# Transportation problem

Transportation problem is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

The transportation problem in linear programming is to find the optimal transportation plan for certain volumes of resources from suppliers to consumers, taking into account the cost of transportation. The plan is a table (matrix), whose rows and columns correspond to the suppliers and consumers, the cells are placed in cargo volume.

Example of the transportation problem:

Consumer 1,
need 20 kg
Consumer 2,
need 30 kg
Consumer 3,
need 10 kg
Supplier 1,
supply 25 kg
\$3 per kg \$5 per kg \$7 per kg
Supplier 2,
supply 35 kg
\$3 per kg \$2 per kg \$5 per kg

The object is to solve the classical transport problem using the method of potentials (with redistributive cycle) with the preparation of the initial transportation plan by the north-west corner of the features to be implemented in this task. The input is the number of suppliers and customers, inventory levels, needs and cost matrix transport cargo. The output of the program is the optimal plan. If necessary, enter a fictitious vendor or customer.

The solution for the above example would be the plan:

Consumer 1 Consumer 2 Consumer 3
Supplier 1 20 kg - 5 kg
Supplier 2 - 30 kg 5 kg

## 1C

```// based on the program of <romix>

перем m,n; // Table size
перем u,v;
перем БазисныеЯчейки;
перем iЦикл, jЦикл;
перем Цены, Спрос, Предложение, Отгрузки; // Arrays of the transportation problem
перем i1, j1;
перем СпросОстаток, ПредложениеОстаток;
перем гл_сч;
перем гсч;

Функция РаспределениеМетодомСевероЗападногоУгла()
Для j=1 по n Цикл
СпросОстаток[j]=Спрос[j];
КонецЦикла;
Для i=1 по m Цикл
ПредложениеОстаток[i]=Предложение[i];
КонецЦикла;
Для i=1 по m Цикл
Для j=1 по n Цикл
БазисныеЯчейки[i][j]=0;
Отгрузки[i][j]=0;
КонецЦикла;
КонецЦикла;
Для i=1 по m Цикл
Для j=1 по n Цикл
Если ПредложениеОстаток[i]=0 Тогда
Прервать;
ИначеЕсли ПредложениеОстаток[i]<0 Тогда
ВызватьИсключение("Error: balance of the offer less than 0");
КонецЕсли;
чОбъем=СпросОстаток[j];
Если чОбъем=0 Тогда
Продолжить;
ИначеЕсли чОбъем<0 Тогда
ВызватьИсключение("Error: balance of the demand less than 0");
КонецЕсли;
Если ПредложениеОстаток[i]<чОбъем Тогда
чОбъем=ПредложениеОстаток[i];
КонецЕсли;
СпросОстаток[j]=СпросОстаток[j]-чОбъем;
ПредложениеОстаток[i]=ПредложениеОстаток[i]-чОбъем;
БазисныеЯчейки[i][j]=1;
Отгрузки[i][j]=чОбъем;
КонецЦикла;
КонецЦикла;
КонецФункции

Функция ПроверкаПравильностиОтгрузок()
Для i=1 по m Цикл
стр="Отгрузки: ";
Для j=1 по n Цикл
стр=стр+Отгрузки[i][j]+" ";
КонецЦикла;
Сообщить(стр);
КонецЦикла;
Для i=1 по m Цикл
чОбъем=0;
Для j=1 по n Цикл
чОбъем=чОбъем+Отгрузки[i][j];
КонецЦикла;
Если чОбъем<>Предложение[i] Тогда
ВызватьИсключение("Error: shipment on the line does not equal the proposal in the row "+i);
КонецЕсли;
КонецЦикла;
Для j=1 по n Цикл
чОбъем=0;
Для i=1 по m Цикл
чОбъем=чОбъем+Отгрузки[i][j];
КонецЦикла;
Если чОбъем<>Спрос[j] Тогда
ВызватьИсключение("Error: shipment by the column does not equal to the demand in the column "+j);
КонецЕсли;
КонецЦикла;
Возврат Истина;
КонецФункции

Функция ВычислениеПотенциалов()
перем i, j;
Для i=1 по m Цикл
u[i]=НеОпределено;
КонецЦикла;
Для j=1 по n Цикл
v[j]=НеОпределено;
КонецЦикла;
u[1]=0;
гл_сч=m*n;
ВычислениеПотенциаловПоГоризонтали(1);
Для i=1 по m Цикл
Если u[i]=НеОпределено Тогда
Сообщить("Failed to evaluate the potential u["+i+"]");
Возврат Ложь;
КонецЕсли;
КонецЦикла;
Для j=1 по n Цикл
Если v[j]=НеОпределено Тогда
Сообщить("Failed to evaluate the potential v["+j+"]");
Возврат Ложь;
КонецЕсли;
КонецЦикла;
Возврат Истина;
КонецФункции

Функция ВычислениеПотенциаловПоВертикали(j)
Если v[j]=НеОпределено Тогда
ВызватьИсключение("Failed to get the potential v["+j+"]");
КонецЕсли;
Для i=1 по m Цикл
Если БазисныеЯчейки[i][j]=0 Тогда
Продолжить;
КонецЕсли;
Если u[i]<>НеОпределено Тогда
Продолжить;
Иначе
u[i]=Цены[i][j]-v[j];
ВычислениеПотенциаловПоГоризонтали(i);
КонецЕсли;
КонецЦикла;
КонецФункции

Функция ВычислениеПотенциаловПоГоризонтали(i)
гл_сч=гл_сч-1;
Если гл_сч=0 Тогда
ВызватьИсключение("Looping in the calculation of potential");
КонецЕсли;
Если u[i]=НеОпределено Тогда
ВызватьИсключение("Failed to get potential u["+i+"]");
КонецЕсли;
Для j=1 по n Цикл
Если БазисныеЯчейки[i][j]=0 Тогда
Продолжить;
КонецЕсли;
Если v[j]<>НеОпределено Тогда
Продолжить;
Иначе
v[j]=Цены[i][j]-u[i];
ВычислениеПотенциаловПоВертикали(j);
КонецЕсли;
КонецЦикла;
КонецФункции

Функция ПроверкаОптимальности()
перем чРешениеОптимально, чМинимальнаяДельта, i, j, Дельта;
чРешениеОптимально=Истина;
чМинимальнаяДельта=НеОпределено;
Для i=1 по m Цикл
стр="Дельта=";
Для j=1 по n Цикл
Если БазисныеЯчейки[i][j]=1 Тогда
Дельта=0;
Иначе
Дельта = Цены[i][j]-u[i]-v[j];
КонецЕсли;
стр=стр+Дельта+" ";
Если Дельта<0 Тогда
чРешениеОптимально=Ложь;
КонецЕсли;
Если чМинимальнаяДельта=НеОпределено Тогда
чМинимальнаяДельта=Дельта;
i1=i;
j1=j;
Иначе
Если Дельта<чМинимальнаяДельта Тогда
чМинимальнаяДельта=Дельта;
i1=i;
j1=j;
КонецЕсли;
КонецЕсли;
КонецЦикла;
КонецЦикла;
Возврат чРешениеОптимально;
КонецФункции

Функция СтоимостьПеревозки()
чСумма=0;
Для i=1 по m Цикл
Для j=1 по n Цикл
чСумма=чСумма+(Отгрузки[i][j]*Цены[i][j]);
КонецЦикла;
КонецЦикла;
Возврат чСумма;
КонецФункции

Функция ПоискНулевойЯчейкиДляВводаВБазис()
ок=0;
Для i=1 по m Цикл
Для j=1 по n Цикл
Если БазисныеЯчейки[i][j]=0 Тогда
ок=1;
Прервать;
КонецЕсли;
КонецЦикла;
Если ок=1 Тогда
Прервать;
КонецЕсли;
КонецЦикла;
Если ок=0 Тогда
ВызватьИсключение("There is no nonbasic (zero) cell entry into the basis");
КонецЕсли;
Пока 1=1 Цикл
i=ГСЧ.СлучайноеЧисло(1, m);
j=ГСЧ.СлучайноеЧисло(1, n);
Если БазисныеЯчейки[i][j]=1 Тогда
Продолжить;
КонецЕсли;
Если Отгрузки[i][j]<>0 Тогда
ВызватьИсключение("Nonzero shipment for nonbasic cell");
КонецЕсли;
БазисныеЯчейки[i][j]=1;
Сообщить("В базис введена ячейка "+i+" "+j);
Возврат Истина;
КонецЦикла;
КонецФункции

Функция НайтиЦикл(i0, j0)
гл_сч = m*n;
iЦикл.Очистить();
jЦикл.Очистить();
Если НайтиЦикл_ПоГоризонтали(i0, j0) Тогда
Возврат Истина;
КонецЕсли;
Возврат Ложь;
КонецФункции

Функция НайтиЦикл_ПоГоризонтали(i0, j0)
гл_сч=гл_сч-1;
Если гл_сч=0 Тогда
ВызватьИсключение("Too many iterations in the cycle search");
КонецЕсли;
Для j=1 по n Цикл
Если j=j0 Тогда
Продолжить;
КонецЕсли;
Если БазисныеЯчейки[i0][j]=0 Тогда
Продолжить;
КонецЕсли;
Если НайтиЦикл_ПоВертикали(i0, j) Тогда
iЦикл.Добавить(i0);
jЦикл.Добавить(j);
Возврат Истина;
КонецЕсли;
КонецЦикла;
Возврат Ложь;
КонецФункции

Функция НайтиЦикл_ПоВертикали(i0, j0)
Для i=1 по m Цикл
Если (j0=j1) и (i=i1) Тогда
iЦикл.Добавить(i);
jЦикл.Добавить(j0);
Возврат Истина;
КонецЕсли;
Если i=i0 Тогда
Продолжить;
КонецЕсли;
Если БазисныеЯчейки[i][j0]=0 Тогда
Продолжить;
КонецЕсли;
Если НайтиЦикл_ПоГоризонтали(i, j0) Тогда
iЦикл.Добавить(i);
jЦикл.Добавить(j0);
Возврат Истина;
КонецЕсли;
КонецЦикла;
Возврат Ложь;
КонецФункции

Функция ПерераспределениеПоЦиклу()
Сообщить("Redistribution by the cycle "+iЦикл.Количество());
Если jЦикл.Количество()<>iЦикл.Количество() Тогда
ВызватьИсключение("Unequal dimension for the cycle coordinates");
КонецЕсли;
Если iЦикл.Количество()<4 Тогда
ВызватьИсключение("Cycle is less than 4 items");
КонецЕсли;
Тета=НеОпределено;
Знак="+";
Для й=0 по iЦикл.ВГраница() Цикл
i=iЦикл[й];
j=jЦикл[й];
Если Знак="-" Тогда
Объем=Отгрузки[i][j];
Если Тета=НеОпределено Тогда
Тета=Объем;
Иначе
Если Объем<Тета Тогда
Тета=Объем;
КонецЕсли;
КонецЕсли;
Знак="+";
Иначе
Знак="-";
КонецЕсли;
КонецЦикла;
Если Тета=НеОпределено Тогда
ВызватьИсключение("Failed to evaluate variable theta.");
КонецЕсли;
Сообщить("Тета="+Тета);
Если Тета=0 Тогда
Возврат Ложь;
КонецЕсли;
Знак="+";
Для й=0 по iЦикл.ВГраница() Цикл
i=iЦикл[й];
j=jЦикл[й];
Если Знак="-" Тогда
Отгрузки[i][j]=Отгрузки[i][j]-Тета;
Знак="+";
Иначе
Отгрузки[i][j]=Отгрузки[i][j]+Тета;
Знак="-";
КонецЕсли;
КонецЦикла;
Возврат Истина;
КонецФункции

Функция РешениеТранспортнойЗадачи()
ГСЧ = Новый ГенераторСлучайныхЧисел();
БазисныеЯчейки = Новый Массив(m+1,n+1);
Отгрузки = Новый Массив(m+1,n+1);
СпросОстаток=Новый Массив(n+1);
ПредложениеОстаток=Новый Массив(m+1);
u=Новый Массив(m+1);
v=Новый Массив(n+1);
iЦикл = Новый Массив;
jЦикл = Новый Массив;
чСпрос=0;
Для j=1 по n Цикл
чСпрос=чСпрос+Спрос[j];
КонецЦикла;
чПредложение=0;
Для i=1 по m Цикл
чПредложение=чПредложение+Предложение[i];
КонецЦикла;
Если чПредложение>чСпрос Тогда
Сообщить("Offering more than the demand for "+(чПредложение-чСпрос)+" units of cargo. Create a fictitious user.");
Возврат Ложь;
ИначеЕсли чПредложение<чСпрос Тогда
Сообщить("Offering less than the demand for "+(чСпрос-чПредложение)+" units of cargo. Create a fictitious vendor.");
Возврат Ложь;
КонецЕсли;
РаспределениеМетодомСевероЗападногоУгла();
чСумма=СтоимостьПеревозки();
Сообщить("The cost of transportation by the north-west corner: "+чСумма);
Пока 1=1 Цикл
ПроверкаПравильностиОтгрузок();
счБазисных=0;
Для i=1 по m Цикл
Для j=1 по n Цикл
Если Отгрузки[i][j]>0 Тогда
БазисныеЯчейки[i][j]=1;
счБазисных=счБазисных+1;
ИначеЕсли Отгрузки[i][j]<0 Тогда
ВызватьИсключение("Shipments should not be negative");
Иначе
БазисныеЯчейки[i][j]=0;
КонецЕсли;
КонецЦикла;
КонецЦикла;
Пока счБазисных<(m+n-1) Цикл
Сообщить("Решение вырождено");
ПоискНулевойЯчейкиДляВводаВБазис();
счБазисных=счБазисных+1;
КонецЦикла;
Если ВычислениеПотенциалов()=Ложь Тогда
Продолжить;
КонецЕсли;
Если ПроверкаОптимальности()=Истина Тогда
Сообщить("Solution is optimal.");
Прервать;
КонецЕсли;
Сообщить("Solution is not optimal.");
Если НайтиЦикл(i1, j1)= Ложь Тогда
ВызватьИсключение("Unable to find a cycle");
КонецЕсли;
ПерераспределениеПоЦиклу();
чСумма=СтоимостьПеревозки();
Сообщить("***");
Сообщить("The cost of transport: "+чСумма);
КонецЦикла;
Возврат Истина;
КонецФункции

&НаКлиенте
Процедура КомандаРассчитать(Команда)
РешениеТранспортнойЗадачи();
КонецПроцедуры
```

## C#

Translation of: Java
```using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;

namespace TransportationProblem {
class Shipment {
public Shipment(double q, double cpu, int r, int c) {
Quantity = q;
CostPerUnit = cpu;
R = r;
C = c;
}

public double CostPerUnit { get; }

public double Quantity { get; set; }

public int R { get; }

public int C { get; }
}

class Program {
private static int[] demand;
private static int[] supply;
private static double[,] costs;
private static Shipment[,] matrix;

static void Init(string filename) {
string line;
var numArr = line.Split();
int numSources = int.Parse(numArr[0]);
int numDestinations = int.Parse(numArr[1]);

List<int> src = new List<int>();
List<int> dst = new List<int>();

numArr = line.Split();
for (int i = 0; i < numSources; i++) {
}

numArr = line.Split();
for (int i = 0; i < numDestinations; i++) {
}

// fix imbalance
int totalSrc = src.Sum();
int totalDst = dst.Sum();
if (totalSrc > totalDst) {
} else if (totalDst > totalSrc) {
}

supply = src.ToArray();
demand = dst.ToArray();

costs = new double[supply.Length, demand.Length];
matrix = new Shipment[supply.Length, demand.Length];

for (int i = 0; i < numSources; i++) {
numArr = line.Split();
for (int j = 0; j < numDestinations; j++) {
costs[i, j] = int.Parse(numArr[j]);
}
}
}
}

static void NorthWestCornerRule() {
for (int r = 0, northwest = 0; r < supply.Length; r++) {
for (int c = northwest; c < demand.Length; c++) {
int quantity = Math.Min(supply[r], demand[c]);
if (quantity > 0) {
matrix[r, c] = new Shipment(quantity, costs[r, c], r, c);

supply[r] -= quantity;
demand[c] -= quantity;

if (supply[r] == 0) {
northwest = c;
break;
}
}
}
}
}

static void SteppingStone() {
double maxReduction = 0;
Shipment[] move = null;
Shipment leaving = null;

FixDegenerateCase();

for (int r = 0; r < supply.Length; r++) {
for (int c = 0; c < demand.Length; c++) {
if (matrix[r, c] != null) {
continue;
}

Shipment trial = new Shipment(0, costs[r, c], r, c);
Shipment[] path = GetClosedPath(trial);

double reduction = 0;
double lowestQuantity = int.MaxValue;
Shipment leavingCandidate = null;

bool plus = true;
foreach (var s in path) {
if (plus) {
reduction += s.CostPerUnit;
} else {
reduction -= s.CostPerUnit;
if (s.Quantity < lowestQuantity) {
leavingCandidate = s;
lowestQuantity = s.Quantity;
}
}
plus = !plus;
}
if (reduction < maxReduction) {
move = path;
leaving = leavingCandidate;
maxReduction = reduction;
}
}
}

if (move != null) {
double q = leaving.Quantity;
bool plus = true;
foreach (var s in move) {
s.Quantity += plus ? q : -q;
matrix[s.R, s.C] = s.Quantity == 0 ? null : s;
plus = !plus;
}
SteppingStone();
}
}

static List<Shipment> MatrixToList() {
List<Shipment> newList = new List<Shipment>();
foreach (var item in matrix) {
if (null != item) {
}
}
return newList;
}

static Shipment[] GetClosedPath(Shipment s) {
List<Shipment> path = MatrixToList();

// remove (and keep removing) elements that do not have a
// vertical AND horizontal neighbor
int before;
do {
before = path.Count;
path.RemoveAll(ship => {
var nbrs = GetNeighbors(ship, path);
return nbrs[0] == null || nbrs[1] == null;
});
} while (before != path.Count);

// place the remaining elements in the correct plus-minus order
Shipment[] stones = path.ToArray();
Shipment prev = s;
for (int i = 0; i < stones.Length; i++) {
stones[i] = prev;
prev = GetNeighbors(prev, path)[i % 2];
}
return stones;
}

static Shipment[] GetNeighbors(Shipment s, List<Shipment> lst) {
Shipment[] nbrs = new Shipment[2];
foreach (var o in lst) {
if (o != s) {
if (o.R == s.R && nbrs[0] == null) {
nbrs[0] = o;
} else if (o.C == s.C && nbrs[1] == null) {
nbrs[1] = o;
}
if (nbrs[0] != null && nbrs[1] != null) {
break;
}
}
}
return nbrs;
}

static void FixDegenerateCase() {
const double eps = double.Epsilon;
if (supply.Length + demand.Length - 1 != MatrixToList().Count) {
for (int r = 0; r < supply.Length; r++) {
for (int c = 0; c < demand.Length; c++) {
if (matrix[r, c] == null) {
Shipment dummy = new Shipment(eps, costs[r, c], r, c);
if (GetClosedPath(dummy).Length == 0) {
matrix[r, c] = dummy;
return;
}
}
}
}
}
}

static void PrintResult(string filename) {
Console.WriteLine("Optimal solution {0}\n", filename);
double totalCosts = 0;

for (int r = 0; r < supply.Length; r++) {
for (int c = 0; c < demand.Length; c++) {
Shipment s = matrix[r, c];
if (s != null && s.R == r && s.C == c) {
Console.Write(" {0,3} ", s.Quantity);
totalCosts += (s.Quantity * s.CostPerUnit);
} else {
Console.Write("  -  ");
}
}
Console.WriteLine();
}
Console.WriteLine("\nTotal costs: {0}\n", totalCosts);
}

static void Main() {
foreach (var filename in new string[] { "input1.txt", "input2.txt", "input3.txt" }) {
Init(filename);
NorthWestCornerRule();
SteppingStone();
PrintResult(filename);
}
}
}
}
```
Output:
```Optimal solution input1.txt

20   -     5
-    30    5

Total costs: 180

Optimal solution input2.txt

-    -    -    12
20   -    10   10
-    30   -     3

Total costs: 130

Optimal solution input3.txt

-    -    -    14
-     9   -     1
10   -     5   -
-     5    7   -
-     1   -    -

Total costs: 1000```

## C++

Translation of: Kotlin
```#include <algorithm>
#include <iomanip>
#include <iostream>
#include <fstream>
#include <numeric>
#include <string>
#include <vector>
#include <cfloat>

using namespace std;

class Shipment {
public:
double costPerUnit;
int r, c;
double quantity;

