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 Post subject: HS 24 X
PostPosted: Mon Apr 04, 2016 1:15 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Human Solvable 24 X

This started off as assassin 335 (I'll post 334 this week) but then I came across the HS bit.

Quite a few trials later: SS gives it 3.4 and JS uses 15 fishes.


After you solve it: adding the middle cage the SS score increases to 4.4 and JS still uses 10 fishes. With the HS JS uses about 6 fishes - in my solution I just used two.

Image

JS Code:
3x3:d:k:3856:3348:7689:7689:7957:10251:10251:3348:1295:3856:7689:7689:7957:7957:7953:10251:10251:1295:7689:7689:7957:7957:2585:7953:7953:10251:10251:7689:7957:7957:1051:2585:28:7953:7953:10251:7702:7702:4119:4119:29:2328:2328:7953:7953:10252:7702:7702:30:2842:1051:10002:10002:7690:10252:10252:7702:7702:2842:10002:10002:7690:7690:1294:10252:10252:7702:10002:10002:7690:7690:3853:1294:3859:10252:10252:10002:7690:7690:3859:3853:

Solution:
651427983
978635142
324819765
517396428
269784531
843251697
735968214
492173856
186542379


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 Post subject: Re: HS 24 X
PostPosted: Sun Apr 10, 2016 4:58 am 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Thanks HATMAN. Not sure if this is the HS you intended - but it worked for me :cheesey:

HS bit:
1. In n1 whatever is in r1c1 is not in 30(7)r1c3. Since the missing numbers from 30(7) are a +15(2) -> r2c1 not in 30(7)
-> Whatever is in r1c4 is in n1 in r3c3
-> Whatever is in r4c1 is in r1c2 (and is Min 4)

Similarly in n3
Whatever is in r1c6 is in n3 in r3c7 and whatever is in r4c9 is in n3 in r1c8

Similarly in n7
Whatever is in r9c4 is in n7 in r7c3 and whatever is in r6c1 is in n7 in r9c2

Similarly in n9
Whatever is in r9c6 is in n9 in r7c7 and whatever is in r6c9 is in n9 in r9c8

2. 4(3)n5 = {13}
16(2)r5 = {79}

!Innies - Outies r6789 -> r5c1 + r5c2 = r6c4 + r6c6 + 5
-> Given (79) already in r5 and (13) already in n5 there is no way to put whatever is in r6c6 into r5c12!

Suppose we try.
This would imply the other one in r5c12 is 5 more than r6c4.
But none of the pairs [61], [72], [83], [94] is possible.

-> Whichever of (13) is in r6c6 goes in n4 in r4c23
-> it can only go in n2 in r3c5
-> it cannot go anywhere in 39(7)r6c7
But since 39(7) must contain a 3
-> r6c6 = 1, r4c4 = 3
-> 1 in r4c12
-> r3c5 = 1
-> r4c5 = 9
-> 16(2)r5 = [97]
Also 1 in 31(7)r2c6 in r5c89

Straightforward from here


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 Post subject: Re: HS 24 X
PostPosted: Fri Apr 29, 2016 3:02 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for a great possible with lots of interesting interactions! :)

Wow! What a powerful logic chain by wellbeback! ;clapclap;

I also used the same initial HS steps, which I also assume were the intended ones. After that my solving path was very different. Some of my steps using multiple rows or columns also felt like HS; I'm not sure whether SudokuSolver can do that type of steps. Even though my solving path was longer, it didn't have any fish or forcing chains. I'm not formally rating my walkthrough but, if I did, I think it would be somewhere in the 1.5s for the multiple row/column steps.

Here is my walkthrough for Human Solvable 24X:
Prelims

a) R12C1 = {69/78}
b) R1C28 = {49/58/67}, no 1,2,3
c) R12C9 = {14/23}
d) R34C5 = {19/28/37/46}, no 5
e) R4C4 + R6C6 = {13}
f) R5C34 = {79}
g) R5C67 = {18/27/36/45}, no 9
h) R67C5 = {29/38/47/56}, no 1
i) R89C1 = {14/23}
j) R89C9 = {69/78}
k) R9C28 = {69/78}

Steps resulting from Prelims
1a. R4C4 + R6C6 = {13}, locked for N5 and D\, clean-up: no 7,9 in R3C5, no 6,8 in R5C7, no 8 in R7C5
1b. R5C34 = {79}, locked for R5, clean-up: no 2 in R5C67

