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//
// Copyright 2012 Hakan Kjellerstrand
//
// Licensed under the Apache License, Version 2.0 (the "License");
// you may not use this file except in compliance with the License.
// You may obtain a copy of the License at
//
// http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.
using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using Google.OrTools.ConstraintSolver;
public class WeddingOptimalChart
{
/**
*
* Finding an optimal wedding seating chart.
*
* From
* Meghan L. Bellows and J. D. Luc Peterson
* "Finding an optimal seating chart for a wedding"
* http://www.improbable.com/news/2012/Optimal-seating-chart.pdf
* http://www.improbable.com/2012/02/12/finding-an-optimal-seating-chart-for-a-wedding
*
* """
* Every year, millions of brides (not to mention their mothers, future
* mothers-in-law, and occasionally grooms) struggle with one of the
* most daunting tasks during the wedding-planning process: the
* seating chart. The guest responses are in, banquet hall is booked,
* menu choices have been made. You think the hard parts are over,
* but you have yet to embark upon the biggest headache of them all.
* In order to make this process easier, we present a mathematical
* formulation that models the seating chart problem. This model can
* be solved to find the optimal arrangement of guests at tables.
* At the very least, it can provide a starting point and hopefully
* minimize stress and arguments…
* """
*
*
* Also see http://www.hakank.org/minizinc/wedding_optimal_chart.mzn
*
*/
private static void Solve()
{
Solver solver = new Solver("WeddingOptimalChart");
//
// Data
//
// Easy problem (from the paper)
int n = 2; // number of tables
int a = 10; // maximum number of guests a table can seat
int b = 1; // minimum number of people each guest knows at their table
/*
// Sligthly harder problem (also from the paper)
int n = 5; // number of tables
int a = 4; // maximum number of guests a table can seat
int b = 1; // minimum number of people each guest knows at their table
*/
// j Guest Relation
// -------------------------------------
// 1 Deb mother of the bride
// 2 John father of the bride
// 3 Martha sister of the bride
// 4 Travis boyfriend of Martha
// 5 Allan grandfather of the bride
// 6 Lois wife of Allan
// 7 Jayne aunt of the bride
// 8 Brad uncle of the bride
// 9 Abby cousin of the bride
// 10 Mary Helen mother of the groom
// 11 Lee father of the groom
// 12 Annika sister of the groom
// 13 Carl brother of the groom
// 14 Colin brother of the groom
// 15 Shirley grandmother of the groom
// 16 DeAnn aunt of the groom
// 17 Lori aunt of the groom
// Connection matrix: who knows who, and how strong
// is the relation
int[,] C = {
{ 1,50, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0},
{50, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0},
{ 1, 1, 1,50, 1, 1, 1, 1,10, 0, 0, 0, 0, 0, 0, 0, 0},
{ 1, 1,50, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0},
{ 1, 1, 1, 1, 1,50, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0},
{ 1, 1, 1, 1,50, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0},
{ 1, 1, 1, 1, 1, 1, 1,50, 1, 0, 0, 0, 0, 0, 0, 0, 0},
{ 1, 1, 1, 1, 1, 1,50, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0},
{ 1, 1,10, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0},
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,50, 1, 1, 1, 1, 1, 1},
{ 0, 0, 0, 0, 0, 0, 0, 0, 0,50, 1, 1, 1, 1, 1, 1, 1},
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1},
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1},
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1},
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1},
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1},
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1}
};
// Names of the guests. B: Bride side, G: Groom side
String[] names = {"Deb (B)",
"John (B)",
"Martha (B)",
"Travis (B)",
"Allan (B)",
"Lois (B)",
"Jayne (B)",
"Brad (B)",
"Abby (B)",
"Mary Helen (G)",
"Lee (G)",
"Annika (G)",
"Carl (G)",
"Colin (G)",
"Shirley (G)",
"DeAnn (G)",
"Lori (G)"};
int m = C.GetLength(0); // number of quests
IEnumerable<int> NRANGE = Enumerable.Range(0, n);
IEnumerable<int> MRANGE = Enumerable.Range(0, m);
//
// Decision variables
//
IntVar[] tables = solver.MakeIntVarArray(m, 0, n-1, "tables");
IntVar z = (from j in MRANGE
from k in MRANGE
where j < k
select C[j,k] * tables[j] == tables[k]
).ToArray().Sum().VarWithName("z");
//
// Constraints
//
foreach(int i in NRANGE) {
solver.Add((from j in MRANGE
from k in MRANGE
where j < k && C[j, k] > 0
select (tables[j] == i) * (tables[k] == i)
).ToArray().Sum() >= b);
solver.Add((from j in MRANGE select tables[j] == i).ToArray().Sum() <= a);
}
// Symmetry breaking
solver.Add(tables[0] == 0);
//
// Objective
//
OptimizeVar obj = z.Maximize(1);
//
// Search
//
DecisionBuilder db = solver.MakePhase(tables,
Solver.INT_VAR_DEFAULT,
Solver.INT_VALUE_DEFAULT);
solver.NewSearch(db, obj);
while (solver.NextSolution()) {
Console.WriteLine("z: {0}",z.Value());
Console.Write("Table: ");
foreach(int j in MRANGE) {
Console.Write(tables[j].Value() + " ");
}
Console.WriteLine();
foreach(int i in NRANGE) {
Console.Write("Table {0}: ", i);
foreach(int j in MRANGE) {
if (tables[j].Value() == i) {
Console.Write(names[j] + " ");
}
}
Console.WriteLine();
}
Console.WriteLine();
}
Console.WriteLine("\nSolutions: {0}", solver.Solutions());
Console.WriteLine("WallTime: {0}ms", solver.WallTime());
Console.WriteLine("Failures: {0}", solver.Failures());
Console.WriteLine("Branches: {0} ", solver.Branches());
solver.EndSearch();
}
public static void Main(String[] args)
{
Solve();
}
}