Discussion between Qwertiy and tuskiomi

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19:38

1

Javascript, more than 10^(16*2^2718281828459046) / 54^3 for(a=b=(Math.E+'').replace(".","");a--;b+=b);alert(b) Description: (Math.E+'') is "2.718281828459045" The dot is dropped, a and b are "2718281828459045" Loop executes 2718281828459045+1 = 2718281828459046 times On every iteration b (an...

tuskiomi

You're score would be 2718281828459045*2718281828459046, no?

Qwertiy

@tuskiomi, in b+=b i'm concatenating string from b with itself, so its length doubles.

tuskiomi

So the above number times 2.

Qwertiy

@tuskiomi, nope, the output is concatenated string, not its length. And that's not addition, that's concatenation. Try following program for(a=4,b=(Math.E+'').replace(".","");a--;b+=b)console.log(b‌);console.log(b) - it makes only 4 steps instead of 2718281828459045 and outputs b to the console on each step and the last value that is alerted in original code.

@tuskiomi, only with 4 iterations value becomes

271828182845904527182818284590452718281828459045271828182845‌90452718281828459‌045‌27182818284590452718‌28182845904527182818‌28459045271828182845‌9045271828‌1828459045‌27182818284590452718‌28182845904527182818‌28459045271828182845‌904‌52718281828459045‌2718281828459045

that is definitely larger than 2718281828459045*2718281828459046 = 7389056098930651665961070771070 :)

tuskiomi

so it's 16*(1+2+3+4 ... 2718281828459046).

Qwertiy

19:38

@tuskiomi why??

tuskiomi

because the string 2718281828459045 is 16 characters long, and the first time you print it out once, then twice, then three times, 1*16+2*16+3*16.... until it ends

simplify to 16*(1+2+3+4...)

Qwertiy

@tuskiomi nope

i print only resulting value

the first time b is

2718281828459045

then i concatenate it with itself and it becomes

27182818284590452718281828459045

now i'm again concatenating it with itself

with current value of it, not original

2718281828459045271828182845904527182818284590452718281828459045

and again:
27182818284590452718281828459045271828182845904527182818284590452718281828459045271828182845904527182818284590452718281828459045

and i'm doing so 2718281828459045+1 times

@tuskiomi

tuskiomi

so it's \sum_{n=0}^{2718281828459045} 16*2^{n}

Qwertiy

@tuskiomi why are you talking about the sum?

only the last value is printed

tuskiomi

only the last value is scored?

Qwertiy

19:46

@tuskiomi in my solution i'm outputting only the last value

its length is 16*2^(n+1)

tuskiomi

so that's more accurate?

Qwertiy

n+1 as the value is printed outside of the loop, so it was increased one more time

@tuskiomi ..^2718281828459045+1

tuskiomi

Qwertiy

@tuskiomi yep, that's the length

the number itself is somewhere near 10^(...)

tuskiomi

2↑10↑↑1.18848665

or thereabout.

Qwertiy

19:51

@tuskiomi what do ↑ and ↑↑ signs mean?

tuskiomi

and reducing gives 10↑↑↑3.552153

Qwertiy

@tuskiomi sorry, i don't understand semantics of this sign, could you give a link, please?

tuskiomi

2↑2 = 2*2
2↑3 = 2*2*2
2↑↑2 = 2^2 = 2↑(2)
2↑↑3 = 2^2^2 = 2↑(2↑(2))
2↑↑↑2 = ...
https://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation

Qwertiy

@tuskiomi interesting

tuskiomi

yeah. I got that number by taking 10↑↑log(log(2718281828459045))

then since that was raised to the 2 I put 2↑10↑↑log(log(2718281828459045))

then simplified from there

Qwertiy

20:01

i'm thinking...

