From 52c34f2cdcc2543cd864694e94c95a9b5e3f9410 Mon Sep 17 00:00:00 2001
From: Hakan Kjellerstrand <hakank@gmail.com>
Date: Sun, 13 Jul 2014 09:40:36 +0200
Subject: [PATCH] MiniZinc: added kidney_exchange_model.mzn
 self_referential_sentence.mzn

---
 minizinc/index.html                    |   4 +-
 minizinc/kidney_exchange_model.mzn     | 109 +++++++++++++++++++++++++
 minizinc/self_referential_sentence.mzn |  50 ++++++++++++
 3 files changed, 162 insertions(+), 1 deletion(-)
 create mode 100644 minizinc/kidney_exchange_model.mzn
 create mode 100644 minizinc/self_referential_sentence.mzn

diff --git a/minizinc/index.html b/minizinc/index.html
index f45c870e..46a5cede 100644
--- a/minizinc/index.html
+++ b/minizinc/index.html
@@ -631,6 +631,7 @@ <h3>Puzzles, small, and large</h3>
 <li> <a href="secret_santa.mzn">secret_santa.mzn</a>: Secret Santa problem
 <li> <a href="secret_santa2.mzn">secret_santa2.mzn</a>: Another Secret Santa problem
 <li> <a href="self_referential_quiz.mzn">self_referential_quiz.mzn</a>: Self-referential quiz
+<li> <a href="self_referential_sentence.mzn">self_referential_sentence.mzn</a>: Self-referential sentence
 <li> <a href="send_more_money.mzn">send_more_money.mzn</a>: Alphametic puzzle: SEND + MORE = MONEY
 <li> <a href="send_more_money2.mzn">send_more_money2.mzn</a>: Alphametic puzzle: SEND + MORE = MONEY, alternative model
 <li> <a href="send_more_money_any_base.mzn">send_more_money_any_base.mzn</a>: Alphametic puzzle: SEND + MORE = MONEY, in any base
@@ -1038,7 +1039,8 @@ <h3>Combinatorial problems</h3>
 <li> <a href="k_consecutive_integers.mzn">k_consecutive_integers.mzn</a>: k consecutive integers (Stack Overflow: <a href=" http://stackoverflow.com/questions/18550915/k-consecutive-integers-constraint">k consecutive integers constraint</a>)
 <li> <a href="kidney_exchange.mzn">kidney_exchange.mzn</a>: Simple kidney exchange model (inspired by Pascal Van Hentenryck's introduction to the Coursera Course "Discrete Optimization")
 <a name="kidney_exchange">
-<li> <a href="kidney_exchange2.mzn">kidney_exchange2.mzn</a>: Kidney exchange model (for .dzn files)<br>
+<li> <a href="kidney_exchange2.mzn">kidney_exchange2.mzn</a>: Kidney exchange model<br>
+     Faster version: <a href="kidney_exchange_model.mzn">kidney_exchange_model.mzn</a>:<br>
    Data files:
    <ul>
      <li> <a href="kidney_exchange_pascal.dzn">kidney_exchange_pascal.dzn</a>
diff --git a/minizinc/kidney_exchange_model.mzn b/minizinc/kidney_exchange_model.mzn
new file mode 100644
index 00000000..a580501b
--- /dev/null
+++ b/minizinc/kidney_exchange_model.mzn
@@ -0,0 +1,109 @@
+% 
+% Kidney exchange in MiniZinc.
+% 
+% A simple model of kidney exchange, inspired by Pascal Van Hentenryck's
+% introduction of the Coursera Course Discrete Optimization.
+%
+% The objective is to maximize the number of kidney exchanges given
+% the compatible 
+%
+% Note that we are looking for cycles, which means that a person receiving
+% a kidney must be able to give a kidney.
