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Python 3, 189 173 bytes lambda s:(f:=lambda i,a,b,c,d:l[i]&1 and(exec('l[i]=0')or a/b+b/c+c/d+d/a+f(i-7,c,a,d,b)+f(i-1,d,c,a,b)+f(i+1,c,d,b,a)+f(i+7,b,d,a,c)))((l:=3*s).index(35,41),4,8,63,7...

posted 7d ago by wizzwizz4‭ · edited 7d ago by wizzwizz4‭

Answer

#3: Post edited by

wizzwizz4‭ · 2026-09-10T11:23:56Z (7 days ago)
Restored attribution, elaborated slightly.

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# [Python 3], <del>189</del> <ins>173</ins> bytes
lambda s:(f:=lambda i,a,b,c,d:l[i]&1 and(exec('l[i]=0')or a/b+b/c+c/d+d/a+f(i-7,c,a,d,b)+f(i-1,d,c,a,b)+f(i+1,c,d,b,a)+f(i+7,b,d,a,c)))((l:=3*s).index(35,41),4,8,63,72)==111
~~[Try it online!][TIO-mtvea3d6]~~
[Python 3]: https://docs.python.org/3/
[TIO-mtvea3d6]: https://tio.run/##pVlNj9NIEL3nV1h@EtjE9Gx2gEGRcmHFFQ4Lpw0a2XFGRARnlASJ0YrfPjjf7nZ31eshF4897epyfb1X1fcP26@r5vrt/fqxmUwfl@X3qi6TzTi7G0@ON4uiLKpiVtTj5X@LL89GSdnU2fznfJY93z2Y/PU8X62T8qoaVlez4eyqHtZX5fAuW7y8ad8qi7qo8v3tqP1z9@BwOxztZLaSy8PtTftn3f53lud5li3Hk@sXm9wsmnr@M7t@Xbwa5cWr4m3x5rq4@TufTEaj0eM/n9@9v/3w/tO/ySSp0jSdDmB2vwHa3@5yuDO@y8DgcGtd9u/1lxqDi9TjnV/qSYBPeE8B@BQwqgLiZ8G7s/ezbGMpn3WSg9D39KXqFtg/5pZaCgC0Y4/CYQIKoOt1REr1vuF@lgFjV32psQJEUQCXNbKxjgKspRBDe@8Dv7iuAghkU@CznK/zhII/U@z3naVRkeV7I@RYkOECImF6IYU4b/l1dTKWCm0iBrxBCCK3uCBkk5vJgkBIQdU1qr5KaWh9upJbtrcM2PrKhwuDMJQLHDQUIwtaVgcqIZGxFBqa2DQMBJhIHSDZFSzG@sURFpBdYCsQn1vQdZUjC3waesWBT0MQMcAbS6aPdhZwBV6JgemgJbFmc79cbLMqnTbTJs0H7e/Dx0@3fa7bUVbD4/MaHY8DaeDDYy90h/AYhgQD6EDsR05EgKxATLq1mLHr0aUgeY6fnvdCC914iW5ifC5wvGXCOSYyrL5jD7oaNrIgdFp@jIMcLmq59EeWgvLQ6kogtLnyogahCwbQdTU8IbD0AKGr1/lBRiJSIiiAEaouRL9jGGM5uAEtuSNC241wmRbjSa0RIIY2TbQoBfROq1cJGakWZDKNgUui2NCWGwOjDwi8azQLxM8SwJYMKbfOrAWRpVikb9Bt74c4B2GkNNS4Lp0w58iiQ9vovdmFsGl69CkeG69S6Qo5X1gqQbbCB6iE0YumsHOoORT5QCx9NNGcUMqbXn2luYsM8jbG8iwjwq7gjcWwYm5UCtYFLrbpxophRISu0IdeVhMFsazIueW7uBOXmHiVahZ4SuatQFAmwPT8U@dZNnGl@gKjz2ph6ISBoRm8ywdopomIOmCY0DZsyWAUcOfkTMIQo32dDvcGFDtTaGcL@zURYw8wGQs1VcWe2@sQuQCAObeT5i6MBU5b0kVTom29pXwjbXiEMYbXle@5CQvYeEEiDGLHHvTpisKK1WoZPGNU6eOli6P5qzgD92FsCLe6aMiN9olO3j/zgkT2dTC6MJHYzggMcjMz8O4xLxcD9FGUNCDsLAWdBSfHEsaiQ9s/qIRcMiK99ZTDjdBUN7Zqqz1qd4aAP5gVuwhzFvfEpt8LnCepxEANJqLpV3G435ziaXwA/JnN3WqdzH5U89tmvk0WTXI@vxkPkvZXbjbz9TZpsuphOy/X6/IhO63O8yK5S7P/T/e/8mTzdfVjWSfVPCn3QpP2cbrfolltb7vbWEdF9lYre7vum6ctu88u2@7edLa@Xy@abZZ@/Jbmj78B "Python 3 – Try It Online"
