Mutation

Reasons why women live longer then men It is an accepted fact of life that men enjoy certain physical advantages over women. Men are stronger, taller, faster and less likely to be overweight, but they have then tendency to die before their female counterparts. Their date rate is higher than women in many different societies. Women, as a group, live longer than men. In all developed countries and most undeveloped ones, women outlive men, sometimes by a margin of as much as 10 years. In the U. S.  life expectancy at birth is about 79 years for women and about 72 years for men. The gender discrepancy is most pronounced in the very old: among centenarians worldwide, women outnumber men nine to one. The gender gap has widened in this century as gains in female life expectancy have exceeded those for males. The death rates for women are lower than those for men at all ages–even before birth. Although boys start life with some numerical leverage about 115 males are conceived for every 100 females. Their numbers are preferentially whittled down thereafter.

Just 104 boys are born for every 100 girls because of the disproportionate rate of spontaneous abortions, stillbirths and miscarriages of male fetuses. More boys than girls die in infancy. And during each subsequent year of life, mortality rates for males exceed those for females, so that by age 25 women are in the majority. The reason women live longer than men can be explained from a biological, psychological and social view point. There are physiological, psychological and social factors that can account for longevity of women.

The reason women live longer than men is because they have better cardiovascular endurance, they have two x chromosomes which further delays aging, men are more engaged in risky behaviors that can cause death, and another reason is that men are more likely to die from depression women. Women have an advantage over men in terns of cardiovascular endurance; disease such a heart attack and stroke are less likely to affect women than men. Woman develop these diseases later in life at about their 70s and 80s while on the other hand male normally get these types of disease about ten years younger in their 50s and 60s.

A woman’s heart can become more active than men. A major contributor to biological age is one’s heart activity; when the heart is more active and healthier a person’s biological age can be reduced because of how the heart is still working in a proper way. A woman’s heart tends to be more active primarily because of how the menstruation process causes the heart to exercise and to work at a greater level. One proposed reason for this is the menstruation process which causes a woman’s heart to become active by the release estradiol hormone in the woman’s body during the process.

Estradiol will work to give the woman’s heart a greater amount of energy because this is an especially powerful form of estrogen that can cause her heart to be more active. It is more powerful than other types of estrogen that can be found in a woman’s body. Because of the workout that is given to the heart during the menstruation period the heart will be able to work at a better rate and as a result to help with getting one’s biological age reduced. Another reason for the delay in cardiovascular disease in women is that women are relatively iron deficient in comparison to men.

This is due to the loss of blood during menstruation; it is especially active in young females. Iron play a very active role in reaction within our cells that produce damaging free radicals which attach unto the cell membrane and DNA, and may translate into aging the cell. A second possible reason for why women live longer than men is the difference in genetic make up. The existence of a second X chromosome in women while men have an X and a Y chromosome is a proposed reason for the longevity of women. When there is damage caused to the cell variations, women have back-up chromosome variations that men do not have.

The assumption is that since men only have one X chromosome, they have a greater chance of aging more rapidly. Chromosomal differences between men and women may also affect their mortality rates due to diseases. Chromosomes carry genetic mutations that cause a number of life-threatening diseases, including muscular dystrophy and hemophilia. Because women have two X chromosomes, a female with an abnormal gene on one of her X chromosomes can use the normal gene on the other and thereby avoid the expression of disease even though she is still a carrier of the defect.

Men, in contrast, have one X chromosome and one Y chromosome, and so they cannot rely on an alternative chromosome if a gene on one of the sex chromosomes is defective. If a man has a defect in this gene, his body’s ability to repair the mutations that arise during cell division could be severely compromised. The accumulation of such mutations is thought to contribute to aging and disease. The second X chromosome as a longevity factor in and of itself. Although one of the two Xs is randomly inactivated early in life, the second X seems to become more active with increasing age.

It may be that genes on the second X activates and compensate for genes on the first X that have been lost or damaged with age. A third reason for the mortality rate be higher for males is the fact that men are more likely to engage in risky life-threatening behavior thus raising the mortality rate among them. While this is a more social reasoning it can be explained using biological terms. The male specie experience what is known as “testosterone storm” in their late teens and 20s. At this stage the level of this hormone is in high quantity and it induce some dangerous behavior and bring out an aggressive nature in men.

Some male dominated sports such as drag racing, motorcycling, sky diving, mountain climbing and even boxing are dangerous sports that have caused death of the participants. They may also engage in reckless behavior such as smoking, abusing alcohol, poor dieting and driving without a seat belt. These reckless behavior can cause death in the long run by death or immediately by accidents. The homicide rate of men is much higher than that of females. Men are more involved in violent and illegal activities that most of the time result is death due to violence.

