int calculateFullEntry(int mat[SIZE1][SIZE2],struct fullentry * bob)
{
int n1,n2;
int autocor[(2*SIZE1-1)*(2*SIZE2-1)];
//int disps[SIZE1*(2*SIZE2-1)];
calculateEntry(mat,&(bob[0].intrep)); //calculate entry part of full entry
calculateAutocor((autocor),mat,&(bob[0].ones));
calculateSpectrum(autocor,&(bob[0].spectrum[0]),bob[0].ones,&(bob[0].peaksidelobe),2*SIZE1-1,2*SIZE2-1);
//display info for debugging
if(bob[0].ones>=25)
{
printf("ones %d \n",bob[0].ones);
printf("calculating full entry \n");
displaymatrix(mat,NULL);
printf("calculating autocor \n");
displaygeneral(autocor,NULL,2*SIZE1-1,2*SIZE2-1);
printf("peak sidelobe %d \n",bob[0].peaksidelobe);
}
/*printf("calculating displacements \n");
calculateDisplacements(mat,disps);
displaygeneral(disps,NULL,SIZE1,2*SIZE2-1);
printf("calculating displacement spectrum \n");
calculateSpectrum((disps),&(bob[0].dispspectrum[0]),&(bob[0].peakdisp),SIZE1,2*SIZE2-1); */
return 0;
}
//take input file from fortran code and convert into format which can be processed by this code, so we can check if equivalent
int convert(int mat[SIZE1][SIZE2],char *infile,char *outfile,int * numbers)
{
printf("reading in data file \n");
int swit2,count,n,val;
struct entry bob;
FILE * fpin;
FILE * fpout;
char text [1000];
fpin=fopen(infile,"r");
fpout=fopen(outfile,"w");
(*numbers)=0; //we don't know how many results we are reading in
fgets(&text[0],1000,fpin); //get first two lines
fgets(&text[0],1000,fpin);
swit2=0;
while(1)
{
//code for locating set of values from line and extracting
count=0;
while(count<SIZE1)
{
//second method for simpler output format
fscanf(fpin,"%d",&val);
if(feof(fpin)) //end of file indicator, read in all words which are part of matrices
{
swit2=1;
break;
}
//printf("%d \n",val);
bob.rows[count]=val;
//printf("%d \n",bob.rows[count]);
count++;
}
if(swit2==1)
{
break;
}
if(count==SIZE1) //if there are SIZE1 components read in, print out matrix to outfile
{
calculateMatrix(bob,mat);
//displayentry(bob,NULL);
displaymatrix(mat,fpout);
(*numbers)++;
fgets(&text[0],1000,fpin); //go to next line
}
}
//printdata(data,size1,(*numbers));
printf("read data in \n");
printf("numbers %d \n",(*numbers));
fclose(fpin);
fclose(fpout);
return 0;
}
int main()
{
char done;
printf("This is the game of the Tic Tac Toe.\n ");
printf("You will be playing against the comuter.\n");
done= ' ';
void initmatrix();
do
{
displaymatrix();
getplayermove();
done=check();
if(done!= ' ') break;
getcomputermove();
done=check();
}while(done== ' ');
if(done=='x') printf("You won!\n");
else printf("I won!!!\n");
displaymatrix();
return 0;
}
int calculateDisplacements(int mat[SIZE1][SIZE2],int * disps)
{
int n1,n2,x1,y1,count;
struct coordinate listones[MAXONES];
count=0;
for(n1=0; n1<SIZE1; n1++)
{
for(n2=0; n2<SIZE2; n2++)
{
if(mat[n1][n2]==1)
{
listones[count].x=n1;
listones[count].y=n2;
count++;
}
}
}
displaymatrix(mat,NULL);
displaylist(listones,count);
for(n1=0; n1<SIZE1; n1++)
{
for(n2=0; n2<(2*SIZE2-1); n2++)
{
disps[n1*(2*SIZE1-1)+n2]=0;
}
}
for(n2=0; n2<count; n2++)
{
for(n1=0; n1< n2; n1++) //entry in counter-1 position is newest 1, we need to verify its displacement with the older ones is not already in the matrix
{
y1=listones[n2].y-listones[n1].y;
x1=listones[n2].x-listones[n1].x;
if(x1<0) //only half of displacements are unique, we use this convention to pick half
{
x1=x1*-1; //just inverting lines, b/c displacements unique up to inversion
