Auto-detect a wider range of RP2040 devices

[rtyley]

At the moment, the device-path detection for Raspberry Pi USB devices only recognises one specific id: 2E8A:0005, the vanilla Raspberry Pi Pico running MicroPython firmware:

Plethora of RP2040 devices...

However, there are many devices using the RP2040 that are similar to the Raspberry Pi Pico and run MicroPython in the same way (Adafruit Feather RP2040, Pimoroni Badger 2040, Raspberry Breadstick, etc) - these all use the same vendor id as the Pico (2E8A), but have different product ids from the large 0x1000 - 0x1fff range which Raspberry Pi sub-license.

Not all of the devices in the 0x1000 - 0x1fff range actually run MicroPython (eg 0x1069, the Zoid Technology Matrix, runs QMK firmware and quite possibly doesn't have a MicroPython build), but if it's plugged into a developers machine and they're trying to access MicroPython on it, it probably does.

Changes made in this PR

• The findusb.py Python script is now much shorter, and no longer takes target usb ids as an argument - it just outputs the full list of attached usb devices (the list is unlikely to be very long - say, fewer than 100 devices) and allows the Kotlin code to do the filtering
• In MicroPythonDeviceProvider, the fixed list of usbIds is replaced with replaced with a checkUsbId() method - this is used to filter the list of usb ids returned by findusb.py. This lets us accept much wider ranges of usb ids without having to define 100s of 'known' MicroPythonUsbIds.

See https://github.com/raspberrypi/usb-pid#assignment - Raspberry Pi have allocated the productId range 0x1000 - 0x1fff for Commercial RP2040 devices, of which they have allocated over 100 ids as of 2023.

[rtyley]

Thanks for the thumbs up @lancelote - I've now signed the JetBrains Contributor License Agreement (CLA) - if the change looks good to you that's great, please merge!

I've tested the changes locally with a Pimoroni Pico LiPo, and they work as intended.

[lancelote]

@rtyley Thank you for the excellent contribution 👍

[rtyley]

Ah wonderful, thank you!