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Numerical Logic (Posted on 2013-05-06) Difficulty: 3 of 5
 
    a    b    c    d    e    f
  +----+----+----+----+----+----+
A |    |    |    |    |    |    |18
  |    |    |    |    |    |    |
  +----+----+----+----+----+----+
B |    |    |    |    |    |    |17
  |    |    |    |    |    |    |
  +----+----+----+----+----+----+
C |    |    |    |    |    |    |22
  |    |    |    |    |    |    |
  +----+----+----+----+----+----+
D |    |    |    |    |    |    |22
  |    |    |    |    |    |    |
  +----+----+----+----+----+----+
E |    |    |    |    |    |    |19
  |    |    |    |    |    |    |
  +----+----+----+----+----+----+
F |    |    |    |    |    |    |28
  |    |    |    |    |    |    |
  +----+----+----+----+----+----+
    18   23   19   18   22   26
The diagram above should contain the numbers one to six, six times each; the numbers next to the grid itself are the row and column number totals. Using the following 12 clues complete the diagram. (NOP means no other pairs of numbers.)

ACROSS

A: Contains two 1's, NOP, no 4's and the two numbers in cells Ae and Af total 11.

B: Contains two 2's, no 6's and NOP.

C: Contains two 4's, two adjacent 3's and the two numbers in cells Cc and Cd total 5.

D: Contains two 1's and two 5's.

E: Contains two 3's, NOP and no 5's.

F: Contains two 5's, two 6's, no 1's or 3's, the numbers in cells Fd to Ff run consecutively, and the numbers in cells Fa and Fb total 7.

DOWN

a: Contains two 1's and no 4's.

b: Contains two 4's, NOP, and the two numbers in cells Ab and Bb total 7.

c: Does not contain a 1, 3 or 4. Contains four 2's, three of which are adjacent. The two numbers in cells Ec and Fc total 8.

d: Contains two adjacent 1's, two 3's and NOP.

e: Contains two 5's, NOP and no 4's.

f: Contains three 4's but no 1's or 2's.

No Solution Yet Submitted by K Sengupta    
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Solution First pencil and paper, then computer Comment 1 of 1

The farthest I got with pencil and paper was:

    a    b    c    d    e    f
  +----+----+----+----+----+----+
A |    |    |    |    |  6 | 5  |18     (two in a box indicating
  |  1 |  3 |  2 |  1 |  5 | 6  |       possible interchange with
  +----+----+----+----+----+----+        the adjacent cell)
B |    |    |    |    |    |    |17
  |    |  4 |  2 |  1 |    |    |
  +----+----+----+----+----+----+
C |    |    |    |    |    |    |22
  |  6 |  4 |  2 |  3 |  3 |  4 |
  +----+----+----+----+----+----+
D |    |    |    |    |    |    |22
  |    |    |    |    |    |  4 |
  +----+----+----+----+----+----+
E |    |    |    |    |    |    |19
  |    |    |  2 |    |    |  4 |
  +----+----+----+----+----+----+
F |   2|5   |    |    |    |    |28
  |   5|2   |  6 |  4 |  5 |  6 |
  +----+----+----+----+----+----+
    18   23   19   18   22   26
   

with the blanks in row B being filled by 2, 3 and 5 in some order;
row D by 1, 1, 5, 5 and 6; row E by 1, 3, 3 and 6.
   
This was enough to make the remaining possibilities manageable for a computer program:

DECLARE SUB permute (a$)
DATA 132100
DATA 042100
DATA 642334
DATA 000004
DATA 002004
DATA 006456
DIM grid(6, 6)
FOR row = 1 TO 6
  READ r$(row)
  FOR col = 1 TO 6
    grid(row, col) = VAL(MID$(r$(row), col, 1))
  NEXT
NEXT row
CLS
FOR Ae = 5 TO 6
  grid(1, 5) = Ae: grid(1, 6) = 11 - Ae
  rowB$ = "235": hB$ = rowB$
  DO
    good = 1
    IF RIGHT$(rowB$, 1) = "2" OR Ae = 5 AND MID$(rowB$, 2, 1) = "5" THEN good = 0
    IF good THEN
      grid(2, 1) = VAL(MID$(rowB$, 1, 1))
      grid(2, 5) = VAL(MID$(rowB$, 2, 1))
      grid(2, 6) = VAL(MID$(rowB$, 3, 1))
  
