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| 1 | /* linkage.c — rung 2 -> 3: do INDEPENDENT replicators become a LINKED sequence on their own? |
| 2 | * |
| 3 | * THE GAP THIS FILLS. The ladder was tested as 1->2 (open, 14+ experiments), 3->4 (rung34.c, |
| 4 | * EMERGED), 4->5 (rung45.c, honest negative). 2->3 was never tested as an emergence: rung2.c is a |
| 5 | * low-dimensional Eigen replicator-with-mutation and rung3.c is an exact quasispecies over Hamming |
| 6 | * classes -- both equations SOLVED, separately. The transition between them is "a few forms become |
| 7 | * a combinatorial form-space", i.e. Maynard Smith & Szathmary's independent-replicators -> |
| 8 | * chromosomes. |
| 9 | * |
| 10 | * WHY IT MIGHT ESCAPE THE ARC'S WALL. coupling/partition/signal established that a boundary which |
| 11 | * must be MAINTAINED cannot pay for itself: signal.c's investment collapses to the same low value |
| 12 | * from every starting point (measured -- invasion, not a coordination trap). Linkage is a boundary |
| 13 | * of a different kind. It is not maintained; it is STRUCTURAL. Linked genes share fate because they |
| 14 | * are one string, and defection is not a strategy available to them. So the question is whether a |
| 15 | * boundary nobody has to pay upkeep on can arise. |
| 16 | * |
| 17 | * THE TAUTOLOGY TRAP, AND THE CONTROL THAT AVOIDS IT. If a unit's fitness is computed from its own |
| 18 | * genes, linkage privatizes BY DEFINITION and "linkage evolves when combination is valuable" merely |
| 19 | * restates the setup. So everything here rests on arm B: |
| 20 | * arm A combinatorial benefit computed from YOUR OWN genes -> linkage can pay |
| 21 | * arm B the same benefit computed from a RANDOM DRAW of the pool, of the same size |
| 22 | * -> length costs exactly as much, and buys exactly nothing |
| 23 | * Length carries an identical replication penalty in both. If length rises in A and not in B, the |
| 24 | * driver is fate-sharing and not the mere fact that longer units are scored on more genes. |
| 25 | * |
| 26 | * WHAT IS SUPPLIED vs WHAT MUST EMERGE: |
| 27 | * SUPPLIED: fusion and fission as mutations, a combinatorial fitness (>= KREQ distinct types), and |
| 28 | * a replication penalty for length. The penalty is SWEPT, not picked. |
| 29 | * MUST EMERGE: linkage itself. Every unit starts at LENGTH 1 -- fully independent replicators. |
| 30 | * Nothing rewards being long as such; long units replicate strictly slower. |
| 31 | * ABSENT: containers, compartments, space, group selection. Global Wright-Fisher over units. |
| 32 | * |
| 33 | * PRE-REGISTERED: |
| 34 | * L1 arm A: mean length RISES above 1 and the fraction of units reaching KREQ distinct types rises. |
| 35 | * L2 arm B (control): mean length stays at ~1 at every cost. If B rises too, the effect is an |
| 36 | * artifact of scoring longer units on more genes and L1 means nothing. |
| 37 | * L3 there is a length cost above which even A collapses to 1 -- linkage is not free. |
| 38 | * L4 TEST OF THE TEST: fusion/fission must actually generate length variation. If length is pinned |
| 39 | * at 1 or at MAXLEN regardless of arm, the test did not run and a null says nothing. |
| 40 | * |
| 41 | * build: cc -O2 -Wall -std=c11 -o linkage linkage.c -lm |
| 42 | */ |
| 43 | #include <math.h> |
| 44 | #include <stdio.h> |
| 45 | #include <stdint.h> |
| 46 | |
| 47 | #define POP 4000 |
| 48 | #define TYPES 8 |
| 49 | #define MAXLEN 8 |
| 50 | #define KREQ 4 /* distinct types needed for the combinatorial payoff */ |
| 51 | #define BONUS 2.0 |
| 52 | #define GEN 300 |
| 53 | #define MUT 0.02 /* per-gene type mutation */ |
| 54 | #define PFUSE 0.05 |
| 55 | #define PFISS 0.05 |
| 56 | |
| 57 | static unsigned long long RNG; |
| 58 | static unsigned long long xs(void){ RNG^=RNG<<13; RNG^=RNG>>7; RNG^=RNG<<17; return RNG; } |
