PB Post 2014 01 11 Lvl 6 (Repeated lower down)
PB Post 2014 01 11 Same as above, except the blue boxes from there are filled in.
A chain starting with the 9 in B9 ended with the 8 in I6, allowing elimination of 8 from I9. Not sure how useful.
From F2 to E1 and also from F2 to F7.
This allowed me to eliminate 9's from B2, A7, and E9.
And thus F7 had the only 9 in the 3x3 group.
Got stuck after filling in the three blue boxes.
A chain beginning with the 8 in C1 and ending with the 6 in A7 let me eliminate 6 from A1 and from C8. This left C1 containing the only 6 in the row. FINIS
PB Post 2014 01 18 Lvl 6 (Repeated lower down)
Great Problem for Color (or chains)
PB Post 2014 01 18 Same as above, except the green boxes are filled in.
This puzzle can be solved if you color 6's.
PB Post 2014 01 25 Lvl 6 (Repeated lower down)
PB Post 2014 01 25 w31 Same as above, except the green boxes are filled in.
 |
| Underlined 4's in row two eliminated due to 6/7 pairs. Coloring 9's eliminates four 9's from column G; H4=9. Done! |
PB Post 2014 02 01 Lvl 6 (Repeated lower down)
Grey boxes are filled in the next picture:
PB Post 2014 02 01 w35 Same as above, except the two blue boxes are filled in.
I'm still working on this puzzle. Despite arduous advances, still not solved!
(w33:Ch8:C1 C7 f7 f1 f2 e2 d3 b3 b5 »b1,c2ø2. Gave more answers, but no solution)
STUCK
PB Post 2014 02 08 Lvl 6 Even the "simple" part of this puzzle is pretty rigorous. Lots of cleverness with two pairs and "doubles"
Many hints for filling in the pink are listed, although other ways exist.
PB Post 2014 02 08 Same as above, except the pink ones there are now filled in.
FINIS
(Note: Using "simple" methods, the underlined numbers can be removed as choices.
Hints for underlined:
Two pairs of 5/7 in the left, middle group; Two pairs of 1/3 along row 5;
Two pairs of 3/8 in row 6 7's along column H
Two 5/6 pairs in column D Two 6/8 pair or two 2/9 doubles in row 1
Two 6/7 pair or 2/4 doubles in column I right, middle group: Note pairs/doubles
To proceed, use color with 9's. It proves D5 and F2 cannot contain 9. Therefore D5=4. Later H4=4.
Got (temporarily) stuck again with only those two new answers. More to come:
PB Post 2014 02 08 (3rd showing) Same as above, except the two blue boxes are filled in.
(Note: Using "simple" methods, the underlined numbers can be removed as choices. Hints are given with the puzzle higher up.)
To proceed, I used a long, boring chain starting with the 8 at F4, then to A4 B5 B2 C1 D1 D3 D7 which proves that F7 cannot be 9. Soon afterward, E5 cannot be 9.
Pretty disappointing results for so much labor. Spent an hour before finding it!
But E5 is now a "double" and it's the beginning of a chain that solves all.
Start with 8 in E5, then H5 H7 H8 proves that E8 cannot be a 6. Therefore E8=7.
Two 5/6 pairs in row 9 helped me, but I saw someone finish without that.
PB Post 2014 02 09 Lvl (5) 6 is labeled a "Level 5" in the Sunday paper. But I needed chains to get unstuck two times, and I still cannot solve! It is reproduced here 4 times, the final time with 33 empty boxes.
One GOOD thing about this puzzle: as dull as it is seeking chains, it is very good for teaching chains because in most cases (if you start at top left) you quickly see to not bother, due to inability to "get out".
+Sun.jpg)
PB Post 2014 02 09 w39. Same as above, except pink boxes from above now filled in:
+Sun+w39.jpg)
(Note: Using "simple" methods, the underlined numbers can be removed as choices.)