Input is given as an ASCII-encoded `bytearray`.
## Explanation
We're solving this one _properly_. A cube has 6 faces, each of which can have 4 orientations, for a total of 24 rotational states, each corresponding to a permutation of the cube's diagonals. Labelling these diagonals:
| | |
| - | - |
| 3 | 2 |
| 4 | 1 |
where they touch the front face, a north rotation is $(a,b,c,d)\mapsto(c,a,d,b)$, and an east rotation is $(a,b,c,d)\mapsto(c,d,b,a)$. The rotation $(a,b,c,d)\mapsto(b,c,d,a)$ does not change which face is shown, so is useful for determining which face that is. If we represent the diagonals by $4$, $8$, $63$ and $72$, the hash function $(a,b,c,d)\mapsto\frac ab+\frac bc+\frac cd+\frac da$ assigns a unique double to each face, with sum $111$ iff a 6-face net is a valid cube.
Now we just need a flood fill algorithm. First, we need to ensure that it won't go out of bounds. The input is "represented as 6 newline _separated_ strings of 6 characters", therefore tripling the string will cause the columns to be slightly offset, meaning that the vertical line input won't take the flood out of bounds. To find our starting `#`, we'll start our search at the beginning of the second copy, at index $36+5=41$. East-west are offset by 1, and north-south are offset by 7. From here, we can simply recurse.
Since `.` and `\n` are both even characters, while `#` is odd, we can take input as a bytestring, and then the test is simply `&1`. To make sure we don't double-count a square, we zero it out once we've visited it. The walrus operator `:=` doesn't work for item assignment, so we use `exec` to turn it into an expression: expensive, but still shorter than `__setitem__`. This returns a falsy value, which requires an extra pair of brackets in the short-circuiting `and` / `or` expression.

