Count with PSS expressions of limited length

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PsiCubed2
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Count with PSS expressions of limited length

Postby PsiCubed2 » Mon May 27, 2019 12:39 pm UTC

This is a hybrid game that combines a counting thread with a "my number is bigger" game.

The rules (concise version):
(1) We write numbers in PSS notation. Don't worry if you have no idea what that is. We'll learn it as we go. For now, you
just need to know that (0,0) adds 1 to the number after it.
(2) We start with (0,0)[10] = 11
(3) The final number in our expressions will always be 10 (for example: (0,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10] = 16)
(4) No double posts
(5) Never use more than 9 (a,b)'s in your expression.
(6) This is a counting thread. Do not skip numbers unless rule #5 forces you to.

The rules (full version):
Spoiler:
(1) We write numbers in PSS notation. Don't worry if you have no idea what that is. We'll learn it as we go. For now, you just need to know that (0,0) adds 1 to the number after it.
(2) We start with (0,0)[10] = 11
(3) The final number in our expressions will always be 10 (for example: (0,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10])
(4) No double posts.
(5) We only use strings of length 9 or less here. In other words, strings of the form:
(a1,b1)...((ak,bk)[10] with 1<=k<=9
(6) This is a counting thread. You are not allowed to skip numbers unless you are forced to by rule 5. For example, the proper reply to (0,0)(0,0)(0,0)[10]=13 is (0,0)(0,0)(0,0)(0,0)[10]=14.
(7) It is recommended (but not required) that you write the actual number along with the PSS expression. When the numbers get too large (which will take quite a while) we'll use approximations.
(8) Two notes for the experts (everybody else can ignore these for now):
(i) We're using the Hardy Hierarchy version of PSS
(ii) We only use standard sequences (so the proper response to 20 will be 22 rather than 21)


Of-course, if you have any questions about the rules, don't hesitate to ask.

I'll start:
(0,0)[10] = 11

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Fri May 31, 2019 10:36 am UTC

PsiCubed2 wrote:(1) We write numbers in PSS notation. Don't worry if you have no idea what that is. We'll learn it as we go.
Yeah, let's learn this. Also hi PsiCubed! :D
(0,0)(0,0)[10] = 12

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Fri May 31, 2019 2:15 pm UTC

Well, hello there. Long time, no see :)

(0,0)(0,0)(0,0)[10]=13

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Fri May 31, 2019 2:16 pm UTC

True, true.

(0,0)(0,0)(0,0)(0,0)[10]=14

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Fri May 31, 2019 2:22 pm UTC

(0,0)(0,0)(0,0)(0,0)(0,0)[10]=15

(Don't worry. Things will start getting interesting pretty soon)

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Fri May 31, 2019 2:36 pm UTC

Yeah I hope so. :)

(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10] = 16

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Fri May 31, 2019 2:38 pm UTC

(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10] = 17

If you're wondering why the length limit is so big (which makes things a bit boring in the beginning): That's to keep the numbers from exploding way too quickly later. Believe me, you will thank me for this when the time comes :)

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Fri May 31, 2019 2:53 pm UTC

I'll take your word for it.

(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10] = 18

I hope you give the next set of rules soon. :wink:

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Fri May 31, 2019 3:17 pm UTC

As much as I'd love to bore everyone forever, the length limit is 9, so I'm forced to give new rules after the current post. :wink:

(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10] = 19

Which is the maximum valid number with only zeros.

New rules:
1. If an expression ends with (0,0)(1,0)[n], erase the "(0,0)(1,0)" and double n.
2. The string "(0,0)(0,0)(1,0)" is taboo and may not be used.
3. Reminder: For ANY expression that ends with (0,0)[n], we erase the "0,0" and add 1 to n

Hint for what to do with the next entry (don't look if you want to figure it out yourself):
Spoiler:
The next expression should evaluate to exactly 20.
Last edited by PsiCubed2 on Fri May 31, 2019 3:32 pm UTC, edited 1 time in total.

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Fri May 31, 2019 3:23 pm UTC

Not looking cause it looks easy so far.

