193  Graph Coloring
Moderator: Board moderators
problem no 193
what is the output of the following input
1
3 2
1 2
1 3
1
3 2
1 2
1 3
193 something is strange
My algorithem is ...
Form the minimum degree node to the maxmum degree node,
if its all neighbors are not black,it is going to be colored.
Is it wrong ...??...
I can got correct results in many cases,but I got WA,WA,and WA...
Form the minimum degree node to the maxmum degree node,
if its all neighbors are not black,it is going to be colored.
Is it wrong ...??...
I can got correct results in many cases,but I got WA,WA,and WA...
Sorry,I can't proof.....
This method broke into my heart occasionally.
I just don't know why it's wrong...^^....
Ex.
1
6 8
1 2
1 3
2 4
2 5
3 4
3 6
4 6
5 6
Step 1. Construct all neighbors for each node
node neighbors degree can_be_color
1 2 3 2 true
2 1 4 5 3 true
3 1 4 6 3 true
4 2 3 6 3 true
5 2 6 2 true
6 3 4 5 3 true
Step 2. Start from the node which has lowest degree.
The lowest degree in this case is "2"
So,we search from node 1 to node 6 to find the node which can be
colored and has degree "2"....we get node "1" , so all neighbors of
node 1 can not be colored...
node neighbors degree can_be_color
1 2 3 2 "colored"
2 1 4 5 3 "false forever"
3 1 4 6 3 "false forever"
4 2 3 6 3 true
5 2 6 2 true
6 3 4 5 3 true
Then we get node 5 which can be colored and has degree "2",so all
neighbors of node 5 can not be colored....
node neighbors degree can_be_color
1 2 3 2 "colored 1 "
2 1 4 5 3 "false forever"
3 1 4 6 3 "false forever"
4 2 3 6 3 true
5 2 6 2 "colored 2 "
6 3 4 5 3 "false forever"
Step 3. Now we have finished all node having degree 2,so we go to
degree 3..
We find node 4 can be colored and has degree "3",so we select it and
find that all node have been considered..............
node neighbors degree can_be_color
1 2 3 2 "colored 1 "
2 1 4 5 3 "false forever"
3 1 4 6 3 "false forever"
4 2 3 6 3 "colored 3 "
5 2 6 2 "colored 2 "
6 3 4 5 3 "false forever"
So the answer is
3
1 4 5
This method broke into my heart occasionally.
I just don't know why it's wrong...^^....
Ex.
1
6 8
1 2
1 3
2 4
2 5
3 4
3 6
4 6
5 6
Step 1. Construct all neighbors for each node
node neighbors degree can_be_color
1 2 3 2 true
2 1 4 5 3 true
3 1 4 6 3 true
4 2 3 6 3 true
5 2 6 2 true
6 3 4 5 3 true
Step 2. Start from the node which has lowest degree.
The lowest degree in this case is "2"
So,we search from node 1 to node 6 to find the node which can be
colored and has degree "2"....we get node "1" , so all neighbors of
node 1 can not be colored...
node neighbors degree can_be_color
1 2 3 2 "colored"
2 1 4 5 3 "false forever"
3 1 4 6 3 "false forever"
4 2 3 6 3 true
5 2 6 2 true
6 3 4 5 3 true
Then we get node 5 which can be colored and has degree "2",so all
neighbors of node 5 can not be colored....
node neighbors degree can_be_color
1 2 3 2 "colored 1 "
2 1 4 5 3 "false forever"
3 1 4 6 3 "false forever"
4 2 3 6 3 true
5 2 6 2 "colored 2 "
6 3 4 5 3 "false forever"
Step 3. Now we have finished all node having degree 2,so we go to
degree 3..
We find node 4 can be colored and has degree "3",so we select it and
find that all node have been considered..............
node neighbors degree can_be_color
1 2 3 2 "colored 1 "
2 1 4 5 3 "false forever"
3 1 4 6 3 "false forever"
4 2 3 6 3 "colored 3 "
5 2 6 2 "colored 2 "
6 3 4 5 3 "false forever"
So the answer is
3
1 4 5
Sorry,I can't proof.....