10↑(16*2↑2718281828459046)

Math.log10(2) == 0.3010299956639812
Math.log10(2)*2718281828459046 == 818284367034505.5

10↑(16*10↑818284367034505.5)

Math.log10(16) == 1.2041199826559248

10↑(10↑818284367034506.7)

@tuskiomi, what should i do next?

tuskiomi

Looks like I misunderstood them

well shoot

ehm

Let's see

from
10↑(16*2↑2718281828459046)

we can say it's equal to 10↑(16*2↑2↑log2(2718281828459046))

10↑(16*2↑2↑51.2716)

Qwertiy

@tuskiomi yep, by the way, we can move 16 as +4 to the degree

tuskiomi

20:16

mmhm 16= 2↑4

thus 10↑(2↑4+2↑2↑51.2716)

Qwertiy

@tuskiomi yep, so 10↑(2↑2718281828459050)

@tuskiomi why? o_O

tuskiomi

sorry

Qwertiy

10↑(2↑2↑log2(2718281828459050)) = 10↑(2↑2↑51.271616464199404)

@tuskiomi you are still doing smth strange...

tuskiomi

hmmm...

and If I continue.. 10↑(2↑4*2↑2↑2↑5.6800)

Qwertiy

@tuskiomi why are you using +?

there was *

tuskiomi

20:25

oh yes, my bad

Qwertiy

and it could've been easily transferred to the degree on the first step

10↑(2↑2↑51.271616464199404) = 10↑(2↑2↑2↑5.680088477859977)

are you going to continue until we can get smth near 2 as a tail?

tuskiomi

10↑(2↑4*2↑2↑2↑2↑2.505913) ...
10↑(2↑4*2↑↑4.2) ...

yeah

and then you collapse it like that

I made a mistake

should be
10↑(2↑2↑2*2↑↑5.2) ...
10↑(2↑↑3*2↑↑5.2) ...

Qwertiy

@tuskiomi near 2 or near 1?

tuskiomi

2. you want it near the base number.

Qwertiy

var n = 3, deg = 5.680088477859977;
console.log(`10↑(${"2↑".repeat(n)}${deg})`) // 10↑(2↑2↑2↑5.680088477859977)
for (; deg > 2.00000001; deg = Math.log2(deg)) ++n
console.log(n, deg, 2**deg) // 5 1.3253365599586389 2.5059134025425713
console.log(`10↑(2↑↑${n})`) // 10↑(2↑↑5)

@tuskiomi, is that right?

@tuskiomi, fixed... but seems i missed the difference between 2.5 and 2

tuskiomi

20:35

what does deg represent?

Qwertiy

@tuskiomi degree - the tail

10↑({"2↑" repeated n times}{deg})

tuskiomi

and n is the count of repitions

okay

Qwertiy

@tuskiomi yep

tuskiomi

That will give you precision to an integer, yes.

Qwertiy

oh

and how can i get it more correctly?

seems like you've calculated 5.2

@tuskiomi?

tuskiomi

20:42

that's a top to the head guess, but you would take it down to the lowest without going over, then say... sorry. am thinking.

I am spitballing, but you would say that logBaseV(deg) = 2, and then your n+=1-V, Which adds the remainder base, or very close to

that means than v^deg = 2. v↑deg = 2 which would incur that 2↑2↑2↑2↑....↑v↑deg

but v↑deg = 2

to then it would be 2↑2↑2↑.... ↑2

Thoughts, @Qwertiy ?

Qwertiy

@tuskiomi hope that understood...

but v would be more than 1

tuskiomi

yeah, which is why you add 1-v.

oh wait... v-1*

Qwertiy

so n+=1-V doesn't seem to be correct

@tuskiomi :)

tuskiomi

is that making sense? or did I have a hole somewhere?

Qwertiy

@tuskiomi it seems reasonable

Simply Beautiful Art

20:52

I'm here

And I can help comprehend the ↑

tuskiomi

Hello, SBA knows about all of this stuff

Simply Beautiful Art

So we should start with ↑'s and whole numbers

tuskiomi

sure.

Qwertiy

d = .5059134025425713

for (l=1, r=2; r-l>1e-6; ) {
  mid = (l+r) / 2
  if (mid**d > 2) r = mid; else l = mid;
}

mid = (l+r) / 2
console.log(l, mid, r) // 1.9999990463256836 1.9999995231628418 2
console.log(d**l, d**mid, d**r) // 0.25594853719362926 0.255948454032902 0.25594837087220185

Simply Beautiful Art

are you trying to calculate these?

