It can't be. Because in H2 and H7 the only possible digits are 2 and 7. So if one of them are 2 or 7, the other one must be oposite. That's why all other 2 and 7 in the H row can be eliminated.
Yeah exactly what this person has said: the software hasn’t picked up that the 2s and 7s form a “pair”, which means that if one is a 7, the other is a 2 and visa versa, therefore no other 2s or 7s can be present in that row, so the 7 there shouldn’t have been visible in the first place It confused me too at first
I really appreciate the added explanation on this “next level” tool to solving puzzles. I know it may be too draw out for some, but there it maybe the only way I and others will get it.
Thanks Bob, I really appreciate your feedback! It also takes me longer to edit into the video all the circles and arrows, hopefully that helps illustrate what I am doing. Thanks for the comment!
Thank you for the nice explanation. I liked the second example better: some of the logic you applied at the end of the first could have been applied before you invoked the skyscraper.
I had to do further research for this as i was also confused and most tutorials don't specify. Sudoku Guy has a good tutorial on Skyscrapers., he said the rows/columns must be in separate blocks!
1st example: both top-7's could be true (A7 and B2), with row H having a 7 in column 6. The argument holds when reasoning from False > True > F > T, but fails in the T>F>T>F case. I.e. if A7 is False then B2 is True, but the reverse doesn't hold. Or am I missing something? The problem is that H2 and H7 don't have a strong link because of H6. EDIT: there *is* a strong link once you identify the matched 2-7 pair in H2/H7 (which eliminates all other 7's from row H). But you do that only afterwards. So yes, the skyscraper argument eventually holds, but only if you had presented it in the right order.
Hi. Just found your channel, your stuff is EXCELLENT! Thanks! Just starting sudoku can do easy ones - watched your episode 1-6. Is there a sequence of your videos I should watch ?
Key thing to remember in finding X-wings or Skyscrapers: 2 "Sides" (2 columns or 2 rows parallel to each other) have to consist of Conjugate Pairs (only 2 Cells in Separate Blocks in a Column or Row contain the Same Number ie in video: b2,h2 and a7,h7) and that on the Third "Side" (in this case, the "Base") the Ends (h2 & h7) just be lined up in a Row or Column. And No, the Base DOESN'T have to be a Conjugate Pair or Matching Pairs. REPEAT: The Ends in the Third Side Doesn't have to be a Conjugate Pair or a Matching pair, they just have to line up. Sounds weird I know, but in a Skyscraper it's on the "Roof" (b2 & a7) where any Eliminations take place. Btw, same rules apply in an X-wing too! The Ends of the Roof (say if was b2 & b7) Doesn't need to be Conjugate Pairs or Matching Pairs (just like in the Base) they just need to Line up. Of course, the difference here is that (as opposed to a Skyscraper) you can also make eliminations in the Base too. Whew! Hope that helps guys! ✌🏼
I see. So it wouldn’t work if the squares were not aligned on the cell with an offset/nonalignment since the only intersections would come from cells that were already guaranteed not to have that number
Great video, thank you! Just a question regarding the “roof” of the pattern. n first example, why couldn’t c7r1 & c2r2 both be sevens? They don’t share the same row? In the second one, c4r5 & c5r3 could both be threes?
I’m just learning this technique myself, but aside from the presence of linked pairs, I think it might not matter if both points of the roof are true. The skyscraper seems like it is only used as a way to eliminate candidates that can see both points of the roof. If either point is true, or if both are true, it still eliminates any candidates that see both.
Um. No, c4r5 & c5r3 (the Roof) can't both be true. The logic here is that there are only two 3s in r3 & only two 3s in r5. If the 3 in c5r3 is true, then the 3 in c9r3 has to be false (and vice versa) because there can be only one 3 in a row. Likewise, if the 3 in c4r5 is true, then the 3 in c9r5 has got to be false (and vice versa) because of that Sudoku rule mentioned. What determines all that is by looking at the Base of the Skyscraper: c9r3, c9r5. The only 3s in c9. Same rule applies. If the 3 in c9r3 is true, then the 3 in c9r5 is false (and vice versa). Soo we've finally come Full Circle 😌 hope that helps you guys 🤙🏼 (why the 3s in c4r5 & c5r3 can't both be true: if the 3 in c9r3 is true, the 3 in c5r3 is false or vice versa. If the 3 in c9r5 is true, then the 3 in c4r5 has got to be false and vice versa)
Yeah. Ideally it's great if the Base is made up of a matched pair (in separate blocks) but the Principle is that it's a Congugated Pair or more simply put, only Two of the Same Number in the Row (or Column) in Separate Blocks. Hope that helps 🤙🏼
H2 being false does imply H7 is true, but only because they're matching pairs which is separate from this technique. Other 2, 7 candidates on row H should have been removed already.