Shipment() : quantity(0), costPerUnit(0), r(-1), c(-1) {
// empty
}

Shipment(double q, double cpu, int r, int c) : quantity(q), costPerUnit(cpu), r(r), c(c) {
// empty
}

friend bool operator==(const Shipment &lhs, const Shipment &rhs) {
return lhs.costPerUnit == rhs.costPerUnit
&& lhs.quantity == rhs.quantity
&& lhs.r == rhs.r
&& lhs.c == rhs.c;
}

friend bool operator!=(const Shipment &lhs, const Shipment &rhs) {
return !(lhs == rhs);
}

static Shipment ZERO;
};
Shipment Shipment::ZERO = {};

vector<int> demand, supply;
vector<vector<double>> costs;
vector<vector<Shipment>> matrix;

void init(string filename) {
ifstream ifs;

ifs.open(filename);
if (!ifs) {
return;
}

size_t numSources, numDestinations;
ifs >> numSources >> numDestinations;

vector<int> src, dst;
int t;

for (size_t i = 0; i < numSources; i++) {
ifs >> t;
src.push_back(t);
}

for (size_t i = 0; i < numDestinations; i++) {
ifs >> t;
dst.push_back(t);
}

// fix imbalance
int totalSrc = accumulate(src.cbegin(), src.cend(), 0);
int totalDst = accumulate(dst.cbegin(), dst.cend(), 0);
if (totalSrc > totalDst) {
dst.push_back(totalSrc - totalDst);
} else if (totalDst > totalSrc) {
src.push_back(totalDst - totalSrc);
}

supply = src;
demand = dst;

costs.clear();
matrix.clear();

double d;
for (size_t i = 0; i < numSources; i++) {
size_t cap = max(numDestinations, demand.size());

vector<double> dt(cap);
vector<Shipment> st(cap);
for (size_t j = 0; j < numDestinations; j++) {
ifs >> d;
dt[j] = d;
}
costs.push_back(dt);
matrix.push_back(st);
}
for (size_t i = numSources; i < supply.size(); i++) {
size_t cap = max(numDestinations, demand.size());

vector<Shipment> st(cap);
matrix.push_back(st);

vector<double> dt(cap);
costs.push_back(dt);
}
}

void northWestCornerRule() {
int northwest = 0;
for (size_t r = 0; r < supply.size(); r++) {
for (size_t c = northwest; c < demand.size(); c++) {
int quantity = min(supply[r], demand[c]);
if (quantity > 0) {
matrix[r][c] = Shipment(quantity, costs[r][c], r, c);

supply[r] -= quantity;
demand[c] -= quantity;

if (supply[r] == 0) {
northwest = c;
break;
}
}
}
}
}

vector<Shipment> matrixToVector() {
vector<Shipment> result;
for (auto &row : matrix) {
for (auto &s : row) {
if (s != Shipment::ZERO) {
result.push_back(s);
}
}
}
return result;
}

vector<Shipment> getNeighbors(const Shipment &s, const vector<Shipment> &lst) {
vector<Shipment> nbrs(2);
for (auto &o : lst) {
if (o != s) {
if (o.r == s.r && nbrs[0] == Shipment::ZERO) {
nbrs[0] = o;
} else if (o.c == s.c && nbrs[1] == Shipment::ZERO) {
nbrs[1] = o;
}
if (nbrs[0] != Shipment::ZERO && nbrs[1] != Shipment::ZERO) {
break;
}
}
}
return nbrs;
}

vector<Shipment> getClosedPath(const Shipment &s) {
auto path = matrixToVector();
path.insert(path.begin(), s);

// remove (and keep removing) elements that do not have a
// vertical AND horizontal neighbor
size_t before;
do {
before = path.size();
path.erase(
remove_if(
path.begin(), path.end(),
[&path](Shipment &ship) {
auto nbrs = getNeighbors(ship, path);
return nbrs[0] == Shipment::ZERO || nbrs[1] == Shipment::ZERO;
}),
path.end());
} while (before != path.size());

// place the remaining elements in the correct plus-minus order
vector<Shipment> stones(path.size());
fill(stones.begin(), stones.end(), Shipment::ZERO);
auto prev = s;
for (size_t i = 0; i < stones.size(); i++) {
stones[i] = prev;
prev = getNeighbors(prev, path)[i % 2];
}
return stones;
}

void fixDegenerateCase() {
double eps = DBL_MIN;
if (supply.size() + demand.size() - 1 != matrixToVector().size()) {
for (size_t r = 0; r < supply.size(); r++) {
for (size_t c = 0; c < demand.size(); c++) {
if (matrix[r][c] == Shipment::ZERO) {
Shipment dummy(eps, costs[r][c], r, c);
if (getClosedPath(dummy).empty()) {
matrix[r][c] = dummy;
return;
}
}
}
}
}
}

void steppingStone() {
double maxReduction = 0;
vector<Shipment> move;
Shipment leaving;
bool isNull = true;

fixDegenerateCase();

for (size_t r = 0; r < supply.size(); r++) {
for (size_t c = 0; c < demand.size(); c++) {
if (matrix[r][c] != Shipment::ZERO) {
continue;
}

Shipment trial(0, costs[r][c], r, c);
vector<Shipment> path = getClosedPath(trial);

double reduction = 0;
double lowestQuantity = INT32_MAX;
Shipment leavingCandidate;

bool plus = true;
for (auto &s : path) {
if (plus) {
reduction += s.costPerUnit;
} else {
reduction -= s.costPerUnit;
if (s.quantity < lowestQuantity) {
leavingCandidate = s;
lowestQuantity = s.quantity;
}
}
plus = !plus;
}
if (reduction < maxReduction) {
isNull = false;
move = path;
leaving = leavingCandidate;
maxReduction = reduction;
}
}
}

if (!isNull) {
double q = leaving.quantity;
bool plus = true;
for (auto &s : move) {
s.quantity += plus ? q : -q;
matrix[s.r][s.c] = s.quantity == 0 ? Shipment::ZERO : s;
plus = !plus;
}
steppingStone();
}
}

void printResult(string filename) {
ifstream ifs;
string buffer;

ifs.open(filename);
if (!ifs) {
return;
}

cout << filename << "\n\n";
while (!ifs.eof()) {
getline(ifs, buffer);
cout << buffer << '\n';
}
cout << '\n';

cout << "Optimal solution " << filename << "\n\n";
double totalCosts = 0.0;
for (size_t r = 0; r < supply.size(); r++) {
for (size_t c = 0; c < demand.size(); c++) {
auto s = matrix[r][c];
if (s != Shipment::ZERO && s.r == r && s.c == c) {
cout << ' ' << setw(3) << s.quantity << ' ';
totalCosts += s.quantity * s.costPerUnit;
} else {
cout << "  -  ";
}
}
cout << '\n';
}
cout << "\nTotal costs: " << totalCosts << "\n\n";
}

void process(string filename) {
init(filename);
northWestCornerRule();
steppingStone();
printResult(filename);
}

int main() {
process("input1.txt");
process("input2.txt");
process("input3.txt");

return 0;
}
```
Output:
```input1.txt

2 3
25 35
20 30 10
3 5 7
3 2 5

Optimal solution input1.txt

20   -     5
-    30    5

Total costs: 180

input2.txt

3 3
12 40 33
20 30 10
3 5 7
2 4 6
9 1 8

Optimal solution input2.txt

-    -    -    12
20   -    10   10
-    30   -     3

Total costs: 130

input3.txt

4 4
14 10 15 12
10 15 12 15
10 30 25 15
20 15 20 10
10 30 20 20
30 40 35 45

Optimal solution input3.txt

-    -    -    14
-     9   -     1
10   -     5   -
-     5    7   -
-     1   -    -

Total costs: 1000```

## D

Translation of: Java
```import std.stdio, std.range, std.algorithm, std.conv, std.math, std.traits;

final class Shipment {
double quantity;
immutable double costPerUnit;
immutable size_t r, c;

this(in double q, in double cpu, in size_t r_, in size_t c_)
pure nothrow @safe @nogc {
quantity = q;
costPerUnit = cpu;
this.r = r_;
this.c = c_;
}
}

alias ShipmentMat = Shipment[][];
alias CostsMat = double[][];

void init(in string fileName, out uint[] demand, out uint[] supply,
out CostsMat costs, out ShipmentMat matrix) {
auto inParts = fileName.File.byLine.map!splitter.joiner;

immutable numSources = inParts.front.to!uint;
inParts.popFront;
immutable numDestinations = inParts.front.to!uint;
inParts.popFront;

foreach (immutable i; 0 .. numSources) {
supply ~= inParts.front.to!uint;
inParts.popFront;
}

foreach (immutable i; 0 .. numDestinations) {
demand ~= inParts.front.to!uint;
inParts.popFront;
}

// Fix imbalance.
immutable totalSrc = supply.sum;
immutable totalDst = demand.sum;

if (totalSrc > totalDst)
demand ~= totalSrc - totalDst;
else if (totalDst > totalSrc)
supply ~= totalDst - totalSrc;

costs = new CostsMat(supply.length, demand.length);
foreach (row; costs)
row[] = 0.0;
matrix = new ShipmentMat(supply.length, demand.length);

foreach (immutable i; 0 .. numSources)
foreach (immutable j; 0 .. numDestinations) {
costs[i][j] = inParts.front.to!double;
inParts.popFront;
}
}

void northWestCornerRule(uint[] demand, uint[] supply, in CostsMat costs,
ShipmentMat matrix) pure nothrow @safe {
size_t northwest = 0;
foreach (immutable r; 0 .. supply.length) {
foreach (immutable c; northwest .. demand.length) {
immutable quantity = min(supply[r], demand[c]);
if (quantity > 0) {
matrix[r][c] = new Shipment(quantity, costs[r][c], r, c);

supply[r] -= quantity;
demand[c] -= quantity;

if (supply[r] == 0) {
northwest = c;
break;
}
}
}
}
}

void steppingStone(in uint[] demand, in uint[] supply,
in CostsMat costs, ShipmentMat matrix) pure @safe {
double maxReduction = 0;
Shipment[] move;
Shipment leaving = null;

fixDegenerateCase(demand, supply, costs, matrix);

foreach (immutable r; 0 .. supply.length) {
foreach (immutable c; 0 .. demand.length) {
if (matrix[r][c] !is null)
continue;

auto trial = new Shipment(0, costs[r][c], r, c);
auto path = getClosedPath(trial, matrix);

double reduction = 0;
double lowestQuantity = uint.max;
Shipment leavingCandidate = null;

bool plus = true;
foreach (s; path) {
if (plus) {
reduction += s.costPerUnit;
} else {
reduction -= s.costPerUnit;
if (s.quantity < lowestQuantity) {
leavingCandidate = s;
lowestQuantity = s.quantity;
}
}
plus = !plus;
}
if (reduction < maxReduction) {
move = path;
leaving = leavingCandidate;
maxReduction = reduction;
}
}
}

if (move !is null) {
auto q = leaving.quantity;
auto plus = true;
foreach (s; move) {
s.quantity += plus ? q : -q;
matrix[s.r][s.c] = (s.quantity == 0) ? null : s;
plus = !plus;
}
steppingStone(demand, supply, costs, matrix);
}
}

auto matrixToSeq(ShipmentMat matrix) pure nothrow @nogc @safe {
return matrix.joiner.filter!(s => s !is null);
}

Shipment[] getClosedPath(Shipment s, ShipmentMat matrix) pure @safe
in {
assert(s !is null);
} out(result) {
assert(result.all!(sh => sh !is null));
} body {
Shipment[] stones = chain([s], matrixToSeq(matrix)).array;

// Remove (and keep removing) elements that do not have
// a vertical AND horizontal neighbor.
while (true) {
auto stones2 = stones.remove!((in e) {
const nbrs = getNeighbors(e, stones);
return nbrs[0] is null || nbrs[1] is null;
});

if (stones2.length == stones.length)
break;
stones = stones2;
}

// Place the remaining elements in the correct plus-minus order.
auto stones3 = stones.dup;
Shipment prev = s;
foreach (immutable i, ref si; stones3) {
si = prev;
prev = getNeighbors(prev, stones)[i % 2];
}
return stones3;
}

Shipment[2] getNeighbors(ShipmentsRange)(in Shipment s, ShipmentsRange seq)
pure nothrow @safe @nogc
if (isForwardRange!ShipmentsRange && is(ForeachType!ShipmentsRange == Shipment))
in {
assert(s !is null);
assert(seq.all!(sh => sh !is null));
} body {
Shipment[2] nbrs;

foreach (o; seq) {
if (o !is s) {
if (o.r == s.r && nbrs[0] is null)
nbrs[0] = o;
else if (o.c == s.c && nbrs[1] is null)
nbrs[1] = o;
if (nbrs[0] !is null && nbrs[1] !is null)
break;
}
}

return nbrs;
}

void fixDegenerateCase(in uint[] demand, in uint[] supply,
in CostsMat costs, ShipmentMat matrix) pure @safe {
immutable eps = double.min_normal;

if (supply.length.signed + demand.length.signed - 1 != matrixToSeq(matrix).walkLength) {
foreach (immutable r; 0 .. supply.length) {
foreach (immutable c; 0 .. demand.length) {
if (matrix[r][c] is null) {
auto dummy = new Shipment(eps, costs[r][c], r, c);
if (getClosedPath(dummy, matrix).length == 0) {
matrix[r][c] = dummy;
return;
}
}
}
}
}
}

void printResult(in string fileName, in uint[] demand, in uint[] supply,
in CostsMat costs, in ShipmentMat matrix) @safe /*@nogc*/ {
writefln("Optimal solution %s", fileName);
double totalCosts = 0;

foreach (immutable r; 0 .. supply.length) {
foreach (immutable c; 0 .. demand.length) {
const s = matrix[r][c];
if (s !is null && s.r == r && s.c == c) {
writef(" %3d ", cast(uint)s.quantity);
totalCosts += s.quantity * s.costPerUnit;
} else
write("  -  ");
}
//writeln; // Not @safe?
write('\n');
}
writefln("\nTotal costs: %s\n", totalCosts);
}

void main() {
foreach (fileName; ["transportation_problem1.txt",
"transportation_problem2.txt",
"transportation_problem3.txt"]) {
uint[] demand, supply;
CostsMat costs;
ShipmentMat matrix;
init(fileName, demand, supply, costs, matrix);
northWestCornerRule(demand, supply, costs, matrix);
steppingStone(demand, supply, costs, matrix);
printResult(fileName, demand, supply, costs, matrix);
}
}
```
Output:
```Optimal solution transportation_problem1.txt
20   -     5
-    30    5

Total costs: 180

Optimal solution transportation_problem2.txt
-    -    -    12
20   -    10   10
-    30   -     3

Total costs: 130

Optimal solution transportation_problem3.txt
-    -    -    14
-     9   -     1
10   -     5   -
-     5    7   -
-     1   -    -

Total costs: 1000```

## Glagol

```ОТДЕЛ Транспорт+;
ИСПОЛЬЗУЕТ
Вывод ИЗ "...\Отделы\Обмен\",
Приём;

ПЕР
Поставщиков, Потребителей: ЦЕЛ;
Запасы, Потребности: ДОСТУП К РЯД ИЗ ЦЕЛ;
Расходы, План: ДОСТУП К РЯД ИЗ РЯД ИЗ ЦЕЛ;
U, V: ДОСТУП К РЯД ИЗ ЦЕЛ;
оцСв: ДОСТУП К РЯД ИЗ НАБОР значение: ЦЕЛ; поставщик, потребитель: ЦЕЛ КОН;
начQ_поставщик, начQ_потребитель: ЦЕЛ;
Q: ДОСТУП К РЯД ИЗ РЯД ИЗ УЗКЦЕЛ;
Поправка, Разница: ЦЕЛ;