After filling in the pink boxes from previous illustration, I got stuck, as illustrated above. To get unstuck, I eventually found chain: Start with 9 in H3. Then H7 H8 I8 G7 G6. This proves that G1 cannot contain 3, and so G6 =3. Three more answers came soon. Then, with 35 remaining I was stuck again. This is shown in the next illustration.
PB Post 2014 02 09 w35 (3rd showing) Same as above, except 4 blue boxes from above filled in:
+Sun+w35.jpg)
(Note: Using "simple" methods, the underlined numbers can be removed as choices.)
Above, with 35 boxes empty, I found a chain beginning with 7 in B6. Then E6 F4 I4 which proves that neither A4 nor C4 nor I6 can contain 9. Therefore C4 = 8. Then I4 = 9.
And I was stuck again with 33 empty boxes.
PB Post 2014 02 09 w33 (4th showing) Same as above, except 2 green boxes from above are now filled in:
+Sun+w33.jpg)
(Note: Using "simple" methods, the underlined numbers can be removed as choices.)
Above, with 33 boxes empty, I'm stuck. If anyone can get more....please tell me how!
STUCK
PB Post 2014 02 14 Lvl 5 (6) This was rated a Level 5, but I cannot solve it.
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The green boxes in the puzzle above should fill in without "chains" or "colors." Then, I got permanently stuck with 24 remaining, as shown in the next puzzle.
PB Post 2014 02 14 Same as above except answers for the green boxes above are now filled in:
+w24.jpg)
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+w+24.jpg)
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+w33.jpg)
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+w35.jpg)
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+w38.jpg)
The results are that C6 cannot contain 5, and so must be 6. 30 naked singles followed!
At least two other chains might have been used:
Chain starting with 5 in B5 ending with 8 in A7 proves that B7=9
Chain starting with 2 in A6 ending with 6 in E7 proves that A7=8
(Note: Using "simple" methods, the 3 underlined numbers in column I can be removed as choices.)
I can't find any more answers!! If YOU can, please let me know how.
STUCK
PB Post 2014 02 15 Lvl 6
Fast puzzle except for one chain needed with 28 remaining:
The pink boxes in the puzzle above should fill in without "chains" or "colors." Then, I got (temporarily) stuck with 28 remaining, as shown in the next puzzle.
PB Post 2014 02 15 w28 Same as above, except the pink boxes are filled in.
(Note: Using "simple" methods, the 8 underlined numbers can be removed as choices.)
To proceed, I found a chain beginning with 5 in C1. Then D1 D7 B7 B2 B6 A6. This proves that B2 cannot contain 3 and so it must equal 6. The rest was easy.
OR... you could have found a chain beginning with the 3 in C1. Then A2, I2, B2, B7. This proved that C7 and C9 cannot contain 5. Therefore C9=6.
OR... you could have found a chain beginning with the 3 in C1. Then A2, I2, B2, B7. This proved that C7 and C9 cannot contain 5. Therefore C9=6.
PB Post 2014 02 20 Lvl 4 (6) (Repeated lower down)
Listed as Level 4, but I can't solve.
PB Post 2014 02 20 w24 Same as above, except the blue boxes are filled in.
(Note: Using "simple" methods, the two underlined 8's can be removed as choices.)
I can't get any more answer to this puzzle (2014 02 20)!
STUCK
PB Post 2014 02 22 Lvl 6 (Repeated lower down)
PB Post 2014 02 22 w22 Same as above, except the gray boxes are filled in.
(Note: Using "simple" methods, the 8 underlined numbers can be removed as choices.)
To proceed, I found a chain beginning with 3 in A2. Then G2, H3, H7. This proves that A7 cannot contain 6, and must equal 9. Finis!
PB Post 2014 03 01 Lvl 6 (Repeated lower down)
Many pairs were required to get the gray boxes filled in.
PB Post 2014 03 01 w17 Same as above, except the gray boxes are filled in.
