Draw a path made by direction changers
25
$begingroup$
This challenge takes place on a grid.
+----------+
| |
| |
| |
| |
| |
| |
| |
| |
| |
+----------+
This one's 10 x 10, but it can be any rectangular shape.
There are four directions on this grid. Up, down, left and right.
The task is to draw a path starting with an upper case direction initial. In this example, will go directly upward from the U.
+----------+
| |
| |
| |
| |
| |
| |
| |
| |
| U |
+----------+
The path will go upwards and be comprised of full-stop characters (.), until it hits a wall, when it will terminate with an asterisk (*).
+----------+
| * |
| . |
| . |
| . |
| . |
| . |
| . |
| . |
| U |
+----------+
In addition to path starts, there's also direction changers, represented by a lower case direction initial.
+----------+
| |
| |
| |
| r.....*|
| . |
| . |
| . |
| . |
| U |
+----------+
Also, an upper case X us an obstacle which will terminate the path.
+----------+
| |
| |
| |
| |
| r...*X |
| . |
| . |
| . |
| U |
+----------+
Rules
- The input is a string consisting of a frame, (consisting of |, - and + characters) containing characters denoting path starts, direction changers, and obstacles.
- Your code should add full stop characters to follow the path described by starts and direction changers, and an asterisk when/if the path meets a wall or obstacle.
- There can be multiple path starts.
- The code will still terminate without error if the path describes a loop.
- If a path meets a path start, it will act as a direction changer.
- It's code golf, low-byte code and no standard loopholes, please.
- I always prefer links to an on-line interpreter.
Test Cases
1: Simple
+----------+
| |
| |
| |
| |
| |
| |
| |
| |
| U |
+----------+
+----------+
| * |
| . |
| . |
| . |
| . |
| . |
| . |
| . |
| U |
+----------+
2: Right turn
+----------+
| |
| |
| |
| r |
| |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| r.....*|
| . |
| . |
| . |
| . |
| U |
+----------+
3: Crossroads
+----------+
| |
| |
| |
| r d |
| |
| u l |
| |
| |
| U |
+----------+
+----------+
| * |
| . |
| . |
| . r..d |
| . . . |
| u....l |
| . |
| . |
| U |
+----------+
4: 4 Crossing paths
+----------+
| D |
| |
| |
|R |
| |
| L|
| |
| |
| U |
+----------+
+----------+
| * D |
| . . |
| . . |
|R........*|
| . . |
|*........L|
| . . |
| . . |
| U * |
+----------+
5: First Loop
+----------+
| |
| |
| |
| r d |
| |
| u l |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| r..d |
| . . |
| u..l |
| . |
| . |
| U |
+----------+
6: Starter as changer
+----------+
| |
| |
| |
| L |
| |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
|*..L |
| . |
| . |
| . |
| . |
| U |
+----------+
7: Straight Loop
+----------+
| |
| |
| |
| |
| r l |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| |
| r..l |
| . |
| . |
| . |
| U |
+----------+
8: Tight Knot
+----------+
| |
| |
| |
| d l |
| r u |
| r u |
| |
| |
| U |
+----------+
+----------+
| * |
| . |
| . |
| d..l |
| .r.u |
| r.u |
| . |
| . |
| U |
+----------+
9: An Obstacle
+----------+
| |
| |
| |
| |
| r X |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| |
| r...*X |
| . |
| . |
| . |
| U |
+----------+
10: S Shape
+----------+
|r d |
| |
| XXXXXXXX|
| d l |
|ul |
|XXXXXXX |
| |
|R u |
| |
+----------+
+----------+
|r.....d |
|. * |
|. XXXXXXXX|
|.d......l |
|ul . |
|XXXXXXX . |
| . |
|R.......u |
| |
+----------+
11: 4-Way Knot
+----------+
| D |
| |
| r |
|R d |
| |
| u L|
| l |
| |
| U |
+----------+
+----------+
| * D |
| . . |
| r.....*|
|R....d. |
| .... |
| .u....L|
|*.....l |
| . . |
| U * |
+----------+
12: Busy Junctions
+----------+
|rrrrr rrrd|
| rlrl |
|ul rrd |
|ruX X |
|udl ll |
|ull |
|rlr |
|rdr d |
|Uruull |
+----------+
+----------+
|rrrrr.rrrd|
|.rlrl .|
|ul rrd .|
|ruX.X. .|
|udl.ll .|
|ull. .|
|rlr. .|
|rdr..d .|
|Uruull *|
+----------+
13: Starts Into Edge
+----------+
| U |
| |
| |
| |
| |
| |
| |
| |
| |
+----------+
+----------+
| U |
| |
| |
| |
| |
| |
| |
| |
| |
+----------+
14: Crossing Dead Paths
+----------+
| |
| |
| |
| R |
| |
| |
| |
| |
| U|
+----------+
+----------+
| *|
| .|
| .|
| R..*|
| .|
| .|
| .|
| .|
| U|
+----------+
code-golf
edited Jan 26 at 11:57
Erik the Outgolfer
32k429103
asked Jan 24 at 13:46
AJFaradayAJFaraday
3,47643159
$endgroup$
|
show 6 more comments
25
$begingroup$
This challenge takes place on a grid.
+----------+
| |
| |
| |
| |
| |
| |
| |
| |
| |
+----------+
This one's 10 x 10, but it can be any rectangular shape.
There are four directions on this grid. Up, down, left and right.
The task is to draw a path starting with an upper case direction initial. In this example, will go directly upward from the U.
+----------+
| |
| |
| |
| |
| |
| |
| |
| |
| U |
+----------+
The path will go upwards and be comprised of full-stop characters (.), until it hits a wall, when it will terminate with an asterisk (*).
+----------+
| * |
| . |
| . |
| . |
| . |
| . |
| . |
| . |
| U |
+----------+
In addition to path starts, there's also direction changers, represented by a lower case direction initial.
+----------+
| |
| |
| |
| r.....*|
| . |
| . |
| . |
| . |
| U |
+----------+
Also, an upper case X us an obstacle which will terminate the path.
+----------+
| |
| |
| |
| |
| r...*X |
| . |
| . |
| . |
| U |
+----------+
Rules
- The input is a string consisting of a frame, (consisting of |, - and + characters) containing characters denoting path starts, direction changers, and obstacles.
- Your code should add full stop characters to follow the path described by starts and direction changers, and an asterisk when/if the path meets a wall or obstacle.
- There can be multiple path starts.
- The code will still terminate without error if the path describes a loop.
- If a path meets a path start, it will act as a direction changer.
- It's code golf, low-byte code and no standard loopholes, please.
- I always prefer links to an on-line interpreter.
