Modified Intersection Puzzle
5
$begingroup$
This is a slightly modified version of the Intersection Puzzle introduced by user477343 here.
In this case, you are given a $7 times 8$ grid with circles and boxes provided in some of the cells.
The aim is to construct a set of rectilinear paths inside the grid with the following rules:
Standard rules
- Paths must consist of horizontal and vertical lines only. For any cell not containing a circle, a path must enter each cell it passes through from the centre of one of its four sides and exit from a different side. (like in Masyu)
- All paths must begin at a circle and end at a circle but no path can begin and end at the same circle.
- Each cell of the grid must contain a segment of some path. Cells which are originally empty must contain exactly one path segment.
- No path may intersect with itself or another path with the exception of intersection in cells which contain boxes.
- Each cell with a box must contain a single path intersection with one path segment passing through horizontally and the other passing through vertically (a crossover).
- The grid is to be treated as a torus. This means that a path may exit a cell at the edge of the grid and "wrap around", continuing in the cell at the end of the row or column, as appropriate. The only exception to this rule is that a path may not wrap around when either of the edge cells contains a circle.
Additional Rules
- Each circle must either have 2 or 3 paths protruding from it.
- For any two circles there is at most one path connecting them.
If any of those rules are unclear, please refer to the solutions of the original puzzle to get an idea of what paths should look like. Users familiar with Flow Free: Bridges or Numberlink may find it easier to think of this as a modified version of those puzzles.
Given all those rules, here is the puzzle
Under the full rule set above, this has a unique solution.
Can you determine it?
grid-deduction
asked Apr 18 at 13:10
hexominohexomino
47.5k4143223
$endgroup$
add a comment |
5
$begingroup$
This is a slightly modified version of the Intersection Puzzle introduced by user477343 here.
In this case, you are given a $7 times 8$ grid with circles and boxes provided in some of the cells.
The aim is to construct a set of rectilinear paths inside the grid with the following rules:
Standard rules
- Paths must consist of horizontal and vertical lines only. For any cell not containing a circle, a path must enter each cell it passes through from the centre of one of its four sides and exit from a different side. (like in Masyu)
- All paths must begin at a circle and end at a circle but no path can begin and end at the same circle.
- Each cell of the grid must contain a segment of some path. Cells which are originally empty must contain exactly one path segment.
- No path may intersect with itself or another path with the exception of intersection in cells which contain boxes.
- Each cell with a box must contain a single path intersection with one path segment passing through horizontally and the other passing through vertically (a crossover).
- The grid is to be treated as a torus. This means that a path may exit a cell at the edge of the grid and "wrap around", continuing in the cell at the end of the row or column, as appropriate. The only exception to this rule is that a path may not wrap around when either of the edge cells contains a circle.
Additional Rules
- Each circle must either have 2 or 3 paths protruding from it.
- For any two circles there is at most one path connecting them.
If any of those rules are unclear, please refer to the solutions of the original puzzle to get an idea of what paths should look like. Users familiar with Flow Free: Bridges or Numberlink may find it easier to think of this as a modified version of those puzzles.
Given all those rules, here is the puzzle
Under the full rule set above, this has a unique solution.
Can you determine it?
grid-deduction
asked Apr 18 at 13:10
hexominohexomino
47.5k4143223
$endgroup$
1
$begingroup$
Yay! Oh, and thanks for the mention in your post! +1 :)
$endgroup$
– user477343
Apr 18 at 13:18
$begingroup$
Regarding the two boxes diagonally adjacent to each other, how would that work? If two lines intersect the top left box of the pair, then if the same happens for the right bottom box, wouldn't that imply that both sets of lines intersect at at least one of the cells orthogonally adjacent to both the boxes? (And how did you verify there is a unique solution? That's cool!)
$endgroup$
– user477343
Apr 18 at 13:32
1
$begingroup$
@user477343 You're beginning to get into the logic now I think. So the lines will meet in the orthogonally adjacent cells as you said and, since intersections of paths are prohibited in these cells, these lines must belong to the same path. Does that make sense?
$endgroup$
– hexomino
Apr 18 at 13:38
$begingroup$
Oh, wait, I just needed to be a little more creative ;)
$endgroup$
– user477343
Apr 18 at 13:39
add a comment |
5
5
5
2
$begingroup$
This is a slightly modified version of the Intersection Puzzle introduced by user477343 here.
