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Very tricky nonogram - where to go next?
Is there a algorithm to decide that the nonogram puzzle is uniqueA Minesweeper CrosswordA simple nonogramCircuit DiagramAn Amazing NonogramWhere did my uncle go?A fortified nonogramHow many possible starting positions are uniquely solvable for a nonogram puzzle?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I've become somewhat addicted to Simon Tatham's "Pattern" (nonogram) puzzles recently. I thought I was becoming fairly adept with them, but this unusually difficult one has me stumped. I've got as far as this by using the usual tricks:
But now I can't figure out how to make any further progress. What am I missing?
How can I make the next step to solve this puzzle?
grid-deduction nonogram
asked 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
$endgroup$
add a comment |
$begingroup$
I've become somewhat addicted to Simon Tatham's "Pattern" (nonogram) puzzles recently. I thought I was becoming fairly adept with them, but this unusually difficult one has me stumped. I've got as far as this by using the usual tricks:
But now I can't figure out how to make any further progress. What am I missing?
How can I make the next step to solve this puzzle?
grid-deduction nonogram
asked 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
$endgroup$
add a comment |
$begingroup$
I've become somewhat addicted to Simon Tatham's "Pattern" (nonogram) puzzles recently. I thought I was becoming fairly adept with them, but this unusually difficult one has me stumped. I've got as far as this by using the usual tricks:
But now I can't figure out how to make any further progress. What am I missing?
How can I make the next step to solve this puzzle?
grid-deduction nonogram
asked 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
$endgroup$
I've become somewhat addicted to Simon Tatham's "Pattern" (nonogram) puzzles recently. I thought I was becoming fairly adept with them, but this unusually difficult one has me stumped. I've got as far as this by using the usual tricks:
But now I can't figure out how to make any further progress. What am I missing?
How can I make the next step to solve this puzzle?
grid-deduction nonogram
grid-deduction nonogram
asked 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
asked 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
asked 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
asked 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
asked 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
73.7k15240488
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
[Edit] Darn, you got it as I was typing this.
There are the ones I saw immediately that can be filled in (the red letters, the blue ones are the blocks that show that the red ones need to be filled in).
]1
answered 8 hours ago
TedTed
536
$endgroup$
add a comment |
$begingroup$
You can use the
1,5,1 column (number 9), because the bottom row block cannot house the 5, which limits it at the top of the column.
Then:
the five block must start in at least row 3 and at most row 5, and this means rows 5,6,7 of column 9 can be filled.
answered 8 hours ago
JonMark PerryJonMark Perry
22.4k643103
$endgroup$
$begingroup$
OK, so the 5 must be in the upper block ... but then what? We don't know exactly where that 5 must be.
$endgroup$
– Rand al'Thor
8 hours ago
add a comment |
$begingroup$
Duh, I got it.
Bottom row: the lone square must be part of the 6-block, but there's not enough space for it to go all the way to the right: it must extend at least two more cells to the left.
Then
edge cells are always useful because we can start from there to fill in whole blocks: in this case, the 5 and 3 at the bottom of the fourth and fifth columns.
I'm guessing the deductions will fall like dominoes from there ...
answered 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
$endgroup$
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
8 hours ago
$begingroup$
You can infer even more about the 6. The center column is (7 2), flanked on either side by a 1 in the bottom position. Nothing can be filled in in the bottom row center column, otherwise you'll have an isolated square in the second-to-bottom row that violates (1 4 4). You can fill in two more squares to the left of the partial 6 block you show in the image, white out the bottom row center column and directly to the right of it, and place the 2 of (6 2) in the hole on the bottom right..
$endgroup$
– Nuclear Wang
8 hours ago
add a comment |
$begingroup$
I’m thinking
Based on your 1 4 4 row (second from bottom), that you have a loner black square on the left side. If you continue that to the right, that should be your second set of 4.
answered 8 hours ago
El-GuestEl-Guest
23.6k35496
$endgroup$
2
$begingroup$
But why would it continue to the right necessarily? It could equally well continue to the left, as far as I can see.
