Harder Fibonacci Puzzles

The Fibonacci puzzles are split into two sections: those with fairly straight-forward and simple explanations as to why the answer is the Fibonacci numbers are on the Easier Fibonacci Puzzles page.

CONTENTS of this Page

• More coin puzzles
  • Pennies for your thoughts - Part 1
  • Pennies for your thoughts - Part 2
• Games and strategies
  • Water Treatment Plants Puzzle
  • Wythoff's game
• Permutation puzzles
  • Non-neighbour Groups
  • Another one!
  • That's Odd!
  • Steven and Todd
• Electrical resistance puzzles
  • A ladder of resistors
    • Series and Parallel
    • Rungs
    • Rungs and One Side
    • One Side only
    • Have you the Capacity for another puzzle?
• Jigsaw and Geometry puzzles
  • A Fibonacci Jigsaw puzzle or How to Prove 64=65!
  • The same jigsaw puzzle proves 64=63!!
  • Yet another Fibonacci Jigsaw Puzzle
• More Links and References

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Pennies for your thoughts - Part 1

The puzzle here is that only one of these two puzzles involves the Fibonacci number series! The other puzzle does not but just begins with a few of the Fibonacci numbers and then becomes something different. One of these puzzles is a fraud, a Fibonacci forgery. So which is the real Fibonacci puzzle?

1. each penny must touch the next in its row
2. each penny except ones on the bottom row touches two pennies on the row below.

The first condition means that there are no gaps in any row and the second means that upper rows are smaller than lower ones.
The following arrangements are not proper combinations for 6 pennies
because the first has a gap in one row and the second has a penny which is not on the bottom row and is not touching two beneath it.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Pennies for your thoughts - Part 2

         i : 0 1 2 3 4 5  6  7  8
     Fib(i): 1 1 2 3 5 8 13 21 34...
       P(n): 1   2   5   13     ?
         n : 1   2   3    4     5

Which one of these two Pennies puzzles is the forgery (it does not continue with a pattern of Fibonacci numbers after some point) and which one genuinely always has Fibonacci numbers of arrangements?

References

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Water Treatment Plants puzzle

• either process any water it may receive from one neighbouring city, together with its own dirty water, discharging the cleaned-up water into the river;
• or send its own dirty water, plus any from its downstream neighbour, along to the upstream neighbouring city's treatment plant (provided that city is not already using the pipe to send it's dirty water downstream);
• or send its own dirty water, plus any from the upstream neighbour, to the downstream neighbouring city's plant, if the pipe is not being used.

CITY A		CITY B
~~RIVER~~

• every city must have its water treated somewhere and
• at least one city must discharge the cleaned water into the river.

Let's represent a city discharging water into the river as "V" (a downwards flow), passing water onto its neighbours as ">" (to the next city on its right) or else "<" (to the left). When we have several cities along the river bank, we assign a symbol to each (V, < or >) and list the cities symbols in order.
For example, two cities, A and B, can

• each treat their own sewage and each discharges clean water into the river. So A's action is denoted V as is B's and we write "VV" ;
• or else city A can send its sewage along the pipe (to the right) to B for treatment and discharge, denoted ">V" ;
• or else city B can send its sewage to (the left to) A, which treats it with its own dirty water and discharges (V) the cleaned water into the river. So A discharges (V) and B passes water to the left (<), and we denote this situation as "V<".

So we have just 3 possible set-ups that fit the conditions:-

Pipe Flow A B RIVER ~~~~~~ Notation VV

A B ~~~~~~ >V

A B ~~~~~~ V<

CITY A		CITY B		CITY C		...
~ ~ ~ ~ ~ ~ R I V E R ~ ~ ~ ~ ~

1. What are the eight set-ups possible for 3 cities?
2. If S(n) is the number of set-ups for n cities, then S(1)=1 and we have just shown that S(2)=3 and S(3)=8. But this does not look like the Fibonacci numbers!
   What is S(4)? What is S(5)?
3. What is the relationship between the S-numbers here and the Fibonacci series!?

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Wythoff's game

Players take it in turns to remove some matches (at least one) from EITHER only one pile OR ELSE an equal number from both piles. The players can decide how large each heap will be before the game starts and the winner is the one who takes the LAST match. A complete heap can be removed as your move if you like. This is not to be recommended however, since your opponent can do the same on the next move and so will win by taking the last match! This leads to the idea of "safe configurations", that is, ones from which it is possible to force a win, no matter what your opponent does.

