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==Discussion of complexity== | ==Discussion of complexity== | ||
− | The corresponding decision problem, which simply asks us to determine whether or not making change is ''possible'' with the given denominations (which it might not be, if we are missing the denomination 1) is known to be [[NP-complete]].<ref>G. S. Lueker. (1975). ''Two NP-complete problems in nonnegative integer programming.'' Technical Report 178, Computer Science Laboratory, Princeton University | + | The corresponding decision problem, which simply asks us to determine whether or not making change is ''possible'' with the given denominations (which it might not be, if we are missing the denomination 1) is known to be [[NP-complete]].<ref>G. S. Lueker. (1975). ''Two NP-complete problems in nonnegative integer programming.'' Technical Report 178, Computer Science Laboratory, Princeton University''</ref> It follows that the optimization and counting problems are both [[NP-hard]] (''e.g.'', because the result of 0 for the counting problem answers the decision problem in the negative, and any nonzero value answers it in the affirmative). |
However, as we shall see, a simple <math>O(nT)</math> solution exists for both versions of the problem. Why then are these problems not in P? The answer is that the ''size'' of the input required to represent the number <math>T</math> is actually the ''length'' of the number <math>T</math>, which is <math>\Theta(\log T)</math> when <math>T</math> is expressed in binary (or decimal, or whatever). Thus, the time and space required by the algorithm is actually <math>O(n 2^{\lg T})</math>, that is, exponential in the size of the input. (This simplified analysis does not take into account the sizes of the denominations, but captures the essence of the argument.) This algorithm is then said to be ''pseudo-polynomial''. No true polynomial-time algorithm is known (and, indeed, none will be found unless it turns out that P = NP). | However, as we shall see, a simple <math>O(nT)</math> solution exists for both versions of the problem. Why then are these problems not in P? The answer is that the ''size'' of the input required to represent the number <math>T</math> is actually the ''length'' of the number <math>T</math>, which is <math>\Theta(\log T)</math> when <math>T</math> is expressed in binary (or decimal, or whatever). Thus, the time and space required by the algorithm is actually <math>O(n 2^{\lg T})</math>, that is, exponential in the size of the input. (This simplified analysis does not take into account the sizes of the denominations, but captures the essence of the argument.) This algorithm is then said to be ''pseudo-polynomial''. No true polynomial-time algorithm is known (and, indeed, none will be found unless it turns out that P = NP). | ||
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The optimization problem exhibits optimal substructure, in the sense that if we remove any coin of value <math>D_i</math> from the optimal means of changing <math>T</math>, then the set of coins remaining is an optimal means of changing <math>T - D_i</math>. This is because if this were ''not'' so; that is, there existed a means of changing <math>T - D_i</math> that used fewer coins than what we obtained by removing the coin <math>D_i</math> from our supposed optimal change for <math>T</math>, then we could just add the coin <math>D_i</math> back in and get change for the original amount <math>T</math> in fewer coins, a contradiction. Therefore, if we let <math>f(x)</math> denote the minimal number of coins required to change amount <math>x</math>, then we can write <math>f(x) = 1 + \min_i f(x - D_i)</math>; we consider all possible minimal solutions to <math>x</math> minus one coin, and take the best one and add that coin back in to get minimal change for <math>x</math>. The base case is <math>f(0) = 0</math>; obviously, 0 coins are required to make change for 0. See [[Dynamic_programming#Optimization_example:_Change_problem|the DP article]] for details and an implementation. | The optimization problem exhibits optimal substructure, in the sense that if we remove any coin of value <math>D_i</math> from the optimal means of changing <math>T</math>, then the set of coins remaining is an optimal means of changing <math>T - D_i</math>. This is because if this were ''not'' so; that is, there existed a means of changing <math>T - D_i</math> that used fewer coins than what we obtained by removing the coin <math>D_i</math> from our supposed optimal change for <math>T</math>, then we could just add the coin <math>D_i</math> back in and get change for the original amount <math>T</math> in fewer coins, a contradiction. Therefore, if we let <math>f(x)</math> denote the minimal number of coins required to change amount <math>x</math>, then we can write <math>f(x) = 1 + \min_i f(x - D_i)</math>; we consider all possible minimal solutions to <math>x</math> minus one coin, and take the best one and add that coin back in to get minimal change for <math>x</math>. The base case is <math>f(0) = 0</math>; obviously, 0 coins are required to make change for 0. See [[Dynamic_programming#Optimization_example:_Change_problem|the DP article]] for details and an implementation. | ||
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==Notes and References== | ==Notes and References== | ||
<references/> | <references/> |