sbv-3.4: SMT Based Verification: Symbolic Haskell theorem prover using SMT solving.

Copyright(c) Levent Erkok
LicenseBSD3
Maintainererkokl@gmail.com
Stabilityexperimental
Safe HaskellNone
LanguageHaskell2010

Data.SBV.Examples.CodeGeneration.GCD

Contents

Description

Computing GCD symbolically, and generating C code for it. This example illustrates symbolic termination related issues when programming with SBV, when the termination of a recursive algorithm crucially depends on the value of a symbolic variable. The technique we use is to statically enforce termination by using a recursion depth counter.

Synopsis

Computing GCD

sgcd :: SWord8 -> SWord8 -> SWord8 Source

The symbolic GCD algorithm, over two 8-bit numbers. We define sgcd a 0 to be a for all a, which implies sgcd 0 0 = 0. Note that this is essentially Euclid's algorithm, except with a recursion depth counter. We need the depth counter since the algorithm is not symbolically terminating, as we don't have a means of determining that the second argument (b) will eventually reach 0 in a symbolic context. Hence we stop after 12 iterations. Why 12? We've empirically determined that this algorithm will recurse at most 12 times for arbitrary 8-bit numbers. Of course, this is a claim that we shall prove below.

Verification

We prove that sgcd does indeed compute the common divisor of the given numbers. Our predicate takes x, y, and k. We show that what sgcd returns is indeed a common divisor, and it is at least as large as any given k, provided k is a common divisor as well.

sgcdIsCorrect :: SWord8 -> SWord8 -> SWord8 -> SBool Source

We have:

>>> prove sgcdIsCorrect
Q.E.D.

Code generation

Now that we have proof our sgcd implementation is correct, we can go ahead and generate C code for it.

genGCDInC :: IO () Source

This call will generate the required C files. The following is the function body generated for sgcd. (We are not showing the generated header, Makefile, and the driver programs for brevity.) Note that the generated function is a constant time algorithm for GCD. It is not necessarily fastest, but it will take precisely the same amount of time for all values of x and y.

/* File: "sgcd.c". Automatically generated by SBV. Do not edit! */

#include <inttypes.h>
#include <stdint.h>
#include <stdbool.h>
#include "sgcd.h"

SWord8 sgcd(const SWord8 x, const SWord8 y)
{
  const SWord8 s0 = x;
  const SWord8 s1 = y;
  const SBool  s3 = s1 == 0;
  const SWord8 s4 = (s1 == 0) ? s0 : (s0 % s1);
  const SWord8 s5 = s3 ? s0 : s4;
  const SBool  s6 = 0 == s5;
  const SWord8 s7 = (s5 == 0) ? s1 : (s1 % s5);
  const SWord8 s8 = s6 ? s1 : s7;
  const SBool  s9 = 0 == s8;
  const SWord8 s10 = (s8 == 0) ? s5 : (s5 % s8);
  const SWord8 s11 = s9 ? s5 : s10;
  const SBool  s12 = 0 == s11;
  const SWord8 s13 = (s11 == 0) ? s8 : (s8 % s11);
  const SWord8 s14 = s12 ? s8 : s13;
  const SBool  s15 = 0 == s14;
  const SWord8 s16 = (s14 == 0) ? s11 : (s11 % s14);
  const SWord8 s17 = s15 ? s11 : s16;
  const SBool  s18 = 0 == s17;
  const SWord8 s19 = (s17 == 0) ? s14 : (s14 % s17);
  const SWord8 s20 = s18 ? s14 : s19;
  const SBool  s21 = 0 == s20;
  const SWord8 s22 = (s20 == 0) ? s17 : (s17 % s20);
  const SWord8 s23 = s21 ? s17 : s22;
  const SBool  s24 = 0 == s23;
  const SWord8 s25 = (s23 == 0) ? s20 : (s20 % s23);
  const SWord8 s26 = s24 ? s20 : s25;
  const SBool  s27 = 0 == s26;
  const SWord8 s28 = (s26 == 0) ? s23 : (s23 % s26);
  const SWord8 s29 = s27 ? s23 : s28;
  const SBool  s30 = 0 == s29;
  const SWord8 s31 = (s29 == 0) ? s26 : (s26 % s29);
  const SWord8 s32 = s30 ? s26 : s31;
  const SBool  s33 = 0 == s32;
  const SWord8 s34 = (s32 == 0) ? s29 : (s29 % s32);
  const SWord8 s35 = s33 ? s29 : s34;
  const SBool  s36 = 0 == s35;
  const SWord8 s37 = s36 ? s32 : s35;
  const SWord8 s38 = s33 ? s29 : s37;
  const SWord8 s39 = s30 ? s26 : s38;
  const SWord8 s40 = s27 ? s23 : s39;
  const SWord8 s41 = s24 ? s20 : s40;
  const SWord8 s42 = s21 ? s17 : s41;
  const SWord8 s43 = s18 ? s14 : s42;
  const SWord8 s44 = s15 ? s11 : s43;
  const SWord8 s45 = s12 ? s8 : s44;
  const SWord8 s46 = s9 ? s5 : s45;
  const SWord8 s47 = s6 ? s1 : s46;
  const SWord8 s48 = s3 ? s0 : s47;
  
  return s48;
}