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Archive for the ‘Puzzles’ Category

Tricks, Traps and Tips for better Problem Solving

Cost: Nada!
Although you’re more than welcome to send me a note of appreciation and/or order some books as a way to support the costs of providing this service.

Where? Head here for all the details.

Summary:
If you work for a living, then you solve problems of all types.This session will explore some simple PS concepts and explain how we can use formal, and informal, PS techniques in every day Life.

Prerequisites:
Peter believes very much in the idea that we learn by doing, more precisely? That we learn by failing at doing.

So??? This presentation WILL be interactive.
1) Make sure you have a deck of playing cards handy.
2) Peter will take ‘questions’ via e-mail as he presents.

Feedback from a similar live session:
Peter … We reviewed the feedback forms this week – of the 90 collected for the Problem Solving sessions, overwhelmingly the ratings were 5s (highest) We summarized the feedback as follows, included member quotes:

‘Over the top’ successful. Very dynamic, excellent speaker.
Members wanted more from Peter; many felt his sessions were too short.
“..energizing & interesting. Very helpful.”
“..too short – was enjoyable & thought provoking!”
“Fabulous, awesome presentation. Great interaction & exercises.
Could have spent the whole afternoon in his session.”
“Excellent session, extremely dynamic presenter! Useful for any level of team member.”
“Peter was the “BEST” part of the day.”
“Please bring Peter back to talk to us.”

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Well, I first stumbled across this puzzle in Charles Barry Townsend’s collection of puzzles, “World’s Toughest Puzzles” and here’s the answer he provided.

The minimum required is 4 men as follows:
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An explorer and financial backer are discussing an upcoming treasure hunt. They’re examining a map in front of them and getting more and more confused.

Here, right in the middle of this desolate land, is where we think they buried the treasure – It’s a single gem, fine enough to be worth a King’s ransom.

You have to travel from this port and there is neither water nor food on the only route inland.

Each person in the party can only travel enough for 5 days, and they can only travel 30 miles per day.

The Port is 120 miles from the location of the treasure.

So, what we need to calculate is, what is the fewest number of men, yourself included, that you will need in your party so that you alone can get to the treasure, stay overnight and return to the port, without dying.

—- There you go… a little bit of project planning for the week. See you next Saturday.

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(This posting is answering the problem posed in Jiggler #3)

The logically and statistically correct answer is – you switch from your original choice to the remaining un-opened box.

When you do this, you’ll increase your chance of finding the $1M cheque by 100% – Your initial chance of success was 1/3… ie. the chance that you initially selected the wrong box was 2/3 – therefore you switch to the last remaining un-opened box.
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Okay – this one is going to start out as an old Chestnut and then take a twist. The starting situation is known as the ‘Monty Hall’ Problem and the answer is easy enough to Google if that’s what moves you. But, don’t do that without at least giving this a try.

I present you with three closed unmarked identical boxes.
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Last week I posted a thinly disguised Nim problem. The goal was to both provide an answer and, even more importantly, speak to how you arrived at the solution.

We had two answers from Jim and Hackz0r

Jim? Close but no cigar… you suggested that we leave multiples of 4… I think you’re so close that you might have misunderstood the goal. The objective was NOT to have to take that last peanut. If you leave multiples of 4, you’re forcing yourself into the losing position.

Consider what happens if you leave 4 peanuts on the table. Since the opponent can take 1,2 or 3… he’ll take three – leaving you that last stinking peanut. You lose.
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Good Morning!

Here’s an easier one than last week’s offering.

You’re sitting across from a friend. There are 23 peanuts between the two of you.

You go first, you’re allowed to remove (and eat if you wish) either one, two or three of the peanuts.

Then your friend can do the same thing, she can take 0ne, two or three of the remaining peanuts.

Then it’s your turn again, and so on and so on.

The person who takes the LAST peanut… pays for dinner.
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