# [Python 3], <del>189</del> <ins>173</ins> bytes
lambda s:(f:=lambda i,a,b,c,d:l[i]&1 and(exec('l[i]=0')or a/b+b/c+c/d+d/a+f(i-7,c,a,d,b)+f(i-1,d,c,a,b)+f(i+1,c,d,b,a)+f(i+7,b,d,a,c)))((l:=3*s).index(35,41),4,8,63,72)==111
[Try it online!][TIO-mtvea3d6] (test harness adapted from [this answer](https://codegolf.codidact.com/posts/292173/294030#answer-294030))
[Python 3]: https://docs.python.org/3/
[TIO-mtvea3d6]: https://tio.run/##pVlNj9NIEL3nV1h@EtjE9Gx2gEGRcmHFFQ4Lpw0a2XFGRARnlASJ0YrfPjjf7nZ31eshF4897epyfb1X1fcP26@r5vrt/fqxmUwfl@X3qi6TzTi7G0@ON4uiLKpiVtTj5X@LL89GSdnU2fznfJY93z2Y/PU8X62T8qoaVlez4eyqHtZX5fAuW7y8ad8qi7qo8v3tqP1z9@BwOxztZLaSy8PtTftn3f53lud5li3Hk@sXm9wsmnr@M7t@Xbwa5cWr4m3x5rq4@TufTEaj0eM/n9@9v/3w/tO/ySSp0jSdDmB2vwHa3@5yuDO@y8DgcGtd9u/1lxqDi9TjnV/qSYBPeE8B@BQwqgLiZ8G7s/ezbGMpn3WSg9D39KXqFtg/5pZaCgC0Y4/CYQIKoOt1REr1vuF@lgFjV32psQJEUQCXNbKxjgKspRBDe@8Dv7iuAghkU@CznK/zhII/U@z3naVRkeV7I@RYkOECImF6IYU4b/l1dTKWCm0iBrxBCCK3uCBkk5vJgkBIQdU1qr5KaWh9upJbtrcM2PrKhwuDMJQLHDQUIwtaVgcqIZGxFBqa2DQMBJhIHSDZFSzG@sURFpBdYCsQn1vQdZUjC3waesWBT0MQMcAbS6aPdhZwBV6JgemgJbFmc79cbLMqnTbTJs0H7e/Dx0@3fa7bUVbD4/MaHY8DaeDDYy90h/AYhgQD6EDsR05EgKxATLq1mLHr0aUgeY6fnvdCC914iW5ifC5wvGXCOSYyrL5jD7oaNrIgdFp@jIMcLmq59EeWgvLQ6kogtLnyogahCwbQdTU8IbD0AKGr1/lBRiJSIiiAEaouRL9jGGM5uAEtuSNC241wmRbjSa0RIIY2TbQoBfROq1cJGakWZDKNgUui2NCWGwOjDwi8azQLxM8SwJYMKbfOrAWRpVikb9Bt74c4B2GkNNS4Lp0w58iiQ9vovdmFsGl69CkeG69S6Qo5X1gqQbbCB6iE0YumsHOoORT5QCx9NNGcUMqbXn2luYsM8jbG8iwjwq7gjcWwYm5UCtYFLrbpxophRISu0IdeVhMFsazIueW7uBOXmHiVahZ4SuatQFAmwPT8U@dZNnGl@gKjz2ph6ISBoRm8ywdopomIOmCY0DZsyWAUcOfkTMIQo32dDvcGFDtTaGcL@zURYw8wGQs1VcWe2@sQuQCAObeT5i6MBU5b0kVTom29pXwjbXiEMYbXle@5CQvYeEEiDGLHHvTpisKK1WoZPGNU6eOli6P5qzgD92FsCLe6aMiN9olO3j/zgkT2dTC6MJHYzggMcjMz8O4xLxcD9FGUNCDsLAWdBSfHEsaiQ9s/qIRcMiK99ZTDjdBUN7Zqqz1qd4aAP5gVuwhzFvfEpt8LnCepxEANJqLpV3G435ziaXwA/JnN3WqdzH5U89tmvk0WTXI@vxkPkvZXbjbz9TZpsuphOy/X6/IhO63O8yK5S7P/T/e/8mTzdfVjWSfVPCn3QpP2cbrfolltb7vbWEdF9lYre7vum6ctu88u2@7edLa@Xy@abZZ@/Jbmj78B "Python 3 – Try It Online"
