Few questions about minesweeper

31201558 · **Posted:** Mon Jan 03, 2011 1:35 am

I know I should not......

But what I need to do ?

Tony · **Posted:** Mon Jan 03, 2011 1:41 am

You need to make sure those values are in the same range as the array. Trace back to where the value comes from, and change your expressions to guarantee that the answer will be in a particular range of values.

31201558 · **Posted:** Mon Jan 03, 2011 1:45 am

add one to both?

cells (minex (i) div 40+1, miney (i) div 40+1) := "mine"

Tony · **Posted:** Mon Jan 03, 2011 1:50 am

That will guarantee the low bound of 1, yes. It will also increase your upper bound value by 1 as well. Coincidentally this might still work, but I'll leave the prove of that up to you.

31201558 · **Posted:** Mon Jan 03, 2011 2:01 am

Turing:

var mx, my, mb, font : int
var dead := false
var cells : array 1 .. 10, 1 .. 10 of string
var value : array 1 .. 9, 1 .. 9 of int
var minex, miney : array 1 .. 10 of int
buttonchoose ("multibutton")
for a : 0 .. 320 by 40
for b : 0 .. 320 by 40
drawfillbox (a, b, a + 38, b + 38, 7)
end for
end for
for i : 1 .. 9
for j : 1 .. 9
cells (i, j) := "not mine"
value (i, j) := 0
end for
end for
%mines
for i : 1 .. 10
minex (i) := Rand.Int (0, 359)
miney (i) := Rand.Int (0, 359)
minex (i) := minex (i) - minex (i) mod 40
miney (i) := miney (i) - miney (i) mod 40
drawfillbox (minex (i), miney (i),minex (i)+38,miney (i)+38 ,green)
end for
for i : 1 .. 10

cells (minex (i) div 40+1, miney (i) div 40+1) := "mine"
end for
for i : 1 .. 9
for j : 1 .. 9
if cells (i, j) = "mine" then
if j + 1 <= 9 then
if cells (i, j + 1) not= "mine" then
value (i, j + 1) += 1
end if
end if
if i + 1 <= 9 and j + 1 <= 9 then
if cells (i + 1, j + 1) not= "mine" then
value (i + 1, j + 1) += 1
end if
end if
if i + 1 <= 9 then
if cells (i + 1, j) not= "mine" then
value (i + 1, j) += 1
end if
end if
if i + 1 <= 9 and j - 1 > 0 then
if cells (i + 1, j - 1) not= "mine" then
value (i + 1, j - 1) += 1
end if
end if
if j - 1 > 0 then
if cells (i, j - 1) not= "mine" then
value (i, j - 1) += 1
end if
end if

if i - 1 > 0 and j - 1 > 0 then
if cells (i - 1, j - 1) not= "mine" then
value (i - 1, j - 1) += 1
end if
end if
if i - 1 > 0 then
if cells (i - 1, j) not= "mine" then
value (i - 1, j) += 1
end if
end if
if i - 1 > 0 and j + 1 <= 9 then
if cells (i - 1, j + 1) not= "mine" then
value (i - 1, j + 1) += 1
end if
end if
end if
end for
end for
for i : 1 .. 9
for j : 1 .. 9
if value (i, j) > 0 then
cells (i, j) := "num cells" + intstr (value (i, j))
end if
end for
end for

loop
mousewhere (mx, my, mb)
mx := mx - mx mod 40
my := my - my mod 40
for i : 1 .. 10
if mx = minex (i) and my = miney (i) and mb = 1 then
dead := true
end if
end for
var a : int
var b : int
a := mx div 40
b := my div 40
if cells (a, b) = "num cells" and mb = 1 then
font := Font.New ("serif:36")
Draw.Text (intstr (value (a, b)), mx, my, font, red)
end if
if cells (a, b) = "not mine" and mb = 1 then
drawfillbox (mx + 1, my + 1, mx + 37, my + 37, white)
end if
exit when dead
delay (100)
end loop

Yes, that problem has solved.

But my minesweeper is still a mess

Can you run this to see where my problem is?

Something gotta be wrong with the mousewhere...

Tony · **Posted:** Mon Jan 03, 2011 2:04 am

Same problem, "Array subscript is out of range". Turing tells you what line the problem is at. You can use the same trick with printing the values being used for indexes, to see when and with what value the bound is broken.

31201558 · **Posted:** Mon Jan 03, 2011 2:05 am

You'll have to use F1 to run this because of the mousewhere.

Just in case you havn't noticed this

31201558 · **Posted:** Mon Jan 03, 2011 2:12 am

I fixed it again.

