sjs34 wrote:

I agree that this puzzle was an unusually difficult iteration of its ilk. On the other hand, it was a subscriber-only puzzle. I would wonder what the numbers of solvers are for other subscriber-only Wednesday 7x7 difficult puzzles.

If expressed again as a percentage of solved puzzles compared to the number of solved easy 4x4's that day,

it's number 3 on the all time list:

1. 2019-04-10: 5.05% (45)

2. 2014-03-12: 5.25% (45)

3. 2019-07-24: 5.37% (45)

So it's up there

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1. e2 must be odd (parity)

2. c3 must be even (parity)

3. e1=4 (only cell in col. 4 with the value 4)

4. one of the cages d4,d5 or d7 must be 3,6, therefore none of those cages can be 6,3 (this would create a situation where there is more than 1 solution)

therefore d1=6

5. cg a1 =2&7

6. one of the cages d4,d5 or d7 must be 2,5 therefore e2=3

7. using parity rule in row 4 either a4 or a5 must be even (this seems of no benefit by itself, but it helps the next step)

8. columns 1&2 must have a total of 8 odd values,

with step 7 in mind we can count the odd values and we get 6 odd values and 1 cage (b2) that is unknown,

therefore cage b2 must have 2 odd values to give a total of 8 odd values in the first 2 columns

therefore cage b2 =5&7

9. there is a cross pair in cells a1,b1 & a2,b2 (value 7), remove 7 from all other cells in columns 1&2

10 from here it is all basic steps to solve the puzzle.

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The percentage below is compared to the number of 4x4 easy puzzles solved (to "normalize" the number somewhat).

rank. [date] [percentage] [nr times]

1. [2017-10-27] [7.6] [71]

2. [2019-07-05] [10.7] [88]

3. [2018-11-02] [11.66] [101]

4. [2017-06-02] [12.01] [97]

5. [2017-03-03] [12.12] [100]

6. [2017-12-15] [13.36] [125]

7. [2017-12-01] [13.49] [124]

8. [2017-10-13] [13.73] [125]

9. [2017-06-30] [13.91] [117]

10. [2018-03-02] [14.34] [133]

1. [2017-10-27] [7.6] [71]

2. [2019-07-05] [10.7] [88]

3. [2018-11-02] [11.66] [101]

4. [2017-06-02] [12.01] [97]

5. [2017-03-03] [12.12] [100]

6. [2017-12-15] [13.36] [125]

7. [2017-12-01] [13.49] [124]

8. [2017-10-13] [13.73] [125]

9. [2017-06-30] [13.91] [117]

10. [2018-03-02] [14.34] [133]

So it's up there alright

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paulv66 wrote:

Be interesting to see how many people in total solve it. To date there are only 77.

The total number of solvers for the previous four Fridays were 229, 233, 268 and 338 respectively.

The final tally is 98, which is significantly less than the previous 4 weeks. I presume this number doesn't include anyone who solved it outside the time limit (I.e. after the solution became available), although I guess someone who was determined enough might have kept going after the time limit was up.

Either way, it shows that I certainly wasn't the only one who struggled with this particular puzzle.

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I had been concentrating on solving one side of the puzzle, so I decided to switch over to the other side. Once I did this, the solution fell into place after a few false starts. Be interesting to see how many people in total solve it. To date there are only 77.

The total number of solvers for the previous four Fridays were 229, 233, 268 and 338 respectively.

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I don't know how long I've spent on this so far, but I think it must add up to several hours. I made several attempts at it yesterday and thought I had narrowed down the number of possibilities by a process of elimination. I went back to it this morning and thought I had finally cracked it until I got stuck again with a few cells to go.

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If anybody would like a more detailed description of the steps taken to solve the puzzle I am happy to provide more information.

d7=5, d8=3

cage g8(7/) = 1,7

>> f8=8

cage c6(72x) = 8,9

cage e4(324x) : row 5 in cage must have 9

>> i4=9

i9 ≠ 7 (row 6 has no 7)

i9 ≠ 1 (row 9 has no 4)

i9 ≠ 2 (cage a7 has no possible solution)

g9 ≠ 3 (row 8 has no 4)

a8 ≠ 2 (row 8 has no 9)

a9 ≠ 9 (col b has no 8)

b9 ≠ 9 (col b has no 8)

c9 ≠ 9 (row 2 has no 8)

>> h9=9

a8 ≠ 6 (row 9 has no 7)

>> a8=9

>> cage b8 = 5,6

cage e8(120x) must have 5 in row 9

>> i9 ≠ 5

i9 ≠ 3 (cage h6(13+) has no solution )

i9 ≠ 6 (row 2 has no 8)

>> i9=8

g9 ≠ 7 (cage c6 (72x) has no solution)

From here the puzzle can be completed easily with basic level steps.

