Homework 4. KenKen solver Solution

Motivation
KenKen is an arithmetical-logical puzzle whose goal is to fill in an N×N grid with integers, so that every row and every column contains all the integers from 1 through N, and so that certain extra constraints can be met. These extra constraints are specified by cages that say that 1 or more contiguous cells must add up to a certain value, or must yield a certain value when multiplied; or that two cells must yield a certain value when divided or subtracted. For example:
 Image credit
A human doing this puzzle can reason it out with arguments like the following. The "11+" at upper left must contain a 5 and a 6, as must the "30×" at medium right. Therefore, there cannot be another 5 in either column 1 or row 4. The "240×" at medium left must contain a 5 somewhere, and since it can't be in either column 1 or row 4, the entry in row 3, column 2 must be where the 5 is; we have now deduced an entry. For fun, you might try filling out the rest of the puzzle, using similar reasoning.
A computer solving this problem doesn't need to be that smart. It can rely on a small list of solution strategies rather than the ad hoc approach taken by humans.
Assignment
Three things. First, write a predicate kenken/3 that accepts the following arguments:
    1. N, a nonnegative integer specifying the number of cells on each side of the KenKen square.
    2. C, a list of numeric cage constraints as described below.
    3. T, a list of list of integers. All the lists have length N. This represents the N×N grid.
Each constraint in C is of the following form:
+(S, L)
means the integer S is the sum of integers in the list L of squares.
*(P, L)
means the integer P is the product of the integers in the list L of squares.
−(D, J, K)
means the integer D is the difference between the integer j in square J and the integer k in square K; D could be equal to either j−k or to k−j.
/(Q, J, K)
means the integer Q is the quotient of the integer j in square J and the integer k in square K; Q could be equal to either j÷k or to k÷j. The remainder must be zero.
In the above description, a square is a term [i|j] where i and j are row and column indexes in the range 1 through N, inclusive. The indexes identify the square in the KenKen diagram that is affected by the constraint in question. The bottom left square is [N|1] and the top right square is [1|N].
Preconditions. N and C must be ground terms, that is, they cannot be logical variables or terms containing any logical variables. N must also be a nonnegative integer. The first argument to each constraint, as described below, must be a nonnegative integer that is less than the current value of vector_max in the GNU Prolog finite domain solver. Your code need not check these preconditions; it can assume that the the preconditions hold.
T may contain logical variables, which can represent integers that kenken/3 should fill in, or can represent entire rows, or the entire grid.
Second, write a predicate plain_kenken/3 that acts like kenken/3 but does not use the GNU Prolog finite domain solver. Instead, plain_kenken/3 should enumerate the possible integer solutions using standard Prolog primitives such as member/2 and is/2. Although plain_kenken/3 should be simpler than kenken/3 and should not be restricted to integers less than vector_max, the tradeoff is that it may have worse performance. Illustrate the performance difference on an example of your choice, measuring performance with statistics/0 or statistics/2.
Third, consider the problem of solving no-op Kenken. No-op Kenken is like Kenken, but the puzzle's cages lack operations, and contain only target numbers. The puzzle solver must deduce the operations, and the choice of operations is part of the solution. This set of five no-op KenKen puzzles can help you get started.
Suppose you want to solve no-op KenKen using GNU Prolog. Design a good application programming interface for your solver, i.e., specify what Prolog terms should be passed to your solver and what the caller should expect should happen to those terms after successful and unsuccessful calls. Also, give an example call, and its behavior, using the style shown below. You do not need to implement the solver, only specify its API and implement a predicate noop_kenken_testcase that performs your example call.
Examples
As a trivial example, kenken(1,[],T) should generate just one answer, T=[[1]]. Slightly less trivially, kenken(2,[],T) should generate two answers, T = [[1,2],[2,1]] and T = [[2,1],[1,2]]. With no constraints, the number of answers grows rapidly with N: for example, kenken(3,[],T) should generate 12 answers, and kenken(4,[],T) should generate 576 answers.
Real KenKen puzzles use constraints to narrow the number of solutions to one. Suppose you have the following fact; it represents the above KenKen diagram.
kenken_testcase(
  6,
  [
   +(11, [[1|1], [2|1]]),
   /(2, [1|2], [1|3]),
   *(20, [[1|4], [2|4]]),
   *(6, [[1|5], [1|6], [2|6], [3|6]]),
   -(3, [2|2], [2|3]),
   /(3, [2|5], [3|5]),
   *(240, [[3|1], [3|2], [4|1], [4|2]]),
   *(6, [[3|3], [3|4]]),
   *(6, [[4|3], [5|3]]),
   +(7, [[4|4], [5|4], [5|5]]),
   *(30, [[4|5], [4|6]]),
   *(6, [[5|1], [5|2]]),
   +(9, [[5|6], [6|6]]),
   +(8, [[6|1], [6|2], [6|3]]),
   /(2, [6|4], [6|5])
  ]
).
Then, the query:
?- fd_set_vector_max(255), kenken_testcase(N,C), kenken(N,C,T).
should output this (reindented to fit):
C = [11+[[1|1],[2|1]], /(2,[1|2],[1|3]), 20*[[1|4],[2|4]],
     6*[[1|5],[1|6],[2|6],[3|6]], -(3,[2|2],[2|3]), /(3,[2|5],[3|5]),
     240*[[3|1],[3|2],[4|1],[4|2]], 6*[[3|3],[3|4]], 6*[[4|3],[5|3]],
     7+[[4|4],[5|4],[5|5]], 30*[[4|5],[4|6]], 6*[[5|1],[5|2]],
     9+[[5|6],[6|6]], 8+[[6|1],[6|2],[6|3]], /(2,[6|4],[6|5])]
N = 6
T = [[5,6,3,4,1,2],
     [6,1,4,5,2,3],
     [4,5,2,3,6,1],
     [3,4,1,2,5,6],
     [2,3,6,1,4,5],
     [1,2,5,6,3,4]] ?
and if you respond with a ";" the next result should be "no".
Here's another example, of a puzzle that is not valid for strict KenKen because it has multiple solutions. Your solver should be able to generate all the solutions:
kenken(
  4,
  [
   +(6, [[1|1], [1|2], [2|1]]),
   *(96, [[1|3], [1|4], [2|2], [2|3], [2|4]]),
   -(1, [3|1], [3|2]),
   -(1, [4|1], [4|2]),
   +(8, [[3|3], [4|3], [4|4]]),
   *(2, [[3|4]])
  ],
  T
), write(T), nl, fail.
This should output:
[[1,2,3,4],[3,4,2,1],[4,3,1,2],[2,1,4,3]]
[[1,2,4,3],[3,4,2,1],[4,3,1,2],[2,1,3,4]]
[[2,1,3,4],[3,4,2,1],[4,3,1,2],[1,2,4,3]]
[[2,1,4,3],[3,4,2,1],[4,3,1,2],[1,2,3,4]]
[[3,1,2,4],[2,4,3,1],[4,3,1,2],[1,2,4,3]]
[[3,2,4,1],[1,4,2,3],[4,3,1,2],[2,1,3,4]]

no
Submit
Submit a file named kenken.pl containing all the requested code. If any extra text information is needed, other than what's in the comments, please submit it as a separate text file.