ЗАДАЧА ПринятьДанные;
ПЕР
сч1, сч2: ЦЕЛ;
сумма1, сумма2, разница: ЦЕЛ;
памЗап, памПотр: ДОСТУП К РЯД ИЗ ЦЕЛ;
УКАЗ
Вывод.Цепь("Number of suppliers: ");
Поставщиков := Приём.Число();
Вывод.Цепь(".^Number of consumers: ");
Потребителей := Приём.Число();
СОЗДАТЬ(памЗап, Поставщиков);
СОЗДАТЬ(памПотр, Потребителей);
Вывод.Цепь(".^Inventories^^");
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
памЗап[сч1] := Приём.Число();
Вывод.Цепь(" ")
КОН;
Вывод.Цепь("^Requirements:^");
ОТ сч1 := 0 ДО Потребителей-1 ВЫП
памПотр[сч1] := Приём.Число();
Вывод.Цепь(" ")
КОН;
сумма1 := 0;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП УВЕЛИЧИТЬ(сумма1, памЗап[сч1]) КОН;
сумма2 := 0;
ОТ сч1 := 0 ДО Потребителей-1 ВЫП УВЕЛИЧИТЬ(сумма2, памПотр[сч1]) КОН;
ЕСЛИ сумма1 > сумма2 ТО
разница := сумма1 - сумма2;
Вывод.ЧЦел("^Introduced a fictitious consumer.", сумма1, сумма2, разница, 0);
УВЕЛИЧИТЬ(Потребителей);
СОЗДАТЬ(Потребности, Потребителей);
ОТ сч1 := 0 ДО Потребителей-2 ВЫП Потребности[сч1] := памПотр[сч1] КОН;
Потребности[Потребителей-1] := разница;
СОЗДАТЬ(Запасы, Поставщиков);
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП Запасы[сч1] := памЗап[сч1] КОН
АЕСЛИ сумма2 > сумма1 ТО
разница := сумма2 - сумма1;
Вывод.ЧЦел("^Introduced a fictitious supplier.", сумма2, сумма1, разница, 0);
УВЕЛИЧИТЬ(Поставщиков);
СОЗДАТЬ(Запасы, Поставщиков);
ОТ сч1 := 0 ДО Поставщиков-2 ВЫП Запасы[сч1] := памЗап[сч1] КОН;
Запасы[Поставщиков-1] := разница;
СОЗДАТЬ(Потребности, Потребителей);
ОТ сч1 := 0 ДО Потребителей-1 ВЫП Потребности[сч1] := памПотр[сч1] КОН
ИНАЧЕ
СОЗДАТЬ(Запасы, Поставщиков);
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП Запасы[сч1] := памЗап[сч1] КОН;
СОЗДАТЬ(Потребности, Потребителей);
ОТ сч1 := 0 ДО Потребителей-1 ВЫП Потребности[сч1] := памПотр[сч1] КОН
КОН;
СОЗДАТЬ(Расходы, Поставщиков, Потребителей);
Вывод.Цепь("^The matrix of costs:^");
ЕСЛИ сумма1 > сумма2 ТО
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-2 ВЫП
Расходы[сч1, сч2] := Приём.Число();
Вывод.Цепь(" ")
КОН;
Расходы[сч1, Потребителей-1] := 0;
Вывод.Цепь("^")
КОН
АЕСЛИ сумма2 > сумма1 ТО
ОТ сч1 := 0 ДО Поставщиков-2 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
Расходы[сч1, сч2] := Приём.Число();
Вывод.Цепь(" ")
КОН;
Вывод.Цепь("^")
КОН;
ОТ сч1 := 0 ДО Потребителей-1 ВЫП Расходы[Поставщиков-1, сч1] := 0 КОН
ИНАЧЕ
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
Расходы[сч1, сч2] := Приём.Число();
Вывод.Цепь(" ")
КОН;
Вывод.Цепь("^")
КОН
КОН;
СОЗДАТЬ(План, Поставщиков, Потребителей);
СОЗДАТЬ(U, Поставщиков);
СОЗДАТЬ(V, Потребителей);
СОЗДАТЬ(оцСв, Потребителей*Поставщиков-(Потребителей+Поставщиков-1));
СОЗДАТЬ(Q, Поставщиков, Потребителей)
КОН ПринятьДанные;

ЗАДАЧА ВывестиПлан;
ПЕР
сч1, сч2: ЦЕЛ;
УКАЗ
ОТ сч1 := 1 ДО Потребителей ВЫП Вывод.Цепь("-----") КОН;
Вывод.Цепь("^");
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] = -1 ТО Вывод.Цепь("  -  ") ИНАЧЕ
Вывод.ЧЦел("%4d ", План[сч1, сч2], 0, 0, 0);
КОН
КОН;
Вывод.Цепь("^")
КОН;
ОТ сч1 := 1 ДО Потребителей ВЫП Вывод.Цепь("-----") КОН
КОН ВывестиПлан;

ЗАДАЧА ПосчитатьПоправку;
ПЕР
сч1, сч2: ЦЕЛ;
УКАЗ
Поправка := -1;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] # -1 ТО
ЕСЛИ Q[сч1, сч2] = -1 ТО
ЕСЛИ Поправка = -1 ТО Поправка := План[сч1, сч2]
АЕСЛИ Поправка > План[сч1, сч2] ТО Поправка := План[сч1, сч2] КОН
КОН
КОН
КОН
КОН;
Разница := Разница * Поправка
КОН ПосчитатьПоправку;

ЗАДАЧА РасставитьНули(недостаток: ЦЕЛ);
ПЕР
Связь: ДОСТУП К РЯД ИЗ РЯД ИЗ УЗКЦЕЛ;
сч1, сч2: ЦЕЛ;
естьБезСвязи: КЛЮЧ;

ЗАДАЧА ЕстьНапротив(строка, столбец: ЦЕЛ): КЛЮЧ;
ПЕР сч: ЦЕЛ;
УКАЗ
ОТ сч := 0 ДО Поставщиков-1 ВЫП
ЕСЛИ (сч # строка) И (Связь[сч, столбец] = 1) ТО ВОЗВРАТ ВКЛ КОН
КОН;
ОТ сч := 0 ДО Потребителей-1 ВЫП
ЕСЛИ (сч # столбец) И (Связь[строка, сч] = 1) ТО ВОЗВРАТ ВКЛ КОН
КОН;
ВОЗВРАТ ОТКЛ
КОН ЕстьНапротив;

ЗАДАЧА СтолбецБезСвязи(номер: ЦЕЛ): КЛЮЧ;
ПЕР сч: ЦЕЛ;
УКАЗ
ОТ сч := 0 ДО Поставщиков-1 ВЫП
ЕСЛИ Связь[сч, номер] = 1 ТО ВОЗВРАТ ОТКЛ КОН
КОН;
ВОЗВРАТ ВКЛ
КОН СтолбецБезСвязи;

УКАЗ
СОЗДАТЬ(Связь, Поставщиков, Потребителей);
естьБезСвязи := ОТКЛ;
ОТ сч1 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[0, сч1] = -1 ТО Связь[0, сч1] := -1 ИНАЧЕ Связь[0, сч1] := 1 КОН
КОН;
ОТ сч1 := 1 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] = -1 ТО Связь[сч1, сч2] := -1 ИНАЧЕ Связь[сч1, сч2] := 0 КОН
КОН
КОН;
ОТ сч1 := 1 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ Связь[сч1, сч2] = 0 ТО
ЕСЛИ ЕстьНапротив(сч1, сч2) ТО Связь[сч1, сч2] := 1
АЕСЛИ НЕ естьБезСвязи ТО естьБезСвязи := ВКЛ КОН
КОН
КОН
КОН;
ЕСЛИ естьБезСвязи ТО
ОТ сч1 := Поставщиков-1 ДО 1 ПО -1 ВЫП
ОТ сч2 := Потребителей-1 ДО 0 ПО -1 ВЫП
ЕСЛИ Связь[сч1, сч2] = 0 ТО
ЕСЛИ ЕстьНапротив(сч1, сч2) ТО Связь[сч1, сч2] := 1
АЕСЛИ НЕ естьБезСвязи ТО естьБезСвязи := ВКЛ КОН
КОН
КОН
КОН
КОН;
ЕСЛИ естьБезСвязи ТО
ОТ сч1 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ СтолбецБезСвязи(сч1) ТО План[0, сч1] := 0; УМЕНЬШИТЬ(недостаток) КОН;
ЕСЛИ недостаток = 0 ТО сч1 := Потребителей КОН
КОН
КОН;
КОЛЬЦО
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ недостаток = 0 ТО ВЫХОД ИНАЧЕ
ЕСЛИ План[сч1, сч2] = -1 ТО
План[сч1, сч2] := 0; УМЕНЬШИТЬ(недостаток);
КОН
КОН
КОН
КОН;
ВЫХОД
КОН
КОН РасставитьНули;

ЗАДАЧА ЗаполнитьОтУгла;
ПЕР
ОсталосьВНаличии, ОсталосьПотребным: ДОСТУП К РЯД ИЗ ЦЕЛ;
занято, недостаток: ЦЕЛ;
сч1, сч2: ЦЕЛ;
УКАЗ
СОЗДАТЬ(ОсталосьВНаличии, Поставщиков);
СОЗДАТЬ(ОсталосьПотребным, Потребителей);
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП ОсталосьВНаличии[сч1] := Запасы[сч1] КОН;
ОТ сч1 := 0 ДО Потребителей-1 ВЫП ОсталосьПотребным[сч1] := Потребности[сч1] КОН;
занято := 0;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ ОсталосьВНаличии[сч1] = 0 ТО План[сч1, сч2] := -1 ИНАЧЕ
ЕСЛИ ОсталосьВНаличии[сч1] > ОсталосьПотребным[сч2] ТО
ЕСЛИ ОсталосьПотребным[сч2] # 0 ТО План[сч1, сч2] := ОсталосьПотребным[сч2]; УВЕЛИЧИТЬ(занято)
ИНАЧЕ План[сч1, сч2] := -1 КОН;
УМЕНЬШИТЬ(ОсталосьВНаличии[сч1], ОсталосьПотребным[сч2]);
ОсталосьПотребным[сч2] := 0
ИНАЧЕ
ЕСЛИ ОсталосьВНаличии[сч1] # 0 ТО План[сч1, сч2] := ОсталосьВНаличии[сч1]; УВЕЛИЧИТЬ(занято)
ИНАЧЕ План[сч1, сч2] := -1 КОН;
УМЕНЬШИТЬ(ОсталосьПотребным[сч2], ОсталосьВНаличии[сч1]);
ОсталосьВНаличии[сч1] := 0
КОН
КОН
КОН
КОН;
недостаток := (Поставщиков+Потребителей-1) - занято;
ЕСЛИ недостаток > 0 ТО РасставитьНули(недостаток) КОН
КОН ЗаполнитьОтУгла;

ЗАДАЧА ОценитьБазисныеКлетки;
ПЕР
сч1, сч2, сч3: ЦЕЛ;
суммы: ДОСТУП К РЯД ИЗ РЯД 3 ИЗ ЦЕЛ;
известно: ДОСТУП К РЯД ИЗ РЯД 2 ИЗ КЛЮЧ;
УКАЗ
СОЗДАТЬ(суммы, Поставщиков+Потребителей-1);
СОЗДАТЬ(известно, Поставщиков+Потребителей-1);
известно[0][0] := ВКЛ; известно[0][1] := ОТКЛ;
ОТ сч1 := 1 ДО (Поставщиков+Потребителей-1)-1 ВЫП известно[сч1][0] := ОТКЛ; известно[сч1][1] := ОТКЛ КОН;
сч3 := 0;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] # -1 ТО
суммы[сч3][0] := сч1; суммы[сч3][1] := сч2; суммы[сч3][2] := Расходы[сч1, сч2];
УВЕЛИЧИТЬ(сч3)
КОН
КОН
КОН;
U[0] := 0;
ОТ сч1 := 1 ДО (Поставщиков+Потребителей-1)-1 ВЫП
ЕСЛИ суммы[сч1][0] = 0 ТО известно[сч1][0] := ВКЛ КОН
КОН;
сч3 := 0;
ПОВТОРЯТЬ
сч1 := 0;
ПОКА НЕ (известно[сч1][0] # известно[сч1][1]) ВЫП
УВЕЛИЧИТЬ(сч1)
КОН;
ЕСЛИ известно[сч1][0] ТО
V[суммы[сч1][1]] := суммы[сч1][2] - U[суммы[сч1][0]];
известно[сч1][1] := ВКЛ;
ОТ сч2 := 0 ДО (Поставщиков+Потребителей-1)-1 ВЫП
ЕСЛИ (суммы[сч2][1] = суммы[сч1][1]) И (НЕ известно[сч2][1]) ТО известно[сч2][1] := ВКЛ КОН
КОН
ИНАЧЕ
U[суммы[сч1][0]] := суммы[сч1][2] - V[суммы[сч1][1]];
известно[сч1][0] := ВКЛ;
ОТ сч2 := 0 ДО (Поставщиков+Потребителей-1)-1 ВЫП
ЕСЛИ (суммы[сч2][0] = суммы[сч1][0]) И (НЕ известно[сч2][0]) ТО известно[сч2][0] := ВКЛ КОН
КОН
КОН;
УВЕЛИЧИТЬ(сч3)
ДО сч3 = Поставщиков+Потребителей-1
КОН ОценитьБазисныеКлетки;

ЗАДАЧА ОценитьСвободныеКлетки(): КЛЮЧ;
ПЕР
сч1, сч2, сч3: ЦЕЛ;
естьПолож: КЛЮЧ;
УКАЗ
естьПолож := ОТКЛ;
сч3 := 0;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] = -1 ТО
оцСв[сч3].значение := U[сч1]+V[сч2]-Расходы[сч1,сч2];
оцСв[сч3].поставщик := сч1; оцСв[сч3].потребитель := сч2;
ЕСЛИ оцСв[сч3].значение > 0 ТО естьПолож := ВКЛ КОН;
УВЕЛИЧИТЬ(сч3)
КОН
КОН
КОН;
ЕСЛИ естьПолож ТО ВОЗВРАТ ОТКЛ ИНАЧЕ ВОЗВРАТ ВКЛ КОН
КОН ОценитьСвободныеКлетки;

ЗАДАЧА Цикл;
ПЕР
сч1, сч2, сч3: ЦЕЛ;
максЗн: ЦЕЛ;
начало, циклНайден: КЛЮЧ;

ЗАДАЧА НаЛинии(наКакой: ЦЕЛ; столб: КЛЮЧ): ЦЕЛ;
ПЕР сч, сколько: ЦЕЛ;
УКАЗ
сколько := 0;
ЕСЛИ столб ТО
ОТ сч := 0 ДО Поставщиков-1 ВЫП
ЕСЛИ (План[сч, наКакой] # -1) ИЛИ ((сч = начQ_поставщик) И (наКакой = начQ_потребитель)) ТО
УВЕЛИЧИТЬ(сколько)
КОН
КОН
ИНАЧЕ
ОТ сч := 0 ДО Потребителей-1 ВЫП
ЕСЛИ (План[наКакой, сч] # -1) ИЛИ ((наКакой = начQ_поставщик) И (сч = начQ_потребитель)) ТО
УВЕЛИЧИТЬ(сколько)
КОН
КОН
КОН;
ВОЗВРАТ сколько
КОН НаЛинии;

ЗАДАЧА^ ИскатьВСтолбце(номер, строка: ЦЕЛ): КЛЮЧ;

ЗАДАЧА ИскатьВСтроке(номер, столбец: ЦЕЛ): КЛЮЧ;
ПЕР
сч: ЦЕЛ;
УКАЗ
ЕСЛИ (НЕ начало) И (номер = начQ_поставщик) И (столбец = начQ_потребитель) ТО циклНайден := ВКЛ КОН;
ЕСЛИ начало ТО начало := ОТКЛ КОН;
ЕСЛИ циклНайден ТО ВОЗВРАТ ВКЛ КОН;
ОТ сч := 0 ДО Потребителей-1 ВЫП
ЕСЛИ
(сч # столбец) И
((План[номер, сч] # -1) ИЛИ ((номер = начQ_поставщик) И (сч = начQ_потребитель))) И
(НаЛинии(сч, ВКЛ) > 1) И
(Q[номер, сч] = 0)
ТО
Q[номер, сч] := -1;
ЕСЛИ НЕ ИскатьВСтолбце(сч, номер) ТО Q[номер, сч] := 0 ИНАЧЕ ВОЗВРАТ ВКЛ КОН
КОН
КОН;
ВОЗВРАТ ОТКЛ
КОН ИскатьВСтроке;

ЗАДАЧА ИскатьВСтолбце(номер, строка: ЦЕЛ): КЛЮЧ;
ПЕР
сч: ЦЕЛ;
УКАЗ
ЕСЛИ (НЕ начало) И (строка = начQ_поставщик) И (номер = начQ_потребитель) ТО циклНайден := ВКЛ КОН;
ЕСЛИ начало ТО начало := ОТКЛ КОН;
ЕСЛИ циклНайден ТО ВОЗВРАТ ВКЛ КОН;
ОТ сч := 0 ДО Поставщиков-1 ВЫП
ЕСЛИ
(сч # строка) И
((План[сч, номер] # -1) ИЛИ ((сч = начQ_поставщик) И (номер = начQ_потребитель))) И
(НаЛинии(сч, ОТКЛ) > 1) И
(Q[сч, номер] = 0)
ТО
Q[сч, номер] := 1;
ЕСЛИ НЕ ИскатьВСтроке(сч, номер) ТО Q[сч, номер] := 0 ИНАЧЕ ВОЗВРАТ ВКЛ КОН
КОН
КОН;
ВОЗВРАТ ОТКЛ
КОН ИскатьВСтолбце;

УКАЗ
максЗн := 0;
ОТ сч1 := 0 ДО Потребителей*Поставщиков-(Потребителей+Поставщиков-1)-1 ВЫП
ЕСЛИ оцСв[сч1].значение > максЗн ТО максЗн := оцСв[сч1].значение КОН
КОН;
сч3 := 0;
ПОКА оцСв[сч3].значение # максЗн ВЫП УВЕЛИЧИТЬ(сч3) КОН;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
Q[сч1, сч2] := 0
КОН
КОН;
Разница := оцСв[сч3].значение;
начQ_поставщик := оцСв[сч3].поставщик; начQ_потребитель := оцСв[сч3].потребитель;
начало := ВКЛ; циклНайден := ОТКЛ;
ЕСЛИ ИскатьВСтроке(начQ_поставщик, начQ_потребитель) ТО КОН
КОН Цикл;

ЗАДАЧА ИзменитьПлан;
ПЕР
сч1, сч2: ЦЕЛ;
занято, недостаток: ЦЕЛ;
УКАЗ
ЕСЛИ Поправка = 0 ТО
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] = 0 ТО План[сч1, сч2] := -1; сч2 := Потребителей; сч1 := Поставщиков КОН
КОН
КОН;
План[начQ_поставщик, начQ_потребитель] := 0
ИНАЧЕ
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ Q[сч1, сч2] = 1 ТО
ЕСЛИ План[сч1, сч2] = -1 ТО План[сч1, сч2] := 0 КОН;
УВЕЛИЧИТЬ(План[сч1, сч2], Поправка);
АЕСЛИ Q[сч1, сч2] = -1 ТО УМЕНЬШИТЬ(План[сч1, сч2], Поправка)
КОН
КОН
КОН;
занято := 0;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] > 0 ТО УВЕЛИЧИТЬ(занято) КОН
КОН
КОН;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] = 0 ТО План[сч1, сч2] := -1 КОН
КОН
КОН;
недостаток := (Поставщиков+Потребителей-1) - занято;
ЕСЛИ недостаток > 0 ТО РасставитьНули(недостаток) КОН
КОН
КОН ИзменитьПлан;

