Can you write a program to find all swastikas in the binary representation of pi?
Rules:
you have to find a 5x5 swastika like pic rel
wrap the image after 5 columns
post the offset for verification.
use this hex data converted to binary ignoring the dot. https://pi2e.ch/blog/2017/03/10/pi-digits-download/#download
>I found 2 in the first billion digits
ez pz
whats the catch?
Pi is infinite, so the answer is infinity.
>Pi is infinite
Proof?
>pi is infinite
>therefore there are infinite swastikas
What
Welcome to infinities. They're not exactly intuitive.
No one is talking about Hitler, except you.
An infinite string isn't guaranteed to have infinitely many occurrences of a given nonempty substring.
An infinite RANDOM string is, though.
Pi isn't random.
It's also not known to be normal.
It's also very likely that God didn't engineer pi not not have swastika patterns. But I don't know if it can be proven.
Any mathematicians in the house? IQfy?
*to not have
Based & Godpilled.
ah, true
you're such a fricking moron lmao
>someone who actually doesn't understand infinity
LAUGH AT THIS MAN
pi is infinite and if pi were simply normal in base 2 there would be infinite swastikas. it is not known if pi is simply normal in base 2.
That being said, the first 31 trillion digits (base 10)
or 3.42*31 trillion digits (base 2) are uniformly distributed.
most intelligent code-monkey
there's only a few million digits which was proven back in 2007 on that google talk
>Can you write a program to find all swastikas in the binary representation of pi?
Not one that halts
All sequences from 0 to n decimal digits.
There is no catch.
>all
lol.
HAHA SO FUNNY
now let's see if you can do it for the first 1 billion digits, smartass.
Yeah it's incredibly easy, I would just check every 5th digit if the proceeding digits were 1011110100111110010111101 and if they were you had a swastika.
I wouldn't have any idea how to do this efficiently but that approach should be fast enough for just a billion digits...
write it.
the fun begins when you realize one hex number is 4 bits, but one line of the swastika is 5.
in theory it sounds easy, but in practice its more interesting
especially if your fav language doesnt support binary operations
>the fun begins when you realize one hex number is 4 bits, but one line of the swastika is 5.
Then check 4*5 = 20 bits at a time? How is this interesting again?
Then drop your solution and we will point why it's wrong.
If you think it's so trivial then you are misunderstanding the challenge.
It can be solved with a fricking for loop advancing 5 digits and checking the next 25 digits. This is so obviously trivial that I have no desire to paste my own solution just to humour you. I honestly don't know why you would think the program not advancing in hex digits is such a large problem.
As somebody else pointed out you can literally fricking grep for the swastika.
Does grep support bit patterns? How?
Anyway I'm not the other guy you were arguing with. What you are not understanding is that you not only have to find the right bit pattern. The bit pattern needs to be ALIGNED to generate the swastika image.
Especially if the initial "3." part of pi generates only 3 bits and the rest of the bits for the fractional part are appended to those first 3 bits.
bits can be offset anon
then you check for the beginning of the swastika once every 20 bits.
but you need to check once every 5 bits
You're presuming he's only doing one check, he can simply do four checks for the swastika within that one check.
I really think you've overthinking something that isn't hard OP. I'm a braindead programmer OP and this is the most brainless tier leetcode babby tier problem I've ever heard of. It's in the realm of difficulty of fizz buzz.
No, you're not understanding the problem or you're trolling. There's a reason nobody is posting blocks of code. This isn't fizzbuzz tier at all.
>There's a reason nobody is posting blocks of code.
oh what possible reason could that be
I bet its globohomo
The reason is that it's not as trivial as it sounds.
>The reason is that it's not as trivial as it sounds.
this means YOURE globohomo, trying to keep me from my fricking swastikas
actually, now that i think of it, without binary operations it is still possible and would involve less writing, albeit the code will run slower
i think ima drop the bin op myself
The tricky part is checking which ones of those instances of the pattern found will have the right 25 bit alignment to form the correct pattern when arranged on a 5x5 grid.
multiline grep
It can be done with standard shell utils but it'd be very slow.
