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Problem: The most creative part of inventing dynamic programming solution is defining recurrent relations. The recurrent relations consist of two parts: state domain and transitions. State domain is a set of states (subproblems) in dynamic programming. For each state the subresult will be calculated eventually. Transitions are the relations between different states which help calculate the subresults.
This recipe covers frequently used state domain types. The general approaches of dealing with them and real SRM examples are given. Also few optimizations specific to particular domains are mentioned here.
Solution Code of DP solution usually contains an array representing subresults on the state domain. For example, classic knapsack problem solution will be like:
int maxcost[items+1][space+1];
memset(maxcost, 63, sizeof(maxcost)); //fill with negative infinity
maxcost[0][0] = 0; //base of DP
for (int i = 0; i<items; i++) //iterations over states in proper order
for (int j = 0; j<=space; j++) {
int mc = maxcost[i][j]; //we handle two types forward transitions
int ni, nj, nmc; //from state (i,j)>mc to state (ni,nj)>nmc
ni = i + 1; //forward transition: do not add ith item
nj = j;
nmc = mc;
if (maxcost[ni][nj] < nmc) //relaxing result for new state
maxcost[ni][nj] = nmc;
ni = i + 1; //forward transition: add ith item
nj = j + size[i];
nmc = mc + cost[i];
if (nj <= space && maxcost[ni][nj] < nmc)
maxcost[ni][nj] = nmc;
}
int answer = 1000000000; //getting answer from state results
for (j = 0; j<=space; j++)
if (maxcost[items][j] > answer)
answer = maxcost[items][j];
return answer;
Here (i,j) is state of DP with result equal to maxcost[i][j]. The result here means the maximal cost of items we can get by taking some of first i items with overall size of exactly j. So the set of (i,j) pairs and concept of maxcost[i][j] here comprise a state domain. The forward transition is adding or not adding the ith item to the set of items we have already chosen.
The order of iterations through all DP states is important. The code above iterates through states with pairs (i,j) sorted lexicographically. It is correct since any transition goes from set (i,*) to set (i+1,*), so we see that i is increasing by one. Speaking in backward (recurrent) style, the result for each state (i,j) directly depends only on the results for the states (i1,*).
To determine order or iteration through states we have to define order on state domain. We say that state (i1,j1) is greater than state (i2,j2) if (i1,j1) directly or indirectly (i.e. through several other states) depends on (i2,j2). This is definition of order on the state domain used. In DP solution any state must be considered after all the lesser states. Else the solution would give incorrect result.
Multidimensional array The knapsack DP solution described above is an example of multidimensional array state domain (with 2 dimensions). A lot of other problems have similar state domains. Generally speaking, in this category states are represented by k parameters: (i1, i2, i3, ..., ik). So in the code we define a multidimensional array for state results like: int Result[N1][N2][N3]...[Nk]. Of course there are some transition rules (recurrent relations). These rules themselves can be complex, but the order of states is usually simple.
In most cases the states can be iterated through in lexicographical order. To do this you have to ensure that if I = (i1, i2, i3, ..., ik) directly depends on J = (j1, j2, j3, ..., jk) then I is lexicographically greater that J. This can be achieved by permuting parameters (like using (j,i) instead of (i,j)) or reversing them. But it is usually easier to change the order and direction of nested loops. Here is general code of lexicographical traversion:
for (int i1 = 0; i1<N1; i1++)
for (int i2 = 0; i2<N1; i2++)
...
for (int ik = 0; ik<Nk; ik++) {
//get some states (j1, j2, j3, ..., jk) > jres by performing transitions
//and handle them
}
Note: changing order of DP parameters in array and order of nested loops can noticably affect performance on modern computers due to CPU cache behavior.
This type of state domain is the easiest to understand and implement, that's why most DP tutorials show problems of this type. But it is not the most frequently used type of state domain in SRMs. DP over subsets is much more popular.


Problem: The most creative part of inventing dynamic programming solution is defining recurrent relations. The recurrent relations consist of two parts: state domain and transitions. State domain is a set of states (subproblems) in dynamic programming. For each state the subresult will be calculated eventually. Transitions are the relations between different states which help calculate the subresults.
This recipe covers frequently used state domain types. The general approaches of dealing with them and real SRM examples are given. Also few optimizations specific to particular domains are mentioned here.
Solution Code of DP solution usually contains an array representing subresults on the state domain. For example, classic knapsack problem solution will be like:
int maxcost[items+1][space+1];
memset(maxcost, 63, sizeof(maxcost)); //fill with negative infinity
maxcost[0][0] = 0; //base of DP
for (int i = 0; i<items; i++) //iterations over states in proper order
for (int j = 0; j<=space; j++) {
int mc = maxcost[i][j]; //we handle two types forward transitions
int ni, nj, nmc; //from state (i,j)>mc to state (ni,nj)>nmc
ni = i + 1; //forward transition: do not add ith item
nj = j;
nmc = mc;
if (maxcost[ni][nj] < nmc) //relaxing result for new state
maxcost[ni][nj] = nmc;
ni = i + 1; //forward transition: add ith item
nj = j + size[i];
nmc = mc + cost[i];
if (nj <= space && maxcost[ni][nj] > nmc)
maxcost[ni][nj] = nmc;
}
int answer = 1000000000; //getting answer from state results
for (j = 0; j<=space; j++)
if (maxcost[items][j] > answer)
answer = maxcost[items][j];
return answer;
Here (i,j) is state of DP with result equal to maxcost[i][j]. The result here means the maximal cost of items we can get by taking some of first i items with overall size of exactly j. So the set of (i,j) pairs and concept of maxcost[i][j] here comprise a state domain. The forward transition is adding or not adding the ith item to the set of items we have already chosen.
