[Question]
  [
A Geiger counter is a device that is used to detect radiation.
We will be making a Geiger counter program.
As we all know, when radiation hits a computer program it removes exactly 1 byte at random. So a Geiger counter program is a program that itself does nothing, but when any byte is removed the modified program prints `beep`, to indicate the presence of radiation.
Answers will be scored in bytes with fewer bytes being better. Answers must be at least 1 byte.
Your program may print `beep` with trailing a newline or print a single newline for empty output, so long as it does so consistently. Your program may also use a different case for `beep` such as `BEEP`, `bEEP`, or `Beep` so long as it does so consistently.
  
[Answer]
# [Lost](https://github.com/Wheatwizard/Lost), ~~303 293 263 253 238~~ 228 bytes
```
v^"peeb"<\>"beepvv"((>@@>>%%>>(((((([[[[[[\
>>>>>>>>>//>>>>>>>>>>>>>>/>>/>>>>>>>>>>>>>\\
>>>>>>>>//>>>>\>>>>>>>>>>/>>>>>>>>>>>>>>>>>\\
>/>>>>>>>/>>>>>>>>>>>>\>>>>>>>>>>>>>>>>>>>>>\\
>>>>>>>>>>>>>>>>>>>>>>\\>>>>\>>>>>>>>>>>>>>>>\
```
[Try it online!](https://tio.run/##y8kvLvn/vyxOqSA1NUnJJsZOKSk1taCsTElDw87Bwc5OVdXOTgMMosEghssOBvT17VCAPhghgRiEYojaGBTFaACkWh@bbIwdNoBkOLoEVk0x////13UMBAA "Lost – Try It Online")
[Verification script](https://tio.run/##hVPbbtswDH22voJzUMhG2lzahwHeYnTr3odhj3HWKjZdC3UkQZIT5OszykmWpMg2wjBE8/DwcmSz9Y1WDzu5Mtp60I4Zq1@tWMEMLOd8t/4VG8Rl/LnI4yWiWa/jJMkfH/P85ibPk97mvRUsP9p4nF/YuH/OrDiB99jiAvzOAnp8LVrk1@yM/H3galKxo0Hn0@xuuqDppfIJ/7r1CKXulM/4LbSoksNaRqhKXWHChSul5GmaMlZhDbZTz612/vmAO@LTjEUb6RvQhkh4yCVCvuEpCAc1RaN6tLHS458MFln0nVUkxsj0aXVsepnuYayNH4c6/WtktnD35QcE1jgdWRRVQv3ozgf1/tbRyJlW0oyF4gQWzmEQvvPzu/sFzGbAv6FHu5JKOi9Lfo6Y7AGc0Z5w/dQIC8H9xNgAngR1rNoteHS@b4lchEa0NQgPArxcIeG@K3BYalWFDm@hbIR6RXjpjNcv4DV8nDAWHGKesMCFVX8Mqb5Bu5EOwTfS9YQu9AXCklRVX2sjtiAVq7UFSQewgT4JhBc6pkEYWcPBncsFfJjBcaoMYAA/36QBWaGiLYiWJlLUd@flGkPTVpS0JUf6/Xvd84yoh6c6w2m2uJQgiv6vwQXmoEL4G0PkTIrTNPT9sLvhDKbkDfY3W4Z6NPchmMPDJIMl3Zy3iO1@Aw) (borrowed from [user 202729's answer](https://codegolf.stackexchange.com/a/169964/76162)). Unfortunately this can only test half of the code at a time, but rest assured I've tested the whole program.
Ouch, this was a tough one. I'll quote WW's deleted answer:
> 
> Lost is perhaps the most interesting language for this challenge. In
>  Lost the start location and direction of the pointer is entirely
>  random, thus to make deterministic programs you must account for every
>  possible start location and direction. At the same time, by the nature
>  of this challenge you must also account for any single byte being
>  removed.
> 
> 
> 
Unfortunately, his answer didn't take into account removing newlines, which screwed everything up.
### Explanation:
(note that a few bytes might be off here and there)
First let's talk about the general structure of the code:
```
v^^"peeb"<<\/"beepvv"((>>>@@>>%%>>(((((([[[[[[[\       Processing line
>>>>>>>>>>>//>>>>>>>>>>>>>>>>>>>/>>>>>>>>>>>>>>\\      Beep line
>>>>>>>>>//>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>\\     Back-up beep line
>//>>>>>>>>>>>>>>>>>>>>\\>>>>>>>>>>>>>>>>>>>>>>>>\\    Back-up return line
>>>>>>>>>>>>>>>>>>>>>>>>\\>>>>>>\>>>>>>>>>>>>>>>>>\    Return line
```
Everything but the processing line must be entirely composed of either `>` or one of `\/`. Why? Well, as an example, let's remove a newline:
```
v^^"peeb"<<\/"beepvv"((>>>@@>>%%>>(((((([[[[[[[\>>>>>>>>>>>//>>>>>>>>>>>>>>>>>>>/>>>>>>>>>>>>>>\\
>>>>>>>>>//>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>\\
>//>>>>>>>>>>>>>>>>>>>>\\>>>>>>>>>>>>>>>>>>>>>>>>\\
>>>>>>>>>>>>>>>>>>>>>>>>\\>>>>>>\>>>>>>>>>>>>>>>>>\
```
The first line is now *way* longer than the rest of the block. If a pointer were to spawn on a non-`>\/` character with vertical movement, then it would get stuck in an infinite loop. 
---
The biggest radiation detector part is the section at the end of each line.
```
 \
 \\
 >\\
 >>\\
 >>>\
```
Normally an IP passing through this from the first line would exit out the last line. However, if any character on the line is removed, then that line shifts down one, e.g:
```
 \
 \\
 >\\
 >\\
 >>>\
```
And the IP instead exits out the line that is missing a byte (with the exception of the last line, where it exits out the second to last).
