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Hi MathsIsFun;

"bobbym" should read "bob bundy", but it looks fine otherwise.

Hi MathsIsFun;

The colour in the solution should be changed to green, but other than that I like it as it is. Figuring out what Melanie knows is the solution key, which the wording says.

Of course, some people would be hoping for the solution to also reveal how Judith was able to figure out the colour of her hat. I'm not sure if the reasoning should be revealed, but if it were, here's an option (wintersolstice's, slightly adjusted):

About crediting, well, I don't really know, just that it would be nice for readers to see the activity, I suppose. So if you were to say something about that, wintersolstice should receive the credit for the discovery, and Bob & I could be mentioned as confirming it.

Btw, when reading "as shown below" in the puzzle I expected to see a diagram, but there isn't one and "one behind the other" might be better.

Hi wintersolstice;

I really like your reasoning in post #2!

Hi wintersolstice and Bob;

I thought you'd never ask, Bob!

Btw, I have one of those, and below is a pic I took of it tonight. It's never worked for me, though!

And...I agree with both of you.

Nice puzzle!

Hi salem_ohio;

All I know is that it can be done with 3.

As thickhead showed in post #12, your {8,8,1} group can be done with 3 (see also my previous post), by what I suspect is the intended method: binary division, as thickhead said in post #17.

Anyway, here's my 'trial and error' solution for {5,5,7}, done in 2 weighings in much the same way as for {7,7,3}:

Groups {A,B,C} = {5,5,7}.

Weigh A:B (5:5):

1. If not balanced, the bag has the rotten apples.

2. If balanced, transfer 2 from A to B and weigh B:C (7:7).

(a) If also balanced, the bag has all good apples.

(b) If not balanced, the bag has the rotten apples.

The 2(b) result is possible without any further weighings. That is because, after returning the 2 transferred apples to A, all other weighing combinations would yield an unbalanced result...shown as follows:

We have {A,B,C} = {5,5,7}.

That gives 3 options of rotten apple combinations: {0,0,5}, {1,1,3} and {2,2,1}.

Transferring 2 apples from A to B yields the following rotten apple options:

For {0,0,5}, it's {0,0,5}.

For {1,1,3}, it's {1,1,3} and {0,2,3}.

For {2,2,1}, it's {2,2,1}, {1,3,1} and {0,4,1}.

Weighing A:C is as follows:

For {0,0,5}, it's {0:5}.

For {1,1,3}, it's {1:3} and {2:3).

For {2,2,1}, it's {2:1}, {3:1} and {4:1}.

None of these weighings give a balanced result.

By this (laborious) method, the minimum number of weighings required for the greengrocer to determine which bag he has is 2.

This image may help:

thickhead wrote:

Salem_Ohio's start of 8,8,1 is to be admired as it is less taxing and shows the strength of

binary divisionto the core(8 vs 8 ;4 vs 4; 2 vs 2 when rotten apples break down) even though it requires 3 weighings.

Yes, salem_ohio's {8,8,1} method is easily the most elegant one here, with its simpler, more logical approach:

Rotten apple numbers from the following weighings:

Split 16: W1 (8:8) = 0 or 2

Split 8: W2 (4:4) = 0 or 1

Split 4: W3 (2:2) is the decider: if there's 1 rotten apple left, it can't be split, and therefore a balanced weigh = all good apples, and unbalanced = rotten.

The final (2:2) weigh is either of these two: 2 good vs 2 good...for a balanced result; or 2 good vs (1 good + 1 bad)...for an unbalanced result.

Done in 3 weighings.

My 2-weigh solution of the {7,7,3} is very laborious, being just trial and error. It would be a pity if it was the winner as it would only win on minimum weighs, not technique. However, I've looked at it pretty closely and can't fault it yet.

And it looks like my {7,7,3} solution isn't the only 2-weigh one, either, because that strategy also seems to work for {5,5,7}. That would further sour the {7,7,3) result, as that means there are multiple solutions. I haven't tried my T&E strategy on other groups yet.

You're right..I'd overlooked that.

I've added two red entries to my post #14. I've checked it a few times and I think it will now work.

I haven't looked at salem_ohio's {8,8,1} yet, but I will soon.

This was my reasoning for all possible rotten apple combinations and their respective weighings:

We have {A,B,C} = {7,7,3}.

That gives 2 options of rotten apple combinations: {1,1,3} and {2,2,1}.

Transferring 2 apples from A to C yields the following rotten apple options:

For {1,1,3}, it's {1,1,3} and {0,1,4}.

For {2,2,1}, it's {2,2,1}, {1,2,2} and {0,2,3}.

Weighing A:C is as follows:

For {1,1,3}, it's {1:3} and {0:4).

For {2,2,1}, it's {2:1}, {1:2} and {0:3}.

None of these weighings give a balanced result.

This image may help:

salem_ohio wrote:

So, just to clarify:

After moving the 2 apples from A to B, it is ALWAYS unbalanced, unless we are in the bag with the good apples, right?

So, in essence, we can do it in 2 weighings?

I don't think we've accounted properly for the possibility of the greengrocer having the bag of good apples...which I suspect would add a minimum of one weighing to each of the earlier results (but I haven't checked them all). That possibility should be eliminated before looking at the rotten apple options...unless it can be accounted for some other way.

I've tried to allow for this as follows:

Groups {A,B,C} = {7,7,3}.

Weigh A:B (7:7):

If not balanced, the bag has the rotten apples.

If balanced, remove 4 from A and weigh A:C (3:3).

EDIT: That line should read: If balanced, transfer 2 from A to C and weigh A:C (5:5).

