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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Constraint Satisfaction Problems
Heuristics for Arc-Consistency Algorithms
Introduction
A Constraint Satisfaction Problem is a triple $(X,D,C)$ where
Step1:  Domain-Heuristics for Arc-Consistency Algorithms
In <a name="ref-1"/>[1] are investigated the effects of a domain-heuristic based on the notion of a double-support check by studying its average time-complexity.
The objective of arc-consistency algorithms is to resolve some uncertainty; it has to be know, for each $v_i \in D_i$ and for each $v_j \in D_j$, whether it is supported.
A single-support check, $(v_i, v_j) \in C_{ij}$, is one in which, before the check is done, it is already known that either $v_i$ or $v_j$ are supported. 
A double-support check $(v_i, v_j) \in C_{ij}$, is one in which there is still, before the check, uncertainty about the support-status of both $v_i$ and $v_j$. 
If a double-support check is successful, two uncertainties are resolved. If a single-support check is successful, only one uncertainty is resolved. A good arc-consistency algorithm, therefore, would always choose to do a double-support check in preference of a single-support check, because the cormer offers the potential higher payback.
The improvement with double-support check is that, where possible, consistency-checks are used to find supports for two values, one value in the domain of each variable, which were previously known to be unsupported. It is motivated by the insight that in order to minimize the number of consistency-checks it is necessary to maximize the number of uncertainties which are resolved per check.
AC-3b
Step2:  At any stage in the process of making 2-variable CSP arc-consistent in AC3b
Step3:  AC3b is a refinement of the AC3 algorithm which consists of the fact that if, when arc $(i,j)$ is being processed and the reverse arc $(j,i)$ is also in the queue, then consistency-checks can be saved because only support for the elements in $S_j^?$ has to be found (as opposed to support for all the elements in $D_j$ in the
AC3 algorithm). <br>
AC3b inherits all its properties like $\mathcal{O}(ed^3)$ time-complexity and $\mathcal{O}(e + nd)$ space-complexity fron AC3 and where $n$ denotes the number of variables in the CSP, $e$ denotes the number of binary constraints and $d$ denotes the maximum domain-size of the variables.
Arc-Heuristics for Arc-Consistency Algorithms
Many arc-heuristics can be devised, based on three major features of CSPs
Step4:  Experimental Results
For the experiments below on binary CSPs, in addition to the two arc-consistency algorithms already cited above, AC3 and AC3b, the AC4 algorithm was used. <br>
The AC4 algorithm runs in $\mathcal{O}(ed^2)$ worst-case time but can be slower than AC3 on average cases.
Step5:  Sudoku
Easy Sudoku
Step6:  Harder Sudoku
Step7:  8 Queens
Step8:  For the experiments below on n-ary CSPs, due to the n-ary constraints, the GAC algorithm was used. <br>
The GAC algorithm has $\mathcal{O}(er^2d^t)$ time-complexity and $\mathcal{O}(erd)$ space-complexity where $e$ denotes the number of n-ary constraints, $r$ denotes the constraint arity and $d$ denotes the maximum domain-size of the variables.
Step9:  Crossword
Step10:  Kakuro
Easy Kakuro
Step11:  Medium Kakuro
Step12:  Harder Kakuro
Step13:  Cryptarithmetic Puzzle
$$
\begin{array}{@{}r@{}}
     S E N D \
{} + M O R E \
   \hline
   M O N E Y
\end{array}
$$
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Python Code:
from csp import *
Explanation: Constraint Satisfaction Problems
Heuristics for Arc-Consistency Algorithms
Introduction
A Constraint Satisfaction Problem is a triple $(X,D,C)$ where: 
- $X$ is a set of variables $X_1, …, X_n$;
- $D$ is a set of domains $D_1, …, D_n$, one for each variable and each of which consists of a set of allowable values $v_1, ..., v_k$;
- $C$ is a set of constraints that specify allowable combinations of values.
A CSP is called arc-consistent if every value in the domain of every variable is supported by all the neighbors of the variable while, is called inconsistent, if it has no solutions. <br>
Arc-consistency algorithms remove all unsupported values from the domains of variables making the CSP arc-consistent or decide that a CSP is inconsistent by finding that some variable has no supported values in its domain. <br> 
Heuristics significantly enhance the efficiency of the arc-consistency algorithms improving their average performance in terms of consistency-checks which can be considered a standard measure of goodness for such algorithms. Arc-heuristic operate at arc-level and selects the constraint that will be used for the next check, while domain-heuristics operate at domain-level and selects which values will be used for the next support-check.