From there, each of the first four lines will redirect to the second line:
```
v
>>>>>>>>>>
>>>>>>>>//
>/
```
Which will then lead into either one of the two `beep`ers.
```
v^"peeb"<<\/"beepvv"((>
>>>>>>>>>>//
```
If any of the bytes in the first `beep`er have been removed, then it instead goes to the second:
```
v^^"peb"<<\/"beepvv"((>
>>>>>>>>>>>//
```
Both `beep`ers then lead back to the first line and the terminating `@`.
Some other miscellaneous parts:
The `(((((([[[[[[[` is used to clear the stack when the pointer starts inside a pair of quotes and ends up pushing the whole first line to stack. It has to be so unfortunately long because the first newline can be removed to make the first line twice the size. Experimenting in generating the `beep` using arithmetic instead of quotes ended up longer.
The `\`s and `/`s scattered across the lines are there to golf bytes in the top line of code by redirecting the pointer to the correct lines. As most of the bottom lines are just filler, only the top line is able to be golfed. If anyone has any ideas for a shorter radiation proof stack clearer then what I've got now, feel free to comment.
[Answer]
# [Hexagony](https://github.com/m-ender/hexagony), 38 bytes
```
.....;p;<>b;e;/<b;e;;p...@@.......;@..
```
[Try it online!](https://tio.run/##y0itSEzPz6v8/18PBKwLrG3skqxTrfVtQKR1AVDMwUEPAqyBjP//AQ "Hexagony – Try It Online")
[Verification program.](https://tio.run/##fVBBboMwEDzjV6xyMVZaI9QbTqqo34gQMrAES41tLUaU11PTkLTqIXuw1vbM7Oz4OfTOvi3m6h0FcAPz5C6kr3AEzvki11JeHd5rhSo7rKfy8e10krdSsVki9JwXr3kZ6caGlH/MAaFxow0Ff4FPtOmmK9E2rsWU66ExhgshGGuxAxpt1eOXvjg7Vxv2zhEFSyYTenA@CvGVH0X5xAXoAbr4m3RyIhPwwWAJYRjJxo2k/6F1OxrrGTLnQ3YflEWvSD5CkSTVsCrvhCTUbRp96WHAGMpTa3Bcg2KdIzBgLJC2F0z/LCxW96RbowO21W@6W3cuTAn7x83s86JkybPJ/7VuFmpEz5dv "Python 3 – Try It Online")
---
## Explanation
We make use of Hexagony's auto-detection of hexagon side-length here.
If no bytes are removed, the program has side length 4 and looks like this:
[![Program without any bytes removed](https://i.stack.imgur.com/vnjiH.png)](https://i.stack.imgur.com/vnjiH.png)
If however, a byte is removed. There are 2 cases.
1. The removed byte is after the second `<`.
The execution flow would be:
[![Program with last byte removed](https://i.stack.imgur.com/Kmuth.png)](https://i.stack.imgur.com/Kmuth.png)
There are 2 consecutive `@` on the 5th line, so even if one of them are removed, the IP will safely hit a `@`.
2. The removed byte is at or before the second `<`.
Then the second half will stay intact, and the IP is no longer redirected upwards by that `<`. Image of execution flow:
[![Program with the second <code><</code> removed](https://i.stack.imgur.com/Hs7dg.png)](https://i.stack.imgur.com/Hs7dg.png)
[Answer]
# [Klein](https://github.com/Wheatwizard/Klein) XXX, ~~144~~ ~~136~~ ~~128~~ ~~122~~ 114 bytes
This answer completes [JoKing's challenge](https://codegolf.stackexchange.com/a/170976) to make a Geiger counter that works in every Klein topology.
```
\..//!?......\
!>>>.>>>>>>>\\\
!.//@"peeb"(<.\.\
!\\\@"peeb"(<..//
>>>>>>>>>>>>////
\\>.>....(.////
>\....?!/1!!./
```
[Try it online!](https://tio.run/##TYwxDoAwDAP3/iKdupCUHaU8xBNVBwRC/H8JaRngtrMtH2fbLzMwi1DhAQKpKusL4O7tGu/WtpgWRl94/CVeB/0hTgD8o/8lHq7oUkhm8j8zyznbVB8 "Klein – Try It Online") [Verification script](https://tio.run/##nVCxboMwEN39FYfLQNTW2IyoIV06V8padyDEKSiRTYMjNYr4dnq2E0KkdOlJYL13d8/Pb1V29TB0ttzbOSedOewrNadxUpV2RomqagM0DjSFAiqzVmy702Rj9tBAoyEWZG0IQHu0tdEZpKa16XanGh3@rD2OWxA38FyhjDlYZn8srjUb@IBw34WdUYjmQCl8Yt/WSuMBEKy8LZfvyxx1nkAbqOpSfynq@xMBcMSmQV9aeaPaG0069Q2xfyrEp4fwqH4W7J99nAKb61z0wcaVehSevONqnOG57qcL/U1mbmGam8P@clf/zM/V3xmulGqDY1ej67t5xhq/m/ePsxPhM4fpBveY8Jj1MEjG0jRaMF@SREVRsCKUlIix@0pbpVY0eWHSTSB9ZbBNikmlWERK1HB6CfO4kA4solREqDdwzgnngnCBpxBEIBaIBWKBOEOcIc7E5RS/ "Bash – Try It Online")
I don't believe it is really possible to make a perfectly satisfactory explanation. There are definitely aspects of this that *I'm* not aware of. It's just got too much going on. So the explanation below is a bit rough. I just want to get the *idea* across even if I gloss over a lot of the details.
The obvious next step is to create a version of Lost that can run on different Klein topologies, and create a Geiger counter in Lost that works in every topology. I don't know if really should be encouraging people to spend there time on such a awful and frustrating task, but I would certainly be willing to give a 500 rep bounty to someone who does the above.
