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1. Apa yang dimaksud Adversarial Search & Constraint Satisfaction Problems? berikan contoh

Pengertian Adversarial Search
Adversarial Search bekerja dengan cara menghitung langkah terbaik dalam dua pertandingan pemain di mana semua informasi tersedia.  Secara teoritis, algoritma pencarian ini didasarkan teorema minimax von Neumann yang menyatakan bahwa dalam jenis permainan selalu ada serangkaian strategi yang mengarah ke kedua pemain untuk mendapatkan nilai yang sama.
CONTOH: PERMAINAN CATUR, TIC TAC TOE

Pengertian & contoh Constraint Satisfaction Problems
CSP atau Constraint Satisfaction Problem adalah permasalahan  yang tujuannya adalah mendapatkan suatu kombinasi variabel-variabel tertentu yang memenuhi aturan-aturan (constraints) tertentu.
CONTOH: MAP-COLORING

2. Apa itu Propositional Logic? Berikan contoh?

Pengertian
-Propositional logic merupakan salah satu bentuk (bahasa) representasi logika yang paling tua dan paling sederhana.

-Dengan cara ini beberapa fakta dapat digambarkan dan dimanipulasi dengan menggunakan aturan-aturan aljabar Boolean.

-Propositional logic membentuk statement sederhana atau statement yang kompleks dengan menggunakan propositional connective, dimana mekanisme ini menentukan kebenaran dari sebuah statement kompleks dari nilai kebenaran yang direpresentasikan oleh statement lain yang lebih sederhana.

-Contoh operator logika pada Propositional Logic:

Tabel Kebenaran:

Contoh Propositional Logic

Jika Anto adalah mahasiswa yang baik maka ia pasti tidak menyontek di ujian
Anto menyontek dalam ujian
___________________________________________________________
Anto bukan mahasiswa yang baik

Dilambangkan:

3. Buat coding (Boleh C, C++ atau Java) untuk  Algoritma A & Algoritma A* (A Star)?

Contoh menggunakan C++ (jujur pak belom ngerti, ini saya dapet hasil googling)

#include <iostream>
#include <iomanip>
#include <queue>
#include <string>
#include <math.h>
#include <ctime>
using namespace std;

const int n=60; // horizontal size of the map
const int m=60; // vertical size size of the map
static int map[n][m];
static int closed_nodes_map[n][m]; // map of closed (tried-out) nodes
static int open_nodes_map[n][m]; // map of open (not-yet-tried) nodes
static int dir_map[n][m]; // map of directions
const int dir=8; // number of possible directions to go at any position
// if dir==4
//static int dx[dir]={1, 0, -1, 0};
//static int dy[dir]={0, 1, 0, -1};
// if dir==8
static int dx[dir]={1, 1, 0, -1, -1, -1, 0, 1};
static int dy[dir]={0, 1, 1, 1, 0, -1, -1, -1};

class node
{
// current position
int xPos;
int yPos;
// total distance already travelled to reach the node
int level;
// priority=level+remaining distance estimate
int priority; // smaller: higher priority

public:
node(int xp, int yp, int d, int p)
{xPos=xp; yPos=yp; level=d; priority=p;}

int getxPos() const {return xPos;}
int getyPos() const {return yPos;}
int getLevel() const {return level;}
int getPriority() const {return priority;}

void updatePriority(const int & xDest, const int & yDest)
{
priority=level+estimate(xDest, yDest)*10; //A*
}

// give better priority to going strait instead of diagonally
void nextLevel(const int & i) // i: direction
{
level+=(dir==8?(i%2==0?10:14):10);
}

// Estimation function for the remaining distance to the goal.
const int & estimate(const int & xDest, const int & yDest) const
{
static int xd, yd, d;
xd=xDest-xPos;
yd=yDest-yPos;

// Euclidian Distance
d=static_cast<int>(sqrt(xd*xd+yd*yd));

// Manhattan distance
//d=abs(xd)+abs(yd);

// Chebyshev distance
//d=max(abs(xd), abs(yd));

return(d);
}
};

// Determine priority (in the priority queue)
bool operator<(const node & a, const node & b)
{
return a.getPriority() > b.getPriority();
}

// A-star algorithm.
// The route returned is a string of direction digits.
string pathFind( const int & xStart, const int & yStart,
const int & xFinish, const int & yFinish )
{
static priority_queue<node> pq[2]; // list of open (not-yet-tried) nodes
static int pqi; // pq index
static node* n0;
static node* m0;
static int i, j, x, y, xdx, ydy;
static char c;
pqi=0;

// reset the node maps
for(y=0;y<m;y++)
{
for(x=0;x<n;x++)
{
closed_nodes_map[x][y]=0;
open_nodes_map[x][y]=0;
}
}

// create the start node and push into list of open nodes
n0=new node(xStart, yStart, 0, 0);
n0->updatePriority(xFinish, yFinish);
pq[pqi].push(*n0);
open_nodes_map[x][y]=n0->getPriority(); // mark it on the open nodes map

// A* search
while(!pq[pqi].empty())
{
// get the current node w/ the highest priority
// from the list of open nodes
n0=new node( pq[pqi].top().getxPos(), pq[pqi].top().getyPos(),
pq[pqi].top().getLevel(), pq[pqi].top().getPriority());

x=n0->getxPos(); y=n0->getyPos();

pq[pqi].pop(); // remove the node from the open list
open_nodes_map[x][y]=0;
// mark it on the closed nodes map
closed_nodes_map[x][y]=1;

