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Date: Mon, 25 May 2015 16:06:06 -0400
From: felix <felix@epepm.cupet.cu>
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Subject: Re: a more precise distance algorithm
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El 25/05/15 15:21, ravas escribió:
> I read an interesting comment:
> """
> The coolest thing I've ever discovered about Pythagorean's Theorem is an alternate way to calculate it. If you write a program that uses the distance form c = sqrt(a^2 + b^2) you will suffer from the lose of half of your available precision because the square root operation is last. A more accurate calculation is c = a * sqrt(1 + b^2 / a^2). If a is less than b, you should swap them and of course handle the special case of a = 0.
> """
>
> Is this valid? Does it apply to python?
> Any other thoughts? :D
>
> My imagining:
>
> def distance(A, B):
>      """
>      A & B are objects with x and y attributes
>      :return: the distance between A and B
>      """
>      dx = B.x - A.x
>      dy = B.y - A.y
>      a = min(dx, dy)
>      b = max(dx, dy)
>      if a == 0:
>          return b
>      elif b == 0:
>          return a
>      else:
>          return a * sqrt(1 + (b / a)**2)
I don't know if precision lose fits here but the second way you gave to 
calculate c is just Math. Nothing extraordinary here.

c = a * sqrt(1 + b^2 / a^2)
c = sqrt(a^2(1 + b^2 / a^2)) applying the inverse function to introduce 
a inside the square root
c = sqrt(a^2 + a^2*b^2/a^2) then just simplify
c = sqrt(a^2 + b^2)


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    <div class="moz-cite-prefix">El 25/05/15 15:21, ravas escribi&oacute;:<br>
    </div>
    <blockquote class=" cite"
      id="mid_b2e66a94_7a89_4be9_bbf7_9434396cc178_googlegroups_com"
      cite="mid:b2e66a94-7a89-4be9-bbf7-9434396cc178@googlegroups.com"
      type="cite">
      <pre wrap="">I read an interesting comment:
"""
The coolest thing I've ever discovered about Pythagorean's Theorem is an alternate way to calculate it. If you write a program that uses the distance form c = sqrt(a^2 + b^2) you will suffer from the lose of half of your available precision because the square root operation is last. A more accurate calculation is c = a * sqrt(1 + b^2 / a^2). If a is less than b, you should swap them and of course handle the special case of a = 0.
"""

Is this valid? Does it apply to python?
Any other thoughts? :D

My imagining:

def distance(A, B):
    """
    A &amp; B are objects with x and y attributes
    :return: the distance between A and B
    """
    dx = B.x - A.x
    dy = B.y - A.y
    a = min(dx, dy)
    b = max(dx, dy)
    if a == 0:
        return b
    elif b == 0:
        return a
    else:
        return a * sqrt(1 + (b / a)**2)
</pre>
    </blockquote>
    I don't know if precision lose fits here but the second way you gave
    to calculate c is just Math. Nothing extraordinary here.<br>
    <br>
    c = a * sqrt(1 + b^2 / a^2) <br>
    c = sqrt(a^2(1 + b^2 / a^2)) applying the inverse function to
    introduce a inside the square root<br>
    c = sqrt(a^2 + a^2*b^2/a^2) then just simplify<br>
    c = sqrt(a^2 + b^2)<br>
    <br>
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