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 Post subject: Assassin 293
PostPosted: Fri May 30, 2014 9:12 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
A very quick first placement possible, then lots of places to explore. Enjoyed this as a nice challenge without having to go complex. SudokuSolve score, 1.45

Assassin 293: [edit after Andrew's post: some cages are diagonally connected which the coloured pic below shows clearly]: note that 1-9 cannot repeat on the diagonals
Image
Colour pic:
Image
code: paste into solver:
3x3:d:k:6656:5633:5633:5633:4866:4611:4611:4100:6405:2566:6656:5633:4866:4866:4611:4611:4100:6405:2566:3335:6656:4866:2824:2824:4100:6405:6405:4617:3335:3335:6656:4100:4100:3082:3082:2571:4617:1548:1548:6656:8973:3854:3854:3854:2571:4617:3343:3343:8973:5136:3601:3601:4882:4115:788:3343:8973:4885:5136:3601:4882:4882:4115:788:8973:4885:4885:5136:5398:2583:2583:4115:8973:3608:3608:3608:5136:5398:5398:5398:5398:
solution:
+-------+-------+-------+
| 5 8 7 | 4 9 3 | 2 6 1 |
| 6 9 3 | 1 2 8 | 5 4 7 |
| 4 2 1 | 7 6 5 | 3 9 8 |
+-------+-------+-------+
| 7 5 6 | 8 1 2 | 9 3 4 |
| 8 4 2 | 3 7 9 | 1 5 6 |
| 3 1 9 | 5 4 6 | 7 8 2 |
+-------+-------+-------+
| 2 3 8 | 6 5 1 | 4 7 9 |
| 1 6 4 | 9 3 7 | 8 2 5 |
| 9 7 5 | 2 8 4 | 6 1 3 |
+-------+-------+-------+
Cheers
Ed


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PostPosted: Sat May 31, 2014 8:08 am 
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Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
Thanks for this week's Assassin, Ed! It was quite hard to solve but that was more my fault since I had to start over twice because of some wrong moves.
A293 Walkthrough:
1. C1235+D/
a) 35(5) = {56789} locked for D/
b) 3(2) = {12} locked for C1+N7
c) 10(2) <> 8,9
d) 18(3) = 8{37/46} -> 8 locked for C1+N4 because 6{39/57} blocked by Killer pairs (36,67) of 10(2) and {459} blocked by Killer pair (45) of 6(2)
e) Innies C1 = 14(2) = {59} -> CPE: R9C9+R5C5 <> 5,9
f) Outies C6789 = 7(2): R3C5 <> 2,7,8,9; R4C5 <> 6
g) 16(5) = {12346} -> R1C8 = 6
h) Outies N4 = 5(2) = [14/23]

2. N3+C9
a) 25(4) = 9{178/358/457} <> 2 -> 9 locked for N3
b) 25(4): R23C9+R3C8 <> 1,3,4 since R1C9 = (134)
c) Innies+Outies C9: 6 = R3C8 - R9C9 -> R3C8 <> 5; R9C9 = (123)

3. R456+D/
a) Outies R6789 = 22(2+1) = [787]/[868] -> 8 locked for R5
b) 15(3) <> 6 because {456} blocked by Killer pair (45) of 6(2) and {267} blocked by R5C15 = (678)
c) Killer pairs (25,45) locked in 6(2) + 15(3) for R5
d) 2 locked in R2C8+R3C7+R4C6 @ D/ for 16(5)

4. R456+C9 !
a) ! Hidden Killer quad (1236) in 16(3) for C9 since 10(2) can only have one of (1236) -> 16(3) <> {457}
b) 16(3) <> 4 because {349} blocked by Killer quad (1234) in R19C1 + 10(2)
c) ! Consider placement of 4 in C9 -> 4 locked in R4C59 for R4:
- i) R1C9 = 4 -> R4C5 = 4 (HS @ 16(5))
- ii) R4C9 = 4
d) 12(2) <> 8
e) Killer pairs (35,59) locked in 13(3) + 12(2) for R4
f) 10(2) <> 3,7 since (37) is a Killer pair of 12(2); 10(2) = [19/46]
g) Naked pair (14) locked in R4C59 for R4
h) R4C6 = 2