Such activities include scamming, drug dealing, gang involvement, gun trafficking, among others. In Jamaica murder is one of the leading cause of death for men in early adulthood, and some of the time these victims are perpetrators of this same crime. As the saying goes “live by the gun, die by the gun”. A forth and final reason is that men are more likely to die from depression that females. Depressing is higher among male that females. While many females will go to seek help for their depression, male are more reluctant to actively seek help from a counselor or even a friend.

They would prefer to keep it inside and they may hold the motion that as long as it is not visible then its fine. Many men are harboring emotional distress and torment that cause depression. Depression and grief and affect one’s health and further cause death. Depression can lead to suicide; men are more likely to die from attempted suicide than females. This is because male tend to use more fatal methods to carry out the action, for example, a woman would probably try to kill herself by overdosing on pills or medication while a man would jump off a ten story building.

Overdose of pills can be remedied if detected early enough but if when a person jumps off a building there is no turning back or pause, they are heading at full speed ahead to the ground causing immediate death. In conclusion, the mortality rate for men is higher at all ages than female; it is almost as if women have a natural adaptation for survival. There are biological, psychological and social reasoning for this phenomenon which are hormonal difference, difference in genetic make up, psychological healthiness and the social on look on risk taking and dangerous activities.

I do not support radon gas because of its harmful effects that can affect everyone around or using the gas. Radon is second leading cause of lung cancer, the primary adverse health effect associated with chronic exposure to radon is lung cancer typically bronchogenic, squamous cell carcinoma, small cell carcinoma, adenocarcinoma, large cell carcinoma. Radon also has the potential to generate genotoxic effects higher incidence of chromosomal aberrations. Radon has been linked with erotogenic effects. Radon and smoking displayed a synergistic effect.

Increased levels of exposure to radon increase the probability of observing an effect; however, increased levels of exposure do not necessarily influence the type or severity of the effect. There are no zero risk thresholds for radon exposure, but instead, risk increases proportionate to the exposure time. Pipelines can explode at any time for many different reasons, and those reasons can be the cause of why shopping malls, ouses, and schools could be the victims of a fatal fire after the pipelines explode.

You state that the pipelines would have periodic checks and be monitored 24 hours a day, every year, but tell that to the campers of Floyd Benet Park, the trees there are prone to fire, and I sincerely doubt that you wish to spend all your profits fixing on the reparation needed to replenish all the trees. Radon decays to equally radioactive and dangerous particles, like polonium and radioactive lead, In addition, before being broken down to regular, non-radioactive lead.

When people breathe in, radon is exhaled, but the radon particles ends up inside the lungs, which then will cause cancer. Only because the radon is a heavy, gas and usually gravitates towards the open floor, making in harmful for children, pets and adults The Radon particles and the progeny can plate out the sides ot gas pipelines, which makes the pipelines hot and radioactive, then making them hazardous. I recently found out that in 1986, EPA had set a limit for exposure to radon in air at 4 picocuries per liter.

Being that there is increased exposure to many inds of radiation in the world today. Universities and Organizations like John Hopkins Universities and the World Health Organization have said that 2. 7 picocuries per liter would be a much better standard, but in reality, there is really know safe level of exposure. To Whom is reading this Letter I sincerely hope you take account all of the facts and examples that I have written and explained to you this day, and I hope the Federal Energy Regulatory Commission understands the dangers or this gas and how it can affect our community,

practaSAGAR INSTITUTE OF RESEARCH AND TECHNOLOGY SOFT COMPUTING PRACTICAL FILE (CS-801) Subject Guide: Submitted by: INDEX S. No. | List of Experiments| Signature| 1. | Implement Perceptron network with binary input and output. | | 2. | Using Madaline net, generate XOR function with bipolar inputs and targets. | | 3. | Calculation of new weights for a back propagation network, given the values of input pattern, output pattern, target output, learning rate and activation function. | | 4. | Use of ART algorithm to cluster vectors. | 5. | Implement traveling salesman problem using genetic algorithm. | | 6. | Implement various laws associated with fuzzy sets. | | 7. | Implement fuzzy sets. | | 8. | Implement word matching using GA. | | Experiment 1: Implement Perceptron network with binary input and output. Program: /*PERCEPTRON*/ #include<stdio. h> #include<conio. h> main() { signed int x[4][2],tar[4]; float w[2],wc[2],out=0; int i,j,k=0,h=0; float s=0,b=0,bc=0,alpha=0; float theta; clrscr(); printf(“Enter the value of theta & alpha”); scanf(“%f%f”,&theta,&alpha); for(i=0;i<=3;i++) printf(“Enter the value of %d Inputrow & Target”,i); for(j=0;j<=1;j++) { scanf(“%d”,&x[i][j]);} scanf(“%d”,&tar[i]); w[i]=0; wc[i]=0;} printf(“Net Target Weight changes New weights Bias changes Bias
“); printf(“—————————————————————————–
“); mew: printf(“ITERATION %d

