Add original unix v6
This commit is contained in:
memnoth
@@ -0,0 +1,463 @@
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/* diff - differential file comparison
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*
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* Uses an algorithm due to Harold Stone, which finds
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* a pair of longest identical subsequences in the two
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* files.
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*
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* The major goal is to generate the match vector J.
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* J[i] is the index of the line in file1 corresponding
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* to line i file0. J[i] = 0 if there is no
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* such line in file1.
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*
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* Lines are hashed so as to work in core. All potential
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* matches are located by sorting the lines of each file
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* on the hash (called value_____). In particular, this
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* collects the equivalence classes in file1 together.
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* Subroutine equiv____ replaces the value of each line in
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* file0 by the index of the first element of its
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* matching equivalence in (the reordered) file1.
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* To save space equiv_____ squeezes file1 into a single
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* array member______ in which the equivalence classes
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* are simply concatenated, except that their first
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* members are flagged by changing sign.
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*
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* Next the indices that point into member______ are unsorted_______ into
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* array class_____ according to the original order of file0.
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*
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* The cleverness lies in routine stone______. This marches
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* through the lines of file0, developing a vector klist_____
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* of "k-candidates". At step i a k-candidate is a matched
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* pair of lines x,y (x in file0 y in file1) such that
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* there is a common subsequence of lenght k
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* between the first i lines of file0 and the first y
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* lines of file1, but there is no such subsequence for
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* any smaller y. x is the earliest possible mate to y
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* that occurs in such a subsequence.
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*
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* Whenever any of the members of the equivalence class of
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* lines in file1 matable to a line in file0 has serial number
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* less than the y of some k-candidate, that k-candidate
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* with the smallest such y is replaced. The new
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* k-candidate is chained (via pred____) to the current
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* k-1 candidate so that the actual subsequence can
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* be recovered. When a member has serial number greater
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* that the y of all k-candidates, the klist is extended.
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* At the end, the longest subsequence is pulled out
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* and placed in the array J by unravel_______.
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*
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* With J in hand, the matches there recorded are
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* check_____ed against reality to assure that no spurious
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* matches have crept in due to hashing. If they have,
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* they are broken, and "jackpot " is recorded--a harmless
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* matter except that a true match for a spuriously
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* mated line may now be unnecessarily reported as a change.
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*
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* Much of the complexity of the program comes simply
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* from trying to minimize core utilization and
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* maximize the range of doable problems by dynamically
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* allocating what is needed and reusing what is not.
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* The core requirements for problems larger than somewhat
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* are (in words) 2*length(file0) + length(file1) +
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* 3*(number of k-candidates installed), typically about
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* 6n words for files of length n. There is also space for buf1
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* used which could, by moving data underfoot and reallocating
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* buf1 together with buf2, be completely overlaid.
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*/
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struct buf {
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int fdes;
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char data[516];
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} *buf1, *buf2;
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struct cand {
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int x;
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int y;
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struct cand *pred;
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} cand;
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struct line {
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int serial;
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int value;
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} *file[2], line;
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int len[2];
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int *class; /*will be overlaid on file[0]*/
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int *member; /*will be overlaid on file[1]*/
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struct cand **klist; /*will be overlaid on file[0] after class*/
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int *J; /*will be overlaid on class*/
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int *ixold; /*will be overlaid on klist*/
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int *ixnew; /*will be overlaid on file[1]*/
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char *area;
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char *top;
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alloc(n)
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{
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register char *p;
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p = area;
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n = (n+1) & ~1;
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area =+ n;
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while(area > top) {
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if(sbrk(1024) == -1) {
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mesg("Out of space\n");
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exit(1);
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}
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top =+ 1024;
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}
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return(p);
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}
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mesg(s)
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char *s;
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{
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while(*s)
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write(2,s++,1);
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}
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sort(a,n) /*shellsort CACM #201*/
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struct line *a;
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{
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struct line w;
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register int j,m;
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struct line *ai;
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register struct line *aim;
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int k;
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for(j=1;j<=n;j=* 2)
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m = 2*j - 1;
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for(m=/2;m!=0;m=/2) {
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k = n-m;
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for(j=1;j<=k;j++) {
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for(ai = &a[j]; ai > a; ai =- m) {
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aim = &ai[m];
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if(aim->value > ai[0].value ||
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aim->value == ai[0].value &&
