Conquer Club

Husband A rows husband B across. drops him off then rows back.
picks up husband C. rows him across, drops him off and rows back.
Husband A gets out and wife C and B row across. wife B drops off wife C, picks up her husband (B) and rows back.
drops off her husband and picks up wife A. rows her across they both get out and husband C rows back picks up husband B and rows across. they both get out and wife A rows across to get her man.

porkenbeans wrote:Husband A rows husband B across. drops him off then rows back.
picks up husband C. rows him across, drops him off and rows back.
Husband A gets out and wife C and B row across. wife B drops off wife C, picks up her husband (B) and rows back.
drops off her husband and picks up wife A. rows her across they both get out and husband C rows back picks up husband B and rows across. they both get out and wife A rows across to get her man.

This solution meets all requirements (and I believe it to be the shortest possible solution)- the ball is now in Porkenbean's court to propose our next puzzle!

I'm going to jump in here with a fairly simple one... How can you get exactly four gallons of water into a 5 gallon jug using only a 3 gallon jug and the aforementioned 5 gallon jug?

pimphawks70 wrote:I'm going to jump in here with a fairly simple one... How can you get exactly four gallons of water into a 5 gallon jug using only a 3 gallon jug and the aforementioned 5 gallon jug?

Fill the 3 gallon jug, then pour it into the 5. Fill the 3 again, then fill the 5 to capacity from it. This will leave 1 gallon in the 3. Empty the 5, pour the 1 gallon from the 3 into the 5, refill the 3, then add its contents to the 5, for a total of 4 gallons in the 5 gallon jug.

Can I just say here that I had finally found time this afternoon to start work on a spreadsheet for the jealous wives/river crossing problem, which was set up to depict legal intermediate states when the boat was empty, and from which I was going to determine legal and useful transfers leading from the initial to the final state? And that I mailed that spreadsheet home from work, and it never arrived, and I had to start over, and then I had to go out, and I was going to solve it tonight?!?!??!!! Honest, I was...

Ah, heck, well done, porkenbeans, give us a new problem.

pimphawks70 wrote:I'm going to jump in here with a fairly simple one... How can you get exactly four gallons of water into a 5 gallon jug using only a 3 gallon jug and the aforementioned 5 gallon jug?

A classic diophantine problem. And infinately reusable- that is to say, there are infinite ordered triples (a,b,c) where a and b are the size of the jugs and c is the amount of water to be measured for which there is a solution. Of course not every such triple works- measuring 7 gallons of water cannot be done with a 4 gallon jug and 6 gallon jug! So while we wait for porken, what conditions must a,b, and c meet for the puzzle to be solvable?

Yeah, it was head scratcher.

OK, here is one I remember-

Get yourself 6 toothpicks, or sticks, of equal length. Now arrange them to make 2 equilateral triangles.

Now, move as many sticks as you want to make 4 equilateral triangles.

The triangles must be the same size as the original first 2, so, no overlapping, or breaking any sticks.

porkenbeans wrote:Yeah, it was head scratcher.

OK, here is one I remember-

Get yourself 6 toothpicks, or sticks, of equal length. Now arrange them to make 2 equilateral triangles.

Now, move as many sticks as you want to make 4 equilateral triangles.

The triangles must be the same size as the original first 2, so, no overlapping, or breaking any sticks.

Tetrahedron .

Someone else post a new one, my head's spinning right now...

YoursFalsey wrote:

pimphawks70 wrote:I'm going to jump in here with a fairly simple one... How can you get exactly four gallons of water into a 5 gallon jug using only a 3 gallon jug and the aforementioned 5 gallon jug?

A classic diophantine problem. And infinately reusable- that is to say, there are infinite ordered triples (a,b,c) where a and b are the size of the jugs and c is the amount of water to be measured for which there is a solution. Of course not every such triple works- measuring 7 gallons of water cannot be done with a 4 gallon jug and 6 gallon jug! So while we wait for porken, what conditions must a,b, and c meet for the puzzle to be solvable?

