二维数组中长度8的所有可能组合

Question

I've been trying to solve a problem in combinations. I have a matrix 6X6 i'm trying to find all combinations of length 8 in the matrix.

I have to move from neighbor to neighbor form each row,column position and i wrote a recursive program which generates the combination but the problem is it generates a lot of duplicates as well and hence is inefficient. I would like to know how could i eliminate calculating duplicates and save time.

int a={{1,2,3,4,5,6},
   {8,9,1,2,3,4},
   {5,6,7,8,9,1},
   {2,3,4,5,6,7},
   {8,9,1,2,3,4},
   {5,6,7,8,9,1},
  }

 void genSeq(int row,int col,int length,int combi)
 {
    if(length==8)
    {
      printf("%d\n",combi);
      return;
    }
    combi = (combi * 10) + a[row][col];
    if((row-1)>=0)

    genSeq(row-1,col,length+1,combi);

if((col-1)>=0)

    genSeq(row,col-1,length+1,combi);

if((row+1)<6)

    genSeq(row+1,col,length+1,combi);

if((col+1)<6)

    genSeq(row,col+1,length+1,combi);

if((row+1)<6&&(col+1)<6)

    genSeq(row+1,col+1,length+1,combi);

if((row-1)>=0&&(col+1)<6)

    genSeq(row-1,col+1,length+1,combi);

if((row+1)<6&&(row-1)>=0)

    genSeq(row+1,col-1,length+1,combi);

if((row-1)>=0&&(col-1)>=0)

    genSeq(row-1,col-1,length+1,combi);
   }

I was also thinking of writing a dynamic program basically recursion with memorization. Is it a better choice?? if yes than I'm not clear how to implement it in recursion. Have i really hit a dead end with approach???

Thankyou

Edit Eg result 12121212,12121218,12121219,12121211,12121213.

the restrictions are that you have to move to your neighbor from any point, you have to start for each position in the matrix ie each row,col. you can move one step at a time, ie right, left, up, down and the both diagonal positions. Check the if conditions. ie if your in (0,0) you can move to either (1,0) or (1,1) or (0,1) ie three neighbors. if your in (2,2) you can move to eight neighbors.
so on...

Answer 1

To eliminate duplicates you can covert 8 digit sequences into 8-digit integers and put them in a hashtable.

Memoization might be a good idea. You can memoize for each cell in the matrix all possible combinations of length 2-7 that can be achieved from it. Going backwards: first generate for each cell all sequences of 2 digits. Then based on that of 3 digits etc.

UPDATE: code in Python

# original matrix
lst = [
   [1,2,3,4,5,6],
   [8,9,1,2,3,4],
   [5,6,7,8,9,1],
   [2,3,4,5,6,7],
   [8,9,1,2,3,4],
   [5,6,7,8,9,1]]

# working matrtix; wrk[i][j] contains a set of all possible paths of length k which can end in lst[i][j]
wrk = [[set() for i in range(6)] for j in range(6)]

# for the first (0rh) iteration initialize with single step paths
for i in range(0, 6):
    for j in range(0, 6):
        wrk[i][j].add(lst[i][j])

# run iterations 1 through 7
for k in range(1,8):
    # create new emtpy wrk matrix for the next iteration
    nw = [[set() for i in range(6)] for j in range(6)]

    for i in range(0, 6):
        for j in range(0, 6):
            # the next gen. wrk[i][j] is going to be based on the current wrk paths of its neighbors
            ns = set()
            if i > 0:
                for p in wrk[i-1][j]:
                    ns.add(10**k * lst[i][j] + p)
            if i < 5:
                for p in wrk[i+1][j]:
                    ns.add(10**k * lst[i][j] + p)
            if j > 0:
                for p in wrk[i][j-1]:
                    ns.add(10**k * lst[i][j] + p)
            if j < 5:
                for p in wrk[i][j+1]:
                    ns.add(10**k * lst[i][j] + p)

            nw[i][j] = ns
    wrk = nw

# now build final set to eliminate duplicates
result = set()

for i in range(0, 6):
    for j in range(0, 6):
        result |= wrk[i][j]

print len(result)
print result

Answer 2

There are LOTS of ways to do this. Going through every combination is a perfectly reasonable first approach. It all depends on your requirements. If your matrix is small, and this operation isn't time sensitive, then there's no problem.

I'm not really an algorithms guy, but I'm sure there are really clever ways of doing this that someone will post after me.

Also, in Java when using CamelCase, method names should start with a lowercase character.

Answer 3

int a={{1,2,3,4,5,6}, {8,9,1,2,3,4}, {5,6,7,8,9,1}, {2,3,4,5,6,7}, {8,9,1,2,3,4}, {5,6,7,8,9,1}, }

By length you mean summation of combination of matrix elements resulting 8.
ie, elements to sum up 8 with in row itself and with the other row elements.
From row 1 = { {2,6}, {3,5}, } and now row 1 elements with row 2 and so on. Is that what you are expecting ?

Answer 4

You can think about your matrix like it is one-dimension array - no matter here ("place" the rows one by one). For one-dimension array you can write a function like (assuming you should print the combinations)

f(i, n) prints all combinations of length n using elements a[i] ... a[last].

It should skip some elements from a[i] to a[i + k] (for all possible k), print a[k] and make a recursive call f(i + k + 1, n - 1).