Test Cases
1: Simple
+----------+
| |
| |
| |
| |
| |
| |
| |
| |
| U |
+----------+
+----------+
| * |
| . |
| . |
| . |
| . |
| . |
| . |
| . |
| U |
+----------+
2: Right turn
+----------+
| |
| |
| |
| r |
| |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| r.....*|
| . |
| . |
| . |
| . |
| U |
+----------+
3: Crossroads
+----------+
| |
| |
| |
| r d |
| |
| u l |
| |
| |
| U |
+----------+
+----------+
| * |
| . |
| . |
| . r..d |
| . . . |
| u....l |
| . |
| . |
| U |
+----------+
4: 4 Crossing paths
+----------+
| D |
| |
| |
|R |
| |
| L|
| |
| |
| U |
+----------+
+----------+
| * D |
| . . |
| . . |
|R........*|
| . . |
|*........L|
| . . |
| . . |
| U * |
+----------+
5: First Loop
+----------+
| |
| |
| |
| r d |
| |
| u l |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| r..d |
| . . |
| u..l |
| . |
| . |
| U |
+----------+
6: Starter as changer
+----------+
| |
| |
| |
| L |
| |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
|*..L |
| . |
| . |
| . |
| . |
| U |
+----------+
7: Straight Loop
+----------+
| |
| |
| |
| |
| r l |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| |
| r..l |
| . |
| . |
| . |
| U |
+----------+
8: Tight Knot
+----------+
| |
| |
| |
| d l |
| r u |
| r u |
| |
| |
| U |
+----------+
+----------+
| * |
| . |
| . |
| d..l |
| .r.u |
| r.u |
| . |
| . |
| U |
+----------+
9: An Obstacle
+----------+
| |
| |
| |
| |
| r X |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| |
| r...*X |
| . |
| . |
| . |
| U |
+----------+
10: S Shape
+----------+
|r d |
| |
| XXXXXXXX|
| d l |
|ul |
|XXXXXXX |
| |
|R u |
| |
+----------+
+----------+
|r.....d |
|. * |
|. XXXXXXXX|
|.d......l |
|ul . |
|XXXXXXX . |
| . |
|R.......u |
| |
+----------+
11: 4-Way Knot
+----------+
| D |
| |
| r |
|R d |
| |
| u L|
| l |
| |
| U |
+----------+
+----------+
| * D |
| . . |
| r.....*|
|R....d. |
| .... |
| .u....L|
|*.....l |
| . . |
| U * |
+----------+
12: Busy Junctions
+----------+
|rrrrr rrrd|
| rlrl |
|ul rrd |
|ruX X |
|udl ll |
|ull |
|rlr |
|rdr d |
|Uruull |
+----------+
+----------+
|rrrrr.rrrd|
|.rlrl .|
|ul rrd .|
|ruX.X. .|
|udl.ll .|
|ull. .|
|rlr. .|
|rdr..d .|
|Uruull *|
+----------+
13: Starts Into Edge
+----------+
| U |
| |
| |
| |
| |
| |
| |
| |
| |
+----------+
+----------+
| U |
| |
| |
| |
| |
| |
| |
| |
| |
+----------+
14: Crossing Dead Paths
+----------+
| |
| |
| |
| R |
| |
| |
| |
| |
| U|
+----------+
+----------+
| *|
| .|
| .|
| R..*|
| .|
| .|
| .|
| .|
| U|
+----------+
code-golf
edited Jan 26 at 11:57
Erik the Outgolfer
32k429103
asked Jan 24 at 13:46
AJFaradayAJFaraday
3,47643159
$endgroup$
$begingroup$
@TFeld Added, thanks!
$endgroup$
– AJFaraday
Jan 24 at 14:49
1
$begingroup$
It seems like all direction changers are always reached in your test cases, which could allow to simplify the algorithm. I'd suggest to add a test case where it's not true.
$endgroup$
– Arnauld
Jan 24 at 16:49
$begingroup$
@Arnauld I'm pretty sure there's some unused direction changers in case 12.
$endgroup$
– AJFaraday
Jan 24 at 16:54
1
$begingroup$
suggested testcase
$endgroup$
– tsh
Jan 25 at 3:37
3
$begingroup$
It is stated that the grid can be any rectangular shape, but all test cases seem to be identical in size and shape.
$endgroup$
– gastropner
Jan 25 at 4:18
|
show 6 more comments
25
25
25
3
$begingroup$
This challenge takes place on a grid.
+----------+
| |
| |
| |
| |
| |
| |
| |
| |
| |
+----------+
This one's 10 x 10, but it can be any rectangular shape.
There are four directions on this grid. Up, down, left and right.
The task is to draw a path starting with an upper case direction initial. In this example, will go directly upward from the U.
+----------+
| |
| |
| |
| |
| |
| |
| |
| |
| U |
+----------+
The path will go upwards and be comprised of full-stop characters (.), until it hits a wall, when it will terminate with an asterisk (*).
+----------+
| * |
| . |
| . |
| . |
| . |
| . |
| . |
| . |
| U |
+----------+
In addition to path starts, there's also direction changers, represented by a lower case direction initial.
+----------+
| |
| |
| |
| r.....*|
| . |
| . |
| . |
| . |
| U |
+----------+
Also, an upper case X us an obstacle which will terminate the path.
+----------+
| |
| |
| |
| |
| r...*X |
| . |
| . |
| . |
| U |
+----------+
Rules
- The input is a string consisting of a frame, (consisting of |, - and + characters) containing characters denoting path starts, direction changers, and obstacles.
- Your code should add full stop characters to follow the path described by starts and direction changers, and an asterisk when/if the path meets a wall or obstacle.
- There can be multiple path starts.
- The code will still terminate without error if the path describes a loop.
- If a path meets a path start, it will act as a direction changer.
- It's code golf, low-byte code and no standard loopholes, please.
- I always prefer links to an on-line interpreter.
Test Cases
1: Simple
+----------+
| |
| |
| |
| |
| |
| |
| |
| |
| U |
+----------+
+----------+
| * |
| . |
| . |
| . |
| . |
| . |
| . |
| . |
| U |
+----------+
2: Right turn
+----------+
| |
| |
| |
| r |
| |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| r.....*|
| . |
| . |
| . |
| . |
| U |
+----------+
3: Crossroads
+----------+
| |
| |
| |
| r d |
| |
| u l |
| |
| |
| U |
+----------+
+----------+
| * |
| . |
| . |
| . r..d |
| . . . |
| u....l |
| . |
| . |
| U |
+----------+
4: 4 Crossing paths
+----------+
| D |
| |
| |
|R |
| |
| L|
| |
| |
| U |
+----------+
+----------+
| * D |
| . . |
| . . |
|R........*|
| . . |
|*........L|
| . . |
| . . |
| U * |
+----------+
5: First Loop
+----------+
| |
| |
| |
| r d |
| |
| u l |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| r..d |
| . . |
| u..l |
| . |
| . |
| U |
+----------+
6: Starter as changer
+----------+
| |
| |
| |
| L |
| |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
|*..L |
| . |
| . |
| . |
| . |
| U |
+----------+
7: Straight Loop
+----------+
| |
| |
| |
| |
| r l |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| |
| r..l |
| . |
| . |
| . |
| U |
+----------+
8: Tight Knot
+----------+
| |
| |
| |
| d l |
| r u |
| r u |
| |
| |
| U |
+----------+
+----------+
| * |
| . |
| . |
| d..l |
| .r.u |
| r.u |
| . |
| . |
| U |
+----------+
9: An Obstacle
+----------+
| |
| |
| |
| |
| r X |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| |
| r...*X |
| . |
| . |
| . |
| U |
+----------+
10: S Shape
+----------+
|r d |
| |
| XXXXXXXX|
| d l |
|ul |
|XXXXXXX |
| |
|R u |
| |
+----------+
+----------+
|r.....d |
|. * |
|. XXXXXXXX|
|.d......l |
|ul . |
|XXXXXXX . |
| . |
|R.......u |
| |
+----------+
11: 4-Way Knot
+----------+
| D |
| |
| r |
|R d |
| |
| u L|
| l |
| |
| U |
+----------+
+----------+
| * D |
| . . |
| r.....*|
|R....d. |
| .... |
| .u....L|
|*.....l |
| . . |
| U * |
+----------+
12: Busy Junctions
+----------+
|rrrrr rrrd|
| rlrl |
|ul rrd |
|ruX X |
|udl ll |
|ull |
|rlr |
|rdr d |
|Uruull |
+----------+
+----------+
|rrrrr.rrrd|
|.rlrl .|
|ul rrd .|
|ruX.X. .|
|udl.ll .|
|ull. .|
|rlr. .|
|rdr..d .|
|Uruull *|
+----------+
13: Starts Into Edge
+----------+
| U |
| |
| |
| |
| |
| |
| |
| |
| |
+----------+
+----------+
| U |
| |
| |
| |
| |
| |
| |
| |
| |
+----------+
14: Crossing Dead Paths
+----------+
| |
| |
| |
| R |
| |
| |
| |
| |
| U|
+----------+
+----------+
| *|
| .|
| .|
| R..*|
| .|
| .|
| .|
| .|
| U|
+----------+
code-golf
edited Jan 26 at 11:57
Erik the Outgolfer
32k429103
asked Jan 24 at 13:46
AJFaradayAJFaraday
3,47643159
$endgroup$
This challenge takes place on a grid.
