# Can you solve this football league table problem?

Each team (A,B,C,D,E,F) plays each other once (on a neutral ground). Maximum number of goals scorable is 4. A team gets $$3$$ points for a win, $$1$$ point for a draw and $$0$$ points for a loss.

At the end of the season, the scores are added, and the table is sorted by points, then goal difference, goals for, wins, losses, name.

The scorecard (on the left) lists each match, and can be divided into $$5$$ blocks of $$3$$ matches each, where each team plays exactly once.

It was a rainy day, and some of the entries have become blurred.

Can you replace the missing entries?

• This isn't really a [grid-deduction] puzzle in the sense of that tag. Feb 21, 2020 at 17:23
• I take it that the left is the game scores with a team names in columns one and four and their scores in columns two and three? Feb 21, 2020 at 20:02
• Can I just check: Is there definitely a logically deducible path through this? I've just spent over an hour on this and have only managed to fill 11 trivial league table spaces and 1 space in the results table... It is entirely possible I am missing something obvious or am just far too tired (it's getting late), but just wanted to confirm that you know there's a logical way to get the solution and haven't just deleted random entries from the tables in the hopes that someone 'might be able to find a solution'?? Thanks!
– Stiv
Feb 22, 2020 at 22:58
• My only thought is whether you have assumed that each subsequent block of 3 matches contains each team only once, i.e. 5 matchdays on which each team plays one match. Nowhere is this stated in the puzzle, so I have not assumed it, instead assuming the teams may have played each other in staggered fashion, much like the Premier League works these days due to TV rights...
– Stiv
Feb 23, 2020 at 8:54
• @shoover; It's all mine AFAIK, I wrote in it Excel.
– JMP
Feb 23, 2020 at 18:45

Solution as follows (write-up complete at last)...

Before the original post was edited to explain that the scorecard could be split into 5 matchdays (blocks of 3 matches in which each team plays only once) it did not have a unique deducible solution. However, with this new information it is now entirely possible, and the final table and results look like this:

There is a single ambiguity which cannot be ironed out, but ultimately it doesn't matter - that regarding which of teams C and E are placed in the left- and right-hand columns for their 3-3 draw.

Logical deduction process:

Before the edit to the post (i.e. without knowing the list could be split by matchday), it was possible to note that there are 11 trivial observations in the league table to start (dependent on W+D+L=5 and 3W+D=PTS), and 1 easily deducible score for Team B:

- 1st place lost 0 - cannot be D (known to have lost 2-4 to F); must have scored at least 8 goals to achieve goal difference of +8.
- 2nd place drew 2, with 8 pts (2 wins + 2 draws).
- 3rd place won 2 and GD must be -2 (GD always sums to 0 across all teams); thus GF must be 9.
- B (in 4th place) must have drawn 2 to get 5 pts total. We know 3 of B's scores, totalling 5 goals; since they scored 13 (and 4 max per game), B must have scored 4 in each of their other 2 matches (making the game vs F a 4-4 draw).
- F (in 5th place) won 1 and lost 2 (to get 5 pts total). Known to have scored 8 goals in 3 games and conceded 6 in 2 games; with a GD of -2 they thus must have conceded at least 10 goals.
- 6th place cannot be E, since we know E won a match (2-1 in match 3).

Overall, this means that 9 games ended in a win for one side; the remaining 6 were draws.

Now, with the matchday clarification outlined above, it is possible to infer all but 2 of the teams in the results list and begin filling in some other gaps:

- Since F's matches against B, C and D are already identified, they must have played E in match 3 and A in match 13.
- B's match on matchday 5 cannot be against D, since we know they must score 4 goals and D's opponents here score only 2; thus B play C in match 14 and E play D in match 15.
- Match 1 must now be between A and B, while C must play D in match 2.
- Thus, match 5 must now be B vs E, making match 4 A vs C.
- In match 7, A must play D, since C and E play each other in match 9 on the same matchday.
- Thus, match 10 must now be A vs E while match 11 is B vs D.
- C has lost vs B in match 14, so cannot finish 1st (since the first-placed team loses 0).
- F now has 9 goals for and 10 against (with a GD of -1), with one game's GF and one game's GA still to identify; to get a net total GD of -2 the difference between these unknown values must be -1. F must therefore lose to C, since C must score at least one goal. This must mean their match vs A is F's last remaining draw, so they score 2 against A to draw and concede 3 against C to get the target additional -1 GD. F's GF and GA are now known.
- B's win against C is their only one, so match 1 must be an A win or a draw. A thus cannot lose 2 matches and finish last; coupled with the knowledge that B, C, E and F have all won at least one match (and thus cannot finish last), D must be in 6th place.
- This also means that the result between B and D in match 11 must end in a 2-2 draw, since B have had their one win already, and D never win. D must also score 2 against E to bring their GF total up to the known 8 goals.

Next:

- B's one win comes against C in match 14. Match 1 must therefore end in an A win or a 3-3 draw. Similarly, since D win 0, match 7 must result in an A win or draw. A therefore win or draw all of their matches and must be the team who finishes 1st.
- B must concede 5 goals in total against A in match 1 and E in match 5. These means they must concede in both matches and they must lose to E. Since E come 2nd or 3rd and both places contain teams who won exactly 2 matches, we now have both of E's wins pinpointed (B and F in match 3).
- For the match between C and E, we know that C must concede 2 or 3 goals (they concede 8 in their other 4 games and must concede 10 or 11 overall) and that E must concede 3 or 4 goals (they concede 7 in their other 4 games and must concede 10 or 11 overall) - i.e. C must score at least 3 goals vs E. This means the most C can score against A in match 4 is 1 goal (as they only score 9 in total, regardless of whether they finish 2nd or 3rd) and they must lose to A.
- This in turn means we now know C loses 2 games in total - against A and B (in match 14), which means they must finish 3rd and E 2nd. This means they concede 11 goals in total and must let in 3 against E. E concede 10 goals in total and must also let in 3 against C - the game ends in a draw.
- C's match vs D must thus be a 1-0 win (to get their 2 wins) and they score 0 vs A.
- E's two draws are against C and D, they have beaten B and F, so must lose to A.

Finally:

- D must concede 2 against A to end up conceding 11 in total, so A vs D is a 2-2 draw.
- We have now identified all 6 draws in the tournament, so A must beat B 4-3. This means A score 16 goals in total, and thus must concede 8 in total to get a net GD of 8 - they must therefore concede 1 against E.
- This means E must score 1 in their victory against B. This makes B's GF and GA both 13 as required also.

And the puzzle is solved, albeit with the ambiguity in the C-E draw... Phew!