V6/usr/source/s1/diff1.c
/* diff - differential file comparison
*
* Uses an algorithm due to Harold Stone, which finds
* a pair of longest identical subsequences in the two
* files.
*
* The major goal is to generate the match vector J.
* J[i] is the index of the line in file1 corresponding
* to line i file0. J[i] = 0 if there is no
* such line in file1.
*
* Lines are hashed so as to work in core. All potential
* matches are located by sorting the lines of each file
* on the hash (called value_____). In particular, this
* collects the equivalence classes in file1 together.
* Subroutine equiv____ replaces the value of each line in
* file0 by the index of the first element of its
* matching equivalence in (the reordered) file1.
* To save space equiv_____ squeezes file1 into a single
* array member______ in which the equivalence classes
* are simply concatenated, except that their first
* members are flagged by changing sign.
*
* Next the indices that point into member______ are unsorted_______ into
* array class_____ according to the original order of file0.
*
* The cleverness lies in routine stone______. This marches
* through the lines of file0, developing a vector klist_____
* of "k-candidates". At step i a k-candidate is a matched
* pair of lines x,y (x in file0 y in file1) such that
* there is a common subsequence of lenght k
* between the first i lines of file0 and the first y
* lines of file1, but there is no such subsequence for
* any smaller y. x is the earliest possible mate to y
* that occurs in such a subsequence.
*
* Whenever any of the members of the equivalence class of
* lines in file1 matable to a line in file0 has serial number
* less than the y of some k-candidate, that k-candidate
* with the smallest such y is replaced. The new
* k-candidate is chained (via pred____) to the current
* k-1 candidate so that the actual subsequence can
* be recovered. When a member has serial number greater
* that the y of all k-candidates, the klist is extended.
* At the end, the longest subsequence is pulled out
* and placed in the array J by unravel_______.
*
* With J in hand, the matches there recorded are
* check_____ed against reality to assure that no spurious
* matches have crept in due to hashing. If they have,
* they are broken, and "jackpot " is recorded--a harmless
* matter except that a true match for a spuriously
* mated line may now be unnecessarily reported as a change.
*
* Much of the complexity of the program comes simply
* from trying to minimize core utilization and
* maximize the range of doable problems by dynamically
* allocating what is needed and reusing what is not.
* The core requirements for problems larger than somewhat
* are (in words) 2*length(file0) + length(file1) +
* 3*(number of k-candidates installed), typically about
* 6n words for files of length n. There is also space for buf1
* used which could, by moving data underfoot and reallocating
* buf1 together with buf2, be completely overlaid.
*/
struct buf {
int fdes;
char data[516];
} *buf1, *buf2;
struct cand {
int x;
int y;
struct cand *pred;
} cand;
struct line {
int serial;
int value;
} *file[2], line;
int len[2];
int *class; /*will be overlaid on file[0]*/
int *member; /*will be overlaid on file[1]*/
struct cand **klist; /*will be overlaid on file[0] after class*/
int *J; /*will be overlaid on class*/
int *ixold; /*will be overlaid on klist*/
int *ixnew; /*will be overlaid on file[1]*/
char *area;
char *top;
alloc(n)
{
register char *p;
p = area;
n = (n+1) & ~1;
area =+ n;
while(area > top) {
if(sbrk(1024) == -1) {
mesg("Out of space\n");
exit(1);
}
top =+ 1024;
}
return(p);
}
mesg(s)
char *s;
{
while(*s)
write(2,s++,1);
}
sort(a,n) /*shellsort CACM #201*/
struct line *a;
{
struct line w;
register int j,m;
struct line *ai;
register struct line *aim;
int k;
for(j=1;j<=n;j=* 2)
m = 2*j - 1;
for(m=/2;m!=0;m=/2) {
k = n-m;
for(j=1;j<=k;j++) {
for(ai = &a[j]; ai > a; ai =- m) {
aim = &ai[m];
if(aim->value > ai[0].value ||
aim->value == ai[0].value &&
aim->serial > ai[0].serial)
break;
w.value = ai[0].value;
ai[0].value = aim->value;
aim->value = w.value;
w.serial = ai[0].serial;
ai[0].serial = aim->serial;
aim->serial = w.serial;
}
}
}
}
unsort(f, l, b)
struct line *f;
int *b;
{
int *a;
int i;
a = alloc((l+1)*sizeof(a[0]));
for(i=1;i<=l;i++)
a[f[i].serial] = f[i].value;
for(i=1;i<=l;i++)
b[i] = a[i];
area = a;
}
prepare(i, arg)
char *arg;