If A:C (3:3) is also balanced, the greengrocer has the bag of good apples, otherwise he has the bag with the rotten apples (because, after having returned the 2 transferred apples to A, all other weighing combinations would yield an unbalanced result - which can be shown without any further weighing/s).

By this method, the minimum number of weighings required for the greengrocer to determine which bag he has is 2.

After first weighing A:B in a 6:6:5 combination and then transferring 1 apple from A to C, the second weighing, this time with B:C in a 5:6:6 combination, identifies that the transferred apple is bad.

I tried to use that information to solve the puzzle at the third weighing, but was unsuccessful. It might be possible, but I gave up looking when I saw thickhead's solution.

*EDIT: Hi, thickhead. I just read your last post, where you gave the solution to the place where I'd got stuck with the 6:6:5, with the outcome that this combination can be solved in 3 weighings. Well done! I hadn't seen that the opportunity was there to increase a group's bad apple count to 3, which would always result in an imbalance. *

salem_ohio wrote:

So, just to clarify:

After moving the 2 apples from A to B, it is ALWAYS unbalanced, unless we are in the bag with the good apples, right?So, in essence, we can do it in 2 weighings?

Exactly! To both questions.

Btw, thickhead's the genius, not me!

Here's my spreadsheet image of thickhead's answer, just looking at it a little differently (but essentially the same).

Xlnt, thickhead! Can't see anything wrong with that!

I was still at the stage of moving just 1 apple over, and then saw your post.

I'd got 4 min in nothing flat and it all seemed so open, so I was pretty sure there'd be a lower min...but not right down to 2!

Hi salem_ohio;

How have you gone with this so far?

I'm down to 4 weighings (hope they're right!), but feel that it can be lowered further. Not too sure about my prospects, though, because I've tried for a while...

Hi, sirluke4, and welcome to the forum!

I have an answer, and it is one of the six possibilities you listed.

Here are a couple of hints to help you solve the puzzle:

This morning the OneDrive links I referred to in my previous post broke again, but this afternoon most - not all - worked!

Anyway, I've now changed them all over to Imgur.

Since the index rebuild, I've noticed that:

1. the Search results display incredibly quickly;

2. clicking the post link at the top of the post on the Search results page works reliably now (only sometimes before).

Well...looks like that index rebuild has fixed all my OneDrive links! None are broken now.

So I've only had to replace the Kiwi6 links, which I know for sure are broken.

And for me, too...busy changing the links now.

Thanks, MIF!

Yes, some are old - from 2015 and before - but some are from 2016, with one as recent as October last year. Furthermore, a post from a couple of weeks earlier - in September - was successful on the same image that refused to change in the October post!!

Riddle me that one, Batman!

Anyway, it's 'good' to hear that I'm not the only one this is happening to. I'll stop pulling my hair out, then.

**phrontister**- Replies: 9

Hi;

I now use Imgur as my image file host, but previously it was OneDrive and Kiwi6 before that. All my links to Kiwi6 are broken due to account inactivity, and a lot to OneDrive broke approx last year when M$ changed their hotlinking method.

During the last few days I've tried to replace many of these broken links with Imgur's, but while some were successful, most failed...and I haven't been able to work out why.

The failures get as far as displaying Imgur's image after clicking 'Review', but 'Submit' hangs, no matter how much time I give it. Also, I've noticed that I can't make any non-image changes (eg, text) to posts where the attempted image change failed.

Does anyone have any ideas about this? I'd like to replace broken links for future reference purposes.

Thanks!

Hi Beach Coach;

Sorry, but I can't answer your ranking question. Maybe someone else here can.

I googled around a bit and came across quite a range of coverage about this topic on the net. Here are some search suggestions:

"Sports teams' ranking"

"Sports rating system methods"

"Sports power ranking"

Hopefully a ready-made system exists for your model!

Hi chen.aavaz;

If for some reason you'd like to hide the image (eg, image too large for main post page, or if it's a spoiler) you can put it into a hidebox, like this:

[hyde][img]your image link[/img][/hyde]

Note: Change the 'y' in 'hyde' (both instances) to 'i' (if I'd typed 'i', then my instruction would have disappeared and become a hidebox...hence the 'trick').

Example:

The word "example" gives a name to the hidebox, and can be something you choose. If left blank, "Hidden Text" will appear in the box.

You can always learn from how others do things by clicking on the 'Quote' button in the bottom right-hand corner of their post (try it in mine).

Hi all;

bobbym wrote:

From the point of view of EM, the problem is solved.

I'm quite sure of that too, so here's the full list of 945 possible combinations of 5 pairs from 10 players, condensed down to 4 image files:

Hi iamaditya

After getting more details i found that the total no. Of possibilities are 10C2 or C(10,2) =10!/(2!*8!)=45

bobbym and I both got 945 possibilities. The full list is a bit large to post, but in the hidebox below is a random sample of 50 from the list. The entries are in strict numerical order.

In post #14, my image in the 'level 5' hidebox shows the first 60 and the last 60 possibilities from the 945 (entries are in numerical order). 4 from the list of 50 random samples are in there too, so that makes 166 possibilities that I've posted.

They all seem to be valid to me, and if you check them I think you'll agree that there are many more possibilities than your answer of 45.

So I suppose there's an error in your formula, but I'm sorry, I don't know enough about this kind of maths to be able to show where it is.

That's how I see it too.

Understanding the problem, finding the logic required to solve it, applying that logic correctly with EM, using that logic for a mathematical approach that confirms the EM results, having two people from opposite sides of world who weren't looking over each other's shoulder arrive at identical results...I think we've done that.

I just hope that one of us didn't misunderstand the problem and influence the other ESP-ly...we are cousins, after all!