guide to solving bohliao puzzles
stayed onboard on monday and my boss was feeling boh liao and got all the officers together for supper. now is uni vacation, so got a lot of young officers who are attached to ship to entertain my boss. somehow, they started asking each other stupid puzzles which seem to make my boss relive his younger days. dunno if it's "i'm a maths guy" or it's "i got a super memory for puzzles". i seem to have heard most of the puzzles and recall the solutions readily.
anyway, while it seems the puzzle require some great ingenuity or wat pple call "think out of the box", it's not difficult if one attempt to use some method. (maybe different pple got different sizes of boxes to think out of) here's a simple tip to solving most maths-related problems.
reduce to a simpler/smaller case.
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example 1. there are 10 boxes, each containing 10 marbles. in 9 boxes, the marble each weighed 1 gram and in one box, the marble each weighed 1.1 gram. we have an electronic weighing machine which can give the exact weight.
aim: in one weighing, find the box with the heavier marbles.
some analysis: obviously weighing everything together will not help and ppl get stuck here. Let's reduce to 2 boxes with 2 marbles each. One has marbles weighing 1 gram each and one has marbles weighing 1.1 grams each.
how many marbles from each box should i choose to weigh? if i take 2 marbles from each box, i'll definitely get the reading 4.2 gram. similarly, if i take 1 marble from each box, the reading will always be 2.1 gram. ie. no conclusion.
now, let's name the boxes B1 and B2. if i take 1 marble from B1 and 2 marble from B2.
if B1 contains the heavier marbles, the reading will be 3.1 gram. if B2 contains the heavier marbles, the reading will be 3.2 gram. there's a different reading for each possibility. that is, we are able to conclude from the reading which box contains the heavier marbles.
hmmm... not v difficult to guess the solution for 10 boxes and 10 marbles right?
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example 2. there are 8 marbles with 1 lighter than the rest. there's a balancing weigh (ie those see-saw type which can tell which marble is heavier).
aim: in two weighing, find the lighter marble.
let's reduce the problem. in one weighing, how can i find a lighter marble?
let's call the scales, left and right. in one weighing, how many possible outcomes can there be? three. (A) the left being heavier, (B) the right being heavier, (C) both just as heavy. hmmmm. so, if i got three marbles with one lighter, i just weigh any one against another. (A) would imply that the right is lighter. (B) imply the left is lighter. (C) would imply the one not weighed is lighter.
it's also clear with only 3 possible outcomes, there's no possible way to find a ligher marble from a pile of 4 marbles with just one weighing.
so, returning to the question, let's say we completed the first weighing. we must conclude that the lighter marble is one in a pile of 3 marbles. So, that means, in our first weighing, we must weigh a pile of 3 against another pile of 3. the rest of the details can be worked out easily.
as a maths guy, i like to generalise the problem. so, in N weighings, it's possible to find a lighter marble from a pile of at most 3N marbles. So, if some guy ask you to find the lighter marble from a pile of 30 with just 3 weighings, you can tell him, mighty mao say impossible.
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hmmm. so where's the box to think out from?
stayed onboard on monday and my boss was feeling boh liao and got all the officers together for supper. now is uni vacation, so got a lot of young officers who are attached to ship to entertain my boss. somehow, they started asking each other stupid puzzles which seem to make my boss relive his younger days. dunno if it's "i'm a maths guy" or it's "i got a super memory for puzzles". i seem to have heard most of the puzzles and recall the solutions readily.
anyway, while it seems the puzzle require some great ingenuity or wat pple call "think out of the box", it's not difficult if one attempt to use some method. (maybe different pple got different sizes of boxes to think out of) here's a simple tip to solving most maths-related problems.
reduce to a simpler/smaller case.
===================================================
example 1. there are 10 boxes, each containing 10 marbles. in 9 boxes, the marble each weighed 1 gram and in one box, the marble each weighed 1.1 gram. we have an electronic weighing machine which can give the exact weight.
aim: in one weighing, find the box with the heavier marbles.
some analysis: obviously weighing everything together will not help and ppl get stuck here. Let's reduce to 2 boxes with 2 marbles each. One has marbles weighing 1 gram each and one has marbles weighing 1.1 grams each.
how many marbles from each box should i choose to weigh? if i take 2 marbles from each box, i'll definitely get the reading 4.2 gram. similarly, if i take 1 marble from each box, the reading will always be 2.1 gram. ie. no conclusion.
now, let's name the boxes B1 and B2. if i take 1 marble from B1 and 2 marble from B2.
if B1 contains the heavier marbles, the reading will be 3.1 gram. if B2 contains the heavier marbles, the reading will be 3.2 gram. there's a different reading for each possibility. that is, we are able to conclude from the reading which box contains the heavier marbles.
hmmm... not v difficult to guess the solution for 10 boxes and 10 marbles right?
==============================================
example 2. there are 8 marbles with 1 lighter than the rest. there's a balancing weigh (ie those see-saw type which can tell which marble is heavier).
aim: in two weighing, find the lighter marble.
let's reduce the problem. in one weighing, how can i find a lighter marble?
let's call the scales, left and right. in one weighing, how many possible outcomes can there be? three. (A) the left being heavier, (B) the right being heavier, (C) both just as heavy. hmmmm. so, if i got three marbles with one lighter, i just weigh any one against another. (A) would imply that the right is lighter. (B) imply the left is lighter. (C) would imply the one not weighed is lighter.
it's also clear with only 3 possible outcomes, there's no possible way to find a ligher marble from a pile of 4 marbles with just one weighing.
so, returning to the question, let's say we completed the first weighing. we must conclude that the lighter marble is one in a pile of 3 marbles. So, that means, in our first weighing, we must weigh a pile of 3 against another pile of 3. the rest of the details can be worked out easily.
as a maths guy, i like to generalise the problem. so, in N weighings, it's possible to find a lighter marble from a pile of at most 3N marbles. So, if some guy ask you to find the lighter marble from a pile of 30 with just 3 weighings, you can tell him, mighty mao say impossible.
=========================================================
hmmm. so where's the box to think out from?
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