This. Why bother with C or even python when we have hundreds of mature languages who's entire syntax is built around solving a certain class of problems elegantly and quickly.
This is why python is essentially useless and C should be getting smaller (i.e. people move back to ANSI) and its useful domain will be much more clearly defined as a result.
On a related note, teaching mathematicians python is downright shameful, a mathematician won't be scared by a language like ivy or R
For a list that most mathematicians should know
so in order of importance: TeX/LaTeX, a bit of shell, noq, more shell, ivy, R, and some CAS. If they really need a generalist language then Haskell as it's generally very good at doing useless shit.
Grep is a standard, and has hundreds of implementation, with all the documentation and repos with varying levels of coverage it's much easier to see how to strip out the cruft and only use what you need. Besides, even gnu grep can search 10 GB files for lines in a tenth of a second; this is way more than enough for a billion digits binary or decimal.
You also don't get it. This is not a search bytes in a file problem. This is a search bits in a file that are aligned to a 25 bit grid. You will need to iterate over the bytes, grep wont save you.
I bet any shell solution you can come up with will be at least 10 times slower than a naive C solution.
>You also don't get it.
No, you don't get it
Compute 1 billion digits of pi
convert to 5 column binary and put in a tmp file
multi line grep for '10111n10100n11111n00101n11101' and maybe rotations and inversion
done. line count is free.
And again, grep can do 10GB in a tenth of a second.
Why do you think it's required to be aligned to your chosen columns?
The conversion is going to be the part that is much slower than C. What standard util would you even use for converting a big file to its binary representation in ascii.
Also, it's going to involve lots of disk IO because you're expanding each bit to a byte, which again is gonna be slow.
>What standard util would you even use for converting a big file to its binary representation in ascii
xxd -b
hmm fair enough
or bc or a dozen other things; he has to be trolling.
Are you stupid? save 1 billion digits of pi /tmp convert it in /tmp it never touches disk.
>you're expanding each bit to a byte, which again is gonna be slow.
wow .1 seconds so slow. and the conversion isn't even part of the program.
thats even worse, you're eating memory like crazy and will be limited by ram+swap
sub 10GB
Okay ramlet
What if you want to find more digits? It's not scalable.
Sure it is. Just buy more ram to scale it
That's what google does. Besides, NVMe drives aren't slow either.
>87238019(You) ultra gey, C is literally made for this shit
I literally found them all before I made that post.
>I literally found them all before I made that post.
then why no post code and results?
its a little late for that now...
01110
00001
10100
1
Nope. No swastikas.
>shut it down
>you have to find a 5x5 swastika like pic rel
>wrap the image after 5 columns
>post the offset for verification.
Your rules are unclear Black person. What am I generating? An image? The digits? The offset?
You are generating the offset which would display a swastika if you generated the image.
The digits are already available to be downloaded but you can generate your own if you want.
So literally finding 0x17A7CBD in the number?
This is my code https://pastebin.com/pLSZjfYR
It only finds one at bit offset 2127667875.
I haven't fully checked the code, I think the pattern matching works but I'm not sure if I don't have an off by one error in the modulo 25 bit filter.
Output:
detected input size 1000000002
searching...
found sequence at vector1_copy: 8d2fe97c, vector2_copy: bc8f8c43, byte_offset: 531916969, local bit offset: 3, global bit offset: 2127667875
Pointer in the hex/string input
Buffer of binary data
Add to the buffer from the hex pointer until buffer is above 25
check for target string
pop first 5 values in buffer
loop
very nice
Fixed my program. https://pastebin.com/C6HNstSs
Now I assume the initial "3." part generates 8 bits but also added a line that when uncommented changes the assumption to 2 bits.
