Change coins, it just doesn’t work out right

Question

Hello, I'm trying to create an algorithm that finds out how many ways to revert changes. But I just can’t get the edit, I keep getting 4, where I have to get 6, and I just don’t understand why.

This is my implementation in C #, this is creating from pseudocode from http://www.algorithmist.com/index.php/Coin_Change

     private static int[] S = { 1, 2, 5 };
        private static void Main(string[] args)
        {
            int amount = 7;
            int ways = count2(amount, S.Length);
            Console.WriteLine("Ways to make change for " + amount + " kr: " + ways.ToString());
            Console.ReadLine();
        }    
static int count2(int n, int m)
        {
        int[,] table = new int[n,m];

        for (int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                // Rules
                // 1: table[0,0] or table[0,x] = 1
                // 2: talbe[i <= -1, x] = 0
                // 3: table[x, j <= -1] = 0

                int total = 0;

                // first sub-problem
                // count(n, m-1)
                if (i == 0) // rule 1
                    total += 1;
                else if (i <= -1) // rule 2
                    total += 0;
                else if (j - 1 <= -1)
                    total += 0;
                else
                    total += table[i, j-1];

                // second sub-problem
                // count(n-S[m], m)
                if (j - 1 <= -1) // rule 3
                    total += 0;
                else if (i - S[j - 1] == 0) // rule 1
                    total += 1;
                else if (i - S[j - 1] <= -1) // rule 2
                    total += 0;
                else
                    total += table[i - S[j-1], j];

                table[i, j] = total;
            }
        }
        return table[n-1, m-1];
    }

Answer 1

Here the algorithm is operational. Click the green "play" arrow to launch it. http://www.exorithm.com/algorithm/view/make_change

, , -1. , , , .

Answer 2

( , )... , linq , , ...

    static IEnumerable<List<int>> GetChange(int target, IQueryable<int> coins)
    {
        var availableCoins = from c in coins where c <= target select c;
        if (!availableCoins.Any())
        {
            yield return new List<int>();
        }
        else
        {
            foreach (var thisCoin in availableCoins)
            {
                foreach (var result in GetChange(target - thisCoin, from c in availableCoins where c <= thisCoin select c))
                {
                    result.Add(thisCoin);
                    yield return result;
                }
            }
        }
    }

Answer 3

, , . , n, m, i, . , , coinType .

, , . , i j 1 .. m 1 .. n - . , 2 3 :

// ...
  else if (i <= -1)     // <- can never be 'true'
    total += 0; 
  else if (j - 1 <= -1) // <- 'true' when 'j == 0'
// ...
  if (j - 1 <= -1)      // <- can never be 'true'
// ...

i -1 j .

Answer 4

.

1), count , ( ). - ( )

func count(table, i, j)
  if ( i == 0 ) 
    return 1
  if ( i < 0 )
    return 0
  if ( j <= 0 and i >= 1 )
    return 0

  return table[i][j]
end

table[i][j] = count(table, i, j - 1) + count(table, i - S[j], j)

.

2) count2 n - 1 . , n == 1 . : , .

Answer 5

, [i, j], , {0, 1, 2,..., j - 1}.

, while , [i, j] - j, S [j]. , , [i, j - 1] + table [i - S [j], j]; - S [j], - i, S [j].

:

        // second sub-problem
        // count(n-S[m], m)
        if (j - 1 <= -1) // rule 3
            total += 0;
        else if (i - S[j - 1] == 0) // rule 1
            total += 1;
        else if (i - S[j - 1] <= -1) // rule 2
            total += 0;
        else
            total += table[i - S[j-1], j];

        // second sub-problem
        // count(n-S[m], m)
        if (i - S[j] == 0) // rule 1
            total += 1;
        else if (i - S[j] < 0) // rule 2
            total += 0;
        else
            total += table[i - S[j], j];

FYI, - (j < 0), (j <= -1).