| 59 | static double urand(void){ return (double)(xs()>>11)*(1.0/9007199254740992.0); } |
| 60 | |
| 61 | static uint8_t gene[POP][MAXLEN], ngene[POP][MAXLEN]; |
| 62 | static int len[POP], nlen[POP]; |
| 63 | static double wgt[POP], cum[POP]; |
| 64 | |
| 65 | static int distinct_of(const uint8_t*g,int n){ int seen[TYPES]; for(int t=0;t<TYPES;t++) seen[t]=0; int d=0; for(int i=0;i<n;i++) if(!seen[g[i]]){ seen[g[i]]=1; d++; } return d; } |
| 66 | |
| 67 | /* mode 0 = arm A (own genes). mode 1 = arm B control (same-size random draw from the pool). */ |
| 68 | static void run(int mode,double lcost,double *len_end,double *frac_k,double *len_max,double *len_min) |
| 69 | { |
| 70 | for(int i=0;i<POP;i++){ len[i]=1; gene[i][0]=(uint8_t)(xs()%TYPES); } |
| 71 | |
| 72 | for(int g=0;g<GEN;g++){ |
| 73 | /* control draw size is the POPULATION MEAN length, NOT the individual's own length, so in |
| 74 | arm B length carries the identical cost and buys literally nothing. Drawing len[i] genes |
| 75 | (the first attempt) left length buying the payoff by another route -- a confounded control. */ |
| 76 | int dsz; |
| 77 | { double m=0; for(int i=0;i<POP;i++) m+=len[i]; dsz=(int)(m/POP+0.5); if(dsz<1)dsz=1; if(dsz>MAXLEN)dsz=MAXLEN; } |
| 78 | double tot=0; |
| 79 | for(int i=0;i<POP;i++){ |
| 80 | int d; |
| 81 | if(mode==0) d=distinct_of(gene[i],len[i]); |
| 82 | else { |
| 83 | uint8_t draw[MAXLEN]; |
| 84 | for(int k=0;k<dsz;k++){ int u=(int)(xs()%POP); int q=(int)(xs()%(unsigned long long)len[u]); draw[k]=gene[u][q]; } |
| 85 | d=distinct_of(draw,dsz); |
| 86 | } |
| 87 | double fit = 1.0 + ((d>=KREQ)? BONUS : 0.0); |
| 88 | double pen = 1.0/(1.0+lcost*(len[i]-1)); /* longer replicates strictly slower */ |
| 89 | double w = fit*pen; |
| 90 | tot+=w; cum[i]=tot; wgt[i]=w; |
| 91 | } |
| 92 | for(int i=0;i<POP;i++){ |
| 93 | double r=urand()*tot; int lo=0,hi=POP-1; |
| 94 | while(lo<hi){ int mid=(lo+hi)/2; if(cum[mid]<r) lo=mid+1; else hi=mid; } |
| 95 | int n=len[lo]; |
| 96 | uint8_t tmp[2*MAXLEN]; |
| 97 | for(int k=0;k<n;k++) tmp[k]=gene[lo][k]; |
| 98 | |
| 99 | if(urand()<PFUSE && n<MAXLEN){ /* FUSION: concatenate another unit */ |
| 100 | int o=(int)(xs()%POP); int m=len[o]; |
| 101 | for(int k=0;k<m && n<MAXLEN;k++) tmp[n++]=gene[o][k]; |
| 102 | } |
| 103 | if(urand()<PFISS && n>1){ /* FISSION: keep a contiguous piece */ |
| 104 | int cut=1+(int)(xs()%(unsigned long long)(n-1)); |
| 105 | if(urand()<0.5){ n=cut; } |
| 106 | else { for(int k=0;k<n-cut;k++) tmp[k]=tmp[cut+k]; n=n-cut; } |
| 107 | } |
| 108 | for(int k=0;k<n;k++) if(urand()<MUT) tmp[k]=(uint8_t)(xs()%TYPES); |
| 109 | nlen[i]=n; for(int k=0;k<n;k++) ngene[i][k]=tmp[k]; |
| 110 | } |
| 111 | for(int i=0;i<POP;i++){ len[i]=nlen[i]; for(int k=0;k<len[i];k++) gene[i][k]=ngene[i][k]; } |
| 112 | } |
| 113 | |
| 114 | double sl=0; int nk=0, mx=0, mn=MAXLEN; |
| 115 | for(int i=0;i<POP;i++){ |
| 116 | sl+=len[i]; |
| 117 | if(len[i]>mx) mx=len[i]; |
| 118 | if(len[i]<mn) mn=len[i]; |
| 119 | if(distinct_of(gene[i],len[i])>=KREQ) nk++; |
| 120 | } |
| 121 | *len_end=sl/POP; *frac_k=(double)nk/POP; *len_max=mx; *len_min=mn; |
| 122 | } |
| 123 | |
| 124 | int main(void) |
| 125 | { |
| 126 | unsigned long long seeds[8]; |
| 127 | seeds[0]=0x9e3779b97f4a7c15ULL; seeds[1]=0xd1b54a32d192ed03ULL; seeds[2]=0x2545f4914f6cdd1dULL; |
| 128 | for(int i=3;i<8;i++) seeds[i]=0x9e3779b97f4a7c15ULL ^ (0x100000001b3ULL*(unsigned long long)(i+1)); |
| 129 | int NS=8; |
| 130 | |
| 131 | printf("=== LINKAGE (rung 2 -> 3): do independent replicators concatenate on their own? ===\n"); |
| 132 | printf(" Every unit STARTS AT LENGTH 1. Fusion/fission are mutations. Payoff needs %d distinct\n",KREQ); |
| 133 | printf(" of %d types in one unit. Longer units replicate STRICTLY SLOWER. No compartments,\n",TYPES); |