In this case, you are given a $7 times 8$ grid with circles and boxes provided in some of the cells.
The aim is to construct a set of rectilinear paths inside the grid with the following rules:
Standard rules
- Paths must consist of horizontal and vertical lines only. For any cell not containing a circle, a path must enter each cell it passes through from the centre of one of its four sides and exit from a different side. (like in Masyu)
- All paths must begin at a circle and end at a circle but no path can begin and end at the same circle.
- Each cell of the grid must contain a segment of some path. Cells which are originally empty must contain exactly one path segment.
- No path may intersect with itself or another path with the exception of intersection in cells which contain boxes.
- Each cell with a box must contain a single path intersection with one path segment passing through horizontally and the other passing through vertically (a crossover).
- The grid is to be treated as a torus. This means that a path may exit a cell at the edge of the grid and "wrap around", continuing in the cell at the end of the row or column, as appropriate. The only exception to this rule is that a path may not wrap around when either of the edge cells contains a circle.
Additional Rules
- Each circle must either have 2 or 3 paths protruding from it.
- For any two circles there is at most one path connecting them.
If any of those rules are unclear, please refer to the solutions of the original puzzle to get an idea of what paths should look like. Users familiar with Flow Free: Bridges or Numberlink may find it easier to think of this as a modified version of those puzzles.
Given all those rules, here is the puzzle
Under the full rule set above, this has a unique solution.
Can you determine it?
grid-deduction
asked Apr 18 at 13:10
hexominohexomino
47.5k4143223
$endgroup$
This is a slightly modified version of the Intersection Puzzle introduced by user477343 here.
In this case, you are given a $7 times 8$ grid with circles and boxes provided in some of the cells.
The aim is to construct a set of rectilinear paths inside the grid with the following rules:
Standard rules
- Paths must consist of horizontal and vertical lines only. For any cell not containing a circle, a path must enter each cell it passes through from the centre of one of its four sides and exit from a different side. (like in Masyu)
- All paths must begin at a circle and end at a circle but no path can begin and end at the same circle.
- Each cell of the grid must contain a segment of some path. Cells which are originally empty must contain exactly one path segment.
- No path may intersect with itself or another path with the exception of intersection in cells which contain boxes.
- Each cell with a box must contain a single path intersection with one path segment passing through horizontally and the other passing through vertically (a crossover).
- The grid is to be treated as a torus. This means that a path may exit a cell at the edge of the grid and "wrap around", continuing in the cell at the end of the row or column, as appropriate. The only exception to this rule is that a path may not wrap around when either of the edge cells contains a circle.
Additional Rules
- Each circle must either have 2 or 3 paths protruding from it.
- For any two circles there is at most one path connecting them.
If any of those rules are unclear, please refer to the solutions of the original puzzle to get an idea of what paths should look like. Users familiar with Flow Free: Bridges or Numberlink may find it easier to think of this as a modified version of those puzzles.
Given all those rules, here is the puzzle
Under the full rule set above, this has a unique solution.
Can you determine it?
grid-deduction
grid-deduction
asked Apr 18 at 13:10
hexominohexomino
47.5k4143223
asked Apr 18 at 13:10
hexominohexomino
47.5k4143223
asked Apr 18 at 13:10
hexominohexomino
47.5k4143223
asked Apr 18 at 13:10
hexominohexomino
47.5k4143223
asked Apr 18 at 13:10
hexominohexomino
47.5k4143223
47.5k4143223
1
$begingroup$
Yay! Oh, and thanks for the mention in your post! +1 :)
$endgroup$
– user477343
Apr 18 at 13:18
$begingroup$
Regarding the two boxes diagonally adjacent to each other, how would that work? If two lines intersect the top left box of the pair, then if the same happens for the right bottom box, wouldn't that imply that both sets of lines intersect at at least one of the cells orthogonally adjacent to both the boxes? (And how did you verify there is a unique solution? That's cool!)
$endgroup$
– user477343
Apr 18 at 13:32
1
$begingroup$
@user477343 You're beginning to get into the logic now I think. So the lines will meet in the orthogonally adjacent cells as you said and, since intersections of paths are prohibited in these cells, these lines must belong to the same path. Does that make sense?