Qwertiy

20:54

so we got 10↑(2↑↑5.25594837087220185) - кшпре,

@tuskiomi v-1 or 2↑v?

Simply Beautiful Art

I believe 10↑(2↑↑5.25594837087220185) = 10↑2↑2↑2↑2↑2↑2↑0.25594837087220185

Qwertiy

wait.. seems like i've made smth wrong

@SimplyBeautifulArt really?

Simply Beautiful Art

@Qwertiy yeah

5 towers of 2↑ with a 2↑frac() at the top.

frac() = fractional part/decimal

Qwertiy

@tuskiomi v^deg = 2 by deg you meant fraction part d = .5059134025425713?

if so, smth like sqrt(v) should be 2, so v is somewhere near 4 - that seems wrong, isn't it?

@tuskiomi?

tuskiomi

Sorry am here

No, hold on

Simply Beautiful Art

21:01

So...
3*2 = 3+3 = 6
3^2 = 3*3 = 3+3+3 = 9
3↑↑2 = 3^3 = 3*3*3 = ... = 27
3↑↑↑2 = 3↑↑3 = 3^3^3 = 3^27 ≈ 7.6e12
3↑↑↑↑2 = 3↑↑↑3 = 3↑↑3↑↑3 ≈ 3↑↑7.6e12 = 3^3^3^... with 7.6e12 amount of 3's.

tuskiomi

Deg is close but not equal to 2.

Qwertiy

@SimplyBeautifulArt fine, but how should i use it?

Simply Beautiful Art

90

Q: Largest Number Printable

Before opening this I've done a little search and I've found there is a similar question, but it has been closed because it was ambiguous. I hope this won't. So, your goal is to write a program that prints a number. The bigger is the number, the more points you'll get. But be careful! Code lengt...

tuskiomi

Yes yes. But the algebra of the notation

Simply Beautiful Art

Comparing numbers here

@tuskiomi It is right to left notation. Start on the right and work to the left

tuskiomi

21:04

Ooh. So stuff distributes, etc to the right

Simply Beautiful Art

Quite messily so

Qwertiy

@SimplyBeautifulArt ?

Simply Beautiful Art

@Qwertiy Hard to compare numbers without these ↑'s

tuskiomi

Huh. Is there a reduction method? Eg reducing 10↑(16*2↑2↑51.2716)

Simply Beautiful Art

Yes, there is a reduction rule

Qwertiy

21:06

@SimplyBeautifulArt yep) but it's hard to convert too)))

Simply Beautiful Art

You would start on the right, so....

Qwertiy

@SimplyBeautifulArt could you show it?

tuskiomi

:) this is so cool

Simply Beautiful Art

2↑51.2716 = 2^51.2716 ≈ e*e15

where the first e is Euler's number and the second is scientific notation

just thought it cool that it came out like that

2↑e*e15 = 2^(e*10^15) = something pretty big

tuskiomi

2.61..?

Simply Beautiful Art

21:08

Then multiply by 16

Then do 10^all that

@tuskiomi ?

tuskiomi

That first reduce it though

Simply Beautiful Art

I get 2.7182...

tuskiomi

Doesn't* sorry. I'm on mobile

Simply Beautiful Art

2↑51.2716 = 2^51.2716 ≈ 2.7182∙10^15

Qwertiy

@SimplyBeautifulArt Math.exp(1) // 2.718281828459045 - is it Euler's number?

yep, see.. thinking

Simply Beautiful Art

21:09

@Qwertiy It's Euler's number out 5 places, coincidentially

now 2↑↑pi, just as an example:

= 2^2^2^2^0.1415926
= 2^2^2^1.103122228
= 2^2^2.148190935
= 2^4.432716006
= 21.5963561

3 amount of 2's, followed by one 2^frac(pi)

and always right to left operation

Qwertiy

but how to get the frac part when you have only a number?

so, i had

console.log(n, deg, 2**deg) // 5 1.3253365599586389 2.5059134025425713
console.log(`10↑(2↑↑${n})`) // 10↑(2↑↑5)

tuskiomi

Be right back

Simply Beautiful Art

Is this what you wanted to calculate?

10↑(2↑↑5)

Qwertiy

what should i add to 5 to get it?