5. C456+N3
a) Outies C6789 = 7(2) = [34/61]
b) 11(2): R3C6 = (58)
c) Innies+Outies N23: -1 = R4C56 - R1C4 -> R1C4 = (47)
d) 2 locked in 19(4) = {1279} @ N2 locked for N2 since other combos blocked by Killer pairs (35,36,58) of 11(2) and {2467} blocked by R1C4 = (47)
e) Innies+Outies N23: -3 = R4C5 - R1C4 -> R1C4 = 4, R4C5 = 1

6. C456
a) Outie C6789 = R3C5 = 6
b) Cage sum: R3C6 = 5
c) Innies C5 = 18(3) = {279} -> R5C5 = 7; 2,9 locked for C5+N2
d) R4C9 = 4 -> R5C9 = 6, R5C1 = 8, R4C1 = 7 -> R6C1 = 3, R3C1 = 4, R2C1 = 6

7. C123+N9
a) 13(3) @ R3 = {256} -> R3C2 = 2; 5,6 locked for R4+N4
b) 26(5) = {13589} -> R4C4 = 8; 1,5 locked for N1+D\
c) 22(4) = {3478} -> 3 locked for N1
d) 26(5) = {13589} -> R5C4 = 3; 9 locked for D\
e) R3C7 = 3, R2C8 = 4, R4C7 = 9, R4C8 = 3, R8C8 = 2 -> R8C7 = 8, R9C9 = 3, R1C9 = 1
f) 16(3) = {259} -> R6C9 = 2; 5,9 locked for C9+N9
g) Hidden Single: R7C3 = 8 @ D/, R7C8 = 7
h) 19(3) = {478} @ N9 -> R6C8 = 8, R7C7 = 4

8. Rest is Naked Singles.

Rating:
Easy 1.5. I used two Killer quads and a small forcing chain.


Last edited by Afmob on Wed Jun 04, 2014 5:22 am, edited 3 times in total.

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 Post subject: Re: Assassin 293
PostPosted: Sat May 31, 2014 5:10 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
I echo Afmob and thanks Ed. Once I got my key step it was not too difficult. Not as hard as the rating suggested.

Hidden Text:
1. 35(5)@r5c5 = {56789}
-> Remaining D/ = {1234}
16(5)@r1c8 = {12346}
-> Whatever goes in r1c9 (from 1234) also goes in r4c5
-> r1c8 = 6

2. 3(2)@c1 = {12}
Innies c1 -> r19c1 = +14 = {59} or {68}
-> 18(3)@n4 cannot be {567}
Also 6(2)@n4 prevents 18(3)@n4 being {459}
-> 5 in c1 can only be in r19c9 -> r19c9 = {59}

3. r1c9 = {59} -> r5c5 from (678) (D/ and D\)
Innies - Outies r6789 -> r5c5 = r6c1 + 4
Since 2 already in c1 -> r6c1 from (34) and r5c5 from (78)
-> Possibilties for 18(3)@n4 and r5c5 are [783][7] and [864][8]

4. r5c9?
Either (a) r5c15 = [87]
-> Innies r5 -> r5c49 = +9
-> Can only be {36} since 6(2)@r5 prevents it from being {45}

or (b) r5c15 = [68] and r6c1 = 4
-> 6(2)@r5 = {15}
-> Innies r5 -> r5c49 = +10 = {37}

-> r5c9 from (367)

5. -> 10(2)@n6 = [46] or {37}
The latter puts 12(2)@n6 = {48}
-> 4 placed in r4 and n6 in r4c789
-> (Since whatever goes in r4c5 also goes in r1c9) -> 4 placed in D/ and n3 in (r2c8, r3c7)

6. Innies - Outies c9 -> r3c8 = r9c9 + 6
-> r3c8 from (789) and r9c9 from (123)

7! Putting 10(2)@n6 = {37} leaves no place for 4 in c9!
-> 10(2)@n6 = [46]
-> 18(3)@n4 = [783] and r5c5 = 7 and r5c4 = 3
-> 12(2)@n6 = {39}
-> 15(3)@r5c6 = [9{15}]
-> 6(2)@n4 = {24}
Also 13(3)@r6c2 = [{19}3]
-> 13(3)@r3c2 = [2{56}]
Also r4c56 = {12}
-> r4c4 = 8