+----------+
| |
| |
| |
| |
| |
| |
| |
| |
| |
+----------+
This one's 10 x 10, but it can be any rectangular shape.
There are four directions on this grid. Up, down, left and right.
The task is to draw a path starting with an upper case direction initial. In this example, will go directly upward from the U.
+----------+
| |
| |
| |
| |
| |
| |
| |
| |
| U |
+----------+
The path will go upwards and be comprised of full-stop characters (.), until it hits a wall, when it will terminate with an asterisk (*).
+----------+
| * |
| . |
| . |
| . |
| . |
| . |
| . |
| . |
| U |
+----------+
In addition to path starts, there's also direction changers, represented by a lower case direction initial.
+----------+
| |
| |
| |
| r.....*|
| . |
| . |
| . |
| . |
| U |
+----------+
Also, an upper case X us an obstacle which will terminate the path.
+----------+
| |
| |
| |
| |
| r...*X |
| . |
| . |
| . |
| U |
+----------+
Rules
- The input is a string consisting of a frame, (consisting of |, - and + characters) containing characters denoting path starts, direction changers, and obstacles.
- Your code should add full stop characters to follow the path described by starts and direction changers, and an asterisk when/if the path meets a wall or obstacle.
- There can be multiple path starts.
- The code will still terminate without error if the path describes a loop.
- If a path meets a path start, it will act as a direction changer.
- It's code golf, low-byte code and no standard loopholes, please.
- I always prefer links to an on-line interpreter.
Test Cases
1: Simple
+----------+
| |
| |
| |
| |
| |
| |
| |
| |
| U |
+----------+
+----------+
| * |
| . |
| . |
| . |
| . |
| . |
| . |
| . |
| U |
+----------+
2: Right turn
+----------+
| |
| |
| |
| r |
| |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| r.....*|
| . |
| . |
| . |
| . |
| U |
+----------+
3: Crossroads
+----------+
| |
| |
| |
| r d |
| |
| u l |
| |
| |
| U |
+----------+
+----------+
| * |
| . |
| . |
| . r..d |
| . . . |
| u....l |
| . |
| . |
| U |
+----------+
4: 4 Crossing paths
+----------+
| D |
| |
| |
|R |
| |
| L|
| |
| |
| U |
+----------+
+----------+
| * D |
| . . |
| . . |
|R........*|
| . . |
|*........L|
| . . |
| . . |
| U * |
+----------+
5: First Loop
+----------+
| |
| |
| |
| r d |
| |
| u l |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| r..d |
| . . |
| u..l |
| . |
| . |
| U |
+----------+
6: Starter as changer
+----------+
| |
| |
| |
| L |
| |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
|*..L |
| . |
| . |
| . |
| . |
| U |
+----------+
7: Straight Loop
+----------+
| |
| |
| |
| |
| r l |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| |
| r..l |
| . |
| . |
| . |
| U |
+----------+
8: Tight Knot
+----------+
| |
| |
| |
| d l |
| r u |
| r u |
| |
| |
| U |
+----------+
+----------+
| * |
| . |
| . |
| d..l |
| .r.u |
| r.u |
| . |
| . |
| U |
+----------+
9: An Obstacle
+----------+
| |
| |
| |
| |
| r X |
| |
| |
| |
| U |
+----------+
+----------+
| |
| |
| |
| |
| r...*X |
| . |
| . |
| . |
| U |
+----------+
10: S Shape
+----------+
|r d |
| |
| XXXXXXXX|
| d l |
|ul |
|XXXXXXX |
| |
|R u |
| |
+----------+
+----------+
|r.....d |
|. * |
|. XXXXXXXX|
|.d......l |
|ul . |
|XXXXXXX . |
| . |
|R.......u |
| |
+----------+
11: 4-Way Knot
+----------+
| D |
| |
| r |
|R d |
| |
| u L|
| l |
| |
| U |
+----------+
+----------+
| * D |
| . . |
| r.....*|
|R....d. |
| .... |
| .u....L|
|*.....l |
| . . |
| U * |
+----------+
12: Busy Junctions
+----------+
|rrrrr rrrd|
| rlrl |
|ul rrd |
|ruX X |
|udl ll |
|ull |
|rlr |
|rdr d |
|Uruull |
+----------+
+----------+
|rrrrr.rrrd|
|.rlrl .|
|ul rrd .|
|ruX.X. .|
|udl.ll .|
|ull. .|
|rlr. .|
|rdr..d .|
|Uruull *|
+----------+
13: Starts Into Edge
+----------+
| U |
| |
| |
| |
| |
| |
| |
| |
| |
+----------+
+----------+
| U |
| |
| |
| |
| |
| |
| |
| |
| |
+----------+
14: Crossing Dead Paths
+----------+
| |
| |
| |
| R |
| |
| |
| |
| |
| U|
+----------+
+----------+
| *|
| .|
| .|
| R..*|
| .|
| .|
| .|
| .|
| U|
+----------+
code-golf
code-golf
edited Jan 26 at 11:57
Erik the Outgolfer
32k429103
asked Jan 24 at 13:46
AJFaradayAJFaraday
3,47643159
edited Jan 26 at 11:57
Erik the Outgolfer
32k429103
asked Jan 24 at 13:46
AJFaradayAJFaraday
3,47643159
edited Jan 26 at 11:57
Erik the Outgolfer
32k429103
edited Jan 26 at 11:57
Erik the Outgolfer
32k429103
edited Jan 26 at 11:57
Erik the Outgolfer
32k429103
32k429103
asked Jan 24 at 13:46
AJFaradayAJFaraday
3,47643159
asked Jan 24 at 13:46
AJFaradayAJFaraday
3,47643159
asked Jan 24 at 13:46
AJFaradayAJFaraday
3,47643159
3,47643159
$begingroup$
@TFeld Added, thanks!
$endgroup$
– AJFaraday
Jan 24 at 14:49
1
$begingroup$
It seems like all direction changers are always reached in your test cases, which could allow to simplify the algorithm. I'd suggest to add a test case where it's not true.
$endgroup$
– Arnauld
Jan 24 at 16:49
$begingroup$
@Arnauld I'm pretty sure there's some unused direction changers in case 12.
$endgroup$
– AJFaraday
Jan 24 at 16:54
1
$begingroup$
suggested testcase
$endgroup$
– tsh
Jan 25 at 3:37
3
$begingroup$
It is stated that the grid can be any rectangular shape, but all test cases seem to be identical in size and shape.
$endgroup$
– gastropner
Jan 25 at 4:18
|
show 6 more comments
$begingroup$
@TFeld Added, thanks!
$endgroup$
– AJFaraday
Jan 24 at 14:49
1
$begingroup$
It seems like all direction changers are always reached in your test cases, which could allow to simplify the algorithm. I'd suggest to add a test case where it's not true.
$endgroup$
– Arnauld
Jan 24 at 16:49
$begingroup$
@Arnauld I'm pretty sure there's some unused direction changers in case 12.
$endgroup$
– AJFaraday
Jan 24 at 16:54
1
$begingroup$
suggested testcase
$endgroup$
– tsh
Jan 25 at 3:37
3
$begingroup$
It is stated that the grid can be any rectangular shape, but all test cases seem to be identical in size and shape.
$endgroup$
– gastropner
Jan 25 at 4:18
$begingroup$
@TFeld Added, thanks!
$endgroup$
– AJFaraday
Jan 24 at 14:49
$begingroup$
@TFeld Added, thanks!
$endgroup$
– AJFaraday
Jan 24 at 14:49
1
1
$begingroup$
It seems like all direction changers are always reached in your test cases, which could allow to simplify the algorithm. I'd suggest to add a test case where it's not true.
$endgroup$
– Arnauld
Jan 24 at 16:49
$begingroup$
It seems like all direction changers are always reached in your test cases, which could allow to simplify the algorithm. I'd suggest to add a test case where it's not true.