$endgroup$
– Rand al'Thor
8 hours ago
add a comment |
$begingroup$
At least
on the bottom row, we know the 4th and 5th cell have to be part of the 6.
answered 8 hours ago
jafejafe
30.5k487312
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
[Edit] Darn, you got it as I was typing this.
There are the ones I saw immediately that can be filled in (the red letters, the blue ones are the blocks that show that the red ones need to be filled in).
answered 8 hours ago
TedTed
536
$endgroup$
add a comment |
$begingroup$
[Edit] Darn, you got it as I was typing this.
There are the ones I saw immediately that can be filled in (the red letters, the blue ones are the blocks that show that the red ones need to be filled in).
answered 8 hours ago
TedTed
536
$endgroup$
add a comment |
$begingroup$
[Edit] Darn, you got it as I was typing this.
There are the ones I saw immediately that can be filled in (the red letters, the blue ones are the blocks that show that the red ones need to be filled in).
answered 8 hours ago
TedTed
536
$endgroup$
[Edit] Darn, you got it as I was typing this.
There are the ones I saw immediately that can be filled in (the red letters, the blue ones are the blocks that show that the red ones need to be filled in).
answered 8 hours ago
TedTed
536
answered 8 hours ago
TedTed
536
answered 8 hours ago
TedTed
536
answered 8 hours ago
TedTed
536
536
add a comment |
add a comment |
$begingroup$
You can use the
1,5,1 column (number 9), because the bottom row block cannot house the 5, which limits it at the top of the column.
Then:
the five block must start in at least row 3 and at most row 5, and this means rows 5,6,7 of column 9 can be filled.
answered 8 hours ago
JonMark PerryJonMark Perry
22.4k643103
$endgroup$
$begingroup$
OK, so the 5 must be in the upper block ... but then what? We don't know exactly where that 5 must be.
$endgroup$
– Rand al'Thor
8 hours ago
add a comment |
$begingroup$
You can use the
1,5,1 column (number 9), because the bottom row block cannot house the 5, which limits it at the top of the column.
Then:
the five block must start in at least row 3 and at most row 5, and this means rows 5,6,7 of column 9 can be filled.
answered 8 hours ago
JonMark PerryJonMark Perry
22.4k643103
$endgroup$
$begingroup$
OK, so the 5 must be in the upper block ... but then what? We don't know exactly where that 5 must be.
$endgroup$
– Rand al'Thor
8 hours ago
add a comment |
$begingroup$
You can use the
1,5,1 column (number 9), because the bottom row block cannot house the 5, which limits it at the top of the column.
Then:
the five block must start in at least row 3 and at most row 5, and this means rows 5,6,7 of column 9 can be filled.
answered 8 hours ago
JonMark PerryJonMark Perry
22.4k643103
$endgroup$
You can use the
1,5,1 column (number 9), because the bottom row block cannot house the 5, which limits it at the top of the column.
Then:
the five block must start in at least row 3 and at most row 5, and this means rows 5,6,7 of column 9 can be filled.
answered 8 hours ago
JonMark PerryJonMark Perry
22.4k643103
edited 8 hours ago
answered 8 hours ago
JonMark PerryJonMark Perry
22.4k643103
answered 8 hours ago
JonMark PerryJonMark Perry
22.4k643103
answered 8 hours ago
JonMark PerryJonMark Perry
22.4k643103
22.4k643103
$begingroup$
OK, so the 5 must be in the upper block ... but then what? We don't know exactly where that 5 must be.
$endgroup$
– Rand al'Thor
8 hours ago
add a comment |
$begingroup$
OK, so the 5 must be in the upper block ... but then what? We don't know exactly where that 5 must be.
$endgroup$
– Rand al'Thor
8 hours ago
$begingroup$
OK, so the 5 must be in the upper block ... but then what? We don't know exactly where that 5 must be.
$endgroup$
– Rand al'Thor
8 hours ago
$begingroup$
OK, so the 5 must be in the upper block ... but then what? We don't know exactly where that 5 must be.