Output now with the 8 bit assumption is:
detected input size 1000000002
searching...
found sequence at vector1_copy: 7d2fe970, vector2_copy: c187ff30, byte_offset: 614406368, local_bit_offset: 3, global_bit_offset: 2457625475
found sequence at vector1_copy: e97f4bd3, vector2_copy: 7aa2d131, byte_offset: 906489200, local_bit_offset: 0, global_bit_offset: 3625956800
Oh, and also fixed an out of range read.
Can anyone make their own program to confirm the output is correct?
I don't want to check by hand.
cool
did you only check for the first 2 because thats a good enough estimate? and because every 1st and 2nd number on each check would be an odd number (always 6 * something?)
i wonder if there would be more of them if you checked for them on every line instead of every 1-5 being a swastika or not
it checks every possible 25 bit substring of a 64 bit long segment based on an offset that advances 32 bits each iteration.
this ensures that if the 25 bit pattern is across boundaries, in the next iteration it will be in the middle and thus found.
again this is very suboptimal, ideally you would calculate on which bytes each 25 bit segment falls on and only check those bytes, and only check the bits at that bit offset on which the 25 bit segment falls on.
Wouldn't that double-count matches that fit entirely in one of your chosen 32-bit words?
yes good point. i suppose it would be an easy fix it by ignoring the last 8 bits (so only 24 bits are checked in the second 32 bit half).
That would still double count matches aligned perfectly to your 32-bit boundary.
no because a match needs 25 bits so if you're only checking up to 24 there is no way to generatd a match.
Oh wait it's a 25-bit pattern so that would work fine.
You using co-pilot?
No, what makes you think that?
>in the binary representation of pi
in WHICH binary representation of pi?
there is literally a file with the binary data linked in the op
the problem is that the subset of people who could solve this and people who are willing to solve this == null
nil* (nil)
holy shit an actual post, is IQfy finally healing
---swastika found at char no 3180096 offset 3
---swastika found at char no 9297182 offset 0
---swastika found at char no 16742682 offset 1
---swastika found at char no 19126475 offset 3
---swastika found at char no 26808175 offset 2
---swastika found at char no 34052436 offset 3
---swastika found at char no 38149656 offset 2
---swastika found at char no 40905358 offset 1
---swastika found at char no 45781783 offset 2
---swastika found at char no 48814915 offset 1
---swastika found at char no 51059660 offset 3
---swastika found at char no 51431539 offset 3
---swastika found at char no 65933960 offset 3
yay now i only have to write a program which can get me a series of a dozen caracters starting at position 3180096
looks like im getting close.
offsets are messed up, but the pattern checks out
first pattern occurs at character nr. 3180094.
bit offset is correct. (3)
checked it by hand.
still gotta debug the difference of 2 between the result and the actual offset
taking a break.
will upload code once ill debug it or i give up
also its certain theres a lot more than just two occurences of said pattern
I'm gonna do my implementation when I get home.
I must be doing something wrong because I'm finding a lot more swastickers than others.
And I'm finding it rather odd that every has offsets that are less than 4.
Each byte you read has 8 bits, and you have to read into columns of 5.
So if the first two bytes are 11110000 10101010
You'd get
11110
00010
10101
0
Correct? So if at this position you don't have a swasticker you read from the next line instead
00010
10101
0
Which would be byte 1 (or 0, whatever), offset 5
Then byte 2 offset 2, then byte 2 offset 7.
Found my second problem, padding out to 8 characters instead of 4, because each byte you're reading in is actually a half byte.