Well clearly c <= a + b, and I think another necessary condition would be that a and b are mutually prime. With that, I think you can always find a solution for the triple (a, b, 1), and intuition tells me that from there, you can solve with problem with any value of c, but I can't be sure that you can retain that 1 while working towards another value. Do I get part marks?

Nice matchstick puzzle!

Regarding the jug question, ender, a and b don't have to be mutually prime- however c must be a multiple of their greatest common divisor. (Which of course it is if they are mutually prime) So just as 3 and 5 can be used to measure out 4, 6 and 10 can be used to measue out 8, 9 and 15 to measure out 12, and so on, and so on.

The corespondance to diophantine equations (ax + by = c) where a,b,c are integer constants and x and y are integer solutions is that positive values of x or y refer to filling a jug from the tap and negative values to emptying the other jug out, with pouring the water from one jug to the other a "free" action. Witness also there are generally 2 solutions- for example with our 3 jug and our 5 jug, (3)3+(-1)5=4. This was ender's solution- he filled the 3-jug three times, emptied the 5-jug once and had 4 gallons remaining. But also (-2)3+(2)5=4. My solution would have been to fill the 5-jug. Pour from the 5-jug into the three jug, and empty that. Pour the remaining 2 gallons into the 3-jug. Fill the 5-jug again, and top off the 3-jug from the 5 gallon, leaving 4 gallons in the 5 jug. (And empty the 3-jug again because we don't need the extra 3 gallons.)
(There are infinite solutions to the diaphantine equation, but the others just involve repetion- so you could fill the 5-jug five times and empty the 3-jug seven times, or fill the 5-jug eight times and empty the 3-jug twelve times, etc, but that is because filling the 5-jug three times and using it to empty the 3-jug five times is a net zero-effect set of actions...)

YoursFalsey wrote:Nice matchstick puzzle!

Regarding the jug question, ender, a and b don't have to be mutually prime- however c must be a multiple of their greatest common divisor. (Which of course it is if they are mutually prime) So just as 3 and 5 can be used to measure out 4, 6 and 10 can be used to measue out 8, 9 and 15 to measure out 12, and so on, and so on.

The corespondance to diophantine equations (ax + by = c) where a,b,c are integer constants and x and y are integer solutions is that positive values of x or y refer to filling a jug from the tap and negative values to emptying the other jug out, with pouring the water from one jug to the other a "free" action. Witness also there are generally 2 solutions- for example with our 3 jug and our 5 jug, (3)3+(-1)5=4. This was ender's solution- he filled the 3-jug three times, emptied the 5-jug once and had 4 gallons remaining. But also (-2)3+(2)5=4. My solution would have been to fill the 5-jug. Pour from the 5-jug into the three jug, and empty that. Pour the remaining 2 gallons into the 3-jug. Fill the 5-jug again, and top off the 3-jug from the 5 gallon, leaving 4 gallons in the 5 jug. (And empty the 3-jug again because we don't need the extra 3 gallons.)
(There are infinite solutions to the diaphantine equation, but the others just involve repetion- so you could fill the 5-jug five times and empty the 3-jug seven times, or fill the 5-jug eight times and empty the 3-jug twelve times, etc, but that is because filling the 5-jug three times and using it to empty the 3-jug five times is a net zero-effect set of actions...)

Very interesting analysis. I did not know that such puzzles could be solved by an equation. I always approached them in a mostly ad hoc manner.

The matchstick problem has NOT been solved. I believe that "slo" was being facetious.

Tetrahedron: 6 edges, 4 sides, each an equilateral triangle.

explanation enough?

slowreactor wrote:Tetrahedron: 6 edges, 4 sides, each an equilateral triangle.

explanation enough?

