Ask HN: Whatever happened to Wolfram Alpha?

They're just serving up answers which is boring to HN readers. Where's the drama in collecting data privately? Where's the drama from censoring results? No drama == No interest? Gawd, I have become cynical.

You should search comments, rather than stories. It's very regularly referenced in HN comments, often for calculations, sometimes in other contexts.

https://hn.algolia.com/?dateRange=all&page=0&prefix=false&qu...

This is for two pairs of R/C in series ( R||C + R||C ).

Real:

https://www.wolframalpha.com/input/?i=Real%28+%281%2F%281%2F...

Imaginary:

https://www.wolframalpha.com/input/?i=Imaginary%28+%281%2F%2...

Edit: Sorry, I don't know how to make the search query text show up since it has special characters, probably best to just use the links to see the query.

Why would it?

I use Google for those pretty often.

Yeah, it's great for these types of things. It also has a bunch of values built in, so you can do things like:

  (day length of jupiter) * 80

you might want to try units(1).

https://www.gnu.org/software/units/units.html

the input language is less flexible than wolframalpha/google, but i quickly got used to it. it's nice to have something local and reliable. you can also define custom units.

i prefer using it in terse mode:

    $ units -t 0.03$/hr*1month
    21.914532 US$
    $ units -t 10TB/month Mbps
    30.421214

   */ 0.03 24 30

   1e6 %~ (10e12 * 8) % */ 30 24 3600

    141 pounds force 30 mm  # in joules
    1160/4
    log(3)/3/(log(2)/2)  # how much more efficient is one-hot ternary than one-hot binary?
    5V 7 μs / 7.3 A
    .0117% half avogadro mol / 1.251e9 years / (potassium+chlorine)g  # how radioactive is lite salt?
    3.27$/gallon  # in $/liter
    sqrt(2 2000 electronvolt/electronmass)
    18.8 foot pounds force # in joules
    163$/(7.9 g/cc * 1500 mm 3000 mm 3.2 mm)  # cold rolled steel price is higher than steel sold by weight
    m3/4 / 15 cfh
    2 pi sqrt(200 um / gravity)

    $ qalc '0.03$/hr*1month'
    error: "r" is not a valid variable/function/unit.
    (0.03 × (USD / hour)) × (1 × month) = $21.915

    $ qalc '0.03$/h*1month -> CAD'
    (0.03 × (USD / hour)) × (1 × month) ≈ CAD 27.28388011

    $ qalc '10TB/month -> Mbit/s'
    10 × (terabyte / month) ≈ 30.42056430 megabits/s

    $ qalc 'integrate(x+x^2)'
    integrate(x + (x^2)) = x^3 / 3 + x^2 / 2 + C

You also might want to try google search. They display calculations for these particular queries and quite a few more.

Same here, the way it seamlessly wrangles even the most ridiculous combinations of units is insanely useful. Just yesterday I used it to calculate power consumption for a house by timing one of those spinning wheel meter things. Something like "(10 rot / 46 s) / (375 rot / kW*h)" and it gave me straight answer in watts.

I definitely could've worked that out by hand, but it would've taken a minute or a few, mostly on unit conversions. With WA, I can just think in variable relationships and not worry about units at all.

Don't get me wrong, it often returns complete garbage, see all the memes of Siri passing non-math questions to it. It's annoying to figure out or explain to someone because the syntax is very loose and you just kind of need to get a feel for it, but once you do, it's really powerful.

or frink [1], which started off as a tool like the others mentiond here, but is now a full fledged units-based programming language. See some examples of them here [2]

1: https://frinklang.org/ 2: https://frinklang.org/#SampleCalculations

Its been on HN before.

Wolfram isn't reporting how much protein you'll get from eating something; it's reporting how much there is in the bread. Protein assimilation depends on a huge range of factors, and varies significantly between individuals (based on everything from gut microbiome to health factors to how much you chew your food to your saliva production to... Well, it's a long list). There's no way a website could report the amount of protein you will get from bread. Reporting how much is in the bread makes much more sense. It's a shame your friend didn't explain that.

This is something that actually annoys me immensely when people say "you eat too much!" to fat people. Two people can have the exact same diet and the exact same exercise regime, and if one assimilates particular foods more effectively they'll be getting more calories, and put on weight. Food intake is far more complex than many people believe.

It might help to explain what your friend says is “wrong and useless” so others could provide feedback.

It also might help to avoid insinuating things about strangers online in order to promote discussion and not stifle it.

