The first thing to notice is that since each team played 3 games,
the only way to get 7 points is 2 wins and 1 draw, the only way to get 6 points is 2 wins and 1 loss, the only way to get 2 points is 2 draws and 1 loss, and the only way to get 1 point is 1 draw and 2 losses
So we have
A: 2 wins 1 draw
B: 2 wins 1 loss
C: 2 draws 1 loss
D: 1 draw 2 losses
Since the only teams with wins are
A and B, each of them must have defeated D. B can't have lost to itself, so A must have defeated B, and B defeated C.
That leaves the draws:
A drew C, and C drew D.
Summary of win-loss-draw:
A def. B
A drew C
A def. D
B def. C
B def. D
C drew D
For the goals scored, notice that
D has 1 draw and 2 losses but a goal differential of only -2, so each of D's losses must be by 1 goal, to A and to B. A has 2 wins and 1 draw, but a goal differential of only +2, so each of A's wins must be by 1 goal, over B and over D. C has 2 draws and 1 loss, but a goal differential of -2, so C's loss must be by 2 goals, to B. So B defeated D by 1 goal, lost to A by 1 goal, and defeated C by 2 goals.
Then we know
D scored 6 goals, but the opponents of A and B together scored only 5, and at least 1 of B's opponents' goals was scored by A, so C-D must be at least 2-2, but no more than 4-4. A-B and A-D can be 1-0, 2-1, or 3-2 because A's opponents scored 2 total. B-D can be 1-0, 2-1, or 3-2. It can't be 4-3 because B's opponents scored a total of 3, and B lost 1 game. B-C can be 2-0, 3-1, or 4-2. C scored 4 goals, at least 2 of them against D, so A-C is 0-0, 1-1, or 2-2.
Looking at D's games,
we know C-D was 2-2, 3-3, or 4-4; A-D was 1-0, 2-1, or 3-2; and B-D was also 1-0, 2-1, or 3-2. Also, D's total goals scored was 6 and conceded was 8.
This gives the following possibilities:
C-D A-D B-D
--- --- ---
2-2 1-0 5-4 xxx
2-2 2-1 4-3 xxx
2-2 3-2 3-2
3-3 1-0 4-3 xxx
3-3 2-1 3-2
3-3 3-2 2-1
4-4 1-0 3-2
4-4 2-1 2-1
4-4 3-2 1-0
Considering the above combinations that are still possible, look at A's games
and consider A-B, which is 1-0, 2-1, or 3-2, but A's goal totals are only 4 scored and 2 conceded.
C-D A-D B-D A-B
--- --- --- ---
2-2 3-2 3-2 1-0
3-3 2-1 3-2 1-0
3-3 2-1 3-2 2-1
3-3 3-2 2-1 1-0
4-4 1-0 3-2 1-0
4-4 1-0 3-2 2-1
4-4 1-0 3-2 3-2
4-4 2-1 2-1 1-0
4-4 2-1 2-1 2-1
4-4 3-2 1-0 1-0
Now consider B's games
recalling that B-C can be 2-0, 3-1, or 4-2, while B's goal totals are 5 scored and 3 conceded.
This gives the following possibilities:
C-D A-D B-D A-B B-C
--- --- --- --- ---
2-2 3-2 3-2 1-0 2-0
3-3 2-1 3-2 1-0 2-0
3-3 2-1 3-2 2-1 xxx
3-3 3-2 2-1 1-0 3-1
4-4 1-0 3-2 1-0 2-0
4-4 1-0 3-2 2-1 xxx
4-4 1-0 3-2 3-2 xxx
4-4 2-1 2-1 1-0 3-2
4-4 2-1 2-1 2-1 2-0
4-4 3-2 1-0 1-0 4-2
This leaves for the last game and goal totals:
C-D A-D B-D A-B B-C A-C goals: A-x B-x C-x D-x
--- --- --- --- --- --- --- --- --- ---
2-2 3-2 3-2 1-0 2-0 0-0 4-2 5-3 2-4 xxx
3-3 2-1 3-2 1-0 2-0 1-1 4-2 5-3 4-6 6-8 <==
3-3 3-2 2-1 1-0 3-1 0-0 4-2 5-3 4-6 6-8 <==
4-4 1-0 3-2 1-0 2-0 2-2 4-2 5-3 6-8 xxx
4-4 2-1 2-1 1-0 3-1 1-1 4-2 5-3 6-8 xxx
4-4 2-1 2-1 2-1 2-0 0-0 4-2 5-3 4-6 6-8 <==
4-4 3-2 1-0 1-0 4-2 0-0 4-2 5-3 6-8 xxx
So the final solution is
that there are three solutions.
A B C D
A \ 1-0 1-1 2-1
B 0-1 \ 2-0 3-2
C 1-1 0-2 \ 3-3
D 1-2 2-3 3-3 \
----------------------
A B C D
A \ 1-0 0-0 3-2
B 0-1 \ 3-1 2-1
C 0-0 1-3 \ 3-3
D 2-3 1-2 3-3 \
----------------------
A B C D
A \ 2-1 0-0 2-1
B 1-2 \ 2-0 2-1
C 0-0 0-2 \ 4-4
D 1-2 1-2 4-4 \
The restriction of no team scored more than 3 that was edited in late reduces this to one:
A B C D
A \ 1-0 1-1 2-1
B 0-1 \ 2-0 3-2
C 1-1 0-2 \ 3-3
D 1-2 2-3 3-3 \