its given me a brain ache

http://www.ebaumsworld.com/pearl.shtml

I've tried it for long enough an still havn't won , it must be a fix

Mike

The point is, never to let him leave you with two rows of two or three pearls each or with an one-two-three-combination.


So: A good start is to take two pearls from the last row, so there is one pearl left in that row.


No matter what he does next, force him into one of the constellations described above.

Na still can't get it Imust be as stupid as i look

Mike

i get the suspicion there is a way to win. Im going to try again!

Think Binary

Easy

quote:

---

Originally posted by Hellfire
Think Binary

Easy

---

OK, I've worked out one of the more common ones. Let computer go first.

Computer takes 5 pearls on row of 6
I take all pearls on row of three
Computer takes 1 pearl on row of 4
I take 3 pearls on row of 5
Computer takes 1 pearl on row of 4
I take remainder of pearls on row of 6
Computer takes 1 pearl on row of 4
I take remainder of row of 5

I win and he turns around and walks away

[Edited on 27-11-05 by Hellfire]

how do you let him go first?!

its wee easy if you make him go first! minter.

[Edited on 27/11/05 by JoelP]

next challenge, win with you going first!

Before you make a move, just click go

Given that in binary 1= 001, 2= 010, 3= 011, 4= 100, 5= 101 and 6= 110, add up the columns. ie

011
100
101
110
___
322

Every time you counter the computers move, ensure that the total reads in combinations of 0's and/or 2's by removing the necessary pearl/s from the columns.

It makes sense to me but does that make sense to you?

Doesn't matter who goes first, him or you............ it's easy. Beat him every time

[Edited on 27-11-05 by Hellfire]

i noticed a pattern of odds and evens at times, but never persued it. Will have a look and see if i can follow your plan