$endgroup$
– hexomino
Apr 18 at 13:38
$begingroup$
Oh, wait, I just needed to be a little more creative ;)
$endgroup$
– user477343
Apr 18 at 13:39
add a comment |
1
$begingroup$
Yay! Oh, and thanks for the mention in your post! +1 :)
$endgroup$
– user477343
Apr 18 at 13:18
$begingroup$
Regarding the two boxes diagonally adjacent to each other, how would that work? If two lines intersect the top left box of the pair, then if the same happens for the right bottom box, wouldn't that imply that both sets of lines intersect at at least one of the cells orthogonally adjacent to both the boxes? (And how did you verify there is a unique solution? That's cool!)
$endgroup$
– user477343
Apr 18 at 13:32
1
$begingroup$
@user477343 You're beginning to get into the logic now I think. So the lines will meet in the orthogonally adjacent cells as you said and, since intersections of paths are prohibited in these cells, these lines must belong to the same path. Does that make sense?
$endgroup$
– hexomino
Apr 18 at 13:38
$begingroup$
Oh, wait, I just needed to be a little more creative ;)
$endgroup$
– user477343
Apr 18 at 13:39
1
1
$begingroup$
Yay! Oh, and thanks for the mention in your post! +1 :)
$endgroup$
– user477343
Apr 18 at 13:18
$begingroup$
Yay! Oh, and thanks for the mention in your post! +1 :)
$endgroup$
– user477343
Apr 18 at 13:18
$begingroup$
Regarding the two boxes diagonally adjacent to each other, how would that work? If two lines intersect the top left box of the pair, then if the same happens for the right bottom box, wouldn't that imply that both sets of lines intersect at at least one of the cells orthogonally adjacent to both the boxes? (And how did you verify there is a unique solution? That's cool!)
$endgroup$
– user477343
Apr 18 at 13:32
$begingroup$
Regarding the two boxes diagonally adjacent to each other, how would that work? If two lines intersect the top left box of the pair, then if the same happens for the right bottom box, wouldn't that imply that both sets of lines intersect at at least one of the cells orthogonally adjacent to both the boxes? (And how did you verify there is a unique solution? That's cool!)
$endgroup$
– user477343
Apr 18 at 13:32
1
1
$begingroup$
@user477343 You're beginning to get into the logic now I think. So the lines will meet in the orthogonally adjacent cells as you said and, since intersections of paths are prohibited in these cells, these lines must belong to the same path. Does that make sense?
$endgroup$
– hexomino
Apr 18 at 13:38
$begingroup$
@user477343 You're beginning to get into the logic now I think. So the lines will meet in the orthogonally adjacent cells as you said and, since intersections of paths are prohibited in these cells, these lines must belong to the same path. Does that make sense?
$endgroup$
– hexomino
Apr 18 at 13:38
$begingroup$
Oh, wait, I just needed to be a little more creative ;)
$endgroup$
– user477343
Apr 18 at 13:39
$begingroup$
Oh, wait, I just needed to be a little more creative ;)
$endgroup$
– user477343
Apr 18 at 13:39
add a comment |
1 Answer
1
active
oldest
votes
5
$begingroup$
answered Apr 18 at 13:38
noednenoedne
9,56512666
$endgroup$
4
$begingroup$
I was just coming to post the answer, only to see that you've already done it! Great job! +1
$endgroup$
– Eagle
Apr 18 at 13:42
$begingroup$
I might be missing something, but which rule uniquely determines the paths in column 7+8, row 3+4?
$endgroup$
– w l
Apr 18 at 14:34
2
$begingroup$
@wl 8
$endgroup$
– noedne
Apr 18 at 14:42
$begingroup$
@noedne Thanks, I completely overlooked the additional rule.
$endgroup$
– w l
Apr 18 at 14:43
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
$begingroup$
answered Apr 18 at 13:38
noednenoedne
9,56512666
$endgroup$
4
$begingroup$
I was just coming to post the answer, only to see that you've already done it! Great job! +1
$endgroup$
– Eagle
Apr 18 at 13:42
$begingroup$
I might be missing something, but which rule uniquely determines the paths in column 7+8, row 3+4?
$endgroup$
– w l
Apr 18 at 14:34
2
$begingroup$
@wl 8
$endgroup$
– noedne
Apr 18 at 14:42
$begingroup$
@noedne Thanks, I completely overlooked the additional rule.
