Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

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: 7

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!

I have some confidence in the result now.

I'm convinced!

Hi Bobby;

Well, that gets 270270 alright, but I have no idea what that code means.

I fiddled around with it a bit to see if I could get to understand it, and got the first 4 levels {10,105,1260,17325) that we'd got before.

So, including your code, this is the pattern:

```
(1/3840)(m-9)(m-8)(m-7)(m-6)(m-5)(m-4)(m-3)(m-2)(m-1)m/.m->13 = 270270
(10/3840)(m-7)(m-6)(m-5)(m-4)(m-3)(m-2)(m-1)m/.m->11 = 17325
(80/3840)(m-5)(m-4)(m-3)(m-2)(m-1)m/.m->9 = 1260
(480/3840)(m-3)(m-2)(m-1)m/.m->7 = 105
(1920/3840)(m-1)m/.m->5 = 10
```

It looks interesting - like it should mean something - but I don't really know what it was that I did!

Didn't know you could do that!

Excellent!

Does your code solve for 5 pairs out of 13 athletes?

My computer can't handle that number of permutations. M stops working straightaway, giving this message:

"General::nomem: The current computation was aborted because there was insufficient memory available to complete the computation."

I noticed that when I ran the M code for 11 athletes, it used up all available RAM (20GB or so) and then began to use virtual memory from the hard drive...which turned out to be enough to get a result. The extra amount of memory needed to solve for 13 athletes would be absolutely humongous, so I let that one slide and just took OEIS's word for it that 270270 was the next element after 17325 in the sequence.

Their formula, (2n+3)!/(3!*n!*2^n), and their M code, Table[(2n+5)!!/3-(2n+3)!!,{n,1,5}]/2, confirm that. Of course, I don't understand the how/what/why etc of those formulas!

Oops, so I did. I'll try again!

bobbym wrote:

Can you calculate 4 pairs out of 9 people?

That's 17325 possible combinations of 4 chosen pairs and their accompanying threesomes who are omitted.

My image in the 'level 4' hidebox in post #24 covers that one (players in green).

I get 105 possible combinations of 2 chosen pairs and their accompanying threesomes who are omitted.

My image in the 'levels 1&2' hidebox in post #24 covers that one (players in orange)...and also the one for 5 athletes (players in blue).

In my images, the combination totals and the number of athletes involved are shown at the bottom of the combination lists. Each list has its own colour, highlighting the chosen players.

Hi Bobby;

What do you get for 3 pairs from a group of 9 players?

I get 1260 possible combinations of 3 chosen pairs and their accompanying threesomes who are omitted. That totals 9 athletes for each combination.

See my image in the 'level 3' hidebox in post #24. It lists the first and last 60 combinations that I got from my M code. The chosen player groups are highlighted in yellow.

```
n = 9;
s = Permutations[Range[n]];
teams = Partition[#, 2] & /@ s;
DeleteDuplicates[Map[Sort, Most /@ teams, 2]];
Length[%]
%%
```

No pressure on us, then!

I'm enjoying the struggle!

It's nice to know that you're on board, Coach!

Often an OP disappears after posting their problem, despite the fact that a full-on discussion by forum members about its solution may follow!

Hopefully we can get somewhere with yours...

I sure hope that the buzzing sound in your new machine is not from a sick bee, or something more sinister!

Must go to bed now...see you tomorrow.

I believe it is correct.

In the figures below, which are from level 1, the first column lists the 10 valid pairs that can be formed from the five athletes {1,2,3,4,5} on the roster, subject to the constraint that there are 3 more athletes on the roster than the number chosen to play (same as for level 5). So the second column lists those who miss out.

For the STP I had to decide how, for levels 1-4, I should deal with the difference of 3 that exists in level 5 (ie, 13-10), and I opted for maintaining it consistently throughout all levels.

That is why all 10 lines contain just one chosen pair, and a threesome who missed out.

```
1,2 3,4,5
1,3 2,4,5
1,4 2,3,5
1,5 2,3,4
2,3 1,4,5
2,4 1,3,5
2,5 1,3,4
3,4 1,2,5
3,5 1,2,4
4,5 1,3,2
```

But my thinking may be wrong on that, so please convince me otherwise if I've erred.

Hi Bobby;

In the meantime, can I see what you did?

In post #24 I explained what I did (although maybe not fully), but I didn't show my M code that I adapted from yours. This is it:

```
n = 11;
s = Permutations[Range[n]];
teams = Partition[#, 2] & /@ s;
DeleteDuplicates[Map[Sort, Most /@ teams, 2]] // Length
```

It's only good for n=an odd number (your code does even numbers), for which n=11 (answer 17325) is the highest my computer will go...because of the 'combinatorial explosion'.

The following code is from OEIS. I changed the min and max and divided by 2 to suit our problem.

`Table[(2 n + 5)!!/3 - (2 n + 3)!!, {n, 1, 5}]/2`

The adapted code outputs all five levels, the last being the 270270.

My images in post #24 hopefully will help show how I identified the pattern.

Hi Bobby;

Have you had a look yet at my post #24 answer to the 13-athletes question from post #1? Just wondering if you agree with my 270270, or have something else.

Beach Coach wrote:

We actually have 13 athletes on the roster. 10 of them can play in a match (5 doubles pairs). How many possible combinations of 5 pairs can I make from 13 athletes?

270270, I think.

I played STP by hand in Excel for levels 1 & 2, but the size of the problem was escalating enormously and I had to change tack. So then I adapted Bobby's M code from post #21 to fit my levels 1 & 2 results and got levels 3 & 4 (but then ran out of computer memory, preventing further progress that way).

At that stage I had the sequence {10,105,1260,17325}, covering the first 4 levels.

At OEIS I found the last (5th) level and the following formula for the sequence:

(2n+3)!/(3!*n!*2^n)

For our problem, *n* is the number of doubles pairs (5), from which (2*5+3)!/(3!*5!*2^5) = 270270.

Here are images of my work for the STP: