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April 2, 2013 9×9 is considered by many players (sjs34, jaek, pnm, clm, bram, garrybl, picklepep,…) among the more difficult puzzles that appear on this site. (Otherwise, refer to the topic “4/2/13 9×9” in the “Special Puzzles / Personal Puzzles” section). For example, Picklepep (one of those great players) commented (April 8, 2013): “I just found this puzzle worth only 130. It’s a bit ridiculous because it’s the hardest puzzle I have ever made. “This puzzle has been solved by about 60 jigsaw puzzles, but perhaps it ‘s time to come up with a complete solution (step by step) for a puzzle of these characteristics (I bring the solution to the section” Solutions and Tips “) to show you some. Concepts and techniques (actually we will only use the basic rules for this puzzle) that hopefully will be useful to many other players who are always giving up. And to show that at In “analysis” we must always believe in the “old” theorem: “If there is only one solution, it can only be solved by analysis. “

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## Whats 9×9

Step 1. It is clear (apply additional rules to the right of the three columns with the sum of 135) that g4 + g6 = 10 >>> [2, 8] or [4, 6] … but what Interestingly, that is f4 + f6 = even (the balance rule for mixed cages “3-” and “1-” where the sum is equal, in fact, f4 and f6 are both odd); Since (now applying parity law to the left of all three rows) c4 + c6 = odd >>> d4 + d6 = 26 – odd = then (parity rule for middle three columns) “60x” is odd ” 378x “e, which is [1, 6, 7, 9] or [2, 3, 7, 9] (from [3, 3, 6, 7] which is subtracted by 3 doubles) It is always strange, so we Conclude that “60x” = [1, 3, 4, 5] is an incorrect combination of [1, 2, 5, 6] (and obviously [2, 2, 3, 5] because 2 – minus two .)

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We then eliminate [4, 6] for pairs g4-g6, then arrive at [2, 8] as a result f4 = 5.

Graph 1: Status g4 = 4 >>> f4 = 1 [Now there are 7 in “13+” (c4-d4)]. If we look at the graph, we arrive quickly at b4 = 3. In these conditions, on the addition of the three middle rows (135 in total): a + b + c = 19 (the sum of cells b6, f6, and h6); But if b = 5 >>> a + c = 14 does not occur and if b = 7 >>> a + c = 12 and such as c6-d6 = [4, 9] this situation cannot be achieved.

Graph 2: case g4 = 6 >>> f4 = 3 (c4-d4 = [4, 9]) >>> f6 = 5 >>> f7-f8 = [4, 8] >>> f2-f3 = [2, 6] (f2 = 2; f3 = 6). 2 all three in columns “d”, “e” and “f” (three middle columns) force “378x” = [1, 6, 7, 9] by 23 and now by adding both middle columns Three (total 135): a + b = 17 (combination of cells d4 and d6) which is not possible because of the 8 in the “16x” cage.

We first created in step 1 that d4 + d6 is odd, that is (call these cells “a” and “b” (to “base”) them): a + b = [11, 13, 15 ] Because it is clear that this sum odd and different from 17 cannot be 9 or less (that is, if this sum is 9 >>> c4 + c6 = 26 – 9 = 17 since c3 = 8 Impossible and values lower than 19. or more for c4 + c6 are isolated).

### Tis A Fearful Thing (yellow/gray)

At this point we can specify that 6 cannot be part of “378x” according to the following: If f9 = 6 (shown in red) there is no place for 2 in column “f” because f6 is odd and is 2. Not a “60x” operand cage; Also, 6 in d9 (shown in red) will force d2-d3 = [4, 7] and it is immediately clear that a + b cannot be 11 or 13 or 15. So 6 is not part of the board “378x” >>> “378x” = [2, 3, 7, 9] (the only remaining combination for this board) and this means: f9 = 2 And now only the 6th place in the “f” column is f5 (both shown in green).

Note: Note that the previous analysis to determine the set [2, 3, 7, 9] for cell “378x” can be done immediately after step 1, it just depends on the fact that “60x” = [1 , 3, 4, 5] and other arguments to be considered in step 1 (often “steps” or “methods” can be substituted in “analysis”).

Next (in green): If f6 = 1, then 9 of the column “f” must go to cage “12+” (= [3, 9]) and then there is no place for 7 in “f”. Columns. This leads to f1 = 1 >>> d5 = 1 (which is now just the 1st place in the “d” column).

1 of column “e” should be in e4, e5 or e6. Because e4 + e5 + e6 = 25 – 1 – 6 = 18 >>> e4-e5-e6 = [1, 8, 9] (shown in green).

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Blue: A 9 in those places d9 = 9 and then e9 = 3 (because h8 = 3) and e8 = 7. 3 of the cage “60x” to d1 and e1-e2 = [4, 5]. An additional method applied to column “d” indicates that d4 + d6 = 11 and since d6 = 5 is forbidden, so d4-d6 = [4, 7] >>> d2-d3 = [5, 6] is out. Come on. Otherwise c4 + c6 = 26 – 11 = 15 = [6, 9]. And finally find 3 in column “c” it should be c2 = 3 >>> b2 = 9.

Step 4 (Graph 4, last number before or candidate shown in black). What happens if we put the number 2 in g6? >>> f6 = 3 (because f1 = 1) and there is no place for 7 in the “f” column, so: g6 = 8, g4 = 2 (in brown).

We now work with the three leftmost columns, a total of 135, adding cages “1-” (b6-b7) and “1-” (b8-c8) (combinations labeled s1 and s2 as indigo. color) is 18 (135 – 21 – 12 – 8 – 8 – 24 – 15 – 2 – 26 – 1) that is s1 + s2 = 18; Since there are 9 two in the columns “b” and “c” we get: “1-” (b8-c8) = [4, 5], [5, 6] (there are only two possibilities) but [4 , 5] Is this impossible because s1 = 18 – s2 = 18 – 9 = 9 = [4, 5] and two cages at the same time [4, 5] can not take those positions in the form “Perpendicular” so “1-” is necessary (b8-c8) = [5, 6].

There are already 6 in the column “c”: b8 = 6, c8 = 5; s1 = 18 – 11 = 7 that is “1-” (b6-b7) = [3, 4] and from there “8+” (b3-b4) = [1, 7].

### Skyscraper Puzzles (9×9, Very Easy)

At this point, we focus on the following three rows, a total of 135 to get: b7 + h7 = 12 (135 – 26 – 1 – 11 – 10 – 6 – 21 – 12 – 7 – 4 – 25) >>> b7 = 3 (b6 = 4), h7 = 9 (h6 = 2) (shown in blue) (also b7 = 4 >>> h7 = 8 >>> h6 = 3 with double 3 in column “h “).

Step 5 (Continued with Chart 4 in blue): b6 = 4 >>> d6 = 7 (c6 = 6) >>> f6 = 9 >>> e6 = 1. Also f6 = 9 and f4 = 5 >>>> “12+” = [4, 8] >>> “4-” = [3, 7] with f2 = 7 (since c2 = 3), f3 = 3. And finally c4 = 9, d4 = 4; And c4 = 9 >>> e4 = 8, e5 = 9.

Step 6 (Continued with Graph 4 in green): Now we focus on the top three rows to get: b3 + h3 = 11 (135 – 21 – 12 – 8 – 13 – 11 – 2 – 10 – 25 – 13 – 9) Obviously b3 1 (b3 = 1 >>> h3 = 10 = out of order) so b3 = 7 (b4 = 1), h3 = 4 (h4 = 7); b2 = 9 and 7 in g7

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