Wednesday, January 11, 2006


Pearls before swine

Is it possible to win this game ?

Yes! I did it by taking 2 from the top, then 3 from the bottom then 1 from the top, then two from the bottom.
The only chance for winning is if you start. There are 3 rows (3, 4 and 5 pearls). Any move you make will lead to having the following combinations:

1 - 2 - 3
3 - 3
4 - 4
1 - 1 - 1

Then you lose.
Not that I can tell. As long as you're only allowed to pick from an even number of balls, you'll always lose. Plus the computer has too much control. If there was a set number of balls that he had to every time, you'd be able to win.

The end-game scenario always has 4 pearls. There's no way for the player that goes first to win.

I played a trick similar to this on folks at parties (I used dollar bills with "winner takes all" as the hook). It was amazing how angry some people became. It was equally amazing how quickly competition short circuited their brain.

Only a very small number of people actually said, "You go first this time". Of that small number, a large number were female.
yes and no. since you always have to go first you will always lose. After you get rid of your pearls count the number of pearls the guy holding them removes. He alwas makes it an even number total between yours and his. (until the last move) Try playing with someone where THEY go first and use this rule. You will always win!
Well spotted, Alex.

Actually I realized later (before reading later commments) that there was a way of solving it. Except that I wasn't able to solve it. Not persistent enough. Bummer.
Alex is a genius. Try it, his method is foolproof. Well done.
No, I take it back... not quite foolproof, but it does seem to work sometimes.
What about the "Glass Bead Game"?
thank you very much, anonymous
Alex got it..also if you don't take any balls your opponent will be left with all the balls including the last one.
Hey guys, this is actually a form of an ancient game called NIM, and it is completely deterministic. If you know the strategy, you can win every time, even his more advanced games II & III (I located the strategy and beat him continuously up to level 10 in game III and got board). Those of you that are interested, and know how to convert the number of balls in a row to Base 2, take a look here: . Good Luck! My email is if you want to contact me, the user name that I wanted is in use, and decided not to bother :)
you guys are need to use ur head, you can win every single time if u go first. it starts off 3,4,5. if u make it, 1,4,5...there is no possible way for him to win

he would usually make it, 1,3,5. you easily change that to 1,3,2. He makes that..2,2,1.

you change it to 2,2. There is no possible move for him to make to win.
This explains how to win
Yes! I did it by taking 2 from the top, then 3 from the bottom then 1 from the top, then two from the bottom.

certified, this really works, but how it works, still i canĀ“t tell
A lot of good info here.

It is possible to win and the strategy is simple. You must be the first person to set the Nim sum to zero to win (see wikipedia - Nim)and reset that zero on your turn each time.

He who then disturbs the zero loses!

Basic strategy (for any game where the last to draw loses).

Its all done with binary addition with absolutely no carries.

Remember Binary number bits:

64 32 16 8 4 2 0

Set the bits that represent the number of pearls remaining on the row.

Create a binary representation of each row.

Next add the binary rows together to create the Nim Sum (in binary).

If you have an even number of 1's then that bit for the Nim Sum becomes zero. An Odd number of 1's sets that bit to one (this is an XOR function). Zero is always Zero.

To win, the bits representing the Nim sum must be set to zeros on your turn. If you can't make it zero that means you've made a mistake on a previous move and are in danger of losing unless your opponent makes a mistake.

Pull the pearls from a single row that will allow you to set the bits representing the Nim sum to all zeros and don't disturb existing bits already at zero!

Once you set the bits on the pearl row to make the Nim sum zero, the sum of the bits on the pearl row will tell you how many pearls to leave on that row to end your turn.

If the game starts out with the NIM sum already at zero make your opponent goes first (to disturb the zero).

He will then automatically lose unless you make a mistake. If you fail to make him go first in this situation Juan will always win.

There's a little more to the strategy but this info will get you about a 90% win. It can get tricky sometimes times to know which row to pull from but this is rare. A little observation can help here. There's math for that situation too.

Hmmmmm....... what if i win it the loser way?

Open 2 windows and just follow his move in one of the windows. You'll definitely win at one of the two games. Heck, he's playing with himself, but oh wells, i don't really care. \(-__-)/
u also can win if take 3 balls from middle,
then take all from bottom,so the ball has left 1 from middle and 3 from will be stupid because not take all from top..
it take only 1 from top..
so u can take all from top...
Well I have tried this method on my app Pearls Before Swine, and it does not work! I will have to check to see if the computer is any different! :(
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