Suppose we have one N x N table of currency exchange rates. We have to check whether there is some sequence of trades we can make or not. Now starting with some amount A of any currency, we can end up with some amount greater than A of that currency. There are no transaction costs and we can also trade fractional quantities.
The value at entry [I, j] in this matrix represents the amount of currency j we can buy with one unit of currency i. Now consider currency 0 is USD, 1 is CAD and 2 is EUR. We can make an arbitrage with the following −
Sell 1 CAD for 0.65 EUR
Sell 0.65 EUR for 0.7865 USD (0.65 * 1.21)
Sell 0.7865 USD for 1.00672 CAD (0.65 * 1.21 * 1.28)
So, if the input is like
| 1 | 1.28 | 0.82 |
| 0.78 | 1 | 0.65 |
| 1.21 | 1.55 | 1 |
then the output will be True.
To solve this, we will follow these steps −
for i in range 0 to size of matrix, do
for j in range 0 to size of matrix[0], do
matrix[i,j] := −log base 2 value of (matrix[I, j])
v := row count of matrix
for k in range 0 to v, do
for i in range 0 to v, do
for j in range 0 to v, do
matrix[I, j] := minimum of matrix[I, j] and (matrix[I, k] + matrix[k, j])
return True if any of the items in the diagonal of the matrix is non−zero.
Let us see the following implementation to get better understanding −
Python
import math class Solution: def solve(self, matrix): for i in range(len(matrix)): for j in range(len(matrix[0])): matrix[i][j] = −math.log(matrix[i][j], 2) v = len(matrix) for k in range(0, v): for i in range(0, v): for j in range(0, v): matrix[i][j] = min(matrix[i][j], matrix[i][k] + matrix[k][j]) return any(matrix[i][i] < 0 for i in range(len(matrix))) ob = Solution() matrix = [ [1, 1.28, 0.82], [0.78, 1, 0.65], [1.21, 1.55, 1] ] print(ob.solve(matrix))
Input
matrix = [ [1, 1.28, 0.82], [0.78, 1, 0.65], [1.21, 1.55, 1] ]
Output
True