y1=y1*-1;
}
y1=y1+SIZE2-1;
//printf("next displacement added %d %d \n",x1,y1);
disps[x1*(2*SIZE2-1)+y1]++;
}
}
return 0;
}
void GlmTest::displayAnova(void) {
unsigned int nVars = tm->nVars;
unsigned int i, j;
const char *testname[3] // string array, only pointers stored
= {"sqrt(WALD)", "SCORE", "LR"}; // 2, 3, 4
displaymatrix(bootID, "bootID");
printf("\n ========================= \n");
printf("\nAnova Table (resampling under ");
if (tm->resamp == CASEBOOT)
printf("H1):\n");
else
printf("H0):\n");
if (tm->corr == SHRINK)
displayvector(tm->anova_lambda, "Est. shrink.param in anova");
unsigned int test = tm->test - 2;
printf("Hypo\t Alter\t dff\t %s\t P-value \n", testname[test]);
for (i = 0; i < nModels - 1; i++)
printf("M%d\t M%d\t %d\t %.3f %.3f\t\t \n", (int)i + 1, (int)i, dfDiff[i],
gsl_matrix_get(anovaStat, i, 0), gsl_matrix_get(Panova, i, 0));
if (tm->punit != NONE) {
if (tm->punit == FREESTEP)
printf("\nUnivariate Tests (FREESTEP adjusted):\n\t\t");
else
printf("\nUnivariate Tests:\n\t\t");
for (i = 0; i < nModels - 1; i++)
printf("\tM%d v. M%d\t", (unsigned int)i + 1, (unsigned int)i);
printf("\n");
for (j = 1; j < nVars + 1; j++) {
printf("[Response %d]:", (unsigned int)j);
for (i = 0; i < nModels - 1; i++)
printf("\t%.3f (%.3f)", gsl_matrix_get(anovaStat, i, j),
gsl_matrix_get(Panova, i, j));
printf("\n");
}
printf("\n");
}
}
int sort(char * fname, int counts, int mat[SIZE1][SIZE2])
{
FILE * fp;
int n,n1,n2, inti,max,start,maxdiff;
struct fullentry * matrices;
int minsidelobes [MAXONES];
fp=fopen(fname,"r");
matrices=(struct fullentry *)calloc(counts,sizeof(struct fullentry));
printf("Beginning to sort results \n");
max=0;
for(n=0; n<MAXONES; n++)
{
minsidelobes[n]=100;
}
for(n=0; n<counts; n++)
{
for(n2=0; n2<SIZE2; n2++)
{
for(n1=0; n1<SIZE1; n1++)
{
fscanf(fp,"%d ",&inti);
mat[n1][n2]=inti;
}
fscanf(fp,"\n ",&inti);
}
lexleast(mat,1);
calculateFullEntry(mat,&matrices[n]);
//displayentry(matrices[n].intrep,NULL);
if(matrices[n].ones>max)
max=matrices[n].ones;
if(minsidelobes[matrices[n].ones]>matrices[n].peaksidelobe)
minsidelobes[matrices[n].ones]=matrices[n].peaksidelobe;
}
fclose(fp);
printf("Maximum ones reached in search is %d \n",max);
printf("optimal sidelobe for each level \n");
start=-1;
//
//printf("minones %d max %d \n",MINONES,max);
int min=100;
//need to be careful here, get start of range and minimum sidelobe independently
for(n=MINONES; n<=max; n++)
{
//printf("%d \n",minsidelobes[n]);
if(minsidelobes[n]<min)
{
min=minsidelobes[n];
}
if(start==-1 && minsidelobes[n]!=100)
start=n;
if(start!=-1)
printf("%d ",minsidelobes[n]);
}
printf(" \n");
//create matrix for storing sidelobe and length statistics, if find lower sidelobe, need to also break down by sidelobe
int * counttype=(int *)calloc((max-start+1)*(MAXSIDELOBE-min+1),sizeof(int));
qsort((void *)matrices,counts,sizeof(struct fullentry),comparefullentry);
n2=0;
//now print out only the unique entries
fp=fopen(fname,"w");
fprintf(fp,"Matrices for maximum sidelobe %d with dimensions %d by %d \n",MAXSIDELOBE,SIZE1,SIZE2);
//need to print out 1st matrix in list first, code works by printing out 1st unique matrix of sorted group and ignoring the rest
calculateMatrix(matrices[0].intrep,mat);
displaymatrix(mat,fp);
displayentry(matrices[0].intrep,fp);
fprintf(fp,"\n");
//printf("start %d \n",start);
counttype[(max-start+1)*(matrices[0].peaksidelobe-min)+matrices[0].ones-start]++;