The order of iterations through all DP states is important. The code above iterates through states with pairs (i,j) sorted lexicographically. It is correct since any transition goes from set (i,*) to set (i+1,*), so we see that i is increasing by one. Speaking in backward (recurrent) style, the result for each state (i,j) directly depends only on the results for the states (i1,*).
To determine order or iteration through states we have to define order on state domain. We say that state (i1,j1) is greater than state (i2,j2) if (i1,j1) directly or indirectly (i.e. through several other states) depends on (i2,j2). This is definition of order on the state domain used. In DP solution any state must be considered after all the lesser states. Else the solution would give incorrect result.
Multidimensional array The knapsack DP solution described above is an example of multidimensional array state domain (with 2 dimensions). A lot of other problems have similar state domains. Generally speaking, in this category states are represented by k parameters: (i1, i2, i3, ..., ik). So in the code we define a multidimensional array for state results like: int Result[N1][N2][N3]...[Nk]. Of course there are some transition rules (recurrent relations). These rules themselves can be complex, but the order of states is usually simple.
In most cases the states can be iterated through in lexicographical order. To do this you have to ensure that if I = (i1, i2, i3, ..., ik) directly depends on J = (j1, j2, j3, ..., jk) then I is lexicographically greater that J. This can be achieved by permuting parameters (like using (j,i) instead of (i,j)) or reversing them. But it is usually easier to change the order and direction of nested loops. Here is general code of lexicographical traversion:
for (int i1 = 0; i1<N1; i1++)
for (int i2 = 0; i2<N1; i2++)
...
for (int ik = 0; ik<Nk; ik++) {
//get some states (j1, j2, j3, ..., jk) > jres by performing transitions
//and handle them
}
Note: changing order of DP parameters in array and order of nested loops can noticably affect performance on modern computers due to CPU cache behavior.
This type of state domain is the easiest to understand and implement, that's why most DP tutorials show problems of this type. But it is not the most frequently used type of state domain in SRMs. DP over subsets is much more popular.


Problem: The most creative part of inventing dynamic programming solution is defining recurrent relations. The recurrent relations consist of two parts: state domain and transitions. State domain is a set of states (subproblems) in dynamic programming. For each state the subresult will be calculated eventually. Transitions are the relations between different states which help calculate the subresults.
This recipe covers frequently used state domain types. The general approaches of dealing with them and real SRM examples are given. Also few optimizations specific to particular domains are mentioned here.
Solution Code of DP solution usually contains an array representing subresults on the state domain. For example, classic knapsack problem solution will be like:
int maxcost[items+1][space+1];
memset(maxcost, 63, sizeof(maxcost)); //fill with negative infinity
maxcost[0][0] = 0; //base of DP
for (int i = 0; i<items; i++) //iterations over states in proper order
for (int j = 0; j<=space; j++) {
int mc = maxcost[i][j]; //we handle two types forward transitions
int ni, nj, nmc; //from state (i,j)>mc to state (ni,nj)>nmc
ni = i + 1; //forward transition: do not add ith item
nj = j;
nmc = mc;
if (maxcost[ni][nj] > nmc) //relaxing result for new state
maxcost[ni][nj] = nmc;
ni = i + 1; //forward transition: add ith item
nj = j + size[i];
nmc = mc + cost[i];
if (nj <= space && maxcost[ni][nj] > nmc)
maxcost[ni][nj] = nmc;
}
int answer = 1000000000; //getting answer from state results
for (j = 0; j<=space; j++)
if (maxcost[items][j] > answer)
answer = maxcost[items][j];
return answer;
Here (i,j) is state of DP with result equal to maxcost[i][j]. The result here means the maximal cost of items we can get by taking some of first i items with overall size of exactly j. So the set of (i,j) pairs and concept of maxcost[i][j] here comprise a state domain. The forward transition is adding or not adding the ith item to the set of items we have already chosen.
The order of iterations through all DP states is important. The code above iterates through states with pairs (i,j) sorted lexicographically. It is correct since any transition goes from set (i,*) to set (i+1,*), so we see that i is increasing by one. Speaking in backward (recurrent) style, the result for each state (i,j) directly depends only on the results for the states (i1,*).
To determine order or iteration through states we have to define order on state domain. We say that state (i1,j1) is greater than state (i2,j2) if (i1,j1) directly or indirectly (i.e. through several other states) depends on (i2,j2). This is definition of order on the state domain used. In DP solution any state must be considered after all the lesser states. Else the solution would give incorrect result.