End of explanation
%psource AC3
%psource revise
Explanation: Domain-Heuristics for Arc-Consistency Algorithms
In <a name="ref-1"/>[1] are investigated the effects of a domain-heuristic based on the notion of a double-support check by studying its average time-complexity.
The objective of arc-consistency algorithms is to resolve some uncertainty; it has to be know, for each $v_i \in D_i$ and for each $v_j \in D_j$, whether it is supported.
A single-support check, $(v_i, v_j) \in C_{ij}$, is one in which, before the check is done, it is already known that either $v_i$ or $v_j$ are supported. 
A double-support check $(v_i, v_j) \in C_{ij}$, is one in which there is still, before the check, uncertainty about the support-status of both $v_i$ and $v_j$. 
If a double-support check is successful, two uncertainties are resolved. If a single-support check is successful, only one uncertainty is resolved. A good arc-consistency algorithm, therefore, would always choose to do a double-support check in preference of a single-support check, because the cormer offers the potential higher payback.
The improvement with double-support check is that, where possible, consistency-checks are used to find supports for two values, one value in the domain of each variable, which were previously known to be unsupported. It is motivated by the insight that in order to minimize the number of consistency-checks it is necessary to maximize the number of uncertainties which are resolved per check.
AC-3b: an improved version of AC-3 with Double-Support Checks
As shown in <a name="ref-2"/>[2] the idea is to use double-support checks to improve the average performance of AC3 which does not exploit the fact that relations are bidirectional and results in a new general purpose arc-consistency algorithm called AC3b.
End of explanation
%psource AC3b
%psource partition
Explanation: At any stage in the process of making 2-variable CSP arc-consistent in AC3b:
- there is a set $S_i^+ \subseteq D_i$ whose values are all known to be supported by $X_j$;
- there is a set $S_i^? = D_i \setminus S_i^+$ whose values are unknown, as yet, to be supported by $X_j$.
The same holds if the roles for $X_i$ and $X_j$ are exchanged.
In order to establish support for a value $v_i^? \in S_i^?$ it seems better to try to find a support among the values in $S_j^?$ first, because for each $v_j^? \in S_j^?$ the check $(v_i^?,v_j^?) \in C_{ij}$ is a double-support check and it is just as likely that any $v_j^? \in S_j^?$ supports $v_i^?$ than it is that any $v_j^+ \in S_j^+$ does. Only if no support can be found among the elements in $S_j^?$, should the elements $v_j^+$ in $S_j^+$ be used for single-support checks $(v_i^?,v_j^+) \in C_{ij}$. After it has been decided for each value in $D_i$ whether it is supported or not, either $S_x^+ = \emptyset$ and the 2-variable CSP is inconsistent, or $S_x^+ \neq \emptyset$ and the CSP is satisfiable. In the latter case, the elements from $D_i$ which are supported by $j$ are given by $S_x^+$. The elements in $D_j$ which are supported by $x$ are given by the union of $S_j^+$ with the set of those elements of $S_j^?$ which further processing will show to be supported by some $v_i^+ \in S_x^+$.
End of explanation
%psource dom_j_up
%psource sat_up
Explanation: AC3b is a refinement of the AC3 algorithm which consists of the fact that if, when arc $(i,j)$ is being processed and the reverse arc $(j,i)$ is also in the queue, then consistency-checks can be saved because only support for the elements in $S_j^?$ has to be found (as opposed to support for all the elements in $D_j$ in the
AC3 algorithm). <br>
AC3b inherits all its properties like $\mathcal{O}(ed^3)$ time-complexity and $\mathcal{O}(e + nd)$ space-complexity fron AC3 and where $n$ denotes the number of variables in the CSP, $e$ denotes the number of binary constraints and $d$ denotes the maximum domain-size of the variables.
Arc-Heuristics for Arc-Consistency Algorithms
Many arc-heuristics can be devised, based on three major features of CSPs:
- the number of acceptable pairs in each constraint (the constraint size or satisfiability);
- the domain size;
- the number of binary constraints that each variable participates in, equal to the degree of the node of that variable in the constraint graph. 
Simple examples of heuristics that might be expected to improve the efficiency of relaxation are:
- ordering the list of variable pairs by increasing relative satisfiability;
- ordering by increasing size of the domain of the variable $v_j$ relaxed against $v_i$;
- ordering by descending degree of node of the variable relaxed.
In <a name="ref-3"/>[3] are investigated the effects of these arc-heuristics in an empirical way, experimenting the effects of them on random CSPs. Their results demonstrate that the first two, later called sat up and dom j up for n-ary and binary CSPs respectively, significantly reduce the number of consistency-checks.