// quit searching when the goal state is reached
//if((*n0).estimate(xFinish, yFinish) == 0)
if(x==xFinish && y==yFinish)
{
// generate the path from finish to start
// by following the directions
string path=””;
while(!(x==xStart && y==yStart))
{
j=dir_map[x][y];
c=’0’+(j+dir/2)%dir;
path=c+path;
x+=dx[j];
y+=dy[j];
}

// garbage collection
delete n0;
// empty the leftover nodes
while(!pq[pqi].empty()) pq[pqi].pop();
return path;
}

// generate moves (child nodes) in all possible directions
for(i=0;i<dir;i++)
{
xdx=x+dx[i]; ydy=y+dy[i];

if(!(xdx<0 || xdx>n-1 || ydy<0 || ydy>m-1 || map[xdx][ydy]==1
|| closed_nodes_map[xdx][ydy]==1))
{
// generate a child node
m0=new node( xdx, ydy, n0->getLevel(),
n0->getPriority());
m0->nextLevel(i);
m0->updatePriority(xFinish, yFinish);

// if it is not in the open list then add into that
if(open_nodes_map[xdx][ydy]==0)
{
open_nodes_map[xdx][ydy]=m0->getPriority();
pq[pqi].push(*m0);
// mark its parent node direction
dir_map[xdx][ydy]=(i+dir/2)%dir;
}
else if(open_nodes_map[xdx][ydy]>m0->getPriority())
{
// update the priority info
open_nodes_map[xdx][ydy]=m0->getPriority();
// update the parent direction info
dir_map[xdx][ydy]=(i+dir/2)%dir;

// replace the node
// by emptying one pq to the other one
// except the node to be replaced will be ignored
// and the new node will be pushed in instead
while(!(pq[pqi].top().getxPos()==xdx &&
pq[pqi].top().getyPos()==ydy))
{
pq[1-pqi].push(pq[pqi].top());
pq[pqi].pop();
}
pq[pqi].pop(); // remove the wanted node

// empty the larger size pq to the smaller one
if(pq[pqi].size()>pq[1-pqi].size()) pqi=1-pqi;
while(!pq[pqi].empty())
{
pq[1-pqi].push(pq[pqi].top());
pq[pqi].pop();
}
pqi=1-pqi;
pq[pqi].push(*m0); // add the better node instead
}
else delete m0; // garbage collection
}
}
delete n0; // garbage collection
}
return “”; // no route found
}

int main()
{
srand(time(NULL));

// create empty map
for(int y=0;y<m;y++)
{
for(int x=0;x<n;x++) map[x][y]=0;
}

// fillout the map matrix with a ‘+’ pattern
for(int x=n/8;x<n*7/8;x++)
{
map[x][m/2]=1;
}
for(int y=m/8;y<m*7/8;y++)
{
map[n/2][y]=1;
}

// randomly select start and finish locations
int xA, yA, xB, yB;
switch(rand()%8)
{
case 0: xA=0;yA=0;xB=n-1;yB=m-1; break;
case 1: xA=0;yA=m-1;xB=n-1;yB=0; break;
case 2: xA=n/2-1;yA=m/2-1;xB=n/2+1;yB=m/2+1; break;
case 3: xA=n/2-1;yA=m/2+1;xB=n/2+1;yB=m/2-1; break;
case 4: xA=n/2-1;yA=0;xB=n/2+1;yB=m-1; break;
case 5: xA=n/2+1;yA=m-1;xB=n/2-1;yB=0; break;
case 6: xA=0;yA=m/2-1;xB=n-1;yB=m/2+1; break;
case 7: xA=n-1;yA=m/2+1;xB=0;yB=m/2-1; break;
}

cout<<“Map Size (X,Y): “<<n<<“,”<<m<<endl;
cout<<“Start: “<<xA<<“,”<<yA<<endl;
cout<<“Finish: “<<xB<<“,”<<yB<<endl;
// get the route
clock_t start = clock();
string route=pathFind(xA, yA, xB, yB);
if(route==””) cout<<“An empty route generated!”<<endl;
clock_t end = clock();
double time_elapsed = double(end – start);
cout<<“Time to calculate the route (ms): “<<time_elapsed<<endl;
cout<<“Route:”<<endl;
cout<<route<<endl<<endl;

// follow the route on the map and display it
if(route.length()>0)
{
int j; char c;
int x=xA;
int y=yA;
map[x][y]=2;
for(int i=0;i<route.length();i++)
{
c =route.at(i);
j=atoi(&c);
x=x+dx[j];
y=y+dy[j];
map[x][y]=3;
}
map[x][y]=4;

// display the map with the route
for(int y=0;y<m;y++)
{
for(int x=0;x<n;x++)
if(map[x][y]==0)
cout<<“.”;
else if(map[x][y]==1)
cout<<“O”; //obstacle
else if(map[x][y]==2)
cout<<“S”; //start
else if(map[x][y]==3)
cout<<“R”; //route
else if(map[x][y]==4)
cout<<“F”; //finish
cout<<endl;
}
}
getchar(); // wait for a (Enter) keypress
return(0);
}

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