This method broke into my heart occasionally.
I just don't know why it's wrong...^^....
Ex.
1
6 8
1 2
1 3
2 4
2 5
3 4
3 6
4 6
5 6
Step 1. Construct all neighbors for each node
node neighbors degree can_be_color
1 2 3 2 true
2 1 4 5 3 true
3 1 4 6 3 true
4 2 3 6 3 true
5 2 6 2 true
6 3 4 5 3 true
Step 2. Start from the node which has lowest degree.
The lowest degree in this case is "2"
So,we search from node 1 to node 6 to find the node which can be
colored and has degree "2"....we get node "1" , so all neighbors of
node 1 can not be colored...
node neighbors degree can_be_color
1 2 3 2 "colored"
2 1 4 5 3 "false forever"
3 1 4 6 3 "false forever"
4 2 3 6 3 true
5 2 6 2 true
6 3 4 5 3 true
Then we get node 5 which can be colored and has degree "2",so all
neighbors of node 5 can not be colored....
node neighbors degree can_be_color
1 2 3 2 "colored 1 "
2 1 4 5 3 "false forever"
3 1 4 6 3 "false forever"
4 2 3 6 3 true
5 2 6 2 "colored 2 "
6 3 4 5 3 "false forever"
Step 3. Now we have finished all node having degree 2,so we go to
degree 3..
We find node 4 can be colored and has degree "3",so we select it and
find that all node have been considered..............
node neighbors degree can_be_color
1 2 3 2 "colored 1 "
2 1 4 5 3 "false forever"
3 1 4 6 3 "false forever"
4 2 3 6 3 "colored 3 "
5 2 6 2 "colored 2 "
6 3 4 5 3 "false forever"
So the answer is
3
1 4 5
This method broke into my heart occasionally.
I just don't know why it's wrong...^^....
Ex.
1
6 8
1 2
1 3
2 4
2 5
3 4
3 6
4 6
5 6
Step 1. Construct all neighbors for each node
node neighbors degree can_be_color
1 2 3 2 true
2 1 4 5 3 true
3 1 4 6 3 true
4 2 3 6 3 true
5 2 6 2 true
6 3 4 5 3 true
Step 2. Start from the node which has lowest degree.
The lowest degree in this case is "2"
So,we search from node 1 to node 6 to find the node which can be
colored and has degree "2"....we get node "1" , so all neighbors of
node 1 can not be colored...
node neighbors degree can_be_color
1 2 3 2 "colored"
2 1 4 5 3 "false forever"
3 1 4 6 3 "false forever"
4 2 3 6 3 true
5 2 6 2 true
6 3 4 5 3 true
Then we get node 5 which can be colored and has degree "2",so all
neighbors of node 5 can not be colored....
node neighbors degree can_be_color
1 2 3 2 "colored 1 "
2 1 4 5 3 "false forever"
3 1 4 6 3 "false forever"
4 2 3 6 3 true
5 2 6 2 "colored 2 "
6 3 4 5 3 "false forever"
Step 3. Now we have finished all node having degree 2,so we go to
degree 3..
We find node 4 can be colored and has degree "3",so we select it and
find that all node have been considered..............