$endgroup$
– Arnauld
Jan 24 at 16:49
$begingroup$
@Arnauld I'm pretty sure there's some unused direction changers in case 12.
$endgroup$
– AJFaraday
Jan 24 at 16:54
$begingroup$
@Arnauld I'm pretty sure there's some unused direction changers in case 12.
$endgroup$
– AJFaraday
Jan 24 at 16:54
1
1
$begingroup$
suggested testcase
$endgroup$
– tsh
Jan 25 at 3:37
$begingroup$
suggested testcase
$endgroup$
– tsh
Jan 25 at 3:37
3
3
$begingroup$
It is stated that the grid can be any rectangular shape, but all test cases seem to be identical in size and shape.
$endgroup$
– gastropner
Jan 25 at 4:18
$begingroup$
It is stated that the grid can be any rectangular shape, but all test cases seem to be identical in size and shape.
$endgroup$
– gastropner
Jan 25 at 4:18
|
show 6 more comments
5 Answers
5
active
oldest
votes
9
$begingroup$
JavaScript (ES6), 191 183 181 bytes
Thanks to @tsh for helping fix a bug
Takes input as a matrix of characters. Outputs by modifying the input.
f=(a,X,Y,d,n=0)=>a.map((r,y)=>r.map((v,x)=>(a+0)[i=' .*dlurDLUR'.indexOf(v),n]?X?X-x+~-d%2|Y-y+(d-2)%2?0:~i?f(a,x,y,i>2?i&3:d,n+1,r[x]=i?v:'.'):n?a[Y][X]='*':0:i>6&&f(a,x,y,i&3):0))
Try it online!
Commented
f = ( a, // a = input matrix
X, Y, // X, Y = coordinates of the previous cell
d, // d = current direction (0 .. 3)
n = 0 // n = number of iterations for the current path
) => //
a.map((r, y) => // for each row r a position y in a:
r.map((v, x) => // for each character v at position x in r:
(a + 0)[ //
i = ' .*dlurDLUR' // i = index of the character
.indexOf(v), // blocking characters '-', '|' and 'X' gives -1
n // by testing (a + 0)[n], we allow each cell to be
] // visited twice (once horizontally, once vertically)
? // if it is set:
X ? // if this is not the 1st iteration:
X - x + ~-d % 2 | // if x - X is not equal to dx[d]
Y - y + (d - 2) % 2 ? // or y - Y is not equal to dy[d]:
0 // ignore this cell
: // else:
~i ? // if this is not a blocking character:
f( // do a recursive call:
a, // pass a unchanged
x, y, // pass the coordinates of this cell
i > 2 ? i & 3 : d, // update d if v is a direction char.
n + 1, // increment n
r[x] = i ? v : '.' // if v is a space, set r[x] to '.'
) // end of recursive call
: // else (this is a blocking character):
n ? // if this is not the 1st iteration:
a[Y][X] = '*' // set the previous cell to '*'
: // else:
0 // do nothing
: // else (1st iteration):
i > 6 && // if v is a capital letter:
f(a, x, y, i & 3) // do a recursive call with this direction
: // else ((a + 0)[n] is not set):
0 // we must be in an infinite loop: abort
) // end of inner map()
) // end of outer map()
answered Jan 24 at 20:44
ArnauldArnauld
76.9k693323
$endgroup$
$begingroup$
btw,[...""+a].mapcould create an array with at least 2x length of a. I'm not sure if it helps.
$endgroup$
– tsh
Jan 25 at 9:27
$begingroup$
(a+0)[n]does save a byte, even thoughnnow needs to be initialized.
$endgroup$
– Arnauld
Jan 25 at 9:38
add a comment |
8
$begingroup$
Python 2, 283 279 293 288 279 bytes
e=enumerate
def f(M):
s=[(x,y,c)for y,l in e(M)for x,c in e(l)if'A'<c<'X'];v=set(s)
for x,y,C in s:
d=ord(C)%87%5;q=d>1;X,Y=x-d+q*3,y+~-d-q;c=M[Y][X];N=(X,Y,[C,c]['a'<c<'x'])
if'!'>c:M[Y][X]='.'
if(c in'-|X')*('/'>M[y][x]):M[y][x]='*'
if(c in'udlr. *')>({N}<v):v|={N};s+=N,
Try it online!
Takes a list of lists.
Outputs by modifying the input array.
answered Jan 24 at 14:26
TFeldTFeld
15.3k21245
$endgroup$
add a comment |
6
$begingroup$
Perl 5, 203 188 166 bytes
$l='K[ a-z](?=';$t='([-|X])?';$s=$_;/
/;$n='.'x"@-";{$_|=s/(?|R[.*]*$l$t)|$t${l}[.*]*L)|D$n(?:[.*]$n)*$l$n$t)|$t$n$l$n([.*]$n)*U))/$&eq$"?$1?'*':'.':uc$&/es?redo:$s}
TIO
How it works
$s=$_to save input into$sto restore lowercase changers.$_|=$sbecause bitwise or with spacewill not change.and*, lowercase lettersurldwill be restored with bitwise or operation.
/n/;$n='.'x"@-"to get "width" and$nto match any character "width" times
$l='K[ a-z](?=';$t='([-|X])?'to reduce regex length ;$lto match a lowercase letterurldor a space on a path,$tto match a terminator.
After replacement :
(?|
R[.*]*K[ a-z](?=([-|X])?)
|
([-|X])?K[ a-z](?=[.*]*L)
|
D$n(?:[.*]$n)*K[ a-z](?=$n([-|X])?)
|
([-|X])?$nK[ a-z](?=$n([.*]$n)*U)
)
- switches
/eto eval,/sso that.(inside$n) matches also a newline character
$&eq$"?$1?'*':'.':uc$&if matched is a space, if termiator matched*otherwise.otherwise uppercase.
answered Jan 24 at 16:14
Nahuel FouilleulNahuel Fouilleul
2,53529
$endgroup$
1
$begingroup$
@Arnauld, it works if you input one test case at a time.
$endgroup$
– Shaggy
Jan 24 at 18:29
$begingroup$
yes i posted quickly and couldn't check it's fixed reseting$sin footer.$sis used to save the input and to restaure lowercase letters because are switched to uppercase when drawing the path
$endgroup$
– Nahuel Fouilleul
Jan 25 at 8:03
add a comment |
4
$begingroup$
Clean, 409 bytes
import StdEnv,Data.List
q=flatlines
$m=foldl(zipWitha b|a=='*'||b=='*'='*'=max a b)(q m)[q(foldl(m(_,y,x)=[[if(b<>x||a<>y)if(k=='*')'.'k'*'\k<-r&b<-[0..]]\r<-m&a<-[0..]])m(last(takeWhile(not o hasDup)(inits(f 0y 0x)))))\l<-m&y<-[0..],c<-l&x<-[0..]|isUpper c]
where f a y b x=let(u,v)=(a+y,b+x)in(case toLower((m!!u)!!v)of' '=[((a,b),u,v):f a u b v];'r'=f 0u 1v;'l'=f 0u -1v;'u'=f -1u 0v;'d'=f 1u 0v;_=)
Try it online!
answered Jan 25 at 0:02
ΟurousΟurous
7,32111035
$endgroup$
add a comment |
3
$begingroup$
Python 2, 250 bytes
def f(G,e=enumerate):
for i,k in e(G):
for j,l in e(k):
v=X=x=y=m,=l,
while(m in'-X|')<(l in'DLRU')>(X in v):v+=X,;y,x=zip((1,0,0,-1,y),(0,-1,1,0,x))['DLRU dlru'.find(m)%5];G[i][j]=(m,'.*'[G[i+y][j+x]in'-X|'])[m<'!'];i+=y;j+=x;X=x,i,j;m=G[i][j]
Try it online!
Takes a list of lists of 1-char strings, as explicitly allowed by the OP.
