# The Universe of Disco

Fri, 05 Oct 2007

Van der Waerden's problem: program 2
In this series of articles I'm going to analyze four versions of a program that I wrote around 1988, and then another program that does the same thing that I wrote last month without referring to the 1988 code.

If you don't remember what the program does, here's an explanation.

Here is program 1, which was an earlier attempt to do the same thing.

## Program 2

In yesterday's article I wrote about a crappy program to search for "good" strings in van der Waerden's problem. It was crappy because it searched the entire space of all 327 strings, with no pruning.

I can't remember whether I expected this to be practical at the time. Did I really think it would work? Well, there was some sense to it. It does work just fine for the 29 case. I think probably my idea was to do the simplest thing that could possibly work, and get as much information out of it as I could. On my current machine, this method proves that V(3,3) > 19 by finding a witness (RRBRRBBYYRRBRRBBYYB) in under 10 seconds. If we estimate that the computer I had then was 10,000 times slower, then I could have produced the same result in about 28 hours. I was at college, and there was plenty of free computing power available, so running a program for 28 hours was easily done. While I was waiting for it to finish, I could work on a better program.

Excerpts of the better program follow. The complete source code is here.

The idea behind this program is that the strings of length less than V form a tree, with the empty string as the root, and the children of string s are obtained from s by appending a single character to the end of s. If the string at a node is bad, so will be all the strings under it, and we can prune the entire branch at that node. This leaves us with a tree of all the good strings. The ones farthest from the root will be the witnesses we seek for the values of V(n, C), and we can find these by doing depth-first search on the tree,

There is nothing wrong with this idea in principle; that's the way my current program works too. The problem is all in the implementation. You see, this program actually constructs the entire tree in memory:

    #define NEWN		((struct tree *) Malloc(sizeof(struct tree)));\
printf("*")
struct tree {
struct tree *away[MAXCOLORS];
} *root;

struct tree is a tree node structure. It represents a string s, and has a flag to record whether s is bad. It also has pointers to its subnodes, which will represents strings sA, sB, and so on.

MAXCOLORS is a compiled-in limit on the number of different symbols the strings can contain, an upper bound on C. Apparently I didn't know the standard technique for avoiding this inflexibility. You declare the array as having length 1, but then when you allocate the structure, you allocate enough space for the array you are actually planning to use. Even though the declared size of the array is 1, you are allowed to refer to node->away[37] as long as there is actually enough space in the allocated chunk. The implementation would look like this:

        struct tree {
struct tree *away[1];
} ;

struct tree *make_tree_node(char bad, unsigned n_subnodes)
{
struct tree *t;
unsigned i;

t =  malloc(sizeof(struct tree)
+ (n_subnodes-1) * sizeof(struct tree *));

if (t == NULL) return NULL;

for (i=0; i < n_subnodes; i++) t->away[i] = NULL;

return t;
}

(Note for those who are not advanced C programmers: I give you my solemn word of honor that I am not doing anything dodgy or bizarre here; it is a standard, widely-used, supported technique, guaranteed to work everywhere.)

(As before, this code is in a pink box to indicate that it is not actually part of the program I am discussing.)

Another thing I notice is that the NEWN macro is very weird. Note that it may not work as expected in a context like this:

        for(i=0; i<10; i++)
s[i] = NEWN;

This allocates ten nodes but prints only one star, because it expands to:

        for(i=0; i<10; i++)
s[i] = ((struct tree *) Malloc(sizeof(struct tree)));
printf("*");

and the for loop does not control the printf. The usual fix for multiline macros like this is to wrap them in do...while(0), but that is not appropriate here. Had I been writing this today, I would have made NEWN a function, not a macro. Clevermacroitis is a common disorder of beginning C programmers, and I was no exception.

The main business of the program is in the makenodes function; the main routine does some argument processing and then calls makenodes. The arguments to the makenodes function are the current tree node, the current string that that node represents, and an integer howfar that says how deep a tree to construct under the current node.

There's a base case, for when nothing needs to be constructed:

    if (!howfar)
{
for (i=0; i<colors; i++)
n->away[i] = NULL;
return;
}

But in general the function calls itself recursively:

    for (i=0; i<colors; i++)
{
n->away[i] = NEWN;
if (apchk(s,'A'+i))
{
}
else
...

Recall that apchk checks a string for an arithmetic progression of equal characters. That is, it checks to see if a string is good or bad. If the string is bad, the function prunes the tree at the current node, and doesn't recurse further.

Unlike the one in the previous program, this apchk doesn't bother checking all the possible arithmetic progressions. It only checks the new ones: that is, the ones involving the last character. That's why it has two arguments. One is the old string s and the other is the new symbol that we want to append to s.

If s would still be good with symbol 'A'+i appended to the end, the function recurses:

        ...
else
{
ls = strlen(s);
newarg = STRING(ls + 1);
strcpy(newarg,s);
newarg[ls+1] = '\0';
newarg[ls] = 'A' + i;
makenodes(n->away[i],howfar-1,newarg);
Free(newarg,ls+2);
Free(n->away[i],sizeof(struct tree));
}
}
}

The entire string is copied here into a new buffer. A better technique sould have been to allocate a single buffer back up in main, and to reuse that buffer over again on each call to makenodes. It would have looked something like this:

        char *s = String(maxlen);
memset(s, 0, maxlen+1);
makenodes(s, s, maxlen);

void
makenodes(char *start, char *end, unsigned howfar)
{
...
for (i=0; i<colors; i++) {
*end = 'A' + i;
makenodes(start, end+1, howfar-1);
}
*end = '\0';
...
}

This would have saved a lot of consing, ahem, I mean a lot of mallocing. Also a lot of string copying. We could avoid the end pointer by using start+maxlen-howfar instead, but this way is easier to understand.

