[bonanova]

Four young couples go for a stroll and come upon a river.

They find a boat, able to carry two persons at a time.

The girls, of course can row.

But there is a problem: the girls are extremely jealous.

So there is an agreement:

No girl may stay be in the company of another's boy friend unless her own b/f is present as well.

How do they all get across? They can't.

But one of the boys spots a small island in the river,

which can be used for intermediate loading and unloading.

Now is it possible?

[Guest]

yes it is possible

it takes 5 trips without need of the island in the middle. Lets label the couples 1A,1B,2A,2B and lets say that A = girl and B = boy. Starting point is X, and other side is Y.

#1: 1b takes 1a from x to y

#2: 1b goes back to x alone

#3: 1b and 2b go from x to y

#4: 2b goes back to x alone

#5: 2b and 2a go from x to y

neither girl is stuck with other girls boyfriend and all are on the other side. Middle island was not needed.

Edited August 9, 2008 by triciapain

[bonanova]

yes it is possible

it takes 5 trips without need of the island in the middle. Lets label the couples 1A,1B,2A,2B and lets say that A = girl and B = boy. Starting point is X, and other side is Y.

#1: 1b takes 1a from x to y

#2: 1b goes back to x alone

#3: 1b and 2b go from x to y

#4: 2b goes back to x alone

#5: 2b and 2a go from x to y

neither girl is stuck with other girls boyfriend and all are on the other side. Middle island was not needed.

Four couples, not two. Keep trying.

[Guest]

If A,B = couple 1 and so forth where the fist letter is girl and second is boy.

A,B go to island.

A stays, B rows back.

B gets D and rows past island island.

D stays across the river.

B then rows to island, stays and A rows back to pick up C.

A and C row across the river.

and voila! C and D are together

repeat this until all 4 couples are across.

This of course is impossible though because girls can't row.

[Guest]

This of course is impossible though because girls can't row.

careful marv - it says they can in the OP and you have just trod on the eggshells in atopic set by a moderator.

I can back it up that they can - all the scandinavian au pairs I knew cld row very very very well - makes you wonder how the brits do well in the olympiic rowing - go go go UK/NL - actually one of them wass called Brita - from Norway (or did she say No way!)

[unreality]

careful marv - it says they can in the OP and you have just trod on the eggshells in atopic set by a moderator.

I can back it up that they can - all the scandinavian au pairs I knew cld row very very very well - makes you wonder how the brits do well in the olympiic rowing - go go go UK/NL - actually one of them wass called Brita - from Norway (or did she say No way!)

marv was making a joke

[Guest]

marv was making a joke

probably, but no smiley?

I was just reminissing of a time when i was kissing, and rowed with friends but now i row with the wife!

[bonanova]

If A,B = couple 1 and so forth where the fist letter is girl and second is boy.

A,B go to island.

A stays, B rows back.

B gets D and rows past island island.

D stays across the river.

B then rows to island, stays and A rows back to pick up C.

A and C row across the river.

and voila! C and D are together

repeat this until all 4 couples are across.

Nope.

When B gets D and rows past the island,

C is left with boys F and H but without her boyfriend D.

[Guest]

Not sure if there is any other easier solution. But this will work.

Let the couples be denoted as, b1 g1, b2 g2, b3 g3, b4 g4. Let their initial position be at s1 and they are going to s2. Let the island be denoted as ID.

1) g1 g2 goes to ID

2) g1 remains in ID and g2 comes back to s1.

3) g2 g3 goes to ID

4) g2 remains in ID and g3 comes back to s1

5) g3 g4 goes to s2

6) g3 remains in s2 and g4 comes back to s1

7) b2 b3 goes to s2

8) b3 remains with g3 in s2 and b2 comes back to s1

9) b1 b2 goes to s2

10) b2 remains with b3 g3 in s2 and b1 comes back to s1

11) b4 g4 goes to s2

12) b4 g4 remains with b3 g3 in s2 and b2 comes back to s1

13) b1 b2 goes to ID

14) b2 remains with g2 in ID and b1 g1 comes back to s1

15) g1 remains in s1 and b1 goes to ID

16) b1 b2 goes to s2

17) b1 remains in s2 and b2 comes back to ID

18) b2 g2 goes to s2

19) b1 goes to s1

20) b1 g1 goes to s2.