For further details, see
T. H. O'Beirne Puzzles and Paradoxes, Dover press, 1965, chapter 8.
Ball, W.W.R. and Coxeter, H.S.M. Mathematical Recreations and Essays, 13th edition, Dover Publications, 1987. A great classic with plenty to keep you amused and enthused on Maths - definitely worth buying! (You can order it online via the title-link.)

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Non-neighbour Groups

For instance, if there are 3 people in the class, let's label them according to their position when in the alphabetical order, so they are 1, 2 and 3.

So if 1 is in the group, then 2 cannot be and 3 may be or not; so we have the groups:

{1} and {1,3}

If 2 is in the group then, since both 1 and 3 are 2's alphabetical neighbours, then that group will consist of 2 alone!

{2}

If 3 is in the group then 2 cannot be and 1 may be. But remember that the group with 3 and 1 in it has already been included above! So we have the following possible new groups with 3 in:

{3}

The Fibonacci Split D L Silverman, J Rec Maths, 9 (1976/7), page 298 is the origin of the mathematics behind this puzzle.
On The Number of Fibonacci Partitions of a Set Helmut Prodinger Fibonacci Quarterly Vol 19, 1981, pages 463 - 465 generalises this problem.

Another one!

length 1
length 2
length 3
length 4
length 5
...	...

• There are now 2 "lines" of colours which are of length 1:
     
• and 5 ways to make a line of length 2:
                 
  
• Can you find all 13 of length 3?
• How many are there of length 4?
• and how many of length 5?
• Can you guess how many there will be of length N? Rest your mouse here for a hint
• Can you find a general method of generating all solutions of length N using the solutions of smaller length?

That's odd!

• in No One! there were no rods of length 1
• in Ones and Twos the only rods available were of lengths 1 or 2
• the Another one!, the previous puzzle on this page, we had two different colours of rod of length 1

• there are 3 ways to make a line of length 4:
  4 is 1111, 13 and 31
  The last two, 13 and 31 are different since the order of the colours is different.
• Find the 5 ways to make a line of length 5 using only the odd-length rods
• How many ways are there using odd rods to make a line of length 6?
• and of length 7?
• Guess the number of solutions for a line of length N
• Find a method of generating the solutions of length N if you have already found all solutions of smaller lengths

Steven and Todd

1. 1
2. 1 2
3. 3
4. 1 2 3

Now try the same puzzle but with Steven always starting:

Finally put all the results together:

Two challenging problems in Combinatorics section 17 of Mathematical Chestnuts from Around The World, R Honsberger, Mathematical Association of America (2001) where this problem is called Terquem's Problem after Olry Terquem who solved it in 1839.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

A ladder of resistors

Series and Parallel

You'll remember that if we have 2 resistances R and S in parallel: then the combined resistance, T, is given by

                  1   1   1
                  - = - + -
                  T   R   S 
   

Rungs

...

    1/R   = 1/1   + 1/1 = 2         so    R=1/2
    

For the 3 resistor ladder, we have combined the 2 resistor ladder with another resistor of 1-ohm, in parallel, so the combined resistance S here is

     1/S = 1/(1/2) + 1/1 = 2+1 = 3   so   S=1/3
   

What pattern are you getting for the combined resistance?

Can you prove that your pattern always holds?

Rungs and One Side

 1/S = 1/1 + 1/2 = 3/2   i.e. S=2/3

What about the next two ladders? What is the general pattern now?

Again, can you prove that your pattern will always hold?

One Side only

Have you the Capacity for another puzzle?

References on the Resistance Ladders

The Golden Ratio in an Electrical Network, J Wlodarski in The Fibonacci Quarterly Vol 9 (1971) pages 188 and pg 194.

Generalization of Modified Morgan-Voyce Polynomials, Fibonacci Quarterly Vol 38 No 1, 2000, pages 8-16.

An advanced mathematical article dealing with resistors, capacitors and inductors.

The appearance of Fibonacci Numbers and the Q Matrix in electrical network theory S L Basin Mathematics Magazine (1963) pages 84-97

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

A Fibonacci Jigsaw puzzle
or How to Prove 64=65

This puzzle can be repeated with other consecutive Fibonacci numbers,

replace 5, 8 and 13  by  8, 13 and 21  or by 3, 5 and 8

Not convinced?

