# Is there a secret message in the first decimals of PI?

Is there a secret message encoded in the first decimals of $$\pi$$?

Source: Vi Hart

• This legitamately seems like an interesting question — although potentially opinion-based. Dec 19, 2018 at 21:29
• Well, I can't speak for the other downvoters, but if it's a puzzle, it's very dull: the only possible answer is NO, it's a mathematical/physical constant. As a physics question about whether pi is a mathematical or physical constant, it's off topic (the answer is, mathematical). As a theological question about what God may or may not have done when He created the integers, it's off topic. Dec 20, 2018 at 0:07
• Before you silly persons manage to close the question, let me reiterate: this is a puzzle. A good one. It has a puzzle solution. The solution is unique, and it is not too hard to find. This puzzle follows all the good cipher puzzle guidelines. If you think there cannot possibly be any messages in an infinite string of random digits, at least not in any well-known encoding, even if the puzzle creator gets to choose the starting point, then, well, please think again.
– Bass
Dec 20, 2018 at 7:40
• @Bass I share your frustration but you don't really need to get offensive. Dec 20, 2018 at 8:48
• I reopened this; if hexomino's answer, or something at least equally good, is what Bass intended, then I think it's reasonable to consider it a valid puzzle. (While liking hexomino's answer a lot, I confess that I retain a bit of sympathy with the downvoters and close-voters.) I wonder whether we need some standard way for posters to say "no, really, this is better than it looks"; I think Bass has enough credibility that doing so might have saved this puzzle from the frosty reception it got. Dec 20, 2018 at 22:26

Updated: Solution

The first four digits after the decimal point in $$\pi$$ are 1415. If we convert this using an alphabet cipher ($$A=1, B=2, \ldots, Z=26$$) we get $$14|15 \rightarrow NO$$ So we have

Q: "Is there a secret message encoded in the first decimals of $$\pi$$?"
A: "NO"

If we take $$\pi$$ up to eight digits after the decimal point i.e, $$3.14159265$$ and use an alphabet cipher ($$A=1, B=2, \ldots, Z=26$$) we can construct the following $$3.14|15|9|26|5 \rightarrow C.NOIZE$$ and C-Noize is apparently a musical artist with some music on youtube.

So maybe the universe is telling us its favourite type of music?

• I tried to exclude the integer part by mentioning "decimals", but I'm not really certain if English works that way.. Anyway, you are more than 200% of the way there :-)
– Bass
Dec 20, 2018 at 16:42
• @Bass, thanks for the suggestion. I've updated my answer which I think is in line with your hint. Dec 20, 2018 at 17:10
• That's it, exactly :-) Tick will follow in a couple of days.
– Bass
Dec 20, 2018 at 23:02
• haha, ok, I take it back :-P Dec 21, 2018 at 0:00
• It is amusing that the presence of the answer contradicts that very same answer ;-) Dec 22, 2018 at 10:55

Here's my guess

Take the first 100 digits of pi: 1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679

STEP 1: Based on whether a digit is odd and even, convert it to AB format. Result: ABAAA BBAAA BAAAA BABBB BBBAA BABAA ABBBB BAAAA BAAAA AAAAB ABBBA ABABB AABAB ABABB BBBBB BBBAA BBBBB ABBBA ABBAA ABAAA

STEP 2: This obviously relates to Francis Bacon's biliteral cipher (see wikipedia), a long-time favorite way of hiding messages; however, this is apparantly an unpublished version of the cipher, since some combinations are not in Bacon's public alphabet. The letter pattern looks like this, however (i.e. which letters are repeated and where and which letters are not repeated at all):

12345 673(repeats 3rd letter, so the 3rd letter in the message is repeated)38 9(10)(11)(10)(12) 5(12)9(13)1 (the parentheses are there to avoid confusion when there are two digits in the number)

STEP 3: It is important to know that anagrams were a primary form of cryptography in Bacon's day (see wikipedia). Thus, when we produce the following gibbberish through simple monoalphabetic substitution (see wikipedia)...:

hltreknttoaswsieiafh

STEP 4: We can guess that it is in fact an anagram, and not gibberish. Rearranging the letters and using an "s" as a stop(period), we get.....

"I know the earth is flat."

/s

• This solution actually works; the "/s" is because I pretty much pulled it out of my rear end, not pi. That and I'm not a flat-earther. Dec 28, 2018 at 21:31
• Holy sh$\cdot\cdot$ Apr 8, 2019 at 14:51
• "I used the circles to disprove the circles." Dec 6, 2020 at 17:13
• Behold the power of randomness! White noise can be shaped into any signal: a bird's song or Beatles' song… That's even how GAN models work, which make those dreamy paintings. Also part of DALL-E. Generator starts with white noise, like "static" on an old TV; both real and Generator-made images are shown to Arbiter, who says real or fake—also random-seeded, the 1st answer is random; Arbiter's answers and the real image (in DALL-E, also a verbal description) are shown to Generator. Generator's goal is to fool Arbiter, Arbiter's to catch Generator on forgery. Noise contains everything! Apr 13 at 23:10
• How do you produce an h from ABAAA, an l from BBAAA, etc? Nov 16 at 14:34

Does this count?:

Pi music (link only sorry due to being a tune)