PB Post 2013 01 05 Lvl 6 (Repeated lower down)
"Simple" methods solve the grey boxes.
PB Post 2013 01 05 Lvl 6 Same as above, except the gray boxes are filled in.
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Jan 5 puzzle is repeated here, but its grey boxes are now filled in. Advanced techniques are needed to continue. I originally used a 2x2 with 9's (rows 4 and 8 have ONLY two 9's, allowing you to strike OTHER 9's in columns B & D), followed by a chain from D6 to G1 (eliminating 9 from G6).
I later found that the puzzle could be solved by simply "coloring" 6's. !!!
Practice learning chains AFTER 9's eliminated with the 2x2, find chain from B2 to A7 (eliminate 7 from A2), and from B2 to F9. (eliminate 6 from F2)
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PB Post 2013 01 12 Lvl 6 (Repeated lower down)
PB Post 2013 01 12 Lvl 6 Same as above, except the pink boxes are filled in.
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First note that two lonely pairs of 1/2 in row 5 allow removal of 1 and 2 from E5. Then advanced techniques are required.
Chain from A5 to H4 eliminates 8 from H5. Another chain from A6 to H4 eliminates 8 from H6. H4 must equal 8.
Amazingly, puzzle STILL cannot be solved, with only 22 empty boxes. Symmetric pairs of 3's in rows 4 and 7 allow removal of 3's from D6 and F6. Finally Chain from B5 to H6 to remove 3 from H5, meaning H5 must equal 4.
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PB Post 2013 01 19 Lvl 6 (Repeated lower down)
I used "Multi Colors" to eliminate 3 from C4. Possibly not necessary.
In row 1, there are 2 pairs of 2/7 which allows me to strike 9 from B1, and strike 3&5 from C1.
Then I was stuck
PB Post 2013 01 19 Lvl 6 w 39 Same as above, except the blue boxes are filled
Note: Using methods described above the 4 underlined numbers - in Grps 1 & 4 - can be eliminated as choices.
I could not find any chains, and I resorted to "Guessing".
PB Post 2013 01 19 Lvl 6 w 39 Same as above, except the doubles are circled, and initial "guess" is indicated.
Guessing that A4 = 3 leads to conflicts (after lots of analysis, including some of the 3-candidate boxes). This PROVES that A4 cannot equal 3, and A4 = 4. FINIS!
PB Post 2013 01 26 Lvl 6 (Repeated lower down)
PB Post 2013 01 26 w 38 Lvl 6 Same as above, except the green boxes are filled.
(Note: Using "simple" methods, the 3 underlined numbers can be eliminated as choices.)
With 38 filled in I used a chain starting with the 6 in C4. Then F4, F2, I2, I9, G7. This proved that C7 and G4 cannot contain 8.
Another chain started with 7 in F2. Then I2, I9, G7, G6. Proves that G2 cannot contain 3, and therefore H1=3.
Found a chain starting with 8 in C4. Then H4, G4, G6, G7. Proves that C7 cannot contain 6.
STUCK
PB Post 2013 02 09 Lvl 6 (Repeated lower down)
PB Post 2013 02 09 Lvl 6 Same as above, except the green boxes are filled
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| First notice the lonely 6/7 pairs in the bottom row, allowing elimination of 6 from E9 and I9. Then, use colors for 7. FINIS! |
PB Post 2013 02 16 Lvl 6 (Repeated lower down)
PB Post 2013 02 16 Lvl 6 Same as above, except the pink boxes are filled
To proceed, color the 1's. It becomes evident that D3, then E3 cannot = 1. All the 1's are soon solved. FINIS.
PB Post 2013 02 23 Lvl 6 (Repeated lower down)
PB Post 2013 02 23 Lvl 6 Same as above, except the blue boxes are filled
Note simple methods: Consider row values to remove four underlined 7's; 5/6 pairs in row 5 eliminates two 6s.
It is possible - but not necessary, and not sufficient - to use multi-color with 9's and eliminate 9 from A2, C2, and later E3.
Puzzle-solver: Chain starting with 6 in B4, then B9 and A8. Proves that A5 cannot contain 9, and thus A5 must equal 8. FINIS.