“,h); printf(“—————————————————————————-
“); for(i=0;i<=3;i++) {for(j=0;j<=1;j++) {s+=(float)x[i][j]*w[j];} s+=b; printf(“%. 2f “,s); if(s>theta) out=1; else if(s<-theta) ut=-1; else { out=0;} printf(“%d “,tar[i]); s=0; if(out==tar[i]) {for(j=0;j<=1;j++) {wc[j]=0; bc=0; printf(“%. 2f “,wc[j]);} for(j=0;j<=1;j++) printf(“%. 2f “,w[j]); k+=1; b+=bc; printf(“%. 2f “,bc); printf(“%. 2f “,b); } else {for(j=0;j<=1;j++) {wc[j]=x[i][j]*tar[i]*alpha; w[j]+=wc[j]; printf(“%. 2f “,wc[j]); wc[j]=0;} for(j=0;j<=1;j++) printf(“%. 2f “,w[j]); bc=tar[i]*alpha; b+=bc; printf(“%. 2f “,bc); printf(“%. 2f “,b); } printf(“
“); } if(k==4) {printf(“
Final weights
“); for(j=0;j<=1;j++) {printf(“w[%d]=%. 2f “,j,w[j]); } printf(“Bias b=%. 2f”,b); } else k=0;h=h+1; getch(); goto mew;}getch(); } Output: Experiment 2: Using Madaline net, generate XOR function with bipolar inputs and targets. Program: /*MADALINE*/ #include<stdio. h> #include<conio. h> main() { signed int x[4][4],tar[4]; float wc[4],w[4],e=0,er=0,yin=0,alp=0. 5,b=0,bc=0,t=0; int i,j,k,q=1; clrscr(); for(i=0;i<=3;i++) {printf(“
Enter the %d row and target “,i); for(j=0;j<=3;j++) {scanf(“%d”,&x[i][j]);} scanf(“%d”,&tar[i]); printf(“%d”,tar[i]); w[i]=0. 0; wc[i]=0. 0;} mew: er=0;e=0; yin=0; printf(“
ITERATION%d”,q); printf(“
——————“); or(i=0;i<=3;i++) {t=tar[i]; for(j=0;j<=3;j++) {yin=yin+x[i][j]*w[j];} b=b+bc; yin=yin+b; bc=0. 0; printf(“
Net=%f “,yin); e=(float)tar[i]-yin; yin=0. 0; printf(“Error=%f “,e); printf(“Target=%d
“,tar[i]); er=er+e*e; for(k=0;k<=3;k++) {wc[k]=x[i][k]*e*alp; w[k]+=wc[k]; wc[k]=0. 0;} printf(“Weights “); for(k=0;k<=3;k++) {printf(“%f “,w[k]);} bc=e*alp; printf(“b=%. 2f “,b); getch(); printf(“
Error Square=%f”,er); if(er<=1. 000) {printf(“
“); for(k=0;k<=1;k++) printf(“%f “,w[k]); getch();} else {e=0; er=0; yin=0; q=q+1; goto mew;} getch();}} Output:

