## 104 - Arbitrage

**Moderator:** Board moderators

### 104 - WA

Need Help!!!

I've summited it lots of times, but still WA

My algorithm is simple, I save in a table (table

I've summited it lots of times, but still WA

My algorithm is simple, I save in a table (table

*[j][k]) the best way to reach "j", starting from "i" with "k" iterations.*

What's wrong?? Is it precission??

My code

[cpp]#include <iostream>

#include <vector>

#define EPS 0.0000001

using namespace std;

int main()

{

int n;

while(cin >> n){

double best[n][n][n+1];

int path[n][n][n];

double table[n][n];

int i,j,k,z;

for(i=0;i<n;i++){

tableWhat's wrong?? Is it precission??

My code

[cpp]#include <iostream>

#include <vector>

#define EPS 0.0000001

using namespace std;

int main()

{

int n;

while(cin >> n){

double best[n][n][n+1];

int path[n][n][n];

double table[n][n];

int i,j,k,z;

for(i=0;i<n;i++){

table

*=1;*

for(j=0;j<n;j++)

if(i!=j) cin >> tablefor(j=0;j<n;j++)

if(i!=j) cin >> table

*[j];*

}

for(i=0;i<n;i++)

for(j=0;j<n;j++)

for(k=0;k<n+1;k++)

best}

for(i=0;i<n;i++)

for(j=0;j<n;j++)

for(k=0;k<n+1;k++)

best

*[j][k]=0;*

for(i=0;i<n;i++)

bestfor(i=0;i<n;i++)

best

*[0]=1;*

int min=40;

int inimin;

for(i=0;i<n;i++){

bool fin=false;

for(j=1;(j<n+1)&&(j<min);j++){

for(k=0;(k<n)&&(!fin);k++){

for(z=0;(z<n)&&(!fin);z++){

if(bestint min=40;

int inimin;

for(i=0;i<n;i++){

bool fin=false;

for(j=1;(j<n+1)&&(j<min);j++){

for(k=0;(k<n)&&(!fin);k++){

for(z=0;(z<n)&&(!fin);z++){

if(best

*[k][j]<best**[z][j-1]*table[z][k]){*

bestbest

*[k][j]=best[i][z][j-1]*table[z][k];*

path[i][k][j-1]=z;

if((i==k)&&(best[i][k][j]-1.01>EPS)){

min=j-1;

inimin=i;

fin=true;

}

}

}

}

}

}

if(min==40) cout << "no arbitrage sequence exists" << endl;

else{

int list[min+1];

int a=inimin;

for(i=min;0<=i;i--){

list[i]=path[inimin][a][i];

a=path[inimin][a][i];

}

for(i=0;i<=min;i++)

cout << list[i]+1<< " ";

cout << inimin+1<<endl;

}

}

return 0;

}[/cpp]

Thanks in advancepath[i][k][j-1]=z;

if((i==k)&&(best[i][k][j]-1.01>EPS)){

min=j-1;

inimin=i;

fin=true;

}

}

}

}

}

}

if(min==40) cout << "no arbitrage sequence exists" << endl;

else{

int list[min+1];

int a=inimin;

for(i=min;0<=i;i--){

list[i]=path[inimin][a][i];

a=path[inimin][a][i];

}

for(i=0;i<=min;i++)

cout << list[i]+1<< " ";

cout << inimin+1<<endl;

}

}

return 0;

}[/cpp]

Thanks in advance

### hints for 104

I've been trying to solve this problem for sooo long and finally got AC so today is a memorable day for me

So here are some clues for those who haven't solved it yet.

1) There are some posts talking about floating point errors, but I didn't check for any in my solution. I use a simple test
However, I don't use subtractions and they may create errors.

2) My solution is similar to a Floyd-Warshall (which runs in O(n^3)) but I use an additional outer loop, with the number of exchanges involved, so it's O(n^4) and was within the time limit. I use a 3d table (the new dimension is the number of steps); the rest is pretty much like F-W.

If you grow desperate as I did I can further explain how I solved it.

Good luck!

So here are some clues for those who haven't solved it yet.

1) There are some posts talking about floating point errors, but I didn't check for any in my solution. I use a simple test

Code: Select all

`if (profit > 1.01) ...`

2) My solution is similar to a Floyd-Warshall (which runs in O(n^3)) but I use an additional outer loop, with the number of exchanges involved, so it's O(n^4) and was within the time limit. I use a 3d table (the new dimension is the number of steps); the rest is pretty much like F-W.

If you grow desperate as I did I can further explain how I solved it.

Good luck!

Now, you can't just try with brute force (trying all combinations) because it'll be too slow and you'll get Time Limit Exceeded. However, there's a well known algorithm, Floyd-Warshall, which will find all the shortest paths between every node to the others in just O(n^3) time. You can find more info about F-W in the net. In my previous post I said how you have to change the general F-W algorithm to work for this particular problem...

As for floating point errors, most numbers representation isn't totally acurate; for instance, 0.1 is usually stored as 0.10000000000000001. After some operations, the error may influence the final result. Again, search the net for floating point errors. However, you won't have to deal with these errors to solve this problems (at least I didn't).