Changes the list in place.
For easier I/O, use this.
answered Jan 26 at 13:15
Erik the OutgolferErik the Outgolfer
32k429103
$endgroup$
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
9
$begingroup$
JavaScript (ES6), 191 183 181 bytes
Thanks to @tsh for helping fix a bug
Takes input as a matrix of characters. Outputs by modifying the input.
f=(a,X,Y,d,n=0)=>a.map((r,y)=>r.map((v,x)=>(a+0)[i=' .*dlurDLUR'.indexOf(v),n]?X?X-x+~-d%2|Y-y+(d-2)%2?0:~i?f(a,x,y,i>2?i&3:d,n+1,r[x]=i?v:'.'):n?a[Y][X]='*':0:i>6&&f(a,x,y,i&3):0))
Try it online!
Commented
f = ( a, // a = input matrix
X, Y, // X, Y = coordinates of the previous cell
d, // d = current direction (0 .. 3)
n = 0 // n = number of iterations for the current path
) => //
a.map((r, y) => // for each row r a position y in a:
r.map((v, x) => // for each character v at position x in r:
(a + 0)[ //
i = ' .*dlurDLUR' // i = index of the character
.indexOf(v), // blocking characters '-', '|' and 'X' gives -1
n // by testing (a + 0)[n], we allow each cell to be
] // visited twice (once horizontally, once vertically)
? // if it is set:
X ? // if this is not the 1st iteration:
X - x + ~-d % 2 | // if x - X is not equal to dx[d]
Y - y + (d - 2) % 2 ? // or y - Y is not equal to dy[d]:
0 // ignore this cell
: // else:
~i ? // if this is not a blocking character:
f( // do a recursive call:
a, // pass a unchanged
x, y, // pass the coordinates of this cell
i > 2 ? i & 3 : d, // update d if v is a direction char.
n + 1, // increment n
r[x] = i ? v : '.' // if v is a space, set r[x] to '.'
) // end of recursive call
: // else (this is a blocking character):
n ? // if this is not the 1st iteration:
a[Y][X] = '*' // set the previous cell to '*'
: // else:
0 // do nothing
: // else (1st iteration):
i > 6 && // if v is a capital letter:
f(a, x, y, i & 3) // do a recursive call with this direction
: // else ((a + 0)[n] is not set):
0 // we must be in an infinite loop: abort
) // end of inner map()
) // end of outer map()
answered Jan 24 at 20:44
ArnauldArnauld
76.9k693323
$endgroup$
$begingroup$
btw,[...""+a].mapcould create an array with at least 2x length of a. I'm not sure if it helps.
$endgroup$
– tsh
Jan 25 at 9:27
$begingroup$
(a+0)[n]does save a byte, even thoughnnow needs to be initialized.
$endgroup$
– Arnauld
Jan 25 at 9:38
add a comment |
9
$begingroup$
JavaScript (ES6), 191 183 181 bytes
Thanks to @tsh for helping fix a bug
Takes input as a matrix of characters. Outputs by modifying the input.
f=(a,X,Y,d,n=0)=>a.map((r,y)=>r.map((v,x)=>(a+0)[i=' .*dlurDLUR'.indexOf(v),n]?X?X-x+~-d%2|Y-y+(d-2)%2?0:~i?f(a,x,y,i>2?i&3:d,n+1,r[x]=i?v:'.'):n?a[Y][X]='*':0:i>6&&f(a,x,y,i&3):0))
Try it online!
Commented
f = ( a, // a = input matrix
X, Y, // X, Y = coordinates of the previous cell
d, // d = current direction (0 .. 3)
n = 0 // n = number of iterations for the current path
) => //
a.map((r, y) => // for each row r a position y in a:
r.map((v, x) => // for each character v at position x in r:
(a + 0)[ //
i = ' .*dlurDLUR' // i = index of the character
.indexOf(v), // blocking characters '-', '|' and 'X' gives -1
n // by testing (a + 0)[n], we allow each cell to be
] // visited twice (once horizontally, once vertically)
? // if it is set:
X ? // if this is not the 1st iteration:
X - x + ~-d % 2 | // if x - X is not equal to dx[d]
Y - y + (d - 2) % 2 ? // or y - Y is not equal to dy[d]:
0 // ignore this cell
: // else:
~i ? // if this is not a blocking character:
f( // do a recursive call:
a, // pass a unchanged
x, y, // pass the coordinates of this cell
i > 2 ? i & 3 : d, // update d if v is a direction char.
n + 1, // increment n
r[x] = i ? v : '.' // if v is a space, set r[x] to '.'
) // end of recursive call
: // else (this is a blocking character):
n ? // if this is not the 1st iteration:
a[Y][X] = '*' // set the previous cell to '*'
: // else:
0 // do nothing
: // else (1st iteration):
i > 6 && // if v is a capital letter:
f(a, x, y, i & 3) // do a recursive call with this direction
: // else ((a + 0)[n] is not set):
0 // we must be in an infinite loop: abort
) // end of inner map()
) // end of outer map()
answered Jan 24 at 20:44
ArnauldArnauld
76.9k693323
$endgroup$
$begingroup$
btw,[...""+a].mapcould create an array with at least 2x length of a. I'm not sure if it helps.
$endgroup$
– tsh
Jan 25 at 9:27
$begingroup$
(a+0)[n]does save a byte, even thoughnnow needs to be initialized.
$endgroup$
– Arnauld
Jan 25 at 9:38
add a comment |
9
9
9
$begingroup$
JavaScript (ES6), 191 183 181 bytes
Thanks to @tsh for helping fix a bug
Takes input as a matrix of characters. Outputs by modifying the input.
f=(a,X,Y,d,n=0)=>a.map((r,y)=>r.map((v,x)=>(a+0)[i=' .*dlurDLUR'.indexOf(v),n]?X?X-x+~-d%2|Y-y+(d-2)%2?0:~i?f(a,x,y,i>2?i&3:d,n+1,r[x]=i?v:'.'):n?a[Y][X]='*':0:i>6&&f(a,x,y,i&3):0))
Try it online!
Commented
f = ( a, // a = input matrix
X, Y, // X, Y = coordinates of the previous cell
d, // d = current direction (0 .. 3)
n = 0 // n = number of iterations for the current path
) => //
a.map((r, y) => // for each row r a position y in a:
r.map((v, x) => // for each character v at position x in r:
(a + 0)[ //
i = ' .*dlurDLUR' // i = index of the character
.indexOf(v), // blocking characters '-', '|' and 'X' gives -1
n // by testing (a + 0)[n], we allow each cell to be
] // visited twice (once horizontally, once vertically)
? // if it is set:
X ? // if this is not the 1st iteration:
X - x + ~-d % 2 | // if x - X is not equal to dx[d]
Y - y + (d - 2) % 2 ? // or y - Y is not equal to dy[d]:
0 // ignore this cell
: // else:
~i ? // if this is not a blocking character:
f( // do a recursive call:
a, // pass a unchanged
x, y, // pass the coordinates of this cell
i > 2 ? i & 3 : d, // update d if v is a direction char.
n + 1, // increment n
r[x] = i ? v : '.' // if v is a space, set r[x] to '.'
) // end of recursive call
: // else (this is a blocking character):
n ? // if this is not the 1st iteration:
a[Y][X] = '*' // set the previous cell to '*'
: // else:
0 // do nothing
: // else (1st iteration):
i > 6 && // if v is a capital letter:
f(a, x, y, i & 3) // do a recursive call with this direction
: // else ((a + 0)[n] is not set):
0 // we must be in an infinite loop: abort
) // end of inner map()
) // end of outer map()
answered Jan 24 at 20:44
ArnauldArnauld
76.9k693323
$endgroup$
JavaScript (ES6), 191 183 181 bytes
Thanks to @tsh for helping fix a bug
Takes input as a matrix of characters. Outputs by modifying the input.
f=(a,X,Y,d,n=0)=>a.map((r,y)=>r.map((v,x)=>(a+0)[i=' .*dlurDLUR'.indexOf(v),n]?X?X-x+~-d%2|Y-y+(d-2)%2?0:~i?f(a,x,y,i>2?i&3:d,n+1,r[x]=i?v:'.'):n?a[Y][X]='*':0:i>6&&f(a,x,y,i&3):0))
Try it online!