for(n1=0; n1<=MAXSIDELOBE; n1++)
{
fprintf(fp,"%d ",matrices[0].spectrum[n1]);
}
fprintf(fp," \n");
fprintf(fp, "ones %d, peak sidelobe %d \n \n",matrices[0].ones,matrices[0].peaksidelobe);
n2++;
maxdiff=matrices[0].ones-matrices[0].peaksidelobe;
printf("range for table is %d ones to %d, and sidelobe %d to %d \n",max,start,MAXSIDELOBE,min);
printf("entering loop %d counts \n",counts);
for(n=1; n<counts; n++)
{
if(compareentry(&matrices[n],&matrices[n-1])!=0) //do not print out as long as previous entry same as current entry
{
calculateMatrix(matrices[n].intrep,mat);
if(matrices[n].peaksidelobe>MAXSIDELOBE)
continue;
displaymatrix(mat,fp);
displayentry(matrices[n].intrep,fp);
fprintf(fp,"\n");
for(n1=0; n1<=MAXSIDELOBE; n1++)
{
fprintf(fp,"%d ",matrices[n].spectrum[n1]);
}
fprintf(fp," \n");
fprintf(fp, "ones %d, peak sidelobe %d \n \n",matrices[n].ones,matrices[n].peaksidelobe);
//printf("%d %d \n",(max-start+1)*(matrices[n].peaksidelobe-min)+matrices[n].ones-start,(max-start+1)*(MAXSIDELOBE-min+1));
//printf("%d peaksidelobe, %d ones \n",matrices[n].peaksidelobe,matrices[n].ones);
counttype[(max-start+1)*(matrices[n].peaksidelobe-min)+matrices[n].ones-start]++;
n2++;
if((matrices[0].ones-matrices[0].peaksidelobe)>maxdiff)
maxdiff=(matrices[0].ones-matrices[0].peaksidelobe);
}
}
fprintf(fp,"Total Unique matrices are %d \n",n2);
printf("sorted through matrices and eliminated redundant elements, have %d unique matrices out of %d \n",n2,counts);
fprintf(fp,"Maximum ones reached in search is %d \n",max);
fprintf(fp,"optimal sidelobe for each level \n");
for(n=start; n<=max; n++)
fprintf(fp,"%d : %d ",n,minsidelobes[n]);
fprintf(fp," \n");
printf("maxdiff is %d, start is %d, max is %d \n",maxdiff,start,max);
printf("printing out table \n");
//print out table of counts to give breakdown and sidelobe and int counts for given max
for(n1=0; n1<=(max-start); n1++)
{
fprintf(fp," \n");
if(n1==0)
fprintf(fp," ");
else
fprintf(fp,"%d ",n1+start);
for(n2=0; n2<=(MAXSIDELOBE-min); n2++)
{
if(n1==0)
fprintf(fp,"%d ",n2+min);
else
fprintf(fp,"%d ",counttype[(max-start+1)*n2+(n1)]);
}
}
printf("finished loop \n");
//needs to also give sum of lowest diagonal corresponding to highest max
int sum=0;
for(n1=0; n1<=(max-start); n1++)
{
//printf("%d \n",start+n1-maxdiff-minsidelobes[start]);
if((start+n1-maxdiff-min)>=0) //if not lined up properly can give sidelobe we did not include before
sum=sum+counttype[(max-start+1)*(start+n1-maxdiff-min)+n1]; //need to sweep constant diffmax contour
}
printf("\n \n There are %d results with optimal difference of %d \n",sum,maxdiff);
fprintf(fp,"\n \n There are %d results with optimal difference of %d \n",sum,maxdiff);
fclose(fp);
printf("reached end \n");
return 0;
}
//1st thing need to do is add new differences to displacements and confirm that these displacements are new
int addones(int mat[SIZE1][SIZE2], struct coordinate disps [MAXONES][MAXONES], struct coordinate listones[MAXONES], int counter, int displacements[SIZE1][2*SIZE2-1], FILE * fp, int * outcount )
{
int n1, n2, minx, miny,x1,y1;
/*if(counter>=11)
{
printf("\n \n");
printf("level %d \n",counter);
displaylist(listones,counter);
displaymatrix(mat,NULL);
printf("added one %d %d \n",listones[counter-1].x,listones[counter-1].y);
}*/
//1st check whether or not matrix is lexographically least, criteria is if less than half of max ones in upper half