УКАЗ
ПринятьДанные;
ЗаполнитьОтУгла;
Разница := -1;
КОЛЬЦО
ОценитьБазисныеКлетки;
ЕСЛИ ОценитьСвободныеКлетки() ТО ВЫХОД КОН;
Цикл;
ПосчитатьПоправку;
ИзменитьПлан
КОН;
ВывестиПлан

КОН Транспорт.
```

### Input

```Number of suppliers: 3.
Number of consumers: 3.
Inventories:
12 40 33
Requirements:
20 30 10
Introduced a fictitious consumer.
The matrix of costs:
3 5 7
2 4 6
9 1 8
```

### Output

```--------------------
-    -    -    12
20   -    10   10
-    30   -     3
--------------------
```

## Go

Translation of: Java
```package main

import (
"bufio"
"container/list"
"fmt"
"io/ioutil"
"log"
"math"
"os"
"strconv"
)

type shipment struct {
quantity, costPerUnit float64
r, c                  int
}

var shipZero = shipment{}

type transport struct {
filename       string
supply, demand []int
costs          [][]float64
matrix         [][]shipment
}

func check(err error) {
if err != nil {
log.Fatal(err)
}
}

func minOf(i, j int) int {
if i < j {
return i
}
return j
}

func newTransport(filename string) *transport {
file, err := os.Open(filename)
check(err)
defer file.Close()
scanner := bufio.NewScanner(file)
scanner.Split(bufio.ScanWords)
scanner.Scan()
numSources, err := strconv.Atoi(scanner.Text())
check(err)
scanner.Scan()
numDests, err := strconv.Atoi(scanner.Text())
check(err)
src := make([]int, numSources)
for i := 0; i < numSources; i++ {
scanner.Scan()
src[i], err = strconv.Atoi(scanner.Text())
check(err)
}
dst := make([]int, numDests)
for i := 0; i < numDests; i++ {
scanner.Scan()
dst[i], err = strconv.Atoi(scanner.Text())
check(err)
}

// fix imbalance
totalSrc := 0
for _, v := range src {
totalSrc += v
}
totalDst := 0
for _, v := range dst {
totalDst += v
}
diff := totalSrc - totalDst
if diff > 0 {
dst = append(dst, diff)
} else if diff < 0 {
src = append(src, -diff)
}

costs := make([][]float64, len(src))
for i := 0; i < len(src); i++ {
costs[i] = make([]float64, len(dst))
}
matrix := make([][]shipment, len(src))
for i := 0; i < len(src); i++ {
matrix[i] = make([]shipment, len(dst))
}
for i := 0; i < numSources; i++ {
for j := 0; j < numDests; j++ {
scanner.Scan()
costs[i][j], err = strconv.ParseFloat(scanner.Text(), 64)
check(err)
}
}
return &transport{filename, src, dst, costs, matrix}
}

func (t *transport) northWestCornerRule() {
for r, northwest := 0, 0; r < len(t.supply); r++ {
for c := northwest; c < len(t.demand); c++ {
quantity := minOf(t.supply[r], t.demand[c])
if quantity > 0 {
t.matrix[r][c] = shipment{float64(quantity), t.costs[r][c], r, c}
t.supply[r] -= quantity
t.demand[c] -= quantity
if t.supply[r] == 0 {
northwest = c
break
}
}
}
}
}

func (t *transport) steppingStone() {
maxReduction := 0.0
var move []shipment = nil
leaving := shipZero
t.fixDegenerateCase()
for r := 0; r < len(t.supply); r++ {
for c := 0; c < len(t.demand); c++ {
if t.matrix[r][c] != shipZero {
continue
}
trial := shipment{0, t.costs[r][c], r, c}
path := t.getClosedPath(trial)
reduction := 0.0
lowestQuantity := float64(math.MaxInt32)
leavingCandidate := shipZero
plus := true
for _, s := range path {
if plus {
reduction += s.costPerUnit
} else {
reduction -= s.costPerUnit
if s.quantity < lowestQuantity {
leavingCandidate = s
lowestQuantity = s.quantity
}
}
plus = !plus
}
if reduction < maxReduction {
move = path
leaving = leavingCandidate
maxReduction = reduction
}
}
}

if move != nil {
q := leaving.quantity
plus := true
for _, s := range move {
if plus {
s.quantity += q
} else {
s.quantity -= q
}
if s.quantity == 0 {
t.matrix[s.r][s.c] = shipZero
} else {
t.matrix[s.r][s.c] = s
}
plus = !plus
}
t.steppingStone()
}
}

func (t *transport) matrixToList() *list.List {
l := list.New()
for _, m := range t.matrix {
for _, s := range m {
if s != shipZero {
l.PushBack(s)
}
}
}
return l
}

func (t *transport) getClosedPath(s shipment) []shipment {
path := t.matrixToList()
path.PushFront(s)

// remove (and keep removing) elements that do not have a
// vertical AND horizontal neighbor
var next *list.Element
for {
removals := 0
for e := path.Front(); e != nil; e = next {
next = e.Next()
nbrs := t.getNeighbors(e.Value.(shipment), path)
if nbrs[0] == shipZero || nbrs[1] == shipZero {
path.Remove(e)
removals++
}
}
if removals == 0 {
break
}
}

// place the remaining elements in the correct plus-minus order
stones := make([]shipment, path.Len())
prev := s
for i := 0; i < len(stones); i++ {
stones[i] = prev
prev = t.getNeighbors(prev, path)[i%2]
}
return stones
}

func (t *transport) getNeighbors(s shipment, lst *list.List) [2]shipment {
var nbrs [2]shipment
for e := lst.Front(); e != nil; e = e.Next() {
o := e.Value.(shipment)
if o != s {
if o.r == s.r && nbrs[0] == shipZero {
nbrs[0] = o
} else if o.c == s.c && nbrs[1] == shipZero {
nbrs[1] = o
}
if nbrs[0] != shipZero && nbrs[1] != shipZero {
break
}
}
}
return nbrs
}

func (t *transport) fixDegenerateCase() {
eps := math.SmallestNonzeroFloat64
if len(t.supply)+len(t.demand)-1 != t.matrixToList().Len() {
for r := 0; r < len(t.supply); r++ {
for c := 0; c < len(t.demand); c++ {
if t.matrix[r][c] == shipZero {
dummy := shipment{eps, t.costs[r][c], r, c}
if len(t.getClosedPath(dummy)) == 0 {
t.matrix[r][c] = dummy
return
}
}
}
}
}
}

func (t *transport) printResult() {
fmt.Println(t.filename)
check(err)
fmt.Printf("\n%s\n", string(text))
fmt.Printf("Optimal solution for %s\n\n", t.filename)
totalCosts := 0.0
for r := 0; r < len(t.supply); r++ {
for c := 0; c < len(t.demand); c++ {
s := t.matrix[r][c]
if s != shipZero && s.r == r && s.c == c {
fmt.Printf(" %3d ", int(s.quantity))
totalCosts += s.quantity * s.costPerUnit
} else {
fmt.Printf("  -  ")
}
}
fmt.Println()
}
fmt.Printf("\nTotal costs: %g\n\n", totalCosts)
}

func main() {
filenames := []string{"input1.txt", "input2.txt", "input3.txt"}
for _, filename := range filenames {
t := newTransport(filename)
t.northWestCornerRule()
t.steppingStone()
t.printResult()
}
}
```
Output:
```input1.txt

2 3
25 35
20 30 10
3 5 7
3 2 5

Optimal solution for input1.txt

20   -     5
-    30    5

Total costs: 180

input2.txt

3 3
12 40 33
20 30 10
3 5 7
2 4 6
9 1 8

Optimal solution for input2.txt

-    -    -    12
20   -    10   10
-    30   -     3

Total costs: 130

input3.txt

4 4
14 10 15 12
10 15 12 15
10 30 25 15
20 15 20 10
10 30 20 20
30 40 35 45

Optimal solution for input3.txt

-    -    -    14
-     9   -     1
10   -     5   -
-     5    7   -
-     1   -    -

Total costs: 1000
```

## J

The current task description refers to the algorithm by name, but I feel that those names are ambiguous (inadequately descriptive - we need the algorithm specified here on rosettacode, not problems which we expect are so well understood that no one needs to describe them).

So, I will be working with this interpretation:

(*) Assign shipments for the lowest cost unsatisfied need which can be supplied. When breaking ties, pick the first one which would be encountered when scanning left to right, top to bottom (which is the same order you use when reading english text on a page).

(*) Supply however much of the need which can be supplied by that shipment.

(*) Repeat until done.

(If this algorithm is incorrect for this task, that would just underline the need for a better task description. And, probably, also: the need for a more representative task example.)

In other words:

```NB. C's y[m] v= x  implemented as  x m ndxasgn v y
ndxasgn=: conjunction define
:
((m{y)v x) m} y
)

:
need=. x
supl=. y
cost=. m
dims=. supl ,&# need
r=. dims\$0
while. 1 e., xfr=. supl *&*/ need do.
'iS iN'=. ndxs=. dims#:(i. <./), cost % xfr
n=. (iS { supl) <. iN { need
need=. n iN ndxasgn - need
supl=. n iS ndxasgn - supl
r=. n (<ndxs)} r
end.
)
```

```need=: 20 30 10
supply=: 25 35
cost=:3 5 7,:3 2 5
```

```   need cost trans supply
20  0 5
0 30 5
```

## Java

Works with: Java version 8
```import java.io.File;
import java.util.*;
import static java.util.Arrays.stream;
import static java.util.stream.Collectors.toCollection;

public class TransportationProblem {

private static int[] demand;
private static int[] supply;
private static double[][] costs;
private static Shipment[][] matrix;

private static class Shipment {
final double costPerUnit;
final int r, c;
double quantity;

public Shipment(double q, double cpu, int r, int c) {
quantity = q;
costPerUnit = cpu;
this.r = r;
this.c = c;
}
}

static void init(String filename) throws Exception {

try (Scanner sc = new Scanner(new File(filename))) {
int numSources = sc.nextInt();
int numDestinations = sc.nextInt();

List<Integer> src = new ArrayList<>();
List<Integer> dst = new ArrayList<>();

for (int i = 0; i < numSources; i++)

for (int i = 0; i < numDestinations; i++)

// fix imbalance
int totalSrc = src.stream().mapToInt(i -> i).sum();
int totalDst = dst.stream().mapToInt(i -> i).sum();
if (totalSrc > totalDst)
else if (totalDst > totalSrc)

supply = src.stream().mapToInt(i -> i).toArray();
demand = dst.stream().mapToInt(i -> i).toArray();

costs = new double[supply.length][demand.length];
matrix = new Shipment[supply.length][demand.length];

for (int i = 0; i < numSources; i++)
for (int j = 0; j < numDestinations; j++)
costs[i][j] = sc.nextDouble();
}
}

static void northWestCornerRule() {

for (int r = 0, northwest = 0; r < supply.length; r++)
for (int c = northwest; c < demand.length; c++) {

int quantity = Math.min(supply[r], demand[c]);
if (quantity > 0) {
matrix[r][c] = new Shipment(quantity, costs[r][c], r, c);

supply[r] -= quantity;
demand[c] -= quantity;

if (supply[r] == 0) {
northwest = c;
break;
}
}
}
}

static void steppingStone() {
double maxReduction = 0;
Shipment[] move = null;
Shipment leaving = null;

fixDegenerateCase();

for (int r = 0; r < supply.length; r++) {
for (int c = 0; c < demand.length; c++) {

if (matrix[r][c] != null)
continue;

Shipment trial = new Shipment(0, costs[r][c], r, c);
Shipment[] path = getClosedPath(trial);

double reduction = 0;
double lowestQuantity = Integer.MAX_VALUE;
Shipment leavingCandidate = null;

boolean plus = true;
for (Shipment s : path) {
if (plus) {
reduction += s.costPerUnit;
} else {
reduction -= s.costPerUnit;
if (s.quantity < lowestQuantity) {
leavingCandidate = s;
lowestQuantity = s.quantity;
}
}
plus = !plus;
}
if (reduction < maxReduction) {
move = path;
leaving = leavingCandidate;
maxReduction = reduction;
}
}
}

if (move != null) {
double q = leaving.quantity;
boolean plus = true;
for (Shipment s : move) {
s.quantity += plus ? q : -q;
matrix[s.r][s.c] = s.quantity == 0 ? null : s;
plus = !plus;
}
steppingStone();
}
}

return stream(matrix)
.flatMap(row -> stream(row))
.filter(s -> s != null)
}

static Shipment[] getClosedPath(Shipment s) {

// remove (and keep removing) elements that do not have a
// vertical AND horizontal neighbor
while (path.removeIf(e -> {
Shipment[] nbrs = getNeighbors(e, path);
return nbrs[0] == null || nbrs[1] == null;
}));

// place the remaining elements in the correct plus-minus order
Shipment[] stones = path.toArray(new Shipment[path.size()]);
Shipment prev = s;
for (int i = 0; i < stones.length; i++) {
stones[i] = prev;
prev = getNeighbors(prev, path)[i % 2];
}
return stones;
}

static Shipment[] getNeighbors(Shipment s, LinkedList<Shipment> lst) {
Shipment[] nbrs = new Shipment[2];
for (Shipment o : lst) {
if (o != s) {
if (o.r == s.r && nbrs[0] == null)
nbrs[0] = o;
else if (o.c == s.c && nbrs[1] == null)
nbrs[1] = o;
if (nbrs[0] != null && nbrs[1] != null)
break;
}
}
return nbrs;
}

static void fixDegenerateCase() {
final double eps = Double.MIN_VALUE;

if (supply.length + demand.length - 1 != matrixToList().size()) {

for (int r = 0; r < supply.length; r++)
for (int c = 0; c < demand.length; c++) {
if (matrix[r][c] == null) {
Shipment dummy = new Shipment(eps, costs[r][c], r, c);
if (getClosedPath(dummy).length == 0) {
matrix[r][c] = dummy;
return;
}
}
}
}
}

static void printResult(String filename) {
System.out.printf("Optimal solution %s%n%n", filename);
double totalCosts = 0;

for (int r = 0; r < supply.length; r++) {
for (int c = 0; c < demand.length; c++) {

Shipment s = matrix[r][c];
if (s != null && s.r == r && s.c == c) {
System.out.printf(" %3s ", (int) s.quantity);
totalCosts += (s.quantity * s.costPerUnit);
} else
System.out.printf("  -  ");
}
System.out.println();
}
System.out.printf("%nTotal costs: %s%n%n", totalCosts);
}

public static void main(String[] args) throws Exception {

for (String filename : new String[]{"input1.txt", "input2.txt",
"input3.txt"}) {
init(filename);
northWestCornerRule();
steppingStone();
printResult(filename);
}
}
}
```
```input1.txt

2 3
25 35
20 30 10
3 5 7
3 2 5

Optimal solution input1.txt

20   -     5
-    30    5

Total costs: 180.0
```
```input2.txt

3 3
12 40 33
20 30 10
3 5 7
2 4 6
9 1 8

Optimal solution input2.txt

-    -    -    12
20   -    10   10
-    30   -     3

Total costs: 130.0
```
```input3.txt

4 4
14 10 15 12
10 15 12 15
10 30 25 15
20 15 20 10
10 30 20 20
30 40 35 45

Optimal solution input3.txt

-    -    -    14
-     9   -     1
10   -     5   -
-     5    7   -
-     1   -    -

Total costs: 1000.0```

## Julia

Code taken from here using JuMP.

```using JuMP, GLPK

# cost vector
c = [3, 5, 7, 3, 2, 5];
N = size(c,1);

# constraints Ax (<,>,=) b
A = [1 1 1 0 0 0
0 0 0 1 1 1
1 0 0 1 0 0
0 1 0 0 1 0
0 0 1 0 0 1];
b = [ 25,  35,  20,  30,  10];
s = ['<', '<', '=', '=', '='];

# construct model
model = Model(GLPK.Optimizer)
@variable(model, x[i=1:N] >= 0, base_name="traded quantities")
cost_fn = @expression(model, c'*x)                                              # cost function
@constraint(model, C1, A[1:2,:]*x .<= b[1:2])                                   # inequality constraints
@constraint(model, C2, A[3:5,:]*x .== b[3:5])                                   # equality constraints
@objective(model, Min, cost_fn)                                                 # objective function

# solve model
status = JuMP.optimize!(model);
xstar = value.(x);
println("solution vector of quantities = ", xstar)
println("minimum total cost = ", JuMP.objective_value(model))

# recover Lagrange multipliers for post-optimality
λ = [JuMP.dual(C1[1]),JuMP.dual(C1[2])]
μ = [JuMP.dual(C2[1]),JuMP.dual(C2[2]),JuMP.dual(C2[3])]
```
Output:
```solution vector of quantities = [20.0, 0.0, 5.0, 0.0, 30.0, 5.0]
minimum total cost = 180.0```

## Kotlin

Translation of: Java
```// version 1.1.51

import java.io.File
import java.util.Scanner

class Transport(val filename: String) {

private val supply: IntArray
private val demand: IntArray
private val costs : Array<DoubleArray>
private val matrix: Array<Array<Shipment>>

class Shipment(
var quantity: Double,
val costPerUnit: Double,
val r: Int,
val c: Int
)

companion object {
private val ZERO = Shipment(0.0, 0.0, -1, -1) // to avoid nullable Shipments
}

init {
val sc = Scanner(File(filename))
try {
val numSources = sc.nextInt()
val numDestinations = sc.nextInt()
val src = MutableList(numSources) { sc.nextInt() }
val dst = MutableList(numDestinations) { sc.nextInt() }

