results on PC, while interesting, won't be reported in the list (that is missing so far the newest posts) though.
It is about pocket devices with calc apps or calculators (thus also desktop calc like the HP 9100).

For PC stuff there are more reasonable benchmarks out there, I think.

(04-13-2023 05:14 PM)stineylee Wrote: [ -> ]There are several different types of benchmarks that can be used, depending on the specific features and capabilities of the calculator in question. Summation-based benchmarks, which test a calculator's ability to perform basic arithmetic operations such as addition and subtraction, are one common type of benchmark.

Here are a few summation-based benchmarks that are commonly used to test calculators:

The "10-key" test: This benchmark measures a calculator's ability to enter and calculate a series of numbers quickly and accurately, using only the 10 numeric keys (0-9) and the basic arithmetic functions (+, -, x, รท). In this test, the user is given a series of 10-digit numbers to enter and calculate, and the speed and accuracy of their input is measured.

The "addition stress test": This benchmark measures a calculator's ability to perform rapid-fire addition calculations without making errors. In this test, the user is given a series of simple addition problems (e.g. 1+2, 3+4, etc.) to solve as quickly as possible, with a set time limit.

The "accuracy test": This benchmark measures a calculator's ability to perform accurate calculations over a range of values. In this test, the user is given a series of increasingly complex arithmetic problems to solve, and the accuracy of their answers is compared to a known correct result.

The "decimal test": This benchmark measures a calculator's ability to handle decimal numbers accurately and consistently. In this test, the user is given a series of arithmetic problems involving decimal numbers, and the accuracy of their answers is compared to a known correct result.

The "repeated calculation test": This benchmark measures a calculator's ability to perform the same calculation repeatedly without making errors. In this test, the user is given a simple arithmetic problem to solve, and is then asked to solve it again and again as quickly as possible, without making mistakes.

Are you sure those tests are not just to test the operators?

(04-13-2023 05:14 PM)stineylee Wrote: [ -> ]Here are a few summation-based benchmarks that are commonly used to test calculators:

I agree with Ian, these 'benchmarks' test the user's skills and knowledge much more than they test calculators, unless you intended to use the word 'calculators' to mean the actual people being tested...

And having spent considerable time in a factory in Japan, the young lady seen in the video is no stunt for the TV piece, there are plenty of young ladies (and maybe men, but all that I met were ladies) with these skills, and they are highly sought-after and even 'stolen' away from one company or department for another. Really impressive to watch live!

(12-22-2017 09:41 PM)pier4r Wrote: [ -> ]So as said the summation based test will be very similar to the savage benchmark and I wonder how many similar tests were developed on various forum or groups regarding calculators.

Anyway, inspired by the test done in the thread of Eddie linked in the 1st post, I decided to pick randomly some scientific functions to assemble a summation that may give an idea of the performances of a calculator for common math functions.

Your results show
~ 18s - fx9860GIII , micropython result has reduced precision p#199

When I run the following on my fx9860GIII with OS version 3.70.0200, I get < 7 seconds.
I really wish it had an RTC. My reflexes aren't that great!

Code:

from math import *
s=0
for x in range(1,1001):
s+=(exp(sin(atan(x))))**(1/3)
print(s)

Quote:Casio fx-9860GIII, OS version 3.70.0200, MicroPython

N options:
For 100 the computation takes too little time, Result: 100.0
1000 < 1 second Result: 1000.0
10,000 3.9 seconds Result: 10000.0
100,000 62 seconds Result: 100000.0
For 1,000,000 the computation takes too much time, I skipped it

Code:

Code:

from math import *
def fourops(n):
sum = 0
b = 0; c = 0
for a in range(1, n+1):
b = a
c = a
sum += sqrt(b*c)/a
print(sum)

...snip... would find it interesting to see that on a single implementation the value you get does not really converge depending on the underlying precision:

Please post those directly in the thread because if I am practially inactive (as I am more or less since 2019) those (maybe) interesting values could go lost.
Now going to add those too in the main post.