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aim->serial > ai[0].serial)
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break;
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w.value = ai[0].value;
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ai[0].value = aim->value;
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aim->value = w.value;
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w.serial = ai[0].serial;
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ai[0].serial = aim->serial;
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aim->serial = w.serial;
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}
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}
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}
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}
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unsort(f, l, b)
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struct line *f;
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int *b;
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{
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int *a;
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int i;
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a = alloc((l+1)*sizeof(a[0]));
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for(i=1;i<=l;i++)
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a[f[i].serial] = f[i].value;
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for(i=1;i<=l;i++)
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b[i] = a[i];
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area = a;
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}
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prepare(i, arg)
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char *arg;
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{
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register char *temp;
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temp = file[i] = area;
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alloc(sizeof(line));
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input(arg);
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len[i] = (area - temp)/sizeof(line) - 1;
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alloc(sizeof(line));
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sort(file[i], len[i]);
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}
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input(arg)
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{
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register int h, i;
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register struct line *p;
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if(fopen(arg,buf1) == -1) {
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mesg("Cannot open ");
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mesg(arg);
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mesg("\n");
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exit(1);
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}
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for(i=0; h=readhash(buf1);) {
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p = alloc(sizeof(line));
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p->serial = ++i;
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p->value = h;
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}
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close(buf1->fdes);
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}
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equiv(a,n,b,m,c)
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struct line *a, *b;
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int *c;
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{
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register int i, j;
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i = j = 1;
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while(i<=n && j<=m) {
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if(a[i].value <b[j].value)
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a[i++].value = 0;
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else if(a[i].value == b[j].value)
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a[i++].value = j;
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else
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j++;
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}
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while(i <= n)
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a[i++].value = 0;
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b[m+1].value = 0;
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j = 0;
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while(++j <= m) {
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c[j] = -b[j].serial;
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while(b[j+1].value == b[j].value) {
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j++;
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c[j] = b[j].serial;
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}
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}
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c[j] = -1;
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}
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main(argc, argv)
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char **argv;
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{
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int k;
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if(argc>1 && *argv[1]=='-') {
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argc--;
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argv++;
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}
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if(argc!=3) {
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mesg("Arg count\n");
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exit(1);
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}
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area = top = sbrk(0);
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buf1 = alloc(sizeof(*buf1));
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prepare(0, argv[1]);
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prepare(1, argv[2]);
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member = file[1];
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equiv(file[0], len[0], file[1], len[1], member);
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class = file[0];
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unsort(file[0], len[0], class);
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klist = &class[len[0]+2];
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area = &member[len[1]+2];
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k = stone(class, len[0], member, klist);
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J = class;
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unravel(klist[k]);
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ixold = klist;
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ixnew = file[1];
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area = &ixnew[len[1]+2];
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buf2 = alloc(sizeof(*buf2));
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if(check(argv))
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mesg("Jackpot\n");
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output(argv);
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}
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stone(a,n,b,c)
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int *a;
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int *b;
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struct cand **c;
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{
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register int i, k,y;
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int j, l;
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int skip;
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k = 0;
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c[0] = 0;
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for(i=1; i<=n; i++) {
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j = a[i];
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if(j==0)
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continue;
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skip = 0;
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do {
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y = b[j];
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if(y<0) y = -y;
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if(skip)
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continue;
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l = search(c, k, y);
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if(l > k) {
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c[k+1] = newcand(i,y,c[k]);
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skip = 1;
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k++;
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}
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else if(c[l]->y > y && c[l]->x < i)
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c[l] = newcand(i,y,c[l-1]);
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} while(b[++j] > 0);
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}
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return(k);
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}
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struct cand *
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newcand(x,y,pred)
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struct cand *pred;
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{
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struct cand *p;
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p = alloc(sizeof(cand));
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p->x = x;
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p->y = y;
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p->pred = pred;
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return(p);
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}
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search(c, k, y)
