integer and identity code

Can you guys help me with this one?

Thanks

[Roger Stafford]

In case it is a mystery to you why the described percentage is not one hundred percent, it should be realized that if x is any integer other than a power of two, the reciprocal 1/x cannot be achieved exactly in the binary numbers, which your computer uses, with a finite number of binary digits and must therefore approximate that reciprocal by rounding. It is then a matter of chance whether the multiplication x*(1/x) will turn out to be exactly one or off by some tiny amount.

Answer 1

[Star Strider]

It seems straightforward. Use the ‘double equal’ operator == to test for equality.

A 1 means the condition is met, a 0 indicates it was not. To understand the nature of the problem (if it’s not already been explained to you) see Why is 0.3 - 0.2 - 0.1 (or similar) not equal to zero? (link).

[John Jamison]

so how would i get the code to calculate how many = 1 and how many = 0?

[Star Strider]

The equality test will produce a logical 1 if the test is met and a logical 0 if it is not. In MATLAB, logical values become double-precision variables if you perform arithmetic operations on them. The sum function will work to sum the vector, so the result will be the number of times the condition is met. Divide that by 1000 to get the percentage (or by 100000 to get the decimal fraction) of times the condition was met. The number of 0 values is simply 100000 minus the number of 1 values. You can expand on that idea to get the percentage of 0 values in your experiment.

—————

EDIT — I get the same result Image Analyst gets:

N = 100000;
Test = [1:N].*(1./[1:N]);
Passed = sum(Test == 1);
Passed_Pct = Passed/1000

Passed_Pct =
                    86.884

[John Jamison]

Why do you multiply 1:N times the equation? Why can't you just use the equation: ie. (1 ./ 1:N)?

[Stephen23]

@John Jamison: if you consider it mathematically this code:

 [1:N].*(1./[1:N])

will create a vector of ones. But because floating point calculations accumulate errors, not all of the values will be ==1 when that code is calculated numerically by a computer. Therefore that code allows a simple estimate of what proportion of rational fractions within that range can be multiplied by their inverse (i.e. the integer) to be ==1.

[Star Strider]

Thank you, Stephen.

The vector code I used replaces an explicit for loop that would do essentially the same calculation, as the similarity of Image Analyst’s and my code results demonstrate.

[John Jamison]

I still don't get why its not just (1 ./ 1:N)?

thanks

[Star Strider]

The problem in the ‘Screen Shot 2017-02-01 at 5.55.24 PM.png’ that you posted wants to calculate: x*(1/x) and test to see if it equals 1. That is the reason it’s not just 1./[1:N].

Answer 2

[Image Analyst]

How about a simple for loop? Here's a start

equalityCount = 0;
for i = 1 : 100000
  if i * (1/i) == 1

I trust you can finish it. I get 86884 = 86.884% Empirically it seems to be higher than the theoretical number Roger suggests. Not sure why.

>> format long
>> 17 * (1/17)
ans =
     1
>> 17 * (1/17) == 1
ans =
  logical
   1