Calculator Forensics (2002)

(rskey.org)

86 points | by ColinWright 4 days ago ago

38 comments

  • LeoPanthera 10 hours ago

    It's better to start from the parent page otherwise you're missing all context:

    https://www.rskey.org/~mwsebastian/miscprj/forensics.htm

    This is where this post should probably point, IMHO.

  • OskarS 16 hours ago

    Couldn't resist checking what math.h does for floats and doubles on godbolt, it returns slightly above 9, but curiously not exactly equal to any answer in that list [1]. Maybe not that surprising, these transcendental functions are always iffy in the last digits.

    [1]: https://godbolt.org/z/dK85Eq8r6

    • chongli 10 hours ago

      I would be really surprised if any physical calculator used binary floating point numbers instead of binary-coded decimal. The appearance of repeating bits even for relatively simple decimal numbers like 0.1 is likely to confuse and annoy a calculator user expecting an exact answer.

      • OskarS 5 hours ago

        I would assume most basic calculators use regular floating point circuits, and then just round aggressively for the display.

    • masfuerte 14 hours ago

      CORE-MATH provides correctly rounded transcendental functions. (Though that doesn't mean that this calculation will return exactly 9. And I haven't tried it.)

      https://core-math.gitlabpages.inria.fr/

    • avipars 7 hours ago

      chalk it up to floating point accuracy

  • avipars 7 hours ago

    Wolfram alpha result - https://www.wolframalpha.com/input?i=arcsin+%28arccos+%28arc...

    3pi - 9

    = 0.42477796076937971538793014983850865259150819812531746292483377692344921885

    • nneonneo 7 hours ago

      This is the result if you do it in radians, but the test actually requires that you set the calculator to degrees mode first.

      Note that all of the inverse trig functions are multivalued because the trig functions are periodic. Here, Wolfram Alpha is giving you one of the possible answers. The entire family of answers should be +/-9 + n*pi for any integer n; the sign on the 9 is due to cos being an even function.

      • dominicrose an hour ago

        irb(main):012> Math.asin(Math.acos(Math.atan(Math.tan(Math.cos(Math.sin(9*Math::PI/180))))))*180/Math::PI => 9.00000000000001

  • kazinator 17 hours ago

    Forensics? Have these calculators committed a crime (against numerical analysis)?

    • maxbond 13 hours ago

      I think it's forensic in the sense that, should you perform this operation on an unknown calculator, this chart could be used to identify it.

      • maxbond 10 hours ago

        On second thought what you'd probably do is identify the chipset of a calculator.

        • kstrauser 10 hours ago

          I think you'd identify the "OS" of a calculator. For example, HP famously had lots of long, complex, highly accurate algorithms in its firmware. They could reuse large portions of that between different models, regardless of the hardware executing it.

          • maxbond 8 hours ago

            For sure. I was imagining that the software would be pretty thin and the FP behavior would be primarily a function of chipset, but I stand corrected.

    • gs17 16 hours ago

      If they have, this site has their fingerprints.

    • gblargg 8 hours ago

      This is a good way to find out what fugitive chip an unknown calculator is harboring.

    • bobmcnamara 14 hours ago

      Failure to identify an identity! Straight to calculator jail!

  • petters 11 hours ago

    Numerically difficult since intermediate values end up close to a local maximum of cos

  • harshreality 10 hours ago

    It would be nice to see an update. HP Prime? TI Nspire series? Swissmicros? Numworks? (the numworks app gives 9, but are the algorithms hardware-independent?)

  • qiqitori 12 hours ago

    Nice. Checked on a Commodore SR-37, produces the same result as the SR-36 per the site. (9.08210803 Commodore SR-36)

  • quirkot 17 hours ago

    I always wonder... if there was an AGI and it's chipset gave the wrong answer, how would it ever know?

    • GuB-42 17 hours ago

      The neural networks we use today have really terrible accuracy, and we tend to make them worse, not better, as having more neurons is better than having more precision. Human brains are also a mess, but somehow, they work, and we are usually able to correct our own mistakes.

      Since by AGI, we usually mean human-like, that system should be able to self correct the same way we do.