Experiment 3: Calculation of new weights for a back propagation network, given the values of input pattern, output pattern, target output, learning rate and activation function. Program: /*BACK PROPAGATION NETWORK*/ #include<stdio. h> #include<conio. h> #include<math. h> #include<stdlib. h> void main() {float v[2][4],w[4][1],vc[2][4],wc[4][1],de,del[4],bl,bia,bc[4],e=0; float x[4][2],t[4],zin[4],delin[4],yin=0,y,dy,dz[4],b[4],z[4],es,alp=0. 02; int i,j,k=0,itr=0; v[0][0]=0. 1970; v[0][1]=0. 3191; v[0][2]=-0. 1448; v[0][3]=0. 3594; v[1][0]=0. 3099; v[1][1]=0. 1904; v[1][2]=-0. 0347; [1][3]=-0. 4861; w[0][0]=0. 4919; w[1][0]=-0. 2913; w[2][0]=-0. 3979; w[3][0]=0. 3581; b[0]=-0. 3378; b[1]=0. 2771; b[2]=0. 2859; b[3]=-0. 3329; bl=-0. 141; x[0][0]=-1; x[0][1]=-1; x[1][0]=-1; x[1][1]=1; x[2][0]=1; x[2][1]=-1; x[3][0]=1; x[3][1]=1; t[0]=0; t[1]=1; t[2]=1; t[3]=0; clrscr(); for(itr=0;itr<=387;itr++) {e=0; es=0; for(i=0;i<=3;i++) {do { for(j=0;j<=1;j++) {zin[k]+=x[i][j]*v[j][k];} zin[k]+=b[k]; k+=1; }while(k<=4); for(j=0;j<=3;j++) {z[j]=(1-exp(-zin[j]))/(1+exp(-zin[j])); dz[j]=((1+z[j])*(1-z[j]))*0. 5;} for(j=0;j<=3;j++) {yin+=z[j]*w[j][0];} yin+=bl; y=(1-exp(-yin))/(1+exp(-yin)); y=((1+y)*(1-y))*0. 5; de=(t[i]-y)*dy; e=t[i]-y; es+=0. 5*(e*e); for(j=0;j<=3;j++) {wc[j][0]=alp*de*z[j]; delin[j]=de*w[j][0]; del[j]=delin[j]*dz[j];} bia=alp*de; for(k=0;k<=1;k++) {for(j=0;j<=3;j++) {vc[k][j]=alp*del[j]*x[i][k]; v[k][j]+=vc[k][j];}} for(j=0;j<=3;j++) {bc[j]=alp*del[j]; w[j][0]+=wc[j][0]; b[j]+=bc[j];} bl+=bia; for(j=0;j<=3;j++) {zin[j]=0; z[j]=0; dz[j]=0; delin[j]=0; del[j]=0; bc[j]=0;} k=0;yin=0;y=0; dy=0;bia=0;de=0;} printf(“
Epoch %d:
“,itr); for(k=0;k<=1;k++) {for(j=0;j<=3;j++) {printf(“%f “,v[k][j]);} printf(“
“);} printf(“
“); for(k=0;k<=3;k++) {printf(“%f “,w[k][0]);} rintf(“
%f”,bl); printf(” “); for(k=0;k<=3;k++) {printf(“%f “,b[k]);} getch(); } getch(); } Output: Experiment 4: Use of ART algorithm to cluster vectors. Program: /* ART NETWORK TO CLUSTER FOUR VECTORS */ #include<stdio. h> #include<conio. h> main() {float n=4. 0,m=3. 0,o=0. 4,l=2. 0; float b[4][3],t[3][4],s[4],x[4],sin=0,y[3],xin=0; int i,j,k=0,J,c=0; y[0]=0,y[1]=0,y[2]=0; clrscr(); for(i=0;i<=3;i++) {for(j=0;j<=2;j++) {b[i][j]=0. 2;}} for(i=0;i<=2;i++) {for(j=0;j<=3;j++) {t[i][j]=1. 0;}} mew: printf(“Enter the input value:
“); for(i=0;i<=3;i++) {scanf(“%f”,&s[i]); x[i]=s[i]; in+=s[i];} for(i=0;i<=2;i++) {printf(“
Y”); do {y[i]+=s[k]*b[k][i]; k+=1;} while(k<=3); if(y[0]>=y[1]) {if(y[0]>=y[2]) J=0; else J=2;} else {if(y[1]>=y[2]) J=1; else J=2;} for(i=0;i<=3;i++) {x[i]=s[i]*t[J][i]; xin+=x[i];} if(xin/sin>=0. 4) {for(i=0;i<=3;i++) {b[i][J]=(2*x[i])/(1+xin); t[J][i]=x[i];}} else {y[J]=-1;} printf(“
“); for(i=0;i<=3;i++) {for(j=0;j<=2;j++) {printf(“%f “,b[i][j]);} printf(“
“);} for(i=0;i<=2;i++) {for(j=0;j<=3;j++) {printf(“%f “,t[i][j]);} printf(“
“);} getch(); y[0]=y[1]=y[2]=0; sin=xin=0; c+=1; k=0; if(c<=3) goto mew;} getch();} Output:

Experiment 5: Implement traveling salesman problem using genetic algorithm. Program: #include<stdio. h> #include<conio. h> int tsp[10][10]={{999,10,3,2,5,6,7,2,5,4}, {20,999,3,5,10,2,8,1,15,6}, {10,5,999,7,8,3,11,12,3,2}, {3,4,5,999,6,4,10,6,1,8}, {1,2,3,4,999,5,10,20,11,2}, {8,5,3,10,2,999,6,9,20,1}, {3,8,5,2,20,21,999,3,5,6}, {5,2,1,25,15,10,6,999,8,1}, {10,11,6,8,3,4,2,15,999,1}, {5,10,6,4,15,1,3,5,2,999}}; int pa[1000][10]= {{0,1,2,3,4,5,6,7,8,9}, {9,8,6,3,2,1,0,4,5,7}, {2,3,5,0,1,4,9,8,6,7}, {4,8,9,0,1,3,2,5,6,7}}; int i,j,k,l,m,y,loc,flag,row,col,it,x=3,y=3; int count,row=0,res[1][10],row1,col1,z; nt numoff=4; int offspring[1000][10]; int mincost=9999,mc; main() {int gen; clrscr(); printf(“Number of Generation : “); scanf(“%d”,&gen); offcal1(pa); offcal2(pa); printf(“
First Generation
“); for(i=0;i<count;i++) {for(j=0;j<10;j++) printf(“%d “,offspring[i][j]); printf(“
“);} for(y=1;y<=gen-1;y++) {getch(); clrscr(); for(i=0;i<count;i++) for(j=0;j<10;j++) pa[i][j]=offspring[i][j]; numoff=count; offcal1(pa); offcal2(pa); printf(“
%d Generation
“,y+1); for(i=0;i<count;i++) {for(j=0;j<10;j++) printf(“%d “,offspring[i][j]); printf(“
“);} getch(); clrscr();} rintf(“

Minimum Cost Path
“); for(z=0;z<10;z++) printf(“%d “,res[0][z]); printf(“
Minimum Cost %d
“,mincost);} /* finding the offspring using cyclic crossover */ offcal1(pa) int pa[1000][10]; {count=0; for(i=0;i<1000;i++) for(j=0;j<10;j++) offspring[i][j]=-1; for(k=0;k<numoff;k++) {for(l=k+1;l<numoff;l++) {offspring[row][0]=pa[k][0]; loc=pa[l][0]; flag=1; while(flag ! = 0) {for(j=0;j<10;j++) {if(pa[k][j] == loc ) { if (offspring[row][j]==-1) {offspring[row][j]=loc; loc=pa[l][j];} else flag=0;}} }/* end while*/ for(m=0;m<10;m++) {if(offspring[row][m] == -1) offspring[row][m]=pa[l][m];} or(z=0;z<10;z++) {if(z<9) {row1=offspring[row][z]; col1=offspring[row][z+1]; mc=mc+tsp[row1][col1];} else { row1=offspring[row][z]; col1=offspring[row][0]; mc=mc+tsp[row1][col1];}} if(mc < mincost) {for(z=0;z<10;z++) res[0][z]=offspring[row][z]; mincost=mc;} count++; row++; }/* end l*/}} offcal2(pa) int pa[1000][10]; {for(k=0;k<numoff;k++) { for(l=k+1;l<numoff;l++) { offspring[row][0]=pa[l][0]; loc=pa[k][0]; flag=1; while(flag ! = 0) {for(j=0;j<10;j++) {if(pa[l][j] == loc ) {if (offspring[row][j]==-1) {offspring[row][j]=loc; loc=pa[k][j];} else flag=0;}} }/* end while*/ for(m=0;m<10;m++) if(offspring[row][m] == -1) offspring[row][m]=pa[k][m];} for(z=0;z<10;z++) {if(z<9) {row1=offspring[row][z]; col1=offspring[row][z+1]; mc=mc+tsp[row1][col1];} else {row1=offspring[row][z]; col1=offspring[row][0]; mc=mc+tsp[row1][col1]; }} row++; if(mc < mincost) {for(z=0;z<10;z++) res[0][z]=offspring[row][z]; mincost=mc; } count++; }/* end l*/ } } Output: Experiment 6: Implement various laws associated with fuzzy sets. Program: #include<stdio. h> #include<alloc. h> #include<conio. h> #include<stdlib. h> struct SET {float nr[5]; float dr[5]; int n;}; typedef struct SET fuzzy; oid printval(fuzzy *m,char *x) {int i; printf(“
%s={“,x); for(i=0;i<m->n;i++) {printf(” %6. 2f / %6. 2f”,m->nr[i],m->dr[i]); if(i! =m->n-1) putch(‘+’);} printf(” }”);} fuzzy unionset(fuzzy a,fuzzy b) { fuzzy temp; char ch; int i; temp. n=a. n; for(i=0;i<a. n;i++) {if(a. dr[i]! =b. dr[i]) {printf(“
Denominators not equal”); getch(); exit(0);} if(a. nr[i]<b. nr[i]) temp. nr[i]=b. nr[i]; else temp. nr[i]=a. nr[i]; temp. dr[i]=a. dr[i];} return temp;} fuzzy intersect(fuzzy a,fuzzy b) {fuzzy temp; int i; temp. n=a. n; for(i=0;i<a. n;i++) {if(a. dr[i]! =b. dr[i]) {printf(“
Denominators not equal”); etch(); exit(0); } if(a. nr[i]>b. nr[i]) temp. nr[i]=b. nr[i]; else temp. nr[i]=a. nr[i]; temp. dr[i]=a. dr[i];} return temp;} fuzzy complement(fuzzy a) { fuzzy temp; int i; temp. n=a. n; for(i=0;i<a. n;i++) {temp. nr[i]=1-a. nr[i]; temp. dr[i]=a. dr[i];} return temp;} void main() {fuzzy a,b,ans; char ch; clrscr(); a. n=b. n=3; a. nr[0]=0. 1; a. dr[0]=1; a. nr[1]=0. 2; a. dr[1]=2; a. nr[2]=0. 3; a. dr[2]=3; b. nr[0]=0. 4; b. dr[0]=1; b. nr[1]=0. 3; b. dr[1]=2; b. nr[2]=0. 2; b. dr[2]=3; printval(&a,”A”); printval(&b,”B”); printf(“
Menu:
1. AUB
2. A^B
3. A~
4.

