In between the years, a PISA question (nothing essential): The last digit of 7^190

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Alexander 2023-12-28

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评论： Alexander 2023-12-29

采纳的回答： Paul

Last digit of 7^190, I thought a simple question if you have Matlab. But

fprintf('%f\n',7.0^190.0)

reports 4 as last digit, but it must be an odd number. 7^190 is fare away from realmax. Why that? It's not a big issue, only for my understanding.

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Stephen23 2023-12-28

编辑：Stephen23 2023-12-28

"7^190 is fare away from realmax"

REALMAX is irrelevant.

The value is well above FLINTMAX (so, as the documentation states, your value is outside the range of consecutive integer values), so there is no expectation that your value will be represented accurately down to the ones:

https://www.mathworks.com/help/matlab/ref/flintmax.html

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Paul 2023-12-28

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Well, if 7^190 is larger than realmax, then using double precison won't work to just try to compute it. You could try the Symbolic Math Toolbox. Or do some analysis and see the pattern in the last digit of 7^n for increments of n, and then figure out where the pattern would be for n == 190.

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John D'Errico 2023-12-28

编辑：John D'Errico 2023-12-28

在 MATLAB Online 中打开

@Alexander

format long g
flintmax
ans = 
      9.00719925474099e+15

is the largest integer that is exactly, uniquely representable as a double. (There is also a flintmax for singles too.)

flintmax is the number

2^53-1
ans = 
      9.00719925474099e+15

Any number larger than that is not uniquely represented by a double. If we go any further, then we would find the least significant bits of that number are not perfectly known.

Note that flintmax is significantly less than realmax, by many orders of magnitude, so there are many numbers that double can represent, but only imperfectly. And we see that your number lies in the nether region, between the two boundaries.

7^190
ans = 
      3.7036300802816e+160
realmax
ans = 
     1.79769313486232e+308

certaainly falls into that domain, as beyond flintmax, but well below realmax. Here is the exact number, computed as a sym

sym(7)^190
ans = 37036300802816027166397504837023132103935904664589688209993011047970173724580953468862621004041894128193388186915258951321654310122171416116357699953172946967249

And here is what the double precision version sees:

sym(7^190)
ans = 37036300802816026017917201676324198784916119353248748160351881978039946084074545334861220526439049530753562341694871901345447895734391162141803321189399405789184

The two numbers are very different beyond about the 16th decimal digit. So it is certainly true that the double version would have mostly garbage for digits.

John D'Errico 2023-12-28

编辑：John D'Errico 2023-12-28

在 MATLAB Online 中打开

@Stephen23

NO. realmax is NOT the largest integer exactly, uniquely representable as a double. NOT. NOT.

Just because you read that on Stackoverflow does not add any credibility. That claim is flat out wrong.

realmax
ans = 1.7977e+308
realmax == realmax - 1
ans = logical
   1

Do you see that? I can subtract 1 from realmax, and as far as MATLAB is concerned, it sees the same number.

Any number beyond flintmax is not UNIQUELY represntable by a double. That is why it is called flintmax. It is the largest INTEGER that is uniquely representable as a float.

All of this stems from the fact that MATLAB (as well as a huge number of other languages) uses 52 binary bits to represent doubles. (plus a sign bit, and an exponent.)

@Alexander

It is not a bug. A feature is perhaps a better description. But even the word fature suggests something unintended. And all of this is completely understood and accepted by those who wrote MATLAB and any other language that uses floating point arithmetic.

Really, all of this is just a limitation of floating point arithmetic. A float can only encode a fixed number of binary digits. So once you go beyond the flintmax limit, there will be more than one integer that all essentially get assigned to any value encoded in that double.

For example,

x = 1e18;
x == x + (1:10:100)
ans = 1×10 logical array
   1   1   1   1   1   1   1   0   0   0

So I've chosen X as a number that explicitly exceeds flintmax. But there will be many other integers in that region, all of which implicitly get stuffed into the same bin.

Rik 2023-12-29

Though Steven is technically correct that realmax is the largest integer that can be represented by a double, stating this is not helpful, as people will misinterpret it. Realmax does represent an integer, but since multiple numbers will map to realmax, its uses are limited. If you want to guarantee the correct last digit, flintmax is the limit.

There have been efforts to implement the quad datatype in Matlab, though to my knowledge that didn't reach a practical state. That would push flintmax to 2^113-1 (if I understand it correctly). Still not enough for this question, but a lot better.

Alexander 2023-12-29