End of explanation
%psource AC4
Explanation: Experimental Results
For the experiments below on binary CSPs, in addition to the two arc-consistency algorithms already cited above, AC3 and AC3b, the AC4 algorithm was used. <br>
The AC4 algorithm runs in $\mathcal{O}(ed^2)$ worst-case time but can be slower than AC3 on average cases.
End of explanation
sudoku = Sudoku(easy1)
sudoku.display(sudoku.infer_assignment())
%time _, checks = AC3(sudoku, arc_heuristic=no_arc_heuristic)
f'AC3 needs {checks} consistency-checks'
sudoku = Sudoku(easy1)
%time _, checks = AC3b(sudoku, arc_heuristic=no_arc_heuristic)
f'AC3b needs {checks} consistency-checks'
sudoku = Sudoku(easy1)
%time _, checks = AC4(sudoku, arc_heuristic=no_arc_heuristic)
f'AC4 needs {checks} consistency-checks'
sudoku = Sudoku(easy1)
%time _, checks = AC3(sudoku, arc_heuristic=dom_j_up)
f'AC3 with DOM J UP arc heuristic needs {checks} consistency-checks'
sudoku = Sudoku(easy1)
%time _, checks = AC3b(sudoku, arc_heuristic=dom_j_up)
f'AC3b with DOM J UP arc heuristic needs {checks} consistency-checks'
sudoku = Sudoku(easy1)
%time _, checks = AC4(sudoku, arc_heuristic=dom_j_up)
f'AC4 with DOM J UP arc heuristic needs {checks} consistency-checks'
backtracking_search(sudoku, select_unassigned_variable=mrv, inference=forward_checking)
sudoku.display(sudoku.infer_assignment())
Explanation: Sudoku
Easy Sudoku
End of explanation
sudoku = Sudoku(harder1)
sudoku.display(sudoku.infer_assignment())
%time _, checks = AC3(sudoku, arc_heuristic=no_arc_heuristic)
f'AC3 needs {checks} consistency-checks'
sudoku = Sudoku(harder1)
%time _, checks = AC3b(sudoku, arc_heuristic=no_arc_heuristic)
f'AC3b needs {checks} consistency-checks'
sudoku = Sudoku(harder1)
%time _, checks = AC4(sudoku, arc_heuristic=no_arc_heuristic)
f'AC4 needs {checks} consistency-checks'
sudoku = Sudoku(harder1)
%time _, checks = AC3(sudoku, arc_heuristic=dom_j_up)
f'AC3 with DOM J UP arc heuristic needs {checks} consistency-checks'
sudoku = Sudoku(harder1)
%time _, checks = AC3b(sudoku, arc_heuristic=dom_j_up)
f'AC3b with DOM J UP arc heuristic needs {checks} consistency-checks'
sudoku = Sudoku(harder1)
%time _, checks = AC4(sudoku, arc_heuristic=dom_j_up)
f'AC4 with DOM J UP arc heuristic needs {checks} consistency-checks'
backtracking_search(sudoku, select_unassigned_variable=mrv, inference=forward_checking)
sudoku.display(sudoku.infer_assignment())
Explanation: Harder Sudoku
End of explanation
chess = NQueensCSP(8)
chess.display(chess.infer_assignment())
%time _, checks = AC3(chess, arc_heuristic=no_arc_heuristic)
f'AC3 needs {checks} consistency-checks'
chess = NQueensCSP(8)
%time _, checks = AC3b(chess, arc_heuristic=no_arc_heuristic)
f'AC3b needs {checks} consistency-checks'
chess = NQueensCSP(8)
%time _, checks = AC4(chess, arc_heuristic=no_arc_heuristic)
f'AC4 needs {checks} consistency-checks'
chess = NQueensCSP(8)
%time _, checks = AC3(chess, arc_heuristic=dom_j_up)
f'AC3 with DOM J UP arc heuristic needs {checks} consistency-checks'
chess = NQueensCSP(8)
%time _, checks = AC3b(chess, arc_heuristic=dom_j_up)
f'AC3b with DOM J UP arc heuristic needs {checks} consistency-checks'
chess = NQueensCSP(8)
%time _, checks = AC4(chess, arc_heuristic=dom_j_up)
f'AC4 with DOM J UP arc heuristic needs {checks} consistency-checks'
backtracking_search(chess, select_unassigned_variable=mrv, inference=forward_checking)
chess.display(chess.infer_assignment())
Explanation: 8 Queens
End of explanation
%psource ACSolver.GAC
Explanation: For the experiments below on n-ary CSPs, due to the n-ary constraints, the GAC algorithm was used. <br>
The GAC algorithm has $\mathcal{O}(er^2d^t)$ time-complexity and $\mathcal{O}(erd)$ space-complexity where $e$ denotes the number of n-ary constraints, $r$ denotes the constraint arity and $d$ denotes the maximum domain-size of the variables.
