The Road Not Taken (Logic Puzzle)

In Robert Frost’s famous poem, The Road Not Taken (sometimes mistakenly referred to as “The Road Less Travelled”), a traveler is faced with the difficult choice of which road he should follow.

This brings to mind a classic logic puzzle, with mathematical implications.

Here is my poetic interpretation of the puzzle.

THE ROAD NOT TAKEN – Mr. Wagneezy Version

(Line 1 by Robert Frost)

Two roads diverge in a yellow wood

One leads to certain death
The other leads to riches untold
I stop to catch my breath

I soon discover I have no clue

Exactly which road is which
I look to the right, and then to the left
My eyes begin to twitch

Suddenly two gnomes appear

From out of nearby briars
One of them is a truthful gnome
The other one is a liar

In looking at these gnomes, alas

I cannot tell the difference
Which one speaks truth?   Which one speaks lies?
I fight the urge to wince

These seemingly identical gnomes

Both know which road to take
But instead of making it clear to me
They make it nearly opaque

The gnomes agree that one of them

Will answer a single question
Once I get the answer
They will end the conversation

I still can’t tell which one speaks truth

And which one is the liar
They smirk at me, these pesky gnomes
That came out of the briars

I must determine what to say

It is a daunting task
To get the gnomes to reveal the way
What question should I ask?

-----------------------------------------------------------------

Successfully getting these gnomes to show us which road to choose will take careful planning.  Since we have no way of knowing which gnome tells the truth and which gnome lies, we must be able to come up with a question that both gnomes would answer the same way.

Obviously, the direct approach (e.g. “Which road leads to untold riches?”) will not work, because the truth-teller would point to one road while the liar would point to the other road.

Therefore, we must ask an indirect question – one that incorporates both the truth and the lie.  We can accomplish this by asking one gnome to tell us which road the other gnome would point us towards.  (E.g. “If I asked the other gnome which road leads to untold riches, which road would he point to?”)

The logic here is that the truth about a lie gives the same result as a lie about the truth.

More specifically, if we happen to talk to the truth-teller, he will point to the wrong road because that is the road the liar would have pointed to.

On the other hand, if we happen to talk to the liar, he will also point to the wrong road because that is not the road the truth-teller would have pointed to.

In either case, the wrong road will be indicated, and we can choose the other road to travel on.

-----------------------------------------------------------------

For movie buffs – a version of this puzzle appeared in the movie Labyrinth :

-----------------------------------------------------------------

 

Earlier, I mentioned that this puzzle has mathematical implications.

Let’s deconstruct this puzzle into a similar mathematical question.

I hope you agree with me that telling the truth is positive, and lying is negative.

Consider mathematical operations that can be performed on two numbers.  Suppose one of the numbers is positive, and the other number is negative, but we DON’T KNOW WHICH IS WHICH.

Which operations are guaranteed to give us results with the same sign, regardless of which number is positive?

If we arbitrarily choose ±2 and ±3 for our numbers and use them to explore each operation, we get the following:

 

 

As we can see, multiplication and division are the only operations that fit the bill.

(Not coincidentally, multiplication and division have the same priority in the Order of Operations.)

Therefore, the logical argument

“The truth about a lie is equivalent to a lie about the truth”

seems to match up with the mathematical concepts

“A negative times a positive is equivalent to a positive times a negative”

and

“A negative divided by a positive is equivalent to a positive divided by a negative”.

By the way – multiplication and division both exhibit the desired property because

1)      Multiplying is the same thing as dividing by the reciprocal.

2)      Dividing is the same thing as multiplying by the reciprocal.

and

3)      A number’s sign (positive or negative) is the same as the sign of the number’s reciprocal.

 

May all your roads be wisely chosen.

 

-----------------------------------------------------------------

 

For Math Tutorials & Silly Math Songs, visit www.onlinemathpro.com.

 

Affiliate Disclosure

Watermelons and New Years' Parties (Puzzle)

Here’s a fun math puzzle involving percents.  The solution is not terribly difficult – but it is surprising how many people end up getting the answer wrong.  Why not give it a try:

The owner of a farm stand puts a bunch of watermelons on display at the beginning of the day.  The watermelons have a mass of 200 kg.  99% of the watermelons’ mass is water.  It is hot, and during the day water evaporates from the melons.  At the end of the day, 98% of the watermelons’ mass is water.  Assuming that no watermelons were sold during the day, what is the mass of the watermelons at the end of the day?