(0,0)(1,0)[10] = 20

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Fri May 31, 2019 3:32 pm UTC

(0,0)(1,0)(0,0)[10] = 22

And in case you wonder why (0,0)(1,0) results in doubling, here is the actual way the notation works:
Spoiler:
The actual rule for expressions ending with (0,0)(1,0) is:
1. Delete the "(1,0)"
2. Copy the "(0,0)" an additional n-1 times and append the copies just before the [n]

So (0,0)(1,0)[n] = (0,0)...(0,0)[n] with n (0,0)'s = n+1+...+1 with n 1's = n+n = 2n

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Re: Count with PSS expressions of limited length

Postby Sabrar » Fri May 31, 2019 5:21 pm UTC

(0,0)(1,0)(0,0)(0,0)[10] = 24

This notation will get us up to 320 176 I believe.

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sat Jun 01, 2019 6:39 pm UTC

Yup.

(0,0)(1,0)(0,0)(0,0)(0,0)[10]=26

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Sat Jun 01, 2019 7:03 pm UTC

Okay then.

(0,0)(1,0)(0,0)(0,0)(0,0)(0,0)[10] = 28

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sat Jun 01, 2019 7:43 pm UTC

(0,0)(1,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10]=30

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Sat Jun 01, 2019 7:54 pm UTC

(0,0)(1,0)(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10] = 32

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sat Jun 01, 2019 8:01 pm UTC

(0,0)(1,0)(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10]=34

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Sat Jun 01, 2019 8:09 pm UTC

(0,0)(1,0)(0,0)(1,0)[10] = 40

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sat Jun 01, 2019 8:40 pm UTC

(0,0)(1,0)(0,0)(1,0)(0,0)[10]=44

Notice that if we strip away the parens and the commas and the "[10]", the expressions we write here are ordered in a simple lexicographic order. So far we had:

00
0000
000000
00000000
0000000000
000000000000
00000000000000
0000000000000000
000000000000000000
0010
001000
00100000
0010000000
001000000000
00100000000000
0010000000000000
001000000000000000
00100010
0010001000

This pattern will continue all the way to the end (001122334455667788, which would correspond to a HUGE number if we ever reach that far).

Given this information, perhaps someone would like to venture a guess as to what expression will come after 176?
(don't bother trying to guess the actual value of that expression. I haven't given enough information to deduce that yet)

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Sat Jun 01, 2019 9:05 pm UTC

(0,0)(1,0)(0,0)(1,0)(0,0)(0,0)[10] = 48

(0,0)(1,0)(1,0)??

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sat Jun 01, 2019 9:13 pm UTC

Yup.

(0,0)(1,0)(0,0)(1,0)(0,0)(0,0)(0,0)[10]=52

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Sat Jun 01, 2019 9:36 pm UTC

Cool.

(0,0)(1,0)(0,0)(1,0)(0,0)(0,0)(0,0)(0,0)[10] = 56

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sat Jun 01, 2019 10:26 pm UTC

(0,0)(1,0)(0,0)(1,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10] = 60

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Re: Count with PSS expressions of limited length

Postby Sabrar » Sun Jun 02, 2019 3:58 am UTC

(0,0)(1,0)(0,0)(1,0)(0,0)(0,1)[10] = 80

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sun Jun 02, 2019 12:04 pm UTC

(0,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)[10] = 88 (miles per hour)

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Re: Count with PSS expressions of limited length

Postby Sabrar » Sun Jun 02, 2019 1:21 pm UTC

(0,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)(0,0)[10] = 96

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sun Jun 02, 2019 4:24 pm UTC

(0,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)(0,0)(0,0)[10] = 104

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Re: Count with PSS expressions of limited length

Postby Sabrar » Sun Jun 02, 2019 4:38 pm UTC

(0,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)(0,1)[10] = 160

So how about that new rule? :wink:

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sun Jun 02, 2019 5:17 pm UTC

(0,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)[10] = 176

Can't postpone it any longer, can I?