PB Post 2013 03 02 Lvl 6 (Repeated lower down)
PB Post 2013 03 02 Lvl 6 w 26 Same as above, except the green boxes are filled
Note: Using "simple" methods, the 5 underlined numbers (in rows 1 and 3) can be eliminated as choices.
To proceed, use a chain starting with the 8 in A3. Then C3, C8, D8, D3, D1, F1. This proves that F3 cannot contain 6. Therefore, F3 = 7 FINIS.
OR GUESSING that B2 = 6 leads to conflicts, which prove that B2 must equal 4. Hint: You only need to work columns A-D to get the conflict. Include some 3-choice boxes.
PB Post 2013 03 09 Lvl 6 (Repeated TWICE lower down)
PB Post 2013 03 09 Lvl 6 w 38 Same as above, except the pink boxes are filled.
Note: Using "simple" methods, the 6 underlined numbers (in A2, B5,G7 and I8) can be eliminated as choices.
To proceed, I found a chain beginning with the 9 in B8. Then F8, I8, G7, G4. This proved that B4 could not contain a 2. Easily see G5 and H5 cannot contain 2s. But no answers yet!
Chain beginning with 9 in A2. Then A7, B8, F8, I8, I3. This proved that B3, G2, and H2 could not contain 2. Therefore I3 = 2.
In row 2, Two pairs of 7/8 led to info on 7's.
PB Post 2013 03 09 Lvl 6 w 28 THIRD showing. Same as above, except the blue boxes are filled.
Some underlined numbers were found with advanced methods, already mentioned.
With 28, Multicolor with 9's proved that D7 could not be a 9. Not sure if necessary.
The clincher: With 28, a chain beginning with 1 in A1. Then D1, D3, E2, E6, and H6. This proved that A6 could not be a 7, and therefore A6 = 1. FINIS.
PB Post 2013 03 16 Lvl 6 (Repeated lower down)
PB Post 2013 03 16 Lvl 6 w 29 Same as above, except the pink boxes are filled
Note: Using "simple" methods, the underlined 3 in D7 can be eliminated as a choice.
To proceed, I found a chain beginning with 1 in C4. Then C5, G5, G1, D1 and D9. This proves that D4 cannot equal 6, and so it must equal 1. Two more answers came easily (the three blue boxes). Then I got temporarily stuck again.
PB Post 2013 03 16 Lvl 6 w 26 (THIRD TME) Same as above, except the blue boxes are filled
Note: Using "simple" methods, the underlined 3 in D7 can be eliminated as a choice..
To proceed, I found a chain beginning with 8 in D7. Then D5, G4, and G1. This proves that D1 cannot contain 7, and therefore equals 3. FINIS.
PB Post 2013 03 23 Lvl 6 (Repeated lower down)
PB Post 2013 03 23 Lvl 6 w 32 Same as above, except the green boxes are filled
Note: Using "simple" methods, the 5 underlined numbers - in B3&B9, G3&G9, and H9 - can be eliminated as choices.
To proceed,I used MULTICOLOR of 5's, beginning in E1 and G1. It proved that B1, H1, I1, F4, and J4 cannot contain 5. (Also possible with 2x2 analysis of 5's.
Continuing coloring with 5s shows F2 and I2 cannot contain 5, and so F2 must equal 2.
I was temporarily stuck again. The progress is shown in the next display.
PB Post 2013 03 23 Lvl 6 w 31 (THIRD SHOWING) Same as above, except the pink box is filled.
Note: Using "simple" methods, the 5 underlined numbers - in B3&B9, G3&G9, and H9 - can be eliminated as choices. 5 more underlined choices were removed using multicolors, as explained in the PREVIOUS puzzle.
To proceed, I found a chain beginning with the 8 in C1. Then I1, H1, H3, H7. This proves that C7 cannot contain 4, and so it must equal 9.
The rest were easy. FINIS.
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PB Post 2013 03 24 Lvl 5 (6) w 29 Same as above, except the blue boxes are filled.
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Note: Using "simple" methods, the 5 underlined numbers (C4. C9, and D4) can be eliminated as choices.
To proceed, hopefully the 8/9 doubles in Column C allowed crossing out some choices.
