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9 x 9 Tuesday May 3rd https://www.calcudoku.org/forum/viewtopic.php?f=16&t=785 
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Author:  eclipsegirl [ Fri May 06, 2016 4:32 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
A question for firefly (or anyone else) At step 6) you state that 8 can not be in 180x or 168x, and therefore must be in the bottom row, in the 38+ Can you explain to me why 8 is eliminated in the 168x? Thank you for your help. ETA : CLM's earlier post explained it. Sorry to disturb 
Author:  firefly [ Fri May 06, 2016 8:28 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
eclipsegirl wrote: A question for firefly (or anyone else) At step 6) you state that 8 can not be in 180x or 168x, and therefore must be in the bottom row, in the 38+ Can you explain to me why 8 is eliminated in the 168x? Thank you for your help. ETA : CLM's earlier post explained it. Sorry to disturb Oops, yes, I goofed a bit there. I didn't think to note down my original solve order (which was more like 340 steps instead of 34 steps), so the solution post was made by backtracking my solution afterthefact. I'm afraid I piggybacked that first part of the solution a bit off the earlier posts, because I couldn't initially remember exactly how I'd moved past that first part. Honestly, now that I think about it a bit, I don't think I actually solved it quite in that order. I figured out that there must be two nines in the 38+ first, which led me to the 10+ being [3,7], which led me to the fact that the 38+ couldn't have two 7s (due to the 10+ and 168x), which led me to the fact that the 38+ had to have one 8, but obviously couldn't have two. When I was drafting my post, I was trying to clean up and shorten any logic chains to make it more easy to follow, so I think I just accidentally shortcutted to the "must have an 8" part without explaining the process. In my attempt to clean up the logic, I was attempting to focus on the 38+ first and pull the 10+ out to after, but I guess I should have addressed the 10+ in the middle. I will correct my solution post. 
Author:  jaek [ Fri May 06, 2016 9:33 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
firefly wrote: Oops, yes, I goofed a bit there. I didn't think to note down my original solve order (which was more like 340 steps instead of 34 steps), so the solution post was made by backtracking my solution afterthefact. I'm afraid I piggybacked that first part of the solution a bit off the earlier posts, because I couldn't initially remember exactly how I'd moved past that first part. Honestly, now that I think about it a bit, I don't think I actually solved it quite in that order. I figured out that there must be two nines in the 38+ first, which led me to the 10+ being [3,7], which led me to the fact that the 38+ couldn't have two 7s (due to the 10+ and 168x), which led me to the fact that the 38+ had to have one 8, but obviously couldn't have two. When I was drafting my post, I was trying to clean up and shorten any logic chains to make it more easy to follow, so I think I just accidentally shortcutted to the "must have an 8" part without explaining the process. In my attempt to clean up the logic, I was attempting to focus on the 38+ first and pull the 10+ out to after, but I guess I should have addressed the 10+ in the middle. I will correct my solution post. This was still a lovely and well thought out analysis. Thanks for putting in all that hard work. It's much appreciated. 
Author:  clm [ Fri May 06, 2016 11:42 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
firefly wrote: Here's a Full Analytical Solution (FAS) for this puzzle, for those who are interested: ... Steps 1824 were clearly the most difficult part to analyze of the whole puzzle. There were other ways to go at that point, but the ones I saw required too much TAE for my liking. Perhaps clm came up with a more elegant way of getting through that middle section. He mentioned parity in the previous post, and while I toyed with parity during my solve, I ultimately never really used it. So perhaps he was able to see something I missed. … Thank you, firefly, for this FAS, these solutions are really appreciated and they represent a great effort. Perhaps the next days (depending on available time) we may continue developing other sections, but before I would like to comment something more about the three bottom rows and why I sometimes use the parity rule focusing, for instance, in these three bottom rows. From your step 10, instead of identifying/separating cells a7a8e8i7i8 by factoring out the bottom row, I follow a different way: I factorize the full three bottom rows in order to obtain only the product of the four cells b7h7g8h8, this product is: (9!)