Input is given as an ASCII-encoded `bytearray`.
## Explanation
We're solving this one _properly_. A cube has 6 faces, each of which can have 4 orientations, for a total of 24 rotational states, each corresponding to a permutation of the cube's diagonals. Labelling these diagonals:
| | |
| - | - |
| 3 | 2 |
| 4 | 1 |
where they touch the front face, a north rotation is $(a,b,c,d)\mapsto(c,a,d,b)$, and an east rotation is $(a,b,c,d)\mapsto(c,d,b,a)$. The rotation $(a,b,c,d)\mapsto(b,c,d,a)$ does not change which face is shown, so is useful for determining which face that is. If we represent the diagonals by $4$, $8$, $63$ and $72$, the hash function $(a,b,c,d)\mapsto\frac ab+\frac bc+\frac cd+\frac da$ assigns a unique double to each face, with sum $111$ iff a 6-face net is a valid cube.
Now we just need a flood fill algorithm. First, we need to ensure that it won't go out of bounds. The input is "represented as 6 newline _separated_ strings of 6 characters", therefore tripling the string will cause the columns to be slightly offset, meaning that the vertical line input won't take the flood out of bounds. To find our starting `#`, we'll start our search at the beginning of the second copy, at index $36+5=41$. East-west are offset by 1, and north-south are offset by 7. From here, we can simply recurse.
Since `.` and `\n` are both even characters, while `#` is odd, we can take input as a bytestring, and then the test is simply `&1`. To make sure we don't double-count a square, we zero it out once we've visited it. The walrus operator `:=` doesn't work for item assignment, so we use `exec` to turn it into an expression: expensive, but still shorter than `__setitem__`. This returns a falsy value, which requires an extra pair of brackets in the short-circuiting `and` / `or` expression.
This solution does not require that nets are aligned to the top-left, and it also works for several nets that are not hexominoes (though there is a septomino that crashes it, so be warned).

