Number of comparisons made in median of 3 function?

问题

As of right now, my functioin finds the median of 3 numbers and sorts them, but it always makes three comparisons. I'm thinking I can use a nested if statement somewhere so that sometimes my function will only make two comparisons.

int median_of_3(int list[], int p, int r)
{
    int median = (p + r) / 2;

    if(list[p] > list[r])
        exchange(list, p, r);
    if(list[p] > list[median])
        exchange(list, p, median);
    if(list[r] > list[median])
        exchange(list, r, median);

    comparisons+=3;                // 3 comparisons for each call to median_of_3

    return list[r];
}

I'm not sure I see where I can make that nested if statement.

回答1:

If you only need the median value, here's a branch-less solution based on min/max operators:

median = max(min(a,b), min(max(a,b),c));

Intel CPU's have SSE min/max vector instructions, so depending on your or your compiler's ability to vectorize, this can run extremely fast.

回答2:

If we allow extra operations, we could use at most 2 comparisons to find the median. The trick is to use exclusive or to find the relationship among three numbers.

void median3(int A[], int p, int r)
{
    int m = (p+r)/2;
    /* let a, b, c be the numbers to be compared */
    int a = A[p], b = A[m], c = A[r];
    int e = a-b;
    int f = a-c;

    if ((e^f) < 0) {
        med_comparisons += 1;
        /* a is the median with 1 comparison */
        A[m] = a;
        /* b < a < c ? */
        if (b < c) /* b < a < c */ { A[p] = b, A[r] = c; }
        else       /* c < a < b */ { A[p] = c, A[r] = b; }
        comparisons += 2;
    } else {
        med_comparisons += 2;
        int g = b-c;
        if ((e^g) < 0) {
            /* c is the median with 2 comparisons */ 
            A[m] = c;
            /* a < c < b ? */
            if (a < b) /* a < c < b */ { A[p] = a, A[r] = b; }
            else       /* b < c < a */ { A[p] = b, A[r] = a; }
        } else {
            /* b is the median with 2 comparisons */
            A[m] = b;
            /* c < b < a ? */
            if (a > c) /* c < b < a */ { A[p] = c; A[r] = a; }
            else       /* a < b < c */ { /* do nothing */    }
        }
        comparisons += 3;
    }
}

The first exclusive or (e^f) is to find out the difference of the sign bit between (a-b) and (a-c).
If they have different sign bit, then a is the median. Otherwise, a is either the minimum or the maximum. In that case, we need the second exclusive or (e^g).

Again, we are going to find out the difference of the sign bit between (a-b) and (b-c). If they have different sign bit, one case is that a > b && b < c. In this case, we also get a > c because a is the maximum in this case. So we have a > c > b. The other case is a < b && b > c && a < c. So we have a < c < b; In both cases, c is the median.

If (a-b) and (b-c) have the same sign bit, then b is the median using similar arguments as above. Experiments shows that a random input will need 1.667 comparisons to find out the median and one extra comparison to get the order.

回答3:

int m = (p + r) / 2;
if (list[p] < list[m])
    if (list[p] >= list[r])
        return list[p];
    else if (list[m] < list[r])
        return list[m];
else
    if (list[p] < list[r])
        return list[p];
return list[r];

回答4:

To sort 3 items, you need exactly 3 comparisons.

To find the middle one by chance, you need 2.

To find the middle one exactly, you need on average 2+2/3 ~= 2.67 (with uniformly distributed random data)

if (a<b) {
   // partial order = a,b
   if (b<c) {  } // 2 comparisons: order is a,b,c
      else { // order is a,c,b or c,a,b
          if (a<c) { } // order is a,c,b -- 3 comparisons
          else { }     // order is c,a,b -- 3 comparisons
      }
} else {
   // partial order = b,a  
   if (c<b) {  } // 2 comparisons: order is c,b,a
   else {  // order is b,c,a or b,a,c
      if (c>a) { } // order is b,a,c -- 3 comparisons
      else { }   // order is b,c,a -- 3 comparisons
   }
}

As an additional side note: some languages (Fortran, IIRC), as well as some ISAs (VAX, again IIRC) support comparisons, where the next PC address is selected from three choices: LT,EQ,GT. With small enough alphabet this chance reduces slightly the number of needed comparisons.

Also this has probably no practical use, taken, that penalties from wrong branch predictions because of overly complex nested structures can be much larger than gain from a saved comparison.

回答5:

回答6:

Python V2

def bigger(a,b):
    if a > b:
       return a
    else:
    return b

def biggest(a,b,c):
    return bigger(a,bigger(b,c))

def median(a,b,c):
    big = biggest(a,b,c)
    if big == a:
       return bigger(b,c)
    if big == b:
       return bigger(a,c)
    else:
       return bigger(a,b)

to print the Median

print(median(1,18,10)) # => 10

来源：https://stackoverflow.com/questions/12937732/number-of-comparisons-made-in-median-of-3-function

标签

c++

algorithm

median