Found my problem, wasn't forward filling the byte conversion to bits with 0s.
good luck anons!
import time
st = time.time()
target = "1011110100111110010111101"
buffer = ""
step = 0
with open("pi_hex_1b.txt", "rb") as f:
EOF = False
while not EOF:
while len(buffer) < 25:
byte = f.read(1)
if byte == b".":
continue
if byte == b"":
EOF = True
break
buffer += bin(int(byte, base=16))[2:].zfill(4)
if buffer.find(target) > -1:
offset = buffer.find(target)
print("swasticker found at", step, "offset", offset)
for i in range(5):
print(buffer[i*5+offset:i*5+5+offset])
buffer = buffer[16:]
step += 4
else:
buffer = buffer[4:]
step += 1
## Print runtime
et = time.time()
if (et - st) < 1:
rt = str(round((et - st) * 1000,3)) + "ms"
else:
rt = str(round(et - st,3)) + "s"
print("Runtime:> ", rt)
My results are off by 1 in comparison, could be that you are 1 indexed?
i count the dot as legit character in my offsets
(i dont parse it, but i take it into account for the position)
import time
st = time.time()
target = "1011110100111110010111101"
buffer = ""
step = 0
count = 0
with open("pi_hex_1b.txt", "rb") as f:
EOF = False
while not EOF:
while len(buffer) < 30:
byte = f.read(1)
if byte == b".":
continue
if byte == b"":
EOF = True
break
buffer += bin(int(byte, base=16))[2:].zfill(4)
if buffer[:28].find(target) > -1:
offset = buffer.find(target)
print("swasticker found at", step, "offset", offset)
count += 1
for i in range(5):
print(buffer[i*5+offset:i*5+5+offset])
buffer = buffer[4:]
step += 1
print("Total swastickers found: ", count)
## Print runtime
et = time.time()
if (et - st) < 1:
rt = str(round((et - st) * 1000,3)) + "ms"
else:
rt = str(round(et - st,3)) + "s"
print("Runtime:> ", rt)
Some improvements. Ensure overlap in the buffer, but also ensure that overlapping swastickers can be found (if there are any).
I get a total of 114 found. Takes 1280s to run.
>Takes 1280s to run.
the absolute state of python
mine finds 114 swastikas too
>Takes 1280s to run.
yeah... thats a little while...
not to brag, (duh, i didnt write C), but just to evidence that C-like still has its use- especially when you have limited ressources available
mine took 17 secs
and takes like 2Mb on the memory
what did you write in then
if you meant in which language did i write it- it is C.
i meant that i didnt write the language (compiler) itself, so the fact that it runs so fast is not my merit.
i could optimize it further, even thread it, but even without that it still is very fast
if you mean how did i manage buffering the data,
i have a 10kb buffer into which i load the file chunk by chunk
the 2megs footprint come from the fact i use libraries (mostly stdio.h i think)
Mine also does
nice.
did you optimize it?
Yes, -O3
aaaah ok
but thats cheating:p
>ot to brag, (duh, i didnt write C), but just to evidence that C-like still has its use- especially when you have limited ressources available
I highly doubt mine is an optimized way of doing it in python.
It's basically just a naive solution.
All signs point to pi being infinite and random. If you want to offer any evidence to the otherwise, be our guest.
>All signs point to
moron
>pi being infinite
we know it has an infinite decimal expansion because pi is irrational. there is no need for evidence, we have a proof of its irrationality.
>and random
there is absolutely no "evidence" for this claim, and there is no proof that pi's decimal expansion is a normal sequence. whether or not pi has a normal decimal expansion, it is still not "random" and you are the biggest dumbest pseud homosexual on this webzone
Oh, come on. The likelihood of there being an infinite amount of swastikas is very high, even if it hasn't been proven.
>The likelihood of there being an infinite amount of swastikas is very high
Bayes is rolling in his grave seeing how midwits constantly misapply his shit.
Enlighten us then
See Euler's sum of powers. Your moronic misapplication would say that it's very likely true even if it hasn't been proven and your moronic ass would be wrong.
Swastikas appearing in the pi is not evidence of there being infinite appearances of them, trying to apply Bayes here is moronic.
No one said anything about Bayes except you here
http://justinparrtech.com/JustinParr-Tech/pi-does-not-contain-the-universe/
not the anon you are arguing with but here's an interesting read
not visiting your site thanks
If something is infinite and pi is semi-random then it follows that you will encounter that particular arrangement somewhere along the line an infinite number of times.