  rowD$ = "11556": hD$ = rowD$
  DO
    good = 1
    IF (Ae = 5 OR grid(2, 5) = 5) AND MID$(rowD$, 5, 1) = "5" THEN good = 0
    IF good THEN
      grid(4, 1) = VAL(MID$(rowD$, 1, 1))
      grid(4, 2) = VAL(MID$(rowD$, 2, 1))
      grid(4, 3) = VAL(MID$(rowD$, 3, 1))
      grid(4, 4) = VAL(MID$(rowD$, 4, 1))
      grid(4, 5) = VAL(MID$(rowD$, 5, 1))
 
  rowE$ = "1336": hE$ = rowE$
  DO
    good = 1
    IF grid(4, 1) = 1 AND MID$(rowE$, 1, 1) = "1" THEN good = 0
    IF good THEN
      grid(5, 1) = VAL(MID$(rowE$, 1, 1))
      grid(5, 2) = VAL(MID$(rowE$, 2, 1))
      grid(5, 4) = VAL(MID$(rowE$, 3, 1))
      grid(5, 5) = VAL(MID$(rowE$, 4, 1))

      FOR Fa = 2 TO 5 STEP 3
        grid(6, 1) = Fa
        grid(6, 2) = 7 - Fa
        good = 1
        IF grid(4, 2) = 5 AND grid(6, 2) = 5 THEN good = 0
        IF grid(4, 1) <> 1 AND grid(5, 1) <> 1 THEN good = 0
        IF grid(4, 4) = 1 OR grid(5, 4) = 1 THEN good = 0
        IF grid(4, 4) <> 3 AND grid(5, 4) <> 3 THEN good = 0
        FOR row = 1 TO 6
          IF grid(row, 3) = 1 THEN good = 0
          IF grid(row, 3) = 3 THEN good = 0
          IF grid(row, 3) = 4 THEN good = 0
        NEXT
        REDIM nct(6)
        FOR row = 1 TO 6
          nct(grid(row, 5)) = nct(grid(row, 5)) + 1
        NEXT
        IF nct(1) <> 1 OR nct(2) <> 1 OR nct(3) <> 1 OR nct(6) <> 1 THEN good = 0
        REDIM nct(6)
        FOR row = 1 TO 6
          nct(grid(row, 2)) = nct(grid(row, 2)) + 1
        NEXT
        IF nct(1) > 1 OR nct(2) > 1 OR nct(3) <> 1 OR nct(5) <> 1 OR nct(6) <> 1 THEN good = 0
        sum5 = (grid(1, 5) = 5) + (grid(2, 5) = 5) + (grid(3, 5) = 5) + (grid(4, 5) = 5) + (grid(5, 5) = 5) + (grid(6, 5) = 5)
        IF ABS(sum5) <> 2 THEN good = 0
        REDIM cct(6)
        FOR row = 1 TO 6
          FOR col = 1 TO 6
            cct(col) = cct(col) + grid(row, col)
          NEXT
        NEXT
        IF cct(1) <> 18 OR cct(2) <> 23 OR cct(3) <> 19 OR cct(4) <> 18 OR cct(5) <> 22 OR cct(6) <> 26 THEN good = 0
        IF good THEN
          FOR row = 1 TO 6
          FOR col = 1 TO 6
            LOCATE (ct \ 4) * 9 + 1 + row, (ct MOD 4) * 15 + 1 + col * 2
            PRINT grid(row, col);
          NEXT:
          NEXT:
          ct = ct + 1
        END IF
      NEXT Fa

    END IF
    permute rowE$
  LOOP UNTIL rowE$ = hE$

    END IF
    permute rowD$
  LOOP UNTIL rowD$ = hD$

    END IF
    permute rowB$
  LOOP UNTIL rowB$ = hB$
NEXT Ae

which finds

1 3 2 1 6 5
5 4 2 1 2 3
6 4 2 3 3 4
1 1 5 6 5 4
3 6 2 3 1 4
2 5 6 4 5 6

The row totals were not needed in either the analysis or the program. I almost forgot the totals altogether, but without consideration of the bottom (column) totals, there were 8 solutions.


  Posted by Charlie on 2013-05-06 17:10:00
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