B~
5. Print S,A,B
6. Exit”); while(1) {switch((ch=getch())) {case ‘1’: ans=unionset(a,b); printval(&ans,”1. AUB”); break; case ‘2’: ans=intersect(a,b); printval(&ans,”2. A^B”); break; case ‘3’: ans=complement(a); printval(&ans,”3. A~”); break; case ‘4’: ans=complement(b); printval(&ans,”4. B~”); break; case ‘5’: printval(&a,”A”); printval(&b,”B”); break; case ‘6’: exit(0); } } } Output: Experiment 7: Implement fuzzy sets. Program: #include<stdio. h> #include<alloc. h> #include<conio. h> #include<stdlib. h> struct SET {float nr[5]; float dr[5]; int n;}; ypedef struct SET fuzzy; void getval(fuzzy *m,char *x) { int i; float f; clrscr(); printf(“
Enter the %s:
“,x); for(i=0;i<m->n;i++) {printf(” Numerator Element %d :”,i+1); scanf(“%f”,&f); m->nr[i]=f; fflush(stdin); printf(” Denominator Element %d:”,i+1); scanf(“%f”,&f); m->dr[i]=f;}} void printval(fuzzy *m,char *x) {int i; printf(“
%s={“,x); for(i=0;i<m->n;i++) {printf(” %6. 2f / %6. 2f”,m->nr[i],m->dr[i]); if(i! =m->n-1) putch(‘+’);} printf(” }”);} fuzzy unionset(fuzzy a,fuzzy b) {fuzzy temp; char ch; int i; temp. n=a. n; for(i=0;i<a. n;i++) {if(a. dr[i]! =b. dr[i]) printf(“
Denominators not equal”); getch(); exit(0);} if(a. nr[i]<b. nr[i]) temp. nr[i]=b. nr[i]; else temp. nr[i]=a. nr[i]; temp. dr[i]=a. dr[i];} return temp;} fuzzy intersect(fuzzy a,fuzzy b) { fuzzy temp; int i; temp. n=a. n; for(i=0;i<a. n;i++) { if(a. dr[i]! =b. dr[i]) { printf(“
Denominators not equal”); getch(); exit(0);} if(a. nr[i]>b. nr[i]) temp. nr[i]=b. nr[i]; else temp. nr[i]=a. nr[i]; temp. dr[i]=a. dr[i]; } return temp;} fuzzy complement(fuzzy a) {fuzzy temp; int i; temp. n=a. n; for(i=0;i<a. n;i++) { temp. nr[i]=1-a. nr[i]; temp. dr[i]=a. dr[i];} return temp;} void main() fuzzy a,b,ans; char ch; clrscr(); printf(“
Enter the no of componets:”); scanf(“%d”,&a. n); b. n=a. n; getval(&a,”A”); getval(&b,”B”); clrscr(); printval(&a,”A”); printval(&b,”B”); getch(); while(1) { clrscr(); printf(“
Menu:
1. AUB
2. A^B
3. A~
4. B~
5. Print S,A,B
6. Exit”); switch((ch=getch())) {case ‘1’: ans=unionset(a,b); printval(&ans,”AUB”); getch(); break; case ‘2’: ans=intersect(a,b); printval(&ans,”A^B”); getch(); break; case ‘3’: ans=complement(a); printval(&ans,”A~”); getch(); break; case ‘4’: ans=complement(b); printval(&ans,”B~”); getch(); break; ase ‘5’: printval(&a,”A”); printval(&b,”B”); getch(); break; case ‘6’: exit(0);}}} Output: Experiment 8: Implement word matching using GA. Program: #include<stdio. h> #include<conio. h> #include<stdlib. h> #include<dos. h> char input[15],parent[50][15],child[50][15],mating_pool[105][15],mutant[05][15]; int pfit[50],cfit[50],fit[105],mfit[05],gen=0; void get_input() {int i; clrscr(); printf(“