And they are all happy.

Edited August 10, 2008 by Drydung

[Guest]

ok so 1 and A are a couple,

2 and B

3 and C

4 and D

numbers are boys.

1 and A go to the other side, drops off A

1 goes and picks up 2, other side drops off 1

2 goes and gets B, other side drops off 2

B goes and gets C, other side drops off B

C goes and gets 3, other side drops off C

3 goes and gets 4, other side drops off 3

4 goes and gets D, other side 4 and D get off

All over

[bonanova]

ok so 1 and A are a couple,

2 and B

3 and C

4 and D

numbers are boys.

1 and A go to the other side, drops off A

1 goes and picks up 2, other side drops off 1

2 goes and gets B, other side drops off 2

B goes and gets C, other side drops off B

C goes and gets 3, other side drops off C

3 goes and gets 4, other side drops off 3

4 goes and gets D, other side 4 and D get off

All over

When 3 drops off C and returns, C is left with 1 and 2.

This violates the agreement.

[bonanova]

Not sure if there is any other easier solution. But this will work.

Let the couples be denoted as, b1 g1, b2 g2, b3 g3, b4 g4. Let their initial position be at s1 and they are going to s2. Let the island be denoted as ID.

1) g1 g2 goes to ID

2) g1 remains in ID and g2 comes back to s1.

3) g2 g3 goes to ID

4) g2 remains in ID and g3 comes back to s1

5) g3 g4 goes to s2

6) g3 remains in s2 and g4 comes back to s1

7) b2 b3 goes to s2

8) b3 remains with g3 in s2 and b2 comes back to s1

9) b1 b2 goes to s2

10) b2 remains with b3 g3 in s2 and b1 comes back to s1

11) b4 g4 goes to s2

12) b4 g4 remains with b3 g3 in s2 and b2 comes back to s1

13) b1 b2 goes to ID

14) b2 remains with g2 in ID and b1 g1 comes back to s1

15) g1 remains in s1 and b1 goes to ID

16) b1 b2 goes to s2

17) b1 remains in s2 and b2 comes back to ID

18) b2 g2 goes to s2

19) b1 goes to s1

20) b1 g1 goes to s2.

And they are all happy.

That does it.

There is a 17-move solution.

[Guest]

I'll try to find the other solution. Interesting problem though.

[Guest]

I think I have a 17-move solution, using Drydung's nice notation:

1) g1 and g2 go to s2

2) g2 goes to s1

3) g2 and g3 go to ID

4) g2 goes to s1

5) b1 and b2 go to s2

6) b3 goes to s1

7) b2 and g2 go to s2

8) g1 goes to ID

9) g3 goes to s1

10) g3 and b3 go to s2

11) b1 goes to s1

12) b4 and g4 go to s2

13) g2 goes to ID

14) g1 goes to s1

15) g1 and b1 go to s2

16) b2 goes to ID

17) b2 and g2 go to s2

[Guest]

I think I have a 17-move solution, using Drydung's nice notation:

1) g1 and g2 go to s2

2) g2 goes to s1

3) g2 and g3 go to ID

4) g2 goes to s1

5) b1 and b2 go to s2

6) b3 goes to s1

7) b2 and g2 go to s2

8) g1 goes to ID

9) g3 goes to s1

10) g3 and b3 go to s2

11) b1 goes to s1

12) b4 and g4 go to s2

13) g2 goes to ID

14) g1 goes to s1

15) g1 and b1 go to s2

16) b2 goes to ID

17) b2 and g2 go to s2

Check step 5.

[Guest]

Why doesn't this work without an island?

First all the girls row across, in pairs on the way, and single on the way back. Once all four girls are across, each one goes back individually to pick up her boyfriend, rows him across, then stays with him on the far side while the other girls repeat.

[bonanova]

Why doesn't this work without an island?

First all the girls row across, in pairs on the way, and single on the way back. Once all four girls are across, each one goes back individually to pick up her boyfriend, rows him across, then stays with him on the far side while the other girls repeat.

