5 ms·
Binary search three 6 bit symbols, that is 18 multiplications. A bit less if you do not blindly pick the centre but estimate - or guess if you will - what the c
by danbruc 4y ago
Binary search three 6 bit symbols, that is 18 multiplications. A bit less if you do not blindly pick the centre but estimate - or guess if you will - what the correct digit should approximately be. I would assume they could have done this to much larger precision if they had cared enough as I find it hard to imagine that they would have been limited by having to test every candidate. If I asked you to find the square root of two to eight fractional digits, you would probably come up with the idea of skipping ahead more than 0.00000001 each time pretty quickly. I guess my point is that it might be interesting that they knew about the idea of the square root of two but the precision does not really surprise me.
- deleted 4y ago[deleted]
- geysersam 4y agoIt does't sound like a lot of work. However, multiplying two six digit numbers by hand isn't that easy! They might have used Newtons algorithm (not using that name probably), but then they would have had to divide, which is even more horrible by hand. Consider that their notation was probably far worse than what we are used to today, and it all becomes quite a hassle.
- danbruc 4y agoI guess for them it would be more like a three - four with the leading one - digit number, would it not? Did they have a 60 x 60 multiplication table at hand? I am almost tempted to try it out, I would guess that it should be doable in two hours in modern notation.
- saalweachter 4y agoI assume it was in their head. I don't know exactly how large of a times table I have in my head, but it's larger than 10x10, more like 15x15. A 60x60 table is 16 times larger than that, but I also learned my times table in elementary school and being a scribe or whatever the job title was for a guy who multiplied was very likely a full time job for a grown Babylonian. Learning your times tables up to 60x60 seems totally plausible.
- airstrike 4y ago15x15 plus 10xInfinity ;-)
- saalweachter 4y agoThere's some other odd ones like 16x16 but the table gets pretty sparse after the 15's.
- r_hoods_ghost 4y agoWe know that by 300bc at the latest that they had abacuses, so I imagine they used these
- RogerL 4y agoSumerians invented it circa 2800-2500BC. 300bc is a thousand years after the tablet in question. I believe the oldest extant example is 300bc.
- bee_rider 4y agoI would expect that, while it is a hassle, somebody would do it. And you only need to do the computation once. Then, stamp it on a square tablet and you can sell it! Probably make a whole set for different aspect ratios. Carpenters might like them. “I used to buy 1.5 ft cuttings of wood at the market to make the diagonals on my cabinets, but thanks to geysersam’s number blocks, I buy 1.42 ft cuttings of wood. It’s like every 17’th cabinet is free!”
- Dylan16807 4y agoI bet that a single digit, 1+25/60, is already more precise than any materials you're likely to buy. Heck, just 1+2/5 is likely sufficient.
- gleenn 4y agoSomeone else pointed out that they likely were doing astronomy predictions/measurements so you actually need the precision.
- Dylan16807 4y agoThat's cool! I was specifically talking about the carpenter use case.
- Ace322 4y agoThat crazy
- s1artibartfast 4y agoIt's not that hard in base 10. It might take 5 minutes with the long multiplication technique. I'm assuming that there is an equivalent in the Babylonian base 60 system
- crdrost 4y agoNewton's method is in fact what is described in the Twitter thread. x → x – f(x)/f'(x) with f = x² – N does in fact give you x → x – [x² – N]/(2 x) Which quickly simplifies to, x → (x + N/x)/2. With that said I am not sure that they knew this coverages quadratically or any of the machinery needed to make Newton's method work... It's a great way to approximate square roots!
- deleted 4y ago[deleted]
- dormento 4y ago> Newton's method is in fact what is described in the Twitter thread. We might in fact have been looking at Nasir's method this whole time.
- nextaccountic 4y agoWho is Nasir?
- rootusrootus 4y agoI was about to ask exactly this. I did a few iterations and it converges pretty fast, nice to know someone went the extra mile to see exactly how many iterations it would take. Annoying if you were doing it by hand, but not that bad really.
- a9h74j 4y agoAbout one bit (in binary equivalent) per naive iteration? Better with some linear interpolation (pseudo-Newton)? IIRC, before e.g. continued fraction representations/approximations became available, high precision estimates of pi were based upon geometric subdivisions of a circle and computation of hundreds of terms.