troubleshooting process dynamic programming optimal substructure
In general the problem can be using a three step process:
Thursday, May 7, 2009
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Separate the problem into smaller subproblems.
Solving these problems optimally using this process three steps recursively. Use
optimal solutions to construct an optimal solution of original problem.
- The subproblems are themselves determined by dividing these into sub-problems and so on until it reaches the simplest case that is easy to solve.
- say that a problem has overlapping subproblems, meaning that one subproblem is used to solve many different problems larger.
- For example, the Fibonacci sequence F3 = F1 + F2 and F4 = F2 + F3, the calculation of each issue involves the calculation of F2. Since both F3 and F4 F5 necessary to calculate a naive method to calculate F5 may end up making F2 calculating two or more times.
overlapping subproblems When a method can waste time recalculating optimal solutions of subproblems already solved. 
To avoid repetition of calculations, you can save the solutions to the problems addressed in a way that if required the same problem then can be reused calculations made earlier.
- memoización This approach is called (no memorization).
- When you are sure that a particular solution and no longer needed can then be deleted, freeing up space.
- In some cases you can make computations that are known are required in later.
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and overlapping subproblems
optimal substructure means that the substructure
subproblems can be used to find the optimal solution to the problem
complete.
For example, the shortest path to a goal from a corner in an acyclic graph can be found by first calculating the nearest road to the goal from all adjacent vertices
then using this to find the best complete path.
Figure 1.
Finding the nearest road in a graph using optimal substructure
, the dark lines show the way close from the beginning to the end. 
say that a problem has overlapping subproblems ie one subproblem is used to solve many different problems bigger.
Troubleshooting dynamic programming
optimal substructure means that the substructure
subproblems can be used to find the optimal solution to the problem
complete.
For example, the shortest path to a goal from a corner in an acyclic graph can be found by first calculating the nearest road to the goal from all adjacent vertices
then using this to find the best complete path.
Figure 1.
Finding the nearest road in a graph using optimal substructure
, the dark lines show the way close from the beginning to the end.
say that a problem has overlapping subproblems Troubleshooting dynamic programming
Tuesday, May 5, 2009
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The problem of dynamic programming currencies.
solve this dynamic programming problem in a function
Java: public int
Minima_devolucion (cantidad_devuelta int, int [] coins) {/ / create
returns array int [] [] matriz_cambio = new int [monedas.length +1] [cantidad_devuelta +1];
/ / Fill 1st column of zeros for (int i = 0; i
matriz_cambio [i] [0] = 0;
/ / The 1 st row less the 1 st column a high number for (int j = 1, j
/ / If the currency is over the amount to return
solve this dynamic programming problem in a function
Java: public int Minima_devolucion (cantidad_devuelta int, int [] coins) {/ / create
returns array int [] [] matriz_cambio = new int [monedas.length +1] [cantidad_devuelta +1];
/ / Fill 1st column of zeros for (int i = 0; i
matriz_cambio [i] [0] = 0;
/ / The 1 st row less the 1 st column a high number for (int j = 1, j
matriz_cambio [0] [j] = 999999; 
for (int i = 1; i for (int j = 1, j if (coins [i-1]> j) { / / If the currency is over the amount to return
matriz_cambio [i] [j] = matriz_cambio [i-1] [j];} else {
/ / If the currency does not exceed the amount to return
/ / Calculate what is the minimum these two positions
int minimum = 0; / / Save
minimum
here if (matriz_cambio [i-1] [j] = matriz_cambio minimum
[i-1] [j];} else {< monedas.length; i++)
minimum = matriz_cambio [i] [j - coins [i-1]] + 1;}
/ / Save <= cantidad_devuelta; j++)
matriz_cambio least [i] [j] = least;
}} return <= monedas.length ; i++)
<= cantidad_devuelta; j++){
matriz_cambio [monedas.length] [cantidad_devuelta];}
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problem with dynamic programming coins
For the problem of dynamic programming currencies
need to create an algorithm that allows a vending machine to give change using the fewest coins possible. By dynamic programming will solve the case in which the number of coins of each type is unlimited. In the currency problem < matriz_cambio[i][j - monedas[i-1]] + 1){
by the greedy algorithm when the number of coins is unlimited.
Description Suppose you have coins worth 1, 4 and 6 and to be repaid an amount corresponding to the value 8. Following the method of dynamic programming
is populate a table with rows for each value for coins and columns with values \u200b\u200bfrom 1 to 8. Each position (i, j) of the table shows the minimum number of coins required to repay the amount with coins j with value less than or equal to i:
____ 1 2 3 4 5 6 7 8
m1 = 1 1 2 3 4 5 6 7 8 m2 = 4 1 2 3 1 2 3 4 2
Algorithm 1. For each cell of the table do: 2. If the current coin value is greater than the amount, is paid in other currencies, ie take the result of the box superior. 3. If the current coin value is less than or equal to the amount, takes the lower of:
1. Pay with other currencies, whichever results in the upper box.
2. Pay with a coin of the current type and the other with the result that was obtained by paying the actual amount which has been subtracted from the value of existing currency. 4. Take as a result the value of the last cell. Pseudocode
solve this dynamic programming problem in a pseudo
.