{
register char *temp;
temp = file[i] = area;
alloc(sizeof(line));
input(arg);
len[i] = (area - temp)/sizeof(line) - 1;
alloc(sizeof(line));
sort(file[i], len[i]);
}
input(arg)
{
register int h, i;
register struct line *p;
if(fopen(arg,buf1) == -1) {
mesg("Cannot open ");
mesg(arg);
mesg("\n");
exit(1);
}
for(i=0; h=readhash(buf1);) {
p = alloc(sizeof(line));
p->serial = ++i;
p->value = h;
}
close(buf1->fdes);
}
equiv(a,n,b,m,c)
struct line *a, *b;
int *c;
{
register int i, j;
i = j = 1;
while(i<=n && j<=m) {
if(a[i].value <b[j].value)
a[i++].value = 0;
else if(a[i].value == b[j].value)
a[i++].value = j;
else
j++;
}
while(i <= n)
a[i++].value = 0;
b[m+1].value = 0;
j = 0;
while(++j <= m) {
c[j] = -b[j].serial;
while(b[j+1].value == b[j].value) {
j++;
c[j] = b[j].serial;
}
}
c[j] = -1;
}
main(argc, argv)
char **argv;
{
int k;
if(argc>1 && *argv[1]=='-') {
argc--;
argv++;
}
if(argc!=3) {
mesg("Arg count\n");
exit(1);
}
area = top = sbrk(0);
buf1 = alloc(sizeof(*buf1));
prepare(0, argv[1]);
prepare(1, argv[2]);
member = file[1];
equiv(file[0], len[0], file[1], len[1], member);
class = file[0];
unsort(file[0], len[0], class);
klist = &class[len[0]+2];
area = &member[len[1]+2];
k = stone(class, len[0], member, klist);
J = class;
unravel(klist[k]);
ixold = klist;
ixnew = file[1];
area = &ixnew[len[1]+2];
buf2 = alloc(sizeof(*buf2));
if(check(argv))
mesg("Jackpot\n");
output(argv);
}
stone(a,n,b,c)
int *a;
int *b;
struct cand **c;
{
register int i, k,y;
int j, l;
int skip;
k = 0;
c[0] = 0;
for(i=1; i<=n; i++) {
j = a[i];
if(j==0)
continue;
skip = 0;
do {
y = b[j];
if(y<0) y = -y;
if(skip)
continue;
l = search(c, k, y);
if(l > k) {
c[k+1] = newcand(i,y,c[k]);
skip = 1;
k++;
}
else if(c[l]->y > y && c[l]->x < i)
c[l] = newcand(i,y,c[l-1]);
} while(b[++j] > 0);
}
return(k);
}
struct cand *
newcand(x,y,pred)
struct cand *pred;
{
struct cand *p;
p = alloc(sizeof(cand));
p->x = x;
p->y = y;
p->pred = pred;
return(p);
}
search(c, k, y)
struct cand **c;
{
register int i, j, l;
int t;
i = 0;
j = k+1;
while((l=(i+j)/2) > i) {
t = c[l]->y;
if(t > y)
j = l;
else if(t < y)
i = l;
else
return(l);
}
return(l+1);
}
unravel(p)
struct cand *p;
{
int i;
for(i=0; i<=len[0]; i++)
J[i] = 0;
while(p) {
J[p->x] = p->y;
p = p->pred;
}
}
/* check does double duty:
1. ferret out any fortuitous correspondences due
to counfounding by hashing (which result in "jackpot")
2. collect random access indexes to the two files */
check(argv)
char **argv;
{
register int i, j;
int ctold, ctnew;
int jackpot;
char c,d;
fopen(argv[1],buf1);
fopen(argv[2],buf2);
j = 1;
ctold = ctnew = 0;
ixold[0] = ixnew[0] = 0;
jackpot = 0;
for(i=1;i<=len[0];i++) {
if(J[i]==0) {
while(getc(buf1)!='\n') ctold++;
ixold[i] = ++ctold;
continue;
}
while(j<J[i]) {
while(getc(buf2)!='\n') ctnew++;
ixnew[j] = ++ctnew;
j++;
}
while((c=getc(buf1))==(d=getc(buf2))) {
if(c=='\n') break;
ctold++;
ctnew++;
}
while(c!='\n') {
jackpot++;
J[i] = 0;
c = getc(buf1);
ctold++;
}
ixold[i] = ++ctold;
while(d!='\n') {
jackpot++;
J[i] = 0;
d = getc(buf2);
ctnew++;
}
ixnew[j] = ++ctnew;
j++;
}
for(;j<=len[1];j++) {
while(getc(buf2)!='\n') ctnew++;
ixnew[j] = ++ctnew;
}
close(buf1->fdes);
close(buf2->fdes);
return(jackpot);
}
output(argv)
char **argv;
{
int dir;
int m;
int i0,i1,j0,j1;
extern fout;
dir = **argv=='-';
fout = dup(1);
buf1->fdes = open(argv[1],0);
buf2->fdes = open(argv[2],0);
m = len[0];
J[0] = 0;
J[m+1] = len[1]+1;
if(dir==0) for(i0=1;i0<=m;i0=i1+1) {
while(i0<=m&&J[i0]==J[i0-1]+1) i0++;
j0 = J[i0-1]+1;
i1 = i0-1;
while(i1<m&&J[i1+1]==0) i1++;
j1 = J[i1+1]-1;
J[i1] = j1;
change(i0,i1,j0,j1,dir);
} else for(i0=m;i0>=1;i0=i1-1) {
while(i0>=1&&J[i0]==J[i0+1]-1&&J[i0]!=0) i0--;
j0 = J[i0+1]-1;
i1 = i0+1;
while(i1>1&&J[i1-1]==0) i1--;
j1 = J[i1-1]+1;
J[i1] = j1;
change(i1,i0,j1,j0,dir);
}
if(m==0)
change(1,0,1,len[1],dir);
flush();
}
change(a,b,c,d,dir)
{
if(a>b&&c>d) return;
range(a,b);
putchar(a>b?'a':c>d?'d':'c');
if(dir==0) range(c,d);
putchar('\n');
if(dir==0) {
fetch(ixold,a,b,buf1,"* ");
if(a<=b&&c<=d) printf("---\n");
}
fetch(ixnew,c,d,buf2,dir==0?". ":"");
if(dir!=0&&c<=d) printf(".\n");
}
range(a,b)
{
if(a>b) printf("%d",b);
if(a<=b) printf("%d",a);
if(a<b) printf(",%d",b);
}
fetch(f,a,b,lb,pref)
int *f;
struct buf *lb;
char *pref;
{
int i, j;
int nc;
for(i=a;i<=b;i++) {
seek(lb->fdes,f[i-1],0);
nc = read(lb->fdes,lb->data,f[i]-f[i-1]);
printf(pref);
for(j=0;j<nc;j++)
putchar(lb->data[j]);
}
}