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Trophy

cinelli
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Trophy

#417901

Postby cinelli » June 7th, 2021, 12:47 pm

Sixteen teams, A to P, entered a knockout football competition to win the Lemon Trophy. In none of the matches did any team score more than five goals, no two matches had the same score, and extra time ensured that there were no draws.

The table shows the total goals, for and against, for each team.

.         A  B  C  D  E  F  G  H  I  J  K  L  M  N  O  P
For 5 14 15 2 1 0 4 1 1 12 6 0 4 1 2 7
Against 6 0 8 4 3 5 6 4 2 12 5 1 5 5 3 6

My own team, K, was knocked out by team who were the eventual champions.

Who won the competition and what were the scores in every match?

Cinelli

GoSeigen
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Re: Trophy

#417985

Postby GoSeigen » June 7th, 2021, 8:30 pm

cinelli wrote:Sixteen teams, A to P, entered a knockout football competition to win the Lemon Trophy. In none of the matches did any team score more than five goals, no two matches had the same score, and extra time ensured that there were no draws.

The table shows the total goals, for and against, for each team.

.         A  B  C  D  E  F  G  H  I  J  K  L  M  N  O  P
For 5 14 15 2 1 0 4 1 1 12 6 0 4 1 2 7
Against 6 0 8 4 3 5 6 4 2 12 5 1 5 5 3 6

My own team, K, was knocked out by team who were the eventual champions.

Who won the competition and what were the scores in every match?

Cinelli


Nice problem to solve sitting at a table with 15 dominoes...

GS

CryptoPlankton
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Re: Trophy

#418028

Postby CryptoPlankton » June 8th, 2021, 4:19 am

Spoiler:


B won the competition.

Last16:

A3 v 1 E
B 5 v 0 F
C 5 v 1 N
G 3 v 2 O
H 1 v 0 L
J 5 v 4 M
K 2 v 1 I
P 4 v 2 D

Q-Finals:

B 4 v 0 H
C 5 v 2 A
J 4 v 3 P
K 4 v 1 G

S-Finals:

B 3 v 0 K
C 5 v 3 J

Final:

B 2 v 0 C

Thanks for something to do on a sleepless night!

UncleEbenezer
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Re: Trophy

#418082

Postby UncleEbenezer » June 8th, 2021, 11:23 am

Spoiler (I see I'm not the first, but never mind).

B won. No draws, no goals against them, they won every game. That's the easiest bit.

The second easy bit: there were (if I've understood the rules aright) 15 games, and 15 different scores. There are also exactly 15 scores allowed by the rules, so each score occurred exactly once.

C was runner-up: 15 goals implies they won three games but lost the final to B. That's three 5-x victories out of a total of five. J was knocked out by C in a high-scoring semi-final. The other semi-finalist got knocked out for no score by B. Since K got knocked out by B to nil, K's six goals imply two wins, making them the other losing semi-finalist.

Knowing the finalists and semi-finalists, we can infer quite a lot from their scores (not at first the order).

- 14 goals to nil tell us B won 5-0, 4-0, 3-0 and 2-0. The 5-0 was F's first round defeat.
- C lost to nil in the final, so C's 15 goals must come from three 5-x victories in earlier games. Eight goals against, of which at least two were in the final, tell us C's victories were 5-1, 5-2 and 5-3, with a 0-2 defeat in the final.
- J's twelve goals include at least nine in their two victories, with seven goals against them and a 5-3 semi-final loss against C. That means theirs is the only remaining 5-score 5-4, and 4-3. Since no team has totals of +3/-4, the 5-4 must be a first round victory against M.
- J's second round opponent lost 4-3 but won round 1. So they lost >=4, and won >= losses (damn, I typed L, then realised how confusing that gets). Only K and P satisfy those constraints, and we already know it wasn't K, so it was P. So P won the first round 4-2 against D.
- K lost the semi-final either 0-3 or 0-4 to B. All 5-x scores are assigned, so K's two victories were with 4 and 2 or 3 and 3. None was to nil, so we infer the five goals against were one in each victory and a 0-3 semi-final defeat. Thus K's victories are 4-1 and 2-1.
- That means B's 4-0 victory was in round 2. The loser has lost >= 4 and 4 >= won >= losses-3. Teams that satisfy that are D,H and M. But D lost to P and M to J, so H lost to B, having beaten L 1-0 in the first round.
- C won the first two rounds 5-1 and 5-2. No team has aggregate +2/-5, so the first round was a victory against N, and round 2 was 5-2 against A, who in turn defeated E 3-1 in round 1.

Are we there yet? I think we may be near enough to fill the gaps by inspection.

Round 1:
A3 E1
B5 F0
C5 N1
P4 D2
H1 L0
J5 M4
K2 I1
G3 O2

Round 2:
B4 H0
C5 A2
K4 G1
J4 P3

Round 3
B3 K0
C5 J3

Round 4
B2 C0

Phew!

cinelli
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Re: Trophy

#418200

Postby cinelli » June 8th, 2021, 6:26 pm

Well done Crypto. Well done Uncle. Although you weren't the first, your analysis is superb.

