Generate a rand array,please help! solved

alexzheng

I use this to generate a rand rows * columns array filled with random number from 1,2,3.
However I got stuck on how to make sure no 3 or more cells with the same value connected horizontally or vertically.
For example the following array filled with 4 numbers of '1' in row 9.

Any good ideas?

1 2 0 2 3 1 3 2 1 0
3 2 3 1 1 2 3 3 2 1
3 0 2 1 3 3 1 1 3 2
2 2 1 0 0 3 0 0 1 0
3 3 1 3 2 1 1 2 2 0
0 1 3 0 1 2 0 3 0 0
1 0 3 3 3 3 1 0 3 0
1 3 0 3 1 2 1 3 1 1
3 3 1 1 1 1 2 1 3 1
3 3 1 2 1 0 2 1 1 0

---create an rand rows * columns array filled with 1,2,3 in num cells local function randarray(rows, columns, num) math.randomseed(os.time())   local array = {} local tmp = {}   for i = 1, rows do tmp[i] = {} array[i] = {} for j = 1, columns do tmp[i][j] = {i, j} array[i][j] = 0 end end   for k = 1, num do local r = math.random(#tmp) local c = math.random(#tmp[r]) --print(r..'|'..c)   local r1, c1 = tmp[r][c][1], tmp[r][c][2],   table.remove(tmp[r], c) if #tmp[r] == 0 then table.remove(tmp, r) end   array[r1][c1] = math.random(3) end   for i = 1, #array do print(unpack(array[i])) end     end   randarray(10, 10, 80)

techdojo

Once you've generated the array, you need to "post process" it repeatedly until it passes the "no matching tiles" rule.

To do that - run over the array and then for each cell, check it's (potentially) four neighbours, the easiest way is to call a function that just moves along the array in the given direction returning the number of continuous matches, if they are >= 3 then you have to change the value (ideally pick one that isn't one of the four neighbours) and start again at the beginning.

Obviously when you can process the entire table and each function returns < 3 matches your job is done.

It may take a few dozen iterations to get right to the end of the table, but considering the speed of Gideros you won't notice it.

alexzheng

That 's the most straight way that I though of at first,but after replace one with another value, it may produce anorther mathch, and I'm afraid it will be endless in worst case.
May be any clever way can do.

atilim

If you have 4 different values (like 1, 2, 3, 4) instead of 3 (1, 2, 3), you can make sure that, your replacement doesn't produce another match. But I think with 3 values, it's hard to do.

alexzheng

Yes, that's scratch my hairs

alexzheng

finally, I get what I expected，I checked each value before assign it and drop the cell if none of the value can be used.

local function checkChain(array, row, column) local paths = { {{1, 0}, {-1, 0}}, {{0, 1}, {0, -1}}, }   local values = {1, 2, 3}   local rand1, rand2 = math.random(#values), math.random(#values) values[rand1], values[rand2] = values[rand2], values[rand1]   for index = 1, #values do local value = values[index] local canUse = true for i = 1, #paths do local path = paths[i] local set = {} for j = 1, #path do local r, c r, c = row + path[j][1], column + path[j][2]   local value2 = array[r] and array[r][c] while value == value2 do set[#set + 1] = value2 r, c = r + path[j][1], c + path[j][2] value2 = array[r] and array[r][c] end end   if #set > 1 then canUse = false end end   if canUse then return value end   end   return nil   end   local function randarray(rows, columns, num) math.randomseed(os.time())   local array = {} local tmp = {}   for i = 1, rows do tmp[i] = {} array[i] = {} for j = 1, columns do tmp[i][j] = {i, j} array[i][j] = 0 end end   local count = 0   local total = rows * columns for k = 1, total do local r = math.random(#tmp) local c = math.random(#tmp[r]) --print(r..'|'..c)   local r1, c1 = tmp[r][c][1], tmp[r][c][2],   table.remove(tmp[r], c) if #tmp[r] == 0 then table.remove(tmp, r) end     local value = checkChain(array, r1, c1)   if value then array[r1][c1] = value count = count + 1   if count >= num then break end end end   print('---------------------------------------')   for i = 1, #array do print(unpack(array[i])) end   print('---------------------------------------')     end   for i = 1, 10 do randarray(10, 10, 80)     end