(2) 2-circles, so we can use earlier defined procedure to make \(LF^{LF(2)}(2)\) forests. Then, by reducing 4-circle, we have that many 2-circles. After reducing all of them, we have \(LF^{LF^{LF(2)}(2)}(2)\) forests. When we finish with 4-circles, there will be LF(2) exponents! If we continue, we’ll see that almost certainly CLF is at level \(\omega\) in fast-growing hierarchy based on LF. (again, I couldn’t prove it.)

OK, while writing this in MS Word above took one and a half page with font 11. This is probably where I should mention SCG again.

SCG(1)

Here I’ll just describe graphs G2-G9 obeying SCG definition such that later we can proceed with CLF(10)

  • G2 – 2 vertices with triple edge between them
  • G3 – 3 vertices a, b, c with 2 edges between a and b, 1 edge between b and c and loop at c. Let G’3 be same graph without loop at c
  • G4 – G’3 and loop
  • G5 – G’3 and stick
  • G6 – G’3 and 3 dots
  • G7 – G’3 and 2 dots
  • G8 – G’3 and 1 dot
  • G9 – G’3

And that’s it! From here you can proceed with circles, looped paths, looped trees, forests and looped forests!

Appendix

I mentioned in two places “up to division by a constant”. While dealing with integers of size like this division by 2 or 3 makes almost no difference. Problem starts when divisor and dividend become close to each other. I made analysis using more complex looped trees which I didn’t want to put here. I’m still not sure, but results seem appealing.

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