//need mod offset for odd dimensions so can strictly define with bigger half of matrix, otherwise can't gaurantee anything with given class
if(SIZE2*SIZE1/2+MOD <(listones[counter-1].y*SIZE1+listones[counter-1].x) && counter <MINONES/2 )
{
return 0;
}
//2nd check whether or not we can add displacements to list
for(n1=0; n1< (counter-1); n1++) //entry in counter-1 position is newest 1, we need to verify its displacement with the older ones is not already in the matrix
{
y1=listones[counter-1].y-listones[n1].y;
x1=listones[counter-1].x-listones[n1].x;
if(x1<0) //only half of displacements are unique, we use this convention to pick half
{
x1=x1*-1; //just inverting lines, b/c displacements unique up to inversion
y1=y1*-1;
}
y1=y1+SIZE2-1;
//printf("next displacement added %d %d \n",x1,y1);
displacements[x1][y1]++;
disps[n1][counter-1].x=x1; //add new displacements to displacement list
disps[n1][counter-1].y=y1;
if(displacements[x1][y1]>1)
{
mat[listones[counter-1].x][listones[counter-1].y]=0; //need to remove 1 from matrix
for(n2=0; n2<=n1; n2++)
{
displacements[disps[n2][counter-1].x][disps[n2][counter-1].y]--; //remove all previously added displacements
}
return 0;
}
}
//3rd check if satisfies the minimum ones requirement, if so print out
if(counter>=MINONES)
{
//if(lexleast(mat,0)==1) //only display matrix if lex least
//{
(*outcount)++;
displaymatrix(mat,fp);
//}
//displaydisplacements(displacements,fp);
}
minx=listones[counter-1].x;
miny=listones[counter-1].y;
//iterate deeper by adding more ones
for(n2=miny; n2<SIZE2; n2++)
{
for(n1=minx; n1<SIZE1; n1++)
{
listones[counter].x=n1; //put one at spot in matrix
listones[counter].y=n2;
if(mat[n1][n2]==0) //only add 1 to spot if none already there, don't touch ones from other levels
{
mat[n1][n2]=1;
addones(mat,disps,listones,counter+1,displacements,fp,outcount);
mat[n1][n2]=0; //have to conserve ones at a given level, once test with one take it away and add another place
}
}
minx=0; //allow to access any column in row after restriction on row with original entry
}
//needs to remove displacements when returning so it will not effect the operation of the next branch
for(n2=0; n2< (counter-1); n2++)
{
displacements[disps[n2][counter-1].x][disps[n2][counter-1].y]--; //remove all previously added displacements
}
return 0; //shouldn't need to return whether or successful or not for code to contineu
}
//same method as before except in loop form for increased speed, minimize the number of function calls
int imperativeform(int mat[SIZE1][SIZE2], struct coordinate disps[MAXONES][MAXONES], struct coordinate listones[MAXONES], int displacements[SIZE1][2*SIZE2-1],int n1[MAXONES],int n2[MAXONES],int minx[MAXONES], FILE *fp, int *outcount,int counter,int * minsidelobes, int * maxsidelobes,struct fullentry * bob)
{
int inc,inc2,x1,y1;
int matp[SIZE1][SIZE2];
//when 1st comes in counter=1
struct entry ref;
ref.rows[0]=2;
ref.rows[1]=17;
ref.rows[2]=22;
//first loop needs to be of special form because of restriction on search space, but only amounts to Size/4 function calls anyways so don't need here
start: ;
//printf("counter %d %d %d \n",counter,listones[counter].x,listones[counter].y);
/*printf("\n \n");
printf("level %d \n",counter+1);
displaylist(listones,counter+1);
displaymatrix(mat,NULL);
printf("added one %d %d \n",listones[counter].x,listones[counter].y); */
/*if((counter+1)==MINONES)
{
identity(mat,matp,SIZE1,SIZE2);
lexleast(matp,1);
calculateFullEntry(matp,bob);