#2: Post edited by

wizzwizz4‭ · 2026-09-10T10:47:28Z (7 days ago)
Saved 16 bytes by changing the hash function and rewriting to a lambda expression.

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~~# [Python 3], 189 bytes~~
~~def n(s):~~
def f(i,a,b,c,d):
~~if l[i]&1:l[i]=0;l[0]+=10*(a/b+b/c+c/d+d/a);f(i-7,c,a,d,b);f(i-1,d,c,a,b);f(i+1,c,d,b,a);f(i+7,b,d,a,c)~~
l=[0,*3*s];f(l.index(35,42),1,4,5,8);return l[0]==487
~~[Try it online!][TIO-mtujoxev] (test harness adapted from [this answer](https://codegolf.codidact.com/posts/292173/294030#answer-294030))~~
[Python 3]: https://docs.python.org/3/
[TIO-mtujoxev]: https://tio.run/##pVpNb9NAEL3nV6z8JBQ3ZtvQoiKqXEBc4QCc2qrKp4iInCpJJRDitxc7n9717szbkotjZz0ez9d7M5vH35sfy/Ly@XkynZmyu87fd0z9ddadF8NiVIyLSX3JzGdmcTu/f9V/Xx8GFzeL24v73qB/cdYdno96o/Nxb3w@6U3Oh/lNde/r6@rOYTEpRrvTfvW1vrA77fVruZX03eLedfV1Uv06zjtmMbi9KM4uz9b31U8LOy8n01/dy7fF1Zu86BdXxdviXX6zmm6eVqWpdRgMrt5dP3/8/uHTw@dP376agRllWXbXga0/HVSf@rA7s6FDx2J36hy297WXWouT1P1ZWOpBQEh4SwGEFLCqAuJrIfjk4Gu5xlJe6yAHsfdpS9UtsL3MLXUUAGjH7oXDRhRA0@tIlBq8w38tC8au@lLrBIiiAE5rZGPtBThLIYb21gdhcU0FEMmmyGt5bxcIhXCmuPd7S5MiK3RHzLEgwwVEwrRCCmneCuvqZSwV2kQMBIMQRG5xQcgmN5MFkZCCqmtSfZXS0Hl1Jbdcb1mw9ZUPFwZhKBd4aChGFrSsjlRCImMpNLSpaRgJMJE6QLIrWIwNiyMsILvAVSA9t6DrKkcW@DQMigOfhiBigDeWTB/dLOAKvBIDd52KxNr142K@6Y6yu/KuzPJO9fn85dtDm@s2lNXw@LhGx@NIGoTwOAjdMTyGJcEAOhCHkRMJICsQk2YtZuy6dylInhOm563QQjNekpuYkAs8b9l4jokMq@3Yna6WjSwInVYY4yCHi1ouw5GloDy0uhIJba68qEHogwF0XS1PCBw9QOgadH6UkYiUCApgxKoL0e9YxlgebkBL7oTQ9iNcpsV4UWsEiKFNEy1KAb3TalVCRqoDmUxj4JMoNrTlxsDqA4LgGs0C6bMEsCVDyq0ja0FiKRbpG3TbhyHOQxgpDTWuSyfMMbLo0LZ6b3YibJoebYrHxqtUumLOF5ZKkK3wASph9KIpPDnWHIp8IJU@2mROKOVNq77S3EUGeRdjeZaRYFfwxmJYMTcqBesCH9t0Y6UwIkJX6EMvp4mCWFbk3Aod/IlLSrxKNQs8JQtWICgTYHr@qfMsl7hSfYHVZ7WwdMLA0gze5wM000RCHbBMaFu2ZDAK@HNyJmGI0b5Oh1sDitoU2t7Cdk3C2ANMxkJNVbHnDjpELgBg9u2kuQtjgcMj6aIp0bbWUr6RtjzCWMvryvfchAVcvCARBqljD3p3RWHFarWM7jGq9PHUxdH8VZyBhzA2hltNNORG@0QnH555QSL7OhidmEhqZwQGuZkZeHObl4sBeitKGhA2loLOgoNjCWPRoR0eVEIuGYneesnmRmyqm1q11R61OUPAf8yKfYQ5inth0x8EzoNUYqAGm9D0qzjcbk7xMj4Afs9mtlyZ8dNo@lBON2ZemuP@Tf2/KmOG6/V0tTFl97AmL8ws6/45nP7NzfrH8mkxMaOpGW4lmepytpVbLjcPTdnO/pArf1k/o7l@/5zmpdOz6uXe8x5X83LTzb78zPLnfw "Python 3 – Try It Online"
We're solving this one _properly_. A cube has 6 faces, each of which can have 4 orientations, for a total of 24 rotational states, each corresponding to a permutation of the cube's diagonals. Labelling these diagonals:
| | |
| - | - |
| 3 | 2 |
| 4 | 1 |
where they touch the front face, a north rotation is $(a,b,c,d)\mapsto(c,a,d,b)$, and an east rotation is $(a,b,c,d)\mapsto(c,d,b,a)$. The rotation $(a,b,c,d)\mapsto(b,c,d,a)$ does not change which face is shown, so is useful for determining which face that is. If we represent the diagonals by $1$, $4$, $5$ and $8$, the hash function $(a,b,c,d)\mapsto5(\frac ab+\frac bc+\frac cd+\frac da)$ assigns a unique double to each face, with sum $243.5$ iff a 6-face net is a valid cube.
Now we just need a flood fill algorithm. First, we need to ensure that it won't go out of bounds. The input is "represented as 6 newline _separated_ strings of 6 characters", therefore tripling the string will cause the columns to be slightly offset, meaning that the vertical line input won't take the flood out of bounds. To find our starting `#`, we'll start our search at the beginning of the second copy, at index $36+5=41$. East-west are offset by 1, and north-south are offset by 7. From here, we can simply recurse.
Since `.` and `\n` are both even characters, while `#` is odd, we can take input as a bytestring, and then the test is simply `&1`. Of course, we also need a place to store the running total: the `nonlocal` keyword costs a few bytes, so we'll stuff it at the beginning of the list, and add 1 to the indexes.