Hope this helps

Simple F-W goes like this:

Code: Select all

```
// initialization
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
best[i][j] = path[i][j];
path[i][j] = i;
}
}
for (k = 0; k < n; k++) {
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (best[i][k] + best[k][j] < best[i][j]) {
best[i][j] = best[i][k][ + best[k][j];
path[i][j] = k;
}
}
}
}
```

*[j] and path*

*[j] I have best**[j][s] and path**[j][s], which means "best path from i to j in s steps". I also have an outer loop, representing the number of steps.*

Code: Select all

```
// initialization
best[i][j][s] = 0, for all i,j,s
best[i][j][1] = input for the program
best[i][i][1] = 1, for all i
path[i][j][1] = i, for all i, j
for (steps = 2; steps <= n; steps++)
for k...
for i..
for j..
tmp = best[i][k][steps-1] * best[k][j][1]
if (tmp > table[i][j][steps]) {
table[i][j][steps] = tmp;
path[i][j][steps] = k;
}
```

After this, you scan though best

*[s] (best way to go from i to i again, in s steps). Remember that you want the minimum number of steps that yields a profit; so it's this:*

Code: Select all

```
for (steps = 2; steps <= n; steps++)
for (i = 0; i < n; i++)
if (best[i][i][steps] > 1.01) {
// score!
}
```

*[steps], which is the vertex just before the last (i).*

For instance, if i is 1 and steps is 4. Check path[1][1][4]. Suppose it is 2. So the path ends with "... 2 1". Now check path[1][2][3]. Supposing it's 4, the path ends with "... 4 2 1". Now check path[1][4][2]. If it's 5, the path is "... 5 4 2 1". Finally check path[1][5][1], which will be obviously 1. So the complete path is "1 5 4 2 1" (exactly 4 steps).

Just remember that pathFor instance, if i is 1 and steps is 4. Check path[1][1][4]. Suppose it is 2. So the path ends with "... 2 1". Now check path[1][2][3]. Supposing it's 4, the path ends with "... 4 2 1". Now check path[1][4][2]. If it's 5, the path is "... 5 4 2 1". Finally check path[1][5][1], which will be obviously 1. So the complete path is "1 5 4 2 1" (exactly 4 steps).

Just remember that path

*[j][s] is the vertex which is before the last one in a path from i to j in s steps. If s is 1, the path is simply "i j". In this problem you have to start and finish in the same currency; that's why we only search in best**[steps].*

Hope this helps, I almost wrote the complete program!

You could also stop the algorithm as soon as you found a solution. In other posts, some people talked about a O(n^3) solution but I can't figure it out... if someone has such a solution please mail it to me, I would love to learn how to do this in a better way.

Good luck and happy coding!Hope this helps, I almost wrote the complete program!

You could also stop the algorithm as soon as you found a solution. In other posts, some people talked about a O(n^3) solution but I can't figure it out... if someone has such a solution please mail it to me, I would love to learn how to do this in a better way.

Good luck and happy coding!

I think it should be:

Code: Select all

```
for (steps = 2; steps <= n; steps++)
for k...
for i..
for j..
tmp = best[i][k][steps-1] * best[k][j][1]
if (tmp > best[i][j][steps-1] && tmp > best[i][j][steps])
{
best[i][j][steps] = tmp;
path[i][j][steps] = k;
}
```

I stay home. Don't call me out.

Code: Select all

```
for (steps = 2; steps <= n; steps++)
for k...
for i..
for j..
tmp = best[i][k][steps-1] * best[k][j][1]
if (tmp > best[i][j][steps]) {
best[i][j][steps] = tmp;
path[i][j][steps] = k;
}
```

*[j][s] is pre-loaded with 0, for all i,j and s >= 2.*

Are you getting floating point error? If you post your code I may help.

Are you getting floating point error? If you post your code I may help.

Oh, now I have not floating point error, that was just because of a wrong typing.

But I got WA. This is my code which works all right on the sample input.

But I got WA. This is my code which works all right on the sample input.

Code: Select all

```
#include<stdio.h>
int main()
{
int n,/*维数*/
i,j,k,step,
path[20][20][20],/*用来记录最佳路径*/
get,
seq[20];/*最后要输出的数组*/
double best[20][20][20],/*best[i][j][s]表示从i到j花了s步的最大利润*/
temp;
while(scanf("%d",&n)==1)
{
get=0;
for(step=0;step<n;step++)
if(step==0)
{
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
if(i==j)
best[i][j][step]=1.0;
else
scanf("%lf",&best[i][j][step]);
path[i][j][step]=i;
}
}
else
for(i=0;i<n;i++)
for(j=0;j<n;j++)
best[i][j][step]=0;
for(step=1;step<n;step++)/*题目要求最多n步*/
for(k=0;k<n;k++)
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
temp=best[i][k][step-1]*best[k][j][0];
if(temp>best[i][j][step])
{
best[i][j][step]=temp;
path[i][j][step]=k;
}
}
for(step=1;step<n;step++)
{
for(i=0;i<n;i++)
{
if(best[i][i][step]>1.01)
{
seq[step]=path[i][i][step];
for(j=step-1;j>=0;j--)
seq[j]=path[i][seq[j+1]][j];
for(j=0;j<=step;j++)
printf("%d ",seq[j]+1);
printf("1\n");
get=1;
break;
}
}
if(get==1)
break;
}
if(get!=1)
printf("no arbitrage sequence exists\n");
}
return 0;
}
```

Last edited by ImLazy on Tue Feb 15, 2005 12:45 pm, edited 1 time in total.

I stay home. Don't call me out.

Code: Select all

```
6
0.0 0.0 0.0 0.0 0.0
0.0 0.0 0.1 0.0 0.0
0.0 0.0 1.0 0.0 0.0
0.0 5.0 0.0 0.0 0.0
0.0 0.0 0.0 0.0 9.0
0.0 0.0 0.0 0.0 1.0
```

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`5 6 5`

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`5 6 1`

Actually, you have a silly bug check again the line

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`printf("1\n");`

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`printf("%d\n", i+1)`