A chain was found beginning with 5 in C4. Then A5, then I5. This proves that I4 cannot equal 1, and it must equal 7. The rest was pretty easy.
PB Post 2013 03 30 Lvl 6 (Repeated lower down)
PB Post 2013 03 30 Lvl 6 w 26 Same as above, except the blue boxes are filled.
To proceed, first I used Color with 4's. This proved that B9 cannot be a 4. Continuing, B5 also cannot be 4, and soon C2 cannot be 4.
I found a chain beginning with 3 in B5. Then B6, C6, H6, H9, and D9. This proves that B9 cannot be 9.
Next, simple methods indicate that A5 cannot contain 9.
Not too exciting until next discovery: A chain beginning with 4 in B3. Then F3 F7 and E9. This proves that B9 cannot be 7, and thus B9 = 3.
FINIS
PB Post 2013 04 06 Lvl 6 (Repeated lower down)
PB Post 2013 04 06 Lvl 6 w41 Same as above, except the green boxes are filled.
Note: Using "simple" methods, the 10 underlined numbers (A1&8, B3&9, E8, F2&9, I1,2,&3) can be eliminated as choices.
To proceed a little I colored with 4's (columns D and G). This eliminated other 3's in rows 1 and 4.
STUCK!
PB Post 2013 04 13 Lvl 6 (Repeated lower down)
Opened with 13 naked singles. Then, I chose to focus on doubles and got 8 more answers pretty quickly.
Eventually, "simple" techniques failed me with 26 remaining...
PB Post 2013 04 13 Lvl 6 w 26 Same as above, except the blue boxes are filled in.
Note: Using "simple" method, the underlined 8 in F3 can be eliminated as a choice.
To proceed, I first discovered I could Color with 9's, beginning in C1. This proved that F3 and G3 could not contain a 9.
Next, MULTI-COLORING 6 beginning at C1 and at E1 proves that D1, F1, D9, and F9 cannot contain 6. This led to finding D9=8 and C1 = 8.
I next found chains...
PB Post 2013 04 13 Lvl 6 w24 (THIRD SHOWING) Same as above, except pink boxes filled in.
Note: Using "simple" method, the 4 underlined numbers (D1, F3, G3, F9) can be eliminated as choices.
To proceed, first found a chain beginning with the 9 in A7. This might NOT be necessary, so see if you can skip to the second chain I found and solve!
Anyway, the first chain started with 9 in A7. Then A3, C1, D1, D7, D6, F6. This proved that F7 cannot contain a 6. Not sure I need to go to all that trouble.
SECOND chain I found started with the 9 in D1. Then, C1, C9, G9, G3. This proved that G1 and F3 could not equal 4.
The solution came pretty quickly after that.
PB Post 2013 04 20 Lvl 6 (Repeated lower down)
PB Post 2013 04 20 Lvl 6 w 33 Same as above, except the pink boxes are filled.
Note: Using "simple" methods, the 7 underlined numbers (A4&5, D3, E9, F2&8) can be eliminated as choices.
To proceed, I used MULTI-COLORING with 4s. It proved that F8 cannot equal 9. FINIS.
PB Post 2013 04 27 Lvl 6 (Repeated lower down)
PB Post 2013 04 27 Lvl 6 w17 Same as above, except the green boxes are filled in.
Note: Using "simple" methods, the underlined 4 in I3 can be eliminated as a choice.
To proceed, I used "color" with 5's. If you mark A1 as being a 5, you will eventually be forced to mark B2 as a 5 also. (You also get two 5's in row 3). Of course, that is WRONG and you end up concluding NEITHER ONE can equal 5; B3 = 5. All the 5's are revealed.
ONE of the several rules of coloring is that ... if you end up with TWO check marks or with TWO X's in the same row, column, or group, then in fact they MUST both be X's! And their opposites MUST be check marks, whose values are now confirmed.
PB Post 2013 05 04 Lvl 6 (Repeated lower down)
PB Post 2013 05 04 Lvl 6 w21 Same as above, except the gray boxes are filled in.
To proceed, I colored 2s. This easily proved that E3 and F2 cannot have a 2.
