证明数组的平均值

3

你好我想证明对于一维数组中包含的所有值的平均值的计算,以下是我的程序:

#include <stdbool.h>

typedef unsigned int size_t;
typedef struct Average avg;
struct Average
{
    bool success;
    float average;
};
/*@
axiomatic Float_Div{
    logic real f_div(real a,real b) = a/b ;

    axiom div:
        \forall real q,a,b;  0 != b ==>
        (a == b*q <==> q == f_div(a, b));

    axiom split :
        \forall real q,a,b,c;  0 != b ==>
        f_div(a + c , b) == f_div(a,b) + f_div(c,b);

}

axiomatic Average {
    logic real average(int * t, integer start, integer stop, integer size);

    axiom average_0:
        \forall int *t, integer start , integer stop, size;
        start >= stop ==> average(t,start, stop, size) == 0;

    axiom average_n:
        \forall int *t, integer start , integer stop, integer size;
        start < stop && size >0  ==>
        average(t,start, stop, size) ==
        f_div((real)stop-1 ,(real) size) +( average(t,start, stop-1, size) );

    axiom average_split :
        \forall int *t, integer start ,integer middle, integer stop, integer size;
        start < middle < stop && size >0  ==>
        average(t,start, stop, size) ==  average(t,start, middle, size) + average(t,middle, stop, size);

    axiom average_unit :
        \forall int *t, integer start , integer stop, integer size;
        start == stop-1 && size >0  ==>
        average(t,start, stop, size) == f_div((real)stop-1 ,(real) size);

}

*/


/*@
requires \valid(array + (0..size-1));
ensures (!\result.success) ==> size == 0 ;
ensures (\result.success) ==> \result.average == average(array, 0, size, size);
assigns \nothing;
*/
avg average(int * array, size_t size){
    avg ret;
    ret.success = true ;
    ret.average = 0 ;
    if (size == 0){
        ret.success = false;
        return ret;
    }
    float average = 0;
    /*@
    loop assigns i, average;
    loop invariant 0 <= i <= size;
    loop invariant  average(array , 0, i , size) == average;
    */
    for (size_t i = 0 ; i < size ; i ++){
        float value = ((float)array[i] / size);
        average += value;
    }
    ret.average = average ;
    return ret;
}

frama-c无法证明此循环不变式:

loop invariant  average(array , 0, i , size) == average;

我做错了什么吗? 我不知道我的问题是否来自浮点数的精度。 我尝试了很多断言,但仍然不起作用。 在Frama-c中能做些什么吗?

编辑:

最终我证明了我的函数是正确的,因为我之前是在将两个数相加之前先进行除法运算,每次试图先做加法时都会出现溢出。

关键是我需要证明我的求和不会溢出,所以我导入了limits.h,并添加了一个新的循环不变式:INT_MIN * i <= sum <= INT_MAX * i; 所以我的代码现在看起来像这样:

#include <stdbool.h>
#include <limits.h>

typedef unsigned int size_t;
typedef struct Average avg;
struct Average
{
    bool success;
    long long average;
};
/*@
axiomatic Sum{
    logic integer sum(int * t , integer start, integer end);

    axiom sum_false :
        \forall int *t, integer start , integer stop;
        start >= stop ==> sum(t,start,stop) == 0;

    axiom sum_true_start :
        \forall int *t, integer start , integer stop;
        0 <= start < stop ==>
        sum(t,start,stop) == sum(t,start,start+1) + sum(t,start+1,stop);

    axiom sum_true_end :
        \forall int *t, integer start , integer stop;
        0 <= start < stop ==>
        sum(t,start,stop) == sum(t,start,stop-1) + sum(t,stop-1,stop);

    axiom sum_split :
        \forall int *t, integer start , integer stop, integer middle;
        0 <= start<=  middle < stop ==>
        sum(t,start,stop) == sum(t,start,middle) + sum(t,middle,stop);


    axiom sum_alone :
        \forall int *t, integer start;
        (0<=start)
        ==>
        sum(t,start,start+1) == t[start] ;
}