## Explanation
The idea for this is to make a program with 2 parts.
The first is the path taken during normal execution. This path doesn't cross any boundaries, so we can avoid having to deal with the topology, and most importantly is very, very fragile. The idea here being that any disturbance to the program breaks the path and sends the ip flying off randomly.
The second part is a robust core. This part is sort of the opposite, it's role is to catch pointers that have gone off the path and direct them to the beeper. It needs to be impervious to radiation and it needs to catch as many stray pointers as possible.
The general strategy here was to build a program like this and just keep fixing the problems.
### The Path
The path looks like this:
```
\   / ?......\
!   .        \\
    @         \.\
!              //
              //
\\           //
 \....?! 1! ./
```
Some important about this things are
* We mostly use mirrors (`\` and `/`) rather than hard redirects. They are more fragile, and make it harder to get back on the path.
* The last character of every line is part of the path. This is required since removing them needs to break the path.
* It is sort of hexagon shaped. This is because a right angle turn can be simulated by a pole (e.g. the NE corner in 101) in the topology. So if we try to make a right angle turn at the corner there will end up being a topology that doesn't even need the mirror to make the turn.
* We have `?!/1` at the bottom and `?` near the end. This is to make sure that the bottom portion gets executed exactly once. Sometimes the pointer ends up back on the path after getting knocked off. This makes sure that if it hits the bottom portion of the path it doesn't get stuck in a loop there, and if it skips that portion it can't get to the end of the path.
* It covers basically the entire perimeter of the program. Hypothetically we only really need the left edge of the program but a full circuit is more fragile and we have to get to the other side anyway.
I experimented with both clockwise and anticlockwise paths. Ultimately clockwise should be better since you can make a shorter path. But I could not get it to work and some earlier attempts using this approach just never worked out. I managed to make an anticlockwise path so that's what's here. I might take some time with the lessons learned here to try clockwise again.
### The Core
The core is nothing too special. It is based off of the method used by pretty much every 2D Geiger counter we have two lines each which print `beep`, and we use the length of each to figure out which is damaged, so we can execute the other one. We surround this with ton of `>`s to force ips where we want them, this is the classic strategy for Lost programs.
[Answer]
# [Hexagony](https://github.com/m-ender/hexagony), ~~34~~ 29 bytes
```
//..>;e;<b@;p;/|/;e;;\.b@;p<@
```
[Try it online!](https://tio.run/##y0itSEzPz6v8/19fX0/PzjrV2ibJwbrAWr9GH8i2jtED8Wwc/v8HAA "Hexagony – Try It Online") [Verification!](https://tio.run/##fZDBbsIwDIbPzVN4XNyILRHarYUJ7SF2YQil1C2RIIncVKzS3r1LN2DTDtxs@f8//3YY4sG759GegucIvhOBfcvmBCtgRBy1VuqlpHJZrctQ6k@d6vJdTd1yPSbFZlE8LbbJZl3M8XWIBHvfu1jgIxzJ5ReeIrf3NeVour21KKUUQtTUAPdud6AP03o37C7iq0kWIjvbeAAfEgknQKLiGSWYDpo0zRp1Zhvp5hAZU@zZpVNU@LY1M@6rAbQPUV8X6RSWOCQpseIKJvJMKiZT5ymY6TpK37gbDVYrQBSNZ7BgHbBxLeV/LpZTeja1NZHq3e9bL9WmsFuY3zo7XxRbkd3b/J/1E6EiCnh9/5s52voB5fgF "Python 3 – Try It Online")
### Explanation:
Here is the normal code formatted into a proper hexagon using [HexagonyColorer](https://github.com/Timwi/HexagonyColorer):
[![No Cancer...](https://i.stack.imgur.com/49CEI.png)](https://i.stack.imgur.com/49CEI.png)
The double `//` at the beginning ensures that this path is always taken. If any character is removed, the `@` is removed from the path, either being shifted one back or being removed itself:
[![Cancer!](https://i.stack.imgur.com/wVoyR.png)](https://i.stack.imgur.com/wVoyR.png)
In this case, we've removed a character after the `|`, which makes it follow this path, printing `beep`:
[![First beep](https://i.stack.imgur.com/wR1iS.png)](https://i.stack.imgur.com/wR1iS.png)
If we instead remove a character from before the `|` (or the `|` itself), we follow the other beep printer:
[![Second beep](https://i.stack.imgur.com/4v55c.png)](https://i.stack.imgur.com/4v55c.png)
We've then accounted for all possibilities and we only `beep` using the un-irradiated parts of the program.
[Answer]
# [Self-modifying Brainfuck](https://soulsphere.org/hacks/smbf/), ~~73~~ 63 bytes
```
<<[[[[<<]]>[[.>>..>>.[,>]]]]   bbeepp+[<<<]>>[[>]>>>.>>..>>.,+]
```
[Try it online!](https://tio.run/##K85NSvv/38YmGghsbGJj7aKj9ezs9EA4WscuFggYGBiSklJTCwq0gQpsYu2AKuyApB1MmY527P//AA "Self-modifying Brainfuck – Try It Online") [Verification!](https://tio.run/##dZDRboMgFIav5SnOTBokGtuld0a92O73Ao4sVLGStEAA1/XpHVhrl2UjEcTz///xO/rqBiX30zSJs1bGgbJoebNXi5A26mjYGapwza3rhMwNZ11C/KFPrOUJBpzh96/dDpOgF9Il@OXqOLRqlK7AGZy4TJagnMtWdd7EbCsEJoQslqXuIzregxnlx/JlrRQougg3gNI@DocUH40vmACz0Ptq1OcXIxx/ZEWGu9FID5Xr2dbHW6Xd1p4P/bxBiInJgoSQGp1H/av7nexNQTsweeQBzMsJYtbyMLhgrQDjMAT@@TowA7drrwwIEBJM8CU/pkEClOhhuTaCwlMFd3tAunWNNx0kG0uKeJOI7KEmGXDZVdhPPor@/femEDRdTelzQcmaPBNE0S@GA@caz5oV5NEUTWXZ@FWWlNZNk9d1Hp4mq6lfAHAIdp16QUlrr6j9Xt9lWUq/AQ)
The spaces in the middle of the code are actually representing NUL bytes.
