Aaron

Aaron Bediako Grade 9 Math First Draft

Mathematics Coursework

Frogs by Lucas

Aaron Nana KojoBoatengBediako
Type: 1st draft


For my Math coursework I have decided to choose Frogs as my topic to
help me get full marks. Frogs was created by a French mathematician
called Lucas.
Aim: The aim of the game is to swap the positions of the discs so that they end up
the other way round with a space in the middle.

Rules: The rules are simple-
1. A disc can slide one square in either direction onto an empty square.
2. A disc can hop over one adjacent disc of another color provided it can
land on an empty square.

Example 1:

1. Slide B one square to the left. A B

A B
2. A hops over B to the right

3. B slides one square to the left B A

That took three moves. B
A

Slides: 2

Hops: 1

Moves: 3


Example 2: The next game is with two frogs on either side:

1. Slide A one square to the right.
A A B B

2. Hop B over A to the left. A B A B

nd A B A B
3. Slide the 2 B one square to the left.

st nd A B
4. Hop the 1 A over the 2 B to the B A
right.

nd
5. Hop the 2 A to the right.
B
A B A

st
6. Slide the 1 B to the left. B B A
A

nd
7. Hop the 2 B over the A to the left. B B A A

nd B
8. Slide the 2 A one square to the right. B A A
That took eight moves.

Slides: 4

Hops: 4

Moves: 8


Example 3: This has three frogs on each side.

A A A B B B

Slide the 1stA one square to the right.

A A A B B B

Hop the 1st B frog to the left over the A.

A A B A B B

Slide the 2nd B frog to the left.

A A B A B B

Hop the A over the 2nd B to the right.

A A B B A B

Hop the 2nd A over the 1st B.

A B A B A B

Slide the 3rd A to the right.

A B A B A B


Hop the 1st B to the left over the 3rd A

B A A B A B

Hop the 2nd B to the left over the 2nd A

B A B A A B

Hop the 3rd B to the left over the 1st A.

B A B A B A

Slide the 1st A to the right.

B A B A B A

Hop the 2nd A over the B.

B A B B A A

Hop the 3rd A over the 2nd B.

B B A B A A

Slide the 2nd B to the left.

B B A B A A

Hop the 3rd B over the 3rd A.

B B B A A A

Slide the 3rd A to the left.


B B B A A A

That took:

6 slides

9 hops

This makes 15 moves.

I continued doing this and got a results table.

Results:
No. of discs No. of squares No. of slides No. of hops No of moves
made
2 3 2 1 3
4 5 4 4 8
6 7 6 9 15
8 9 8 16 24
10 11 10 25 35
12 13 12 36 48
14 15 14 49 63
16 17 16 64 80

From the results above I can garner that each result has a
pattern to it, when you start from the first results with two
discs. The number of squares is equal to the number of discs


plus one and the number of slides is equal to the number of
discs.
I found many other sub-formulas but this question is about
finding the formula for the least number of moves needed in a
particular game. I mixed around the numbers and pondered
over fake equations until I got it. I used my knowledge of
sequences to make a modified table and using the nth term I got
closer to my answer:
N 1 2 3 4 5 6 7 N
Discs 2 4 6 8 10 12 14
Moves 3 8 15 24 35 48 63

I noticed that the N numbers were half the number of the
number of discs. I also noticed that they were multiples of the
number of moves and could be multiplied to get them. I made a
new modified table with the numbers that the N numbers were
multiplied by at the bottom:
N 1 2 3 4 5 6 7 N
Discs 2 4 6 8 10 12 14
Moves 3 8 15 24 35 48 63
×3 ×4 ×5 ×6 ×7 ×8 ×9


I sensed a pattern in the arrangement of the numbers. I
realized that the N numbers were being multiplied by numbers
two numbers higher than them. I found the formula:
M=N×(N+2)
This translates into:
½ the number of discs × (1/2 the number of discs + 2)
The examples to prove this are:
With 2 discs three moves are made
M= N × (N+2)
M=1× (1+2)
=1×3
=3
The number of moves is 3.

With 10 discs the number of moves is 35:

M= N× (N+2)
M= 5× (5+2)
= 5×7
= 35
The number of moves is 35.

With 14 discs the number of moves is 63:

M= N × (N+2)


M= 7 × (7+2)
=7×9
=63
The result is 63.

My fourth and final example is with 8 discs which has 24 moves:

M= N × (N+2)
M= 4 × (4+2)
=4×6
= 24
The final number of moves is 24.

This equation works for all of the results that I have and all the
others that I tested it on.

I have also found ways to find the formulas for each of the
fields in the table found using the number of discs:

Moves= ½ discs × (1/2 discs + 2)
Slides= discs
Squares= discs + 1
Hops= (1/2 discs × (1/2 discs + 2)) - discs


I have found a second formula to find the least number of
moves needed for any game as I typed the formulas above and
I found it by moving more numbers around then using another
modified table that I made:

N 1 2 3 4 5 6 7 N
Discs 2 4 6 8 10 12 14
Moves 3 8 15 24 35 48 63

I used the same table to find the formula:

M = (N × N) + 2N
Which is the same as:
M = N2 + 2N
This is equal to
Moves = (½ discs × ½ discs) + discs

This formula has been proven to work on all the results in the
table and more. I have provided a few examples below:

The 1st example has 10 discs in the game with 35 moves:

M = (N × N) + 2N
M = (5 × 5) + 10
= 25 + 10
= 35
The formula is proven to work here.