// fix imbalance
val totalSrc = src.sum()
val totalDst = dst.sum()
if (totalSrc > totalDst)
else if (totalDst > totalSrc)
supply = src.toIntArray()
demand = dst.toIntArray()

costs  = Array(supply.size) { DoubleArray(demand.size) }
matrix = Array(supply.size) { Array(demand.size) { ZERO } }
for (i in 0 until numSources) {
for (j in 0 until numDestinations) costs[i][j] = sc.nextDouble()
}
}
finally {
sc.close()
}
}

fun northWestCornerRule() {
var northwest = 0
for (r in 0 until supply.size) {
for (c in northwest until demand.size) {
val quantity = minOf(supply[r], demand[c]).toDouble()
if (quantity > 0.0) {
matrix[r][c] = Shipment(quantity, costs[r][c], r, c)
supply[r] -= quantity.toInt()
demand[c] -= quantity.toInt()
if (supply[r] == 0) {
northwest = c
break
}
}
}
}
}

fun steppingStone() {
var maxReduction = 0.0
var move: Array<Shipment>? = null
var leaving = ZERO
fixDegenerateCase()

for (r in 0 until supply.size) {
for (c in 0 until demand.size) {
if (matrix[r][c] != ZERO) continue
val trial = Shipment(0.0, costs[r][c], r, c)
val path = getClosedPath(trial)
var reduction = 0.0
var lowestQuantity = Int.MAX_VALUE.toDouble()
var leavingCandidate = ZERO
var plus = true
for (s in path) {
if (plus) {
reduction += s.costPerUnit
}
else {
reduction -= s.costPerUnit
if (s.quantity < lowestQuantity) {
leavingCandidate = s
lowestQuantity = s.quantity
}
}
plus = !plus
}
if (reduction < maxReduction) {
move = path
leaving = leavingCandidate
maxReduction = reduction
}
}
}

if (move != null) {
val q = leaving.quantity
var plus = true
for (s in move) {
s.quantity += if (plus) q else -q
matrix[s.r][s.c] = if (s.quantity == 0.0) ZERO else s
plus = !plus
}
steppingStone()
}
}

private fun matrixToList() =
LinkedList<Shipment>(matrix.flatten().filter { it != ZERO } )

private fun getClosedPath(s: Shipment): Array<Shipment> {
val path = matrixToList()

// remove (and keep removing) elements that do not have a
// vertical AND horizontal neighbor
while (path.removeIf {
val nbrs = getNeighbors(it, path)
nbrs[0] == ZERO || nbrs[1] == ZERO
}) ; // empty statement

// place the remaining elements in the correct plus-minus order
val stones = Array<Shipment>(path.size) { ZERO }
var prev = s
for (i in 0 until stones.size) {
stones[i] = prev
prev = getNeighbors(prev, path)[i % 2]
}
return stones
}

private fun getNeighbors(s: Shipment, lst: LinkedList<Shipment>): Array<Shipment> {
val nbrs = Array<Shipment>(2) { ZERO }
for (o in lst) {
if (o != s) {
if (o.r == s.r && nbrs[0] == ZERO)
nbrs[0] = o
else if (o.c == s.c && nbrs[1] == ZERO)
nbrs[1] = o
if (nbrs[0] != ZERO && nbrs[1] != ZERO) break
}
}
return nbrs
}

private fun fixDegenerateCase() {
val eps = Double.MIN_VALUE
if (supply.size + demand.size - 1 != matrixToList().size) {
for (r in 0 until supply.size) {
for (c in 0 until demand.size) {
if (matrix[r][c] == ZERO) {
val dummy = Shipment(eps, costs[r][c], r, c)
if (getClosedPath(dummy).size == 0) {
matrix[r][c] = dummy
return
}
}
}
}
}
}

fun printResult() {
println("\$filename\n\n\$text")
println("Optimal solution \$filename\n")
var totalCosts = 0.0

for (r in 0 until supply.size) {
for (c in 0 until demand.size) {
val s = matrix[r][c]
if (s != ZERO && s.r == r && s.c == c) {
print(" %3s ".format(s.quantity.toInt()))
totalCosts += s.quantity * s.costPerUnit
}
else print("  -  ")
}
println()
}
println("\nTotal costs: \$totalCosts\n")
}
}

fun main(args: Array<String>) {
val filenames = arrayOf("input1.txt", "input2.txt", "input3.txt")
for (filename in filenames) {
with (Transport(filename)) {
northWestCornerRule()
steppingStone()
printResult()
}
}
}
```
Output:
```Same as Java entry
```

## Nim

Translation of: Go
```import fenv, lists, math, sequtils, strformat, strutils

type

Shipment = object
quantity: float
costPerUnit: float
r, c: int

Transport = object
filename: string
supply: seq[int]
demand: seq[int]
costs: seq[seq[float]]
matrix: seq[seq[Shipment]]

const ShipZero = Shipment()

template emitError(msg: string) =
raise newException(ValueError, msg)

proc initTransport(filename: string): Transport =
let infile = filename.open()
let numSources = fields[0]
let numDests = fields[1]
if numSources < 1 or numDests < 1:
emitError "wrong number of sources or destinations."
if src.len != numSources:
emitError "wrong number of sources; got \$1, expected \$2.".format(src.len, numSources)
if dst.len != numDests:
emitError "wrong number of destinations; got \$1, expected \$2.".format(dst.len, numDests)

# Fix imbalance.
let totalSrc = sum(src)
let totalDst = sum(dst)
let diff = totalSrc - totalDst
if diff > 0: dst.add diff
elif diff < 0: src.add -diff

var costs = newSeqWith(src.len, newSeq[float](dst.len))
var matrix = newSeqWith(src.len, newSeq[Shipment](dst.len))

for i in 0..<numSources:
if fields.len > dst.len:
emitError "wrong number of costs; got \$1, expected \$2.".format(fields.len, numDests)
for j in 0..<numDests:
costs[i][j] = fields[j]

result = Transport(filename: filename, supply: move(src),
demand: move(dst), costs: move(costs), matrix: move(matrix))

func northWestCornerRule(tr: var Transport) =
var northWest = 0
for r in 0..tr.supply.high:
for c in northWest..tr.demand.high:
let quantity = min(tr.supply[r], tr.demand[c])
if quantity > 0:
tr.matrix[r][c] = Shipment(quantity: quantity.toFloat, costPerUnit: tr.costs[r][c], r: r, c: c)
dec tr.supply[r], quantity
dec tr.demand[c], quantity
if tr.supply[r] == 0:
northWest = c
break

func getNeighbors(tr: Transport; s: Shipment; list: ShipmentList): array[2, Shipment] =
for o in list:
if o != s:
if o.r == s.r and result[0] == ShipZero:
result[0] = o
elif o.c == s.c and result[1] == ShipZero:
result[1] = o
if result[0] != ShipZero and result[1] != ShipZero:
break

func matrixToList(tr: Transport): ShipmentList =
for m in tr.matrix:
for s in m:
if s != ShipZero:
result.append(s)

func getClosedPath(tr: Transport; s: Shipment): seq[Shipment] =
var path = tr.matrixToList
path.prepend(s)

# Remove (and keep removing) elements that do not have a
# vertical and horizontal neighbor.
while true:
var removals = 0
for e in path.nodes:
let nbrs = tr.getNeighbors(e.value, path)
if nbrs[0] == ShipZero or nbrs[1] == ShipZero:
path.remove(e)
inc removals
if removals == 0:
break

# Place the remaining elements in the correct plus-minus order.
var prev = s
var i = 0
for _ in path:
prev = tr.getNeighbors(prev, path)[i]
i = 1 - i

func fixDegenerateCase(tr: var Transport) =
const Eps = minimumPositiveValue(float)
if tr.supply.len + tr.demand.len - 1 != tr.matrix.len * tr.matrix[0].len:
for r in 0..tr.supply.high:
for c in 0..tr.demand.high:
if tr.matrix[r][c] == ShipZero:
let dummy = Shipment(quantity: Eps, costPerUnit: tr.costs[r][c], r: r, c: c)
if tr.getClosedPath(dummy).len == 0:
tr.matrix[r][c] = dummy
return

func steppingStone(tr: var Transport) =
var maxReduction = 0.0
var move: seq[Shipment]
var leaving = ShipZero
tr.fixDegenerateCase()

for r in 0..tr.supply.high:
for c in 0..tr.demand.high:
if tr.matrix[r][c] != ShipZero:
continue
let trial = Shipment(quantity: 0, costPerUnit: tr.costs[r][c], r: r, c: c)
var path = tr.getClosedPath(trial)
var reduction = 0.0
var lowestQuantity = float(int32.high)
var leavingCandidate = ShipZero
var plus = true
for s in path:
if plus:
reduction += s.costPerUnit
else:
reduction -= s.costPerUnit
if s.quantity < lowestQuantity:
leavingCandidate = s
lowestQuantity = s.quantity
plus = not plus
if reduction < maxReduction:
move = move(path)
leaving = leavingCandidate
maxReduction = reduction

if move.len != 0:
let q = leaving.quantity
var plus = true
for s in move.mitems:
if plus: s.quantity += q
else: s.quantity -= q
tr.matrix[s.r][s.c] = if s.quantity == 0: ShipZero else: s
plus = not plus
tr.steppingStone()

proc printResult(tr: Transport) =
echo tr.filename, '\n'
echo "\nOptimal solution for ", tr.filename, '\n'
var totalCosts = 0.0
for r in 0..tr.supply.high:
for c in 0..tr.demand.high:
let s = tr.matrix[r][c]
if s != ShipZero and s.r == r and s.c == c:
stdout.write &" {int(s.quantity):3} "
totalCosts += s.quantity * s.costPerUnit
else:
stdout.write "  -  "
echo()
echo &"\nTotal costs: {totalCosts:g}\n"

when isMainModule:

const Filenames = ["input1.txt", "input2.txt", "input3.txt"]
for filename in Filenames:
var tr = initTransport(filename)
tr.northWestCornerRule()
tr.steppingStone()
tr.printResult()
```
Output:
```input1.txt

2 3
25 35
20 30 10
3 5 7
3 2 5

Optimal solution for input1.txt

20   -     5
-    30    5

Total costs: 180

input2.txt

3 3
12 40 33
20 30 10
3 5 7
2 4 6
9 1 8

Optimal solution for input2.txt

-    -    -    12
20   -    10   10
-    30   -     3

Total costs: 130

input3.txt

4 4
14 10 15 12
10 15 12 15
10 30 25 15
20 15 20 10
10 30 20 20
30 40 35 45

Optimal solution for input3.txt

-    -    -    14
-     9   -     1
10   -     5   -
-     5    7   -
-     1   -    -

Total costs: 1000```

## Pascal

```Program transport;
{ based on the program of <Svetlana Belashova> }

Uses Crt;

Label l1;

Const N=10;
n1=7; n2=7;
Sa:longint=0;
Sb:longint=0;

Type points=Array [1..N] of longint;
distribution=Array [1..N,1..N] of longint;

Var A,B,alfa,beta,B_d,x:points;
c,p:distribution;
f,f0,x_min,Sp:longint;
Nt,x_p,r,r_min,ki,kj,Na,Nb,h,l,i,j:byte;
d:char;
u:Array[1..N*N] of byte;

Procedure Nul (var a:points);
var i:byte;
Begin
for i:=1 to N do a[i]:=0;
End;

Procedure PrintS (x,y:byte; s:string; c:byte);
Begin
TextColor(c);
GotoXY(x,y);
Write(s);
End;

Procedure Print (x,y:byte; n:byte; a:longint; c:byte);
Begin
TextColor(c);
GotoXY(x,y); Write(' ':n);
GotoXY(x,y); Write(a);
End;

var i:integer;
s:string;
c:char;
j,k:byte;
Begin
s:=''; i:=1;
TextColor(11);
Repeat
Case ord(c) of
48..57:         begin s:=s+c;
Write(c);
inc(i);
end;
8:              if i>1 then begin dec(i);
Delete(s,i,1);
Write(chr(8),' ',chr(8));
end;
end;
j:=WhereX;
GotoXY(60,1); ClrEOL;
if i>y then begin
TextColor(4);
Write('Not more than ');
for k:=1 to y-1 do Write('9');
TextColor(11);
end;
GotoXY(j,1);
Until (ord(c)=13) and (i<y+1);
val(s,x,i);
End;

Procedure horizontal (a,b,c,d,e:char);
var i,j:byte;
Begin
Write(a);
for i:=1 to n2 do Write(b);
Write(c);
for i:=1 to Nb do begin
for j:=1 to n1 do Write(b);
if i<>Nb then Write(d) else Write(c);
end;
for i:=1 to 4 do Write(b);
Write(e);
End;

Procedure vertical;
var i:byte;
Begin
Write('│',' ':n2,'║');
for i:=1 to Nb-1 do Write(' ':n1,'│');
WriteLn(' ':n1,'║',' ' :4,'│');
End;

Procedure Table; { Drawing the table }
Begin
ClrScr;
TextColor(1);
h:=6+Na*3;
l:=14+Nb*7;
GotoXY(1,3);
for i:=3 to h do vertical;
GotoXY(1,2);
horizontal('┌','─','╥','┬','┐');
for i:=1 to Na+1 do begin
GotoXY(1,i*3+2);
if (i=1) or (i=Na+1)
then horizontal('╞','═','╬','╪','╡')
else horizontal('├','─','╫','┼','┤');
end;
GotoXY(1,h+1);
horizontal('└','─','╨','┴','┘');
TextColor(9);
for i:=1 to Na do begin
GotoXY(5,i*3+3);
Write('A',i);
end;
for i:=1 to Nb do begin
GotoXY(i*(n1+1)+n2-2,3);
Write('B',i);
end;
l:=Nb*(n1+1)+n2+3;
h:=Na*3+6;
PrintS(4,3,'\Bj',9);
PrintS(4,4,'Ai\',9);
PrintS(1,1,'Table N1',14);
PrintS(l,4,'alfa',9);
PrintS(3,h,'beta',9);
End;

Procedure EnterIntoTheTable (var a:points; b:byte; c:char); { Entering into the table }
var i,l,m:byte;
Begin
for i:=1 to b do begin
TextColor(3);
GotoXY(32,1);
ClrEOL;
Write(c,i,'=  ');
TextColor(14);
Case c of
'A':     GotoXY(n2-trunc(ln(a[i])/ln(10)),i*3+4);
'B':     GotoXY(n2+i*(n1+1)-trunc(ln(a[i])/ln(10)),4);
end;
Write(a[i]);
end;
End;

Function CalculatingTheCost:longint;        { Calculating the cost of the plan }
var i,j:byte;
f:longint;
Begin
f:=0;
for i:=1 to Na do
for j:=1 to Nb do
if p[i,j]>0 then inc(f,c[i,j]*p[i,j]);
GotoXY(65,Nt+2);
TextColor(10);
Write('F',Nt,'=',f);
CalculatingTheCost:=f;
End;

Function CalculatingThePotentials:boolean;      { Calculating the potentials }
var k,i,j:byte;
Z_a,Z_b:points;
d:boolean;
Begin
Nul(Z_a); Nul(Z_b);
alfa[1]:=0; Z_a[1]:=1; k:=1;
Repeat
d:=1=1;
for i:=1 to Na do
if Z_a[i]=1 then
for j:=1 to Nb do
if (p[i,j]>-1) and (Z_b[j]=0) then begin
Z_b[j]:=1;
beta[j]:=c[i,j]-alfa[i];
inc(k);
d:=1=2;
end;
for i:=1 to Nb do
if Z_b[i]=1 then
for j:=1 to Na do
if (p[j,i]>-1) and (Z_a[j]=0) then begin
Z_a[j]:=1;
alfa[j]:=c[j,i]-beta[i];
inc(k);
d:=1=2;
end;
Until (k=Na+Nb) or d;
if d then begin
i:=1;
While Z_a[i]=1 do inc(i);
j:=1;
While Z_b[j]=0 do inc(j);
p[i,j]:=0;
Print((j+1)*(n1+1)+n2-8,i*3+4,1,p[i,j],7);
end;

CalculatingThePotentials:=d;
End;

Procedure OutputThePlan;         { Output the plan of distribution }
var i,j,h,l,k:byte;
c_max:longint;
Begin
k:=0;
for i:=1 to Na do begin
h:=i*3+4;
for j:=1 to Nb do begin
l:=j*(n1+1)+n2-5;
GotoXY(l,h);
Write(' ':n1);
if p[i,j]>0 then begin
inc(k);
Print(l-trunc(ln(p[i,j])/ln(10))+5,h,1,p[i,j],14);
end
else if p[i,j]=0 then begin
Print(l+n1-2,h,1,p[i,j],14);
inc(k);
end;
end;
end;

While CalculatingThePotentials do inc(k);

if k>Na+Nb-1 then PrintS(40,1,'k > n+m-1',12);
End;

Function CalculatingTheCoefficients(var ki,kj:byte):integer; { Calculation the coefficients in the free cells }
var i,j:byte;
k,k_min:integer;
b:boolean;
Begin
b:=1=1;
for i:=1 to Na do
for j:=1 to Nb do
if p[i,j]=-1 then begin
k:=c[i,j]-alfa[i]-beta[j];
if b then begin
b:=1=2;
ki:=i; kj:=j; k_min:=k;
end else
if k<k_min then begin
k_min:=k;
ki:=i; kj:=j;
end;
TextColor(6);
GotoXY(j*(n1+1)+n2-5,i*3+4);
Write('(',k,')');
end;
if k_min<0 then PrintS(kj*(n1+1)+n2,ki*3+4,'X',12);
CalculatingTheCoefficients:=k_min;
End;

Procedure div_mod(c:byte; var a,b:byte);   { Translate one-dimensional array to two-dimensional }
Begin
b:=c mod Nb; a:=c div Nb +1;
if b=0 then begin
b:=Nb; dec(a);
end;
End;

Procedure Recursive(Xi,Yi:byte; var z:boolean; var c:byte);
var i,j:byte;
Begin
z:=1=2;
Case c of
1:   for i:=1 to Na do
if i<>Xi then
if p[i,Yi]>-1 then begin
if u[(i-1)*Nb+Yi]=0 then begin
u[(Xi-1)*Nb+Yi]:=(i-1)*Nb+Yi;
c:=2;
Recursive(i,Yi,z,c);
if z then exit;
end;
end
else if (i=ki) and (Yi=kj) then begin
u[(Xi-1)*Nb+Yi]:=(ki-1)*Nb+kj;
z:=not z;
exit;
end;
2:   for i:=1 to Nb do
if i<>Yi then
if p[Xi,i]>-1 then begin
if u[(Xi-1)*Nb+i]=0 then begin
u[(Xi-1)*Nb+Yi]:=(Xi-1)*Nb+i;
c:=1;
Recursive(Xi,i,z,c);
if z then exit;
end;
end
else if (Xi=ki) and (i=kj) then begin
u[(Xi-1)*Nb+Yi]:=(ki-1)*Nb+kj;
z:=not z;
exit;
end;
end;
u[(Xi-1)*Nb+Yi]:=0;
c:=c mod 2 +1;
End;

Procedure Contour;       { Determine the contour of displacement }
var i,j,k,mi,mj,l:byte;
z:boolean;
p_m:longint;
Begin
for i:=1 to N*N do u[i]:=0;
l:=1;
Recursive(ki,kj,z,l);
i:=ki; j:=kj;
k:=u[(i-1)*Nb+j];
div_mod(k,i,j);
mi:=i; mj:=j; l:=1;
Repeat
inc(l);
k:=u[(i-1)*Nb+j];
div_mod(k,i,j);
if l mod 2=1 then
if p[i,j]<p[mi,mj] then begin
mi:=i; mj:=j;
end;
Until (i=ki) and (j=kj);

i:=ki; j:=kj; l:=0;
p_m:=p[mi,mj];
Repeat
if l mod 2=0 then begin
inc(p[i,j],p_m);
PrintS((n1+1)*j+n2-1,i*3+3,'(+)',12);
end else begin
dec(p[i,j],p_m);
PrintS((n1+1)*j+n2-1,i*3+3,'(-)',12);
end;
if l=0 then inc(p[i,j]);
k:=u[(i-1)*Nb+j];
div_mod(k,i,j);
inc(l);
Until (i=ki) and (j=kj);
p[mi,mj]:=-1;
End;

Procedure Pause;
var d:char;
Begin
TextColor(6);
GotoXY(40,1);
Write('Press any key');
GotoXY(40,1);
ClrEOL;
End;

BEGIN
Nul(alfa); Nul(beta);
Nt:=1;
ClrScr;
TextColor(10);
Repeat
Write('Enter the number of suppliers (2<=Na<=',N-1,')   ');
Write('Enter the number of consumers (2<=Nb<=',N-1,')   ');
Until (Na>1) and (Na<=N-1) and (Nb>1) and (Nb<=N-1);
Table;

PrintS(1,1,'Enter the production quantity:',3);
EnterIntoTheTable(A,Na,'A');
EnterIntoTheTable(B,Nb,'B');
TextColor(3);
GotoXY(1,1); ClrEOL;
Write('Enter the cost of transportation');
for i:=1 to Na do
for j:=1 to Nb do begin
TextColor(3);
GotoXY(29,1); ClrEOL;
Write('A',i,' - B',j,'  ');
Print((n1+1)*j+n2-4,i*3+3,1,c[i,j],11);
end;

GotoXY(1,1);
ClrEOL;
TextColor(14);
Write('Table N1');

for i:=1 to Na do Sa:=Sa+A[i];
for i:=1 to Nb do Sb:=Sb+B[i];
if Sa<>Sb then begin
PrintS(20,1,'The problem is open (Press any key)',7);
if Sa>Sb then begin
inc(Nb);
B[Nb]:=Sa-Sb;
for i:=1 to Na do c[i,Nb]:=0;
end else begin
inc(Na);
A[Na]:=Sb-Sa;
for i:=1 to Nb do c[Na,i]:=0;
end;
Table;
for i:=1 to Na do
for j:=1 to Nb do Print((n1+1)*j+n2-4,i*3+3,1,c[i,j],11);
for i:=1 to Na do
Print(n2-trunc(ln(A[i])/ln(10)),i*3+4,1,A[i],14);
for i:=1 to Nb do
Print(n2+i*(n1+1)-trunc(ln(B[i])/ln(10)),4,1,B[i],14);
PrintS(20,1,'The problem is open',7);
end
else PrintS(20,1,'The problem is closed',7);

(**************** Drafting the basic plan ******************)
for i:=1 to Nb do B_d[i]:=B[i];
for i:=1 to Na do begin
for j:=1 to Nb do x[j]:=j;
for j:=1 to Nb-1 do begin
x_min:=c[i,x[j]];
r_min:=j;
for r:= j+1 to Nb do
if (x_min>c[i,x[r]]) or
((x_min=c[i,x[r]]) and (B[x[r]]>b[x[r_min]])) then
begin
x_min :=c[i,x[r]];
r_min:=r;
end;
x_p:=x[r_min];
x[r_min]:=x[j];
x[j]:=x_p;
end;
Sp:=0;
for j:=1 to Nb do begin
p[i,x[j]]:=B_d[x[j]];
if p[i,x[j]]>A[i]-Sp then p[i,x[j]]:=A[i]-Sp;
inc(Sp,p[i,x[j]]);
dec(B_d[x[j]],p[i,x[j]]);
end;
end;
(***********************************************************)

for i:=1 to Na do
for j:=1 to Nb do if p[i,j]=0 then p[i,j]:=-1;
OutputThePlan;
f:=CalculatingTheCost; f0:=F;

While CalculatingThePotentials do;
for i:=1 to Na do Print(l+1,i*3+3,3,alfa[i],11);
for i:=1 to Nb do Print(i*(n1+1)+n2-4,h,6,beta[i],11);
Pause;

(******* gradual approach the plan to the optimality ******)
While CalculatingTheCoefficients(ki,kj)<0 do begin
Contour;
pause;
for i:=1 to Na do
for j:=1 to Nb do PrintS((n1+1)*j+n2-1,i*3+3,'   ',14);
inc(Nt);
GotoXY(1,1);
Write('Table N',Nt);
OutputThePlan;
f0:=f; f:=CalculatingTheCost;
if CalculatingThePotentials then Goto l1;
for i:=1 to Na do Print(l+1,i*3+3,3,alfa[i],11);
for i:=1 to Nb do Print(i*(n1+1)+n2-4,h,6,beta[i],11);
Pause;
end;
(***********************************************************)

PrintS(40,1,'Solution is optimal',12);
PrintS(60,1,'(any key)',6);
for i:=1 to Na do
for j:=1 to Nb do if p[i,j]=-1 then begin
h:=i*3+4;
l:=j*(n1+1)+n2-5;
GotoXY(l,h);
Write(' ':n1);
end;
GotoXY(40,1);
END.
```