As input parameters the function takes C, which corresponds to the amount to be refunded and a vector M of coins, which stores the value of each type. Returns num, the number of coins needed to make the return.
fun change (C: nat, M [1 .. n] of nat) dev num: nat var
T [0 .. n] [0 .. C] of nat
begin T [0] [ 1 .. C]: = \\ infty;
T [1 .. n] [0]: = 0;
for i: = 1 to n do
for j: = 1 to C do
if M [i]> j then T [i, j]: = T [i-1, j] if T [i, j]: = min {T [i-1, j], T [i, jM [i]] + 1}
fsi
fto fto num: = T [n, C]
Ffun
/ / The 1st row 1st column unless a high number for (int j = 1, j
matriz_cambio [0] [j] = 999999;
for (int i = 1; i for (int j = 1, j
if (coins [i-1]> j) {
For the problem of dynamic programming currencies
need to create an algorithm that allows a vending machine to give change using the fewest coins possible. By dynamic programming will solve the case in which the number of coins of each type is unlimited. In the currency problem < matriz_cambio[i][j - monedas[i-1]] + 1){
by the greedy algorithm when the number of coins is unlimited.
Description Suppose you have coins worth 1, 4 and 6 and to be repaid an amount corresponding to the value 8. Following the method of dynamic programming
is populate a table with rows for each value for coins and columns with values \u200b\u200bfrom 1 to 8. Each position (i, j) of the table shows the minimum number of coins required to repay the amount with coins j with value less than or equal to i:
____ 1 2 3 4 5 6 7 8
m1 = 1 1 2 3 4 5 6 7 8 m2 = 4 1 2 3 1 2 3 4 2
m3 = 6 1 2 3 1 2 1 2 2 
Example for position i = 2 j = 7, it takes a coin with value 4 type 2 and three coins of a value type 1, therefore the table the number of coins in position (2.7) will be 1 + 3 = 4. Algorithm 1. For each cell of the table do: 2. If the current coin value is greater than the amount, is paid in other currencies, ie take the result of the box superior. 3. If the current coin value is less than or equal to the amount, takes the lower of:
1. Pay with other currencies, whichever results in the upper box.
2. Pay with a coin of the current type and the other with the result that was obtained by paying the actual amount which has been subtracted from the value of existing currency. 4. Take as a result the value of the last cell. Pseudocode
solve this dynamic programming problem in a pseudo
.
As input parameters the function takes C, which corresponds to the amount to be refunded and a vector M of coins, which stores the value of each type. Returns num, the number of coins needed to make the return.
fun change (C: nat, M [1 .. n] of nat) dev num: nat var
T [0 .. n] [0 .. C] of nat
begin T [0] [ 1 .. C]: = \\ infty;
T [1 .. n] [0]: = 0;
for i: = 1 to n do
for j: = 1 to C do
if M [i]> j then T [i, j]: = T [i-1, j] if T [i, j]: = min {T [i-1, j], T [i, jM [i]] + 1}
fsi
fto fto num: = T [n, C]
Ffun
matriz_cambio [i] [0] = 0;
Code Java
solve this dynamic programming problem
in Java code: public
Minima_devolucion int (cantidad_devuelta int, int [] coins)
{/ / Create the array of returns
int [] [] matriz_cambio = new int [monedas.length +1] [cantidad_devuelta +1];
/ / Fill the 1 st column of zeros
for (int i = 0; i
/ / The 1st row 1st column unless a high number for (int j = 1, j
matriz_cambio [0] [j] = 999999;
for (int i = 1; i for (int j = 1, j
if (coins [i-1]> j) {
/ / If the currency is higher than the amount to be refunded
matriz_cambio [i] [j] = matriz_cambio [i-1] [ j];
} else {/ / If the currency does not exceed the amount to return
/ / Calculate what is the minimum of these two positions
int minimum = 0; / / Save the minimum
here if ( matriz_cambio [i-1] [j]
matriz_cambio minimum = [i-1] [j];} else {< monedas.length; i++)
minimum = matriz_cambio [i] [j - coins [i-1]] + 1;}
/ / Save <= cantidad_devuelta; j++)
minimum matriz_cambio [i] [j] = least;
}} return <= monedas.length ; i++)
<= cantidad_devuelta; j++){
matriz_cambio [monedas.length] [cantidad_devuelta];}
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