Simple methods (In row 2, two pairs of 2/6) proved that E2 cannot be an 8, so F2 must equal 8. FINIS
PB Post 2013 05 11 Lvl 6 (Repeated lower down)
PB Post 2013 05 11 Lvl 6 w45 Same as above, except the green boxes are filled in.
This puzzle had many high skilled eliminations. Among others, look in row 5 for two 7/8 pairs, and in row 7 for two 1/4 pairs. This allows several ellims, and leaves C4 = 7. (shown below)
After those, coloring 3s can prove that A8 cannot contain a 3. MULTI-COLORING 8's beginning in B4 and E6 proves that B9 cannot equal 8.
A chain begins with the 6 in E1. Then D2, I2, I5, I8, I4, and G4. This proves that E4 cannot=2.
After all that work......STUCK
PB Post 2013 05 11 Lvl 6 w44 (THIRD SHOWING) Same as above, except the pink box is filled in.
Explanation is given ABOVE this showing. STUCK
PB Post 2013 05 18 Lvl 6 (Repeated lower down)
PB Post 2013 05 18 Lvl 6 w 32 Same as above, except the blue boxes are filled
Note: Using "simple" methods, the underlined numbers () can be eliminated as choices.
To proceed, Coloring 9 beginning at D7 will prove that D6 cannot contain 9.
A chain beginning with 6 in F3 continues to F9, H9, and G7 will prove that G3 cannot be 9.
Simpler methods then show I4, I6, and E6 cannot contain 9.
THEN, another chain begins with 6 in F3. Continue to F9, H9, E9, E6, and E4. This proves that E2 and F4 cannot be 9. Therefore E4 = 9.
Results are shown in the next puzzle. (Another chain was used before solving.)
PB Post 2013 05 18 Lvl 6 w 31 (THIRD SHOWING) Same as above, except the green box is filled in.
See description above about elimination of underlined values.
To proceed, a chain was found beginning with the 5 in F4. Then E6, E9, D7 (or H9, H8), and finally ending at D8. This proves that D4 and F7 cannot contain 7, and so D4 = 4
PB Post 2013 05 25 Lvl 6 (Repeated lower down)
PB Post 2013 05 25 Lvl 6 w 49 Same as above, except the blue boxes are filled in.
Note: Using "simple" methods, the underlined numbers () can be eliminated as choices.
To proceed,
PB Post 2013 05 25 Lvl 6 w 37 (THIRD SHOWING) Same as above, except the boxes are filled.
Note: Using "simple" methods, the 6 underlined numbers (in row 2 and D5) can be eliminated as choices.
To proceed,
PB Post 2013 06 01 Lvl 6 (Repeated lower down)
This puzzle starts out by giving away 16 easy singles.
PB Post 2013 06 01 Lvl 6 w30 Same as above, except the green boxes are filled
Note: Using "simple" methods, the 6 underlined numbers (A1, B1, C7, A9, C9) can be eliminated as choices.
With 30 remaining, to proceed, I found a chain beginning with 6 in E7. Then G7 and H9. This proves that H7 and E9 cannot be 7. And therefore H9 = 7.
After eliminating more 7's, the puzzle finished.
PB Post 2013 06 08 Lvl 6 (Repeated lower down)
PB Post 2013 06 08 Lvl 6 w 41 Same as above, except the blue boxes are filled in.
Note: Using "simple" methods, the 15 underlined numbers (mostly rows 1 & 9; also one each in rows 2,3, and 5) can be eliminated as choices.
To proceed, I found a chain beginning with 2 in C2. Then B3 and B9. This proves that C9 cannnot be 5, and so it must equal 1. The B9 became 5. Then stuck again as shown next:
?? MULTI-COLOR with 1s starting at A1 and at G8 can prove that G5 cannot be 1. Not sure how useful.
PB Post 2013 06 08 Lvl 6 w 39 3rd Showing! Same as above, except the gray boxes are filled
Note: Using "simple" methods, the underlined numbers can be eliminated as choices. The "1" in G5 was eliminated with multicolors beginning in A1 and G8. Probably not necessary.