If you’d like to check your answer, the correct solution is below.

 

If you’re having a hard time, consider this puzzle.  The math involved the same as the watermelon problem, but the way in which the puzzle is presented tends to make the solution easier to find.

Some guys planned a New Years’ party and invited only women.  At the party, there were a total of 200 people, 99% of whom were women.  A lot of the women thought that the party was lame, so they left.  After one hour, 98% of the people left at the party were women.  How many total people were at the party after one hour?

This puzzle is slightly easier to grasp because it explicitly states that there are men and women.  (And, it states that only women leave the party.)

 

See below for the solutions.

 

 

 

Solution: New Years’ party puzzle

It’s not very hard to determine how many men and women there were at the beginning of the party:

Women:  99% of 200 = 198           Men:  1% of 200 = 2

Since only women left the party, there are still two men after one hour.  However, these two men now comprise 2% of the total party crowd.  2 is 2% of 100… therefore, there are 100 people at the party after one hour.

 

Solution: Watermelon puzzle

The watermelon puzzle is a little trickier simply because the problem doesn’t explicitly state that the watermelon is made up of water and solids.  Therefore, many people get hung up focusing on the water.

If 99% of the watermelon mass is water, then 1% is made up of solids.  1% of 200kg is 2 kg – so there are 2 kg of solids.  Solids do not evaporate, so at the end of the day, there are still 2 kg of solids.  This now comprises 2% of the total watermelon mass.  2 kg is 2% of 100 kg.  Therefore, there are 100 kg of watermelons left at the end of the day.

Happy New Years’, Everyone!

 

------------------------------

 

For Math Tutorials and Silly Math Songs, visit   www.onlinemathpro.com

 

 

 

Affiliate Disclosure

Ready, Go, SET!

Riddles and puzzles are a good way to challenge yourself and increase your logic skills.  They also give valuable practice in solving the story problems that seem to haunt so many math students.

Here’s a brain teaser that can be solved with simple math.  (As is the case with many math puzzles, there is more than one way to solve the problem.)  If you think you’re up to it, try to answer the question before reading the solutions and explanations at the bottom of this page.

There are 340 players in a SET* tournament.  Each game in the tournament is played with 4 people.  For each game, there is one winner who moves on to the next round.  Rounds will continue until there is just one winner left – the champion.  In any round, if the number of players is not a multiple of 4, then some players will be chosen randomly to advance automatically so that the number of people playing in that round is divisible by 4.  (This will ensure that each game that is played will have 4 players.)  How many total games will be played in the tournament?

{Scroll down for solutions}

--------------------------------------------------

*I love card games.  I also love activities that help people learn and/or increase their brain power.  SET is a card game that is simple to learn, fun for a wide range of players (recommended for players 6 years old and above), and helps build brain power.  As a teacher and card game enthusiast, I give SET my full recommendation.  (And now SET-Junior is available for younger players.)

--------------------------------------------------

Solutions/Explanations

Method 1 (slow)

To solve the problem, many people get out pencil and paper and begin calculating the results, round by round.  For each round, they keep track of the number of games, and then add these up at the end to find the answer to the question.  This method can be quite time consuming.  One such solution is shown:

Method 2 (fast)

Of course, there is another way to solve this problem – a way that takes much less time.  In the previous solution, the focus was on the WINNERS from each round.  (The number of players in rounds 2 – 5 was determined by figuring out how many winners there were in the previous round and adding this amount to the number of people who were lucky enough to advance automatically.)  A much quicker solution involves focusing on LOSERS:

We know each game that is played will have 4 players.  This means that each game is guaranteed to have 3 losers.  The tournament begins with 340 players, and 339 of those players will be a loser at some point.  All we need to do is divide 339 by 3 to find our answer: There will be 113 games.

When looking at a tough problem, we often hurt ourselves by focusing on the wrong part of the problem.  Sometimes it is helpful to take a step back, reexamine the problem, and figure out if our focus is in the right place.


For math tutorials and silly math songs, visit www.onlinemathpro.com




Affiliate Disclosure