Simply Beautiful Art

frac(Int) = 0

Qwertiy

21:15

no, i want to reduce

i had 10↑(2↑2718281828459050) and i want to get 10↑(2↑↑x)

and i found out that 5<x<6

console.log(n, deg, 2**deg) // 5 1.3253365599586389 2.5059134025425713

Simply Beautiful Art

Ah, I see

2↑2718281828459050 = 2↑↑x → x=?

Yes, 5<x<6

Qwertiy

by the way, shouldn't we switch to base 10 instead of 2 before?

Simply Beautiful Art

Use whatever base is best for scenario

Qwertiy

as we want to get 10↑↑↑y

Simply Beautiful Art

perhaps not so intuitively, the base doesn't matter in the long run

Qwertiy

21:18

@SimplyBeautifulArt yep)

@SimplyBeautifulArt ok, than let's continue with 2 :)

Simply Beautiful Art

2↑2718281828459050 = 2↑↑x = 2^2^2^2^2^2^frac(x)

Qwertiy

@SimplyBeautifulArt yep

Simply Beautiful Art

so frac(x) = log(log(log(log(log(2718281828459050)))))

where we use base two

Qwertiy

interesting

Simply Beautiful Art

I get frac(x) = 0.406358769

so x=5.406358769

Qwertiy

21:20

eval("Math.log2(".repeat(5) + 2718281828459050 + ")".repeat(5)) // 0.40635876852612085

Simply Beautiful Art

DO NOT calculate the numbers explicitly :)

Qwertiy

@SimplyBeautifulArt why?

Simply Beautiful Art

yup

Obviously the numbers get too big

especially once we hit ↑↑↑

Qwertiy

@SimplyBeautifulArt it's only a fraction, not to big :)

Simply Beautiful Art

I mean the rest of the mess gets too big. So don't try evaluating the entire thing explicitly and solving for x that way
>.<

Qwertiy

21:21

so we got 10↑(2↑↑5.40635876852612085)

49 mins ago, by Qwertiy

var n = 3, deg = 5.680088477859977;
console.log(`10↑(${"2↑".repeat(n)}${deg})`) // 10↑(2↑2↑2↑5.680088477859977)
for (; deg > 2.00000001; deg = Math.log2(deg)) ++n
console.log(n, deg, 2**deg) // 5 1.3253365599586389 2.5059134025425713
console.log(`10↑(2↑↑${n})`) // 10↑(2↑↑5)

Simply Beautiful Art

I assume you want to convert the entire thing into 10↑↑x?

Qwertiy

@SimplyBeautifulArt 10↑↑x or 10↑↑↑x?

Simply Beautiful Art

It's nowhere near ↑↑↑ level

Just to hint at the strength of three arrows:

3↑↑↑3 = 3↑↑3↑↑3 = 3↑↑7.6e12 = 3^3^3^.... with 7.6e12 amount of 3's

10↑↑↑x must involve minuscule values of x.

Qwertiy

@SimplyBeautifulArt but it's already 10↑(2↑↑5.40635876852612085)

Simply Beautiful Art

a↑(b↑↑c) is generally way smaller than something you would write in the form x↑↑↑y

10↑↑y is much much more reasonable

Qwertiy

21:27

@SimplyBeautifulArt ok, how can we do that?

Simply Beautiful Art

Well

I'd gander a real quick look and say

10↑(2↑↑5.40635876852612085) ≈ 2↑(2↑↑5.40635876852612085) = 2↑↑6.40635876852612085

As I said before, base hardly matters

For example, note that 10^3 = 1000 ≈ 1024 = 2^10

Qwertiy

don't think so))

Simply Beautiful Art

So 10^x ≈ 2^(10x/3)

But 10∙(2↑↑5.40635876852612085)/3 does not change the number significantly

at most, it will affect like the 10th digit

Hence how one makes quick analysis

Qwertiy

ok, if we switched to base 10 on the first step, we would've get 10↑(10↑↑y) and then 10↑↑(y+1) right?

Simply Beautiful Art

yes

But that would've been harder to analyze

Qwertiy

21:32

i'll try :)

@SimplyBeautifulArt why?

