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#1 Re: Help Me ! » 5 students test » Today 00:33:57

Hmmm...many duplicates got under my guard. sad

I've tried to eliminate them all, and am now getting 2041 as the answer.

Still checking my work, though...

#2 Re: Help Me ! » Geometry Homework » Yesterday 13:46:01

Hi evene;

You use the 'hide' tags.

In the following two examples I've included a space before "hide" in the opening tag. You must remove that space in each example for them to work for you, but their inclusion enabled me to prevent my text from turning into hide boxes. smile

This:

[ hide=Hint]Type your text here...[/hide]

will give you this:

And this:

[ hide]Type your text here...[/hide]

will give you this:

Click on the two boxes to display their contents.

If you click on the "Quote" button in the bottom right-hand corner of my post you'll see exactly how I did it for my two boxes.

You can also just copy my examples into your post and experiment there.

#3 Re: Help Me ! » Password lock » 2016-02-05 03:31:17

Thanks!

I generated the list of 80730 combinations in M and pasted it into an Excel spreadsheet, which I used for the rest. That was the least time-consuming option for me.

#4 Re: Help Me ! » Password lock » 2016-02-04 22:17:20

Hi;

Here are images of the 54 solutions, displaying one solution per row.

The 5 tries were applied in ascending order, and the images display the progress.

#5 Re: Help Me ! » Password lock » 2016-02-04 11:16:18

Thanks! I'll take off my dunce cap, then! smile

download?resid=C20C46B976D069EE!3988&authkey=!APahCPKKF5S2XxY&v=3&ithint=photo%2cjpg

#6 Re: Help Me ! » Password lock » 2016-02-04 02:58:08

Well, surprise, surprise! Here I was thinking that 9 would win, but it looks like 5 wins...as Anna thought it would.

I found 54 different combinations (listed below) that 'will' open the lock on the fifth try.

I say 'will', because I haven't tested them all...although the 10 random combinations I tried all worked. The other 44 should too, because all 54 were found by the same program.

Each set of codes will open the lock irrespective of the order in which they're tried.

111	122	212	221	333
111	122	233	312	321
111	123	213	221	332
111	123	232	313	321
111	132	212	231	323
111	132	223	312	331
111	133	213	231	322
111	133	222	313	331
111	222	233	323	332
111	223	232	322	333
112	121	211	222	333
112	121	233	311	322
112	123	213	222	331
112	123	231	313	322
112	131	211	232	323
112	131	223	311	332
112	133	213	232	321
112	133	221	313	332
112	221	233	323	331
112	223	231	321	333
113	121	211	223	332
113	121	232	311	323
113	122	212	223	331
113	122	231	312	323
113	131	211	233	322
113	131	222	311	333
113	132	212	233	321
113	132	221	312	333
113	221	232	322	331
113	222	231	321	332
121	132	213	322	331
121	132	222	231	313
121	133	212	323	331
121	133	223	231	312
121	212	233	313	332
121	213	232	312	333
122	131	213	321	332
122	131	221	232	313
122	133	211	323	332
122	133	223	232	311
122	211	233	313	331
122	213	231	311	333
123	131	212	321	333
123	131	221	233	312
123	132	211	322	333
123	132	222	233	311
123	211	232	312	331
123	212	231	311	332
131	212	223	313	322
131	213	222	312	323
132	211	223	313	321
132	213	221	311	323
133	211	222	312	321
133	212	221	311	322
anna_gg wrote:

Compose a strategy to find the minimum number of tries to open the lock, under the worst situation and prove that this strategy is optimal.

My strategy was to do a brute force search of the 80,730 possible combinations of 5 choices from 27 (the maximum number of different guesses possible), which was successful. To make sure that 5 was the minimum number, I also tried 3 and 4, but no luck there. Not very elegant; but then, neither am I! smile

I think a good strategy now would be for the lock operator to memorise one of the solutions, and here I'd recommend {112,121,211,222,333} as being one of the easier sets to remember, because it comprises the two triplets {2,2,2} and {3,3,3}, and the three permutations of {1,1,2}.

Having taken so long to properly understand both the problem and Nehushtan's method (and also because there are so many solutions!), I feel a little uneasy about the result, so someone please check a few of the combinations and let me know how they performed.

Thanks!

#7 Re: Help Me ! » Password lock » 2016-02-03 17:17:07

anna_gg wrote:

...it can be done in 5 guesses, if you pick some other sets instead of the triplets. Maybe this way we manage to eliminate all 6 of the remaining combinations.

