Talk:Friedman's SSCG function

Proof?

Can someone corroborate the reference given for "SSCG(3) is not only larger than TREE(3), it is much, much larger than TREE(TREE(…TREE(3)…))"? The reference provided, https://cp4space.wordpress.com/2013/01/13/graph-minors/ is not a formal mathematical paper. It's credited to wordpress user "apgoucher": https://mathoverflow.net/users/39521/adam-p-goucher.Kemery72 (talk) 07:38, 9 October 2017 (UTC)[reply]

Friedman? Which friedman... not THE 'Harvey Friedman'?

92.0.246.54 (talk) 22:26, 14 September 2019 (UTC)[reply]

It's Harvey Friedman, from Ohio State. 67.198.37.16 (talk) 20:54, 5 March 2024 (UTC)[reply]

SSCG(4) vs TREE^SSCG(3)(3) - Which is bigger?

SSCG(0) = 2

TREE^SSCG(0)(3) = TREE(TREE(3))

SSCG(1) = 5

TREE^SSCG(1)(3) = TREE(TREE(TREE(TREE(TREE(3)))))

SSCG(2) < TREE(3)

SSCG(3) > TREE^TREE(3)(3)

So, what happens, if you compare SSCG(4) with TREE^SSCG(3)(3)? — Preceding unsigned comment added by 84.151.250.207 (talk) 19:33, 25 April 2020 (UTC)[reply]

SCG-function vs Loader's number

What kind of SCG(n) would be about as big as Loader's number?

Numerical values?

The values presented for SSCG(2) (without reference) may not be correct. Correct me if I am wrong but when I do modulo arithmetic I find that the final digit should be 0, not 8. And when I compute the decimal approximation by calculating the exponent using extended precision floats and then converting to a base-10 logarithm, the integer part of the exponent appears to end with the digits 65, not 66 (the mantissa seems correct). Where did these numbers come from? Should they even appear on this page without a reliable external reference? Was this original research? vttoth (talk) 05:19, 31 July 2022 (UTC)[reply]

The current text shows the final digit to be 0. The exponent now ends with 65. Edit history shows that the fixes were done on 16:22, 29 April 2023‎ and also in June, by User:Kwékwlos. Both results remains uncited. 67.198.37.16 (talk) 21:04, 5 March 2024 (UTC)[reply]

So is someone going to clarify what fε2*2 means?

2601:58B:4204:B6B0:89C7:F2B0:4321:F034 (talk) 21:23, 16 March 2024 (UTC)[reply]

As far as I am aware the original source of the claim is this blog post.

f_{\varepsilon _{2}2}

is a function in the fast-growing hierarchy, but with a different system of fundamental sequences than any on Wikipedia. There is a definition of the system of fundamental sequences in the blog post, but I think it is not well-defined, as there's not an order-preserving bijection from subcubic graphs under the graph minor relation to the ordinals with their usual order. C7XWiki (talk) 03:21, 21 March 2024 (UTC)[reply]

SSCG(3) vs extended TREE recursion

k(1) = TREE(1), k(2) = TREE(TREE(TREE(2))), k(3) = TREE^TREE(3)(3), k(n) = TREE^TREE(n)(n)

Then, how does SSCG(3) compare to k^k(3)(3)? 2A00:6020:A123:8B00:D1CF:4C5D:D7F1:5F8D (talk) 16:31, 10 October 2024 (UTC)[reply]

Who got rid of the mention of the TREE(3) nesting of the TREE sequence in the article? 174.103.211.175 (talk) 02:12, 2 March 2025 (UTC)[reply]

In Orders of magnitude (numbers), it was Ovinus (https://en.wikipedia.org/w/index.php?title=Orders_of_magnitude_%28numbers%29&diff=1113008463&oldid=1113007655).

In Friedman's SSCG function, SimpleSubCubicGraph wrote the confusing description: "...TREE^TREE(3)(3), that is, you have TREE(3) different unique nodes." (https://en.wikipedia.org/w/index.php?title=Friedman%27s_SSCG_function&diff=1273197541&oldid=1272928933), since TREE(3) different nodes would be TREE(TREE(3)), while TREE^TREE(3)(3) is the much longer TREE nesting TREE(TREE(...TREE(3)...)) with TREE(3) iterations.

But, you just wrote a second question, instead of the answer to the original question, which is why I now included a Tier-2 counter, m(x) = k^k(x)(x).

The original question was about k^k(3)(3) vs SSCG(3), where k(n) is the TREE nesting counter from https://news.ycombinator.com/item?id=16239690.

The TREE nesting counter, k(n), goes like this.: k(1) = TREE(1), k(2) = TREE(TREE(TREE(2))) = TREE(TREE(3)), k(3) = TREE^TREE(3)(3), k(n) = TREE^TREE(n)(n)

The Tier-2 counter, m(x), goes like this.: m(1) = k(1) = TREE(1), m(2) = k^k(2)(2) = k^{TREE(TREE(3))}(2), m(3) = k^k(3)(3), m(x) = k^k(x)(x).

Then, how does SSCG(3) compare to k^k(3)(3) and m^m(3)(3)? 2A00:6020:A123:8B00:2195:A85:C7D2:35ED (talk) 21:44, 8 March 2025 (UTC)[reply]