It seems that Wolfram Alpha also has some difficulty figuring out whether I'm talking about raw oats or cooked oats, even when I use the word raw in my query. As a result, it can be off by a factor of 3. I agree that it's not useful if you have to carefully check the output every time.

>protein assimilability from bread is 40%

This is wrong; the digestibility of gluten is 80-90%. Your friend was probably thinking of the PDCAAS, which is more like 45 for gluten. But this is nutritional quality vs an egg white equivalent as defined by the bioavailability and concentration of essential amino acids (egg = 100 by definition; the score is based on the lowest fraction of any EAA, so gelatin — no tryptophan — has PDCAAS 0), not the fraction absorbed or utilized. For an idea of what utilization looks like see e.g.:

https://mdpi-res.com/d_attachment/nutrients/nutrients-10-001...

> The net protein utilization is profoundly affected by the limiting amino acid content—the essential amino acid found in the smallest quantity in the foodstuff. It is therefore a good idea to mix foodstuffs that have different weaknesses in their essential amino acid distributions.

> The limiting amino acid for wheat is lysine.

From what I gather, you still can process all of the protein from wheat if you get lysine from somewhere else:

> A vegetarian or low animal protein diet can be adequate for protein, including lysine, if it includes both cereal grains and legumes.

This also means that any statements about protein utilization from compound meals are more-or-less bogus if done without calculating the different amino acids.

You can improve the protein assimilability of bread by combining it with a high-lysine protein (and I'm not sure but I think eggs and cheddar might fit the bill) but you may not care if you're looking for low-glycemic-index low-fat calories rather than amino acids specifically.

I don't know what's wrong with the downvoters.

Step by step solutions were free around 9/10 years ago

SymPy Gamma, for example: https://www.sympygamma.com/input/?i=diff%28a**%285*x**2%29%2...

A reminder that GNU Units still exists, e.g.

  $ units
  Currency exchange rates from FloatRates (USD base) on 2021-01-17 
  3677 units, 109 prefixes, 114 nonlinear units
  
  You have: 4 amu * (1000 nm/s)^2
  You want: joules
          * 6.6421563e-39
          / 1.5055352e+38

  You have: ^D

It’s slightly less DWIMish (you have to say “atomicmassunits”, “atomicmassunit”, “amu”, or “u”, not “atomic mass units”) and somewhat awkward as a separate tool, but then resorting to your web browser for unit conversions is awkward in a different way. Non-interactive invocations, like units VALUE-OR-UNIT UNIT, work as well.

[1]: https://www.gnu.org/software/units/

... Seriously, though, if you’re actually need this type of calculation regularly and didn’t just pick a random example, atomic-scale calculations are absolutely miserable to do in SI (and this is not a problem, it’s a human-scale, engineering system, after all; and its metrological aspects, which were the actual advance originally, are completely unimportant here).

If I had to do this in my head or with a desk calculator, I’d just do it in high-energy units (c = ℏ = 1, mass and energy in eV, length and time in eV^-1). So,

  4 amu = 4 × 0.93 GeV (a proton weighs 939 MeV, an amu is slightly smaller due do binding energy, rounding to 1 GeV is good enough for most purposes) ≈ 4 GeV,

  (1000 nm / s)^2 = (1e4 Å / s)^2 = (1e4 / 1.97 keV^-1 s^-1)^2 (an angstrom is a typical atomic size, a keV is a typical [large] atomic energy, a fermi aka femtometer is a typical nuclear size, a MeV is a typical [not so large] nuclear energy, remember any of 197 MeV fm = 1.97 keV Å = 1, though again 200 is almost always good enough) ≈ (1e4 / 2 keV^-1 s^-1)^2 = 25e6 keV^-2 s^-2,

  4 GeV × 25e6 keV^-2 s^-2 = 4e6 keV × 100e6/4 × keV^-2 s^-2 = 1e14 keV^-1 s^-2.

This is slightly inconvenient, we wanted energy in eV, but the seconds don’t seem to want to go away. I don’t remember Planck’s constant in eV s, but I do remember 2 keV Å ≈ 1 and 300e3 km/s = 3e8 m/s = 1, so let’s sprinkle it with those,

  1e14 keV^-1 s^-2 ≈ 1e14 keV^-1 s^-2 × (2 keV Å)^2 / (3e8 m/s)^2 = 4/9 × 1e14 × 1e-16 keV Å^2 m^-2 = 0.44 × 1e14 × 1e-16 keV × (1e-10)^2 ≈ 0.44e-22 keV ≈ 0.44e-19 eV.