node neighbors degree can_be_color
1 2 3 2 "colored 1 "
2 1 4 5 3 "false forever"
3 1 4 6 3 "false forever"
4 2 3 6 3 "colored 3 "
5 2 6 2 "colored 2 "
6 3 4 5 3 "false forever"
So the answer is
3
1 4 5
193 
/*@BEGIN_OF_SOURCE_CODE*/
#include<stdio.h>
#include<stdlib.h>
#define BLACK 2
#define WHITE 1
int vertex[102][102];
int flag[102],x[102],max,vertx;
void Coloring(int i,int count)
{
int j;
if(i==vertx){
if(max<count){
max =count;
for(j=0;j<vertx;j++)
if(flag[j]==BLACK) x[j] =1;
else x[j] = 0;
}
return;
}
for(j=0;j<vertx;j++)
if(vertex[j] == 1 && flag[j] == BLACK) break;
if(j==vertx){
flag =BLACK;
Coloring(i+1,count+1);
flag = 0;
}
else{
flag = WHITE;
Coloring(i+1,count);
flag = 0;
}
}
int main()
{
int edge,n,i,j,a,b,count;
scanf("%d",&n);
while(n>0){
scanf("%d%d",&vertx,&edge);
max = 0;
for(i=0;i<vertx;i++) {
flag = 0,x = 0;
for(j = 0;j<vertx;j++)
vertex[j] = 0;
}
for(i=0;i<edge;i++){
scanf("%d%d",&a,&b);
vertex[a1][b1]=vertex[b1][a1]=1;
}
for(i =0 ;i<vertx;i++)
Coloring(i,0);
printf("%d\n",max);
for(i=0;i<vertx;i++)
if(x == 1) printf("%d ",i+1);
printf("\n");
}
return 0;
}
/*@END_OF_SOURCE_CODE*/
*******
Is the algorithm right?
Can anyone help me?
#include<stdio.h>
#include<stdlib.h>
#define BLACK 2
#define WHITE 1
int vertex[102][102];
int flag[102],x[102],max,vertx;
void Coloring(int i,int count)
{
int j;
if(i==vertx){
if(max<count){
max =count;
for(j=0;j<vertx;j++)
if(flag[j]==BLACK) x[j] =1;
else x[j] = 0;
}
return;
}
for(j=0;j<vertx;j++)
if(vertex[j] == 1 && flag[j] == BLACK) break;
if(j==vertx){
flag =BLACK;
Coloring(i+1,count+1);
flag = 0;
}
else{
flag = WHITE;
Coloring(i+1,count);
flag = 0;
}
}
int main()
{
int edge,n,i,j,a,b,count;
scanf("%d",&n);
while(n>0){
scanf("%d%d",&vertx,&edge);
max = 0;
for(i=0;i<vertx;i++) {
flag = 0,x = 0;
for(j = 0;j<vertx;j++)
vertex[j] = 0;
}
for(i=0;i<edge;i++){
scanf("%d%d",&a,&b);
vertex[a1][b1]=vertex[b1][a1]=1;
}
for(i =0 ;i<vertx;i++)
Coloring(i,0);
printf("%d\n",max);
for(i=0;i<vertx;i++)
if(x == 1) printf("%d ",i+1);
printf("\n");
}
return 0;
}
/*@END_OF_SOURCE_CODE*/
*******
Is the algorithm right?
Can anyone help me?

 Learning poster
 Posts: 63
 Joined: Thu Apr 04, 2002 2:00 am
193  Graph Coloring
Hello,
I tried to solve 193 (and got AC in 0.01s), but I'm using some kind of probabilistic approach: I randomly select a node, remove it and its neighbours until there are no nodes left. I repeat this several times (100 is enough) to get the maximum node count.
I know that this problem (Maximum Independent Set) is NPcomplete, but is there another (maybe a bit more deterministic) solution?
I tried to solve 193 (and got AC in 0.01s), but I'm using some kind of probabilistic approach: I randomly select a node, remove it and its neighbours until there are no nodes left. I repeat this several times (100 is enough) to get the maximum node count.
I know that this problem (Maximum Independent Set) is NPcomplete, but is there another (maybe a bit more deterministic) solution?

 New poster
 Posts: 3
 Joined: Wed May 01, 2002 2:05 am
problem 193
i used the following heuristic to solve the problem:
find the node with minimum degree put it in the solution vector and delete it and all the nodes connected to it and so on till u finish all the nodes.
it generates the right answer for the test cases so can any body points out what is wrong with this algorithm and give me some difficult test cases, thanks.
find the node with minimum degree put it in the solution vector and delete it and all the nodes connected to it and so on till u finish all the nodes.
it generates the right answer for the test cases so can any body points out what is wrong with this algorithm and give me some difficult test cases, thanks.