>pi is semi-random
pi is not random at all how fricking difficult is it for this concept to make it past your thick skull holy shit
i hate you pseuds so much
>then it follows that you will encounter that particular arrangement somewhere along the line an infinite number of times.
this isn't even true if pi were in fact a random sequence (which it is not)
you can have events with probability 1, or sequences of events that converge to 1 in probability, and those events may never occur
god you're so fricking moronic just shut up
>semi-random
Define that.
>it follows that you will encounter that particular arrangement somewhere along the line an infinite number of times
Provide proof.
to be more explicit for your reptile brain:
you can have a random process that generates infinite sequences and you may be able to prove (depending on what kind of random process it is) that some given finite subsequence "eventually" appears with probability 1. this DOES NOT MEAN that your chosen finite subseqeunce actually does appear in any particularly realized infinite sequence generated by this random process
all of this is, of course, irrelevant, because the decimal expansion of pi is not a random sequence, nor is it generated by any known random process, and the fact of whether or not it is normal is unknown.
in conclusion: please do us all a favor and have a nice day in real life unironically
Lmao keep believing that
>you can have a random process that generates infinite sequences and you may be able to prove (depending on what kind of random process it is) that some given finite subsequence "eventually" appears with probability 1. this DOES NOT MEAN that your chosen finite subseqeunce actually does appear in any particularly realized infinite sequence generated by this random process
That's exactly what it means.
that is exactly not what it means. you should learn some measure theory: an event having probability 1 does not mean it actually occurs.
No, you tard. Your sample space is infinite so its subsets with Lebesgue measure 0 don't need to be empty. In infinite sample space, "almost surely" is distinct to "surely".
>cherrypicking
What about all the other times where likely conjectures were proven decades down the line?
Besides, there being an infinite number of instances of some funny pattern in an irrational number's decimal is much more likely, a priori in my opinion, to some inequality holding true for all numbers just because it happens to hold true for a lot of small numbers, since an inequality is the default state of things, as finding arbitrary patterns in irrational numbers is the default state of things.
>The likelihood of there being an infinite amount of swastikas is very high
based on what you fricking moron
>pi is semi-random
oh okay based on your moronic fever dream
that's awesome dude it's so cool when you draw false inferences from false assumptions
pi is irrational, which means there is no repeating sequence
i.e. it's random
>pi is irrational, which means there is no repeating sequence
>i.e. it's random
you MUST be a troll at this point. there is no other explanation.
just in case you are not, consider the following sequence:
0.01011011101111011111...
this is a non-repeating sequence corresoponding to some irrational, real number. nowhere in this sequence does the subsequence 00 appear. this sequence is completely deterministic: give me an index, and i can tell you exactly what digit is in that index in the sequence.
>nowhere in this sequence does the subsequence 00 appear.
this is different than OP's question because he already proved that the subsequence does appear in the sequence
consider the following infinite, non-repeating sequence:
0.00101101110111011110111110...
as we can clearly see, this sequence demonstrates the fact that you gargle wieners you fricking moron
>if I just make something up, it provea you wrong
Can you show any evidence that the decimal expansion of pi will ever enter a sequence that will never produce any more swastickers?
can you show me any evidence that the decimal expansion of pi contains infinitely-many swastikas?
no, neither of those questions currently has an answer, homosexual.
because i can write down for you the generating function for this sequence and i can prove to you that it does not. in fact, this would be an exercise for a first-year undergrad or a junior/senior high-school student learning discrete maths.
>can you show me any evidence that the decimal expansion of pi contains infinitely-many swastikas?
can you show me any evidence that it doesn't?
>can you show me any evidence that the decimal expansion of pi contains infinitely-many swastikas?
All decimal expansions of pi to date have never entered a sequence that won't ever produce swastickers.
This is not proof. It is evidence. Something you have none of.