More simple clean up from here


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 Post subject: Re: Assassin 293
PostPosted: Mon Jun 02, 2014 9:39 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
My solution is quite different though in the same areas. Step 9 is the difference [edit: and steps 10 & 11. No one else used step 10 and Andrew found step 11 to be crucial to his WT also]. So, like the others found, bit of a one-trick pony though interesting to have such different "tricks" in the three walkthroughs.

Start to A293
22 steps:
[edit: thanks to Afmob and Andrew for some additions and corrections]Prelims courtesy of SudokuSolver
Cage 3(2) n7 - cells ={12}
Cage 6(2) n4 - cells only uses 1245
Cage 12(2) n6 - cells do not use 126
Cage 10(2) n1 - cells do not use 5
Cage 10(2) n9 - cells do not use 5
Cage 10(2) n6 - cells do not use 5
Cage 11(2) n2 - cells do not use 1
Cage 19(3) n69 - cells do not use 1
Cage 19(3) n78 - cells do not use 1
Cage 16(5) n35 - cells ={12346}
Cage 35(5) n57 - cells ={56789}

1. 35(5)r5c5 = {56789}: all locked for D/

2. "45" on D/: 2 outies r1c8+r4c5 - 6 = 1 innie r1c9
2a. r1c8 and r1c9 cannot be equal since they are in the same nonet -> the IOD of 6 cannot be in r4c5 (IOU)
2b. no 6 in r4c5
2c. 16(5)r1c8 must have 6 -> r1c8 = 6

3. 3(2)n7 = {12}: both locked for c1 and n7
3a. no 8,9 in 10(2)n1

4. 18(3)n4: {459} blocked by 6(2)n4 = 4/5
4a. {369/567} blocked by 10(2)n1 = 3/6;6/7
4b. = {378/468}(no 5,9)
4c. must have 8: locked for c1 and n4

5. Hidden pair {59} in c1 -> r19c1 = {59} only
5a. r5c5 see both of those cells through D\/ -> no 5,9 in r5c5 [edit: note that the same is true for r9c9 -> no 5,9 in r9c9. This is not essential for this optimised WT.]

6. "45" on r6789: 1 outie r5c5 = 1 innie r6c1 + 4
6a. = [73/84]

7. 18(3)r4c1 = {378/468}: can't have both 3 & 4 -> no 3,4 in r45c1

8. "45" on n4: 2 outies r37c2 = 5 = [14/23]

The key step
9. "45" on r123: 1 innie r3c2 + 12 = 4 outies r4c456+r5c4
9a. r3c2 = (12) -> 4 outies = 13/14
9b. Min. any 3 cells = 6 -> max. any one cell = 8 (no 9 in r45c4)

10. "45" on r5: 4 innies r5c1459 = 24 and must have two of 6,7,8 for r5c15 for r5 = {1689/2679/3678/4578}
10a. 9 in {1689/2679} must be in r5c9 -> no 1,2 in r5c9
10b. no 8,9 in r4c9
[Andrew pointed out that {4578} is blocked, see my step 21. However, I deliberately left it out to see if it was really necessary for this solution. Since step 21 shows it is necessary then a truly optimised solution would have included it here. Nice optimising Andrew!!]

11. 13(3)r3c2 must have 1 or 2 for r3c2 = {139/157/247/256}: ie, if it has 9 must also have 3
11a. 9 in r4 in 13(3) = {39} or 9 in 12(2)n6 = {39} -> 3 locked for r4
11b. no 7 in r5c9

12. 3 must be in 16(5)r1c8, only in n3: 3 locked for n3

13. 25(4)n3 = {1789/4579}(no 2)
13a. must have 7 & 9: both locked for n3

14. 2 on D/ only in 16(5)r1c8 -> no 2 in r4c5

15. "45" on n23: 1 innie r1c4 - 1 = 2 outies r4c56
15a. = [4]+[12]/[7]+[42]
15b. -> r4c6 = 2, r1c4 = (47)