I was thinking this afternoon how it's intersting the way I wrote this. It's written the way it would have been done, had I been using a functional programming language. In a functional language, you would never mutate the same string for each function call; you always copy the old structure and construct a new one, just as I did in this program. This is why C programmers abominate functional languages.

Had I been writing makenodes today, I would probably have eliminated the other argument. Instead of passing it a node and having it fill in the children, I would have had it construct and return a complete node. The recursive call would then have looked like this:

  struct tree *new = NEWN;
...
for (i=0; i<colors; i++) {
new->away[i] = makenodes(...);
...
}
return new;

One thing I left out of all this was the diagnostic printfs; you can see them in the complete code if you want. But there's one I thought was worth mentioning anyway:

    #define TABS	"                                        "
....

#ifdef DIAG
printf("%s makenoding with string %s, depth %d.\n",
TABS+12-maxlen+howfar,s,maxlen-howfar);
#endif

The interesting thing here is the TABS+12-maxlen+howfar argument, which indents the display depending on how far the recursion has progressed. In Perl, which has nonaddressable strings, I usually do something like this:

        my $TABS = " " x (maxlen - howfar); print$TABS, "....";

The TABS trick here is pretty clever, and I'm a bit surprised that I thought of it in 1988, when I had been programming in C for only about a year. It makes an interesting contrast to my failure to reuse the string buffer in makenodes earlier.

(Peeking ahead, I see that in the next version of the program, I did reuse the string buffer in this way.)

TABS is actually forty spaces, not tabs. I suspect I used tabs when I tested it with V(2, 3), where maxlen was only 9, and then changed it to spaces for calculating V(3, 3), where maxlen was 27.

The apchk function checks to see if a string is good. Actually it gets a string, qq, and a character, q, and checks to see if the concatenation of qq and q would be good. This reduces its running time to O(|qq|) rather than O(|qq|2).

  int
apchk(qq,q)
char *qq ,q;
{
int lqq, f, s, t;

t = lqq = strlen(qq);
if (lqq < 2) return NO;

for (f=lqq % 2; f <= lqq - 2; f += 2)
{
s = (f + t) / 2;
if ((qq[f] == qq[s]) && (qq[s] == q))
return YES;
}
return NO;
}

It's funny that it didn't occur to me to include an extra parameter to avoid the strlen, or to use q instead of qq[s] in the first == test. Also, as in the previous program, I seem unaware of the relative precedences of && and ==. This is probably a hangover from my experience with Pascal, where the parentheses are required.

It seems I hadn't learned yet that predicate functions like apchk should be named something like is_bad, so that you can understand code like if (is_bad(s)) { ... } without having to study the code of is_bad to figure out what it returns.

I was going to write that I hated this function, and that I could do it a lot better now. But then I tried to replace it, and wasn't as successful as I expected I would be. My replacement was:

        unsigned
{
size_t qql = strlen(qq);
char *f = qq + qql%2;
char *s = f + qql/2;
while (f < s) {
if (*f == q && *s == q) return 1;
f += 2; s += 1;
}
return 0;
}

I could simplify the initializations of f and s, which are the parts I dislike most here, by making the pointers move backward instead of forward, but then the termination test becomes more complicated:
        unsigned
{
char *s = strchr(qq, '\0')-1;
char *f = s-1;
while (1) {
if (*f == q && *s == q) return 1;
if (f - qq < 2) break;
f -= 2; s -= 1;
}
return 0;
}

Anyway, I thought I could improve it, but I'm not sure I did. On the one hand, I like the f -= 2; s -= 1;, which I think is pretty clear. On the other hand, s = (f + t) / 2 is pretty clear too; s is midway between f and t. I'm willing to give teenage Dominus a passing grade on this one.

Someone probably wants to replace the while loop here with a for loop. That person is not me.

The Malloc and Free functions track memory usage and were presumably introduced when I discovered that my program used up way too much memory and crashed—I think I remember that the original version omitted the calls to free. They aren't particularly noteworthy, except perhaps for this bit, in Malloc:

        if (p == NULL)
{
fprintf(stderr,"Couldn't get %d bytes.\n",c);
fprintf(stderr,"Total get was %d.\n",gotten);
fprintf(stderr,"P\n L\n  O\n   P\n    !\n");
abort();
}

Plop!

It strikes me as odd that I was using void in 1988 (this is before the C90 standard) but still K&R-style function declarations. I don't know what to make of that.

## Behavior

This program works, almost. On my current machine, it can find the length-26 witnesses for V(3, 3) in no time. (In 1998, it took several days to run on a Sequent Balance 21000.) The major problem is that it gobbles memory: the if (!howfar) base case in makenodes forgets to release the memory that was allocated for the new node. I wonder if the Malloc and Free functions were written in an unsuccessful attempt to track this down.

Sometime after I wrote this program, while I was waiting for it to complete, it occurred to me that it never actually used the tree for anything, and I could take it out.

I have this idea that one of the principal symptoms of novice programmers is that they take the data structures too literally, and always want to represent data the way it will appear when it's printed out. I haven't developed the idea well enough to write an article about it, but I hope it will show up here sometime in the next three years. This program, which constructs an entirely unnecessary tree structure, may be one of the examples of this idea.

I'll show the third version sometime in the next few days, I hope.

[ Addendum 20071014: Here is part 3. ]