### Explanation:
The code is split into two sections by 3 NUL bytes in the middle. Both of them basically prints `beep` if the other section is irradiated (with a couple of exceptions).
First, the `<<[[` at the beginning is to ensure that all `]`s are matched at any time. `[`s won't attempt to look for a matching `]` if the cell is positive, while `]`s *do*. If any `]` jumps back to one of these brackets it usually jumps back straight away because the cell is `0`.
The next part, `[[<<]]>` then checks if the length of section 2 is even. If so, it executes the other half of section 1, which prints `beep` using the `bbeepp` at the start of section 2.
```
[[.>>..>>.[,>]]]]
```
It then clears all of section 2 so it doesn't execute.
In section 2, we check if the length of section 1 and the NUL bytes is divisible by `3` with `+[<<<]>>`.
```
[[>]>>>.>>..>>.,+]
```
Similarly, we print `beep`.
[Answer]
# [Z80Golf](https://github.com/lynn/z80golf), ~~53~~ ~~36~~ 34 bytes
*-16 bytes thanks to @Lynn*  
*-2 byte thanks to @Neil*
Since this is just Z80 machine code, there is a lot of unprintables in this one, so have a `xxd -r`-reversible hexdump:
```
00000000: ddb6 2120 10dd b615 280c 003e 62ff 3e65  ..! ....(..>b.>e
00000010: ffff 3e70 ff76 003e 62ff 3e65 ffff 3e70  ..>p.v.>b.>e..>p
00000020: ff76                                     .v
```
[Try it online! (exhaustive tester in Python)](https://tio.run/##fVFNT8MwDD23v8Ibh7QabJ34EKq0HThwRQJudJrSxtkitiRK0q0D8dcpaddtgBCHWHZsPz8/651bKnlZi7VWxoGyoTZqYegaJpDVOckqxrIqv@lBVo2TQ@T96yirksK/ZJpnFedTbGzr6cZu/kmRWhshXUTudg6hUKV0KTmHFcqomx7HYciQgynlvPs6ptIw2Aq3BKV9PSkUQ99LtjmJgVrgPh3w4dYIh8eWMDDoSiP9fkPd9vVHSrvR222yUCs@sqY4@NAA9uOhQcoiz4Jai16Yv4jAZAKEhFwZECAkGCoXGP3YwpMxlAnqkM1PwnbeSypmMDhGYjBOZ2FwtteG95/ROiEXkFTvIk2uqo@@30OVTpfOY3wn9HuErxMcutKeJ5kjatIIc8B@xLXaNOBuiZA3V6AOKGMGrT0NhNyoVwThhvDQgqXwvkftmZZN0KlzT1cWw@6oT2VReJgeietPL7JQ0tYX5Rc "Z80Golf – Try It Online")
## Explanation
z80golf is Anarchy Golf's hypothetical Z80 machine, where `call $8000` is a putchar, `call $8003` is a getchar, `halt` makes the interpreter exit, your program is placed at `$0000`, and all other memory is filled with zeroes. Making programs radiation-proof in assembly is pretty hard, but a generically useful technique is using one-byte idempotent instructions. For example,
```
or c        ; b1    ; a = a | c
```
is just one byte, and `a | c | c == a | c`, so it can be made radiation-proof by just repeating the instruction. On the Z80, an 8-bit immediate load is two bytes (where the immediate is in the second byte), so you can load *some* values into registers reliably too. This is what I originally did at the beginning of the program, so you can analyze the longer variants I archived at the bottom of the answer, but then I realized that there is a simpler way.
The program consists of two independent payloads, where one of them could have been damaged by radiation. I check whether a byte was removed, and whether the removed byte was before the second copy of the payload, by checking the values of some absolute memory addresses.
First, we need to exit if no radiation was observed:
```
    or a, (ix+endbyte) ; dd b6 21 ; a |= memory[ix+0x0021]
    jr nz, midbyte     ; 20 10    ; jump to a halt instruction if not zero
```
If any byte was removed, then all the bytes will shift and `$0020` will contain the last `76`, so `$0021` will be a zero. We can afford radiating the beginning of the program, even though there is virtually no redundancy:
* If the jump offset `$10` is removed, then radiation will correctly be detected, the jump will not be taken, and the offset will not matter. The first byte of the next instruction will be consumed, but since it's designed to be resistant to byte removals, this doesn't matter.
* If the jump opcode `$20` is removed, then the jump offset `$10` will decode as `djnz $ffe4` (consuming the next instruction byte as the offset -- see above), which is a loop instruction -- decrement B, and jump if the result is not zero. Because `ffe4-ffff` is filled with zeroes (`nop`s), and the program counter wraps around, this will run the beginning of the program 256 times, and then finally continue. I am amazed this works.
* Removing the `$dd` will make the rest of the snippet decode as `or (hl) / ld ($1020), hl`, and then slide into the next part of the program. The `or` will not change any important registers, and because HL is zero at this point, the write will also cancel out.
* Removing the `$b6` will make the rest decode as `ld ($1020), ix` and proceed as above.
* Removing the `$21` will make the decoder eat the `$20`, triggering the `djnz` behavior.
Note that using `or a, (ix+*)` saves two bytes over `ld a, (**) / and a / and a` thanks to the integrated check for zero.
