As can be seen in this example, each channel (R, G, B) in a BMP file takes an input. A 24-bit BMP image has 8 bit for-R , 8-bit for G, and 8 bit for B. I saved an image in MS-paint as monochrome(black and white). Its property says the image's depth is 1-bit. The question is who gets this 1 bit: R , G or B? Is it not mandatory that all the three channels must get certain value? I am not able to understand how MS-Paint has drawn this BMP image using 1 bit.
Thanks in advance for your replies.
There's multiple ways to store a bitmap. In this case, the important distinction is RGB versus indexed.
In an RGB bitmap, every pixel is associated with three separate values, one for red, another for green, and another for blue. Depending on the "bitness" (bit depth) and the specific pixel format, the different colour channels can have different amount of bits allocated for them - the simplest case is the typical true-color with 8 bits for each of the channels, and another 8 bits (optional) for the alpha channel. However, some pixel formats allocate a bit differently - the idea is that the human eye has different sensitivity to each of those channels, and you can save up on space and improve visual quality by allocating the bits in a smarter way. For example, one of the more popular pixel formats is BGR-565 - that is, 16 bits total, 5 bits for blue, 6 bits for green and 5 bits for red.
In an indexed bitmap, the value stored with each of the pixels is an index (hence "indexed bitmap") into a palette (a colour table). The palette is usually a simple table of colours, using RGB "pixel" formats to assign each index with some specific colour. For example, index 0 might mean black, 1 might mean turqouise etc.
In this case, the bit-depth doesn't exactly map into colour quality - you're not trying to map the whole colour space, you're focusing on some subset of the possible colours instead. For example, if you have 256 shades of grey (say, from black to white), a true-colour bitmap would need at least three bytes per pixel (and each of those three bytes would have the same value), while you could use an indexed bitmap with a pallete of all the grey colours, requiring only one byte per pixel (plus the cost of the pallete - 256 * 3 bytes). There's a lot of benefits to using indexed bitmaps, and a lot of tricks to improve the visual quality further without using more bits-per-pixel, but that would be way beyond the scope of this question.
This also means that you only need as many possible values as you want to show. If you only need 16 different colours, you only need four bits per pixel. If you only need a monochromatic bitmap (that is, either a pixel is "on", or it's "off"), you only need one bit per pixel - and that's exactly your case. If you have the amount of distinct colours you need, you can easily get the required bit depth by taking a base-2 logarithm (e.g. log 256 = 8).
So let's say you have an image that only uses two colours - black and white. You'll build a pallete with two colours, black and white. And for each of the pixels in the bitmap, you either save 0 if it's black, or 1 if it's white.
Now, when you want to draw a bitmap like this, you simply read the palette (0 -> RGB(0, 0, 0), 1 -> RGB(1, 1, 1) in this case), and then you read one pixel after another. If the bit is zero, paint a black pixel. If it's one, paint a white pixel. Done :)
No, it depends on the type of data you chose to save as. Because you chose to save as monochrome, the RGB mapping is not used here, and the used mapping would go as a one byte per pixel, ranging from white to black.
Each type has its own mapping ways, saving as 24-bit will give you RGB mapping, saving as 256 will map a byte for each pixel, each value represents a color( you can find the table on the internet), as for monochrome, you'll have the same as a 256 bitmap, but the color table will only have white and black colors.
Sorry for the mistake, the way I explained for monochrome is actually used by Gray Scale, the monochrome uses one bit to indicate if the pixel is black or white, depending on the value of each bit, no mapping table is used.