To proceed used I found a chain beginning with 8 in B1. Then G4, I5, and A5. This proves that A1 cannot be 1, and so A1 = 6, then A5 = 1.
With 37 remaining, 1/5 pairs in Group 6 let me eliminate some there, but still stuck.
A Chain begins with 1 in B1. Then B3, I3, I8, H7, H9, and H4. This proves that B4 cannot be 8, and so B4 = 3. FINIS.
* * * * * * * * * * * * * * * * * * * * * * * *
PB Post 2013 06 15 Lvl 6 (Repeated lower down)
Wonderful puzzle. You think you're almost there, but with only 15 empty boxes, simple techniques fail.
PB Post 2013 06 15 Lvl 6 w15 Same as above, except the blue boxes are filled
With only 15 empty boxes, the easiest way for some people is by coloring 9's. IF you begin by assuming G5 is 9, you run into conflicts which in fact reveals the correct position for the 4 remaining 9s. Even if you assumed G5 was NOT 9, you should be conscious to the fact that it could not be otherwise, because you end up with two NON-9's in a single column and in a single row. Obviously, that could NOT be reversed.
An ALTERNATE solution is to begin a chain with the 7 in E5. Then go on to E8, I8, I2, and finally I6. This proves that D6 cannot be 9, and so D6 = 8. (Other ways exist to get from E5 to D6)
PB Post 2013 07 04 Lvl 4 (6) (Repeated lower down)
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PB Post 2013 08 17 Lvl 6 w 29 Same as above, except the blue boxes are filled in.
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PB Post 2013 07 20 Lvl 6 (Repeated lower down)
PB Post 2013 07 20 Lvl 6 w 36 Same as above, except the pink boxes are filled
Note: Using "simple" methods, the 5 underlined numbers (in E4, E5, E6, and F5) can be eliminated as choices.
To proceed, I used Colors with 5s. This proved that E4 could NOT be a 5.
Then, using colors with 7 shows that shows that (again) E4 cannot be a 7. Therefore, E4=2.
I got four more answers easily, then stuck again for awhile, as indicated in the next view.
PB Post 2013 07 20 Lvl 6 w 31 (3rd showing!) Same as above, except the gray boxes are filled
Note: Using "simple" methods, the 5 underlined numbers (in E5, E6, and F5) can be eliminated as choices.
To proceed, first, I found a chain - but I don't think this is necessary!!!
Begin with 1 in F5. Then E5, E1, E2, and I2. This proves that I5 cannot be a 4. (Who cares?)
Then a found a chain beginning with the 7 in F9. Then B9, B5, I8, and H8. This proves that E8 cannot be 1. Therefore E5 must equal 1. Then H8 = 1.
FINIS
PB Post 2013 08 17 Lvl 6 (Repeated lower down)
PB Post 2013 08 17 Lvl 6 w 29 Same as above, except the pink boxes are filled in.
Note: Using "simple" methods, the underlined numbers can be eliminated as choices.
To proceed with 29 remaining, a chain begins with the 3 in A2. Then D2, E3, E4, H4,H5, and B5. This proves that A4 and A5 cannot contain 7, and so A2 = 7.
I eliminated a 7 from I5, but could not solve more.
I admit this is dumb, but I repeat the puzzle again with this little bit of extra progress:
PB Post 2013 08 17 Lvl 6 THIRD SHOWING! Same as above, except the gray box is filled in.
Note: Using "simple" methods, the underlined numbers can be eliminated as choices.
To proceed with 28 remaining, I found a chain beginning with 6 in C3. Then E3, E4, H4, and A4. This proves that A1 and C6 cannot contain 1, and that C6 must equal 2.
The FINISH slowly came after that.
PB Post 2013 0 Lvl 6 (Repeated lower down)
PB Post 2013 0 Lvl 6 Same as above, except the boxes are filled
Note: Using "simple" methods, the underlined numbers () can be eliminated as choices.
To proceed,
PB Post 2013 0 Lvl 6 (Repeated lower down)
PB Post 2013 0 Lvl 6 Same as above, except the boxes are filled
Note: Using "simple" methods, the underlined numbers () can be eliminated as choices.