We now need to decide which of the two copies of the payload to execute:
```
    or (ix+midbyte)  ; dd b6 15
    jr z, otherimpl  ; 28 0c
    nop              ; 00
    ; first payload
    ld a, 'b'        ; 3e 62
    rst $0038        ; ff
    ld a, 'e'        ; 3e 65
    rst $0038        ; ff
    rst $0038        ; ff
    ld a, 'p'        ; 3e 70
    rst $0038        ; ff
midbyte:
    halt             ; 76
otherimpl:
    nop              ; 00
    ld a, 'b'        ; 3e 62
    ; ...            ; ...
    rst $0038        ; ff
endbyte:
    halt             ; 76
```
The two copies are separated by a nop, since a relative jump is used to choose between them, and radiation could have shifted the program in a way that would make the jump skip the first byte after the destination. Additionally, the nop is encoded as a zero, which makes it easy to detect shifted bytes. Note that it doesn't matter which payload is chosen if the switch itself is corrupted, because then both copies are safe. Let's make sure that it will not jump into uninitialized memory, though:
* Deleting `$dd` will make the next two bytes decode as `or (hl) / dec d`. Clobbers D. No big deal.
* Deleting `$b6` will create an undocumented longer encoding for `dec d`. Same as above.
* Deleting `$15` will read the `$28` instead as the offset, and execution will proceed at the `$0c`, as below.
* When `$28` disappears, the `$0c` is decoded as `inc c`. The payload does not care about `c`.
* Deleting `$0c` - that's what the nop is for. Otherwise, the first byte of the payload would have been read as the jump offset, and the program would jump into uninitialized memory.
The payload itself is pretty simple. I think the small size of the string makes this approach smaller than a loop, and it's easier to make position-independent this way. The `e` in `beep` repeats, so I can shave off one `ld a`. Also, because all memory between `$0038` and `$8000` is zeroed, I can fall through it and use a shorter `rst` variant of the `call` instruction, which only works for `$0`, `$8`, `$10` and so on, up to `$38`.
## Older approaches
### 64 bytes
```
00000000: 2e3f 3f2e 3f3f 7e7e a7a7 201f 1e2b 2b1e  .??.??~~.. ..++.
00000010: 2b2b 6b00 7ea7 2814 003e 62cd 0080 3e65  ++k.~.(..>b...>e
00000020: cd00 80cd 0080 3e70 cd00 8076 003e 62cd  ......>p...v.>b.
00000030: 0080 3e65 cd00 80cd 0080 3e70 cd00 8076  ..>e......>p...v
```
### 58 bytes
```
00000000: 2e39 392e 3939 7e7e a7a7 2019 3a25 00a7  .99.99~~.. .:%..
00000010: 2814 003e 62cd 0080 3e65 cd00 80cd 0080  (..>b...>e......
00000020: 3e70 cd00 8076 003e 62cd 0080 3e65 cd00  >p...v.>b...>e..
00000030: 80cd 0080 3e70 cd00 8076                 ....>p...v
```
### 53 bytes
This one has an explanation in the edit history, but it's not too different.
```
00000000: 3a34 00a7 a720 193a 2000 a728 1400 3e62  :4... .: ..(..>b
00000010: cd00 803e 65cd 0080 cd00 803e 70cd 0080  ...>e......>p...
00000020: 7600 3e62 cd00 803e 65cd 0080 cd00 803e  v.>b...>e......>
00000030: 70cd 0080 76                             p...v
```
## What if: any non-empty output was fine instead of beep
### 1 byte
```
v
```
`halt`s the program normally, but if radiation removes it, then the memory will be full of zeroes, making `$8000` execute an infinite number of times, printing a lot of null bytes.
[Answer]
# [><>](https://esolangs.org/wiki/Fish), 23 bytes
```
<|o<"beep"/
>o<"beep"<\
```
[Try it online!](https://tio.run/##S8sszvj/36Ym30YpKTW1QEmfyw7GtIn5/x8A "><> – Try It Online") [Verification.](https://tio.run/##fVHBbsIwDD2Tr/DKIa3YitBuHeywfcAOOzKEstal1iCJEpeq0v6duQwGmrRdIif2e37vxffcOHt/oJ13gcFF5YPbBLODBQSt9WH@6ebJO6JPpurxXM7fDtJbzoq72UoAZDnVTz0jlK61XOhb2KJNT0w52tJVmGoTSyKdZZlSFdYQWrveusjr09x5PivUqCNuwHkh0QNWCHWnMzARaumO6rwLxPiDUKOA3AYr@nN/hNWJ/3YGU@d5WlNsjkfuexgYkywPaKpUtLiWB7N/qVHKxIhDNsPcArRWYhn3z40JMFwflBrDCzcYOooI3FAEph3GI8IEiaSCxmzrzvRAVtUuAEkBwdgNpldJZYN1quF0XdIKbhZwXlYAjOH1gzxQhZapNFuxYiOWLdNespcZUzKGKAn9b2pZCPXksmcyK1YS4uiX0@GvtTxf2b1IU4cv "Python 3 – Try It Online")
It's honestly really surprising how short I got this. Ends with an error.