$endgroup$
– Rand al'Thor
8 hours ago
add a comment |
$begingroup$
Duh, I got it.
Bottom row: the lone square must be part of the 6-block, but there's not enough space for it to go all the way to the right: it must extend at least two more cells to the left.
Then
edge cells are always useful because we can start from there to fill in whole blocks: in this case, the 5 and 3 at the bottom of the fourth and fifth columns.
I'm guessing the deductions will fall like dominoes from there ...
answered 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
$endgroup$
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
8 hours ago
$begingroup$
You can infer even more about the 6. The center column is (7 2), flanked on either side by a 1 in the bottom position. Nothing can be filled in in the bottom row center column, otherwise you'll have an isolated square in the second-to-bottom row that violates (1 4 4). You can fill in two more squares to the left of the partial 6 block you show in the image, white out the bottom row center column and directly to the right of it, and place the 2 of (6 2) in the hole on the bottom right..
$endgroup$
– Nuclear Wang
8 hours ago
add a comment |
$begingroup$
Duh, I got it.
Bottom row: the lone square must be part of the 6-block, but there's not enough space for it to go all the way to the right: it must extend at least two more cells to the left.
Then
edge cells are always useful because we can start from there to fill in whole blocks: in this case, the 5 and 3 at the bottom of the fourth and fifth columns.
I'm guessing the deductions will fall like dominoes from there ...
answered 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
$endgroup$
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
8 hours ago
$begingroup$
You can infer even more about the 6. The center column is (7 2), flanked on either side by a 1 in the bottom position. Nothing can be filled in in the bottom row center column, otherwise you'll have an isolated square in the second-to-bottom row that violates (1 4 4). You can fill in two more squares to the left of the partial 6 block you show in the image, white out the bottom row center column and directly to the right of it, and place the 2 of (6 2) in the hole on the bottom right..
$endgroup$
– Nuclear Wang
8 hours ago
add a comment |
$begingroup$
Duh, I got it.
Bottom row: the lone square must be part of the 6-block, but there's not enough space for it to go all the way to the right: it must extend at least two more cells to the left.
Then
edge cells are always useful because we can start from there to fill in whole blocks: in this case, the 5 and 3 at the bottom of the fourth and fifth columns.
I'm guessing the deductions will fall like dominoes from there ...
answered 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
$endgroup$
Duh, I got it.
Bottom row: the lone square must be part of the 6-block, but there's not enough space for it to go all the way to the right: it must extend at least two more cells to the left.
Then
edge cells are always useful because we can start from there to fill in whole blocks: in this case, the 5 and 3 at the bottom of the fourth and fifth columns.
I'm guessing the deductions will fall like dominoes from there ...
answered 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
answered 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
answered 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
answered 8 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
73.7k15240488
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
8 hours ago
$begingroup$
You can infer even more about the 6. The center column is (7 2), flanked on either side by a 1 in the bottom position. Nothing can be filled in in the bottom row center column, otherwise you'll have an isolated square in the second-to-bottom row that violates (1 4 4). You can fill in two more squares to the left of the partial 6 block you show in the image, white out the bottom row center column and directly to the right of it, and place the 2 of (6 2) in the hole on the bottom right..
$endgroup$
– Nuclear Wang
8 hours ago
add a comment |
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
8 hours ago
$begingroup$
You can infer even more about the 6. The center column is (7 2), flanked on either side by a 1 in the bottom position. Nothing can be filled in in the bottom row center column, otherwise you'll have an isolated square in the second-to-bottom row that violates (1 4 4). You can fill in two more squares to the left of the partial 6 block you show in the image, white out the bottom row center column and directly to the right of it, and place the 2 of (6 2) in the hole on the bottom right..
$endgroup$
– Nuclear Wang
8 hours ago
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
8 hours ago
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
8 hours ago
$begingroup$
You can infer even more about the 6. The center column is (7 2), flanked on either side by a 1 in the bottom position. Nothing can be filled in in the bottom row center column, otherwise you'll have an isolated square in the second-to-bottom row that violates (1 4 4). You can fill in two more squares to the left of the partial 6 block you show in the image, white out the bottom row center column and directly to the right of it, and place the 2 of (6 2) in the hole on the bottom right..