To proceed,
PB Post 2013 09 07 Lvl 6 (Repeated lower down)
PB Post 2013 09 07 Lvl 6 w 37 Same as above, except the green boxes are filled in.
Note: Using "simple" methods, the 11 underlined numbers (Check rows 1-3 and Group 9) can be eliminated as choices.
To proceed, a chain begins with the 2 in A2. Then C2, H2, H7, H9, I9, and I7. This proves that I2 cannot be 9. This left two 4/5 doubles in Row 2 proving that C2 cannot =4. C2 =2. FINIS.
PB Post 2013 09 14 Lvl 6 (Repeated lower down)
PB Post 2013 09 14 Lvl 6 Same as above, except the blue boxes are filled
Note: using "simple" methods, the 4 underlined numbers (in A9, C3, and C7) can be eliminated as choices.
STUCK
PB Post 2013 10 12 Lvl 6 (Repeated lower down)
Here's the same puzzle as above, with the blue boxes filled in. Note several of the tiny possibilities are underlined. You should be able to eliminate them.
PB Post 2013 11 24 Lvl 5 (6) (Repeated lower down)
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PB Post 2013 11 24 Lvl 5 (6) Same as above, except the green boxes are filled in.
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Note: Using "simple" methods, the 2 underlined numbers (in column I) can be eliminated as choices.
To proceed, I first used MULTI-COLORS with 1. It proves that B7 cannot be 1. This was helpful because a chain beginning at B7 led to the solution.
With 33 remaining, a chain begins with the 4 in B7. Then, G7 and H8. This proves that C8 and H7 cannot be 7. Therefore both B7 and H7 = 7.
The rest was easy.
PB Post 2013 11 30 Lvl 6 (Repeated lower down)
PB Post 2013 11 30 Lvl 6 w 25 Same as above, except the blue boxes are filled
Note: Using "simple" methods, the 3 underlined numbers (in G9, H2, and I3) can be eliminated as choices.
To proceed, I found a chain beginning with 1 in B8. Then A9, A3, I3, G3, and G7. This proves that H8 cannot equal 2 and therefore H8 = 1.
Easy after that
Dec 7 - I got stuck. I could fill in all the pink ones. Enlarged results follow are shown lower down.
Here is same puzzle (Dec 7) with pink ones filled in. I can't get any more answers! Stuck!
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| Not stuck. Maybe. Long mixed up history. w 33:MCr 9»e6ø9;Reg D2ø9; Ch4:c 7e7 d9 h9 h6 g4 f4 f8 » d7,b8 ø7; b8=4; w30 Col2 »b4ø2 ;Ch1: b2 b4 g4 h6 h9 d9 d2 e3 » c3,e2 ø 9,c3=4 w 26 MC 2»d2ø2; Ch1:b2 b4 g4 f4 f8 d9 d6 » b6 ø9; 1/7 prs in B»b2=9;c2=1 w24: Ch5:c4…e8 » e4 ø6»d5=6 FINIS Not sure if all given are necessary. |
PB Post 2013 12 14 Lvl 6 (Repeated lower down)
PB Post 2013 12 14 Lvl 6 w 37 Same as above, except the pink boxes are filled in.
Note: Using "simple" methods, the 3 underlined numbers (in G9, H2, and I3) can be eliminated as choices.
To proceed I found a chain beginning with the 2 in B1. Then B4, G4, I5, and I2. This proves that B2 cannot contain 6, and therefore B1 must equal 6.
Seven more answers came easily, filling all the green boxes.
Then stuck again with 29 remaining.
PB Post 2013 12 14 Lvl 6 w 29 (THIRD showing)
Same as above, except the green boxes are filled in.
To proceed, you only need to find the chain beginning with 6 in C5. Then I5 and H6. This proves that C6 cannot be a 4, and so it must equal 3. The rest was easy.
(Along the way of that discovery, I also found a chain beginning with 3 in B4. Then G4, G9, G7. This proved that B7 could not equal 2. It wasn't required.)
PB Post 2013 Lvl 5 (Repeated lower down)
PB Post 2013 Lvl 5 Same as above, except the boxes are filled
Note: Using "simple" methods, the underlined numbers () can be eliminated as choices.
To proceed,
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