To proceed, I found a chain beginning with 6 in B2. Then H2, H6, G4. This proves that G2 cannot contain 4, and must equal 9. Finis!
PB Post 2014 03 02 SundayLvl 5 (6) (Repeated lower down)
Labeled a "5" but I needed a chain to solve it.
PB Post 2014 03 02 Sunday Lvl 5(6) w33 Same as above, except the gray boxes are filled in.
Note: Using "simple" methods, the 10 underlined numbers (in E123 G1 I1 G789 H8) can be removed as choices.
To proceed, I first found - but perhaps not necessary - that multicolors with 2 (beginning at E1 and at H2) prove that E8 cannot equal 2.
The clincher was a chain starting with 9 in C3. Then B3 B1 C1 I1 I6 I7 F7 and ending F2. This proved that E3 and F3 cannot be 6. (Be sure to cross out both!!) Therefore, F3 = 1. FINIS.
PB Post 2014 03 08 Lvl 6 (Repeated lower down)
This level 6 OPENS with 26 naked singles! Followed by 3 hidden singles! Then it gets harder.
PB Post 2014 03 08 Lvl 6 w 22 Same as above, except the pink boxes are filled in.
Note: Using "simple" methods, the 3 underlined numbers (in F1 & D3) can be removed as choices.
To proceed, I first found I could MULTICOLOR 7s beginning at A1 and at E1. It proves that C1, F1, D5, and F5 cannot be 7.
Chain starts with 8 in C3. Then I3 D3 D8 D5 and F6. This proves that C6 cannot be 7, and so C6=5.
Quick finish after that. Notice that D5 was required in this chain, so the previous multicolor with 7 was probably necessary.
PB Post 2014 03 15 Lvl 6 (Repeated lower down)
PB Post 2014 Lvl 6 w Same as above, except the pink boxes are filled in.
Note: Using "simple" methods, the 3 underlined numbers (in F7, F8, & G4) can be removed as choices.
To proceed, a chain begins with 2 in C7. Then F7 and D9. This proves that C9 cannnot be 1, and therefore C9 = 9. FINIS!
PB Post 2014 03 22 Lvl 6 (Repeated lower down)
PB Post 2014 03 22 Lvl 6 w44 Same as above, except the blue boxes are filled in.
Note: Using "simple" methods (hidden pairs), the 5 underlined numbers -in Groups 6 and 8 - can be removed as choices.
To proceed, I found a chain beginning with 4 in A7. Then A5, C5, G5, H4, and H1. This proved that H7 cannot be a 1. Blech!! No help.
STUCK
PB Post 2014 03 29 Lvl 6 (Repeated lower down)
Nice puzzle because you ALMOST finish with simple techniques. With only 18 remaining, a chain is required.
PB Post 2014 03 29 Lvl 6 w18 Same as above, except the gray boxes are filled in.
Note: Using "simple" methods, the underlined 5's (G4 and I5) can be removed as choices.
To proceed, a chain was found, beginning with the 5 in B4. Then B9, G9, and G3. This proves that G4 cannot equal 1, and so G4 must equal 9. FINIS!
PB Post 2014 04 05 Lvl 6 (Repeated lower down)
PB Post 2014 04 05 Lvl 6 w39 Same as above, except the pink boxes are filled in.
STUCK
PB Post 2014 04 11 Lvl 5 (6) (Repeated lower down)
Some notes on author's "journey" to solve. Purposely used only naked and hidden singles until all the pink boxes were filled in. Wrongly believed that some simple techniques would finish the puzzle.
PB Post 2014 04 11 Lvl 5 (6) w 35 Same as above, except the pink boxes are filled in.
Note: Using "simple" methods, the underlined numbers can be removed as choices.
Here are explanations of how numbers became underlined (and thus deleted as possibilities):
Group 1 (row 1): single line of 7's in , causes their elimination in Group 3 (row 1).
Group 6 (row 4): single line of 9's in , causes their elimination in Group 5 (row 4).
