November 13, 200322 yr Ok i figured out a cheap way to win. I cheated! Open 2 windows, you make him go first in window 1, then do what he did in window 2 as the first move, just keep doing what he does and switching windows........i dunno if there is a legit way to win but i still feel fulfilled
November 13, 200322 yr Ok i figured out a cheap way to win. I cheated! Open 2 windows, you make him go first in window 1, then do what he did in window 2 as the first move, just keep doing what he does and switching windows........i dunno if there is a legit way to win but i still feel fulfilled It's not cheating if he doesn't find out...
November 13, 200322 yr ***SOLUTION*** Don't look if you don't want to know. Ok, it's a little complicated but here goes. What you need to do is try to get it down to the point where there is one marble in each of three rows on his turn. First, you need to assign each row a value based on how many pearls are in that row: 6 = 110 5 = 101 4 = 100 3 = 11 2 = 10 1 = 1 So the starting configuration works out like this 3 pearls + 4 pearls + 5 pearls + 6 pearls Which equals 11+100+101+110 Which totals 322 What you need to do is remove pearls so that the digits in the total are all even, zero counts as even. For instance, if you go first and remove the whole row containing four pearls the equation becomes 11+101+110 = 222 Now say he removes two pearls from the row with three, 1+101+110 =212. You can remove two pearls from the bottom row with six, leaving four. The equation is now 101+100+1 = 202. Continue removing pearls leaving totals with even digits in this fashion outlined above. Make sure to leave at least one pearl in three rows, don’t get down to two. Sooner or later he will remove pearls so that there is one row of one and one row of either two or three and one row with >3 pearls. On your turn remove pearls in the >3 row so that there is one row with one pearl, one row with two pearls, and one row with 3 pearls. At this point you should be able to get him down to one pearl in each of three rows pretty easy. If he takes one from the two row then take two from the three row, or vice versa. Once you are down to one pearl in each of three rows on his turn it’s all over. If he happens to take the one pearl row in the one, two, three scenario then just take one from the three row leaving two rows of two. If he takes away one pearl take away the remaining two row. If he takes away a two row take away one pearl from the remaining row. It sounds complicated but once you do it a couple of times it becomes instinctual.
November 15, 200322 yr ***SOLUTION*** Don't look if you don't want to know. Ok, it's a little complicated but here goes. What you need to do is try to get it down to the point where there is one marble in each of three rows on his turn. First, you need to assign each row a value based on how many pearls are in that row: 6 = 110 5 = 101 4 = 100 3 = 11 2 = 10 1 = 1 So the starting configuration works out like this 3 pearls + 4 pearls + 5 pearls + 6 pearls Which equals 11+100+101+110 Which totals 322 What you need to do is remove pearls so that the digits in the total are all even, zero counts as even. For instance, if you go first and remove the whole row containing four pearls the equation becomes 11+101+110 = 222 Now say he removes two pearls from the row with three, 1+101+110 =212. You can remove two pearls from the bottom row with six, leaving four. The equation is now 101+100+1 = 202. Continue removing pearls leaving totals with even digits in this fashion outlined above. Make sure to leave at least one pearl in three rows, don’t get down to two. Sooner or later he will remove pearls so that there is one row of one and one row of either two or three and one row with >3 pearls. On your turn remove pearls in the >3 row so that there is one row with one pearl, one row with two pearls, and one row with 3 pearls. At this point you should be able to get him down to one pearl in each of three rows pretty easy. If he takes one from the two row then take two from the three row, or vice versa. Once you are down to one pearl in each of three rows on his turn it’s all over. If he happens to take the one pearl row in the one, two, three scenario then just take one from the three row leaving two rows of two. If he takes away one pearl take away the remaining two row. If he takes away a two row take away one pearl from the remaining row. It sounds complicated but once you do it a couple of times it becomes instinctual. That makes a little sense, but I'm totally messed up with the pearls having numerical value. That shoots right over my head.
November 15, 200322 yr That makes a little sense, but I'm totally messed up with the pearls having numerical value. That shoots right over my head. There's a method to how the numbers are assigned but you don't need to know it. Just count up the number of pearls in a row and look at the value as I listed it in my post. It works for Pearls Before Swine III too but you need to know how to figure out the values for higher numbers. I got to level 21 last night before I had to quit because it was taking too much time.
November 17, 200322 yr There's a method to how the numbers are assigned but you don't need to know it. Just count up the number of pearls in a row and look at the value as I listed it in my post. It works for Pearls Before Swine III too but you need to know how to figure out the values for higher numbers. I got to level 21 last night before I had to quit because it was taking too much time. OK, I did it. I have no idea why those rows get assigned the numerical values that they do, but thanks anyways, bj!
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