>entered a sequence that won't ever produce swastickers.
this is complete nonsense.
>It is evidence.
oh cool, something that's completely irrelevant to the subject at hand, as EVIDENCED by the innumerable examples of infinite, non-repeating sequences with arbitrarily many occurences of a given subsequence that, nonetheless, do not contain infinitely-many occurences of that sequence
>It is evidence
not evidence of it containing infinite number of them
If pi is infinite, yes it is.
no it's not and there are countless examples, even some posted in this thread, that prove you wrong.
But those examples aren't pi and don't have the same properties as pi.
I ran the algorithm for the 1st billion and it found it.
I ran the algorithm for the 2nd billion and it found it.
I ran the algorithm for the 3rd billion and it found it.
I ran the algorithm for the 4th billion and it found it.
I ran the algorithm for the 5th billion and it found it.
I ran the algorithm for the 6th billion and it found it.
This isn't good enough for you? Do you even believe in inductive evidence?
If we follow your argument to its logical conclusion, if I try to go through a closed door, there's a chance that I will never ever be stopped again. Just because for the last 1000 iterations I collided with it, doesn't mean that at some point my body will not just stop colliding with other objects. After all, we haven't PROVEN that atoms won't just stop interacting with each other at some point in the future.
your inability to comprehend how small "6 billion" itterations is in the face of infinity is why every reasonable person in this thread is ridiculing you.
please fricking have a nice day.
Ah, yes, the "it might eventually enter a sequence where that pattern never appears again" argument.
So is the number of years in recorded history. Maybe tomorrow aliens land on Earth and decide to kill 90% of the human population.
Since there is no proof either way, it's as likely to happen as it is not.
Since time is infinite, it's not proper to draw a probabilistic conclusion from a sample size of only a few billion years.
>0.00101101110111011110111110...
What would be the reason for me to think this doesn't contain other instances of 00 in the sequence?
to expand: give me any finite sequence you want, and i will generate for you an infinite, non-repeating sequence that contains your finite subsequence exactly N (and no more) times for any natural number value of N that you desire, and i can make this sequence arbitrarily "complex"
>and i will generate for you
But we're talking about pi which has a determined way of being generated
and no one knows (it is an open question in mathematics) whether or not pi admits arbitrary subsequences. if you had an answer to this question, you'd be first in line for some sort of award (likely not the Field's, but maybe idk)
the point is that a sequence being infinite and non-repeating doesn't mean it contains arbitrary subsequences, homosexual
that was the whole point of those examples, how did you not see that.
can you show me any evidence that it does?
>can you show me any evidence that it does?
>no u
1/10
that's literally the same game you're playing, congrats
0/10
>no u again
>no u yet again
HOLY
BASED
guess trans aren't so good now huh?
lmao
someone is fricking seething
ITT: chuds
lmao
troon seething bc low level programming is beyond its grasp
>binary representation of Pi
In what format?
Big Decimal? Fractional? IEEE754?
You got the idea from this guy
Actually I reposted a thread posted by another anon that was taken down by the jannitorwaffen.
then the other anon did
/sci/spergs get OUT
>talks out of his ass
>gets btfo on absolute basics
>f-fricking autists!!!1!
lmao
that was my first post in this thread you massive autist, go take your mood stabilizers
never + ligma + ratio
>keep believing proven statements of mathematics while denying crackpot claims made by normie pseuds
yes, i think i will.
>proven
do you think we can find any swastikas in e?
What's the likelihood of ever finding this exact sequence somewhere in pi?
>111101101101101000111010010010111000111100100101111000111100100101111000111100111100111000111101110101101
can't I just search for the decimal 24804541?