Alright. Here are the rules for ANY expression that ends with either (0,0)[n] or (1,0)[n]:

(1) If it ends with (0,0)[n] then we delete the (0,0) and add 1 to n.
(2) if it ends with (1,0)[n] then we:
(2-i) delete the final (1,0)[n] (but remember the value of n)
(2-ii) Copy everything from the last (0,0) in the resulting expression till the end, an additional n-1 times.
(2-iii) reappend the [n] at the end (without changing its value)
For example, to expand (0,0)(1,0)(2,0)(3,0)(0,0)(1,0)(2,0)(1,0)[3]:

(a) We see that it ends with a (1,0) so we'll follow rule 2.
(b) We delete the (1,0)[3] and remember "n=3"
(0,0)(1,0)(2,0)(3,0)(0,0)(1,0)(2,0)
(c) We look for the last (0,0) in there:
(0,0)(1,0)(2,0)(3,0)(0,0)(1,0)(2,0)
(d) For clarity, let's bold everything starting from that (0,0)
(0,0)(1,0)(2,0)(3,0)(0,0)(1,0)(2,0)
(e) Copy the bolded part two more times (since n is 3):
(0,0)(1,0)(2,0)(3,0)(0,0)(1,0)(2,0)(0,0)(1,0)(2,0)(0,0)(1,0)(2,0)
(f) Append the [3] at the end:
(0,0)(1,0)(2,0)(3,0)(0,0)(1,0)(2,0)(0,0)(1,0)(2,0)(0,0)(1,0)(2,0)[3]

Of-course, we don't yet know how to expand that (it ends in a (2,0)) , but I've chosen this example to illustrate that the above rules will continue to be valid even once we add things like (2,0) and beyond.

We also need a new rule for the forbidden substrings:

Let (a,b)n be shorthand for (a,b)...(a,b) (with n (a,b)'s)

Then the following substrings are forbidden:
(a) (0,0)(0,0)(1,0) (we've already mentioned that one)
(b) (0,0)(1,0)n(0,0)(1,0)m for m>n.

And that's it for now. The limit of this is already going to be a very big number (more info on exactly how big it is, will be given as we progress).

At any rate, you now have all the information needed to calculate the actual value of the next expression. For those who are just interested in the end result, here is the general formula for (0,0)(1,0)(1,0)[n] (if you want to find it on your own, don't peek!):
Spoiler:
(0,0)(1,0)(1,0)[n] = n*2n

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Re: Count with PSS expressions of limited length

Postby Sabrar » Sun Jun 02, 2019 5:49 pm UTC

So...

(0,0)(1,0)(1,0)[n] = {(0,0)(1,0)} repeated a total of n times + [n] = n * 2n

Are the substrings only forbidden because otherwise the lexicographic ordering would not hold up? So we can always look for the next number by appending to the end instead of the front?

(0,0)(1,0)(1,0)[10] = 10 * 210 = 10,240

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sun Jun 02, 2019 7:16 pm UTC

(0,0)(1,0)(1,0)(0,0)[10] = 11x211 = 22 528

Are the substrings only forbidden because otherwise the lexicographic ordering would not hold up?


That's the general idea, but the actual reason goes far deeper than that:

(1) The allowed strings (the "standard strings") are precisely those which can be arrived at from (0,0)(1,1)(2,2)...(k,k) for some k, if we allow ourselves to change the value of [n] as we wish at every step.

For example, (0,0)(1,0)(0,0)(0,0)(0,0)[10] is standard because we can choose k=1 and start with n=3:
(0,0)(1,1)[3] = (0,0)(1,0)(2,0)[3]
(change value of n to 2)
(0,0)(1,0)(2,0)[2] = (0,0)(1,0)(1,0)[2]
(keep value of n at 2)
(0,0)(1,0)(1,0)[2] = (0,0)(1,0)(0,0)(1,0)[2]
(change value of n to 3)
(0,0)(1,0)(0,0)(1,0)[3] = (0,0)(1,0)(0,0)(0,0)(0,0)[3]
(change value of n to 10)
(0,0)(1,0)(0,0)(0,0)(0,0)[10] - and we're done.

For a forbidden string like (0,0)(1,0)(0,0)(1,0)(1,0), the above cannot be done.

