I can never remember how to do this. I always have to work it out every time I see it.
You have six frogs trying to cross the stream. On the first three stones, there are three frogs (A)lex, (B)obby, and (C)arl. On the other side of the stream are three more frogs, on three more stones. Their names are (D)eb, (E)nid, and (F)rancine.
So, when they start, they look like this:
A B C _ D E F
There's an empty stone in the middle. A, B and C want to get to the side D, E and F are on and vice-versa. But none of the frogs want to get wet today. Each frog can leap-frog over another frog (if they are of the opposite sex) or can jump to an empty stone that's next to them. How do you get them to switch sides?
Did I explain that well enough?
D to _
ReplyDeleteC ju D
D to _
B ju D
D to _
B to _
E ju C
F to _
C ju F
B ju E
A ju D
D to _
E ju A
F ju B
B to _
A ju F
F to _
And there you go!
Brian, your answer and mine seem to line up. Thanks for answering!
ReplyDeleteC steps, D hops over her. E steps, C hops, B hops, A steps, D hops, E hops, F hops, C steps, B hops, A hops, E steps, F hops, and A steps. That's 15 moves altogether.
I can not seem to understand it
ReplyDeletei have tried all what you said but they do not seem to add up
what could i have possibly done wrong?
I'm not sure what could be wrong. Keep in mind that A, B and C are always moving to the right, while D, E, and F are always moving left.
ReplyDeleteAlso, if they hop, then they are moving two spaces (jumping over the occupied space) and if they step, then they only move one spot.
Mike's answer is right!
ReplyDeleteBrian C -- step 6 (B to __) is impossible, as it tells to move B backwards!
ABC_DEF
ReplyDeleteAB_CDEF
ABDC_EF
ABDCE_F
ABD_ECF
A_DBECF
_ADBECF
DA_BECF
DAEB_CF
DAEBFC_
DAEBF_C
DAE_FBC
D_EAFBC
DE_AFBC
DEFA_BC
DEF_ABC
It's fine, but can u try to write down the algorithmic pseudo code for this.
DeleteThank you!I've an homework to do this kind of algorithm!
i know the answer, now whats the formula?
ReplyDeletei know hot to do the puzzle, but what is the formula is algerbra terms?
ReplyDeleteIT IS EASY
ReplyDeleteBUT WOT IS THE FORMULA!?
easy.
ReplyDeletenumber of frogs on the right * number of frogs on the left, + number of frogs on the right, + number of frogs on the left. this gives you the minimum number of moves for the given situation.
Can you give a general statement which explains how to work out the number of jumps if you know the number of frogs.
ReplyDeletewhy does the minimum number of moves always seem to work out as one less than the square number sequence ie
ReplyDelete2 frogs->8 moves(9-1) , 3 frogs ->15 moves (16-1) 4 frogs ->24 moves (25-1)
Just helped my daughter with this. I think the method is all around the pattern of the moves. The moves can either be slides (S) where the frog moves into the adjacent space or jumps (J) where the frog jumps over its neighbour into the space. If we think of Green (G) frogs on the left and Brown (B) frogs on the right then the pattern is simply an increasing then decreasing number of jumps, separated by slides.
ReplyDeleteSo for four frogs the pattern is
GS
BJ
BS
GJ
GJ
GS
BJ
BJ
BJ
BS
GJ
GJ
GJ
GJ
BS
BJ
BJ
BJ
GS
GJ
GJ
BS
BJ
GS
A symmetrical pattern or palindrome