Try this demonstration

Try cutting out the pieces as shown and rearranging them yourself if you are not sure the puzzle "works".
It works even better as a class demonstration using an overhead projector: photocopy the square with its grid lines onto an overhead projector transparency, cut out the shapes and show them as a square on the screen, then rearrange it into the rectangle, carefully lining up the grid lines to "show I'm not cheating"!

or look at this diagram of the pieces using a 3-by-8 rectangle and a 5-by-5 square where the 5-by-5 square is 1 unit larger than the 3-by-8 rectangle.

Some References on this puzzle:
Edward Wakeling in Rediscovered Lewis Carroll Puzzles Dover, 1995, says that this puzzle was found in Lewis Carroll's papers (the original is now kept at Princeton University) and that this puzzle goes back to Schlomilch, 1868.
Martin Gardner's Mathematics, Magic and Mystery a 1956 Dover book, is a book with magic tricks and how the mathematics behind them makes them work. It has two chapters on such Geometrical Vanishes and has a full explanation of this and other puzzles. He also traces its origins back to Sam Loyd (senior) who presented it to the American Chess congress (using an 8-by-8 chessboard) in 1858, ten years before Wakeling's reference to Schlomilch in the reference above. However this also appears not to be the earliest reference...
David Wells in The Penguin Book of Curious and Interesting Puzzles (Penguin, 1997) in Puzzle 143 traces its origin back to William Hooper in Rational Recreations of 1774.

How to Prove 64=63!!

Here is another arrangement. This time the blue puzzle's pieces have been re-arranged as shown here in green and now it loses a square -- there are two 5-by-6 rectangles + 3 squares in a row joining them, making a total area of 63!

So what's happened this time???
Fibonacci Sequences and a Geometrical Paradox A F Horadam Mathematics Magazine 35 (1962) pages 1-11, gives an extensive analysis of the problem and shows that with the difference of 1 unit of area, the puzzle extends to each triple of Fibonacci numbers.
Ball, W.W.R. and Coxeter, H.S.M. Mathematical Recreations and Essays, 13th edition, Dover Publications, 1987 has only the first version of the puzzle (page 85) and says the earliest reference the authors can find is Zeitschrift für Mathematik und Physik Leipzig, 1868, vol XIII, page 162.
On Fibonacci Sequences and a Geometrical Paradox Santosh Kumar Mathematics Magazine 37 (1964), pages 221-223 analyses both versions.
The second version comes from Henry E Dudeney's 536 Puzzles and Curious Problems (reprinted 1983); Problems 352 and 353 and their answers
Martin Gardner's Mathematics, Magic and Mystery a 1956 Dover book (mentioned in the first version of this puzzle) says that Sam Loyd junior (who adopted his father's name and continued his father's puzzle columns) was the first to discover this new reduced-square version. This book has a good explanation of how the two puzzles work and that the Fibonacci numbers produce other sizes of puzzle with identical variations of an additional and missing single square. He shows how other generalized Fibonacci sequences (i.e. starting with two other numbers rather than 0 and 1) can be used to devise variations where any number of squares can be made to appear and disappear, together with many other kinds of geometrical dissection puzzles. If you like the puzzles on these two Web pages, you'll enjoy this book too with number, handkerchief and card puzzles based on mathematics.

Yet another Fibonacci Jigsaw Puzzle!

It is made up of 4 pieces,

• a smaller green triangle with height 2 and base 5 ;
• a larger red triangle with height 3 and base 8;
• an orange L-shape of height 2 and width 5;
• a blue L-shape of the same width and height but a different shape.

The puzzle will work with a green triangle height 1 base 3 and a red triangle height 2 base 5, and two straight pieces (1-by-3) that make up a 2-by-3 rectangle. Rearranging them this time makes the small rectangle 1 square smaller this time so the two straight pieces have to overlap.
Similarly, using triangles of height 3 base 8 and height 5 base 13 the rectangle again loses one square.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

More Links and References

The Fibonacci Numbers and Golden Section in Nature

Fibonacci Home Page Previous pages on this Topic: The Easier Fibonacci Puzzles Where to now? This is the last page of Fibonacci Puzzles.

The next Topic is... The Mathematical World of Fibonacci and Phi

© 1996-2021 Dr Ron Knott