1987-03-09 19:15:41 +00:00
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/*
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* (c) copyright 1987 by the Vrije Universiteit, Amsterdam, The Netherlands.
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* See the copyright notice in the ACK home directory, in the file "Copyright".
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*/
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1985-01-08 09:59:28 +00:00
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#ifndef NORCSID
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1994-06-24 11:31:16 +00:00
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static char rcsid[]= "$Id$";
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1985-01-08 09:59:28 +00:00
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#endif
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#include "assert.h"
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#include "param.h"
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#include "set.h"
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2013-05-10 11:04:21 +00:00
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#include "extern.h"
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1985-01-08 09:59:28 +00:00
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#include <stdio.h>
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/*
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* This file implements the marriage thesis from Hall.
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* The thesis says that given a number, say N, of subsets from
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* a finite set, it is possible to create a set with cardinality N,
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* that contains one member for each of the subsets,
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* iff for each number, say M, of subsets from 2 to N the union of
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* each M-tuple sets has cardinality >= M.
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*
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* So what, you might say. As indeed I did.
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* But this is actually used here to check the possibility of each
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* code rule. If a code rule has a number of token_sets in the with
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* clause and a number of properties in the uses rule it must be
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* possible to do this from an empty fakestack. Hall helps.
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*/
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#define MAXHALL (TOKPATMAX+MAXALLREG)
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short hallsets[MAXHALL][SETSIZE];
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int nhallsets= -1;
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int hallfreq[MAXHALL][2];
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hallverbose() {
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register i;
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register max;
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fprintf(stderr,"Table of hall frequencies\n # pre post\n");
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for (max=MAXHALL-1;hallfreq[max][0]==0 && hallfreq[max][1]==0;max--)
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;
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for (i=0;i<=max;i++)
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fprintf(stderr,"%3d%6d%6d\n",i,hallfreq[i][0],hallfreq[i][1]);
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}
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inithall() {
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assert(nhallsets == -1);
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nhallsets=0;
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}
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nexthall(sp) register short *sp; {
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register i;
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assert(nhallsets>=0);
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for(i=0;i<SETSIZE;i++)
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hallsets[nhallsets][i] = sp[i];
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nhallsets++;
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}
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card(sp) register short *sp; {
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register sum,i;
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sum=0;
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1987-02-18 10:47:55 +00:00
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for(i=0;i<8*sizeof(short)*SETSIZE;i++)
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1985-01-08 09:59:28 +00:00
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if (BIT(sp,i))
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sum++;
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return(sum);
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}
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checkhall() {
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assert(nhallsets>=0);
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if (!hall())
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error("Hall says: \"You can't have those registers\"");
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}
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hall() {
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register i,j,k;
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int ok;
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hallfreq[nhallsets][0]++;
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/*
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* If a set has cardinality >= nhallsets it can never be the cause
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* of the hall algorithm failing. So it can be thrown away.
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* But then nhallsets is less, so this step can be re-applied.
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*/
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do {
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ok = 0;
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for(i=0;i<nhallsets;i++)
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if (card(hallsets[i])>=nhallsets) {
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for (j=i+1;j<nhallsets;j++)
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for(k=0;k<SETSIZE;k++)
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hallsets[j-1][k] =
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hallsets[j][k];
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nhallsets--;
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ok = 1;
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break;
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}
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} while (ok);
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/*
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* Now all sets have cardinality < nhallsets
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*/
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hallfreq[nhallsets][1]++;
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ok=recurhall(nhallsets,hallsets);
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nhallsets = -1;
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return(ok);
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}
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recurhall(nhallsets,hallsets) short hallsets[][SETSIZE]; {
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short copysets[MAXHALL][SETSIZE];
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short setsum[SETSIZE];
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register i,j,k,ncopys;
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/*
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* First check cardinality of union of all
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*/
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for(k=0;k<SETSIZE;k++)
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setsum[k]=0;
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for(i=0;i<nhallsets;i++)
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unite(hallsets[i],setsum);
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if (card(setsum)<nhallsets)
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return(0);
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/*
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* Now check the hall property of everything but one set,
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* for all sets
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*/
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for(i=0;i<nhallsets;i++) {
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ncopys=0;
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for(j=0;j<nhallsets;j++) if (j!=i) {
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for(k=0;k<SETSIZE;k++)
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copysets[ncopys][k] = hallsets[j][k];
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ncopys++;
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}
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assert(ncopys == nhallsets-1);
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if (!recurhall(ncopys,copysets))
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return(0);
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}
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return(1);
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}
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unite(sp,into) register short *sp,*into; {
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register i;
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for(i=0;i<SETSIZE;i++)
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into[i] |= sp[i];
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}
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/*
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* Limerick (rot13)
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*
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* N zngurzngvpvna anzrq Unyy
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* Unf n urknurqebavpny onyy,
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* Naq gur phor bs vgf jrvtug
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* Gvzrf uvf crpxre'f, cyhf rvtug
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* Vf uvf cubar ahzore -- tvir uvz n pnyy..
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*/
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