Yes, the classic 4-sided dice shape.

slowreactor wrote:Tetrahedron: 6 edges, 4 sides, each an equilateral triangle.

explanation enough?

Oh, yes I see, another word for 4 sided pyramid.
your go slo.

I'm all burnt out... someone else come up with one please.

How about another crossing puzzle?

Three missionaries (Matthew, Mark, and Micheal) and three cannibals (whose actual names are unpronouncable, so call them Clark, Chuck, and Cedric) need to cross the big river, which has a particularly nasty current going today. Once again, the only canoe they could find can hold at most two people. Two constraints exist: Mark and Clark are the only two individuals who can row the canoe without tipping it over, so Mark and/or Clark must be on the canoe for every trip. And the missionaries are concerned about ending up in the cooking pot, so at no time can the cannibals be allowed to outnumber the missionaries on either shore. (For example, if Mark and Cedric were together on one shore, Clark could not row over to put Mark on the canoe for the trip back because Mark would be outnumbered by Clark and Cedric at the transition moment.) How can all six cross the river?

Use the bridge.
Or eat the cannibals. (Turn about is fair play)

Clark rows chuck and returns.
Clark rows Cedric and returns.
Mark rows Matt, picks up Cedric and returns.
Mark rows Clark, picks up Matt, and returns.
Mark rows Mich, Gives the boat to Clark and he picks up his fellow cannibals one by one.

Man, this one had smoke coming out of my ears.

porkenbeans wrote:Clark rows chuck and returns.
Clark rows Cedric and returns.
Mark rows Matt, picks up Cedric and returns.
Mark rows Clark, picks up Matt, and returns.
Mark rows Mich, Gives the boat to Clark and he picks up his fellow cannibals one by one.

Man, this one had smoke coming out of my ears.

damn you beat me to it pork.

porkenbeans wrote:Clark rows chuck and returns.
Clark rows Cedric and returns.
Mark rows Matt, picks up Cedric and returns.
Mark rows Clark, picks up Matt, and returns.
Mark rows Mich,....

As written, you still have steam coming out of your ears At this point Matt and Cedric are on the original side of the river- when Clark heads back with the canoe, we have violation, and poor Matt ends up inside the pot.
Still, given your final line, I think you meant to add this line:
Mark picks up Chuck and returns. There he drops off Chuck, picks up Matt and crosses the river where he

porkenbeans wrote:.... Gives the boat to Clark and he picks up his fellow cannibals one by one.

Man, this one had smoke coming out of my ears.

Good job! Next puzzle is yours to give!

Very busy right now. If someone wants to take my turn, go ahead.

This one is a classic, which means someone will recognize it and knock it down quickly.

Start with a standard 8x8 chess board:

A1 A2 A3 A4 A5 A6 A7 A8
B1 B2 B3 B4 B5 B6 B7 B8
C1 C2 C3 C4 C5 C6 C7 C8
D1 D2 D3 D4 D5 D6 D7 D8
E1 E2 E3 E4 E5 E6 E7 E8
F1 F2 F3 F4 F5 F6 F7 F8
G1 G2 G3 G4 G5 G6 G7 G8
H1 H2 H3 H4 H5 H6 H7 H8

Covering that board with 32 dominos, each large enough to cover 2 squares, is easy, with many possible solutions. (A1-A2) (A3-A4) (A5-A6) (A7-A8) (B1-B2)...blah,blah,blah..(H5-H6)(H7-H8)- Four dominos going across on each row- is one, but there are many more. So make it a little more tricky: Remove squares A1 and H8. Now use 31 dominos to cover the 62 remaining squares.

Impossible.

slowreactor wrote:Impossible.

I know you're right, but if you want to claim the win, you should explain how you know it is impossible.

ender516 wrote:

slowreactor wrote:Impossible.

I know you're right, but if you want to claim the win, you should explain how you know it is impossible.

Yeah, what ender said! (Particularly since I admitted upfront this chestnut has been around the block a time or two...)