You must have one of each digit in a row so if H2 is false (in other words, it's false that the digit 7 occurs in H2) then since you must have a 7 somewhere in row H and the only other possibility of a 7 is in H7, then it must be true that the 7 is in H7.
There are 2-7 pairs in H2 and H7, which means that the 2's and 7's can be eliminated from H1 and H6 before even beginning the skyscraper technique. After eliminating the 2s and 7s from H1 and H6, then the only 7's remaining are in H2 and H7.
Yes her logic of row H was confusing since there were other 7 candidates but the explanation of pairs cleared it up. If they were 2-7 pairs then this technique would work and could be dangerous assumption
Just realised that rather than saying if H2 is false H7 is True (which only works if they are matching pairs ) if you say if H7 is true then H2 is False. Same thing but makes more sense to me. This means they do not need to be matching pairs.
I don't understand the logic of the first example. It's clear that if B2 is a 7, then A7 surely cannot be a 7. But it seems there is nothing that says that if B2 is a 2, then A7 cannot be a 5. How do you arrive at the conclusion that if B2 is 2 then A7 must be 7?
For anyone who stumbles across this, it all makes sense if you remember that the 7s in each of the two columns in question are their column's only candidates for a 7.
2:25 unfortunately this is wrong here. You can't start a skyscraper by stating something is true. The first assumption must be false. Same problem at 6:20
but at 3.06 if the 7 in H2 is false, the 7 in H6 might be true so the 7 in H7 isn't obligated to be true. I don't understand
It can't be. Because in H2 and H7 the only possible digits are 2 and 7. So if one of them are 2 or 7, the other one must be oposite. That's why all other 2 and 7 in the H row can be eliminated.
Yeah exactly what this person has said: the software hasn’t picked up that the 2s and 7s form a “pair”, which means that if one is a 7, the other is a 2 and visa versa, therefore no other 2s or 7s can be present in that row, so the 7 there shouldn’t have been visible in the first place
It confused me too at first
Thanqq... Tried solving a sudoku which was looking impossible to solve, using this method...and my sudoku was solved in few seconds... It's awesome...
I really appreciate the added explanation on this “next level” tool to solving puzzles. I know it may be too draw out for some, but there it maybe the only way I and others will get it.
Thanks Bob, I really appreciate your feedback! It also takes me longer to edit into the video all the circles and arrows, hopefully that helps illustrate what I am doing. Thanks for the comment!
Hello, i've listened and tried to learn this from videos, and even read on internet.
But this video is finally where i understand it. Thank you 🙂
Thank you so much for the feedback, I really appreciate it!
Thank you for the nice explanation. I liked the second example better: some of the logic you applied at the end of the first could have been applied before you invoked the skyscraper.
I learnt a new technique skyscrapers from you Mam Thanks a lot
Thanks for the video? Do the 4 points that make up the skyscraper have to be in separate blocks for this to work?
I had to do further research for this as i was also confused and most tutorials don't specify. Sudoku Guy has a good tutorial on Skyscrapers., he said the rows/columns must be in separate blocks!
Great video! This was the trick i needed to solve my sudoku
Hello Thank you so much for your very clear explanations ! Where can I find all your lessons for classic sudoku please ?
1st example: both top-7's could be true (A7 and B2), with row H having a 7 in column 6. The argument holds when reasoning from False > True > F > T, but fails in the T>F>T>F case. I.e. if A7 is False then B2 is True, but the reverse doesn't hold. Or am I missing something? The problem is that H2 and H7 don't have a strong link because of H6.
EDIT: there *is* a strong link once you identify the matched 2-7 pair in H2/H7 (which eliminates all other 7's from row H). But you do that only afterwards. So yes, the skyscraper argument eventually holds, but only if you had presented it in the right order.
Hi. Just found your channel, your stuff is EXCELLENT! Thanks! Just starting sudoku can do easy ones - watched your episode 1-6. Is there a sequence of your videos I should watch ?
thanks! what is the app use? :)
Key thing to remember in finding X-wings or Skyscrapers: 2 "Sides" (2 columns or 2 rows parallel to each other) have to consist of Conjugate Pairs (only 2 Cells in Separate Blocks in a Column or Row contain the Same Number ie in video: b2,h2 and a7,h7) and that on the Third "Side" (in this case, the "Base") the Ends (h2 & h7) just be lined up in a Row or Column. And No, the Base DOESN'T have to be a Conjugate Pair or Matching Pairs. REPEAT: The Ends in the Third Side Doesn't have to be a Conjugate Pair or a Matching pair, they just have to line up. Sounds weird I know, but in a Skyscraper it's on the "Roof" (b2 & a7) where any Eliminations take place. Btw, same rules apply in an X-wing too! The Ends of the Roof (say if was b2 & b7) Doesn't need to be Conjugate Pairs or Matching Pairs (just like in the Base) they just need to Line up. Of course, the difference here is that (as opposed to a Skyscraper) you can also make eliminations in the Base too. Whew! Hope that helps guys! ✌🏼
Finally got Helpful explanation
Very well explained excellent
Is it only the roof of the skyscraper that causes elimination of a candidate or can this be done with the base as well?