[Now the first Human Solvable steps. I don’t often see them this early, or maybe I mean that they aren’t available so early.]
2. 45 rule on N1 2(1+1) outies R1C4 + R4C1 = 2(1+1) innies R1C2 + R3C3, R1C2 ‘sees’ R1C4 and all cells on N1 -> R1C2 = R4C1 = {456789}, R1C4 = R3C3, no 1,3 in R1C4

3. 45 rule on N3 2(1+1) outies R1C6 + R4C9 = 2(1+1) innies R1C8 + R3C7, R1C8 ‘sees’ R1C6 and all cells of N3 -> R1C8 = R4C9 = {456789}, R1C6 = R3C3

4. 45 rule on N7 2(1+1) outies R6C1 + R9C4 = 2(1+1) innies R7C3 + R9C2, R9C2 ‘sees’ R9C4 and all cells of N7 -> R9C2 = R6C1 = {6789}, R7C3 = R9C4

5. 45 rule on N9 2(1+1) outies R6C9 + R9C6 = 2(1+1) innies R7C7 + R9C8, R9C8 ‘sees’ R9c6 and all cells of N9 -> R9C8 = R6C9 = {6789}, R7C3 = R9C6, no 1,3 in R9C6

6. 45 rule on R1234 2 innies R4C46 = 2 outies R5C89 + 5, min R5C89 = 3 -> min R4C46 = 8 -> min R4C6 = 5
6a. Max R4C46 = 12 -> max R5C89 = 7, no 8 in R5C89

7. 45 rule on R6789 2 outies R5C12 = 2 innies R6C46 + 5, max R5C12 = 14 -> max R6C46 = 9, no 9 in R6C4

8. 31(7) cage at R2C6 must contain 7 in R23C6 + R34C7 + R4C8, CPE no 7 in R4C6
8a. 45 rule on whole grid 3 innies R4C6 + R5C5 + R6C4 = 16 = {259/268/457}
8b. 7 of {457} must be in R6C4 -> no 4 in R6C4

9. R1234 must contain four each of 1,2,3,4, 30(7) cage at R1C3 and 31(7) cage at R1C5 must each contain all of 1,2,3,4, 40(7) cage at R1C6 + R12C9 must contain all of 1,2,3,4, R34C5 must contain one of 1,2,3,4, R4C4 = {13} which leaves two more of 1,2,3,4 in R1234, 31(7) cage at R2C6 must contain all of 1,2,3,4 -> two of them must be in R1234 with R5C89 = {1234} -> no 4 in R1C28, clean-up: no 9 in R1C28, no 4,9 in R4C1 (step 2), no 4,9 in R4C9 (step 3)

10. Similarly R6789 must contain four each of 6,7,8,9, 40(7) cage at R6C1 and 39(7) cage at R6C7 must contain all of 6,7,8,9, 30(7) cage at R6C9 + R89C9 must contain all of 6,7,8,9, R9C28 must contain two of 6,7,8,9, R67C5 must contain one of 6,7,8,9 and 30(7) cage at R5C1 must contain one of 7,9 which must be in R6789 -> no 6,8 in R6C23 + R7C34 + R8C4, no 6,7,8 in R6C4, clean-up: no 6,8 in R9C4 (step 4)
10a. 30(7) cage at R5C1 contains one of 7,9 in R678 -> must contain one of 6,8 in R5C12
10b. R5C12 = R6C46 + 5 (step 7), R6C46 = [21/23/51/53] = 3,5,6,8 -> R5C12 = 8,10,11,13 = {26/28/46/38/56/58}, no 1
10c. 1 in R5 only in R5C789, locked for N6
10d. 1 in R5 only in R5C789, CPE no 1 in R3C7, clean-up: no 1 in R1C6 (step 3)
10e. 1 in R4 only in R4C234, CPE no 1 in R23C4

11. R4C6 + R5C5 + R6C4 (step 8a) = {259/268}, no 4, 2 locked for N5 and D/, clean-up: no 2 in R1C6 (step 3), no 3 in R2C9, no 8 in R3C5, no 9 in R7C5, no 3 in R8C1, no 2 in R9C4 (step 4)
11a. 9 of {259} must be in R4C6 -> no 5 in R4C6

12. 30(7) cage at R5C1 must contain both of 2,5
12a. R6C4 = {25} -> R6C23 + R7C34 + R8C4 cannot contain both of 2,5 (reverse CPE) -> R5C12 must contain one of 2,5, R5C12 must also contain one of 6,8 (step 10a) -> R5C12 (step 10b) = {26/28/56/58}, no 3,4
12b. 3 in R5 only in R5C789, locked for N6
12c. 3 in R5 only in R5C789, CPE no 3 in R3C7, clean-up: no 3 in R1C6 (step 3)
12d. 3 in R4 only in R4C234, CPE no 3 in R23C4

