Scratch

bharvey wrote:

Hardmath123 wrote:

Let the guests be Alan and Alice, Bill and Beatrix, Carl and Cindy, Daniel and Della, and Ethan and Emma. Also, let Alan be the mathematician.

Aww, I was trying so hard to avoid suggesting that the mathematician has to be male!

You really were... 

The mathematician's spouse is one of the nine people who gave distinct answers.  So once you pair off 0-8, 1-7, etc., then you just say "whoever's left (namely 4) must be married to the mathematician."

Having found a solution, I show it to mum, who says "I still don't understand why the mathematician's spouse has to be 4..." — which it turned out I still didn't understand either! I think I get it now, though, so thanks to both of you for explaining 

Good puzzle 

EDIT: Brian, do you think this could this be solved using a program, a la your thing on logic puzzles?

Last edited by blob8108 (2013-02-10 14:04:05)

bharvey wrote:

Hardmath123 wrote:

Let the guests be Alan and Alice, Bill and Beatrix, Carl and Cindy, Daniel and Della, and Ethan and Emma. Also, let Alan be the mathematician.

Aww, I was trying so hard to avoid suggesting that the mathematician has to be male!

And I guess you must live in one of the states that hasn't gotten around to legalizing gay marriage yet. 

Why'd you think I picked Alan? 

We now proceed to prove that Alice (the mathematician's wife) could not have shaken eight or zero hands. If she had shaken zero hands, then the guest who shook eight hands would obviously not be her spouse. Similarly, if she had shaken eight hands, the guest who shook no hands wouldn't be her spouse. As shown above, neither of these cases is possible, so Alice shook neither eight nor zero hands.

I don't think you have to go through all that work.  The mathematician's spouse is one of the nine people who gave distinct answers.  So once you pair off 0-8, 1-7, etc., then you just say "whoever's left (namely 4) must be married to the mathematician."

This is one of my favorite puzzles, because at first glance it seems like there can't possibly be enough information in it, but by the end of the argument you know exactly who shook hands with whom, including btw the mathematician, who (like his/her spouse) shook hands with 8, 7, 6, and 5.

It's hard to sleep with a math problem on your mind, isn't it?

Wait, you weren't supposed to be sleeping; you were supposed to be prepping for a piano lesson! 

Don't tell my mom. 

PS  The UCB implementation of Lisp for Unix, years ago, was called Franz Lisp.

Cool—why is it that though geeks are mocked for their sense of humor, they are usually the epicest when it comes to inside jokes?

bharvey wrote:

blob8108 wrote:

EDIT: Brian, do you think this could this be solved using a program, a la ...missing URL...

Nice project!  Fun to play with.

I was thinking about that myself.  The thing is, once you get that the answers are [0, 8], it could probably be set up easily enough, but the leap from the problem as stated to that point is the hardest thing in the problem, so that's sort of cheating.

EDIT:  Of course, my logic program is really feeble; it handles only propositional logic, not predicate logic.  In principle, if I can solve it, a computer can solve it.

There is a solution to this problem in MiniZinc. Go to Hakank.org and check out his constraint programming blog. A great resource if you are interested in these types of problem.

Hardmath123 wrote:

should I aim to keep the scripts of the project the same, or the behavior consistent?

I'd intuitively say consistent behaviour, but constructing lists seems overkill. For the broadcast hack, I'd go for it.

I'll post a few more converted projects

Try my nodes one! 

Last edited by blob8108 (2013-02-11 06:34:42)

Hardmath123 wrote:

Oh yeah! Does it use any custom blocks?

Several, in fact — to get atomic loops. Is that a problem?

Hardmath123 wrote:

We now proceed to prove that Alice (the mathematician's wife) could not have shaken eight or zero hands. If she had shaken zero hands, then the guest who shook eight hands would obviously not be her spouse. Similarly, if she had shaken eight hands, the guest who shook no hands wouldn't be her spouse. As shown above, neither of these cases is possible, so Alice shook neither eight nor zero hands.

Wait... why not?

Last edited by joefarebrother (2013-02-11 11:37:00)

joefarebrother wrote:

Hardmath123 wrote:

We now proceed to prove that Alice (the mathematician's wife) could not have shaken eight or zero hands. If she had shaken zero hands, then the guest who shook eight hands would obviously not be her spouse. Similarly, if she had shaken eight hands, the guest who shook no hands wouldn't be her spouse. As shown above, neither of these cases is possible, so Alice shook neither eight nor zero hands.

Wait... why not?

Does this help explain? 



The grey one is the mathematician. If the mathematician's spouse (below the mathematician) shakes 8 hands, all the other 8 people must shake hands with her. But we know that one of them must shake no hands for them all to shake a different number (as there are nine of them).

joefarebrother wrote:

Wait... why not?

As I said earlier, just ignore that paragraph in Hardmath's proof; once you establish that 0 and 8 are married, then that 1 and 7 are married, etc., in the end the only people left are 4 and the mathematician.

Okay, I promise to work on the OF block (Scratch-style, not BYOB3 style). Hardmath123, just consider it to be there 

About the CLONE block, I really don't want the present block to take a sprite name as argument. Can you work around that one?

xly wrote:

If with Snap! one can create a TELL custom block[...]

Yes, once we add OOP in 4.1 this will all be easy.  The current discussion is about the minimal kludgery needed to run Scratch 2.0 projects in Snap! 4.0.

@Jens, How about a kludge in which you add to the left pulldown of the OF block a choice CLONE, so the block could say [CLONE OF Sprite3]?  (EDIT: And that choice would have to change the block's shape to Command.  )

Yeah I know it's a kludge, I already said so. 

Last edited by bharvey (2013-02-11 18:38:36)

bharvey wrote:

OldCodger wrote:

There is a solution to this problem in MiniZinc. Go to Hakank.org and check out his constraint programming blog. A great resource if you are interested in these types of problem.

Could you give me a more specific URL?  I couldn't find it.  Tnx.

Sorry but I'm not yet trusted to give URL's.

Try www.Hakank.org/constraint_programming

OldCodger wrote:

Sorry but I'm not yet trusted to give URL's.

Make some (3, but you didn't hear it from me) projects! The system is automated, and I know the secret recipe. 

@bharvey and Jens: I think the ugliest yet most functional way is to have a secret broadcast called "-snap-clone-*" where you specify the sprite name in place of *. I can do that myself already—just make dummy broadcast receptors with "clone" blocks! Are you guys ok with that hack? It can be an optional parameter-option-flag-thing: -clonehack.

Interestingly, this is exactly the inverse of the problem I had with 2.0 cloning a while ago: they didn't let you clone yourself, only someone else. 

OF blocks will default to (timer) until you implement it (remember to give me the blockspec!). But no pressure, I doubt I'll have any serious coding time for a while now.

Now to work on those custom blocks I've evaded for a while now…

Ooh, I didn't think of that. Basically, you need to have a local variable called "original", and set it to 1. Then under a "when cloned" hat set it to 0. Finally, in the -snap-clone-*" broadcast check if "original" is equal to 1.