Nice content, very helpful
I see. So it wouldn’t work if the squares were not aligned on the cell with an offset/nonalignment since the only intersections would come from cells that were already guaranteed not to have that number
Great video, thank you! Just a question regarding the “roof” of the pattern.
n first example, why couldn’t c7r1 & c2r2 both be sevens? They don’t share the same row?
In the second one, c4r5 & c5r3 could both be threes?
I’m just learning this technique myself, but aside from the presence of linked pairs, I think it might not matter if both points of the roof are true. The skyscraper seems like it is only used as a way to eliminate candidates that can see both points of the roof. If either point is true, or if both are true, it still eliminates any candidates that see both.
Um. No, c4r5 & c5r3 (the Roof) can't both be true. The logic here is that there are only two 3s in r3 & only two 3s in r5.
If the 3 in c5r3 is true, then the 3 in c9r3 has to be false (and vice versa) because there can be only one 3 in a row.
Likewise, if the 3 in c4r5 is true, then the 3 in c9r5 has got to be false (and vice versa) because of that Sudoku rule mentioned.
What determines all that is by looking at the Base of the Skyscraper: c9r3, c9r5. The only 3s in c9. Same rule applies.
If the 3 in c9r3 is true, then the 3 in c9r5 is false (and vice versa).
Soo we've finally come Full Circle 😌 hope that helps you guys 🤙🏼
(why the 3s in c4r5 & c5r3 can't both be true: if the 3 in c9r3 is true, the 3 in c5r3 is false or vice versa. If the 3 in c9r5 is true, then the 3 in c4r5 has got to be false and vice versa)
Looking at the comments below is it therefore true that this only works if the "base" is a matching pair ?
Yeah. Ideally it's great if the Base is made up of a matched pair (in separate blocks) but the Principle is that it's a Congugated Pair or more simply put, only Two of the Same Number in the Row (or Column) in Separate Blocks. Hope that helps 🤙🏼
Where can I find these puzzles? I went to Hoduko and there are no puzzles?
@5:28 , after elimination two 5s in cells A1and B1,the puzzle looks still unsolvable. What is the next step, please?
Can you let us know which sudoku solver that is? I haven't run into any that shows you what level of puzzle it is lol
Erm why does H7 needs to be true if H2 is false? 🧐
H2 being false does imply H7 is true, but only because they're matching pairs which is separate from this technique. Other 2, 7 candidates on row H should have been removed already.
You must have one of each digit in a row so if H2 is false (in other words, it's false that the digit 7 occurs in H2) then since you must have a 7 somewhere in row H and the only other possibility of a 7 is in H7, then it must be true that the 7 is in H7.
There are 2-7 pairs in H2 and H7, which means that the 2's and 7's can be eliminated from H1 and H6 before even beginning the skyscraper technique. After eliminating the 2s and 7s from H1 and H6, then the only 7's remaining are in H2 and H7.
Yes her logic of row H was confusing since there were other 7 candidates but the explanation of pairs cleared it up. If they were 2-7 pairs then this technique would work and could be dangerous assumption
Just realised that rather than saying if H2 is false H7 is True (which only works if they are matching pairs ) if you say if H7 is true then H2 is False. Same thing but makes more sense to me. This means they do not need to be matching pairs.
There is a third scenario. The roof cells can both be true. But the rule of elimination is correct.
Exactly..why can't it be B2 true and A7 true
Wrong, it would eliminate all 7s in H row@@Belle-os8if
H6 seven shouldn't be eliminated too?
Yes, it is eliminated.
I don't understand the logic of the first example. It's clear that if B2 is a 7, then A7 surely cannot be a 7. But it seems there is nothing that says that if B2 is a 2, then A7 cannot be a 5. How do you arrive at the conclusion that if B2 is 2 then A7 must be 7?
For anyone who stumbles across this, it all makes sense if you remember that the 7s in each of the two columns in question are their column's only candidates for a 7.
2:33 this is just plain wrong? if top left is true, bottom right is not necesarrily true
you sound like the english voice of naruto
2:25 unfortunately this is wrong here. You can't start a skyscraper by stating something is true. The first assumption must be false. Same problem at 6:20
It literally makes no difference 😂
@@flockadelrey it does if you're smart enough to spot the fault in the logic