# 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:

- The black frogs can only move to the left and the white can only move to the right.
- A frog can slide from one square in to the one next to it, providing the squares are empty.
- 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:

- Slide Black 6. Slide Black 11. Jump Black
- Jump white 7. Jump white 12. Jump Black
- Slide white 8. Jump white 13. Slide white
- Jump Black 9. Jump white 14. Jump white
- 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.

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 n² 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

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…).

This student written piece of work is one of many that can be found in our GCSE T-Total section.

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