//displayentry((*bob).intrep,NULL);
if(compareentry(&ref,&((*bob).intrep))==0)
{
printf("Have match! \n");
}
}*/
//need mod offset for odd dimensions so can strictly define with bigger half of matrix, otherwise can't gaurantee anything with given class
if(SIZE2*SIZE1/2+MOD <(listones[counter].y*SIZE1+listones[counter].x) && counter <MINONES/2 )
{
//will never satisfy this for counter =1, can just goto loop here
counter=counter-1;
goto loop;
}
//printf("2nd part \n");
//2nd check whether or not we can add displacements to list
maxsidelobes[counter+1]=maxsidelobes[counter]; //start with max from previous level
for(inc=0; inc<(counter); inc++) //entry in counter-1 position is newest 1, we need to verify its displacement with the older ones is not already in the matrix
{
y1=listones[counter].y-listones[inc].y;
x1=listones[counter].x-listones[inc].x;
if(x1<0) //only half of displacements are unique, we use this convention to pick half
{
x1=x1*-1; //just inverting lines, b/c displacements unique up to inversion
y1=y1*-1;
}
y1=y1+SIZE1-1;
//printf("next displacement added %d %d \n",x1,y1);
displacements[x1][y1]++;
disps[inc][counter].x=x1; //add new displacements to displacement list
disps[inc][counter].y=y1;
if(displacements[x1][y1]>maxsidelobes[counter+1])
maxsidelobes[counter+1]=displacements[x1][y1];
if(displacements[x1][y1]>MAXSIDELOBE ) //use sidelobe criteria and previous maxdistance to sort results
{
mat[listones[counter].x][listones[counter].y]=0; //need to remove 1 from matrix
for(inc2=0; inc2<=inc; inc2++)
{
displacements[disps[inc2][counter].x][disps[inc2][counter].y]--; //remove all previously added displacements
}
counter=counter-1; //go back up level, again will never satisfy for counter=0 here
goto loop;
}
}
//printf("3rd part \n");
//3rd check if satisfies the minimum ones requirement, if so print out
if( (counter+1)>=MINONES && ((counter+1-maxsidelobes[counter+1])>=MINDISTANCE) )
{
(*outcount)++;
displaymatrix(mat,fp);
//if(minsidelobes[bob[0].ones]>bob[0].peaksidelobe)
// minsidelobes[bob[0].ones]=bob[0].peaksidelobe;
}
minx[counter]=listones[counter].x;
//iterate deeper by adding more ones
for(n2[counter]=listones[counter].y; n2[counter]<SIZE2; n2[counter]++)
{
for(n1[counter]=minx[counter]; n1[counter]<SIZE1; n1[counter]++)
{
if(mat[n1[counter]][n2[counter]]==0) //only add 1 to spot if none already there, don't touch ones from other levels
{
//need to put one more spot in matrix, needs to put list coordinate at counter+1, because loop currently on counter interation
listones[counter+1].x=n1[counter];
listones[counter+1].y=n2[counter];
mat[n1[counter]][n2[counter]]=1;
counter=counter+1;
goto start; //only can go to start from this point
//bad cases, or terminations will return here, decrementing counter as they come back up
loop: ;
mat[n1[counter]][n2[counter]]=0; //have to conserve ones at a given level, once test with one take it away and add another place
}
}
minx[counter]=0; //allow to access any column in row after restriction on row with original entry
}
//needs to remove displacements when returning so it will not effect the operation of the next branch
for(inc=0; inc< counter; inc++)
{
displacements[disps[inc][counter].x][disps[inc][counter].y]--; //remove all previously added displacements
}
//go back up level
counter=counter-1;
if(counter!=-1)
goto loop; //done checking cases needs to go back to place where called from
else
return 0;
}