WORD MATCHING PROBLEM – GENETIC ALGORITHMS ASSIGNMENT”); printf(“
**********************************************************”); printf(“

ENTER THE WORD TO BE MATCHED : “); canf(“%s”,input); printf(“

THE ASCII EQUIVALENT OF THE LETTERS IN THE ENTERED WORD”); printf(“
————————————————————–“); printf(“

LETTERS :”); for(i=0;i<strlen(input);i++) {printf(” %c “,input[i]);} printf(“
ASCII :”); for(i=0;i<strlen(input);i++) {printf(” %3d”,input[i]);} getch();} void initial_pop() {int i,j; randomize(); for(i=0;i<50;i++) {for(j=0;j<strlen(input);j++) {parent[i][j]=random(26)+97; if(parent[i][j]==input[j]) {pfit[i]++;}}}} void display() {int i,j,nexti; clrscr(); printf(“

THE CHROMOSOMES OF PARENTS AND CHILDREN”); rintf(“
——————————————–
“); printf(“
PREVIOUS GENERATION CHILDREN CHROMOSOMES

“); for(i=0;i<50;i++) {if(((i)%4)==0) printf(“
“); for(j=0;j<strlen(input);j++) {printf(“%c”,child[i][j]);} printf(“% 2d “,cfit[i]);} printf(“
MUTANTS OF THIS GENERATION
“); for(i=0;i<05;i++) {if (i==3) printf(“
“); for(j=0;j<strlen(input);j++) {printf(“%c”,mutant[i][j]);} printf(“% 2d “,mfit[i]);} getch(); clrscr(); printf(“

THE CHROMOSOMES OF PARENTS AND CHILDREN”); printf(“
——————————————–
“); rintf(“
NEXT GENERATION PARENTS CHROMOSOMES

“); for(i=0;i<50;i++) {if(((i)%4)==0) printf(“
“); for(j=0;j<strlen(input);j++) {printf(“%c”,parent[i][j]);} printf(“% 2d “,pfit[i]);} getch();} void reproduction() //sorting_based_on_fitness() {char tempc; int temp; int i,j,k; for(i=0;i<50;i++) {for(j=0;j<strlen(input);j++) {mating_pool[i][j]=parent[i][j]; fit[i]=pfit[i];}} for(i=50;i<100;i++) {for(j=0;j<strlen(input);j++) {mating_pool[i][j]=child[i-50][j]; fit[i]=cfit[i-50];}} for(i=100;i<105;i++) {for(j=0;j<strlen(input);j++) {mating_pool[i][j]=mutant[i-100][j]; fit[i]=mfit[i-100];}} //sorting or(i=0;i<105;i++) {for(j=i+1;j<105;j++) {if(fit[i]<fit[j]) {for(k=0;k<strlen(input);k++) {tempc=mating_pool[i][k]; mating_pool[i][k]=mating_pool[j][k]; mating_pool[j][k]=tempc; temp=fit[i]; fit[i]=fit[j]; fit[j]=temp;}}}} for(i=0;i<50;i++) {for(j=0;j<strlen(input);j++) {parent[i][j]=mating_pool[i][j]; pfit[i]=fit[i];}} for(i=50;i<100;i++) {for(j=0;j<strlen(input);j++) {child[i-50][j]=mating_pool[i][j]; cfit[i-50]=fit[i];}}} void crossover() {int xover_pt; int i,j,k; for(i=0;i<50;i++) {xover_pt=random(strlen(input)); cfit[i]=0; cfit[i+1]=0; for(j=0;j<xover_pt;j++) {child[i][j]=parent[i][j]; f (input[j]==child[i][j]) cfit[i]++; child[i+1][j]=parent[i+1][j]; if(input[j]==child[i+1][j]) cfit[i+1]++;} for(j=xover_pt;j<strlen(input);j++) {child[i][j]=parent[i+1][j]; if(input[j]==child[i][j]) cfit[i]++; child[i+1][j]=parent[i][j]; if(input[j]==child[i+1][j]) cfit[i+1]++;} i++;}} void mutation() {int i,mut_pt,j; char mut_val; randomize(); for(i=0;i<05;i++) {mut_pt=random(strlen(input)); mut_val=random(26)+97; mfit[i]=0; for(j=0;j<mut_pt;j++) {mutant[i][j]=parent[1][j]; if (mutant[i][j]==input[j]) {mfit[i]++;}} mut_val=input[j]; mutant[i][mut_pt]=mut_val; if (mutant[i][j]==input[j]) {mfit[i]++;} or(j=mut_pt+1;j<strlen(input);j++) {mutant[i][j]=parent[1][j]; if (mutant[i][j]==input[j]) {mfit[i]++;}}}} void results() {int i; clrscr(); printf(“