16. "45" on c6789: 2 outies r34c5 = 7 = [61/34]
16a. r3c5 = (36) -> r3c6 = (58)

Missing clean-up from here
17. 2 in n2 must be in 19(4)
17a. but {2359/2368/2458} all blocked by 11(2)n2
17b. and {2467} blocked by r1c4 = (47)
17c. = {1279} only: all locked for n2
17d. r1c4 = 4
17e. -> r1c9 = 1 -> r4c5 = 1 (hidden single in 16(5) -> r3c5 = 6 (h7(2)r34c5), r3c6 = 5

18. Naked pair {38} in r12c6: both locked for c6 and not in r12c7
18a. Naked pair {34} in r2c8+r3c7: 4 locked for n3
18b. Naked pair {25} in r12c7: locked for c7 and 5 for n3

19. "45" on c9: 1 outie r3c8 - 6 = 1 innie r9c9 = [82/93]

20. 7 in n3 only in c9: locked for c9
20a. -> 10(2)n6 = {46} only: both locked for c9 and n6

21. h25(r5c1459): {4578} blocked by 6(2)n4 = 4/5
21a. = {3678} only: all locked for r5
21b. -> r45c9 = [46]; r5c4 = 3

This step would have been simpler if I'd seen Afmob's neat step 3a
22. [684] in 18(3)n4 blocked by IODr6789=-4: ie, two 8's in r5c15
22a. 18(3) = {78}[3] only combination
22b. -> r5c5 = 7 (IODr6789=-4); Placed for D/ and D\


Cracked.
Cheers
Ed


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 Post subject: Re: Assassin 293
PostPosted: Wed Jun 04, 2014 2:24 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin.

Ed wrote:
A very quick first placement possible, then lots of places to explore. Enjoyed this as a nice challenge without having to go complex.
Yes, there's a lot to explore. I got stuck but when I came back a couple of days later I quickly found my breakthrough steps. I didn't need anything complex.

Ed wrote:
My solution is quite different though in the same areas. Step 9 is the difference. So, like the others found, bit of a one-trick pony though interesting to have such different "tricks" in the three walkthroughs.
I was surprised to find that four of us found four different reasons for the first placement.

My solving path was more like Ed's than the other two.

It would have been better if this puzzle had been posted with the colour pic as the main diagram. It's much clearer than the white one, for those of us who set up our own coloured diagrams. I was surprised that I was able to set up the diagram with four colours, which didn't touch in any place; that's usually difficult to do when there are diagonally connected cages.

Here is my walkthrough for Assassin 293:
Prelims

a) R23C1 = {19/28/37/46}, no 5
b) R3C56 = {29/38/47/56}, no 1
c) R4C78 = {39/48/57}, no 1,2,6
d) R45C9 = {19/28/37/46}, no 5
e) R5C23 = {15/24}
f) R78C1 = {12}
g) R8C78 = {19/28/37/46}, no 5
h) 19(3) cage at R6C8 = {289/379/469/478/568}, no 1
i) 19(3) cage at R7C4 = {289/379/469/478/568}, no 1
j) 16(5) cage at R1C8 = {12346}
k) 35(5) cage at R5C5 = {56789}

Steps resulting from Prelims
1a. Naked pair {12} in R78C1, locked for C1 and N7, clean-up: no 8,9 in R23C1
1b. Naked quint {56789} in 35(5) cage at R5C5, locked for D/
1c. Min R9C23 = 7 -> max R9C4 = 7

2. Naked quad {1234} in R1C9 + R2C8 + R3C7 + R4C6, CPE no 1,2,3,4 in R1C8 -> R1C8 = 6, clean-up: no 4 in R8C7

3. 45 rule on C1 2 innies R19C1 = 14 = {59}/[86]
3a. 18(3) cage at R4C1 = {378/468} (cannot be {369/567} which clash with R19C1, cannot be {459} which clashes with R5C23), no 5,9, 8 locked for C1 and N4, clean-up: no 6 in R9C1
3b. Naked pair {59} in R19C1, CPE no 5,9 in R5C5 + R9C9 using diagonals