13. 40(7) cage at R1C6 + R12C9 contain 2, 31(7) cage at R2C6 and 30(7) cage at R6C9 each contain 2 -> 39(7) cage at R6C7 must contain 2 in C678 to give 2 in all four columns of C6789 -> no 2 in R89C5
13a. 39(7) cage contains 2 = {2346789}, no 1,5, clean-up: no 5 in R9C6 (step 5)
13b. 1 in C5 only in R123C5, locked for C2

14. 1 in 31(7) cage at R2C6 only in R5C89, locked for R5, clean-up: no 8 in R5C6
14a. Hidden killer pair 3,4 in R5C67 and R5C89 for R5, R5C67 contains one of 3,4 -> R5C89 must contain one of 3,4 -> R5C89 = {13/14}, no 2

15. 30(7) cage at R1C3 and 30(7) cage at R5C1 contain 4, 40(7) cage at R6C1 + R89C1 contain 4 -> 31(7) cage at R1C5 must contain 4 in C234 to give 4 in all four columns of C1234 -> no 4 in R12C5

16. 30(7) cage at R1C3 and 31(7) cage at R1C5 contain 2, 40(7) cage at R1C6 + R12C9 contain 2, 31(7) cage at R2C6 contains 2 in R234 -> no 2 in R3C5, clean-up: no 8 in R4C5

[I ought to have seen this after step 13, but I was still looking for steps like the earlier ones …]
17. R6C6 = 1 (hidden single in C6) -> R4C4 = 3, clean-up: no 3 in R7C3 (step 4)
17a. 31(7) cage at R1C5 must contain 3 in R12C5, locked for C5 and N2, clean-up: no 7 in R4C5, no 8 in R6C5
17b. 31(7) cage at R2C6 must contain 1,3 -> R5C89 = {13}, 3 locked for R5, clean-up: no 6 in R5C6
17c. Naked pair {45} in R5C67, locked for R5
17d. R5C89 = {13} = 4 -> R4C46 = 9 (step 6) = [36], 6 placed for D/, clean-up: no 6 in R1C2 (step 2), no 6 in R1C8 (step 3), no 7 in R1C28, no 4 in R3C5, no 7 in R4C1 (step 2), no 7 in R4C9 (step 3), no 5 in R7C5, no 6 in R7C7 (step 5)

18. R4C6 + R5C5 + R6C4 = 16 (step 8a), R4C6 = 6 -> R5C5 + R6C4 = 10 = [82], 8 placed for both diagonals, clean-up: no 8 in R1C4 (step 2), no 8 in R1C6 (step 3), no 7 in R2C1, no 2 in R3C3 (step 2), no 7 in R8C9, no 8 in R9C6 (step 5)
18a. Naked pair {58} in R1C28, locked for R1, clean-up: no 5 in R3C3 (step 2), no 5 in R3C7 (step 3)
18b. Naked pair {58} in R4C19, locked for R4

19. 31(7) cage at R2C6 = {1234579} (only remaining combination), no 8, 5 locked for C6 and N2 -> R6C67 = [45], R4C19 = [58], R1C2 = 5 (hidden single in N1), R1C8 = 8, R4C5 = 9 -> R3C5 = 1, R5C34 = [97], R6C5 = 5 -> R7C5 = 6, partial clean-up: no 7 in R9C9
19a. Naked pair {69} in R89C9, locked for C9 and N9 -> R6C9 = 7, R9C8 = 7 -> R9C2 = 8, R6C1 = 8 (step 4), R89C5 = [74], partial clean-up: no 7 in R1C1
19b. Naked pair {69} in R12C1, locked for C1 and N1 -> R5C12 = [26]
19c. Clean-up: no 6,9 in R1C4 (step 2), no 7 in R3C3 (step 2), no 4 in R3C7 (step 3), no 4,7 in R7C3 (step 4), no 4 in R7C7 (step 5), no 1 in R8C1, no 3 in R9C1, no 9 in R9C6 (step 5)

20. R1C4 = 4, R3C3 = 4, placed for D\, R7C7 = 2, placed for D\, R9C6 = 2, clean-up: no 1 in R2C9
20a. R2C2 = 7, R3C12 = [32], R3C12 = [18], R1C9 = 3, placed for D/ -> R2C9 = 2, R9C1 = 1, placed for D/, R8C1 = 4
20b. R7C3 + R8C2 = [59], placed for D/, R9C37 = [63], R9C9 = 9, placed for D\

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


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