WORD MATCHING PROBLEM – GENETIC ALGORITHM ASSIGNMENT”); printf(“
**********************************************************”); printf(“

THE MATCHING WORD FOR THE GIVEN INPUT WORD”); printf(“

OBTAINED USING GENETIC ALGORITHM”); printf(“

“); for(i=0;i<strlen(input);i++) {printf(“%c”,parent[0][i]);} printf(“
–“); for(i=0;i<strlen(input);i++) {printf(“-“);} printf(“–

USER INPUT : %s”,input); rintf(“

THE FITNESS OF THE GA GENERATED WORD AND THE USER’S INPUT”); printf(“

%2d/%d”,pfit[0],strlen(input)); printf(“

GENERATIONS COUNT : %d”,gen);} int input_choice() {int choice,i; clrscr(); printf(“

GENEREATION NUMBER : %d”,gen); printf(“
——————————“); printf(“

THE FITTEST INDIVIDUAL TILL THE PREVIOUS GENERATION

“); for(i=0;i<strlen(input);i++) {printf(“%c”,parent[0][i]);} printf(” / “); for(i=0;i<strlen(input);i++) {printf(“%c”,input[i]);} printf(“

WITH A FITNESS OF %d/%d”,pfit[0],strlen(input)); rintf(“

ENTER YOUR CHOICE (TO CONTINUE 1 TO EXIT 0) : “); scanf(“%d”,&choice); return choice;} void main() {int i,choice; clrscr(); get_input(); initial_pop(); //display(); reproduction();//sorting_based_on_fitness(); display(); printf(“
ENTER YOUR CHOICE (TO CONTINUE 1 TO EXIT 0) : “); scanf(“%d”,&choice); while((choice==1)&&(pfit[0]! =strlen(input))) {crossover(); gen++; mutation(); reproduction();//sorting_based_on_fitness(); display(); choice=input_choice();} sound(1000);delay(200); nosound();delay(200);results(); getch(); sound(1000);delay(200);nosound();} Output:

FlyLab will allow you to play the role of a research geneticist. You will use FlyLab to study important introductory principles of genetics by developing hypotheses and designing and conducting matings between fruit flies with different mutations that you have selected. Once you have examined the results of a simulated cross, you can perform a statistical test of your data by chi-square analysis and apply these statistics to accept or reject your hypothesis for the predicted phenotypic ratio of offspring for each cross.With FlyLab, it is possible to study multiple generations of offspring, and perform testcrosses and backcrosses. FlyLab is a very versatile program; it can be used to learn elementary genetic principles such as dominance, recessiveness, and Mendelian ratios, or more complex concepts such as sex-linkage, epistasis, recombination, and genetic mapping.

Objectives The purpose of this laboratory is to: Simulate basic principles of genetic inheritance based on Mendelian genetics by designing and performing crosses between fruit flies.Help you understand the relationship between an organism’s genotype and its phenotype. Demonstrate the importance of statistical analysis to accept or reject a hypothesis. Use genetic crosses and recombination data to identify the location of genes on a chromosome by genetic mapping. Before You Begin: Prerequisites Before beginning FlyLab you should be familiar with the following concepts: Chromosome structure, and the stages of gamete formation by meiosis.The basic terminology and principles of Mendelian genetics, including complete and incomplete dominance, epistasis, lethal mutations, recombination, autosomal recessive inheritance, autosomal dominant inheritance, and sex-linked inheritance. Predicting the results of monohybrid and dihybrid crosses by constructing a Punnett square.

How genetic mutations produce changes in phenotype, and beneficial and detrimental results of mutations in a population.