4. 45 rule on N4 2 outies R37C2 = 5 = [14/23]
4a. 13(3) cage at R3C2 = {139/157/247/256} (cannot be {346} because R3C2 only contains 1,2)
4b. R3C2 = {12} -> no 1,2 in R4C23
4c. Killer triple 3,4,5 in 13(3) cage, 18(3) cage at R4C1 and R5C23, locked for N4
4d. 13(3) cage at R6C2 = {139/247} (cannot be {346} because 3,4 only in R7C2), no 6
4e. 2 in N4 only in R5C23 = {24} or 13(3) cage at R6C2 = {247} -> 13(3) cage at R3C2 = {139/157/256} (cannot be {247} which clashes with R5C23 + 13(3) cage at R6C2, blocking cages), no 4
4f. 9 in N4 only in 13(3) cage at R3C2 = {139} or 13(3) cage at R6C2 = {139} (locking cages), CPE no 1 in R5C2, clean-up: no 5 in R5C3
4g. Min R9C23 = {35} = 8 (R9C23 cannot be {34} which clashes with R7C2) -> max R9C4 = 6

5. 45 rule on C6789 2 outies R34C5 = 7 = [34/43/52/61], clean-up: R3C6 = {5678}

6. 45 rule on R6789 1 outie R5C5 = 1 innie R6C1 + 4, R5C5 = {78}, R6C1 = {34}
6a. Killer pair 3,4 in R23C1 and R6C1, locked for C1

7. 45 rule on C1234 2 innies R23C4 = 1 outie R5C5 + 1
7a. Max R5C5 = 8 -> max R23C4 = 9, no 9 in R23C4

8. 25(4) cage at R1C9 = {1789/3589/4579}, no 2, 9 locked for N3
8a. R1C9 = {134} -> no 1,3,4 in R2C9 + R3C89
8b. 2 on D/ only in R2C8 + R3C7 + R4C6, locked for 16(4) cage at R1C8, no 2 in R4C5, clean-up: no 5 in R3C5 (step 5), no 6 in R3C6

9. 45 rule on N23 1 innie R1C4 = 2 outies R4C56 + 1
9a. Min R4C56 = 3 -> min R1C4 = 4
9b. Max R4C56 = 7 -> max R1C4 = 8

10. R5C5 + R6C1 = [73/84] (step 6)
10a. 18(3) cage at R4C1 (step 3a) = {378/468} = [783/864], no 6 in R4C1, no 7 in R5C1
10b. 18(3) cage + R5C5 = [7837/8648], 8 locked for R5, clean-up: no 2 in R4C9
[Afmob got a similar result with 45 rule on R6789 3(2+1) outies R45C1 + R5C5 = 22 …]
10c. 2 in R4 only in R4C46, locked for N5

11. 45 rule on R1234 2 outies R5C49 = 1 innie R4C1 + 2
11a. R4C1 = {78} -> R5C49 = 9,10
11b. Hidden killer pair 3,9 in R5C49 and 15(3) cage at R5C6 for R5, neither can contain both of 3,9 -> R5C49 and 15(3) must each contain one of 3,9
11c. R5C49 = 9,10 = {36/37/19}, no 2,4,5, 15(3) cage = {159/249/357}, no 6, clean-up: no 6,8 in R4C9
11d. R4C1 + R5C49 = [736]/8{37}/8{19} (cannot be [763] because R45C9 = [73] clashes with R4C1), no 6 in R5C4
[These combinations can be reduced further using a forcing chain, but I’ll leave that for now and hope to find something simpler.]