I tried a dozen or so combinations, and they all worked like the triplets in eliminating 21 options as long as they contained, in total, three 1's, three 2's and three 3's. The six remaining options in my tries didn't appear to be able to eliminated other than singly.

Yes, there are 80,730 (only) combinations of 5 guesses from the total of 27 - and I've saved the list in Excel and M - but I haven't been able to devise a program to do what we need.

Other numbers of guesses and combinations:
  6 guesses from 27 options =   296,010 combinations;
  7 guesses from 27 options =   888,030 combinations;
  8 guesses from 27 options = 2,220,075 combinations;
  9 guesses from 27 options = 4,686,825 combinations...which are very quickly reduced by Nehushtan's strategy and mine.

#8 Re: Help Me ! » Password lock » 2016-02-03 04:13:38

Hi Nehushtan;

Thanks for your explanation...I finally got it! Your method works perfectly well.

I'm very sorry about having dragged you through this for so long before the penny dropped for me! sad

#9 Re: Help Me ! » Password lock » 2016-02-03 01:07:39

Hi Nehushtan;

Leaving aside the theoretical approach for a bit, I'd now like us to put your method to a practical test...just to dispel any doubts I may have about it working.

I happen to have a combination lock identical to anna_gg's, but unfortunately I don't know the code to open it. So I'd like you to post, in this thread, the codes that will open the lock in a number of tries not exceeding the maximum number you mentioned in posts 2 and 11.

Now, I should let you know in advance that this is a tricky little lock that employs strict avoidance tactics to ensure a worst-case scenario - just like the puzzle's lock!

You can post any number of codes at a time, their quantity not exceeding the maximum number you've stated...although, just out of curiosity should that maximum be reached without success, we could continue until the lock is opened to see how many extra tries it may take. Up to you.

I'll try the codes on the lock in the order that you post them, and then respond with progress reports about their success (or otherwise) as we go.

Your turn...

smile

#10 Re: Help Me ! » Password lock » 2016-02-02 22:06:40

Nehushtan wrote:

Yes.

Ok...sorry about that.

Nehushtan wrote:

No.

Oh, I see.

#11 Re: Help Me ! » Password lock » 2016-02-02 21:38:09

Hi Nehushtan;

Nehushtan wrote:

Read the thread!

I may have misunderstood something here.

Are you saying that if the correct combination is 123 (with wild's position unknown), and you entered 111, 112, 131, 132, 211, 212, 213, 221, 222, 231, 232, 233, 311, 312, 313, 321, 322, 331, 332, and 333 (none of which will open the lock, as I understand it) before using one of the seven successful combinations you listed in post #5, that your method still holds?

#12 Re: Help Me ! » Password lock » 2016-02-02 20:01:01

True, but you've got the wild as the third digit even though its actual position is unknown. That leads to many permutations other than the minimum number in your post.

#14 Re: Help Me ! » 5 students test » 2016-02-02 12:42:29

This is my post #9 method:

Subject to the rule "each question must be chosen by exactly 2 students", there are 2220 different sets (permutations) of first-choice questions, and any set chosen determines which question numbers (but not their exact order) form the set of second-choice questions. No student has the same two questions (not exactly stated, but implied, I'd say).

Here's an explanation of the table below:

The first-choice questions only have three varieties (v1, v2, v3), and their second-choice partners the same. Examples are given.

There are different quantities of these varieties: v1=120, v2=1200, v3=900, which total 2220.

The number of permutations of second-choice questions is constant for each of the varieties: v1=44, v2=32, v3=24.

So now we can calculate the total number of different possible ways the questions can be picked:
v1 + v2 + v3 = (120 x 44) + (1200 x 32) + (900 x 24) = 5280 + 38400 + 21600 = 65280

But half of the combinations of first and second-choice sets are duplicates (eg, 1st choice 1,1,2,3,4 & 2nd choice 4,5,5,2,3 = 1st choice 4,5,5,2,3 & 2nd choice 1,1,2,3,4).

Therefore the answer is 65280/2 = 32640.

                             1st choice | 2nd choice   
                              A B C D E | A B C D E   varieties   2nd choice perms   varieties x perms
v1: no repeat digits          1 2 3 4 5 | 2 3 1 5 4      120             44                5280
v2: 1 pair of repeat digits   1 1 2 3 4 | 4 5 3 2 5     1200             32               38400
v3: 2 pairs of repeat digits  1 1 2 2 3 | 5 4 3 4 5      900             24               21600
                                                Totals  2220                              65280...less 50% duplicates = 32640

#15 Re: Help Me ! » Square ceiling » 2016-02-01 22:27:35

From the way I've spaced the circles vertically, the image shows the top row of lights right at the top of the square. However, if spaced at the minimum amount there is a vacant strip of 4√0.75 across the top. Too small to do anything with, I suppose...but I did have a look to see if I could combine it somehow with other unused spaces to conjure up another spotlight, just in case this puzzle came from some sneaky puzzle setter!