When the first boy comes across, the other girls don't have their b/f's with them.

[brainz]

Four young couples go for a stroll and come upon a river.

They find a boat, able to carry two persons at a time.

The girls, of course can row.

But there is a problem: the girls are extremely jealous.

So there is an agreement:

No girl may stay in the company of another's boy friend unless her own b/f is present as well.

How do they all get across? They can't.

But one of the boys spots a small island in the river,

which can be used for intermediate loading and unloading.

Now is it possible?

[Guest]

When the first boy comes across, the other girls don't have their b/f's with them.

Apologies. Revised solution (requires island):

BF1 brings GF1 to the other side, sails back.

BF2 brings GF2 to the island, sails back.

BF1 brings BF2 to the other side, BF2 sails back.

BF3 brings GF3 to the island, sails back.

BF2 brings BF3 to the other side, BF3 sails back.

BF3 brings BF4 to the other side, BF4 sails back.

BF4 brings GF4 to the other side.

BF3 goes and gets GF3.

BF2 goes and gets GF2.

17 steps. More carefully checked.

[Guest]

Check step 5.

Oh yeah I see where I got it wrong

I should have checked it better!

[Guest]

Apologies. Revised solution (requires island):

BF1 brings GF1 to the other side, sails back.

BF2 brings GF2 to the island, sails back.

BF1 brings BF2 to the other side, BF2 sails back.

BF3 brings GF3 to the island, sails back.

BF2 brings BF3 to the other side, BF3 sails back.

BF3 brings BF4 to the other side, BF4 sails back.

BF4 brings GF4 to the other side.

BF3 goes and gets GF3.

BF2 goes and gets GF2.

17 steps. More carefully checked.

When BF3 brings GF3 to the island GF2 is in the island without BF2...violation.

[Guest]

ok...let couple n 1 is a(a1-bf, a2-gf)

so couple 2 is b(b1-bf, b2gf)

=we have couples C and D

1.a2 takes b2 to island

2 a2 takes c2 to island- so she takes all girls

3..

4. a2 takes a1 to another coast.

5. a1 drives and gets b1 to another coast, then they go out of the bout and

6. a2 goes with bout and brings b2-so we have 1 couple on the coast

7. a2 goes to the island and gives the bout to c2

8. c2 goes and takes c1 to the coast

9. then c2 goes to island and gives a bout to d2

10. d2 takes her d1 and brings him to the coast,

11. then comes back to island and takes a2

12.then comes back to island and takes c2 to the coust.....

It shows that girls do most of the work...

[Guest]

When BF3 brings GF3 to the island GF2 is in the island without BF2...violation.

Oh my, that's the second time. I shall have to go and cut out some multi-colored markers or something and play a little game on my floor.

[bonanova]

Oh my, that's the second time. I shall have to go and cut out some multi-colored markers or something and play a little game on my floor.

Use coins.

[bonanova]

ok...let couple n 1 is a(a1-bf, a2-gf)

so couple 2 is b(b1-bf, b2gf)

=we have couples C and D

1.a2 takes b2 to island

2 a2 takes c2 to island- so she takes all girls

3..

4. a2 takes a1 to another coast.

5. a1 drives and gets b1 to another coast, then they go out of the bout and

6. a2 goes with bout and brings b2-so we have 1 couple on the coast

7. a2 goes to the island and gives the bout to c2

8. c2 goes and takes c1 to the coast

9. then c2 goes to island and gives a bout to d2

10. d2 takes her d1 and brings him to the coast,

11. then comes back to island and takes a2

12.then comes back to island and takes c2 to the coust.....

It shows that girls do most of the work...

You say

a2 goes to the island and gives the bout to c2.

At this point haven't you have left b2 with a1 on the coast?

[Guest]

Variation on my first answer, no island.

GF1 brings BF1 across - sails back.

GF2 brings GF1 across - sails back.

GF2 brings BF2 across - sails back.

GF3 brings GF2 across - sails back.

GF3 brings BF3 across - sails back.

GF4 brings GF3 across - sails back.

GF4 brings BF4 across.

I have not yet sunk to little markers.