$endgroup$
– Nuclear Wang
8 hours ago
$begingroup$
You can infer even more about the 6. The center column is (7 2), flanked on either side by a 1 in the bottom position. Nothing can be filled in in the bottom row center column, otherwise you'll have an isolated square in the second-to-bottom row that violates (1 4 4). You can fill in two more squares to the left of the partial 6 block you show in the image, white out the bottom row center column and directly to the right of it, and place the 2 of (6 2) in the hole on the bottom right..
$endgroup$
– Nuclear Wang
8 hours ago
add a comment |
$begingroup$
I’m thinking
Based on your 1 4 4 row (second from bottom), that you have a loner black square on the left side. If you continue that to the right, that should be your second set of 4.
answered 8 hours ago
El-GuestEl-Guest
23.6k35496
$endgroup$
2
$begingroup$
But why would it continue to the right necessarily? It could equally well continue to the left, as far as I can see.
$endgroup$
– Rand al'Thor
8 hours ago
add a comment |
$begingroup$
I’m thinking
Based on your 1 4 4 row (second from bottom), that you have a loner black square on the left side. If you continue that to the right, that should be your second set of 4.
answered 8 hours ago
El-GuestEl-Guest
23.6k35496
$endgroup$
2
$begingroup$
But why would it continue to the right necessarily? It could equally well continue to the left, as far as I can see.
$endgroup$
– Rand al'Thor
8 hours ago
add a comment |
$begingroup$
I’m thinking
Based on your 1 4 4 row (second from bottom), that you have a loner black square on the left side. If you continue that to the right, that should be your second set of 4.
answered 8 hours ago
El-GuestEl-Guest
23.6k35496
$endgroup$
I’m thinking
Based on your 1 4 4 row (second from bottom), that you have a loner black square on the left side. If you continue that to the right, that should be your second set of 4.
answered 8 hours ago
El-GuestEl-Guest
23.6k35496
answered 8 hours ago
El-GuestEl-Guest
23.6k35496
answered 8 hours ago
El-GuestEl-Guest
23.6k35496
answered 8 hours ago
El-GuestEl-Guest
23.6k35496
23.6k35496
2
$begingroup$
But why would it continue to the right necessarily? It could equally well continue to the left, as far as I can see.
$endgroup$
– Rand al'Thor
8 hours ago
add a comment |
2
$begingroup$
But why would it continue to the right necessarily? It could equally well continue to the left, as far as I can see.
$endgroup$
– Rand al'Thor
8 hours ago
2
2
$begingroup$
But why would it continue to the right necessarily? It could equally well continue to the left, as far as I can see.
$endgroup$
– Rand al'Thor
8 hours ago
$begingroup$
But why would it continue to the right necessarily? It could equally well continue to the left, as far as I can see.
$endgroup$
– Rand al'Thor
8 hours ago
add a comment |
$begingroup$
At least
on the bottom row, we know the 4th and 5th cell have to be part of the 6.
answered 8 hours ago
jafejafe
30.5k487312
$endgroup$
add a comment |
$begingroup$
At least
on the bottom row, we know the 4th and 5th cell have to be part of the 6.
answered 8 hours ago
jafejafe
30.5k487312
$endgroup$
add a comment |
$begingroup$
At least
on the bottom row, we know the 4th and 5th cell have to be part of the 6.
answered 8 hours ago
jafejafe
30.5k487312
$endgroup$
At least
on the bottom row, we know the 4th and 5th cell have to be part of the 6.
answered 8 hours ago
jafejafe
30.5k487312
answered 8 hours ago
jafejafe
30.5k487312
answered 8 hours ago
jafejafe
30.5k487312
answered 8 hours ago
jafejafe
30.5k487312
30.5k487312
add a comment |
add a comment |
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