Commented
f = ( a, // a = input matrix
X, Y, // X, Y = coordinates of the previous cell
d, // d = current direction (0 .. 3)
n = 0 // n = number of iterations for the current path
) => //
a.map((r, y) => // for each row r a position y in a:
r.map((v, x) => // for each character v at position x in r:
(a + 0)[ //
i = ' .*dlurDLUR' // i = index of the character
.indexOf(v), // blocking characters '-', '|' and 'X' gives -1
n // by testing (a + 0)[n], we allow each cell to be
] // visited twice (once horizontally, once vertically)
? // if it is set:
X ? // if this is not the 1st iteration:
X - x + ~-d % 2 | // if x - X is not equal to dx[d]
Y - y + (d - 2) % 2 ? // or y - Y is not equal to dy[d]:
0 // ignore this cell
: // else:
~i ? // if this is not a blocking character:
f( // do a recursive call:
a, // pass a unchanged
x, y, // pass the coordinates of this cell
i > 2 ? i & 3 : d, // update d if v is a direction char.
n + 1, // increment n
r[x] = i ? v : '.' // if v is a space, set r[x] to '.'
) // end of recursive call
: // else (this is a blocking character):
n ? // if this is not the 1st iteration:
a[Y][X] = '*' // set the previous cell to '*'
: // else:
0 // do nothing
: // else (1st iteration):
i > 6 && // if v is a capital letter:
f(a, x, y, i & 3) // do a recursive call with this direction
: // else ((a + 0)[n] is not set):
0 // we must be in an infinite loop: abort
) // end of inner map()
) // end of outer map()
answered Jan 24 at 20:44
ArnauldArnauld
76.9k693323
edited Jan 25 at 9:56
answered Jan 24 at 20:44
ArnauldArnauld
76.9k693323
answered Jan 24 at 20:44
ArnauldArnauld
76.9k693323
answered Jan 24 at 20:44
ArnauldArnauld
76.9k693323
76.9k693323
$begingroup$
btw,[...""+a].mapcould create an array with at least 2x length of a. I'm not sure if it helps.
$endgroup$
– tsh
Jan 25 at 9:27
$begingroup$
(a+0)[n]does save a byte, even thoughnnow needs to be initialized.
$endgroup$
– Arnauld
Jan 25 at 9:38
add a comment |
$begingroup$
btw,[...""+a].mapcould create an array with at least 2x length of a. I'm not sure if it helps.
$endgroup$
– tsh
Jan 25 at 9:27
$begingroup$
(a+0)[n]does save a byte, even thoughnnow needs to be initialized.
$endgroup$
– Arnauld
Jan 25 at 9:38
$begingroup$
btw,
[...""+a].map could create an array with at least 2x length of a. I'm not sure if it helps.$endgroup$
– tsh
Jan 25 at 9:27
$begingroup$
btw,
[...""+a].map could create an array with at least 2x length of a. I'm not sure if it helps.$endgroup$
– tsh
Jan 25 at 9:27
$begingroup$
(a+0)[n] does save a byte, even though n now needs to be initialized.$endgroup$
– Arnauld
Jan 25 at 9:38
$begingroup$
(a+0)[n] does save a byte, even though n now needs to be initialized.$endgroup$
– Arnauld
Jan 25 at 9:38
add a comment |
8
$begingroup$
Python 2, 283 279 293 288 279 bytes
e=enumerate
def f(M):
s=[(x,y,c)for y,l in e(M)for x,c in e(l)if'A'<c<'X'];v=set(s)
for x,y,C in s:
d=ord(C)%87%5;q=d>1;X,Y=x-d+q*3,y+~-d-q;c=M[Y][X];N=(X,Y,[C,c]['a'<c<'x'])
if'!'>c:M[Y][X]='.'
if(c in'-|X')*('/'>M[y][x]):M[y][x]='*'
if(c in'udlr. *')>({N}<v):v|={N};s+=N,
Try it online!
Takes a list of lists.
Outputs by modifying the input array.
answered Jan 24 at 14:26
TFeldTFeld
15.3k21245
$endgroup$
add a comment |
8
$begingroup$
Python 2, 283 279 293 288 279 bytes
e=enumerate
def f(M):
s=[(x,y,c)for y,l in e(M)for x,c in e(l)if'A'<c<'X'];v=set(s)
for x,y,C in s:
d=ord(C)%87%5;q=d>1;X,Y=x-d+q*3,y+~-d-q;c=M[Y][X];N=(X,Y,[C,c]['a'<c<'x'])
if'!'>c:M[Y][X]='.'
if(c in'-|X')*('/'>M[y][x]):M[y][x]='*'
if(c in'udlr. *')>({N}<v):v|={N};s+=N,
Try it online!
Takes a list of lists.
Outputs by modifying the input array.
answered Jan 24 at 14:26
TFeldTFeld
15.3k21245
$endgroup$
add a comment |
8
8
8
$begingroup$
Python 2, 283 279 293 288 279 bytes
e=enumerate
def f(M):
s=[(x,y,c)for y,l in e(M)for x,c in e(l)if'A'<c<'X'];v=set(s)
for x,y,C in s:
d=ord(C)%87%5;q=d>1;X,Y=x-d+q*3,y+~-d-q;c=M[Y][X];N=(X,Y,[C,c]['a'<c<'x'])
if'!'>c:M[Y][X]='.'
if(c in'-|X')*('/'>M[y][x]):M[y][x]='*'
if(c in'udlr. *')>({N}<v):v|={N};s+=N,
Try it online!
Takes a list of lists.
Outputs by modifying the input array.
answered Jan 24 at 14:26
TFeldTFeld
15.3k21245
$endgroup$
Python 2, 283 279 293 288 279 bytes
e=enumerate
def f(M):
s=[(x,y,c)for y,l in e(M)for x,c in e(l)if'A'<c<'X'];v=set(s)
for x,y,C in s:
d=ord(C)%87%5;q=d>1;X,Y=x-d+q*3,y+~-d-q;c=M[Y][X];N=(X,Y,[C,c]['a'<c<'x'])
if'!'>c:M[Y][X]='.'
if(c in'-|X')*('/'>M[y][x]):M[y][x]='*'
if(c in'udlr. *')>({N}<v):v|={N};s+=N,
Try it online!
Takes a list of lists.
Outputs by modifying the input array.
answered Jan 24 at 14:26
TFeldTFeld
15.3k21245
edited Jan 25 at 10:27
answered Jan 24 at 14:26
TFeldTFeld
15.3k21245
answered Jan 24 at 14:26
TFeldTFeld
15.3k21245
answered Jan 24 at 14:26
TFeldTFeld
15.3k21245
15.3k21245
add a comment |
add a comment |
6
$begingroup$
Perl 5, 203 188 166 bytes
$l='K[ a-z](?=';$t='([-|X])?';$s=$_;/
/;$n='.'x"@-";{$_|=s/(?|R[.*]*$l$t)|$t${l}[.*]*L)|D$n(?:[.*]$n)*$l$n$t)|$t$n$l$n([.*]$n)*U))/$&eq$"?$1?'*':'.':uc$&/es?redo:$s}
TIO
How it works
$s=$_to save input into$sto restore lowercase changers.$_|=$sbecause bitwise or with spacewill not change.and*, lowercase lettersurldwill be restored with bitwise or operation.
/n/;$n='.'x"@-"to get "width" and$nto match any character "width" times
$l='K[ a-z](?=';$t='([-|X])?'to reduce regex length ;$lto match a lowercase letterurldor a space on a path,$tto match a terminator.