For that I tried diamond shapes and equilateral triangles, but gave up after a while.

#17 Re: Help Me ! » 5 students test » 2016-01-30 05:17:45

Ok...thanks.

This is how I arrived at my answer of 32640. It's a method I thought up just yesterday to verify the result from my post #9 method (which it did), but is easier to explain than that one.

There are 113400 permutations of the two lots of 5 questions, starting with 1,1,2,2,3,3,4,4,5,5 and ending with 5,5,4,4,3,3,2,2,1,1.

Formula: 
, where 'n' is the total number of questions picked, and letters 'p' to 't' denote the quantity of repeat numbers in different sets (in this case there are 5 sets of pairs).

So, 

In each permutation, student A's questions are the first two digits, B's are the next two...and so on for C, D & E.

Of the 113400 permutations, there are 65280 where no student chooses the same two questions.

65280 is comprised 50:50 of valid possibilities and duplicates: eg,

                                    A  |  B  |  C  |  D  |  E
First valid possibility (#2528) :  1,2 | 1,2 | 3,4 | 3,5 | 4,5 
Its duplicate          (#23348) :  2,1 | 2,1 | 4,3 | 5,3 | 5,4
Last valid possibility (#90053) :  4,5 | 4,5 | 2,3 | 1,3 | 1,2 
Its duplicate          (#110873):  5,4 | 5,4 | 3,2 | 3,1 | 2,1

So the answer = 65280/2 = 32640.

That took miles of coding in Excel (also with a bit of help from M), for which you'd need a 64-bit version of Excel running on a 64-bit operating system with a truckload of ram...all of which my computer has, fortunately.

No doubt there's a tiny mathematical formula hiding out there somewhere that could solve this in less than a blink! smile

#18 Re: Help Me ! » 5 students test » 2016-01-27 13:33:09

Hi anna_gg,

anna_gg wrote:

In how many possible ways can this be done?

Is this an example of two different ways?

     1      2
A:  1,3    1,3
B:  1,4    1,3
C:  2,4    2,4
D:  2,5    2,5
E:  3,5    4,5

It seems so to me, and my answer in post #10 is based on this example being valid.

#19 Re: Exercises » Boxes with disks. » 2016-01-26 19:32:15

Thanks for the offer, but I think I'll pass...for a while at least.

I have so many unfinished projects at home that I should be doing at a quicker rate than I have. As is the case with many owner-built houses - which mine is - I've left a lot of jobs till later. And now it's later!

#20 Re: Exercises » Boxes with disks. » 2016-01-26 19:21:03

Wow - hardly gave me time to blink!! smile I don't understand it, of course. Waaay over my head!

Oh, well...coding up a working sim was satisfying enough for me. It didn't take me long to sack Excel and let M take over.

M rolled the die at about 58000rps compared to Excel's 4000rpm, but Excel had to wade through updating spreadsheet cells, whereas M just counted under the table.

#21 Re: Exercises » Boxes with disks. » 2016-01-26 18:05:33

My code for 1 billion rolls took M 4.75 hours. I know a maths way is quicker than a sim, but I couldn't think of one...and I've never heard of a Markov chain.

What's the quick M way?

#22 Re: Exercises » Boxes with disks. » 2016-01-26 17:57:59

So that's a difference of about 6.76% from my last M simulation.

Maybe I should have let M keep rolling the die till now, and I might have got closer to the right answer!

#23 Re: Help Me ! » 5 students test » 2016-01-26 17:24:40

Is the answer 32640?

EDIT: It's based on the post #9 info.

#24 Re: Help Me ! » 5 students test » 2016-01-25 06:00:23

Well, by my reckoning, there are 2220 different sets of first-choice questions, and the set chosen will determine which questions will form the set of second-choice questions...of which there will be varying numbers of permutations.

That's as far as I've got...

EDIT 1: I think the 2220 should be halved to 1110 to avoid choice-order duplicates where second-choice sets have already been used as first-choice sets.
EDIT 2: "...of which there will be varying numbers of permutations." There are only 3 such varieties: 24, 32 and 44.

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