End of explanation
crossword = Crossword(crossword1, words1)
crossword.display()
words1
%time _, _, checks = ACSolver(crossword).GAC(arc_heuristic=no_heuristic)
f'GAC needs {checks} consistency-checks'
crossword = Crossword(crossword1, words1)
%time _, _, checks = ACSolver(crossword).GAC(arc_heuristic=sat_up)
f'GAC with SAT UP arc heuristic needs {checks} consistency-checks'
crossword.display(ACSolver(crossword).domain_splitting())
Explanation: Crossword
End of explanation
kakuro = Kakuro(kakuro2)
kakuro.display()
%time _, _, checks = ACSolver(kakuro).GAC(arc_heuristic=no_heuristic)
f'GAC needs {checks} consistency-checks'
kakuro = Kakuro(kakuro2)
%time _, _, checks = ACSolver(kakuro).GAC(arc_heuristic=sat_up)
f'GAC with SAT UP arc heuristic needs {checks} consistency-checks'
kakuro.display(ACSolver(kakuro).domain_splitting())
Explanation: Kakuro
Easy Kakuro
End of explanation
kakuro = Kakuro(kakuro3)
kakuro.display()
%time _, _, checks = ACSolver(kakuro).GAC(arc_heuristic=no_heuristic)
f'GAC needs {checks} consistency-checks'
kakuro = Kakuro(kakuro3)
%time _, _, checks = ACSolver(kakuro).GAC(arc_heuristic=sat_up)
f'GAC with SAT UP arc heuristic needs {checks} consistency-checks'
kakuro.display(ACSolver(kakuro).domain_splitting())
Explanation: Medium Kakuro
End of explanation
kakuro = Kakuro(kakuro4)
kakuro.display()
%time _, _, checks = ACSolver(kakuro).GAC()
f'GAC needs {checks} consistency-checks'
kakuro = Kakuro(kakuro4)
%time _, _, checks = ACSolver(kakuro).GAC(arc_heuristic=sat_up)
f'GAC with SAT UP arc heuristic needs {checks} consistency-checks'
kakuro.display(ACSolver(kakuro).domain_splitting())
Explanation: Harder Kakuro
End of explanation
cryptarithmetic = NaryCSP(
    {'S': set(range(1, 10)), 'M': set(range(1, 10)),
     'E': set(range(0, 10)), 'N': set(range(0, 10)), 'D': set(range(0, 10)),
     'O': set(range(0, 10)), 'R': set(range(0, 10)), 'Y': set(range(0, 10)),
     'C1': set(range(0, 2)), 'C2': set(range(0, 2)), 'C3': set(range(0, 2)),
     'C4': set(range(0, 2))},
    [Constraint(('S', 'E', 'N', 'D', 'M', 'O', 'R', 'Y'), all_diff),
     Constraint(('D', 'E', 'Y', 'C1'), lambda d, e, y, c1: d + e == y + 10 * c1),
     Constraint(('N', 'R', 'E', 'C1', 'C2'), lambda n, r, e, c1, c2: c1 + n + r == e + 10 * c2),
     Constraint(('E', 'O', 'N', 'C2', 'C3'), lambda e, o, n, c2, c3: c2 + e + o == n + 10 * c3),
     Constraint(('S', 'M', 'O', 'C3', 'C4'), lambda s, m, o, c3, c4: c3 + s + m == o + 10 * c4),
     Constraint(('M', 'C4'), eq)])
%time _, _, checks = ACSolver(cryptarithmetic).GAC(arc_heuristic=no_heuristic)
f'GAC needs {checks} consistency-checks'
%time _, _, checks = ACSolver(cryptarithmetic).GAC(arc_heuristic=sat_up)
f'GAC with SAT UP arc heuristic needs {checks} consistency-checks'
assignment = ACSolver(cryptarithmetic).domain_splitting()
from IPython.display import Latex
display(Latex(r'\begin{array}{@{}r@{}} ' + '{}{}{}{}'.format(assignment['S'], assignment['E'], assignment['N'], assignment['D']) + r' \\ + ' + 
              '{}{}{}{}'.format(assignment['M'], assignment['O'], assignment['R'], assignment['E']) + r' \\ \hline ' + 
              '{}{}{}{}{}'.format(assignment['M'], assignment['O'], assignment['N'], assignment['E'], assignment['Y']) + ' \end{array}'))
Explanation: Cryptarithmetic Puzzle
$$
\begin{array}{@{}r@{}}
     S E N D \
{} + M O R E \
   \hline
   M O N E Y
\end{array}
$$
End of explanation
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