*/
/*@
requires \valid(array + (0..size-1));
ensures (!\result.success) ==> size == 0 ;
ensures (\result.success) ==> (\result.average == sum(array,0,size)/size) ;
assigns \nothing;
*/
avg average(int * array, size_t size){
    //we use a structure to be sure that the function finish without error
    avg ret;
    ret.success = true ;
    ret.average = 0 ;
    if (size == 0){
        //if the size == 0 the function will fail
        ret.success = false;
        return ret;
    }
    else{
        /*
        the average is the sum of all the element of the array divided by the size
        An int is between - 2^15-1 and 2^15-1 that imply that the sum of
        all the element of an array is between
        -2^15 * size and 2^15 * size as size is between 0 and 2^16
        the sum is between -2^31 and 2^31
        a long long is between -2^63 and 2^63

        the sum of all the element can be inside a long long.
        */
        long long sum = 0;

        /*@
        loop assigns i, sum ;
        loop invariant 0 <= i <= size;
        loop invariant sum == sum(array,0,i);
        loop invariant INT_MIN * i <= sum <= INT_MAX * i;
        */
        for (size_t i = 0 ; i < size ; i ++){
            //@assert INT_MIN * i <= sum <= INT_MAX * i;

            sum += array[i];
            //@assert  i+1 <= size;

            //@assert INT_MIN * (i+1) <= sum <= INT_MAX * (i+1);
            //@assert ((LLONG_MIN < INT_MIN * size ) && (LLONG_MAX > INT_MAX* size));
            //@assert LLONG_MIN <= sum <= LLONG_MAX;

            //@assert sum == sum(array,0,i) + array[i];

        }
        ret.average = sum/size ;
        return ret;
    }
}

我保留了assert,但我确定其中很多都是无用的。


2
尝试给它一个小的容差值。fabs(average(...) - average) < epsilon - Eugene Sh.
也许它不能处理您同时将符号“average”用作函数和变量名称的情况?如果您不使用不同的名称来为变量命名(这也将使您的代码更易于阅读和理解),那会发生什么呢? - Some programmer dude
4
使用浮点类型进行数学计算通常是不精确的,会导致一些你期望相等的值因为舍入误差而微小地不相等。很可能它正是因为这个原因在抱怨,你应该能够进行“足够接近”而非精确相等的比较。 - Fire Lancer
每次循环都进行除法运算,这是有意为之吗?看起来会带来比必要更多的不精确性。除非你的数组足够大以至于会溢出float - Guillaume Petitjean
1
关于:typedef unsigned int size_t; 类型size_t在头文件stdio.h中定义,并被定义为:long unsigned int 你为什么要尝试重新定义它? - user3629249
显示剩余2条评论
1个回答

3

我想证明计算一维数组中所有值的平均值的计算。

为了确保精确的数学计算,请避免使用浮点数。

由于array[]int,请坚持使用整数计算。

建议重写代码。
伪代码用于测试“一维数组中所有值的平均值”

// Compute sum of all elements of the array
wide_integer_type sum = 0
for (i=0; i<n; i++) 
  sum += array[i]

for (i=0; i<n; i++) 
  // below incurs no rounding like `array[i] == (double)sum/n` might
  if ((cast to wide_integer_type)array[i] * n == sum) 
    print "average found!" sum/n

那是我尝试做的第一件事,但我卡在了int溢出上。在每个步骤进行除法是我找到的唯一解决方案。 - Shinbly
1
我遇到了一个整数溢出的问题。建议使用_wide_integer_type_,因此请使用long long sum = 0; ... if ((long long)array[i] * n == sum)的代码。 - chux - Reinstate Monica
即使是 long long sum,我仍然会溢出。 /*@ assert rte: signed_overflow: -9223372036854775808 ≤ current_average + (long long)*(array + i); */ /*@ assert rte: signed_overflow: current_average + (long long)*(array + i) ≤ 9223372036854775807; - Shinbly
我尝试了使用 double(因为和的最大值是 2^63),并且我没有溢出,但证明仍然不完整,我也不记得 frama-c 是否处理双精度浮点数。 - Shinbly
这里中,我期望的是sum + (long long)*(array + i)。不清楚current_average的类型和作用是什么。 - chux - Reinstate Monica

网页内容由stack overflow 提供, 点击上面的
可以查看英文原文,
原文链接