(2) What's ordered lexicographically are not the numbers (for a specific n such as 10), but the functions (for arbitrary n):

(0,0)[n]=n+1
(0,0)(0,0)[n]=n+2
.
.
(0,0)(0,0)...(0,0)[n] = n+k
.
(an infinity of these)
.
(0,0)(1,0)[n] = 2n
(0,0)(1,0)(0,0)[n] = 2n+2
.
.
(an infinity of these)
.
(0,0)(1,0)(0,0)(1,0)[n] = 4n
.
.
(an infinity of these)
.
(0,0)(1,0)(0,0)(1,0)(0,0)(1,0)[n] = 8n
.
(an infinity of these)
.

... (an infinity of these) ...

.

(0,0)(1,0)(1,0)[n] = n*2n
.
.

and so on.

As we go down the list, the functions always grow faster as n goes to infinity. But for a given n, this won't always be the case (n+1000000 is bigger than 2n for n=10, even though the latter function grows faster).

At any rate, what we have here is an hierarchy of functions which are ordered lexicographically. In fact, every function in this list can be given a unique ordinal (remember those?), and for this beautiful structure to hold we need to ignore the nonstandard strings.

(EDITED to fix some mistakes in the examples)

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Sabrar
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Re: Count with PSS expressions of limited length

Postby Sabrar » Sun Jun 02, 2019 7:38 pm UTC

Thanks, very nice explanation (as usual).

(0,0)(1,0)(1,0)(0,0)(0,0)[10] = 12 * 212 = 49,152

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sun Jun 02, 2019 7:39 pm UTC

(0,0)(1,0)(1,0)(0,0)(0,0)(0,0)[10]=13x213=106 496

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Re: Count with PSS expressions of limited length

Postby Sabrar » Sun Jun 02, 2019 7:47 pm UTC

(0,0)(1,0)(1,0)(0,0)(0,0)(0,0)(0,0)[10] = 14 * 214 = 229,376

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sun Jun 02, 2019 8:23 pm UTC

(0,0)(1,0)(1,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10] = 15x215 = 491 520 ≈ 4.9×105

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Re: Count with PSS expressions of limited length

Postby Sabrar » Sun Jun 02, 2019 8:43 pm UTC

(0,0)(1,0)(1,0)(0,0)(0,0)(0,0)(0,0)(0,0)(0,0)[10] = 16 * 216 = 220 ~ 106

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Sun Jun 02, 2019 9:26 pm UTC

I'll define this shorthand:

(0,0)(1,0)n → {n}

So then the next entry is:
(0,0)(1,0)(1,0)(0,0)(1,0)[10] = {2}{1}10 = {2}20 = 20×220 = 20 971 520 ≈ 2.1×107

(if this looks familiar and you're wondering where you've seen this before: The above shorthand is exactly equivalent to Emlightened's {n}-notation from the "My Number is Bigger" everymen thread! Beautiful, isn't it?)

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Re: Count with PSS expressions of limited length

Postby Sabrar » Mon Jun 03, 2019 5:06 am UTC

At first I was going to ask why you would need a shorthand if the sequence is limited to 9 but of course the expansion will be much larger.
Linking for my own benefit.

(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)[10] = {2}{1}{0}10 = {2}22 = 22 * 222 ~ 9,2 * 107

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Re: Count with PSS expressions of limited length

Postby PsiCubed2 » Mon Jun 03, 2019 2:00 pm UTC

(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)(0,0)[10] = {2}{1}{0}{0}10 = {2}{1}12 = {2}24=2×224 = 402 653 184 ≈ 4.0×108

(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)(0,0)[10] = {2}{1}{0}{0}10 = {2}{1}12 = {2}24=24×224 = 402 653 184 ≈ 4.0×108
Last edited by PsiCubed2 on Mon Jun 03, 2019 4:15 pm UTC, edited 1 time in total.

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Re: Count with PSS expressions of limited length

Postby Sabrar » Mon Jun 03, 2019 2:58 pm UTC

You missed a 4 somewhere in there. :)

(0,0)(1,0)(1,0)(0,0)(1,0)(0,0)(0,0)(0,0)[10] = {2}{1}13 = 26 * 226 ~ 1.745 * 109


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