^3 / (5*7*8*9*9) / 8 / 6 / 21 / 4 / 168 / 18 / 8 / 180 = 120, so we have four combinations: 5324, 5381, 5622 and 5641, and taking into account that g8h8 must be consecutive numbers it’s easy to see that 5381, 5622 and 5641 are not valid, this last because 45 in g8h8 force a second 6 in row 7 and 56 force a second 4 in row 7, so we are left with [2,3,4,5] (with a sum of 14) and this means that h7 = 2 (>>> f7 = 9, f8 = 2, a8 = 9); then, as in your solution, b7 is either 3 or 5 and g8h8 is [4,5] or [3,4] with a 4 certainly in that cage “1”. At the same time a double 2 in the three bottom rows >>> 53322 is not a valid combination for “180x” (there are already two 2’s in the three bottom rows). Considering that “180x” has six possible combinations: 53322, 53341, 53621, 56611, 59221 and 59411 (the last three automatically eliminated due to the double 1 or the 9) we are only left with 53341 (which sum is 16) and 53621 (which sum is 17). Cages “168x” and “180x” have a sum of 35 (the difference between the total 135 of the three bottom rows and the rest of cages found), so if “168x” is even, “180x” must be odd = [1,2,3,5,6] and necessarily “168x” = [1,4,6,7] (which sum is 18), here, for instance, the parity rule has been applied in a very simple way, though it can be argued that this could have been seen directly. However, the parity rule will help again thru the solution, for instance, being “180x” odd >>> g4 + g6 is even, … , 
Author:  firefly [ Sat May 07, 2016 1:09 am ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
clm wrote: From your step 10, instead of identifying/separating cells a7a8e8i7i8 by factoring out the bottom row, I follow a different way: I factorize the full three bottom rows in order to obtain only the product of the four cells b7h7g8h8, this product is: (9!)^3 / (5*7*8*9*9) / 8 / 6 / 21 / 4 / 168 / 18 / 8 / 180 = 120, so we have four combinations: 5324, 5381, 5622 and 5641, and taking into account that g8h8 must be consecutive numbers it’s easy to see that 5381, 5622 and 5641 are not valid, this last because 45 in g8h8 force a second 6 in row 7 and 56 force a second 4 in row 7, so we are left with [2,3,4,5] (with a sum of 14) and this means that h7 = 2 (>>> f7 = 9, f8 = 2, a8 = 9); then, as in your solution, b7 is either 3 or 5 and g8h8 is [4,5] or [3,4] with a 4 certainly in that cage “1”. Two different (though similar) ways of getting to the same information so far. The next part is the part that differs. clm wrote: At the same time a double 2 in the three bottom rows >>> 53322 is not a valid combination for “180x” (there are already two 2’s in the three bottom rows). Considering that “180x” has six possible combinations: 53322, 53341, 53621, 56611, 59221 and 59411 (the last three automatically eliminated due to the double 1 or the 9) we are only left with 53341 (which sum is 16) and 53621 (which sum is 17). Cages “168x” and “180x” have a sum of 35 (the difference between the total 135 of the three bottom rows and the rest of cages found), so if “168x” is even, “180x” must be odd = [1,2,3,5,6] and necessarily “168x” = [1,4,6,7] (which sum is 18), here, for instance, the parity rule has been applied in a very simple way, though it can be argued that this could have been seen directly. Indeed, the use of parity here seems to me to be the long way around, since you could have just simply eliminated the [5,3,3,4,1] option for 180x at the same time you discarded the others due to the fact that two 3s were already spoken for in rows 79 (one in the 10+ at 7d and the other in the [7b,8g,8h] leftover set you identified in your previous step). clm wrote: However, the parity rule will help again thru the solution, for instance, being “180x” odd >>> g4 + g6 is even, … , I'm definitely more curious to see if you used parity in the following steps as an alternative to my somewhat bruteforce approach in steps 1824. To be honest, I don't end up using parity all that often (except on certain puzzle types, like the all minus cage puzzles, for example). I find that parity is great for single unknown blocks, like a single multiplication cage in an otherwiseknown parity set, but when it starts extending to two or more unknown blocks, it starts to feel more like a TAE method than hard logic. I'm definitely doing parity a bit of a disservice by thinking that, but most (not all!) situations in which parity is useful can be solved in the same or fewer steps by using other rules of logic, which is what I would typically consider more "elegant". But that's just the snob in me. :P 
Author:  clm [ Sat May 07, 2016 8:58 am ] 
Post subject:  9 x 9 August 18, 2015 