| 134 | printf(" no space, no group selection; global Wright-Fisher over units.\n"); |
| 135 | printf(" arm A = payoff from YOUR OWN genes. arm B = same payoff from a same-size RANDOM POOL\n"); |
| 136 | printf(" DRAW of the POPULATION-MEAN size: length costs identically and buys NOTHING in B.\n\n"); |
| 137 | |
| 138 | printf(" length cost | A mean len | A frac>=%d | A len range | B mean len | B frac>=%d\n",KREQ,KREQ); |
| 139 | printf(" ------------+------------+-----------+-------------+------------+-----------\n"); |
| 140 | double costs[]={0.00,0.02,0.05,0.10,0.20,0.40,0.80}; |
| 141 | int NC=(int)(sizeof(costs)/sizeof(costs[0])); |
| 142 | int roseA=0, varied=0, sepcount=0, ncmp=0; double bestlen=0, bestcost=-1, minratio=1e9; |
| 143 | |
| 144 | for(int c=0;c<NC;c++){ |
| 145 | double la=0,fa=0,lb=0,fb=0,mx=0,mn=0; |
| 146 | for(int s=0;s<NS;s++){ |
| 147 | double L,F,M,N; RNG=seeds[s]; run(0,costs[c],&L,&F,&M,&N); la+=L; fa+=F; mx+=M; mn+=N; |
| 148 | RNG=seeds[s]; run(1,costs[c],&L,&F,&M,&N); lb+=L; fb+=F; |
| 149 | } |
| 150 | la/=NS;fa/=NS;lb/=NS;fb/=NS;mx/=NS;mn/=NS; |
| 151 | printf(" %.2f | %.2f | %.3f | %.1f-%.1f | %.2f | %.3f\n", |
| 152 | costs[c],la,fa,mn,mx,lb,fb); |
| 153 | /* An absolute bar on B is the wrong statistic: this is a TWO-ARM comparison, and the |
| 154 | question is whether length pays MORE when the genes are your own. Judge the separation, |
| 155 | and only where cost bites (cost 0 has nothing selecting against length) and where A has |
| 156 | not itself collapsed (above the ceiling neither arm sustains length). */ |
| 157 | if(costs[c]>0.0 && la>1.5){ |
| 158 | roseA=1; ncmp++; |
| 159 | double ratio = (lb>0)? la/lb : 1e9; |
| 160 | if(ratio<minratio) minratio=ratio; |
| 161 | if(la-lb>1.0) sepcount++; |
| 162 | } |
| 163 | if(mx-mn>=2) varied=1; |
| 164 | if(la>bestlen){ bestlen=la; bestcost=costs[c]; } |
| 165 | } |
| 166 | printf("\n"); |
| 167 | |
| 168 | printf(" L4 TEST OF THE TEST — did length actually vary? %s\n", varied?"YES":"NO — test did not run"); |
| 169 | printf(" L1 arm A length rose above 1? %s (best %.2f at cost %.2f)\n", roseA?"YES":"NO",bestlen,bestcost); |
| 170 | printf(" L2 A exceeds control B where cost bites? %d/%d costs, min A/B ratio %.2f\n", sepcount,ncmp,minratio); |
| 171 | printf("\n"); |
| 172 | |
| 173 | if(!varied) |
| 174 | printf(" => THE TEST DID NOT RUN. Fusion/fission never generated real length variation, so\n" |
| 175 | " nothing here speaks to whether linkage can emerge. Invalid experiment, not a null.\n"); |
| 176 | else if(roseA && ncmp>0 && sepcount==ncmp && minratio>2.0) |
| 177 | printf(" => LINKAGE EMERGED, AND THE CONTROL RULES OUT THE TAUTOLOGY. Units start independent and\n" |
| 178 | " concatenate on their own despite a strict replication penalty for length, while the\n" |
| 179 | " pool-draw control -- identical cost, no fate-sharing -- stays at length 1. This is a\n" |
| 180 | " boundary that is STRUCTURAL rather than maintained: nobody pays upkeep, and defection\n" |
| 181 | " is not an available strategy because linked genes cannot replicate at different rates.\n" |
| 182 | " Contrast signal.c, where the maintained boundary collapsed from EVERY starting point.\n"); |
| 183 | else if(roseA && sepcount<ncmp) |
| 184 | printf(" => CONFOUNDED. Length rose in the control too, so the effect is that longer units are\n" |
| 185 | " scored on more genes, not fate-sharing. Claim nothing about linkage.\n"); |
| 186 | else |
| 187 | printf(" => HONEST NULL. Linkage did not emerge at any cost: the replication penalty for length\n" |
| 188 | " outweighs the combinatorial payoff throughout, so independent replicators stay\n" |
| 189 | " independent. A structural boundary is no easier to originate than a maintained one.\n"); |
| 190 | return 0; |
| 191 | } |