Group 8: Hidden 2/8 pair allows eliminating 5's in E8 and F8
Column E: (Hidden) 2/8 pair allows eliminating 4 in E3.
"Simple" techniques failed me after those eliminations.
To proceed, the first chain I found was pretty long, but it solved:
Begin with the 6 in C2. Then C6, A6, A3, A9, G9, and I7. This proves that C7 cannot equal 4. That resulted in determining I7 = 4.
No major obstacles to FINIS.
PB Post 2014 04 12 Lvl 6 (Repeated lower down)
PB Post 2014 04 12 Lvl 6 w31 Same as above, except the grayboxes are filled in.
Note: Using "simple" methods, the underlined numbers can be removed as choices.
To proceed, I used "coloring" of 4's, beginning at B1. Even though "multi-coloring" might be applied, the single coloring was enough to reveal FOUR DIFFERENT boxes that must contain 4!!
Pretty easy after that to finish.
PB Post 2014 04 19 Lvl 6 (Repeated lower down)
PB Post 2014 04 19 Lvl 6 w23 Same as above, except the blue boxes are filled in.
GUESS A2 and look for contradiction. If none found, guess opposite just to be sure.
PB Post 2014 04 19 Lvl 6 w23 Same as above, except I GUESSED that A2 = 9.
I GUESSED that A2 = 9, then marked consequences in Groups 1 and 2.
From Group 1 I went down to Group 7.
From Group 2 I went down to Group 8.
The contradiction is the two 6's in row 8.
The LOCATION of the contradiction is not important. It varies depending on how the user moved around marking the consequences of the first guess.
Just to be SURE, the user should erase all the marks, then try the opposite guess. No contradictions should appear!
IF NEITHER GUESS GIVES A CONTRADICTION, THE METHOD FAILS.
PB Post 2014 04 26 Lvl 6 (Repeated lower down)
Kickoff: You can do all the 3s and all the 7's!
After filling in the pinks, simpler methods fail. Here's results:
PB Post 2014 04 26 Lvl 6 w27 Same as above, except the pink boxes are filled in.
I was able to solve by coloring 8's. I ASSUMED that F1 = 8. This eventually led to a conflict, proving - in fact - that G1 = 8, and also three other 8's were determined.
FINIS
I never tried chains, but did try the "GUESS" method with doubles. It worked:
PB Post 2014 04 26 Lvl 6 w27 Same as above, except the doubles are circled.
Interestingly, guessing that D2 = 5 or D2=8 produces no results, but guessing that C8=4 produces a conflict.
The conflict varies, depending upon the order of making conclusions. You might get that both D2 and G2 = 5. Or that D2 and D9=8, or other. No matter. Any conflict proves that C8 cannot equal 4, and so it must equal 5.
FINIS.
DETAILS FOR CONTRADICTIONS OF THIS PUZZLE ARE GIVEN ON THE "GUESS" PAGE.
PB Post 2014 05 03 Lvl 6 (Repeated lower down)
Here is an account of the order that I followed in solving. Users may have a different experience.
1. There were 6 naked singles.
2. Two hidden 1's, then a hidden 5.
3. In Column A, I used single-line elimination to cross out some 5s.
4. In Column E, single-line eliminations got rid of some 9s.
5. In Column F, 4/7 doubles allowed me to make several eliminations and prove that F1 = 8
6. Found a hidden 8
7. In Column D, 3/4 doubles means D4 is not 4; D4 = 5
8. In Column E, single-line eliminations of 4s in groups 2 & 8.
FINALLY - with 39, simple techniques failed me. See next puzzle:
PB Post 2014 05 03 Lvl 6 w39 Same as above, except the green boxes are filled in.
Note: Using "simple" methods, the 10 underlined numbers can be eliminated as choices.
To proceed, I colored 4's, beginning with B1. This PROVED that B1 and other could NOT equal 4; it revealed four boxes that MUST equal 4.