My 2nd example is of a game with 2 discs with 3 moves:

M = (N × N) + 2N
M = (1 × 1) + 2
=1+2
=3
It is also proven here that the formula works this example.

My 3rd example is of a game with 6 discs with 15 moves:

M = (N × N) + 2N
M = (3 × 3) + 6
=9+6
= 15
The formula I derived from the table works here also.

My 4th example is of a game with 14 discs which uses 63 moves:

M = (N × N) + 2N
M = (7 × 7) + 14
= 49 + 14
= 63
My formula is also proven to work here.


My 5th and final example is of a game with 8 discs which uses 24
moves:

M = (N × N) + 2N
M = (4 × 4) + 8
= 16 + 8
= 24
The formula I found is proven to work on all my examples and I
can assure you that it works on any game you would try it on.

With the new formula I found I can make a new list of my
results to find the least number of moves needed for any game
by using the number of discs:

• Moves= ½ discs × (1/2 discs + 2)
= (½ discs × ½ discs) + discs
• Slides= discs
• Squares= discs + 1
• Hops= (1/2 discs × (1/2 discs + 2)) - discs

I can conclude the first question by saying that I have found two
formulas to find the least number of moves needed in any
game (above) and that they both work excellently in any
situation. There might be more but I have not found them yet.


Extension
The question for the extension is to find the formula for the
least number of moves needed for x discs of one color and y
discs of another. Such as having 3 blues and 2 reds.

I tried a game or two then played more, up to a game with 9
discs on one side and 8 discs on the other. I have an example
below.

Example 1:

A A B

A A B

Hop the B over the 1st A to the right.
A B A

A B A

Hop the 2nd A over the B.
B A A


Slide the B to the left.
B A A

That took:
5 moves
2 hops
3 slides

Example 2:

A A A B B

A A A B B

Hop the 1st B over the A to the left.
A A B A B

A A B A B

Hop the 1st A to the right.
A A B B A

Hop the 2nd A to the right.
A B A B A


Slide the 3rd A to the right.
A B A B A

Hop the 1st B to the left.
B A A B A

Hop the 2nd B to the left.
B A B A A

Slide the 2nd A to the right.
B A B A A

Hop the 3rd A to the right.
B B A A A

B B A A A

That took:
5 slides
6 hops
11 moves


After playing more of these games with the reds being more
than the blues I could finally make a table with the reds in it as
X and the blues as Y.
The named table is below:

Discs X Y Squares Slides Hops Moves

3 2 1 4 3 2 5
5 3 2 6 5 6 11
7 4 3 8 7 12 19
9 5 4 10 9 20 29
11 6 5 12 11 30 41
13 7 6 14 13 42 55
15 8 7 16 15 56 71
17 9 8 18 17 72 89

This table shows the number of discs in the game which is
followed by the number of X discs in the game and the number
of Y discs in the game. The rest of the table is similar to the
table for the first question.


I noticed that there were a lot if sequences in the table and so
mixed a few numbers around and thought about it. I found a
formula to find the least number of moves needed by focusing
my mathematical abilities solely on the number of discs X and Y
and trying variant forms of math to check.

I eventually found that the way to find it was to add them
together then multiply them together and found the formula:

M = (X + Y) + (X × Y)

I found the formula by randomly placing numbers together. I
couldn’t find a way with normal methods so I focused on the
number of discs.

I checked this formula on the table and it worked.
A few examples are shown below:

A game with 3 reds (X) on one side and 2 blues (Y) on the other
side has 11 moves:

M = (X + Y) + (X × Y)
M = (3 + 2) + (3 × 2)
=5+6
= 11
The formula is proven to work here on this game.


My 2nd example is of a game with 5 reds on one side and 4
blues on the other which is a game which is supposed to use 29
moves:

M = (X + Y) + (X × Y)
M = (5 + 4) + (5 × 4)
= 9 + 20
= 29
The formula is also proven to work on this game.

My 3rd example is of a game with 9 reds on one side and 8 blues
on the other which is a game which is supposed to use 89
moves:

M = (X + Y) + (X × Y)
M = (9 + 8) + (9 × 8)
= 17 + 72
= 89
My formula is also proven to work on the third example.

My 4th example is of a game with 8 reds on one side and 7 blues
on the other which is supposed to have a total of 71 moves:

M = (X + Y) + (X × Y)
M = (8 + 7) + (8 × 7)
= 15 + 56
=71


The formula is proven to work on this example also.

My final example is of a game with 6 reds on one side and 5
blues on the other which is supposed to have a total of 41
moves:

M = (X + Y) + (X × Y)
M = (6 + 5) + (5 × 6)
= 11 + 30
= 41
My formula is proven to work on my final example.

With this formula I can conclude by saying that I have found a
formula for finding both the least number of moves needed for
a game with an equal number of discs on each side and the
formula for a game where the number of X discs is one more
than the Y discs.

I conclude my math coursework by saying that the formulas for
the least number of moves needed in a game with:

1. An equal number of discs: Moves= ½ discs × (1/2 discs + 2)
= (½ discs × ½ discs) + discs
2. And with X being 1 more than Y: Moves = (X + Y) + (X × Y)

Aaron

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