Cinelli

9873210
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Re: Trophy

#418214

Postby 9873210 » June 8th, 2021, 7:41 pm

If anyone would care to comment.

How do you organize all this? I made pretty much all the deductions above but I kept getting to the end and realizing I'd assigned 2-0 twice or some such, and I haven't got my notes organized enough to unwind back to the point of the error.

GoSeigen
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Re: Trophy

#418233

Postby GoSeigen » June 8th, 2021, 9:25 pm

9873210 wrote:If anyone would care to comment.

How do you organize all this? I made pretty much all the deductions above but I kept getting to the end and realizing I'd assigned 2-0 twice or some such, and I haven't got my notes organized enough to unwind back to the point of the error.


See my earlier post.

GS

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Re: Trophy

#418341

Postby Gengulphus » June 9th, 2021, 11:47 am

9873210 wrote:If anyone would care to comment.

How do you organize all this? I made pretty much all the deductions above but I kept getting to the end and realizing I'd assigned 2-0 twice or some such, and I haven't got my notes organized enough to unwind back to the point of the error.

I'd recommend setting up a main table showing how the knockout competition progresses and a subsidiary table showing which of the 15 scorelines are still freely available, like this:

? ? \
> ? ?
? ? / \
> ? ?
? ? \ / \
> ? ? \
? ? / \
> ? ?
? ? \ / \
> ? ? / \
? ? / \ / \
> ? ? \
? ? \ / \
> ? ? \
? ? / \
> ?
? ? \ /
> ? ? /
? ? / \ /
> ? ? /
? ? \ / \ /
> ? ? \ /
? ? / \ /
> ? ?
? ? \ /
> ? ? /
? ? / \ /
> ? ?
? ? \ /
> ? ?
? ? /

5-0 4-0 3-0 2-0 1-0
5-1 4-1 3-1 2-1
5-2 4-2 3-2
5-3 4-3
5-4

The main table should have the winner of each match on the upper branch and the loser on the lower branch, and the subsidiary table should have all scorelines that have been located (either precisely or to within just a few possibilities) struck out. Repeatedly make a copy of the table so far, make some deductions from what you've got so far and fill those deductions in on the copy (but not the original). If you find you've made an incorrect deduction, you only have to back off to before the incorrect deduction, not all the way to the start.

So if e.g. your first set of deductions is from B's 14 For / 0 Against totals, that B must win the tournament with scorelines of 5-0, 4-0, 3-0 and 2-0 in its four matches, and that the scoreline in its first match must be 5-0 against F because no teams have 0 For / 4 Against, 0 For / 3 Against or 0 For / 2 Against totals, you update that to:

B 5 \
> B *
F 0 / \
> B *
? ? \ / \
> ? 0 \
? ? / \
> B *
? ? \ / \
> ? ? / \
? ? / \ / \
> ? 0 \
? ? \ / \
> ? ? \
? ? / \
> B
? ? \ /
> ? ? /
? ? / \ /
> ? ? /
? ? \ / \ /
> ? ? \ /
? ? / \ /
> ? 0
? ? \ /
> ? ? /
? ? / \ /
> ? ?
? ? \ /
> ? ?
? ? /

5-0 4-0 3-0 2-0 1-0
5-1 4-1 3-1 2-1
5-2 4-2 3-2
5-3 4-3
5-4

in which the asterisks must be 2, 3 and 4 in some order.

And if your next set of deductions is that C's 15 For / 8 Against totals must be the result of three of the scorelines 5-1, 5-2, 5-3 and 5-4 in the first three rounds and one of the scorelines 0-2, 0-3 and 0-4 in the final, the only way to keep that down to 8 Against goals is for those scorelines to be 5-1, 5-2, 5-3 and 0-2, and that the scoreline for C's first round match must be 5-1 against N because no teams have 2 For / 5 Against or 3 For / 5 Against totals, you update that to:

B 5 \
> B *
F 0 / \
> B *
? ? \ / \
> ? 0 \
? ? / \
> B 2
? ? \ / \
> ? ? / \
? ? / \ / \
> ? 0 \
? ? \ / \
> ? ? \
? ? / \
> B
C 5 \ /
> C 5 /
N 1 / \ /
> C 5 /
? ? \ / \ /
> ? # \ /
? ? / \ /
> C 0
? ? \ /
> ? ? /
? ? / \ /
> ? #
? ? \ /
> ? ?
? ? /

5-0 4-0 3-0 2-0 1-0
5-1 4-1 3-1 2-1
5-2 4-2 3-2
5-3 4-3
5-4

in which the asterisks must be 3 and 4 in some order, and the hashes must be 2 and 3 in some order.