The rest were naked singles!
An alternate solution was found:
PB Post 2014 05 03 Lvl 6 w39 Same as above, except doubles are circled.
ALTERNATE STRATEGY: Circle doubles, then Guess A2 = 4:
Clashing results are shown lower down.("Clash" is good!)
Note: Using "simple" methods, the 10 underlined numbers can be removed as choices.
The order of proceeding gives different results, but if you assume that A2 = 4, you will eventually run into a conflict. Below is one example that illustrates this.
PB Post 2014 05 03 Lvl 6 w39 Same as above, except clash illustrated when assuming A2=4.
The path above results in two 4's in row 2. This PROVES that the assumption (A2=4) must be wrong! Therefore A2 must equal 3.
The remaining 38 boxes were filled using naked singles!
PB Post 2014 05 10 Lvl 6 or 7! (Repeated lower down)
Opened with seven naked singles.
Then, two 4's, a 5, three 7's, two 8's.
With 38, notice in Row 3: 5/8 pairs eliminate some choices.
Looking across groups 1-3, notice double-line elimination of 2's.
No more info from "simple" techniques.
PB Post 2014 05 10 Lvl 6 OR 7 w 38 Same as above, except the gray boxes are filled in.
Note: Using "simple" methods, the 7 underlined numbers (in groups 1 and 2 ) can be removed as choices.
Still with 38, coloring of 3 proved that H9 could not equal 3. Not too exciting!
Found a chain beginning with the 3 in B2. Then D2, D6, and finally A6. This proves that A1 and B5 cannot contain 9.
Another chain begins with the 3 in B2, then B7, A9, F9, E7, and finally E3. This proves F2 cannot equal 9, and so B2 = 9.
In row 1, note single line elimination of 3's proves that H3 = 3.
Down in groups 7-9, double-line elimination of 3's.
A few more answers came (pink boxes). then, stuck again without advanced technques.
PB Post 2014 05 10 Lvl 6 OR 7 w 33 (3RD SHOWING) PINK boxes are filled in.
Note: Using "simple" methods, the 5 underlined numbers (in groups 2 and 8 ) can be removed as choices.
With 33, A chain begins with the 2 in D6. Then G6, H5, H9, and finally H8. This proved that D8 cannot equal 1. Bleh!
Another chain started with 9 in I1. Then I7, B7, A9, A1, A6, and G6. This proves that G3 cannot be 2, and therefore G3 = 6.
Next, 5 naked singles were found. You'd think it would be over, but no!
I gave up looking for chains. I GUESSED that A1 = 1. It leads to conflicts, which proves that A1 cannot equal 1, and A1 = 3.
FINALLY...FINIS with little trouble.
For what it's worth, guessing would not have worked back when stuck with 38 remaining. Not sure about with 33.
PB Post 2014 05 16 Lvl 5 (6) (Repeated lower down)
PB Post 2014 05 16 Lvl 5 (6) w 38 Same as above, except the green boxes are filled in.
To proceed, note row 3 has a ?semi-naked? triple of 5,7,9. This allows you to cross those numbers out of the OTHER boxes in that row, leaving E3=6. Eventually, you can finish.
PB Post 2014 05 24 Lvl 6 (Repeated lower down)
Chronicle: Found three 3s, then a 5, 6;
PB Post 2014 05 24 Lvl 6 w 50 Same as above, except the blue boxes are filled in.
Note: Using "simple" methods, the 31 underlined numbers can be removed as choices. Details given below
With 50: Row 2: Horizontally, single-line elimination of 7s.
Row 9: 3/4 pair
Column E: 5/9 pair
Group 9: Horizontal double-line elimination of 2
Group 2: Vertically, double-line elimination of 4
Group 6: Triple 2/8/9
Group 7: Horizontally single-line elimination of 1
Group 9: Horizontal double-line elimination of 2
Row 4: 1/7 doubles
MULTICOLOR 9 beginning at B1 and at B8 proves that G2 cannot equal 9. NOT SURE IF THIS IS NECESSARY!!!