12. 45 rule on C9 1 outie R3C8 = 1 innie R9C9 + 6, R3C8 = {789}, R9C9 = {123}

13. 45 rule on R9 2 innies R9C15 = 1 outie R8C6 + 10
13a. Max R9C15 = 17 -> max R8C6 = 7
13b. Min R9C15 = 11, no 1 in R9C5

14. 45 rule on C1234 3 outies R125C5 = 18 = {189/279/378/567} (cannot be {369/459} because R5C5 only contains 7,8, cannot be {468} which clashes with R34C5), no 4

15. 45 rule on N2 3 innies R1C46 + R2C6 = 15 = {159/168/249/258/267/348/456} (cannot be {357} which clashes with R3C56)
15a. 7 of {267} must be in R1C4 -> no 7 in R12C6

16. 45 rule on N2 2 outies R12C7 = 1 innie R1C4 + 3, IOU no 3 in R2C7

17. 45 rule on R123 4 outies R4C456 + R5C4 = 1 innie R3C2 + 12
17a. Max R3C2 = 2 -> max R4C456 + R5C4 = 14, no 9 in R45C4, clean-up: no 1 in R5C9 (step 11c), no 9 in R4C9

[I was slow to spot the next two steps even though I’d used the 45 earlier; it’s more powerful now.]
18. 9 in R4 only in 13(3) cage at R3C2 = {139} or R4C78 = {39} -> 3 locked for R4 (locking cages), clean-up: no 4 in R3C5 (step 5), no 7 in R3C6, no 7 in R5C9
18a. 16(5) cage at R1C8 = {12346}, 3 locked for N3

19. R1C4 = R4C56 + 1 (step 9)
19a. R4C56 = [12/42] (cannot be {14} = 5 because no 6 in R1C4) -> R4C6 = 2
19b. R4C56 = [12/42] = 3,6 -> R1C4 = {47}
19c. 2 in N3 only in R12C7, locked for C7, clean-up: no 8 in R8C8
19d. Naked triple {134} in R1C9 + R2C8 + R3C7, locked for N3
[Cracked. The rest is straightforward.]

20. R1C46 + R2C6 (step 15) = {348/456} (cannot be {159/168} because no 1,5,6,8,9 in R1C4) -> R1C4 = 4, R12C6 = [38/56/83]
20a. 19(4) cage at R1C5 = {1279} (hidden quad in N2), 9 locked for C5
20b. R125C5 (step 14) contains 9 = {189/279}
20c. R5C5 = {78} -> no 7 in R12C5
20d. 19(4) = {1279}, 7 locked for C4, clean-up: no 3 in R5C9 (step 11d), no 7 in R4C9

21. R1C9 = 1, R4C9 = 4 -> R5C9 = 6, R4C5 = 1, R3C5 = 6 (step 5) -> R3C6 = 5, R5C4 = 3, R45C1 = [78], R6C1 = 3 (cage sum), R23C1 = [64], clean-up: no 7 in R3C8 (step 12), no 5,8 in R4C78
21a. R3C7 = 3, R2C8 = 4, R4C78 = [93], clean-up: no 6,7 in R8C7, no 1,7 in R8C8
21b. Naked pair {56} in R4C23, locked for R4 and N4 -> R4C4 = 8, placed for D\
21c. R4C23 = {56} -> R3C2 = 2 (cage sum), R5C2 = 4 -> R5C3 = 2, R7C2 = 3
21d. R5C5 = 7, placed for both diagonals
21e. Naked pair {29} in R12C5, locked for C5 and N2
21f. Naked pair {17} in R23C4, locked for C4

22. R1C9 = 1 -> 25(4) cage at R1C9 = {1789}, locked for N3, 7 also locked for C9
22a. Naked pair {25} in R12C7, locked for C7 -> 15(3) cage at R5C6 = [915], R8C7 = 8 -> R8C8 = 2

23. R6C7 = 7 -> R67C6 = 7 = [61], 6 placed for D/

and the rest is naked singles, without using the diagonals.

Rating Comment:
I'll rate my walkthrough for A293 at Easy 1.5. I used locking cages and blocking cages, which I always rate at least Easy 1.5. Maybe I could have written those steps as combined cages and given my walkthrough a lower rating, but I preferred to use the more elegant locking and blocking cages.


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 Post subject: 4 colour theorem
PostPosted: Wed Jun 04, 2014 8:37 am 
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Grand Master
Grand Master

Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
Andrew wrote:
I was surprised that I was able to set up the diagram with four colours, which didn't touch in any place; that's usually difficult to do when there are diagonally connected cages.

While it maybe hard to find such a colouring it is always doable (excluding remote cages) as the Four color theorem shows.


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