Then your next set of deductions might be that J has to play in at least three rounds because of its 12 Against goals (and indeed also its 12 For goals) and cannot play in all four rounds because we already know which teams are the two finalists and they're not J, so it must be a losing semi-finalist. Furthermore, it cannot be the semi-finalist that loses against B, since that one gets at most 5 For goals from each of the first two rounds and none from its semi-final, for at most 10 in total. So it must be the semi-finalist that loses against C, and its scorelines must be two of those we haven't struck out in the first two rounds and either 2-5 or 3-5 in the semi-final. To get those up to 12 Against goals, we must have 7 Against goals from the first two rounds, which can only come from the 5-4 and 4-3 scorelines, and then the total of 12 For goals requires the semi-final scoreline to be 3-5. Furthermore, as no team has totals of 3 For and 4 Against goals, the 4-3 scoreline must be in the second round and the 5-4 scoreline in the first round, which means that J knocks M out in the first round. So we update the tables to:

B 5 \
> B *
F 0 / \
> B *
? ? \ / \
> ? 0 \
? ? / \
> B 2
? ? \ / \
> ? ? / \
? ? / \ / \
> ? 0 \
? ? \ / \
> ? ? \
? ? / \
> B
C 5 \ /
> C 5 /
N 1 / \ /
> C 5 /
? ? \ / \ /
> ? 2 \ /
? ? / \ /
> C 0
J 5 \ /
> J 4 /
M 4 / \ /
> J 3
? ? \ /
> ? 3
? ? /

5-0 4-0 3-0 2-0 1-0
5-1 4-1 3-1 2-1
5-2 4-2 3-2
5-3 4-3
5-4

in which the asterisks must be 3 and 4 in some order.

And so it goes on - I won't complete the solution because the point of this post is to answer your question about how to organise the process of solving the puzzle, not to provide a complete spoiler (though I am leaving it as an unconcealed partial spoiler, because I reckon spoiler concealment produces too much visual clutter for the tables to be at all easy to read). One other point to make is that this way of organising it is much better done electronically than with paper and pencil, because of the far greater ease of making a copy of your work so far and doing further work in the copy only, leaving the possibility of reverting to the current state of your solution if you make a mistake in that further work.

Completing the solution does require about four or five further similar-sized sets of deductions, so one does end up with a pretty long write-up of the complete solution if one does it all this way. But it can be written up quite quickly once one has designed and written the first set of tables, because each stage of copying and editing in the further deductions doesn't involve much typing - most of the time is spent looking for and finding good sets of deductions to make, not on writing them up. And once one has the complete solution and has checked it, one does have the option of shortening the write-up by combining groups of deductions into larger groups, leaving out the intermediate tables between them. (Doing that too much does make it hard for readers to keep track mentally of where one has got to, but one can certainly use larger groups of deductions than those I've used above to illustrate the process - there's basically a balance to be struck between conciseness and easy readabiility here.)

Gengulphus

UncleEbenezer
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Re: Trophy

#418356

Postby UncleEbenezer » June 9th, 2021, 12:10 pm

Gengulphus wrote:I'd recommend setting up a main table showing how the knockout competition progresses and a subsidiary table showing which of the 15 scorelines are still freely available, like this:

That makes sense for a paper solution, but I hate to think how much effort you must've taken to do it in a purely ASCII medium! I thought my bullet points (of which you presented the first three) a far less laborious approach.

Mine was basically to make the same deductions as you until I had most of the games. Then I jotted down remaining unused scores to fill in the final blanks. A composition box here serves for jottings as well as what one actually posts!

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Re: Trophy

#418373

Postby Gengulphus » June 9th, 2021, 12:49 pm

UncleEbenezer wrote:
Gengulphus wrote:I'd recommend setting up a main table showing how the knockout competition progresses and a subsidiary table showing which of the 15 scorelines are still freely available, like this:

That makes sense for a paper solution, but I hate to think how much effort you must've taken to do it in a purely ASCII medium! I thought my bullet points (of which you presented the first three) a far less laborious approach.

I don't find it all that much effort, as long as I display it in a fixed-pitch font like Courier. I must admit to having had a lot of practice at doing ASCII diagrams in my working life, though - mostly 25+ years ago, but I still use them often enough that I haven't lost the knack.

But I agree that your bullet points are a less laborious approach, provided one doesn't encounter the problem 9873210 did: "I made pretty much all the deductions above but I kept getting to the end and realizing I'd assigned 2-0 twice or some such, and I haven't got my notes organized enough to unwind back to the point of the error." The point of my post was to answer his question about how to organise things to avoid that problem: record the information one needs systematically, and take regular 'checkpoints' of its state that one can easily unwind to if one finds that one has made an error. (And actually, I don't think my suggestion does make much sense for a paper solution, unless one makes a good deal of use of a photocopier or scanner to take those checkpoints!)

Gengulphus

9873210
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Re: Trophy

#418409

Postby 9873210 » June 9th, 2021, 3:46 pm

Thanks for the discussion.

I tried something like both those approaches. Using a spreadsheet rather than ascii for the brackets.

One of the issues I had was that there are a load of deductions don't fit neatly into these schemes. For example you can immediately identify 7 of the 8 first round winners. As you show the last two rounds are easily solved. If you can only identify the last first round winner ... .

If I was being paid to solve this I'd just do a depth first search at that point. There's at worst 9 cases, but most drop out easily. However I was not being paid so I kept looking for the bright shiny object as I worked through the solution. Unfortunately this meant I had a couple of supplementary tables and they got out of synch.


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