### Non-erroring version, 29 bytes
```
<<;oooo"beep"/
 ;oooo"beep"<\
```
[Try it online!](https://tio.run/##S8sszvj/38bGOh8IlJJSUwuU9LkUkHg2Mf//AwA "><> – Try It Online")
[Answer]
# [Klein](https://github.com/Wheatwizard/Klein), one of each topology, totaling 291 bytes
After seeing [WW's answer](https://codegolf.stackexchange.com/a/170741/76162) using the `001` topology, I decided to see how hard it would be to do a Geiger Counter for each topology. (Spoiler: very hard. It's difficult to figure out where the pointer will go without hand gestures that make me look like I'm figuring out which hand is my left)
[Verification!](https://tio.run/##jZPNbqMwFIXX@CkumYVBydiQ2UVEymjeYTZ1VTnFJNYQbBm3UVT12TM2wYBCVHXDj@93rs89GH2xR9X8usqTVsaCapE26mD4qYUtfFxxlmV4gwzGuCh2Cy3EfkER6Z8KBq6wQghn@fewLO@xmDwACSkIGTp@G80Hj5T2ZYYCV9CByu8pOlJxoMZBSNiVMUTYjvQoKwY0v5@5cCQJFmnYeT3PcPQXj1RoxxhhQ3aUuKHp2HUMcz1YpYwNOEN0R8icnTS/FZOkoCimu10xhdH1E6FSVGDempf@KCRWaVWrw2UF/Uq6QdFZ2iMoLZoEv6pS4BXgM06Bt1C5alSRs5FWJEGBIiPsm2ncESO6k1UL3R2@NVClLf1XC9ncrkRfwPeEj7DzJ/z8vUiJEbxMUoSssrx2BzRDqFIGZv5ANuGxJc7FqU285VDdhiJprZHadYxuHZdbqJ2z0fMPbWRjk6/DcBxvW@F@n4dcwGC7Bcwa7H2I9z9HbpwRl3jkR5DesuHNQSRTB951JKuw1ZN8hti7v@l9MTK8lNyK8mU23dPG4ctRu8w3z14RhroTptPao0Fmgk7x1egzRZfBXgjd5RBNkxhH9AF5E0Mb/JfXsozxahrNCu8vVrQEpyjg7hP2q9C9OEV3p/k6rPN3YfhBONX1Pw "Python 3 – Try It Online")
(I also thought about writing a program that is a valid Geiger counter on all topologies, but that might have to wait. If anyone else wants to try though, I'm offering a 500 rep bounty)
### 000 and 010, 21 bytes
```
<<@"peeb"/
.@"peeb"<\
```
[Try 000 online!](https://tio.run/##y85Jzcz7/9/GxkGpIDU1SUmfSw/Kson5//@/gYHBf11HAA "Klein – Try It Online") and [Try 010 online!](https://tio.run/##y85Jzcz7/9/GxkGpIDU1SUmfSw/Kson5//@/gaHBf11HAA "Klein – Try It Online")
This is ported from my [`><>` solution](https://codegolf.stackexchange.com/a/169995/76162). This obviously works in `000`, since that's the default topology for most 2D languages, but I was surprised that it also works in `010`.
### 001 and 011, 26 bytes
```
!.<<@"peeb"/
.@"peeb"..<..
```
[Try 001 online!](https://tio.run/##y85Jzcz7/19Rz8bGQakgNTVJSZ9LD8rS07PR0/v//7@BgeF/XUcA "Klein – Try It Online") and [Try 011 online!](https://tio.run/##y85Jzcz7/19Rz8bGQakgNTVJSZ9LD8rS07PR0/v//7@BoeF/XUcA "Klein – Try It Online")
This one is copied directly from [WW's answer](https://codegolf.stackexchange.com/a/170741/76162). Thanks!
### 100, 21 bytes
```
//@"peeb"\
@"peeb".</
```
[Try it online!](https://tio.run/##y85Jzcz7/19f30GpIDU1SSmGC8rQs9H///@/oYHBf11HAA "Klein – Try It Online")
### 101, 21 bytes
```
//@"peeb"/
@"peeb".<!
```
[Try it online!](https://tio.run/##y85Jzcz7/19f30GpIDU1SUmfC8rQs1H8//@/oYHhf11HAA "Klein – Try It Online")
### 110, 26 bytes
```
<.<@"peeb"\\
.\@."peeb".\<
```
[Try it online!](https://tio.run/##y85Jzcz7/99Gz8ZBqSA1NUkpJoZLL8ZBD8LRi7H5//@/oaHBf11HAA "Klein – Try It Online")
### 111, 24 bytes
```
<<@"peeb"<\
...@"peeb"//
```
[Try it online!](https://tio.run/##y85Jzcz7/9/GxkGpIDU1SckmhktPTw/K0df///@/oaHhf11HAA "Klein – Try It Online")
### 200, 21 bytes
```
<<@"peeb"\
@"peeb".!/
```
[Try it online!](https://tio.run/##y85Jzcz7/9/GxkGpIDU1SSmGC8rQU9T///@/kYHBf11HAA "Klein – Try It Online")
### 201, 31 bytes
```
\\.\.@"peeb"</./
./...@"peeb"<\
```
[Try it online!](https://tio.run/##y85Jzcz7/z8mRi9Gz0GpIDU1SclGX0@fS09fTw8uEPP//38jA8P/uo4A "Klein – Try It Online")
By far the most annoying.
### 210, 26 bytes
```
/\\@"peeb"</\
/@.."peeb"<\
```
[Try it online!](https://tio.run/##y85Jzcz7/18/JsZBqSA1NUnJRj@GS99BTw/Ki/n//7@RocF/XUcA "Klein – Try It Online")
### 211, 27 bytes
```
\\."peeb"((</
!/@@<"peeb"<\
```
[Try it online!](https://tio.run/##y85Jzcz7/z8mRk@pIDU1SUlDw0afS1HfwcEGwreJ@f//v5Gh4X9dRwA "Klein – Try It Online")
The only one where I had to handle entering the beeper through the right side.
[Answer]
# [Klein](https://github.com/Wheatwizard/Klein) 0X1, 26 bytes
*Works in both 011, and 001.*
```
!.<<@"peeb"/
.@"peeb"..<..
