+++++++++++++++++++++++++
A + 2 | 0 | 3 | 0 | 2 | 2 +
+---+---+---+---+---+---+
B + 1 | 3 | 0 | 3 | 0 | 2 +
+---+---+---+---+---+---+
C + 1 | 2 | 2 | 1 | 3 | 0 +
+---+---+---+---+---+---+
A B C D + 2 | 1 | 1 | 2 | 1 | 2 +
+++++++++++++++++++++++++++++++++++++++++
+ 1 | 1 | 2 | 2 + A | B | C | D | C | D +
+---+---+---+---+---+---+---+---+---+---+
+ 2 | 2 | 1 | 1 + D | C | A | B | A | B +
+---+---+---+---+---+---+---+---+---+---+
+ 0 | 1 | 3 | 2 + C | B | C | D | C | D +
+---+---+---+---+---+---+---+---+---+---+
+ 3 | 1 | 0 | 2 + A | D | A | B | D | A +
+---+---+---+---+---+---+---+---+---+---+
+ 1 | 2 | 1 | 2 + D | B | D | C | A | B +
+---+---+---+---+---+---+---+---+---+---+
+ 2 | 2 | 2 | 0 + B | C | A | B | C | A +
+++++++++++++++++++++++++++++++++++++++++
Let r = row and c = column; let A, B, C and D refer to the entries in the grid.
In r3, the three Cs must be in c1, c3 and c5, because C cannot appear in c6. This places B/D in c2, c4 and c6.
In r4, two of the As must appear in c1 and c3, which places B/D in c2 and c4; A/D in c5 and A/B/D in c6.
In c3, there must be an A in r6, which places A/C/D in r1; A/D in r2 and C/D in r5.
In c5, there must be a C in r1, which places A/D in r2; A/C/D in r5 and A/C in r6.
Now let us move to the rest of the options.
In c1, r1 and r2 are A/B/D; r5 is B/D and r6 is A/B. As there are two Ds in the column, r5 must be a D.
In c2, r1 and r2 are B/C/D; r5 and r6 are B/C.
In c4, r1 is B/D; r2 and r5 are B/C/D and r6 is B/C.
In c6, r1, r2 and r5 are A/B/D and r6 is A/B.
As there must be two Cs in r6, c2 has to be one of them. This makes r5c2 a B; r4c2 a D; r3c2 a B; r1c2 a B and r2c2 a C.
As the only B has now been placed in r3, the two remaining cells must be Ds. Now, we can place a B in r4c4. This makes r4c6 an A and r4c5 a D. From this, we can deduce that r2c5 is A, leaving r5c5 and r6c5 as A/C.
Now, r2c6 has to be a B, leaving A/D in r1c6 and B/D in r5c6.
In c4, there has to be a B in r6, which places a C in r6c5 and an A in r5c5. The only place for the other B in r5 is in c6, which makes r6c6 an A; r1c6 a D and r6c1 a B.
Now, look at r1. The only place for the other C is in c3, which puts a D in r5c3; a C in r5c4 and an A in r2c3.
As the only B has already been placed in r1, r1c4 must be D and r1c1 must be A. This puts a D in r2c1 and a B in r2c4.
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