# [Python 3], <del>189</del> <ins>173</ins> bytes
lambda s:(f:=lambda i,a,b,c,d:l[i]&1 and(exec('l[i]=0')or a/b+b/c+c/d+d/a+f(i-7,c,a,d,b)+f(i-1,d,c,a,b)+f(i+1,c,d,b,a)+f(i+7,b,d,a,c)))((l:=3*s).index(35,41),4,8,63,72)==111
[Try it online!][TIO-mtvea3d6]
[Python 3]: https://docs.python.org/3/
[TIO-mtvea3d6]: https://tio.run/##pVlNj9NIEL3nV1h@EtjE9Gx2gEGRcmHFFQ4Lpw0a2XFGRARnlASJ0YrfPjjf7nZ31eshF4897epyfb1X1fcP26@r5vrt/fqxmUwfl@X3qi6TzTi7G0@ON4uiLKpiVtTj5X@LL89GSdnU2fznfJY93z2Y/PU8X62T8qoaVlez4eyqHtZX5fAuW7y8ad8qi7qo8v3tqP1z9@BwOxztZLaSy8PtTftn3f53lud5li3Hk@sXm9wsmnr@M7t@Xbwa5cWr4m3x5rq4@TufTEaj0eM/n9@9v/3w/tO/ySSp0jSdDmB2vwHa3@5yuDO@y8DgcGtd9u/1lxqDi9TjnV/qSYBPeE8B@BQwqgLiZ8G7s/ezbGMpn3WSg9D39KXqFtg/5pZaCgC0Y4/CYQIKoOt1REr1vuF@lgFjV32psQJEUQCXNbKxjgKspRBDe@8Dv7iuAghkU@CznK/zhII/U@z3naVRkeV7I@RYkOECImF6IYU4b/l1dTKWCm0iBrxBCCK3uCBkk5vJgkBIQdU1qr5KaWh9upJbtrcM2PrKhwuDMJQLHDQUIwtaVgcqIZGxFBqa2DQMBJhIHSDZFSzG@sURFpBdYCsQn1vQdZUjC3waesWBT0MQMcAbS6aPdhZwBV6JgemgJbFmc79cbLMqnTbTJs0H7e/Dx0@3fa7bUVbD4/MaHY8DaeDDYy90h/AYhgQD6EDsR05EgKxATLq1mLHr0aUgeY6fnvdCC914iW5ifC5wvGXCOSYyrL5jD7oaNrIgdFp@jIMcLmq59EeWgvLQ6kogtLnyogahCwbQdTU8IbD0AKGr1/lBRiJSIiiAEaouRL9jGGM5uAEtuSNC241wmRbjSa0RIIY2TbQoBfROq1cJGakWZDKNgUui2NCWGwOjDwi8azQLxM8SwJYMKbfOrAWRpVikb9Bt74c4B2GkNNS4Lp0w58iiQ9vovdmFsGl69CkeG69S6Qo5X1gqQbbCB6iE0YumsHOoORT5QCx9NNGcUMqbXn2luYsM8jbG8iwjwq7gjcWwYm5UCtYFLrbpxophRISu0IdeVhMFsazIueW7uBOXmHiVahZ4SuatQFAmwPT8U@dZNnGl@gKjz2ph6ISBoRm8ywdopomIOmCY0DZsyWAUcOfkTMIQo32dDvcGFDtTaGcL@zURYw8wGQs1VcWe2@sQuQCAObeT5i6MBU5b0kVTom29pXwjbXiEMYbXle@5CQvYeEEiDGLHHvTpisKK1WoZPGNU6eOli6P5qzgD92FsCLe6aMiN9olO3j/zgkT2dTC6MJHYzggMcjMz8O4xLxcD9FGUNCDsLAWdBSfHEsaiQ9s/qIRcMiK99ZTDjdBUN7Zqqz1qd4aAP5gVuwhzFvfEpt8LnCepxEANJqLpV3G435ziaXwA/JnN3WqdzH5U89tmvk0WTXI@vxkPkvZXbjbz9TZpsuphOy/X6/IhO63O8yK5S7P/T/e/8mTzdfVjWSfVPCn3QpP2cbrfolltb7vbWEdF9lYre7vum6ctu88u2@7edLa@Xy@abZZ@/Jbmj78B "Python 3 – Try It Online"
Input is given as an ASCII-encoded `bytearray`.
## Explanation
We're solving this one _properly_. A cube has 6 faces, each of which can have 4 orientations, for a total of 24 rotational states, each corresponding to a permutation of the cube's diagonals. Labelling these diagonals:
| | |
| - | - |
| 3 | 2 |
| 4 | 1 |
where they touch the front face, a north rotation is $(a,b,c,d)\mapsto(c,a,d,b)$, and an east rotation is $(a,b,c,d)\mapsto(c,d,b,a)$. The rotation $(a,b,c,d)\mapsto(b,c,d,a)$ does not change which face is shown, so is useful for determining which face that is. If we represent the diagonals by $4$, $8$, $63$ and $72$, the hash function $(a,b,c,d)\mapsto\frac ab+\frac bc+\frac cd+\frac da$ assigns a unique double to each face, with sum $111$ iff a 6-face net is a valid cube.
Now we just need a flood fill algorithm. First, we need to ensure that it won't go out of bounds. The input is "represented as 6 newline _separated_ strings of 6 characters", therefore tripling the string will cause the columns to be slightly offset, meaning that the vertical line input won't take the flood out of bounds. To find our starting `#`, we'll start our search at the beginning of the second copy, at index $36+5=41$. East-west are offset by 1, and north-south are offset by 7. From here, we can simply recurse.
Since `.` and `\n` are both even characters, while `#` is odd, we can take input as a bytestring, and then the test is simply `&1`. To make sure we don't double-count a square, we zero it out once we've visited it. The walrus operator `:=` doesn't work for item assignment, so we use `exec` to turn it into an expression: expensive, but still shorter than `__setitem__`. This returns a falsy value, which requires an extra pair of brackets in the short-circuiting `and` / `or` expression.