## Perl

Just re-using the code from Vogel's approximation method, tweaked to handle specific input:

```use strict;
use warnings;
use feature 'say';
use List::AllUtils qw( max_by nsort_by min );

my \$data = <<END;
A=20 B=30 C=10
S=25 T=35
AS=3 BS=5 CS=7
CT=3 BT=2 CT=5
END

my \$table = sprintf +('%4s' x 4 . "\n") x 3,
map {my \$t = \$_; map "\$_\$t", '', 'A' .. 'C' } '' , 'S' .. 'T';

my (\$cost, %assign) = (0);
while( \$data =~ /\b\w=\d/ ) {
my @penalty;
for ( \$data =~ /\b(\w)=\d/g ) {
my @all = map /(\d+)/, nsort_by { /\d+/ && \$& }
grep { my (\$t, \$c) = /(.)(.)=/; \$data =~ /\b\$c=\d/ and \$data =~ /\b\$t=\d/ }
\$data =~ /\$_\w=\d+|\w\$_=\d+/g;
push @penalty, [ \$_, (\$all[1] // 0) - \$all[0] ];
}
my \$rc = (max_by { \$_->[1] } nsort_by
{ my \$x = \$_->[0]; \$data =~ /(?:\$x\w|\w\$x)=(\d+)/ && \$1 } @penalty)->[0];
my @lowest = nsort_by { /\d+/ && \$& }
grep { my (\$t, \$c) = /(.)(.)=/; \$data =~ /\b\$c=\d/ and \$data =~ /\b\$t=\d/ }
\$data =~ /\$rc\w=\d+|\w\$rc=\d+/g;
my (\$t, \$c) = \$lowest[0] =~ /(.)(.)/;
my \$allocate = min \$data =~ /\b[\$t\$c]=(\d+)/g;
\$table =~ s/\$t\$c/ sprintf "%2d", \$allocate/e;
\$cost += \$data =~ /\$t\$c=(\d+)/ && \$1 * \$allocate;
\$data =~ s/\b\$_=\K\d+/ \$& - \$allocate || '' /e for \$t, \$c;
}

say my \$result = "cost \$cost\n\n" . \$table =~ s/[A-Z]{2}/--/gr;
```
Output:
```cost 170

A   B   C
S  20  --   5
T  --  30   5```

## Phix

The simplest solution I could think of.
Assumes 0 cost is not allowed, but using say -1 as the "done" cost instead should be fine.

```with javascript_semantics
procedure solve(sequence needs, avail, costs)
-- the costs parameter should be length(avail/*aka suppliers*/) rows
--                            of length(needs/*aka customers*/) cols
assert(length(costs)==length(avail))
assert(apply(costs,length)==repeat(length(needs),length(avail)))
sequence res = repeat(repeat(0,length(needs)),length(avail))
while true do
integer best = 0, supplier, customer
for s=1 to length(costs) do
for c=1 to length(costs[s]) do
integer csc = costs[s][c]
if csc!=0 and (best=0 or csc<best) then
best = csc
supplier = s
customer = c
end if
end for
end for
if best=0 then exit end if -- all costs examined
integer amt = min(avail[supplier],needs[customer])
-- obviously amt can be 0, in which case this just
-- removes cost entry from further consideration.
avail[supplier] -= amt
needs[customer] -= amt
res[supplier,customer] = amt
costs[supplier,customer] = 0
end while
pp(res,{pp_Nest,1})
end procedure

solve({20,30,10},{25,35},{{3,5,7},{3,2,5}})
```
Output:
```{{20,0,5},
{0,30,5}}
```

### stepping stones

Obviously I did not really quite understand the problem when I rattled out the above... this does much better.

Translation of: Go
```-- demo\rosetta\Transportation_problem.exw
with javascript_semantics
enum QTY, COST, R, C -- (a shipment)
constant eps = 1e-12

function print_matrix(sequence matrix)
atom total_costs = 0.0
for r=1 to length(matrix) do
for c=1 to length(matrix[r]) do
object s = matrix[r][c]
string st = "  -  "
if s!=0 and s[R]==r and s[C]==c then
atom qty = round(s[QTY]) -- (remove +/-eps)
if qty!=0 then
st = sprintf(" %3d ", qty)
total_costs += qty * s[COST]
end if
end if
puts(1,st)
end for
printf(1,"\n")
end for
end function

procedure print_result(sequence transport, atom expected)
sequence matrix = transport[4]
printf(1,"Optimal solution\n\n")
atom total_costs = print_matrix(matrix)
printf(1,"\nTotal costs: %g (expected %g)\n\n", {total_costs,expected})
end procedure

function get_neighbors(sequence shipment, lst)
sequence nbrs = {0,0}
for e=1 to length(lst) do
sequence o = lst[e]
if o!=shipment then
if o[R]==shipment[R] and nbrs[1]==0 then
nbrs[1] = o
elsif o[C]==shipment[C] and nbrs[2]==0 then
nbrs[2] = o
end if
if nbrs[1]!=0 and nbrs[2]!=0 then
exit
end if
end if
end for
return nbrs
end function

function matrix_to_list(sequence matrix)
sequence l = {}
for r=1 to length(matrix) do
for c=1 to length(matrix[r]) do
if matrix[r,c]!=0 then
l = append(l,matrix[r,c])
end if
end for
end for
return l
end function

function get_closed_path(sequence matrix, shipment)
sequence path = matrix_to_list(matrix)
path = prepend(path,shipment)

-- remove (and keep removing) elements that do not have a
-- vertical AND horizontal neighbor
while true do
integer removals = 0
for e=length(path) to 1 by -1 do
sequence nbrs = get_neighbors(path[e], path)
if nbrs[1]==0 or nbrs[2]==0 then
path[e..e] = {}
removals += 1
end if
end for
if removals==0 then exit end if
end while

-- place the remaining elements in the correct plus-minus order
sequence stones = repeat(0,length(path)),
prev = shipment
for i=1 to length(stones) do
stones[i] = prev
prev = get_neighbors(prev, path)[mod(i,2)+1]
end for
return stones
end function

function fix_degenerate_case(sequence matrix, costs)
if length(matrix)+length(matrix[1])-1 != length(matrix_to_list(matrix)) then
printf(1,"fixing degenerate case...\n")
for r=1 to length(matrix) do
for c=1 to length(matrix[r]) do
if matrix[r][c] == 0 then
sequence dummy = {eps, costs[r][c], r, c}
if length(get_closed_path(matrix,dummy)) == 0 then
matrix[r][c] = dummy
return matrix
end if
end if
end for
end for
?9/0 -- ??
end if
return matrix
end function

function initialise(sequence tests, integer t)
sequence {src,dst,costs} = deep_copy(tests[t])
string cs = ppf(costs,{pp_Nest,1,pp_StrFmt,3,pp_IntCh,false,pp_Indent,7})
printf(1,"test %d:\nsrc: %v,\ndst: %v,\ncosts: %s\n",{t,src,dst,cs})

-- check for and fix any imbalance
atom totalSrc = sum(src),
totalDst = sum(dst),
diff = totalSrc-totalDst
if diff>0 then
dst = append(dst, diff)
for i=1 to length(costs) do
--DEV...
--          costs[i] &= 0
costs[i] = deep_copy(costs[i]) & 0
end for
elsif diff<0 then
src = append(src, -diff)
costs = append(costs,repeat(0,length(dst)))
end if

printf(1,"generating initial feasible solution using northwest corner method...\n")
sequence matrix = repeat(repeat(0,length(dst)),length(src))
integer northwest = 1
for r=1 to length(src) do
for c=northwest to length(dst) do
atom qty = min(src[r],dst[c])
if qty>0 then
matrix[r][c] = {qty,costs[r,c],r,c}
src[r] -= qty
dst[c] -= qty
if src[r]=0 then
northwest = c
exit
end if
end if
end for
end for
printf(1,"\nTotal costs: %g\n\n", print_matrix(matrix))

return {src,dst,costs,matrix}
end function

function stepping_stone(sequence transport)
sequence {src, dst, costs, matrix} = deep_copy(transport)
atom maxReduction = 0
object move = NULL, leaving
matrix = fix_degenerate_case(matrix, costs)
for r=1 to length(src) do
for c=1 to length(dst) do
if matrix[r][c] = 0 then
sequence trial_shipment = {0, costs[r][c], r, c},
path = get_closed_path(matrix,trial_shipment)
atom reduction = 0.0,
lowestQuantity = 1e308
object leavingCandidate = 0
bool plus = true
for i=1 to length(path) do
sequence s = path[i]
if plus then
reduction += s[COST]
else
reduction -= s[COST]
if s[QTY] < lowestQuantity then
leavingCandidate = s
lowestQuantity = s[QTY]
end if
end if
plus = not plus
end for
if reduction < maxReduction then
move = path
leaving = leavingCandidate
maxReduction = reduction
end if
end if
end for
end for

if move!=NULL then
atom q = leaving[QTY]
bool plus = true
for i=1 to length(move) do
sequence s = deep_copy(move[i])
if plus then
s[QTY] += q
else
s[QTY] -= q
end if
if s[QTY] == 0 then
matrix[s[R]][s[C]] = 0
else
matrix[s[R]][s[C]] = s
end if
plus = not plus
end for
{src, dst, costs, matrix} = stepping_stone({src, dst, costs, matrix})
end if
return {src, dst, costs, matrix}
end function

--              --  source           dest            costs          expected total
constant tests = {{{25,35},         {20,30,10},     {{3,5,7},
{3,2,5}},           180},
{{12,40,33},      {20,30,10},     {{3,5,7},
{2,4,6},
{9,1,8}},           130},
{{14,10,15,12},   {10,15,12,15},  {{10,30,25,15},
{20,15,20,10},
{10,30,20,20},
{30,40,35,45}},    1000},
{{100,300,300},   {300,200,200},  {{50,40,30},
{80,40,30},
{90,70,50}},      39000},
{{40,60,50},      {20,30,50,50},  {{4,6,8,8},
{6,8,6,7},
{5,7,6,8}},         920},
{{12,1,5},        {10,8},         {{ 2, 4},
{ 8,12},
{12, 6}},            68},
{{7,9,18},        {5,8,7,14},     {{19,30,50,10},
{70,30,40,60},
{40, 8,70,20}},     743},
{{12,11,14,8},    {10,11,15,5,4}, {{ 7,12, 1, 5, 6},
{15, 3,12, 6,14},
{ 8,16,10,12, 7},
{18, 8,17,11,16}},  259},
{{50,60,50,50},  {30,20,70,30,60},{{16,16,13,22,17},
{14,14,13,19,15},
{19,19,20,23,50},
{50,12,50,15,11}}, 3100},
{{50,75,25},      {20,20,50,60},  {{3,5,7,6},
{2,5,8,2},
{3,6,9,2}},         610}}

--for i=1 to length(tests) do
for i=3 to 3 do
print_result(stepping_stone(initialise(tests,i)),tests[i][4])
end for

?"done"
{} = wait_key()
```
Output:

(Obviously the other eight tests all work fine and produce similar output.)

```test 3:
src: {14,10,15,12},
dst: {10,15,12,15},
costs: {{10,30,25,15},
{20,15,20,10},
{10,30,20,20},
{30,40,35,45}}

generating initial feasible solution using northwest corner method...
10    4   -    -
-    10   -    -
-     1   12    2
-    -    -    12
-    -    -     1

Total costs: 1220

fixing degenerate case...
Optimal solution

-    -    -    14
-     9   -     1
10   -     5   -
-     5    7   -
-     1   -    -

Total costs: 1000 (expected 1000)
```

Note that Vogel's_approximation_method#Phix gets a few of the others wrong and loops on #2, then again that is unbalanced (needs a dummy customer), and I'm not sure whether throwing such at VAM is fair or not.

## R

Using `lpSolve::lp.transport`.

```library(lpSolve)

# cost matrix
costs <- matrix(c(3, 5, 7,
3, 2, 5), nrow = 2, byrow = TRUE)

# constraints for suppliers
row.signs <- rep("<=", 2)
row.rhs <- c(25, 35)
# constraints for customers
col.signs <- rep("=", 3)
col.rhs <- c(20, 30, 10)

# minimum cost (objective value)
lp.transport(costs, "min", row.signs, row.rhs, col.signs, col.rhs)
# solution matrix
sol = lp.transport(costs, "min", row.signs, row.rhs, col.signs, col.rhs)\$solution
rownames(sol) <- c("Supplier 1", "Supplier 2")
colnames(sol) <- c("Customer 1", "Customer 2", "Customer 3")
sol
```
Output:
```Success: the objective function is 180

Customer 1 Customer 2 Customer 3
Supplier 1         20          0          5
Supplier 2          0         30          5
```

## Racket

Translation of: Java

(I understand the letters in Java!)

Using `typed/racket`, to keep track of Vectors of Vectors of data.