>infinite
>never repeats
>somehow this doesn't mean that there will be all possible combinations/arrangements of numbers
If this isn't literal bait y'all need to have a nice day right this moment
see
that's the same bait I was mentioning kys
>being filtered by undergrad baby maths that literal codemonkeys are able to pass
i feel bad for you, anon
see
how is
namecalling? lmao
you're just calling me a troll
open an undergrad textbook on probability/random processes/measure theory, troglodyte
>namecalling
great bait, you've already got some bites
samegay
>and don't have the same properties as pi
how do you know? (fun fact: you don't, because this is an open question in mathematics)
oh no 🙁
>how do you know? (fun fact: you don't, because this is an open question in mathematics)
Signs point to it. No reason to assume otherwise until reasonable evidence appears to show contrary.
can someone explain me how the argument is bait?
there are already multiple explanations of varying degree of autism itt
if you can't follow those, try
>multiple explanations
None relevant to this btw
What's with all the psueds in this thread pretending pi isn't infinite or that it isn't random.
>more lazy bait
>Signs point to it.
they don't, as demonstrated by the counterexamples.
>No reason to assume otherwise until reasonable evidence appears to show contrary.
no one (reasonable) in this thread is assuming the negation, they are denying your claim that it does contain infinitely-many occurances.
>they don't, as demonstrated by the counterexamples
The counter examples that are not examples of sequences found in pi or used in calculating pi.
i'm sorry, you are actually too dumb to speak to. if you can't grasp the point that's being made by now, then remain content being a normalgay and a pseud.
The point being made is that there are some irrational numbers that have sequences that will never appear or repeat.
But pi hasn't been shown to be one of these numbers, nor that it converges towards being one.
Some irrational numbers having property Y doesn't mean all irrational numbers have property Y.
yeah no shit moron
so what don't you understand about the fact that the statement:
"pi has infinitely many swastikas"
is completely unfounded, has zero basis in reality, and can not be claimed as being anywhere near adjacent to True?
>is completely unfounded, has zero basis in reality, and can not be claimed as being anywhere near adjacent to True?
The same can be said about the opposite.
yes, correct. so what is your fricking problem?
did you literally just make up a strawman to fight against?
It's more strongly evidenced that the opposite.
>It's more strongly evidenced that the opposite.
no it is not lmao
you fundamentally don't understand the counterexamples in this thread nor why they are presented as counterexamples
the counterexamples show that you can't use your "signs" as evidence of there being infinite occurrences of some subsequence in a sequence
outstanding b8 thread though
They quite literally don't.
Yeah, that's basically what you're arguing.
This is true in taxicab space
which real life is btw, due to the discrete nature of molecules
wtf the math checks out
>they don't because.... THEY JUST DON'T OKAY!!!
kys psued
no u
I only found 24.
do i compare the pi hexes with the target hex string? Or do i convert the pi file to binary and compare it with the binary target string?
Second one. Otherwise you won't get the sequences in the middle.
I did that one in c but my times are overwhelming. How do you even reach 10-20 seconds?
I wrote a function that maps a hex char to a binary string the followed this
logic.
post the code
the code
Comparing strings is slow
Picrel is how I've done it (
)
sheesh bro maffs be dam racist fr no cap
>Find all
There are infinitely many
we can't keep letting the nazis win
#TakeBackTheSwastika
It should be noted that due to the nature of the problem, binary string representation of a hex expansion of pi, there are multiple sequences that produce a swasticker.
All sequences would have to stop appearing for the number of swastickers to be finite.
watcha doin' there, rabbi?
i can't believe that nazis hid swastikas in pi... guess this means pi is racist now...
Not just swastikas, the entirety of Mein Kampf is embedded in there if you know the super secret offset.
>hexadecimal fraction
lol
hard mode: start at the end and work backwards
Impossible mode.
it's A whirlpool you idiot
nazis copycat japan japan copycat nature and nature copy cat atoms and atoms copycat wave moves and wave moves copycat diff potential between nothing and the absolute nothing = exotic reaction due the coldest temp
And my name is tumpool
You tump my pool,, again..
USA = faint-hearted politics
look at germany as eample
trump looks smart drinking a little bit water
On this point, there are 105 Japanese style swastickers in the input.
bump