What does the for(h=1;h<=(r-1)/9;h=h*3+1) loop mean in 《Algorithms in C part 1-4》p172 shellSort?

Question

I'm studying the Shell sort from Sedgewick Algorithms in C part 1-4 on p172.

I use size (the length of array), not l and r (start and end); so my code is

int i,j,h;
int key;
for( h=1;h<=(size-1)/9;h=h*3+1);
for(;h>0;h/=3)
{
    for(i=h;i<size;i++)
    {
        key=num[i];
        j=i;
        while(j>=h&&key>num[j-h];j-=h)
        {
            num[j]=num[j-h];
        } 
        num[j]=key;
    }
}

I know all of this. I read TAOCP. I know 1, 4, 13, … is the best sequence (comparable). But in this position, my code has

for(h=1;h<size;h=h*3+1);

My question is: why did he write h<(size-1)/9?

What does the '/9' mean?

Answer 1

The loop:

for (h = 1; h < size; h = h * 3 + 1)
    ;

overshoots the size of the array most of the time. The alternative loop keeps the gap within range.

You can see this for yourself with a trivial test program like this:

#include <stdio.h>

static inline int hs_gap9(int size)
{
    int h;
    for (h = 1; h <= (size - 1) / 9; h = h * 3 + 1)
        ;
    return h;
}

static inline int hs_gap3(int size)
{
    int h;
    for (h = 1; h < size; h = h * 3 + 1)
        ;
    return h;
}

int main(void)
{
    for (int i = 1; i < 100; i++)
        printf("Size: %3d; gap9 = %d; gap3 = %d\n", i, hs_gap9(i), hs_gap3(i));
    return 0;
}

Sample output:

Size:   1; gap9 =   1; gap3 =   1
Size:   2; gap9 =   1; gap3 =   4
Size:   3; gap9 =   1; gap3 =   4
Size:   4; gap9 =   1; gap3 =   4
Size:   5; gap9 =   1; gap3 =  13
Size:   6; gap9 =   1; gap3 =  13
Size:   7; gap9 =   1; gap3 =  13
Size:   8; gap9 =   1; gap3 =  13
Size:   9; gap9 =   1; gap3 =  13
Size:  10; gap9 =   4; gap3 =  13
Size:  11; gap9 =   4; gap3 =  13
Size:  12; gap9 =   4; gap3 =  13
Size:  13; gap9 =   4; gap3 =  13
Size:  14; gap9 =   4; gap3 =  40
Size:  15; gap9 =   4; gap3 =  40
Size:  16; gap9 =   4; gap3 =  40
…
Size:  34; gap9 =   4; gap3 =  40
Size:  35; gap9 =   4; gap3 =  40
Size:  36; gap9 =   4; gap3 =  40
Size:  37; gap9 =  13; gap3 =  40
Size:  38; gap9 =  13; gap3 =  40
Size:  39; gap9 =  13; gap3 =  40
Size:  40; gap9 =  13; gap3 =  40
Size:  41; gap9 =  13; gap3 = 121
Size:  42; gap9 =  13; gap3 = 121
Size:  43; gap9 =  13; gap3 = 121
…
Size:  97; gap9 =  13; gap3 = 121
Size:  98; gap9 =  13; gap3 = 121
Size:  99; gap9 =  13; gap3 = 121

As you can see, the 'gap3' algorithm returns an initial value of h that is larger than the size of the array. The 'gap9' algorithm returns an initial value of h that is smaller than the size of the array. This saves a little overhead on the loops (saves one iteration of the outer loop, where the middle loop exits on the first cycle without touching the inner loop.