Classroom narrative: Frogs

submitted by: Nathalie Sinclair
on Tue Jun 10 16:14:59 2003

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Course: Math 7
Topic: Pre-algebra
Resource type: Tool
Catalogue entry:
http://mathforum.org/mathtools/tool.html?id=955
Resource location:
http://hydra.educ.queensu.ca/cgi-bin/Maths/maths.cgi?do=activity&activity=jam
Story:
 

I ask the students to read the directions for Frogs and to start playing the game. Almost immediately, different students announce that they have succeeded in 23 moves, in 21 moves, in 19 moves, and finally in 17 moves. Many think that's the best one can do and take quite a bit more time trying to figure out how to succeed in less moves. One student notices after awhile that it's a "bad thing" to move a frog so that it's behind one of the same colour. Another student suggests that moving frogs from the opposite teams in alternation should be avoided. Finally, after about 10 minutes of playing with Frogs, one students hits the magic number 15 moves.

As news makes it around the class, other students try to replicate the achievement, paying special attention to the two suggestions mentioned above. When another student succeeds, he describes the trick to succeeding as "always slide one more over than you think you should." Since the sequences of moves also generates a sequence of colours, I ask the students what they notice about those colours. They tell me that the sequence seems symmetric, if you can get the minimum number of moves. They also show that you can actually get two opposite sequences, depending on whether you start with the red team or the blue team. I ask them to predict what the sequence of colours would look like if there were four frogs on each team and they quickly deduce the pattern and propose that since three frogs per team gives 1blue, then 2red, then 3blue, 3red, 3blue, then back to 2red, then 1blue, then four frogs per team would give 1blue, 2red ,3blue ,4red, 4blue, 4red, 3blue, 2red,1blue. In other words the length of the 3 middle blocks of colour equals the number of frogs on each side and the number of colours in a block decreases symmetrically on each side. Many of the students continue playing the game, some just to make sure they have "got it" and other taking pleasure in the rhythmic solution they have discovered.

At this point, the students are ready to write their notes of their discoveries. Then we move onto the next investigation: Different numbers of players on each side. Here the students take off on their own without my having to give directions. And they are soon shouting out minimum moves for different combinations: "4 and 4 takes 24," "6 and 1 takes 13," "2 and 5 takes 17," etc. After we have completed all the possible combinations (excluding of course the symmetries) I ask the students how the number of frogs on each side is related to the minimum number of moves. The students look a little blankly at the blackboard and keep trying different combinations on their computers. So I ask them to focus on the even-strength teams. We circle the 1-1, 2-2, 3-3, 4-4, 5-5, 6-6 entries and I challenge them to tell what the minimum number of moves would be if there were 7 frogs on each side. Soon a student notices the following relationship:

Lulu's team Other team Moves
1 1 3 = 1 * 3
2 2 8 = 2 * 4
3 3 15 = 3 * 5
4 4 24 = 4 * 6
5 5 35 = 5 * 7
6 6 48 = 6 * 8

When this is proposed, and we write it on the blackboard, another student notices that the second number is always 2 greater than the first. In order to encourage algebraic expression, I asked the students how they could express this more generally. They propose that if there are L frogs on each side, then the minimum number of moves will be L*(L+2).

But then another student shows that the formula doesn't work for the unevenly matched teams. However, it doesn't take long for a student, who's been concentrating on the numbers for awhile, to notice a pattern, as exemplified with a few examples in this table:

Lulu's team Other team Moves
1 6 13 = 1 * 6 + 1 + 6
4 3 19 = 4 * 3 + 4 + 3
2 1 5 = 2 * 1 + 2 + 1
5 6 41 = 5 * 6 + 5 + 6
5 5 35 = 5 * 5 + 5 + 5

Again, we develop an algebraic expression for this relationship: If there are L frogs on one side and O on the other, then the minimum number of moves will be L*O+L+O. We check this relationship out for many of the entries and are satisfied that it holds. Instead of asking the students just why it holds at this point, I ask them to tell me what they notice about the patterns in the colours and letters of the following investigation: Patterns of colours, slides, and jumps, hoping that this might help them gain insight into the relationship.

The students are quick to notice many things that they are eager to note in their albums. Some of the patterns are: the sequence of colour always starts and ends with the same colour. If there is more frogs on one side than on the other the middle block of colours will be the same length as the number of frogs on the smaller team. For each frog extra that one team has, there will be a repetition of the middle string of j's and S's. The string of j's and S's is also symmetric. At this point, the bell is about to ring and I ask them to finish writing down their discoveries. We will investigate the unanswered questions next class!

Send comments to Nathalie Sinclair at nathsinc@educ.queensu.ca

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Tue Jun 10 16:14:59 2003