#1: Initial revision by

wizzwizz4‭ · 2026-09-10T08:36:18Z (7 days ago)

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# [Python 3], 189 bytes

<!-- language-all: lang-python -->

    def n(s):
     def f(i,a,b,c,d):
      if l[i]&1:l[i]=0;l[0]+=10*(a/b+b/c+c/d+d/a);f(i-7,c,a,d,b);f(i-1,d,c,a,b);f(i+1,c,d,b,a);f(i+7,b,d,a,c)
     l=[0,*3*s];f(l.index(35,42),1,4,5,8);return l[0]==487

[Try it online!][TIO-mtujoxev] (test harness adapted from [this answer](https://codegolf.codidact.com/posts/292173/294030#answer-294030))

[Python 3]: https://docs.python.org/3/
[TIO-mtujoxev]: https://tio.run/##pVpNb9NAEL3nV6z8JBQ3ZtvQoiKqXEBc4QCc2qrKp4iInCpJJRDitxc7n9717szbkotjZz0ez9d7M5vH35sfy/Ly@XkynZmyu87fd0z9ddadF8NiVIyLSX3JzGdmcTu/f9V/Xx8GFzeL24v73qB/cdYdno96o/Nxb3w@6U3Oh/lNde/r6@rOYTEpRrvTfvW1vrA77fVruZX03eLedfV1Uv06zjtmMbi9KM4uz9b31U8LOy8n01/dy7fF1Zu86BdXxdviXX6zmm6eVqWpdRgMrt5dP3/8/uHTw@dP376agRllWXbXga0/HVSf@rA7s6FDx2J36hy297WXWouT1P1ZWOpBQEh4SwGEFLCqAuJrIfjk4Gu5xlJe6yAHsfdpS9UtsL3MLXUUAGjH7oXDRhRA0@tIlBq8w38tC8au@lLrBIiiAE5rZGPtBThLIYb21gdhcU0FEMmmyGt5bxcIhXCmuPd7S5MiK3RHzLEgwwVEwrRCCmneCuvqZSwV2kQMBIMQRG5xQcgmN5MFkZCCqmtSfZXS0Hl1Jbdcb1mw9ZUPFwZhKBd4aChGFrSsjlRCImMpNLSpaRgJMJE6QLIrWIwNiyMsILvAVSA9t6DrKkcW@DQMigOfhiBigDeWTB/dLOAKvBIDd52KxNr142K@6Y6yu/KuzPJO9fn85dtDm@s2lNXw@LhGx@NIGoTwOAjdMTyGJcEAOhCHkRMJICsQk2YtZuy6dylInhOm563QQjNekpuYkAs8b9l4jokMq@3Yna6WjSwInVYY4yCHi1ouw5GloDy0uhIJba68qEHogwF0XS1PCBw9QOgadH6UkYiUCApgxKoL0e9YxlgebkBL7oTQ9iNcpsV4UWsEiKFNEy1KAb3TalVCRqoDmUxj4JMoNrTlxsDqA4LgGs0C6bMEsCVDyq0ja0FiKRbpG3TbhyHOQxgpDTWuSyfMMbLo0LZ6b3YibJoebYrHxqtUumLOF5ZKkK3wASph9KIpPDnWHIp8IJU@2mROKOVNq77S3EUGeRdjeZaRYFfwxmJYMTcqBesCH9t0Y6UwIkJX6EMvp4mCWFbk3Aod/IlLSrxKNQs8JQtWICgTYHr@qfMsl7hSfYHVZ7WwdMLA0gze5wM000RCHbBMaFu2ZDAK@HNyJmGI0b5Oh1sDitoU2t7Cdk3C2ANMxkJNVbHnDjpELgBg9u2kuQtjgcMj6aIp0bbWUr6RtjzCWMvryvfchAVcvCARBqljD3p3RWHFarWM7jGq9PHUxdH8VZyBhzA2hltNNORG@0QnH555QSL7OhidmEhqZwQGuZkZeHObl4sBeitKGhA2loLOgoNjCWPRoR0eVEIuGYneesnmRmyqm1q11R61OUPAf8yKfYQ5inth0x8EzoNUYqAGm9D0qzjcbk7xMj4Afs9mtlyZ8dNo@lBON2ZemuP@Tf2/KmOG6/V0tTFl97AmL8ws6/45nP7NzfrH8mkxMaOpGW4lmepytpVbLjcPTdnO/pArf1k/o7l@/5zmpdOz6uXe8x5X83LTzb78zPLnfw "Python 3 – Try It Online"


We're solving this one _properly_. A cube has 6 faces, each of which can have 4 orientations, for a total of 24 rotational states, each corresponding to a permutation of the cube's diagonals. Labelling these diagonals:

|   |   |
| - | - |
| 3 | 2 |
| 4 | 1 |

where they touch the front face, a north rotation is $(a,b,c,d)\mapsto(c,a,d,b)$, and an east rotation is $(a,b,c,d)\mapsto(c,d,b,a)$. The rotation $(a,b,c,d)\mapsto(b,c,d,a)$ does not change which face is shown, so is useful for determining which face that is. If we represent the diagonals by $1$, $4$, $5$ and $8$, the hash function $(a,b,c,d)\mapsto5(\frac ab+\frac bc+\frac cd+\frac da)$ assigns a unique double to each face, with sum $243.5$ iff a 6-face net is a valid cube.

Now we just need a flood fill algorithm. First, we need to ensure that it won't go out of bounds. The input is "represented as 6 newline _separated_ strings of 6 characters", therefore tripling the string will cause the columns to be slightly offset, meaning that the vertical line input won't take the flood out of bounds. To find our starting `#`, we'll start our search at the beginning of the second copy, at index $36+5=41$. East-west are offset by 1, and north-south are offset by 7. From here, we can simply recurse.

Since `.` and `\n` are both even characters, while `#` is odd, we can take input as a bytestring, and then the test is simply `&1`. Of course, we also need a place to store the running total: the `nonlocal` keyword costs a few bytes, so we'll stuff it at the beginning of the list, and add 1 to the indexes.

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