CHAIN beginning with 1 at A4. Then A9 and B8. This proves that B4, B5, and B6 cannot contain 7.
Group 7: Vetical double-line elimination of 7 proves that C8 = 2.
COLOR 8, beginning at D7, proves that D4 cannot equal 8, and so D4 = 4.
More answers came, stubbornly.
But one, (maybe two) chains still needed with 32 remaining:
PB Post 2014 05 24 Lvl 6 w 32 (THIRD showing) Same as above, except pink boxes are filled in.
Note: Using "simple" methods, the 3 underlined numbers can be removed as choices. Details were given above.
With 32 remaining, I first found this chain. Perhaps not needed:
Chain begins with 8 in B4. Then G4, I6, and I7. It proves that B7 cannot equal 5. Perhaps useless!
Then a chain begins with 5 in I1. It goes to I7 and ends at H8. This proves that H1 cannot be a 9, and therefore H1 = 5.
Eventually, FINIS
PB Post 2014 05 31 Lvl 6 (Repeated lower down)
PB Post 2014 05 31 Lvl 6 w34 Same as above, except the pink boxes are filled in.
Note: Using "simple" methods, the 8 underlined numbers can be removed as choices.
PB Post 2014 05 31 Lvl 6 w31 (3rd view) Same as above, except the blue boxes are filled in.
Note: Using "simple" methods, the 8 underlined numbers can be removed as choices.
To proceed, a chain was found starting with 9 in C2 and ending with 5 in I6. The results are that C6 cannot contain 5, and so must be 6. 30 naked singles followed! TWO MORE VIEWS BELOW SHOW (1) THE PUZZLE WITH SIMPLE ELIMINATIONS GONE AND DOUBLES CIRCLED. THEN (2) THE CHAIN
PB Post 2014 05 31 Lvl 6 w31 (4th view) Same as above, except "simple" eliminations made and doubles circled.
PB Post 2014 05 31 Lvl 6 w31 (5th view) Same as above, except one of the chain paths is shown.
At least two other chains might have been used:
Chain starting with 5 in B5 ending with 8 in A7 proves that B7=9
Chain starting with 2 in A6 ending with 6 in E7 proves that A7=8
PB Post 2014 06 07 Lvl 6 (Repeated lower down)
Remarkable puzzle because stuck for advanced methods with only 16 remaining.
PB Post 2014 Lvl 6 w 16 Same as above, except the pink boxes are filled in.
Note: Using "simple" methods, the 2 underlined 8s can be removed as choices. In both cases, single-line elimination along a row or column was adequate. Solution (next sentence) may not even require their removal this way.
To proceed, I colored 9s, beginning at A3. You end up proving that A3 CANNOT be a 9, and all four remaining 9s are revealed. The puzzle is soon solved.
PB Post 2014 06 14 Lvl 6 (Repeated lower down)
Opens with 24 naked singles.
Then in group 3, a 1, 5, and 8 are revealed.
Next... with 24, a level 5 skill is required. (See next view)
PB Post 2014 06 14 Lvl 6 w24 Same as above, except the blue boxes are filled in.
To proceed, in group 4 (and column C), the hidden 8/9 pairs allow eliminations, soon revealing two 7s and a 4.
Then, with 21 remaining - Color 2 to prove A7 cannot be 2.
In Column B, single-line elimination of 6.
Finally.. a chain starts with the 5 in A4. Then A7, D7, and C7. This proves that A8 and C5 cannot contain 2; A4=2, &C5 = 4.
Finish soon afterwards.
PB Post 2014 Lvl 6 (Repeated lower down)
PB Post 2014 Lvl 6 w Same as above, except the gray boxes are filled in.
Note: Using "simple" methods, the underlined numbers (in ) can be removed as choices.
To proceed,
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