+%
+
+%
+% This MiniZinc model was created by Hakan Kjellerstrand, hakank@gmail.com
+% See also my MiniZinc page: http://www.hakank.org/minizinc/
+%
+
+include "globals.mzn"; 
+
+int: num_people;
+array[1..num_people] of set of int: compatible;
+
+% people that has no no potential donors (and can't get a kidney)
+set of 1..num_people: non_compatible = { p | p in 1..num_people where card(compatible[p]) = 0 };
+
+% The domains for each person
+array[1..num_people] of set of int: 
+   compatible_pruned = [ 
+            if card(compatible[p]) = 0 then {p} else (compatible[p] diff non_compatible) union {p} endif
+   | p in 1..num_people ];
+
+%
+% decision variables
+%
+
+% which kidney does person p get (or p if he/she gets no kidney)
+array[1..num_people] of var 1..num_people: x;
+
+var 0..num_people: z = sum([bool2int(x[i] != i) | i in 1..num_people]);
+
+% solve satisfy;
+% solve maximize z;
+solve :: int_search(
+      x,
+      first_fail, 
+      indomain_median,
+      complete) 
+    maximize z;
+
+% Just allow the compatible people (+ p) in the domains
+% and remove uncompatible people.
+constraint
+  forall(p in 1..num_people) (
+     x[p] in compatible_pruned[p]    
+  )
+  /\
+  alldifferent(x)
+
+  % experimental: first test if all are donors (-> circuit)
+  % 
+  % works in MiniZinc v2.0, at least for some solvers.
+  /\ (circuit(x) \/ true)
+;
+
+output [
+  "z: " ++ show(z) ++ "\n" ++
+  "x: " ++ show(x) ++ "\n" 
+]
+++
+[
+  "person: donor\n"
+]
+++
+[
+  if fix(x[i] = i) then 
+     show_int(3, i) ++ ":   -\n"
+  else 
+     show_int(3, i) ++ ": " ++  show_int(3, x[i]) ++ "\n"
+  endif
+  | i in 1..num_people
+]
+++
+[
+  show(x) ++ "\n"
+  ++ "z: " ++ show(z) ++ "\n" 
+];
+
+
+%
+% data
+%
+
+% The compatibility matrix 
+% (from Pascal's introduction lecture)
+% who can give a kidney to person p
+% This is a directed graph
+% num_people = 8;
+% compatible = 
+%   [
+%     {2,3}, % 1
+%     {1,6}, % 2
+%     {1,4,7}, % 3
+%     {2}, % 4  
+%     {2}, % 5
+%     {5}, % 6
+%     {8}, % 7
+%     {3}, % 8
+%   ];
diff --git a/minizinc/self_referential_sentence.mzn b/minizinc/self_referential_sentence.mzn
new file mode 100644
index 00000000..64b68f4b
--- /dev/null
+++ b/minizinc/self_referential_sentence.mzn
@@ -0,0 +1,50 @@
+% 
+% Self referential sentence in MiniZinc.
+% 
+% E.g. from
+%  http://puzzling.stackexchange.com/questions/1901/write-a-true-self-reflective-statement-about-the-digits-from-0-to-9-or-prove-it
+%  
+%  "This statement contains 1 0's, 7 1's, 3 2's, 2 3's, 1 4's, 1 5's, 1 6's, 2 7's, 1 8's, and 1 9's."
+%
+%  Note: It happens that the number of occurrences of the digits are between 1..9 
+%        so we don't have to worry about converting 10 to 1,0 etc.
+
+% 
+% This MiniZinc model was created by Hakan Kjellerstrand, hakank@gmail.com
+% See also my MiniZinc page: http://www.hakank.org/minizinc/
+%
+
+include "globals.mzn"; 
+
+int: b = 9; % base
+int: n = 2*(b+1);
+
+% decision variables
+array[1..n] of var 0..b: x;
+
+
+solve satisfy;
+% solve :: int_search(x, first_fail, indomain_min, complete) satisfy;
+
+constraint
+  % The digits 0..b are placed in positions x[i*2+1].
+  % Ensure that it is x[i*2+2] occurrences of the digit i
+  forall(i in 0..b) (
+    x[i*2+1] = i 
+    /\
+    count(x,i,x[i*2+2])
+  )
+;
+
+output 
+[
+  "x: ", show(x), "\n",
+  "This statement contains " 
+]
+++
+[
+ show(x[i*2+2]) ++ " " ++ show(x[i*2+1]) ++ "'s, " ++  
+      if i == b-1 then "and " else "" endif
+  | i in 0..b
+];
+