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I am investigating what the rule would be if I had black frogs and white frogs each on the opposite sides of a grid. I want to find out how many moves it would take me to get them to swap sides.

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Introduction

Janine SmithInvestigating Frogs5/1/2007

I am investigating what the rule would be if I had black frogs and white frogs each on the opposite sides of a grid. I want to find out how many moves it would take me to get them to swap sides without having to go through the whole process and just find a rule.

These are the rules I have to follow:

  1. The black frogs can only move to the left and the white can only move to the right.
  2. A frog can slide from one square in to the one next to it, providing the squares are empty.
  3. A frog can Jump over exactly one of the other colour’s to land in an empty space.

Plan

  • Experiment which sequence will allow the black and white frogs to swap positions in the least amount of moves possible.
  • I will then do the same for different amounts of frogs.
  • I will record this data into a table.
  • I will try to find a pattern within my results in order to find a rule for being able to find the number of moves for any number of frogs.
  • I will then test this rule to prove it works with any number of frogs.
  • I will plot this data onto a graph.
  • I will analyze the graph and explain what the graph shows.
  • I will try to further my rule to find the number of frogs used in a certain amount of total moves.
  • I will prove that the rule works if I find one.
  • I will plot the data onto a graph and explain what it shows.
  • I will evaluate my investigation.
  • First I shall Experiment which sequence has the least amount of moves in order for the two colour’s to change position.

To help me with this I used counters in place of the frogs and drew a grid.

I had to get the three black counters into the position the white counters are currently in on the diagram and vice versa for the white counters.

Here are the moves I used in which to achieve this:

  1. Slide Black                        6. Slide Black                11. Jump Black
  2. Jump white                        7. Jump white                12. Jump Black
  3. Slide white                        8. Jump white                13. Slide white
  4. Jump Black                        9. Jump white                14. Jump white
  5. Jump Black                        10. Slide Black                15. Slide Black

This was for three black frogs and three white frogs.

Then I carried on doing this for

4 black and 4 white

5 black and 5 white

6 black and 6 white

                                                        etc…

I looked at the relationship between the black and white moves and found a formula from that which would have been adequate for use.

This was the formula:

(n/2)²+n

I then looked over the results I had and saw that there was symmetry between the jumps and slides so I found a new formula which I found easier to use and I could prove that it worked using a graph too.

So, whilst I did this I noticed that the patterns of slides and jumps were symmetrical for example:

S  J  S  J  J  S  J J J  S  J  J  S  J  S

This will now help me to predict the other results accurately.

  • Next I will draw a table to present my results.
...read more.

Middle

1

1

2

3

5

2

4

4

8

7

3

9

6

15

9

4

16

8

24

11

5

25

10

35

13

6

36

12

48

15

7

49

14

63

17

From this table of results you can see that the number of jumps goes up in squared numbers. The number of slides goes up in 2’s.

  • From the results in my table I will try to find a formula that will enable me to find the total number of moves for any number of frogs on a side.

I looked through my results and saw that there could be a relationship between the jumps and slides for each one. I saw that the jumps went up in squared numbers so I used for the first part of my formula. In the slide column I saw that the data went up in 2’s so I made this 2n and used this for the second part of my formula by adding them together

So my formulae is:                n²+2n

  • Now I have to prove that this formula works.

For 1 frog on each side

SJ S

1²+(2x1) = 3

As you can see referring back to the table of results that the answer is correct and the sequence is symmetrical.

I will do this again for another number of frogs

For 4 frogs on each side:

S  J  S  J  J  S  J  J  J  S  J  JJ  J  S  J  J  J  S  J  J  S  J  S

4²+(2X4) = 24

...read more.

Conclusion

I tried lots of different rules but I found it very difficult to find a relationship between them.

The best I came up with was 2n+2 which would find you the number of slides when you know the number of white frogs on a side.

To try and find another way to do this I added in to the table the number of black moves and the number of white moves.

Number of white frogs

Number of black frogs

Number of slides

Number of Jumps

Number of black move

Number of white moves

Total number of moves

1

2

3

2

3

2

5

2

3

5

6

6

5

11

3

4

7

12

10

9

19

4

5

9

20

15

14

29

The two new columns are in red.

I noticed straight away that there was always one more black move than the white, the difference between both is 3 then 4 then 5, etc.

Evaluation

In this investigation I think that I did reasonable well with what I had managed to find. I saw the connection between the Black moves and white moves first and came up with a different formula, but then I re-looked over the results and saw that there was symmetry between the jumps and slides. I thought from this I could find an easier way to prove that the formula did actually work. I was then able to show on a graph that formula worked as the line crossed the x axis at exactly the right point. After this I tried to find the formulas for different numbers of frogs on each side (2 white, 3 black, etc…).

...read more.

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