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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 11 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 11 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 11 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 11 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
asked 11 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
asked 11 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
asked 11 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
asked 11 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 11 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 11 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
11 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 11 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
$endgroup$
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
11 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
11 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 11 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
11 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 11 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 11 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 11 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 11 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 11 hours ago
TedTed
536
answered 11 hours ago
TedTed
536
answered 11 hours ago
TedTed
536
answered 11 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 11 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
11 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 11 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
11 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 11 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 11 hours ago
JonMark PerryJonMark Perry
22.4k643103
edited 11 hours ago
answered 11 hours ago
JonMark PerryJonMark Perry
22.4k643103
answered 11 hours ago
JonMark PerryJonMark Perry
22.4k643103
answered 11 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
11 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
11 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
11 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
11 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 11 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
$endgroup$
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
11 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
11 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 11 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
$endgroup$
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
11 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
11 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 11 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 11 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
answered 11 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
answered 11 hours ago
Rand al'ThorRand al'Thor
73.7k15240488
answered 11 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
11 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
11 hours ago
add a comment |
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
11 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
11 hours ago
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
11 hours ago
$begingroup$
And yep, I've now solved it completely. facepalm
$endgroup$
– Rand al'Thor
11 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
11 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
11 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 11 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
11 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 11 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
11 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 11 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 11 hours ago
El-GuestEl-Guest
23.6k35496
answered 11 hours ago
El-GuestEl-Guest
23.6k35496
answered 11 hours ago
El-GuestEl-Guest
23.6k35496
answered 11 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
11 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
11 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
11 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
11 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 11 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 11 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 11 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 11 hours ago
jafejafe
30.5k487312
answered 11 hours ago
jafejafe
30.5k487312
answered 11 hours ago
jafejafe
30.5k487312
answered 11 hours ago
jafejafe
30.5k487312
30.5k487312
add a comment |
add a comment |
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