firefly wrote: ... Indeed, the use of parity here seems to me to be the long way around, since you could have just simply eliminated the [5,3,3,4,1] option for 180x at the same time you discarded the others due to the fact that two 3s were already spoken for in rows 79 (one in the 10+ at 7d and the other in the [7b,8g,8h] leftover set you identified in your previous step). Correct. Let's say that I was simply "introducing" the parity in this FAS . firefly wrote: clm wrote: However, the parity rule will help again thru the solution, for instance, being “180x” odd >>> g4 + g6 is even, … , I'm definitely more curious to see if you used parity in the following steps as an alternative to my somewhat bruteforce approach in steps 1824. To be honest, I don't end up using parity all that often (except on certain puzzle types, like the all minus cage puzzles, for example). I find that parity is great for single unknown blocks, like a single multiplication cage in an otherwiseknown parity set, but when it starts extending to two or more unknown blocks, it starts to feel more like a TAE method than hard logic. I'm definitely doing parity a bit of a disservice by thinking that, but most (not all!) situations in which parity is useful can be solved in the same or fewer steps by using other rules of logic, which is what I would typically consider more "elegant". But that's just the snob in me. :P In order to convince you of the wonderful and powerful use of the parity rule, as well as its elegance, let's then go to practically finish the solution (specially steps 1824, ... ) with a very concise way (and few steps). First: Let’s observe the three rightmost columns: g4 + g6 + 19 + “1”(g2) + 1 + “2”(h3) + 7 + 28 + 8 + 3 + “1”(g8) + 17 (“180x”) = 135 that is: g4 + g6 + “1”(g2) + “2”(h3) + “1”(g8) = 52. To satisfy the parity in this equation, g4 + g6 must be even, so this sum must value 4, 6, 8, 10, 12, 14 or 16. Clearly 4 is not possible (g3 = 1) and 6 is not possible (a 2 or a 4 cann’t go in g4); 16 is not possible (g5 = 7) and 14 is not possible, since g7 = 8 >>> [5,9] but none of them can be in g6, so we are left with 10 and 12. Now the rule of “maximums and minimums in the sum of cages” means that, being the maximum of the three subtraction cages = 42 (maximum of each cage, that is: [8,9] + [7,9] + [4,5] = 42), g4 + g6 has a minimum value of 10, which is congruent with our previous result. But, if g4 + g6 = 12 >>> (g6 = 3, g4 = 9), the rest of the cages would sum 40, but now the maximum of “1”(g2) would be 13 ([6,7]), so: “1”(g8) = 40 – 13 – 16 = 11 which is impossible (it would require 56), then: g4 + g6 = 10 >>> g4 = 6, f4 = 8, g6 = 4, f6 = 3, c6 = 7, d6 = 5. And being g4 + g6 = 10, the rest of cages must sum 42, which can only be accomplished with the maximums for all cages, that is, 89, 79 and 45, so: g2 = 9, h2 = 8, h3 = 7, h4 = 9, g8 = 5, h8 = 4. And obviously, b7 = 3, b6 = 6, d7 = 7, d8 = 3, a7 = 5 and i7 = 1. Also d2 = 6, d3 = 8. Let’s finish the three rightmost columns. “28+” has the “brute force” combinations: 9982, 9973, 9964, 9955, 9883, 9874, 9865, 9775, 9766, 8875, 8866 and 8776. Double 9 is not possible (h4 = 9), doble 8 is not possible (h2 = 8), double 7 is not possible (g5 = 7 or h3 = 7), double 6 is not possible (two 6’s already in the three rightmost columns) and 9874 is not possible since none of them can be placed in h5, so "28+" = [5,6,8,9]. Finally, “19+” contains the rest of numbers in the three rightmost columns, so "19+" = [2,3,3,4,7]. Second: We will use again the parity rule to solve the three top rows. b3 is even. Beautyful. Why?: g3 = 1, b7 = 3, e3 = 5, h3 = 7 and c3 = 9, all odd numbers used. And being even, it can not be 4, because of b9 = 9, 6 (f3 = 6) or 8 (d3 = 8) so b3 = 2 (b4 = 7). Now, “240x” must be even, and having the five combinations: 53224 (which sum is 16), 53281 (19), 53441 (17), 56241 (18) and 56811 (21), we see that only 53224 and 56241 are even, but 53224 is impossible due to two 2’s already in the three rows (one in b3 and the other inside “19+”), then “240x” = [1,2,4,5,6]. Now “1”(b2) + “1”(f2) = 18 (135 – 18 – 2 – 9 – 25 – 14 – 5 – 19 – 17 – 1 – 7). This can only be satisfied (considering that no more 2’s in the three rows and that d2 = 6, g2 = 9, h2 = 8) with 34 + 56 (since 45 + 45 is impossible). Then: b2 = 4, c2 = 3, f2 = 5, f3 = 6. In this point I think the puzzle is finished, the rest of the puzzle is easy, straight forward. Still: c4 + c6 is odd , so c4 is even, and since it can not be 4 (in c5), 6(in g4) or 8 (in f4) >>> c4 = 2, d4 = 4 (or simply because c4 is the rest to 135 of the other cages and cells in the three leftmost columns ). Additionally, now, the central cross weights 17 so 36611 or 39221 (among a total of five possible combinations: 33621, 36611, 39221, 39411 and 69211) but two 6's are not possible there, furthermore no 6 is possible since there are three 6's in the three central columns, so "108x" = [1,2,2,3,9], with obviously the two 2's in d5 and e6 due to c4 = 2 and f8 = 2 (d5 = 2, e6 = 2). In summary, without the parity rule, everything is more complex and either a lot of TAE is required or many and more complicated hyphotesis are required, while with the parity rule we can solve this puzzle in a very reduced number of steps. And a final comment about the beauty of Calcudokus, I like those patterns where the big cages (4 cells or more) have all different sums or different products. 
