Does anyone want to flex the old grey matter?

[flobadob]

This is a puzzle where you have to try and swap the male and female frogs to the opposite rocks. Took me a while to get it. The best I could do it in was 15 moves. http://www.sonnyradio.com/leapfrog.htm

[boxtops]

I've seen that before, but I couldn't do it at all. well done you!

here's something else to stretch the grey matter, and donate to charity at the same time without spending money.. http://www.freerice.com/index.php

[ttfn]

Sorry flob I thought it was going to be about me for a minute!

[mollykins]

I did it in fifteen aswell, you have to get them alternating witha a gap at one end that you can move to the middle and you end up with them swapped.

[cazzz1975]

just did it!

[linedancer3]

Perhaps its because I should be in bed by now but please what is the solution -detailed!- to the frogs puzzle?!!!

[Buenchico]

That puzzle (and the extensions of it, by introducing 4 frogs to each side, then 5, then 6, etc) was nearly always the first thing I did with a new class of 12-year-olds learning maths. There are some great number patterns to be studied by examining it.

As you've suggested, Flobadob, it should take 15 moves. (I just did it in 23 seconds but I've probably got an unfair advantage!)

[flobadob]

Chris, is there a mathematical equation which would show that with 6 frogs and 7 rocks it should take 15? Likewise would the same equation apply to as you suggested 8 frogs with 9 rocks etc?

[Buenchico]

To get to the mathematics of it, Flobadob, you need to break down 'moves' into 'slides' (where a frog just moves to an adjacent rock) and 'jumps' (where a frog hops over a frog of the opposite colour).

Doing it with 1 frog on each side (with 3 stones) takes 2 slides & 1 jump = moves
Doing it with 2 frogs on each side (with 5 stones) takes 4 slides & 4 jumps = 8 moves
Doing it with 3 frogs on each side (with 7 stones) takes 6 slides & 9 jumps = 15 moves
Doing it with 4 frogs on each side (with 9 stones) takes 8 slides & 16 jumps = 24 moves

So the number of slides is equal to double the number of frogs on one side, whereas the number of jumps is found by squaring the number of frogs on one side. So the total number of moves = 2n + n-squared (where n is the number of frogs on one side).

So with 100 frogs on each side (of 201 stones) it would take 200 + 10000 = 10200 moves.

[Buenchico]

Er, I seem to have missed a number out there, but I'm sure that you don't need me to add 2 + 1 for you anyway ;-)

[flobadob]

I wouldn't like to chance that last one. The 100 frogs that is, not the 2+1. I got an online calculator to solve that little problem x-)

[mike11111]

One solution is:

Imagine the stones numbered from 1 to 7 from left to right then move as follows:
1. 3 - 4
2. 5 - 3
3. 6 - 5
4. 4 - 6
5. 2 - 4
6. 1 - 2
7. 3 - 1
8. 5 - 3
9. 7 - 5
10. 6 - 7
11. 4 - 6
12. 2 - 4
13. 3 - 2
14. 5 - 3
15. 4 - 5

[linedancer3]

Thanks Mike11111 !! Just what I needed! I can sleep easy tonight now!