Create a chess/checker board with for loop?

cindy.h1986 · November 8, 2013, 10:27am

This code create a checker board style grid, I just wonder if there’s a better way to set the colors or create this style of pattern?

&nbsp; &nbsp; local group = display.newGroup()&nbsp; &nbsp; &nbsp; local blocks = {}&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;local blockSize = display.contentWidth/5 &nbsp;&nbsp;&nbsp;&nbsp;for i = 1, 5, 1 do &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;for j = 1, 5, 1 do &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;blocks[i] = display.newRect(group, &nbsp;0, 0, blockSize, blockSize) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;blocks[i]:setReferencePoint(display.CenterReferencePoint) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;blocks[i].x = ((blockSize/2) + (i-1) \* blockSize) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;blocks[i].y = (100 +(j-1) \* blockSize) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;blocks[i]:setFillColor(0, 0, 0, 255) &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;if (i == 1) and ( j == 1) or (i == 1) and (j == 3) or (i == 1) and (j == 5) &nbsp;then &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp;blocks[i]:setFillColor(255, 255, 255, 255) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;end &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;if (i == 2) and ( j == 2) or (i == 2) and (j == 4) then &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp;blocks[i]:setFillColor(255, 255, 255, 255) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;end &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;if (i == 3) and ( j == 1) or (i == 3) and (j == 3) or (i == 3) and (j == 5) &nbsp;then &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp;blocks[i]:setFillColor(255, 255, 255, 255) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;end &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;if (i == 4) and ( j == 2) or (i == 4) and (j == 4) then &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp;blocks[i]:setFillColor(255, 255, 255, 255) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;end &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;if (i == 5) and ( j == 1) or (i == 5) and (j == 3) or (i == 5) and (j == 5) &nbsp;then &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp;blocks[i]:setFillColor(255, 255, 255, 255) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;end &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;end &nbsp;&nbsp;&nbsp;&nbsp;end

aukStudios · November 8, 2013, 12:35pm

Hi Cindy,

Yes, the better way would be to do it like this:

[lua]

blocks[i]:setFillColor(0, 0, 0, 255)

if (i+j)%2 == 0 then

blocks[i]:setFillColor(255, 255, 255, 255)

end

[/lua]

The expression (i+j)%2 uses the % (modulo operator, which gives you the remainder after dividing. So (i+j)%2==0 tests whether (i+j) is divisible by 2 (remainder is 0).

Andrew

aukStudios · November 8, 2013, 12:35pm