Documentation.SBV.Examples.BitPrecise.BrokenSearch

The classic "binary-searches are broken" example: http://ai.googleblog.com/2006/06/extra-extra-read-all-about-it-nearly.html

Documentation

midPointBroken :: SInt32 -> SInt32 -> SInt32

Model the mid-point computation of the binary search, which is broken due to arithmetic overflow. Note how we use the overflow checking variants of the arithmetic operators. We have:

>>> checkArithOverflow midPointBroken
Documentation/SBV/Examples/BitPrecise/BrokenSearch.hs:35:28:+!: SInt32 addition overflows: Violated. Model:
  low  = 2147483647 :: Int32
  high = 2147483647 :: Int32

Indeed:

>>> (2147483647 + 2147483647) `div` (2::Int32)
-1

giving us a negative mid-point value!

midPointFixed :: SInt32 -> SInt32 -> SInt32

The correct version of how to compute the mid-point. As expected, this version doesn't have any underflow or overflow issues:

>>> checkArithOverflow midPointFixed
No violations detected.

As expected, the value is computed correctly too:

>>> checkCorrectMidValue midPointFixed
Q.E.D.

midPointAlternative :: SInt32 -> SInt32 -> SInt32

Show that the variant suggested by the blog post is good as well:

mid = ((unsigned int)low + (unsigned int)high) >> 1;

In this case the overflow is eliminated by doing the computation at a wider range:

>>> checkArithOverflow midPointAlternative
No violations detected.

And the value computed is indeed correct:

>>> checkCorrectMidValue midPointAlternative
Q.E.D.

Helpers

checkArithOverflow :: (SInt32 -> SInt32 -> SInt32) -> IO ()

A helper predicate to check safety under the conditions that low is at least 0 and high is at least low.

checkCorrectMidValue :: (SInt32 -> SInt32 -> SInt32) -> IO ThmResult

Another helper to show that the result is actually the correct value, if it was done over 64-bit integers, which is sufficiently large enough.