```
[Try it online!](https://tio.run/##y85Jzcz7/19Rz8bGQakgNTVJSZ9LD8rS07PR0/v//7@BgeF/XUcA "Klein – Try It Online")
[Verify!](https://tio.run/##bVHLbsIwEDzHX7Gkh41FMYTeIpDafkQvFCFDNrAq2JZjhPj61IEUUMnFWmsemtl157Cz5q3hg7M@gK2F83br9QHm4BGxGajZ7D11ROt0LFQ3KTVTqonwIi9G@TJq2IQMP8@BYGOPJhT4CnsyWWemyGxsSRnqesOMUkohREkV@KNZdZw/rixEcuKwA@uiAba6aIYnlKBrqCKaVOrkOdBNIRJP4ehNjK/cRVal7lJsCmPrwvhnT2yur3JnaD1hMslh9JFK5UmXWQx0LdGXSApd1xTX0wfCfA74bVBU1gMDG/DabCl7qC/bTl6XrAOVq/uCu2lR8BKGtx8P82Ipkpcuzz@ZvCMPYZ5YkdaT@Yl2Cb8mcm2B7opfes/lAGXzCw "Python 3 – Try It Online")
## Explanation
This program takes advantage of Klein's unique topologies in particular the **001** and **011** topology, which are the Klein bottle and the real projective plane. This explanation will focus on **001**, the klein bottle.
Unedited the program follows the execution path:
[![Orange path](https://i.stack.imgur.com/YSbr0.png)](https://i.stack.imgur.com/YSbr0.png)
Removing a byte from the program can effect the program in 4 ways (each in a different color):
[![Program sections](https://i.stack.imgur.com/ElVmw.png)](https://i.stack.imgur.com/ElVmw.png)
The first thing to note is that `<<` will always deflect the ip to the left of the origin at the start. If one of the `<`s is deleted the other takes its place. So if any byte is removed from the red section the following execution path will be followed:
[![Red path](https://i.stack.imgur.com/G41cW.png)](https://i.stack.imgur.com/G41cW.png)
If the blue byte is removed we get the very simple path:
[![enter image description here](https://i.stack.imgur.com/hi14q.png)](https://i.stack.imgur.com/hi14q.png)
If the newline is removed we get the path:
[![Green path](https://i.stack.imgur.com/l9HIp.png)](https://i.stack.imgur.com/l9HIp.png)
The yellow path is a bit more complex. Since the bottom line is one longer than the top line, when the program is squared at the beginning of execution a virtual character is added to the end of the first line to make them the same size. If any byte on the second line is removed the line is shortened and that virtual character is not added. This is important because `!` normally jumps over the virtual character, but in its absence it jumps over `/` instead.
[![Yellow path](https://i.stack.imgur.com/DBr7q.png)](https://i.stack.imgur.com/DBr7q.png)
[Answer]
# [Self-modifying Brainfuck](https://soulsphere.org/hacks/smbf/), ~~144~~ 102 bytes
Unprintables are displayed as escape sequence (for example `\x01`).
```
\xa8<<[[<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<[[[.<-]>[>-]\x01qffb\x00\x00beep\x00]]]<[>]<<[>[-<+>]+<<]>[[>]>>[.>]]\x01
```
[Verification!](https://tio.run/##dVLLTsMwEDzXX7EqEk5IE@Jyq9Ic@ALuxkJJalMLYhvbUeHryzo0wAEOfmh2ZjK7jvuIR2vuzuezHp31EcLUO28HGQIhVzBYE6KfhggIPvtuJMRtYQ89fXyv67R6KV06hRANb0WDGy@bohVF04iWI9S2vGqFQBKjxLFZjTROoUg3epO9SpO5bV5u8xninFdNieK2nFVvSvWUXKQfUYaMl2EaM8fy6/pdKZFkjpEuBIkd1GBsBG1@QY7xWixwT4tyUzUtFxjnqys0dqxwW/yI1yZm9OEL39HN0ni@lO4xAc5lMjFV5@gXRk7IQSrwk3m6QN@lHVmddDyCdcingz1I1NL@RHPoAigsr1R18jrKbwlZdSZgsJ8HqYajHF6e7BTdFDO@vrUu3oaxV/O23sA6Ga/FBqIeJdL2DF28jJM3gGZkGcdfCWGf3oUSZT3oNCbfmWeZ/e6vZKmP/z34TotiueuC7cTFNf0j9PwJ "Python 3 – Try It Online")
[Answer]
# [Backhand](https://github.com/GuyJoKing/Backhand), ~~25~~ 21 bytes
```
vv""ppeeeebb""jjHH@
```
[Try it online!](https://tio.run/##S0pMzs5IzEv5/7@sTEmpoCAVCJKSRESUlLKyPDwc/v8HAA "Backhand – Try It Online") [Verification!](https://tio.run/##dZDBbsIwDIbP7ZkH8CKhNAIVTbtVqjSxC7vsBRBCaePSsJFESQrq03cJlDJNW0527P@3P5vet1q9DMMgT0ZbD9qlY@R6l6bG6oPlJyhjmjsvpMotcpGxWJPKZ3Tde4Rad8oXdAlfqLJRlKOqtcCMcldLSRljo2SsBwuBDdhO7cefqVKkyUX6FrQJdjS6BGt6oQy4gyZUkya/WOnx4ZVY9J1VASA3V1lDzA0OVtr4VcXrz5YrsVqPwf5debQmyNDmpoc4hbCJTnc@UP@13B38Q0MdjA4YuUM7S7lzGG8YpSVQGm@E57eWW7iljbYgQSqwUZf9OBaLzLKBMd3KHTyVcJdH4ttUMheQzR0ryDyTy0c3WwIqUVIaLpH8u/u2kLvFJFo8Fzs2OV8JkuQXQ4Vo6LVnAnkMTYfzmRBjMLyqms0IOR43m9dv)
This uses Backhand's ability to change the pointer step value to skip an instruction every step and neatly solve the redunancy issue. It then uses the `j` command to check if the code is irradiated by jumping to the last character (`@`, halt) if not, and jumping to the second last (`H`, halt and output stack) if so.