And I noticed that anywhere that's suppose to have a number don't have one.

Why is this happening?

Tony · **Posted:** Mon Jan 03, 2011 2:18 am

Probably because "if cells (a, b) = "num cells" and mb = 1 then" is not true.

31201558 · **Posted:** Mon Jan 03, 2011 2:26 am

Oh my god!!!!!!!!!!!!

It worked!!!!!!!!!!!

Thank you so much!!!!!!!!!!!!

Tony!!!!!!!!!!!!!!!!!!

I'll donate all my bits to you!!!!!!!!!!!!!!!

Well, some for token too......

Tony · **Posted:** Mon Jan 03, 2011 2:32 am

And with 13 hours to spare! Well done Smile

31201558 · **Posted:** Thu Jan 06, 2011 6:33 pm

Well, our teacher has lated the due date because he is so nice......

Therefore I now want to make the open multiboxes feature to work.

I just notice that there's a flood fill algorithm to make it.

Here's my code:

Turing:

var mx, my, mb, font : int
var dead := false
var cells : array 1 .. 10, 1 .. 10 of string
var value : array 1 .. 9, 1 .. 9 of int
var minex, miney : array 1 .. 10 of int
var totalTime : real := 0
buttonchoose ("multibutton")
for a : 0 .. 320 by 40
for b : 0 .. 320 by 40
drawfillbox (a, b, a + 38, b + 38, 7)
end for
end for
for i : 1 .. 9
for j : 1 .. 9
cells (i, j) := "not mine"
value (i, j) := 0
end for
end for
%mines
for i : 1 .. 10
minex (i) := Rand.Int (0, 359)
miney (i) := Rand.Int (0, 359)
minex (i) := minex (i) - minex (i) mod 40
miney (i) := miney (i) - miney (i) mod 40
drawfillbox (minex (i), miney (i), minex (i) + 38, miney (i) + 38, green)
end for
for i : 1 .. 10
cells (minex (i) div 40 + 1, miney (i) div 40 + 1) := "mine"
end for
for i : 1 .. 9
for j : 1 .. 9
if cells (i, j) = "mine" then
if j + 1 <= 9 then
if cells (i, j + 1) not= "mine" then
value (i, j + 1) += 1
end if
end if
if i + 1 <= 9 and j + 1 <= 9 then
if cells (i + 1, j + 1) not= "mine" then
value (i + 1, j + 1) += 1
end if
end if
if i + 1 <= 9 then
if cells (i + 1, j) not= "mine" then
value (i + 1, j) += 1
end if
end if
if i + 1 <= 9 and j - 1 > 0 then
if cells (i + 1, j - 1) not= "mine" then
value (i + 1, j - 1) += 1
end if
end if
if j - 1 > 0 then
if cells (i, j - 1) not= "mine" then
value (i, j - 1) += 1
end if
end if

if i - 1 > 0 and j - 1 > 0 then
if cells (i - 1, j - 1) not= "mine" then
value (i - 1, j - 1) += 1
end if
end if
if i - 1 > 0 then
if cells (i - 1, j) not= "mine" then
value (i - 1, j) += 1
end if
end if
if i - 1 > 0 and j + 1 <= 9 then
if cells (i - 1, j + 1) not= "mine" then
value (i - 1, j + 1) += 1
end if
end if
end if
end for
end for
for i : 1 .. 9
for j : 1 .. 9
if value (i, j) > 0 then
cells (i, j) := "num cells" + intstr (value (i, j))
end if
end for
end for
loop
mousewhere (mx, my, mb)
mx := mx - mx mod 40
my := my - my mod 40
if mx < 0 then
mx := 0
elsif mx > 359 then
mx := 359
end if
if my < 0 then
my := 0
elsif my > 359 then
my := 359
end if
for i : 1 .. 10
if mx = minex (i) and my = miney (i) and mb = 1 then
dead := true
end if
end for
var a : int
var b : int
a := mx div 40
b := my div 40
if cells (a + 1, b + 1) not= "num cells" and mx < 359 and my < 359 and mb = 1 then
font := Font.New ("serif:36")
Draw.Text (intstr (value (a + 1, b + 1)), mx, my, font, red)
end if
if cells (a + 1, b + 1) = "not mine" and mb = 1 then
drawfillbox (mx + 1, my + 1, mx + 37, my + 37, white)
end if
locate (1, 1)
totalTime := totalTime + 0.1
put totalTime
if totalTime = 999 then
dead := true
end if
exit when dead
delay (100)
end loop

The problem is, I don't think I got a reveal procedure. Will I still be able to make the flood fill without an reveal procedure?

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