    • nenenejej 15 hours ago

      How do humans know? usually someone corrects someone else. we have repeatability in physics, or we wait 30 years and quash convictions etc. etc.

    • gs17 16 hours ago

      I'd presume it could reason around the wrong answer, at least to realize something was off. Current LLMs will sometimes hallucinate that this has happened when they're "thinking".

  • _trampeltier 17 hours ago

    Didn't expect to see this site here. From time to time I show this site to young guys.

  • mysteria 16 hours ago

    So what's the correct answer?

    • gs17 16 hours ago

      9 degrees. arcsin(arccos(arctan(tan(cos(sin(9)))))) basically makes a set of sin-cos-tan layers that arctan-arccos-arcsin unwrap one-by-one, which should result in nothing having changed, unless the functions used weren't accurate.

      • measurablefunc 16 hours ago

        That's incorrect, you have to choose the proper inverse branch if you want the answer to be 9.

        • madars 15 hours ago

          There is no choice here - each inverse is uniquely determined. That's similar to how 3 and -3 are both square roots of 9 (i.e., solutions to x^2=9), but sqrt(9)=3 as it denotes the principal square root, which by convention is always the non-negative value. Of course, in a different context we might design functions to have multi-valued properties, like atan2(x,y) != atan(y/x) in general (atan2 takes quadrant in account and returns full range [-pi, pi], atan only returns principal values in [-pi/2, pi/2]) as practical applications benefit from preserving quadrant beyond just the principal inverse (or not failing when x=0!)

          • measurablefunc 15 hours ago

            The inverse branches are not unique, you might think there is no choice being made but picking the standard branch is a choice b/c I can always shift the result by 2π by picking a different branch of the inverse. The answer is not unique & the assumption is that the calculators are using the standard branch.

            • madars 15 hours ago

              Of course, but the choice is standard and thus the answer is 9. I can define a non-standard sqrt(x) which sometimes gives the positive root and sometimes the negative one, and then sqrt(sqrt(16)) could be -2 or undefined (if I defined sqrt(16)=-4) but that's just silly - the only reasonable interpretation for what the calculator should show for sqrt(sqrt(16)) is simply 2.

        • deruta 3 hours ago

          I was with you until I remembered the default unit for angles in calculators is degrees, not radians.

        • saagarjha 13 hours ago

          Yes, that's what those functions do.

    • measurablefunc 16 hours ago

      You can assume that sin(9) is within the range of all the functions that are post-composed w/ it so what you end up w/ in the end is arcsin(sin(9)). Naively you might think that's 9 but you have to be careful b/c the standard inverse branch of sin is defined to be [-1, 1] → [-π/2, π/2].

      Edit: The assumption is that the calculators are using specific branches of the inverse functions but that's still a choice being made b/c the functions are periodic there are no unique choices of inverse functions. You have to pick a branch that is within the domain/range of periodicity.

    • madars 16 hours ago

      arcsin(arccos(arctan(tan(cos(sin(9)))))) = 9 (in degrees mode - when regular trig functions output pure numbers, those numbers get interpreted as degrees for the next function and similar for inverses - calculator style), because each intermediate lands in the principal-value domain of the next inverse (e.g., arctan(tan(x)) = x when x \in (-90°, 90°) and the intermediates happen to be in those ranges). Specifically, sin(9°) ≈ 0.156434, cos(0.156434°) ≈ 0.999996, arctan(tan(0.999996°)) = 0.999996°, arccos(0.999996)≈0.156434°, arcsin(0.156434)≈9°.

  • ani17 13 hours ago

    thanks for sharing!

  • tiahura 14 hours ago

    Disappointed in the HP-48.

  • tomhow 16 hours ago

    Previously:

    Calculator Forensics (2002) - https://news.ycombinator.com/item?id=42757455 - Jan 2025 (1 comment)

    Calculator Forensics (2002) - https://news.ycombinator.com/item?id=28561298 - Sept 2021 (19 comments)

    Calculator Forensics (2002) - https://news.ycombinator.com/item?id=7682045 - May 2014 (2 comments)