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First (basic) eliminations/thoughts:

0 a) CD4 ≠ [8,9] (due to CD6 and “rule” of unique solution; it would already be eliminated after step 1) (later) though);

b) interaction between FG4 and FG6 –

c)

d) the right 1512× cage has combinations [3,7,8,9] and [4,6,7,9];

e) rule of parity says that three rows or columns in a 9×9 calcudoku must get an odd result (3 × 45). So if the right 1512× cage is [3,7,8,9], G4 = [3,7]; if is [4,6,7,9], G4 = [4,6,8];

f) the central 324× cage can by composed of digits 1, 2, 3, 4, 6 and 9.

1) (Im)possible combinations for the central 324× cage: [1,1,4,9,9], [1,1,6,6,9], [1,2,2,9,9], [1,2,3,6,9], [1,3,3,4,9], [1,3,3,6,6], [2,2,3,3,9], [2,3,3,3,6] and [3,3,3,3,4].

Of these, only two are valid: [1,2,3,6,9] and [1,3,3,4,9] (number 4 can only be in E5!). It means that the column E contains 1 in E4 or E6, the row 5 contains 9 in D5 or F5. Thus 9 in the right 1512× cage must be placed in the row 4 –

2) Needed a long chain on 7 in H6, but it was very useful.

If H6 = 7 → H8 = 1, B7 = 7 (due to no other place for the third 7 in lower three rows), B6 = 3, G6 = 1, E6 = 2, A6 = 5, I6 = 6, H5 = 4, DEF5 = [3,6,9], AB5 = [2,8], A4 = 6; H5678 = [1,4,6,7], thus H34 = [2,3].

Now the sum of known numbers in the columns G, H and I is 117, so G2 + H2 + G4 = 18. G4 ≠ 6; if G4 = 8, GH2 is blocked; if G4 = 4, GH2 = [6,8].

This with three 7's in the columns A, B and C means that BC2 = [2,9]. The “rule” of unique solution is again on the scene: BC8 ≠ [2,9] = [5,6].

Now factorial counting the three left columns: 9!^3 ÷ (cages by rows: 2160 × 18 × 7 × 480 × 21 × 1512 × 3 × 30) = 128.

So the cells B3, B4, C4 and C6 must get product of 128 = 2^7; C6 = 8, C4 = 2, B34 = product of 8, which is impossible →

It gives

3) If B6 = 2 → H6 = 6, FG6 = [1,4], A6 = 5, E6 = 3, I6 = 7; if HI5 = [4.6], the 324× cage is blocked; if [3,8], DEF5 = [2,6,9], AB5 = [4,7], A4 = 5. A4 = A6 – contradiction, therefore

4) If H6 = 6 → G8 = 7, G6 = 3; so G4 is even and the right 1512× cage must be [4,6,7,9]; H5 = 4, H8 = 1, H34 = again [2,3]. And again the sum of the columns G, H and I: this time 119, so G2 + H2 + G4 = 16.

All possibilities of numbers 5, 8, and 9 in H2 lead to contradiction, thus

5) Another long chain:

If the right 1512× cage = [3,7,8,9] → in cooperation with B5 = [3,7,8], DEF5 must be [2,6,9], A5 = 4; E46 = [1,3], therefore HI5 = [3,8], B5 = 7, I6 = 7, A6 = [2,4,6]; A4 + A6 = 10 and A5 = 4, thus A6 = 2, A4 = 8; G6 = 1, G8 = 7, G4 = 3 (rule of parity), F4 = 2, E4 = 1. But now exists no number for B4, so

6) a)

b)

c)

d) I789 = min. 15 → A)

e) H5 = [6,7] →

7) If E5 = 4 → DF5 = [3,9], HI5 = [6,7], AB5 = [2,8], A4 + A6 = 11, which makes combination [4,7] here; E8 = 6, A8 = 4. A4 = A8 – contradiction, therefore

8) If FG6 = [1,4] → B6 = 6, HI5 = [4,6], but it forces numbers 1 and 6 to one cell (E4).

9) Due to the 324× cage,

10) At the end the rule of parity helps again:

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