Simply Beautiful Art

@Qwertiy Just my opinion

Qwertiy

@SimplyBeautifulArt usually people are using 10, not 2 :)

Simply Beautiful Art

I thought you people did your numbers in bits. :)

I get 10↑↑4.069506124

I'm gonna be out for a bit

@Qwertiy You do Ruby?

Qwertiy

@SimplyBeautifulArt nope

Simply Beautiful Art

Python?

Qwertiy

21:40

don't program it(

them both

Simply Beautiful Art

okay

Qwertiy

why are you asking?

Simply Beautiful Art

5

A: Largest Number Printable

Ruby: score is approximately (much greater than) fω²(fω²(fω²(fω²(fω²(fω²(966))))) where fα(n) is the fast growing hierarchy. Assuming online interpreter where printing the result is automatically done: def f(a,b=a,c=a,y=?!.ord)(c>y)?a.times{a+=f a,b,c-y}&&a:(b<y)?a:f(a,b-y)end;f(f(f(f(f(f ?φ.o...

My number was made using Ruby, which is close to Python

Qwertiy

:)

Simply Beautiful Art

After you comprehend three ↑'s, four ↑'s, and arbitrarily many ↑'s... then comprehend even bigger things...

Qwertiy

21:43

deg = 2718281828459050
console.log(`10↑(2↑${deg})`) // 10↑(2↑2718281828459050)
deg *= Math.log10(2)
console.log(`10↑(10↑${deg})`) // 10↑(10↑818284367034506.8)

n = 1;
console.log(`10↑(${"10↑".repeat(n)}${deg})`) // 10↑(10↑818284367034506.8)
for (d=deg; d>10; d=Math.log10(d)) ++n
console.log(n, d, 10**d) // 3 1.173562229554173 14.912904254248932
console.log(`10↑(10↑↑${n}.?)`) // 10↑(10↑↑3.?)

console.log(n += eval("Math.log10(".repeat(n) + deg + ")".repeat(n))) // 3.06950612353447
console.log(`10↑(10↑↑${n})`) // 10↑(10↑↑3.06950612353447)

@SimplyBeautifulArt, so i got 10↑↑4.06950612353447

@SimplyBeautifulArt yep, nice :)

@tuskiomi, we calculated smth :)

tuskiomi

22:02

:) I'll read up on that

Simply Beautiful Art

@Qwertiy :D

So triple arrows work in the same manner for the most part:

2↑↑↑pi = 2↑↑2↑↑2↑↑2↑↑frac(pi)

Three 2's separated by double arrows with 2↑↑frac(pi) at the end

2↑↑frac(pi) = 2^frac(pi) = 2^0.14159... = 1.103122228

2↑↑1.103122228 = 2^2^0.103122228 = 2.105401626

2↑↑2.105401626 = 2^2^2^0.105401626 = 4.310578809

And lastly,
2↑↑↑pi = 2↑↑4.310578809 = 2^2^2^2^2^0.310578809 = 4.25709608e10

Likewise, more arrows will expand in the same manner with one less arrow in each expansion.

Qwertiy

:)

Simply Beautiful Art

:P

Ackermann function

In computability theory, the Ackermann function, named after Wilhelm Ackermann, is one of the simplest and earliest-discovered examples of a total computable function that is not primitive recursive. All primitive recursive functions are total and computable, but the Ackermann function illustrates that not all total computable functions are primitive recursive. After Ackermann's publication of his function (which had three nonnegative integer arguments), many authors modified it to suit various purposes, so that today "the Ackermann function" may refer to any of numerous variants of the original...

A(m,n)=[2↑↑↑↑...↑(n+3)]-3 with m-3 amount of arrows

The function is pretty easy to write, and provides some intuition into the up-arrow business

(works only for integer m,n > 0)

Simply Beautiful Art

23:14

repl.it/ICAD/2

Simply Beautiful Art

23:30

@tuskiomi If you don't mind, I'm fixing your edits

tuskiomi

23:47

@SimplyBeautifulArt please

Simply Beautiful Art

@tuskiomi they are all pending

since I don't have enough rep *hmph*

@Qwertiy You should update your answer with 10↑↑x notation

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Discussion between Qwertiy and tuskiomi