[Answer]
# [Runic Enchantments](https://github.com/Draco18s/RunicEnchantments/tree/Console), 29 bytes
```
>>yyLL@"peeb"/
     @"peeb"L\
```
[Try it online!](https://tio.run/##KyrNy0z@/9/OrrLSx8dBqSA1NUlJn0sBBKA8n5j//wE "Runic Enchantments – Try It Online")
Essentially the same as the Klein 000 answer or ><> answer (I started with the Klein one). The only change really needed was to turn the `<` into `L` and `.` into  (translation of command symbols), inserting the IP entry points (need 2, otherwise a deletion would result in a non-compiling program) and the insertion of the dela`y` command to get the two IPs to merge (thus printing only one `beep`), again, needing two. Also required inserting additional NOPs to keep line lengths the same. Klein conveniently also uses `@` for "print and terminate."
No ability to utilize the whitespace in the lower left, as any reflectors to change direction inhibits the ability to detect radiation. e.g. (26 bytes, irradiated `y`):
```
/yLL@"peeb"/
\<<  @"peeb"L\
```
Prints no output, due to the bent entry segment causing a re-reflection back to the lower line's terminator.
[Answer]
# [Befunge-93](https://github.com/catseye/Befunge-93), 55 bytes
```
<< >       vv
>,,@#"beep"<"
v  |   _ #-<_
>"peeb",,,,@>
```
[Try it online!](https://tio.run/##S0pNK81LT/3/38ZGwU4BAsrKuOx0dByUlZJSUwuUbJS4yhQUaoDi8QrKujbxXHZKBampSUo6QOBg9/8/AA "Befunge-93 – Try It Online") [Verification!](https://tio.run/##dZDdbsIwDIWvm6fwglAaUYomrla1Fdru9wIIof64NBIkVZIWIe3dWQKlTNOWKzv2Ofbn7mJbJdfX61WcOqUtKEPGyFwMIZ1WB12cIPNpbGwtZKyxqEPua0LakL1fLEKlemkTFsERZTiKYpSVqjFkhamEYJzzUTLWnUWNDehe7sefqZKQ4CxsC6pzdsy7xGXj3NmZcSgMNK4haOKzFhafdoFG22vpGOLupmzoSnV2VWLTywOuSiF9DN4PvsAW4gjLChbrN8onKtVbR/vXUg/gTwVVWzhDz@vaOSmMQX87L82AMX8bHD7aQsM9bZQGAUKC9rrwx5G4ZxUNjOlW7OAlg4fcY96n0nkN4dzwhM5DET27eQQo64wxhx/8u/s2EbvFJFq8Jjs@Od8IguAXQ4nYsVvPBPIcSq5pCjnc3zCQPIo2M@olNKVkAHdcgD3Mlume5LRDLGnk3ib/Bg "Python 3 – Try It Online")
I'd hoped to get it a little smaller than this, but the length of the `beep` printer is the bottleneck.
[Answer]
# [Wumpus](https://github.com/m-ender/wumpus), ~~37 34 32~~ 31 bytes
```
777*7..@ $o&4"beep"|"@peeb"4&o@
```
[Try it online!](https://tio.run/##Ky/NLSgt/v/f3Nxcy1xPz0FBJV/NRCkpNbVAqUbJoSA1NUnJRC3f4f9/AA "Wumpus – Try It Online") [Verification!](https://tio.run/##dZHfasMgFMav41OchbXGtdiOFQKBQNnu9wJdGfljGqFVUbMssHfPtLXpGNuVHs/3ffo7qsG2UjyN48hPSmoL0qCwM4NBSGl50MUJcl9SY2suqGZFnRC3qGNRsQQDXuK3z/UaE6/nwib4ebAMKtkJm@ElHJlIQhBlopK1MxWm4hwTQoIl9F1EzRrQnXgPJ1MnQ1HPbQtSuTjsU1w07jGBwkDjulFDe80tu2VFmtlOCwdF1dnWxLorB1hJZVd9d1KdCQvVJfjImAQ8hGRnHfZfL7lSvkqo2kIcmId0coIKY5gforfmgLEfCPt4aQsNl7KRGjhwAdr7kh@TIR6QNxDKHd/DXQ5Xu8e73BrPakhmhmTxLOHLm5osgYk6x@4Xoujft@8yvl9MpsVjtidT8pkgin4xlIwpfNZMILdL0Zim6UNK6Rbu5XwTe3H8FW8VY2W8mcvtNw "Python 3 – Try It Online")
This solution uses the fact that `.` jumps to a position modulus the length of the program.
Alternatively for the same amount of bytes
```
" @o&4"beep"}@
@o&4"beep"}$}  
```
[Try it online!](https://tio.run/##Ky/NLSgt/v@fS0nBIV/NRCkpNbVAqdaBC4mjUqug8P8/AA "Wumpus – Try It Online") [Verification!](https://tio.run/##dZHRbsIgFIav4SnOmjlKbKrLdtWkidnu9wLOLNhSS6JAgK7rhc/egWI1i7uCw/n/H76DHlyr5Ms4juKglXGgLI47O1iMtVE7ww5QhjK3rhYyN5zVKfWL3rOKpwRIRj5/lktCg15Il5K3wXGoVCddQTLYc5nGoJzLStXexGwlBKGURkvs@4iaN2A6@RVPpk6BUS9cC0r7OBJSfDTpCQVmofFd1OS9EY5fs5DhrjPSQ@X6ZGsS020HWCjtFn130J2NS262ECITGvEwVp3z2PdecqH8UFC1TO54gPRyipm1PAwxWEsgJAyEf7@3zMC5bJQBAUKCCb70ZjI0AIoGYrkWG3go4WIPeOdbk1kN6czSIpmlIruqaQZc1iXxv4DQv29fF2Izn0zz52JDp@QTAUJ/GLaca3LSTCDXS/GIE1ipp9ckyJLjCt8Uj0eAXw "Python 3 – Try It Online")
This one utilises the difference in the direction of the pointer for odd and even line lengths. (I don't really know how the first `"` actually works when the newline is removed)
]