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Mathematics Mathematica Algebra Wolfram Language Operating System and Network Access

Result of Equal with multiple arguments

Martin Guttmann

Posted 6 years ago

In the documentation it is written: "Approximate numbers with machine precision or higher are considered equal if they differ in at most their last seven binary digits (roughly their last two decimal digits)." and "e1==e2==e3 gives True if all the ei are equal." This works fine with an equation of two arguments in all combinations, but it fails with a comparison of three values, but why? a = 0.999999999999991; b = 1.000000000000000; c = 1.000000000000001; Print[a == b, " : a==b is True"] Print[b == c, " : b==c is True"] Print[a == c, " : a==c is also True"] Print[a == b == c, " : a==b==c should be True as well?"] True : a==b is True True : b==c is True True : a==c is also True False : a==b==c should be True as well? I use Mathematica 10.0.2.0 on Windows 7 (x64)

POSTED BY: Martin Guttmann

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Szabolcs Horvát

Posted 6 years ago

I don't know the exact meaning of `$EqualTolerance`, but its value is 7 bits. In[108]:= Internal`$EqualTolerance/Log10[2] Out[108]= 7. What does that mean? It is not sufficient to compare the last 7 (binary) digits. Using a decimal example, the last 1 digit of `999` and `1000` are not the same even though their difference is only 1. Similarly, `a<1` but `c>b>1`. If we offset them, we get rid of this problem. What is the difference between `({a, b, c} + 0.00000005551145)` and `({a, b, c} + 0.00000005551146)` (see my last post)? Let's see their binary digits: Notice that the one that returns False has 7 digits of difference, the other only 6. 7 is exactly the threshold. The behaviour must have something to do with this. But I still do not understand it.

POSTED BY: Szabolcs Horvát

Szabolcs Horvát

Posted 6 years ago

Where can I post a bug report? http://support.wolfram.com is not that hard to find ... I am not going to comment on whether this should be considered a bug or not, but I will put one question to the community: What would be a reasonable implementation in your opinion? Keep in mind that equality with tolerance is not transitive, and that doing all pairwise comparisons is slow ( $O(n^2)$ would not be nice). One could do `Round`ing and then do exact comparison, but what if two very close values fall on two sides of the rounding boundary? I'm quite curious about how this is implemented.

POSTED BY: Szabolcs Horvát

Szabolcs Horvát

Posted 6 years ago

It's weird that it does this: In[91]:= Equal @@ ({a, b, c} - 0.1) Out[91]= False In[92]:= Equal @@ ({a, b, c} - 0.2) Out[92]= True In[93]:= Equal @@ ({a, b, c} - 0.3) Out[93]= False In[94]:= Equal @@ ({a, b, c} + 0.00000005551146) Out[94]= True In[95]:= Equal @@ ({a, b, c} + 0.00000005551145) Out[95]= False

It's weird that it does this:

In[91]:= Equal @@ ({a, b, c} - 0.1)
Out[91]= False

In[92]:= Equal @@ ({a, b, c} - 0.2)
Out[92]= True

In[93]:= Equal @@ ({a, b, c} - 0.3)
Out[93]= False

In[94]:= Equal @@ ({a, b, c} + 0.00000005551146)
Out[94]= True

In[95]:= Equal @@ ({a, b, c} + 0.00000005551145)
Out[95]= False

POSTED BY: Szabolcs Horvát

Arnoud Buzing

Arnoud Buzing, Wolfram Research

Posted 6 years ago

[removed, because I read the question wrong]

POSTED BY: Arnoud Buzing

Szabolcs Horvát

Posted 6 years ago

In my System (Mma 7.0) a == b == c gives True. I checked several versions. v10.0 and later give False. v7, v8 and v9 give True. Our institutional cluster even has v6 installed, but it no longer starts up on modern Linux.

POSTED BY: Szabolcs Horvát

Sander Huisman

Sander Huisman, University of Twente

Posted 6 years ago

The order still doesn't matter fortunately: a == b == c a == c == b b == c == a b == a == c c == a == b c == b == a all give the same Maybe the one-bit-off doesn't work anymore for the general vararg?