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struct cand **c;
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{
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register int i, j, l;
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int t;
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i = 0;
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j = k+1;
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while((l=(i+j)/2) > i) {
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t = c[l]->y;
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if(t > y)
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j = l;
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else if(t < y)
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i = l;
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else
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return(l);
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}
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return(l+1);
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}
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unravel(p)
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struct cand *p;
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{
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int i;
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for(i=0; i<=len[0]; i++)
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J[i] = 0;
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while(p) {
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J[p->x] = p->y;
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p = p->pred;
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}
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}
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/* check does double duty:
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1. ferret out any fortuitous correspondences due
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to counfounding by hashing (which result in "jackpot")
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2. collect random access indexes to the two files */
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check(argv)
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char **argv;
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{
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register int i, j;
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int ctold, ctnew;
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int jackpot;
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char c,d;
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fopen(argv[1],buf1);
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fopen(argv[2],buf2);
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j = 1;
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ctold = ctnew = 0;
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ixold[0] = ixnew[0] = 0;
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jackpot = 0;
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for(i=1;i<=len[0];i++) {
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if(J[i]==0) {
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while(getc(buf1)!='\n') ctold++;
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ixold[i] = ++ctold;
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continue;
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}
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while(j<J[i]) {
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while(getc(buf2)!='\n') ctnew++;
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ixnew[j] = ++ctnew;
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j++;
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}
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while((c=getc(buf1))==(d=getc(buf2))) {
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if(c=='\n') break;
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ctold++;
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ctnew++;
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}
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while(c!='\n') {
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jackpot++;
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J[i] = 0;
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c = getc(buf1);
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ctold++;
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}
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ixold[i] = ++ctold;
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while(d!='\n') {
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jackpot++;
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J[i] = 0;
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d = getc(buf2);
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ctnew++;
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}
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ixnew[j] = ++ctnew;
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j++;
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}
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for(;j<=len[1];j++) {
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while(getc(buf2)!='\n') ctnew++;
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ixnew[j] = ++ctnew;
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}
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close(buf1->fdes);
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close(buf2->fdes);
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return(jackpot);
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}
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output(argv)
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char **argv;
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{
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int dir;
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int m;
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int i0,i1,j0,j1;
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extern fout;
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dir = **argv=='-';
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fout = dup(1);
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buf1->fdes = open(argv[1],0);
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buf2->fdes = open(argv[2],0);
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m = len[0];
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J[0] = 0;
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J[m+1] = len[1]+1;
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if(dir==0) for(i0=1;i0<=m;i0=i1+1) {
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while(i0<=m&&J[i0]==J[i0-1]+1) i0++;
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j0 = J[i0-1]+1;
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i1 = i0-1;
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while(i1<m&&J[i1+1]==0) i1++;
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j1 = J[i1+1]-1;
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J[i1] = j1;
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change(i0,i1,j0,j1,dir);
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} else for(i0=m;i0>=1;i0=i1-1) {
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while(i0>=1&&J[i0]==J[i0+1]-1&&J[i0]!=0) i0--;
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j0 = J[i0+1]-1;
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i1 = i0+1;
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while(i1>1&&J[i1-1]==0) i1--;
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j1 = J[i1-1]+1;
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J[i1] = j1;
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change(i1,i0,j1,j0,dir);
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}
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if(m==0)
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change(1,0,1,len[1],dir);
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flush();
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}
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change(a,b,c,d,dir)
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{
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if(a>b&&c>d) return;
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range(a,b);
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putchar(a>b?'a':c>d?'d':'c');
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if(dir==0) range(c,d);
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putchar('\n');
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if(dir==0) {
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fetch(ixold,a,b,buf1,"* ");
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if(a<=b&&c<=d) printf("---\n");
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}
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fetch(ixnew,c,d,buf2,dir==0?". ":"");
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if(dir!=0&&c<=d) printf(".\n");
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}
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range(a,b)
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{
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if(a>b) printf("%d",b);
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if(a<=b) printf("%d",a);
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if(a<b) printf(",%d",b);
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}
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fetch(f,a,b,lb,pref)
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int *f;
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struct buf *lb;
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char *pref;
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{
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int i, j;
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int nc;
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for(i=a;i<=b;i++) {
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seek(lb->fdes,f[i-1],0);
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nc = read(lb->fdes,lb->data,f[i]-f[i-1]);
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printf(pref);
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for(j=0;j<nc;j++)
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putchar(lb->data[j]);
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}
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}
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