Code: Select all
[cpp]
#include<iostream>
#include<fstream>
using namespace std;
int findmin();
void adjust(int x);
int a[101]={1};
int b[101][101]={0};
int n;
int main()
{
//ifstream ins;
//ins.open("in.txt");
//cin=ins;
int m,k;
int temp1,temp2;
int f,t;
int count;
cin>>m;
for(t=1;t<=m;t++)
{
count=0;
//initialize a,b
cin>>n>>k;
for(f=1; f<=n;f++)
{
//a[f]=1;
for(int c=1;c<=n;c++)
b[f][c]=0;
}
for(f=1;f<=k;f++)
{
cin>>temp1>>temp2;
if(temp1 !=temp2)
{
b[temp1][temp2]=1;
b[temp2][temp1]=1;
}
}
for(f=1;f<=n;f++)
{
int y=findmin();
if(y>n)
break;
a[f]=y;
count++;
for(int t=1;t<=n;t++)
{
if(b[y][t]==1)
adjust(t);
}
adjust(y);
}
cout<<count<<endl;
for(f=1;f<=count;f++)
{if(f==count)
cout<<a[f];
else
cout<<a[f]<<" ";
}
cout<<endl;
}
return 0;
}
int findmin()
{
int index=1000;
int count;
int mincount=1000;
int i,j;
for(i=1;i<=n;i++)
{
count=0;
for(j=1;j<=n;j++)
{
if(b[i][j]==1)
break;
if(b[i][j])
count++;
}
if(j>n && count<mincount)
{
mincount=count;
index=i;
}
}
return index;
}
void adjust(int x)
{
for(int i=1;i<=n;i++)
{
if(b[i][x]==1)
b[i][x]=0;
b[x][i]=1;
}
}
[/cpp]
mizo
so far as i know this algorithm will generate a maximal independent set, not necessarily the one that has maximum size.
maximal set: no other maximal set contains it
maximum set: the set whose cardinality is maximum
maximal set: no other maximal set contains it
maximum set: the set whose cardinality is maximum
K M Hasan
http://www.cs.umanitoba.ca/~kmhasan/
http://www.cs.umanitoba.ca/~kmhasan/
193 solution???
Hi,
Can anyone provide their solution to the graph coloring problem and the source code(c,c++ or java). I'm working on a problem where I need this solution.
Thanks.
Can anyone provide their solution to the graph coloring problem and the source code(c,c++ or java). I'm working on a problem where I need this solution.
Thanks.

 Experienced poster
 Posts: 147
 Joined: Fri Jun 13, 2003 10:46 pm

 Learning poster
 Posts: 94
 Joined: Sat Oct 05, 2002 5:34 pm
 Location: CS  AIUB, Dhaka, Bangladesh.
 Contact:
193 Graph Coloring
My program got WA in 0.039 sec.
Can anyone tell me some inputs and outputs?
Can anyone tell me some inputs and outputs?
Sajid Online: www.sajidonline.com

 Learning poster
 Posts: 94
 Joined: Sat Oct 05, 2002 5:34 pm
 Location: CS  AIUB, Dhaka, Bangladesh.
 Contact:
hmmm
After taking the Graph, in a loop, I took the graph in another adjcency matrix and call the dfs, which call with the root (the root is 1 to n, end of the loop).
The dfs visits every unvisited node, when it visits a node it colors white to its neighbours.
Finally, took the maximum number of Black nodes from the dfs(1>node)
is it, ok?
Code: Select all
for(i=1;i<=node;i++)
{
temp=graph
dfs(i)
}
Finally, took the maximum number of Black nodes from the dfs(1>node)
is it, ok?
Sajid Online: www.sajidonline.com