Any demoscener from Mensa?
category: residue [glöplog]
Here's one, I'm not sure if you know it already, but it's nice:
You have a Minesweeper game of n*m squares in size, and with k mines (k<m*n) in random (but different) positions. You know; the typical Minesweeper game.
You solve the game, and you sum all the numbers you see in the squares, and call the result S.
Then, in the same board, you set up a different game, where there is a mine in the places where there were no mines and there are no mines where there were mines. So, you generate a "negative" or "inverse" mines set-up.
Now you solve this new game and sum all the numbers in the squares at the end. What will be the result?
You have a Minesweeper game of n*m squares in size, and with k mines (k<m*n) in random (but different) positions. You know; the typical Minesweeper game.
You solve the game, and you sum all the numbers you see in the squares, and call the result S.
Then, in the same board, you set up a different game, where there is a mine in the places where there were no mines and there are no mines where there were mines. So, you generate a "negative" or "inverse" mines set-up.
Now you solve this new game and sum all the numbers in the squares at the end. What will be the result?
It's been a long time that I played Minesweeper. Am I right that the numbers also count for diagonal neighbours, i.e. the max. number is 8, or is it only 4? I'll try it...
BTW, is it possible to solve this without luck?
I guess it isn't... which shows what a crappy game Minesweeper is.
I guess it isn't... which shows what a crappy game Minesweeper is.It's clear that diagonals count, i.e. the max. number displayed on a field is 8.
OK. There are up to 8 fields that border to a field with a bomb. So in up to 8 surrounding fields, the fact that there's a bomb in a field counts for 1.
However, there are also fields which are bordered by 5 or even 3 fields only! This means that the sum S isn't simply 8 * k. We only know that S <= 8 * k.
OK. There are up to 8 fields that border to a field with a bomb. So in up to 8 surrounding fields, the fact that there's a bomb in a field counts for 1.
However, there are also fields which are bordered by 5 or even 3 fields only! This means that the sum S isn't simply 8 * k. We only know that S <= 8 * k.
If we have a negative set-up, this means that in every field in which there was a bomb, there's no bomb now, and the other way round.
If a bomb was bordered by 8 fields, and thus counted for 1 in 8 fields, there will be one field instead numbered 8. So this doesn't change the value of S.
Likewise, if a bomb was located in a corner and thus only bordered by 3 fields, there will now be one field numbered 3. It doesn't change the value of S, either, and the same goes for bombs bordered by 5 fields.
If a field had 2 fields with bombs at its borders, then the number of this field was 2. Now this field would have a bomb, and thus increase the number of the two bordering fields by 1.
It seems to me that the value of S remains unaffected by the inversion of the Minesweeper set-up.
If a bomb was bordered by 8 fields, and thus counted for 1 in 8 fields, there will be one field instead numbered 8. So this doesn't change the value of S.
Likewise, if a bomb was located in a corner and thus only bordered by 3 fields, there will now be one field numbered 3. It doesn't change the value of S, either, and the same goes for bombs bordered by 5 fields.
If a field had 2 fields with bombs at its borders, then the number of this field was 2. Now this field would have a bomb, and thus increase the number of the two bordering fields by 1.
It seems to me that the value of S remains unaffected by the inversion of the Minesweeper set-up.
adok why don't you code a program that tries different m,n,k combinations and see if it's true?
Because it doesn't make sense this time
Adok, please :)
tmb, want to show us how brainy you are?
adok btw this is the only solution:
Uhm. No it's not. See that "x" there in the upper right corner of the window? It usually takes people a while to figure out, though.
Oh yes.
1) Figure out why this works:
http://www.glumbert.com/media/multiply
2) If you need less than 3 minutes, use this algorithm to multiply 987x979
Erm... I don't see any difference with the schoolboy multiply algorithm. It is not even beatiful IMHO. So... crap!
http://www.youtube.com/watch?v=KgH1cIqLd4I
I think the video is nice. It sells something very simple as something very esoteric and complicated.
sorry but i agree with texel, it's just multiplication for dummies. it doesn't even show why. better teach modular arithmetic.
my friends son is 14 and a member of mensa
we're trying to get him to join the scene and stop taking so much drugs etc
we're trying to get him to join the scene and stop taking so much drugs etc
"...join the scene and stop taking so much drugs etc"
anyone see the problem here?
anyone see the problem here?
not if ur a goog gurilla
well or join the scene and continue taking drugs etc
atleast he would be excercising his intelligence and put it to good use instead of laming around
atleast he would be excercising his intelligence and put it to good use instead of laming around
what about solving this?
A coder and musician look sadly at the only people who showed up for their demo party; Only 3 people, all friends of the musician.
"You know," says the musician, "If you multiply the ages of these 3 lamers you get 2450 and if you add them together you get twice your age. Can you tell me how old they are?"
"No I cant" says the coder after thinking a while.
"Well OK, consider that I am the oldest of us all."
After thinking a while the coder says. "Ok, I know how old they are."
The problem: How old is the musician?
(If you heard it before dont spoil it)
"You know," says the musician, "If you multiply the ages of these 3 lamers you get 2450 and if you add them together you get twice your age. Can you tell me how old they are?"
"No I cant" says the coder after thinking a while.
"Well OK, consider that I am the oldest of us all."
After thinking a while the coder says. "Ok, I know how old they are."
The problem: How old is the musician?
(If you heard it before dont spoil it)