The hardest part is pretending to be a normal person: you have to remember what an electronvolt actually is in normal units. Good thing this is numerically the same as remembering the charge of an electron in coulombs (1 eV = 1.6e-19 J),

  0.44e-19 eV = 0.44e-19 eV × 1.6e-19 J / eV (turns out converting to a decimal fraction wasn’t a good idea after all, powers of two FTW) ≈ 4/9 × 16 × 1e-1 × 1e-19 × 1e-19 J = 64/9 × 1e-39 J ≈ 63/9 × 1e-39 J = 7e-39 J.

Good enough to a couple percent.

OK, I won’t pretend that this is easy or that I did it flawlessly the first time just now, but I do think this looks like a skill you could plausibly learn, unlike the textbook “SI all the things” calculation. The good news is that you’ve just seen essentially all the relevant constants you’re going to have to remember, except maybe Avogadro’s number if you’re going to have moles somewhere.

(One place where this doesn’t help is first-principles chemistry, things like electrolysis, because you need to subtract large binding energies to get a change that’s hundreds to thousands times smaller. Calculating things to a couple percent just isn’t good enough.)

    You have: 200 ft**3 / (10 mL/min)
    You want: months
     * 12.921493

> I wish I had a really good calendar math utility

Might be a good learning exercise in machine learning: translating natural-language queries from that domain to whatever standard utility.

WTactualF? When has * ever been anything other than multiplication? Why would the resulting links all be discussing division?

I don’t understand the problem. You are asking what is 48*6, and the correct answer is right at the top.

Hope I'm not sounding like a grammar nazi but its "buried the lede" I only recently discovered - the reason is really interesting - https://www.merriam-webster.com/words-at-play/bury-the-lede-...

For most of these, Wolfram Vertical Line Alpha seems to give reasonable results. However, for the third one, because I'm in Argentina, it helpfully converts US$79.80 into Argentine pesos, getting an answer that's off by about a factor of 2: AR$7969.44. As https://preciodolarblue.com.ar/ explains, the current bid and ask prices for the dollar are AR$195 and AR$199. Wolfram Vertical Line Alpha is apparently using the "official" rate of AR$99.45 or so; this is the rate at which the government converts your dollars into pesos if you are an exporter, but you cannot convert your pesos into dollars at this rate without special permission, granted, for example, if you are going on vacation to Disney World.

That's a bit recursive. I'd have needed english translations of the german articles to get to know the english math terms to be able to write about them in english :)

This sounds bad for the state of the world?

I'm talking about both. When I was comparing them to competitors like Symbolab I was using the Alexa ranking for alpha.

I find it faster and more accurate to use a specific package in an interpreter than query Wolfram Alpha or use Mathematica. And for the simpler things a search engine will do!

I don't think so. You'd need to run it in a terminal with something like

    from sympy.plotting import plot3d
    x,y=symbols('x y')
    plot3d(x*y, (x, -10,10), (y, -10,10))

I doubt the bad performance is due to evaluating expressions itself. If I type N[Log[2]] into Mathematica, it evaluates in less than a millisecond. It's probably because Wolfram Alpha is using natural language process to try to process my query and then finally deciding that by N[Log[2]], I mean N[Log[2]]. And it's probably not because of that, but because their grid scheduler isn't optimized for sub-second latency.

> Sure, there's something to be said for implementing logic in high-level code, but without a plan for lowering that high-level logic to machine code in a way that performs well, you're setting yourself up for long-term pain.

Whatever the reason for the performance issue (I don't know enough about WA to speculate what/why/how), I feel like noting the existence of the wolfram compiler[0] and the various language interfaces[1]. Anyone interested in using Mathematica/WL might get a kick out of exploring those more, at the very least.

[0] https://reference.wolfram.com/language/Compile/tutorial/Over...

[1] https://reference.wolfram.com/language/guide/CLanguageInterf... (a lot of the paclets are bindings for C libraries too)

Mathematica is extremely performant for most of the built-ins, the overhead of interpretation is nearly negligible for all but the tiniest operations.

There is also no reason to think that their request-response boilerplate is written in Mathematica, Mathematica is fully integrated with a lot of languages and runtimes.

I think that's the point. "My internet speed is fast enough that it is not the cause of slowness, so any delay is all on Wolfram|Alpha."

If I didn't include that note, someone would say "Is is slow because you're on 56kbps dial-up?"