```#lang typed/racket
;; {{trans|Java}}
(define-type (V2 A) (Vectorof (Vectorof A)))
(define-type VI (Vectorof Integer))
(define-type V2R (V2 Real))
(define-type Q (U 'ε Integer))
(define ε 'ε)
(struct Shipment ([qty : Q] [cost/unit : Real] [r : Integer] [c : Integer]))
(define-type Shipment/? (Option Shipment))
(define-type V2-Shipment/? (V2 Shipment/?))
(define-type Shipment/?s (Listof Shipment/?))
(define-type Shipments (Listof Shipment))

(: Q+ (Q Q -> Q))
(: Q- (Q Q -> Q))
(: Q<? (Q Q -> Boolean))
(: Q-zero? (Q -> Boolean))
(: Q-unary- (Q -> Q))
(: Q*R (Q Real -> Real))

(define Q+ (match-lambda** [('ε 0) ε] [(0 'ε) ε] [('ε 'ε) ε] [('ε x) x] [(x 'ε) x]
[((? integer? x) (? integer? y)) (+ x y)]))

(define Q<? (match-lambda** [('ε 0) #f] [(0 'ε) #t] [('ε 'ε) #f] [('ε x) #t] [(x 'ε) #f]
[((? integer? x) (? integer? y)) (< x y)]))

(define Q- (match-lambda** [('ε 0) ε] [(0 'ε) ε] [('ε 'ε) 0] [('ε (? integer? x)) (- x)] [(x 'ε) x]
[((? integer? x) (? integer? y)) (- x y)]))

(define Q-unary- (match-lambda ['ε ε] [(? integer? x) (- x)]))

(define Q-zero? (match-lambda ['ε #f] [(? integer? x) (zero? x)]))

(define Q*R (match-lambda** [('ε _) 0] [((? integer? x) y) (* x y)]))

(: vector-ref2 (All (A) ((Vectorof (Vectorof A)) Integer Integer -> A)))
(define (vector-ref2 v2 r c) (vector-ref (vector-ref v2 r) c))

(: vector-set!2 (All (A) ((Vectorof (Vectorof A)) Integer Integer A -> Void)))
(define (vector-set!2 v2 r c v) (vector-set! (vector-ref v2 r) c v))

(define (northwest-corner-rule! [supply : VI] [demand : VI] [costs : V2R] [M : V2-Shipment/?]) : Void
(define supply-l (vector-length supply))
(define demand-l (vector-length demand))
(let loop ((r 0) (nw 0) (c 0))
(cond [(= r supply-l) (void)]
[(= c demand-l) (loop (add1 r) nw 0)]
[else
(define quantity (min (vector-ref supply r) (vector-ref demand c)))
(cond
[(positive? quantity)
(define shpmnt (Shipment quantity (vector-ref2 costs r c) r c))
(vector-set!2 M r c shpmnt)
(define supply-- (- (vector-ref supply r) quantity))
(define demand-- (- (vector-ref demand c) quantity))
(vector-set! supply r supply--)
(vector-set! demand c demand--)
(if (zero? supply--) (loop (add1 r) c 0) (loop r nw (add1 c)))]
[else (loop r nw (add1 c))])])))

(define (stepping-stone! [supply : VI] [demand : VI] [costs : V2R] [M : V2-Shipment/?]) : Void
(fix-degenerate-case! supply demand costs M)
(define-values (move leaving max-reduction)
(for*/fold : (Values Shipments Shipment/? Real)
((move : Shipments null) (leaving : Shipment/? #f) (max-reduction : Real 0))
((r (vector-length supply))
(c (vector-length demand))
(m (in-value (vector-ref2 M r c)))
#:unless m)
(define path (let ((trial (Shipment 0 (vector-ref2 costs r c) r c))) (get-closed-path trial M)))
(define-values (+? reduction leaving-cand lowest-quantity)
(for/fold : (Values Boolean Real Shipment/? (Option Q))
((+? #t) (reduction : Real 0) (leaving-cand : Shipment/? #f) (lowest-q : (Option Q) #f))
((s (in-list path)))
(define s.cpu (Shipment-cost/unit s))
(if +?
(values #f (+ reduction s.cpu) leaving-cand lowest-q)
(let ((reduction-- (- reduction s.cpu))
(s.q (Shipment-qty s)))
(if (or (not lowest-q) (Q<? s.q lowest-q))
(values #t reduction-- s s.q)
(values #t reduction-- leaving-cand lowest-q))))))

(if (< reduction max-reduction)
(values path leaving-cand reduction)
(values move leaving max-reduction))))

(unless (null? move)
(define l.q (Shipment-qty (cast leaving Shipment)))
(for/fold ((+? : Boolean #t)) ((s (in-list move)))
(define s.q+ ((if +? Q+ Q-) (Shipment-qty s) l.q))
(define s+ (struct-copy Shipment s [qty s.q+]))
(vector-set!2 M (Shipment-r s+) (Shipment-c s+) (if (Q-zero? s.q+) #f s+))
(not +?))
(stepping-stone! supply demand costs M)))

(: matrix->list (All (T) ((V2 T) -> (Listof T))))
(define (matrix->list m)
(for*/list : (Listof T) ((r (in-vector m)) (c (in-vector r)) #:when c)
c))

(define (fix-degenerate-case! [supply : VI] [demand : VI] [costs : V2R] [M : V2-Shipment/?]) : Void
(define m-list (matrix->list M))
(unless (= (+ (vector-length supply) (vector-length demand) -1) (length m-list))
(let/ec ret : Void
(for* ((r (vector-length supply)) (c (vector-length demand)) #:unless (vector-ref2 M r c))
(define dummy (Shipment ε (vector-ref2 costs r c) r c))
(when (null? (get-closed-path dummy M))
(vector-set!2 M r c dummy)
(ret (void)))))))

(: get-closed-path (Shipment V2-Shipment/? -> Shipments))
(define (get-closed-path s matrix)
; remove (and keep removing) elements that do not have a vertical AND horizontal neighbour
(define path
(let loop : Shipment/?s
((path (cons s (matrix->list matrix))))
(define (has-neighbours [e : Shipment/?]) : Boolean
(match-define (list n0 n1) (get-neighbours e path))
(and n0 n1 #t))
(define-values (with-nbrs w/o-nbrs)
((inst partition Shipment/? Shipment/?) has-neighbours path))
(if (null? w/o-nbrs) with-nbrs (loop with-nbrs))))

;; place the remaining elements in the correct plus-minus order
(define p-len (length path))
(define-values (senots prev)
(for/fold : (Values Shipments Shipment/?)
((senots : Shipments null) (prev : Shipment/? s))
((i p-len))
(values (if prev (cons prev senots) senots)
(list-ref (get-neighbours prev path) (modulo i 2)))))
(reverse senots))

(define (get-neighbours [s : Shipment/?] [lst : Shipment/?s]) : (List Shipment/? Shipment/?)
(define-values (n0 n1)
(for/fold : (Values Shipment/? Shipment/?)
((n0 : Shipment/? #f) (n1 : Shipment/? #f))
((o (in-list lst)) #:when (and o s) #:unless (equal? o s))
(values (or n0 (and (= (Shipment-r s) (Shipment-r o)) o))
(or n1 (and (= (Shipment-c s) (Shipment-c o)) o)))))
(list n0 n1))

(define (print-result [S : VI] [D : VI] [M : V2-Shipment/?] [fmt : String] . [args : Any *]) : Real
(apply printf (string-append fmt "~%") args)
(define total-costs
(for*/sum : Real
((r (vector-length S)) (c (vector-length D)))
(when (zero? c) (unless (zero? r) (newline)))
(define s (vector-ref2 M r c))
(cond
[(and s (= (Shipment-r s) r) (= (Shipment-c s) c))
(define q (Shipment-qty s))
(printf "\t~a" q)
(Q*R q (Shipment-cost/unit s))]
[else (printf "\t-") 0])))
(printf "~%Total costs: ~a~%~%" total-costs)
total-costs)

;; inits from current-input-port --- make sure you set that before coming in
(define (init) : (Values VI VI V2R V2-Shipment/?)
(define srcs. (for/list : (Listof Integer) ((_ n-sources)) (cast (read) Integer)))
(define dsts. (for/list : (Listof Integer) ((_ n-destinations)) (cast (read) Integer)))

(define sum-src--sum-dest (- (apply + srcs.) (apply + dsts.)))

(define-values (supply demand)
(cond [(positive? sum-src--sum-dest) (values srcs. (append dsts. (list sum-src--sum-dest)))]
[(negative? sum-src--sum-dest) (values (append srcs. (list (- sum-src--sum-dest))) dsts.)]
[else (values srcs. dsts.)]))

(define s-l (length supply))
(define d-l (length demand))
(define costs (for/vector : V2R ((_ s-l)) ((inst make-vector Real) d-l 0)))
(define matrix (for/vector : V2-Shipment/? ((_ s-l)) ((inst make-vector Shipment/?) d-l #f)))
(for* ((i n-sources) (j n-destinations)) (vector-set!2 costs i j (cast (read) Real)))
(values (list->vector supply) (list->vector demand) costs matrix))

(: transportation-problem (Input-Port -> Real))
(define (transportation-problem p)
(parameterize ([current-input-port p])
(define-values (supply demand costs matrix) (init))
(northwest-corner-rule! supply demand costs matrix)
(stepping-stone! supply demand costs matrix)
(print-result supply demand matrix "Optimal solutions for: ~s" name)))

(module+ test
(require typed/rackunit)
(define (check-tp [in-str : String] [expected-cost : Real])
(define cost ((inst call-with-input-string Real) in-str transportation-problem))
(check-equal? cost expected-cost))

(check-tp #<<\$
input1
2 3
25 35
20 30 10
3 5 7
3 2 5
\$
180)

(check-tp #<<\$
input2
3 3
12 40 33
20 30 10
3 5 7
2 4 6
9 1 8
\$
130)

(check-tp #<<\$
input3
4 4
14 10 15 12
10 15 12 15
10 30 25 15
20 15 20 10
10 30 20 20
30 40 35 45
\$
1000))
```
Output:

Output of: `raco test Transportation-problem.rkt`:

```raco test: (submod "transportation-problem.rkt" test)
Optimal solutions for: input1
20      -       5
-       30      5
Total costs: 180

Optimal solutions for: input2
-       -       -       12
20      -       10      10
-       30      -       3
Total costs: 130

Optimal solutions for: input3
-       -       -       14
-       9       -       1
10      -       5       -
-       5       7       -
-       1       -       -
Total costs: 1000

3 tests passed
```

## Raku

(formerly Perl 6)

Works with: Rakudo version 2019.03.1
```my  %costs = :S1{:3C1, :5C2, :7C3}, :S2{:3C1, :2C2, :5C3};
my %demand = :20C1, :30C2, :10C3;
my %supply = :25S1, :35S2;

my @cols = %demand.keys.sort;

my %res;
my %g = (|%supply.keys.map: -> \$x { \$x => [%costs{\$x}.sort(*.value)».key]}),
(|%demand.keys.map: -> \$x { \$x => [%costs.keys.sort({%costs{\$_}{\$x}})]});

while (+%g) {
my @d = %demand.keys.map: -> \$x
{[\$x, my \$z = %costs{%g{\$x}[0]}{\$x},%g{\$x}[1] ?? %costs{%g{\$x}[1]}{\$x} - \$z !! \$z]}

my @s = %supply.keys.map: -> \$x
{[\$x, my \$z = %costs{\$x}{%g{\$x}[0]},%g{\$x}[1] ?? %costs{\$x}{%g{\$x}[1]} - \$z !! \$z]}

@d = |@d.grep({ (.[2] == max @d».[2]) }).&min: :by(*.[1]);
@s = |@s.grep({ (.[2] == max @s».[2]) }).&min: :by(*.[1]);

my (\$t, \$f) = @d[2] == @s[2] ?? (@s[1],@d[1]) !! (@d[2],@s[2]);
my (\$d, \$s) = \$t > \$f ?? (@d[0],%g{@d[0]}[0]) !! (%g{@s[0]}[0], @s[0]);

my \$v = %supply{\$s} min %demand{\$d};

%res{\$s}{\$d} += \$v;
%demand{\$d} -= \$v;

if (%demand{\$d} == 0) {
%supply.grep( *.value != 0 )».key.map: -> \$v
{ %g{\$v}.splice((%g{\$v}.first: * eq \$d, :k),1) };
%g{\$d}:delete;
%demand{\$d}:delete;
}

%supply{\$s} -= \$v;

if (%supply{\$s} == 0) {
%demand.grep( *.value != 0 )».key.map: -> \$v
{ %g{\$v}.splice((%g{\$v}.first: * eq \$s, :k),1) };
%g{\$s}:delete;
%supply{\$s}:delete;
}
}

say join "\t", flat '', @cols;
my \$total;
for %costs.keys.sort -> \$g {
print "\$g\t";
for @cols -> \$col {
print %res{\$g}{\$col} // '-', "\t";
\$total += (%res{\$g}{\$col} // 0) * %costs{\$g}{\$col};
}
print "\n";
}
say "\nTotal cost: \$total";
```
Output:
```	C1	C2	C3
S1	20	-	5
S2	-	30	5	```

## REXX

Translation of: Java
```/* REXX ***************************************************************
* Solve the Transportation Problem using the Northwest Corner Method
Default Input
2 3        # of sources / # of demands
25 35      sources
20 30 10   demands
3 5 7      cost matrix
3 2 5
* 20201210 support no input file -courtesy GS
*          Note: correctnes of input is not checked
* 20210103 remove debug code
**********************************************************************/
Signal On Halt
Signal On Novalue
Signal On Syntax

Parse Arg fid
If fid='' Then
fid='input1.txt'
Call init
matrix.=0
ms=0
Do r=1 To rr
Do c=1 To cc
matrix.r.c=r c cost.r.c 0
End
End
r=1
c=1
Do While r<=rr & c<=cc
q=min(source.r,demand.c)
matrix.r.c=r c cost.r.c q
source.r=source.r-q
demand.c=demand.c-q
If source.r=0 Then r=r+1
If demand.c=0 Then c=c+1
End
Call show_alloc 'after NWC application'
Call steppingstone
Exit

/**********************************************************************
* Subroutines for NWC Algorithm
**********************************************************************/

init:
If lines(fid)=0 Then Do
fid='Default input'
in.1=sourceline(4)
Parse Var in.1 numSources .
Do i=2 To numSources+3
in.i=sourceline(i+3)
End
End
Else Do
Do i=1 By 1 while lines(fid)>0
in.i=linein(fid)
End
End
Parse Var in.1 numSources numDestinations . 1 rr cc .
source_sum=0
Do i=1 To numSources
Parse Var in.2 source.i in.2
ss.i=source.i
source_in.i=source.i
source_sum=source_sum+source.i
End
l=linein(fid)
demand_sum=0
Do i=1 To numDestinations
Parse Var in.3 demand.i in.3
dd.i=demand.i
demand_in.i=demand.i
demand_sum=demand_sum+demand.i
End
Do i=1 To numSources
j=i+3
l=in.j
Do j=1 To numDestinations
Parse Var l cost.i.j l
End
End
Do i=1 To numSources
ol=format(source.i,3)
Do j=1 To numDestinations
ol=ol format(cost.i.j,4)
End
End
ol='   '
Do j=1 To numDestinations
ol=ol format(demand.j,4)
End

Select
When source_sum=demand_sum Then Nop  /* balanced */
When source_sum>demand_sum Then Do   /* unbalanced - add dummy demand */
Say 'This is an unbalanced case (sources exceed demands). We add a dummy consumer.'
cc=cc+1
demand.cc=source_sum-demand_sum
demand_in.cc=demand.cc
dd.cc=demand.cc
Do r=1 To rr
cost.r.cc=0
End
End
Otherwise /* demand_sum>source_sum */ Do /* unbalanced - add dummy source */
Say 'This is an unbalanced case (demands exceed sources). We add a dummy source.'
rr=rr+1
source.rr=demand_sum-source_sum
source_in.rr=source.rr
ss.rr=source.rr
Do c=1 To cc
cost.rr.c=0
End
End
End
Say 'Sources / Demands / Cost'
ol='    '
Do c=1 To cc
ol=ol format(demand.c,3)
End
Say ol

Do r=1 To rr
ol=format(source.r,4)
Do c=1 To cc
ol=ol format(cost.r.c,3)
End
Say ol
End
Return

show_alloc: Procedure Expose matrix. rr cc demand_in. source_in.
Return
Say ''
total=0
ol='    '
Do c=1 to cc
ol=ol format(demand_in.c,3)
End
Do r=1 to rr
ol=format(source_in.r,4)
a=word(matrix.r.1,4)
If a>0 Then
ol=format(a,4)
Else
ol='  - '
total=total+word(matrix.r.1,4)*word(matrix.r.1,3)
Do c=2 To cc
a=word(matrix.r.c,4)
If a>0 Then
ol=ol format(a,4)
Else
ol=ol '  - '
total=total+word(matrix.r.c,4)*word(matrix.r.c,3)
End
Say ol
End
Say 'Total costs:' format(total,4,1)
Return

/**********************************************************************
* Subroutines for Optimization
**********************************************************************/

steppingstone: Procedure Expose matrix. cost. rr cc demand_in.,
source_in. fid
maxReduction=0
move=''

Call fixDegenerateCase

Do r=1 To rr
Do c=1 To cc
Parse Var matrix.r.c r c cost qrc
If qrc=0 Then Do
path=getclosedpath(r,c)
reduction = 0
lowestQuantity = 1e10
leavingCandidate = ''
plus=1
pathx=path
Do While pathx<>''
Parse Var pathx s '|' pathx
If plus Then
reduction=reduction+word(s,3)
Else Do
reduction=reduction-word(s,3)
If word(s,4)<lowestQuantity Then Do
leavingCandidate = s
lowestQuantity = word(s,4)
End
End
plus=\plus
End
If reduction < maxreduction Then Do
move=path
leaving=leavingCandidate
maxReduction = reduction
End
End
End
End
if move <> '' Then Do
quant=word(leaving,4)
plus=1
Do While move<>''
Parse Var move m '|' move
Parse Var m r c cpu qrc
Parse Var matrix.r.c vr vc vcost vquant
If plus Then
nquant=vquant+quant
Else
nquant=vquant-quant
matrix.r.c = vr vc vcost nquant
plus=\plus
End
move=''
Call steppingStone
End
Else Do
Call show_alloc 'Optimal Solution' fid
End
Return

getclosedpath: Procedure Expose matrix. cost. rr cc
Parse Arg rd,cd
path=rd cd cost.rd.cd word(matrix.rd.cd,4)
do r=1 To rr
Do c=1 To cc
If word(matrix.r.c,4)>0 Then Do
path=path'|'r c cost.r.c word(matrix.r.c,4)
End
End
End
path=magic(path)
Return stones(path)

magic: Procedure
Parse Arg list
Do Forever
list_1=remove_1(list)
If list_1=list Then Leave
list=list_1
End
Return list_1

remove_1: Procedure
Parse Arg list
cntr.=0
cntc.=0
Do i=1 By 1 While list<>''
parse Var list e.i '|' list
Parse Var e.i r c .
cntr.r=cntr.r+1
cntc.c=cntc.c+1
End
n=i-1
keep.=1
Do i=1 To n
Parse Var e.i r c .
If cntr.r<2 |,
cntc.c<2 Then Do
keep.i=0
End
End
list=e.1
Do i=2 To n
If keep.i Then
list=list'|'e.i
End
Return list

stones: Procedure
Parse Arg lst
tstc=lst
Do i=1 By 1 While tstc<>''
Parse Var tstc o.i '|' tstc
end
stones=lst
o.0=i-1
prev=o.1
Do i=1 To o.0
st.i=prev
k=i//2
nbrs=getNeighbors(prev, lst)
Parse Var nbrs n.1 '|' n.2
If k=0 Then
prev=n.2
Else
prev=n.1
End
stones=st.1
Do i=2 To o.0
stones=stones'|'st.i
End
Return stones

getNeighbors: Procedure
parse Arg s, lst
Do i=1 By 1 While lst<>''
Parse Var lst o.i '|' lst
End
o.0=i-1
nbrs.=''
sr=word(s,1)
sc=word(s,2)
Do i=1 To o.0
If o.i<>s Then Do
or=word(o.i,1)
oc=word(o.i,2)
If or=sr & nbrs.0='' Then
nbrs.0 = o.i
else if oc=sc & nbrs.1='' Then
nbrs.1 = o.i
If nbrs.0<>'' & nbrs.1<>'' Then
Leave
End
End
return nbrs.0'|'nbrs.1

pelems: Procedure
Parse Arg p
Do i=1 By 1 While p<>''
Parse Var p x '|' p
End
Return i

fixDegenerateCase: Procedure Expose matrix. rr cc ms
Call matrixtolist
If (rr+cc-1)<>ms Then Do
Do r=1 To rr
Do c=1 To cc
If word(matrix.r.c,4)=0 Then Do
matrix.r.c=subword(matrix.r.c,1,3) 1.e-10
Return
End
End
End
End
Return

matrixtolist: Procedure Expose matrix. rr cc ms
ms=0
list=''
Do r=1 To rr
Do c=1 To cc
If word(matrix.r.c,4)>0 Then Do
list=list'|'matrix.r.c
ms=ms+1
End
End
End
Return strip(list,,'|')

Novalue:
Say 'Novalue raised in line' sigl
Say sourceline(sigl)
Say 'Variable' condition('D')
Signal lookaround

Syntax:
Say 'Syntax raised in line' sigl
Say sourceline(sigl)
Say 'rc='rc '('errortext(rc)')'

halt:
lookaround:
If fore() Then Do
Say 'You can look around now.'
Trace ?R
Nop
End
Exit 12
```
Output:
```F:\>rexx tpx2 input1.txt
Sources / Demands / Cost
20  30  10
25   3   5   7
35   3   2   5

after NWC application
20    5   -
-    25   10
Total costs:  185.0

Optimal Solution input1.txt
20   -     5
-    30    5
Total costs:  180.0

F:\>rexx tpx2 input2.txt
This is an unbalanced case (sources exceed demands). We add a dummy consumer.
Sources / Demands / Cost
20  30  10  25
12   3   5   7   0
40   2   4   6   0
33   9   1   8   0

after NWC application
12   -    -    -
8   30    2   -
-    -     8   25
Total costs:  248.0

Optimal Solution input2.txt
-    -    -    12
20   -    10   10
-    30   -     3
Total costs:  130.0

F:\>rexx tpx2 input3.txt
This is an unbalanced case (demands exceed sources). We add a dummy source.
Sources / Demands / Cost
10  15  12  15
14  10  30  25  15
10  20  15  20  10
15  10  30  20  20
12  30  40  35  45
1   0   0   0   0

after NWC application
10    4   -    -
-    10   -    -
-     1   12    2
-    -    -    12
-    -    -     1
Total costs: 1220.0

Optimal Solution input3.txt
-    -    -    14
-     9   -     1
10   -     5   -
-     5    7   -
-     1   -    -
Total costs: 1000.0```

## SAS

Use network solver in SAS/OR:

```/* create SAS data sets */
data cost_data;
input from \$ to \$ cost;
datalines;
s1 c1 3
s1 c2 5
s1 c3 7
s2 c1 3
s2 c2 2
s2 c3 5
;

data supply_data;
input node \$ supply;
datalines;
s1  25
s2  35
c1 -20
c2 -30
c3 -10
;

/* call OPTMODEL procedure in SAS/OR */
proc optmodel;
/* declare sets and parameters, and read input data */
set NODES = union {<i,j> in LINKS} {i,j};
num supply {NODES} init 0;
read data supply_data into [node] supply;

/* call network solver */
solve with network /

/* print optimal solution */
print _OROPTMODEL_NUM_['OBJECTIVE'];
print flow;
quit;
```

Output:

```180

flow
c1 c2 c3
s1 20 0 5
s2 0 30 5
```

## Visual Basic .NET

Translation of: C#
```Module Module1

Class Shipment
Public Sub New(q As Double, cpu As Double, rv As Integer, cv As Integer)
Quantity = q
CostPerUnit = cpu
R = rv
C = cv
End Sub

Public ReadOnly Property CostPerUnit() As Double

Public Property Quantity() As Double

Public ReadOnly Property R As Integer

Public ReadOnly Property C As Integer

Public Shared Operator =(s1 As Shipment, s2 As Shipment) As Boolean
Return s1.CostPerUnit = s2.CostPerUnit _
AndAlso s1.Quantity = s2.Quantity _
AndAlso s1.R = s2.R _
AndAlso s1.C = s2.C
End Operator

Public Shared Operator <>(s1 As Shipment, s2 As Shipment) As Boolean
Return s1.CostPerUnit <> s2.CostPerUnit _
OrElse s1.Quantity <> s2.Quantity _
OrElse s1.R <> s2.R _
OrElse s1.C <> s2.C
End Operator
End Class

Class Program
Private demand() As Integer
Private supply() As Integer
Private costs(,) As Double
Private matrix(,) As Shipment

Sub Init(filename As String)
Dim numArr = line.Split
Dim numSources = Integer.Parse(numArr(0))
Dim numDestinations = Integer.Parse(numArr(1))

Dim src As New List(Of Integer)
Dim dst As New List(Of Integer)

numArr = line.Split
For i = 1 To numSources
Next

numArr = line.Split
For i = 1 To numDestinations
Next

REM fix imbalance
Dim totalSrc = src.Sum
Dim totalDst = dst.Sum
If totalSrc > totalDst Then
ElseIf totalDst > totalSrc Then
End If

supply = src.ToArray
demand = dst.ToArray

ReDim costs(supply.Length - 1, demand.Length - 1)
ReDim matrix(supply.Length - 1, demand.Length - 1)

For i = 1 To numSources
numArr = line.Split
For j = 1 To numDestinations
costs(i - 1, j - 1) = Integer.Parse(numArr(j - 1))
Next
Next
End Sub

Sub NorthWestCornerRule()
Dim northwest = 1
For r = 1 To supply.Length
For c = northwest To demand.Length
Dim quantity = Math.Min(supply(r - 1), demand(c - 1))
If quantity > 0 Then
matrix(r - 1, c - 1) = New Shipment(quantity, costs(r - 1, c - 1), r - 1, c - 1)

supply(r - 1) -= quantity
demand(c - 1) -= quantity

If supply(r - 1) = 0 Then
northwest = c
Exit For
End If
End If
Next
Next
End Sub

Sub SteppingStone()
Dim maxReduction = 0.0
Dim move() As Shipment = Nothing
Dim leaving As Shipment = Nothing

FixDegenerateCase()

For r = 1 To supply.Length
For c = 1 To demand.Length
If Not IsNothing(matrix(r - 1, c - 1)) Then
Continue For
End If

Dim trial As New Shipment(0, costs(r - 1, c - 1), r - 1, c - 1)
Dim path = GetClosedPath(trial)

Dim reduction = 0.0
Dim lowestQuanity = Integer.MaxValue
Dim leavingCandidate As Shipment = Nothing

Dim plus = True
For Each s In path
If plus Then
reduction += s.CostPerUnit
Else
reduction -= s.CostPerUnit
If s.Quantity < lowestQuanity Then
leavingCandidate = s
lowestQuanity = s.Quantity
End If
End If
plus = Not plus
Next
If reduction < maxReduction Then
move = path
leaving = leavingCandidate
maxReduction = reduction
End If
Next
Next

If Not IsNothing(move) Then
Dim q = leaving.Quantity
Dim plus = True
For Each s In move
s.Quantity += If(plus, q, -q)
matrix(s.R, s.C) = If(s.Quantity = 0, Nothing, s)
plus = Not plus
Next
SteppingStone()
End If
End Sub

Sub FixDegenerateCase()
Const eps = Double.Epsilon
If supply.Length + demand.Length - 1 <> MatrixToList().Count Then
For r = 1 To supply.Length
For c = 1 To demand.Length
If IsNothing(matrix(r - 1, c - 1)) Then
Dim dummy As New Shipment(eps, costs(r - 1, c - 1), r - 1, c - 1)
If GetClosedPath(dummy).Length = 0 Then
matrix(r - 1, c - 1) = dummy
Return
End If
End If
Next
Next
End If
End Sub

Function MatrixToList() As List(Of Shipment)
Dim newList As New List(Of Shipment)
For Each item In matrix
If Not IsNothing(item) Then
End If
Next
Return newList
End Function

Function GetClosedPath(s As Shipment) As Shipment()
Dim path = MatrixToList()

REM remove (and keep removing) elements that do not have a veritcal AND horizontal neighbor
Dim before As Integer
Do
before = path.Count
path.RemoveAll(Function(ship)
Dim nbrs = GetNeighbors(ship, path)
Return IsNothing(nbrs(0)) OrElse IsNothing(nbrs(1))
End Function)
Loop While before <> path.Count

REM place the remaining elements in the correct plus-minus order
Dim stones = path.ToArray
Dim prev = s
For i = 1 To stones.Length
stones(i - 1) = prev
prev = GetNeighbors(prev, path)((i - 1) Mod 2)
Next
Return stones
End Function

Function GetNeighbors(s As Shipment, lst As List(Of Shipment)) As Shipment()
Dim nbrs() As Shipment = {Nothing, Nothing}
For Each o In lst
If o <> s Then
If o.R = s.R AndAlso IsNothing(nbrs(0)) Then
nbrs(0) = o
ElseIf o.C = s.C AndAlso IsNothing(nbrs(1)) Then
nbrs(1) = o
End If
If Not IsNothing(nbrs(0)) AndAlso Not IsNothing(nbrs(1)) Then
Exit For
End If
End If
Next
Return nbrs
End Function

Sub PrintResult(filename As String)
Console.WriteLine("Optimal solution {0}" + vbNewLine, filename)
Dim totalCosts = 0.0

For r = 1 To supply.Length
For c = 1 To demand.Length
Dim s = matrix(r - 1, c - 1)
If Not IsNothing(s) AndAlso s.R = r - 1 AndAlso s.C = c - 1 Then
Console.Write(" {0,3} ", s.Quantity)
totalCosts += (s.Quantity * s.CostPerUnit)
Else
Console.Write("  -  ")
End If
Next
Console.WriteLine()
Next
Console.WriteLine(vbNewLine + "Total costs: {0}" + vbNewLine, totalCosts)
End Sub
End Class

Sub Main()
For Each filename In {"input1.txt", "input2.txt", "input3.txt"}
Dim p As New Program
p.Init(filename)
p.NorthWestCornerRule()
p.SteppingStone()
p.PrintResult(filename)
Next
End Sub

End Module
```
Output:
```Optimal solution input1.txt

20   -     5
-    30    5

Total costs: 180

Optimal solution input2.txt

-    -    -    12
20   -    10   10
-    30   -     3

Total costs: 130

Optimal solution input3.txt

-    -    -    14
-     9   -     1
10   -     5   -
-     5    7   -
-     1   -    -

Total costs: 1000```

## Wren

Translation of: Java
Library: Wren-dynamic
Library: Wren-ioutil
Library: Wren-math
Library: Wren-seq
Library: Wren-fmt
```import "./dynamic" for Struct
import "./ioutil" for FileUtil, File
import "./math" for Nums
import "./seq" for Lst
import "./fmt" for Fmt

var Shipment = Struct.create("Shipment", ["quantity", "costPerUnit", "r", "c"])

class Transport {
construct new(filename) {
var split = lines[0].split(" ")
var numSources = Num.fromString(split[0])
var numDests   = Num.fromString(split[1])
var src = List.filled(numSources, 0)
var dst = List.filled(numDests, 0)
split = lines[1].split(" ")
for (i in 0...numSources) src[i] = Num.fromString(split[i])
split = lines[2].split(" ")
for (i in 0...numDests) dst[i] = Num.fromString(split[i])

// fix imbalance
var totalSrc = Nums.sum(src)
var totalDst = Nums.sum(dst)
if (totalSrc > totalDst) {
} else if (totalDst > totalSrc) {
}
_supply = src
_demand = dst
_costs  = List.filled(_supply.count, null)
_matrix = List.filled(_supply.count, null)
for (i in 0..._supply.count) {
_costs[i]  = List.filled(_demand.count, 0)
_matrix[i] = List.filled(_demand.count, null)
}
for (i in 0...numSources) {
split = lines[i + 3].split(" ")
for (j in 0...numDests) _costs[i][j] = Num.fromString(split[j])
}
_filename = filename
}

northWestCornerRule() {
var northwest = 0
for (r in 0..._supply.count) {
var c = northwest
while (c < _demand.count) {
var quantity = _supply[r].min(_demand[c])
if (quantity > 0) {
_matrix[r][c] = Shipment.new(quantity, _costs[r][c], r, c)
_supply[r] = _supply[r] - quantity.floor
_demand[c] = _demand[c] - quantity.floor
if (_supply[r] == 0) {
northwest = c
break
}
}
c = c + 1
}
}
}

steppingStone() {
var maxReduction = 0
var move = null
var leaving = null
fixDegenerateCase_()

for (r in 0..._supply.count) {
for (c in 0..._demand.count) {
if (_matrix[r][c] != null) continue
var trial = Shipment.new(0, _costs[r][c], r, c)
var path = getClosedPath_(trial)
var reduction = 0
var lowestQuantity = Num.maxSafeInteger
var leavingCandidate = null
var plus = true
for (s in path) {
if (plus) {
reduction = reduction + s.costPerUnit
} else {
reduction = reduction - s.costPerUnit
if (s.quantity < lowestQuantity) {
leavingCandidate = s
lowestQuantity = s.quantity
}
}
plus = !plus
}
if (reduction < maxReduction) {
move = path
leaving = leavingCandidate
maxReduction = reduction
}
}
}

if (move) {
var q = leaving.quantity
var plus = true
for (s in move) {
s.quantity = s.quantity + ((plus) ? q : -q)
_matrix[s.r][s.c] = (s.quantity == 0) ? null : s
plus = !plus
}
steppingStone()
}
}

matrixToList_() { Lst.flatten(_matrix).where { |s| s != null }.toList }

getClosedPath_(s) {
var path = matrixToList_()
path.insert(0, s)
// remove (and keep removing) elements that do not have a
// vertical AND horizontal neighbor
while (true) {
var removals = 0
for (e in path) {
var nbrs = getNeighbors_(e, path)
if (nbrs[0] == null || nbrs[1] == null) {
path.remove(e)
removals = removals + 1
}
}
if (removals == 0) break
}

// place the remaining elements in the correct plus-minus order
var stones = List.filled(path.count, null)
var prev = s
for (i in 0...stones.count) {
stones[i] = prev
prev = getNeighbors_(prev, path)[i % 2]
}
return stones
}

getNeighbors_(s, lst) {
var nbrs = List.filled(2, null)
for (o in lst) {
if (o != s) {
if (o.r == s.r && nbrs[0] == null) {
nbrs[0] = o
} else if (o.c == s.c && nbrs[1] == null) {
nbrs[1] = o
}
if (nbrs[0] != null && nbrs[1] != null) break
}
}
return nbrs
}

fixDegenerateCase_() {
var eps = Num.smallest
if (_supply.count + _demand.count - 1 != matrixToList_().count) {
for (r in 0..._supply.count) {
for (c in 0..._demand.count) {
if (_matrix[r][c] == null) {
var dummy = Shipment.new(eps, _costs[r][c], r, c)
if (getClosedPath_(dummy).count == 0) {
_matrix[r][c] = dummy
return
}
}
}
}
}
}

printResult() {
System.print("%(_filename)\n\n%(text)")
System.print("Optimal solution %(_filename)\n")
var totalCosts = 0
for (r in 0..._supply.count) {
for (c in 0..._demand.count) {
var s = _matrix[r][c]
if (s != null && s.r == r && s.c == c) {
Fmt.write(" \$3d ", s.quantity.floor)
totalCosts = totalCosts + s.quantity * s.costPerUnit
} else System.write("  -  ")
}
System.print()
}
System.print("\nTotal costs: %(totalCosts)\n")
}
}

var filenames = ["input1.txt", "input2.txt", "input3.txt"]
for (filename in filenames) {
var t = Transport.new(filename)
t.northWestCornerRule()
t.steppingStone()
t.printResult()
}
```
Output:
```input1.txt

2 3
25 35
20 30 10
3 5 7
3 2 5

Optimal solution input1.txt

20   -     5
-    30    5

Total costs: 180

input2.txt

3 3
12 40 33
20 30 10
3 5 7
2 4 6
9 1 8

Optimal solution input2.txt

-    -    -    12
20   -    10   10
-    30   -     3

Total costs: 130

input3.txt

4 4
14 10 15 12
10 15 12 15
10 30 25 15
20 15 20 10
10 30 20 20
30 40 35 45

Optimal solution input3.txt

-    -    -    14
-     9   -     1
10   -     5   -
-     5    7   -
-     1   -    -

Total costs: 1000
```