$endgroup$
– w l
Apr 18 at 14:43
add a comment |
5
$begingroup$
answered Apr 18 at 13:38
noednenoedne
9,56512666
$endgroup$
4
$begingroup$
I was just coming to post the answer, only to see that you've already done it! Great job! +1
$endgroup$
– Eagle
Apr 18 at 13:42
$begingroup$
I might be missing something, but which rule uniquely determines the paths in column 7+8, row 3+4?
$endgroup$
– w l
Apr 18 at 14:34
2
$begingroup$
@wl 8
$endgroup$
– noedne
Apr 18 at 14:42
$begingroup$
@noedne Thanks, I completely overlooked the additional rule.
$endgroup$
– w l
Apr 18 at 14:43
add a comment |
5
5
5
$begingroup$
answered Apr 18 at 13:38
noednenoedne
9,56512666
$endgroup$
answered Apr 18 at 13:38
noednenoedne
9,56512666
answered Apr 18 at 13:38
noednenoedne
9,56512666
answered Apr 18 at 13:38
noednenoedne
9,56512666
answered Apr 18 at 13:38
noednenoedne
9,56512666
9,56512666
4
$begingroup$
I was just coming to post the answer, only to see that you've already done it! Great job! +1
$endgroup$
– Eagle
Apr 18 at 13:42
$begingroup$
I might be missing something, but which rule uniquely determines the paths in column 7+8, row 3+4?
$endgroup$
– w l
Apr 18 at 14:34
2
$begingroup$
@wl 8
$endgroup$
– noedne
Apr 18 at 14:42
$begingroup$
@noedne Thanks, I completely overlooked the additional rule.
$endgroup$
– w l
Apr 18 at 14:43
add a comment |
4
$begingroup$
I was just coming to post the answer, only to see that you've already done it! Great job! +1
$endgroup$
– Eagle
Apr 18 at 13:42
$begingroup$
I might be missing something, but which rule uniquely determines the paths in column 7+8, row 3+4?
$endgroup$
– w l
Apr 18 at 14:34
2
$begingroup$
@wl 8
$endgroup$
– noedne
Apr 18 at 14:42
$begingroup$
@noedne Thanks, I completely overlooked the additional rule.
$endgroup$
– w l
Apr 18 at 14:43
4
4
$begingroup$
I was just coming to post the answer, only to see that you've already done it! Great job! +1
$endgroup$
– Eagle
Apr 18 at 13:42
$begingroup$
I was just coming to post the answer, only to see that you've already done it! Great job! +1
$endgroup$
– Eagle
Apr 18 at 13:42
$begingroup$
I might be missing something, but which rule uniquely determines the paths in column 7+8, row 3+4?
$endgroup$
– w l
Apr 18 at 14:34
$begingroup$
I might be missing something, but which rule uniquely determines the paths in column 7+8, row 3+4?
$endgroup$
– w l
Apr 18 at 14:34
2
2
$begingroup$
@wl 8
$endgroup$
– noedne
Apr 18 at 14:42
$begingroup$
@wl 8
$endgroup$
– noedne
Apr 18 at 14:42
$begingroup$
@noedne Thanks, I completely overlooked the additional rule.
$endgroup$
– w l
Apr 18 at 14:43
$begingroup$
@noedne Thanks, I completely overlooked the additional rule.
$endgroup$
– w l
Apr 18 at 14:43
add a comment |
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1
$begingroup$
Yay! Oh, and thanks for the mention in your post! +1 :)
$endgroup$
– user477343
Apr 18 at 13:18
$begingroup$
Regarding the two boxes diagonally adjacent to each other, how would that work? If two lines intersect the top left box of the pair, then if the same happens for the right bottom box, wouldn't that imply that both sets of lines intersect at at least one of the cells orthogonally adjacent to both the boxes? (And how did you verify there is a unique solution? That's cool!)
$endgroup$
– user477343
Apr 18 at 13:32
1
$begingroup$
@user477343 You're beginning to get into the logic now I think. So the lines will meet in the orthogonally adjacent cells as you said and, since intersections of paths are prohibited in these cells, these lines must belong to the same path. Does that make sense?
$endgroup$
– hexomino
Apr 18 at 13:38
$begingroup$
Oh, wait, I just needed to be a little more creative ;)
$endgroup$
– user477343
Apr 18 at 13:39