After replacement :
(?|
R[.*]*K[ a-z](?=([-|X])?)
|
([-|X])?K[ a-z](?=[.*]*L)
|
D$n(?:[.*]$n)*K[ a-z](?=$n([-|X])?)
|
([-|X])?$nK[ a-z](?=$n([.*]$n)*U)
)
- switches
/eto eval,/sso that.(inside$n) matches also a newline character
$&eq$"?$1?'*':'.':uc$&if matched is a space, if termiator matched*otherwise.otherwise uppercase.
answered Jan 24 at 16:14
Nahuel FouilleulNahuel Fouilleul
2,53529
$endgroup$
1
$begingroup$
@Arnauld, it works if you input one test case at a time.
$endgroup$
– Shaggy
Jan 24 at 18:29
$begingroup$
yes i posted quickly and couldn't check it's fixed reseting$sin footer.$sis used to save the input and to restaure lowercase letters because are switched to uppercase when drawing the path
$endgroup$
– Nahuel Fouilleul
Jan 25 at 8:03
add a comment |
6
$begingroup$
Perl 5, 203 188 166 bytes
$l='K[ a-z](?=';$t='([-|X])?';$s=$_;/
/;$n='.'x"@-";{$_|=s/(?|R[.*]*$l$t)|$t${l}[.*]*L)|D$n(?:[.*]$n)*$l$n$t)|$t$n$l$n([.*]$n)*U))/$&eq$"?$1?'*':'.':uc$&/es?redo:$s}
TIO
How it works
$s=$_to save input into$sto restore lowercase changers.$_|=$sbecause bitwise or with spacewill not change.and*, lowercase lettersurldwill be restored with bitwise or operation.
/n/;$n='.'x"@-"to get "width" and$nto match any character "width" times
$l='K[ a-z](?=';$t='([-|X])?'to reduce regex length ;$lto match a lowercase letterurldor a space on a path,$tto match a terminator.
After replacement :
(?|
R[.*]*K[ a-z](?=([-|X])?)
|
([-|X])?K[ a-z](?=[.*]*L)
|
D$n(?:[.*]$n)*K[ a-z](?=$n([-|X])?)
|
([-|X])?$nK[ a-z](?=$n([.*]$n)*U)
)
- switches
/eto eval,/sso that.(inside$n) matches also a newline character
$&eq$"?$1?'*':'.':uc$&if matched is a space, if termiator matched*otherwise.otherwise uppercase.
answered Jan 24 at 16:14
Nahuel FouilleulNahuel Fouilleul
2,53529
$endgroup$
1
$begingroup$
@Arnauld, it works if you input one test case at a time.
$endgroup$
– Shaggy
Jan 24 at 18:29
$begingroup$
yes i posted quickly and couldn't check it's fixed reseting$sin footer.$sis used to save the input and to restaure lowercase letters because are switched to uppercase when drawing the path
$endgroup$
– Nahuel Fouilleul
Jan 25 at 8:03
add a comment |
6
6
6
$begingroup$
Perl 5, 203 188 166 bytes
$l='K[ a-z](?=';$t='([-|X])?';$s=$_;/
/;$n='.'x"@-";{$_|=s/(?|R[.*]*$l$t)|$t${l}[.*]*L)|D$n(?:[.*]$n)*$l$n$t)|$t$n$l$n([.*]$n)*U))/$&eq$"?$1?'*':'.':uc$&/es?redo:$s}
TIO
How it works
$s=$_to save input into$sto restore lowercase changers.$_|=$sbecause bitwise or with spacewill not change.and*, lowercase lettersurldwill be restored with bitwise or operation.
/n/;$n='.'x"@-"to get "width" and$nto match any character "width" times
$l='K[ a-z](?=';$t='([-|X])?'to reduce regex length ;$lto match a lowercase letterurldor a space on a path,$tto match a terminator.
After replacement :
(?|
R[.*]*K[ a-z](?=([-|X])?)
|
([-|X])?K[ a-z](?=[.*]*L)
|
D$n(?:[.*]$n)*K[ a-z](?=$n([-|X])?)
|
([-|X])?$nK[ a-z](?=$n([.*]$n)*U)
)
- switches
/eto eval,/sso that.(inside$n) matches also a newline character
$&eq$"?$1?'*':'.':uc$&if matched is a space, if termiator matched*otherwise.otherwise uppercase.
answered Jan 24 at 16:14
Nahuel FouilleulNahuel Fouilleul
2,53529
$endgroup$
Perl 5, 203 188 166 bytes
$l='K[ a-z](?=';$t='([-|X])?';$s=$_;/
/;$n='.'x"@-";{$_|=s/(?|R[.*]*$l$t)|$t${l}[.*]*L)|D$n(?:[.*]$n)*$l$n$t)|$t$n$l$n([.*]$n)*U))/$&eq$"?$1?'*':'.':uc$&/es?redo:$s}
TIO
How it works
$s=$_to save input into$sto restore lowercase changers.$_|=$sbecause bitwise or with spacewill not change.and*, lowercase lettersurldwill be restored with bitwise or operation.
/n/;$n='.'x"@-"to get "width" and$nto match any character "width" times
$l='K[ a-z](?=';$t='([-|X])?'to reduce regex length ;$lto match a lowercase letterurldor a space on a path,$tto match a terminator.
After replacement :
(?|
R[.*]*K[ a-z](?=([-|X])?)
|
([-|X])?K[ a-z](?=[.*]*L)
|
D$n(?:[.*]$n)*K[ a-z](?=$n([-|X])?)
|
([-|X])?$nK[ a-z](?=$n([.*]$n)*U)
)
- switches
/eto eval,/sso that.(inside$n) matches also a newline character
$&eq$"?$1?'*':'.':uc$&if matched is a space, if termiator matched*otherwise.otherwise uppercase.
answered Jan 24 at 16:14
Nahuel FouilleulNahuel Fouilleul
2,53529
edited Jan 28 at 12:01
answered Jan 24 at 16:14
Nahuel FouilleulNahuel Fouilleul
2,53529
answered Jan 24 at 16:14
Nahuel FouilleulNahuel Fouilleul
2,53529
answered Jan 24 at 16:14
Nahuel FouilleulNahuel Fouilleul
2,53529
2,53529
1
$begingroup$
@Arnauld, it works if you input one test case at a time.
$endgroup$
– Shaggy
Jan 24 at 18:29
$begingroup$
yes i posted quickly and couldn't check it's fixed reseting$sin footer.$sis used to save the input and to restaure lowercase letters because are switched to uppercase when drawing the path
$endgroup$
– Nahuel Fouilleul
Jan 25 at 8:03
add a comment |
1
$begingroup$
@Arnauld, it works if you input one test case at a time.
$endgroup$
– Shaggy
Jan 24 at 18:29
$begingroup$
yes i posted quickly and couldn't check it's fixed reseting$sin footer.$sis used to save the input and to restaure lowercase letters because are switched to uppercase when drawing the path
$endgroup$
– Nahuel Fouilleul
Jan 25 at 8:03
1
1
$begingroup$
@Arnauld, it works if you input one test case at a time.
$endgroup$
– Shaggy
Jan 24 at 18:29
$begingroup$
@Arnauld, it works if you input one test case at a time.