Multidimensional array The knapsack DP solution described above is an example of multidimensional array state domain (with 2 dimensions). A lot of other problems have similar state domains. Generally speaking, in this category states are represented by k parameters: (i1, i2, i3, ..., ik). So in the code we define a multidimensional array for state results like: int Result[N1][N2][N3]...[Nk]. Of course there are some transition rules (recurrent relations). These rules themselves can be complex, but the order of states is usually simple.
In most cases the states can be iterated through in lexicographical order. To do this you have to ensure that if I = (i1, i2, i3, ..., ik) directly depends on J = (j1, j2, j3, ..., jk) then I is lexicographically greater that J. This can be achieved by permuting parameters (like using (j,i) instead of (i,j)) or reversing them. But it is usually easier to change the order and direction of nested loops. Here is general code of lexicographical traversion:
for (int i1 = 0; i1<N1; i1++)
for (int i2 = 0; i2<N1; i2++)
...
for (int ik = 0; ik<Nk; ik++) {
//get some states (j1, j2, j3, ..., jk) > jres by performing transitions
//and handle them
}
Note: changing order of DP parameters in array and order of nested loops can noticably affect performance on modern computers due to CPU cache behavior.
This type of state domain is the easiest to understand and implement, that's why most DP tutorials show problems of this type. But it is not the most frequently used type of state domain in SRMs. DP over subsets is much more popular.


Problem: The most creative part of inventing dynamic programming solution is defining recurrent relations. The recurrent relations consist of two parts: state domain and transitions. State domain is a set of states (subproblems) in dynamic programming. For each state the subresult will be calculated eventually. Transitions are the relations between different states which help calculate the subresults.
This recipe covers frequently used state domain types. The general approaches of dealing with them and real SRM examples are given. Also few optimizations specific to particular domains are mentioned here.
Solution Code of DP solution usually contains an array representing subresults on the state domain. For example, classic knapsack problem solution will be like:
int maxcost[items+1][space+1];
memset(maxcost, 63, sizeof(maxcost)); //fill with negative infinity
maxcost[0][0] = 0; //base of DP
for (int i = 0; i<items; i++) //iterations over states in proper order
for (int j = 0; j<=space; j++) {
int mc = maxcost[i][j]; //we handle two types forward transitions
int ni, nj, nmc; //from state (i,j)>mc to state (ni,nj)>nmc
ni = i + 1; //forward transition: do not add ith item
nj = j;
nmc = mc;
if (maxcost[ni][nj] > nmc) //relaxing result for new state
maxcost[ni][nj] = nmc;
ni = i + 1; //forward transition: add ith item
nj = j + size[i];
nmc = mc + cost[i];
if (nj <= space && maxcost[ni][nj] > nmc)
maxcost[ni][nj] = nmc;
}
int answer = 1000000000; //getting answer from state results
for (j = 0; j<=space; j++)
if (maxcost[items][j] > answer)
answer = maxcost[items][j];
return answer;
Here (i,j) is state of DP with result equal to maxcost[i][j]. The result here means the maximal cost of items we can get by taking some of first i items with overall size of exactly j. So the set of (i,j) pairs and concept of maxcost[i][j] here comprise a state domain. The forward transition is adding or not adding the ith item to the set of items we have already chosen.
The order of iterations through all DP states is important. The code above iterates through states with pairs (i,j) sorted lexicographically. It is correct since any transition goes from set (i,*) to set (i+1,*), so we see that i is increasing by one. Speaking in backward (recurrent) style, the result for each state (i,j) directly depends only on the results for the states (i1,*).
To determine order or iteration through states we have to define order on state domain. We say that state (i1,j1) is greater than state (i2,j2) if (i1,j1) directly or indirectly (i.e. through several other states) depends on (i2,j2). This is definition of order on the state domain used. In DP solution any state must be considered after all the lesser states. Else the solution would give incorrect result.
Multidimensional array The state domain used in knapsack problem is represented by multidimensional array. Generally speaking, in this category states are represented by k parameters: (i1, i2, i3, ..., ik). So in the code we define a multidimensional array for state results like: int Result[N1][N2][N3]...[Nk]. Of course there are some transition rules (recurrent relations). These rules themselves can be complex, but the order of states is usually simple.
In most cases the states can be iterated through in lexicographical order. To do this you have to ensure that if I = (i1, i2, i3, ..., ik) directly depends on J = (j1, j2, j3, ..., jk) then I is lexicographically greater that J. This can be achieved by permuting parameters (like using (j,i) instead of (i,j)) or reversing them. But it is usually easier to change the order and direction of nested loops. Here is general code of lexicographical traversion:
for (int i1 = 0; i1<N1; i1++)
for (int i2 = 0; i2<N1; i2++)
...
for (int ik = 0; ik<Nk; ik++) {
//get some states (j1, j2, j3, ..., jk) > jres by performing transitions
//and handle them
}
Note: changing order of DP parameters in array and order of nested loops can noticably affect performance on modern computers due to CPU cache behavior.
This type of state domain is the easiest to understand and implement, that's why most DP tutorials show problems of this type. But it is not the most frequently used type of state domain in SRMs. DP over subsets is much more popular.