Author:  beaker [ Sat May 07, 2016 8:06 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
I wonder what a twin of the Tuesday 9x9 would look like or is there not enough room to put them both on one page? 
Author:  firefly [ Sat May 07, 2016 9:22 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
clm wrote: To satisfy the parity in this equation, g4 + g6 must be even, so this sum must value 4, 6, 8, 10, 12, 14 or 16. Clearly 4 is not possible (g3 = 1) and 6 is not possible (a 2 or a 4 cann’t go in g4); 16 is not possible (g5 = 7) and 14 is not possible, since g7 = 8 >>> [5,9] but none of them can be in g6, so we are left with 10 and 12. Actually, you're also left with 8 (could be [6,2]), which would however be eliminated with the next step. clm wrote: Now the rule of “maximums and minimums in the sum of cages” means that, being the maximum of the three subtraction cages = 42 (maximum of each cage, that is: [8,9] + [7,9] + [4,5] = 42), g4 + g6 has a minimum value of 10, which is congruent with our previous result. But, if g4 + g6 = 12 >>> (g6 = 3, g4 = 9), the rest of the cages would sum 40, but now the maximum of “1”(g2) would be 13 ([6,7]), so: “1”(g8) = 40 – 13 – 16 = 11 which is impossible (it would require 56), then: g4 + g6 = 10 >>> g4 = 6, f4 = 8, g6 = 4, f6 = 3, c6 = 7, d6 = 5. Ah, but see, you could solve the same thing without the extra step of parity. The 14+ at 4f can't be [5,9] because that would change the max value of the 1 at 2g to [6,7] and the max value of the 2 at 3h to [8,6], and thus the minimum value of g6 to 7, which isn't possible. With [8,6] proved in the 14+ at 4f, then the only possible fit for 6g would be 4 in order to satisfy the minimum of 10. clm wrote: “28+” has the “brute force” combinations: 9982, 9973, 9964, 9955, 9883, 9874, 9865, 9775, 9766, 8875, 8866 and 8776. Double 9 is not possible (h4 = 9), doble 8 is not possible (h2 = 8), double 7 is not possible (g5 = 7 or h3 = 7), double 6 is not possible (two 6’s already in the three rightmost columns) and 9874 is not possible since none of them can be placed in h5, so "28+" = [5,6,8,9]. Finally, “19+” contains the rest of numbers in the three rightmost columns, so "19+" = [2,3,3,4,7]. I did roughly the same thing in my step 25, though it would be even more simple now that the 7, 8, and 9 are all taken in column h. And from here we're back on track with step 27+ of my solution. clm wrote: Second: We will use again the parity rule to solve the three top rows. b3 is even. Beautyful. Why?: g3 = 1, b7 = 3, e3 = 5, h3 = 7 and c3 = 9, all odd numbers used. And being even, it can not be 4, because of b9 = 9, 6 (f3 = 6) or 8 (d3 = 8) so b3 = 2 (b4 = 7). Once again, parity is an unnecessary step here. Like my step 28 says, the 5 at 3b must be [2,7] because the 6, 8, and 9 are all taken in column b. clm wrote: Now, “240x” must be even, and having the five combinations: 53224 (which sum is 16), 53281 (19), 53441 (17), 56241 (18) and 56811 (21), we see that only 53224 and 56241 are even, but 53224 is impossible due to two 2’s already in the three rows (one in b3 and the other inside “19+”), then “240x” = [1,2,4,5,6]. Now “1”(b2) + “1”(f2) = 18 (135 – 18 – 2 – 9 – 25 – 14 – 5 – 19 – 17 – 1 – 7). This can only be satisfied (considering that no more 2’s in the three rows and that d2 = 6, g2 = 9, h2 = 8) with 34 + 56 (since 45 + 45 is impossible). Then: b2 = 4, c2 = 3, f2 = 5, f3 = 6. c4 + c6 is odd , so c4 is even, and since it can not be 4 (in c5), 6(in g4) or 8 (in f4) >>> c4 = 2, d4 = 4 (or simply because c4 is the rest to 135 of the other cages and cells in the three leftmost columns ). Additionally, now, the central cross weights 17 so 36611 or 39221 (among a total of five possible combinations: 33621, 36611, 39221, 39411 and 69211) but two 6's are not possible there, furthermore no 6 is possible since there are three 6's in the three central columns, so "108x" = [1,2,2,3,9], with obviously the two 2's in d5 and e6 due to c4 = 2 and f8 = 2 (d5 = 2, e6 = 2). Meh I'd say that my steps 29+ are as simple as you can get, really. And even after I leave off in step 34, it's just as simple: 1b=1 ∴ 2b=4 ∴ 2c=3 ∴ 4c=2 ∴ 4d=4 ∴ 9d=1 ∴ 9e=6 ∴ 9f=4 ∴ 3f=6 ∴ 3a=4 & 5f=1. Then 1d=9 ∴ 5d=2 & 1e=8 ∴ 1a=6 & 2e=1 ∴ 2a=2 & 4e=3 ∴ 4a=1 & 5e=9 ∴ 6e=2 & 5a=3 & 5i=8 ∴ 6i=9 ∴ 6a=8. clm wrote: In summary, without the parity rule, everything is more complex and either a lot of TAE is required or many and more complicated hyphotesis are required, while with the parity rule we can solve this puzzle in a very reduced number of steps. See, and I think I've shown that the easiest (and therefore most elegant?) solution will almost never incorporate the use of parity. Now, I should clarify that I quite often attempt to use parity as an aide to determining what the best step is (though not with any terribly frequent success), but once I've found that step, it almost always turns out that parity is an unnecessary step in explaining why that was the right step. Does that make more sense? The use of parity may be instrumental to one's solving the puzzle, but not in the solution to the puzzle, if you catch the distinction. I'll admit that this is part of what makes me use parity as one of the last options that I check, along with other multipart brute force eliminations. All this aside, however, clm did provide a somewhat shorter workaround to the method I used in steps 1824, even if he didn't actually need to use parity to do so. I'll probably edit my solution post with this shortened approach. 
Author:  clm [ Sat May 07, 2016 10:36 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
firefly wrote: clm wrote: To satisfy the parity in this equation, g4 + g6 must be even, so this sum must value 4, 6, 8, 10, 12, 14 or 16. Clearly 4 is not possible (g3 = 1) and 6 is not possible (a 2 or a 4 cann’t go in g4); 16 is not possible (g5 = 7) and 14 is not possible, since g7 = 8 >>> [5,9] but none of them can be in g6, so we are left with 10 and 12. Actually, you're also left with 8 (could be [6,2]), which would however be eliminated with the next step. Yes, thank you, that should say "we are left with 8, 10 and 12", the 8 is eliminated in the next step with the rule of "the use of maximums and minimums in the sum of cages". firefly wrote: clm wrote: In summary, without the parity rule, everything is more complex and either a lot of TAE is required or many and more complicated hyphotesis are required, while with the parity rule we can solve this puzzle in a very reduced number of steps. See, and I think I've shown that the easiest (and therefore most elegant?) solution will almost never incorporate the use of parity. Now, I should clarify that I quite often attempt to use parity as an aide to determining what the best step is (though not with any terribly frequent success), but once I've found that step, it almost always turns out that parity is an unnecessary step in explaining why that was the right step. Does that make more sense? The use of parity may be instrumental to one's solving the puzzle, but not in the solution to the puzzle, if you catch the distinction. I'll admit that this is part of what makes me use parity as one of the last options that I check, along with other multipart brute force eliminations. All this aside, however, clm did provide a somewhat shorter workaround to the method I used in steps 1824, even if he didn't actually need to use parity to do so. I'll probably edit my solution post with this shortened approach. Your solution is enough clear and elegant, though the developper always thinks that something can be improved, but that requires a lot of additional effort, specially if there are graphics, etc. The parity: I understand what you say, sure, it's only a tool, but a useful tool, the main idea of the analysis is to use in each moment the appropriate (or preferred) tool in order to remove possibilities reducing the options so simplifying the way to the solution: parity rule, maximums and minimums in the sum of cages, segmentation, addition, factorization, location of determined prime numbers like 5, 7, 11, ... . None of these rules by itself gives the solution, of course, that is, the final position of the digits, but we should use them, instead of only TAE (well, perhaps some players prefer TAE simply to endjoy their time, it's a respectable option), according to my preferred "theorem": "If a Calcudoku has a unique solution then it can be solve using only analytical means" , the concept of analysis very discussed in the past, btw. 
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