$endgroup$
– Shaggy
Jan 24 at 18:29
$begingroup$
yes i posted quickly and couldn't check it's fixed reseting
$s in footer. $s is used to save the input and to restaure lowercase letters because are switched to uppercase when drawing the path$endgroup$
– Nahuel Fouilleul
Jan 25 at 8:03
$begingroup$
yes i posted quickly and couldn't check it's fixed reseting
$s in footer. $s is used to save the input and to restaure lowercase letters because are switched to uppercase when drawing the path$endgroup$
– Nahuel Fouilleul
Jan 25 at 8:03
add a comment |
4
$begingroup$
Clean, 409 bytes
import StdEnv,Data.List
q=flatlines
$m=foldl(zipWitha b|a=='*'||b=='*'='*'=max a b)(q m)[q(foldl(m(_,y,x)=[[if(b<>x||a<>y)if(k=='*')'.'k'*'\k<-r&b<-[0..]]\r<-m&a<-[0..]])m(last(takeWhile(not o hasDup)(inits(f 0y 0x)))))\l<-m&y<-[0..],c<-l&x<-[0..]|isUpper c]
where f a y b x=let(u,v)=(a+y,b+x)in(case toLower((m!!u)!!v)of' '=[((a,b),u,v):f a u b v];'r'=f 0u 1v;'l'=f 0u -1v;'u'=f -1u 0v;'d'=f 1u 0v;_=)
Try it online!
answered Jan 25 at 0:02
ΟurousΟurous
7,32111035
$endgroup$
add a comment |
4
$begingroup$
Clean, 409 bytes
import StdEnv,Data.List
q=flatlines
$m=foldl(zipWitha b|a=='*'||b=='*'='*'=max a b)(q m)[q(foldl(m(_,y,x)=[[if(b<>x||a<>y)if(k=='*')'.'k'*'\k<-r&b<-[0..]]\r<-m&a<-[0..]])m(last(takeWhile(not o hasDup)(inits(f 0y 0x)))))\l<-m&y<-[0..],c<-l&x<-[0..]|isUpper c]
where f a y b x=let(u,v)=(a+y,b+x)in(case toLower((m!!u)!!v)of' '=[((a,b),u,v):f a u b v];'r'=f 0u 1v;'l'=f 0u -1v;'u'=f -1u 0v;'d'=f 1u 0v;_=)
Try it online!
answered Jan 25 at 0:02
ΟurousΟurous
7,32111035
$endgroup$
add a comment |
4
4
4
$begingroup$
Clean, 409 bytes
import StdEnv,Data.List
q=flatlines
$m=foldl(zipWitha b|a=='*'||b=='*'='*'=max a b)(q m)[q(foldl(m(_,y,x)=[[if(b<>x||a<>y)if(k=='*')'.'k'*'\k<-r&b<-[0..]]\r<-m&a<-[0..]])m(last(takeWhile(not o hasDup)(inits(f 0y 0x)))))\l<-m&y<-[0..],c<-l&x<-[0..]|isUpper c]
where f a y b x=let(u,v)=(a+y,b+x)in(case toLower((m!!u)!!v)of' '=[((a,b),u,v):f a u b v];'r'=f 0u 1v;'l'=f 0u -1v;'u'=f -1u 0v;'d'=f 1u 0v;_=)
Try it online!
answered Jan 25 at 0:02
ΟurousΟurous
7,32111035
$endgroup$
Clean, 409 bytes
import StdEnv,Data.List
q=flatlines
$m=foldl(zipWitha b|a=='*'||b=='*'='*'=max a b)(q m)[q(foldl(m(_,y,x)=[[if(b<>x||a<>y)if(k=='*')'.'k'*'\k<-r&b<-[0..]]\r<-m&a<-[0..]])m(last(takeWhile(not o hasDup)(inits(f 0y 0x)))))\l<-m&y<-[0..],c<-l&x<-[0..]|isUpper c]
where f a y b x=let(u,v)=(a+y,b+x)in(case toLower((m!!u)!!v)of' '=[((a,b),u,v):f a u b v];'r'=f 0u 1v;'l'=f 0u -1v;'u'=f -1u 0v;'d'=f 1u 0v;_=)
Try it online!
answered Jan 25 at 0:02
ΟurousΟurous
7,32111035
answered Jan 25 at 0:02
ΟurousΟurous
7,32111035
answered Jan 25 at 0:02
ΟurousΟurous
7,32111035
answered Jan 25 at 0:02
ΟurousΟurous
7,32111035
7,32111035
add a comment |
add a comment |
3
$begingroup$
Python 2, 250 bytes
def f(G,e=enumerate):
for i,k in e(G):
for j,l in e(k):
v=X=x=y=m,=l,
while(m in'-X|')<(l in'DLRU')>(X in v):v+=X,;y,x=zip((1,0,0,-1,y),(0,-1,1,0,x))['DLRU dlru'.find(m)%5];G[i][j]=(m,'.*'[G[i+y][j+x]in'-X|'])[m<'!'];i+=y;j+=x;X=x,i,j;m=G[i][j]
Try it online!
Takes a list of lists of 1-char strings, as explicitly allowed by the OP.
Changes the list in place.
For easier I/O, use this.
answered Jan 26 at 13:15
Erik the OutgolferErik the Outgolfer
32k429103
$endgroup$
add a comment |
3
$begingroup$
Python 2, 250 bytes
def f(G,e=enumerate):
for i,k in e(G):
for j,l in e(k):
v=X=x=y=m,=l,
while(m in'-X|')<(l in'DLRU')>(X in v):v+=X,;y,x=zip((1,0,0,-1,y),(0,-1,1,0,x))['DLRU dlru'.find(m)%5];G[i][j]=(m,'.*'[G[i+y][j+x]in'-X|'])[m<'!'];i+=y;j+=x;X=x,i,j;m=G[i][j]
Try it online!
Takes a list of lists of 1-char strings, as explicitly allowed by the OP.
Changes the list in place.
For easier I/O, use this.
answered Jan 26 at 13:15
Erik the OutgolferErik the Outgolfer
32k429103
$endgroup$
add a comment |
3
3
3
$begingroup$
Python 2, 250 bytes
def f(G,e=enumerate):
for i,k in e(G):
for j,l in e(k):
v=X=x=y=m,=l,
while(m in'-X|')<(l in'DLRU')>(X in v):v+=X,;y,x=zip((1,0,0,-1,y),(0,-1,1,0,x))['DLRU dlru'.find(m)%5];G[i][j]=(m,'.*'[G[i+y][j+x]in'-X|'])[m<'!'];i+=y;j+=x;X=x,i,j;m=G[i][j]
Try it online!
Takes a list of lists of 1-char strings, as explicitly allowed by the OP.
Changes the list in place.
For easier I/O, use this.
answered Jan 26 at 13:15
Erik the OutgolferErik the Outgolfer
32k429103
$endgroup$
Python 2, 250 bytes
def f(G,e=enumerate):
for i,k in e(G):
for j,l in e(k):
v=X=x=y=m,=l,
while(m in'-X|')<(l in'DLRU')>(X in v):v+=X,;y,x=zip((1,0,0,-1,y),(0,-1,1,0,x))['DLRU dlru'.find(m)%5];G[i][j]=(m,'.*'[G[i+y][j+x]in'-X|'])[m<'!'];i+=y;j+=x;X=x,i,j;m=G[i][j]
Try it online!
Takes a list of lists of 1-char strings, as explicitly allowed by the OP.
Changes the list in place.
For easier I/O, use this.
answered Jan 26 at 13:15
Erik the OutgolferErik the Outgolfer
32k429103
edited Jan 26 at 13:24
answered Jan 26 at 13:15
Erik the OutgolferErik the Outgolfer
32k429103
answered Jan 26 at 13:15
Erik the OutgolferErik the Outgolfer
32k429103
answered Jan 26 at 13:15
Erik the OutgolferErik the Outgolfer
32k429103
32k429103
add a comment |
add a comment |
draft saved
draft discarded
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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$begingroup$
@TFeld Added, thanks!
$endgroup$
– AJFaraday
Jan 24 at 14:49
1
$begingroup$
It seems like all direction changers are always reached in your test cases, which could allow to simplify the algorithm. I'd suggest to add a test case where it's not true.
$endgroup$
– Arnauld
Jan 24 at 16:49
$begingroup$
@Arnauld I'm pretty sure there's some unused direction changers in case 12.
$endgroup$
– AJFaraday
Jan 24 at 16:54
1
$begingroup$
suggested testcase
$endgroup$
– tsh
Jan 25 at 3:37
3
$begingroup$
It is stated that the grid can be any rectangular shape, but all test cases seem to be identical in size and shape.
$endgroup$
– gastropner
Jan 25 at 4:18