The author’s memory is remarkable. I hardly remember my own name that far back, LOL. Back then, I knew I would always struggle with those types of interviews, so I always carried a floppy disk with me to them. The disk contained a few programs I had written, and I would simply tell the interviewers, “Don’t give me a quiz. I’m terrible at it, so if you do, I’m out.” However, if they were willing to look at my capabilities, I would share a few of my programs. That approach actually worked most of the time and got me the jobs. The good old days!
The circle outlining one seems interesting to me. I definitely didn't read the algorithm ahead of time, and I'm probably not as smart as a revolutionary genius computer scientist, but I thought of 2 basic algorithms in a few minutes.
You could iterate over the degrees 0 to 360, use trig functions to figure out where the point at that angle is, and plot the closest point. Might need to be a bit tricky about the step size, but I bet you could compute a decent guess from the radius using more trig functions. Of course that probably doesn't work if you don't have floating point and trig functions.
You could also split it into four quadrants. Plot each of the top, bottom, right, and left points by direct computation of adding/subtracting the radius, then for each quadrant, you start from the computed point, check 3 specific points in the appropriate direction for which way you're going in that quadrant, plot whichever one is closest to the radius away from the center, then repeat from that plotted point until you reach the computed point for the next quadrant. It would be tedious and repetitive, but should be doable. You could probably also avoid computing square roots by comparing the squares instead. So I guess you'd want something based on that idea to do it fast on 90s hardware.
Exactly! I used this question regularly. It wasn’t a gotcha question or even an impossible-without-experience question as the author thinks. It was a show me HOW you think through something question. There are a lot of ways to solve it, from scanline to sin/cos to using the circle equation, but you can _progress_ from naive to advanced solutions, and all solutions above plus the DDA or Bresenham solutions have symmetries (hint: more than four quadrants are symmetrical) that you can use to optimize it. A practiced interviewer can learn a lot about the candidate with this question.
I wonder if there's a class of people who managed to get CS degrees but really aren't that good. To them, it might feel more like either you remember the perfect and optimal but complex solution you were taught in a class, or you don't happen to remember it and are completely stuck and can't make any progress at all. I don't think I'd want to hire or work with somebody who can't come up with some sort of solution after thinking through it for a few minutes.
In fact, coming up with the CS-perfect solution immediately may be a bit of an anti-signal. I want the person who can think their way through to a solution to a problem that's new to them reasonably well. The fact that you happened to have memorized the best algorithm for this and can recite it on command doesn't tell me much useful, because nobody has the perfect algorithm for everything memorized.
This brings back memories! I could easily pass this interview today, because I used to write code like this all the time 25 years ago doing gamedev (and so did everyone else to some extent). But the really interesting thing is that I just realized I haven't written code like this in a long, long time.
Programming has changed over time, but the change has been so gradual I hadn't even realized this until this article. These days I'm pondering how the profession has changed in the last 2 years due to AI. Feels a lot more like a step change. And yet I'm having more fun than I've had in a long time, both at work and at home, throwing Claude at problems. I still don't fully understand why.
The circle one would have gotten me. "There's a neat algo for this. The name starts with B, and you just get an oracle that tells you if next y ==current y or y-1. And then you loop x, and you have to mirror that to do all octants of the circle with mirroring and flipping by -1 for some octants"
Writing that oracle after 20+ years would have been left as an exercise to the reader.
The "Borland flood fill" routine that Microsoft was trying to beat was in the very popular (at the time) "Turbo Graphics for Borland" package which had an immensely useful stack of graphics primitives, some a little more advanced than what was available in Windows at the time, for pure DOS mode.
It was an awesome package - I shipped many products which used it in that era - and it was always a bit amusing to me that it seemed like Borland had almost everything they needed to write an operating system. But, they didn't.
The second one is absolutely trivial if you've ever read K&R (even if you're not allowed to just call strcpy()), while the fourth one is also very straightforward if you know about https://news.ycombinator.com/item?id=15266331 ; but 32 years ago, knowledge definitely did not propagate as quickly as it does today.
Additionally, I was allowed to store Color however I wanted — so if I needed some precomputation, I was allowed to bake it in there.
I believe it can be done in three operations, not including the precomputation.
This is a fun article. As a current Principal at MSFT I've never seen these type of questions being asked in interviews. I don't think it's fair or accurate to say "If you’re an experienced programmer, you already know how to do all of them". So many of the SWEs and candidates at Microsoft are just studying leetcode using python, joining the company and writing managed C# code.
It's pretty amazing to me that, if your goal is to check that the intern candidate can write plain C, the questions still look pretty reasonable to me even in 2026, maybe except for the question related to colors which will probably confuse the majority of the interns (2 bits per color? how is that possible).
For the circle drawing exercise, it just seems that the interviewer did not do a good job hinting the author. Fun fact: a person I know got this question on their Microsoft interview in around 2016. I guess, it the question works, why bother changing it!
this is probably a rhetorical question, but lemme answer anyways: By packing the colour channels into a single byte. So, for example, you'd have RRGGBBAA within a single byte, for each pixel. Giving you 64 possible colours, with 4 steps of alpha.
Or if you don't need to have alpha, you could pack it even further down to RRGGBB in a byte, which leaves 2 bits left over for the next pixel. Via that, you can pack four pixels worth of colour data into 3 bytes: RRGGBBRR|GGBBRRGG|BBRRGGBB (italics for delineting pixels, vertical bars for delineting bytes)
The latter is a tradeoff between compression and a more complex accessing pattern.
A bit of a tangent, some system used RRRGGGBB for colours, because the eyes are the least sensitive to differentiating the amount of blue, so that's another way to use up a full byte per pixel.
Oh right. Guess the " (2 bits per color? how is that possible)" is what threw me off there, because I read it as 2 bits per colour channel, rather than cga colour. Of course, "indexed" colours can get away with much fewer bits.
It seems the first two questions are coding ones, and the second two ("flood fill" and circle drawing) are more thinking ones.
The flood fill color detection one would be fastest with a look up table returning a bitmask of colors contained in the byte, with the Color param also as a bitmask, then the result is just lut[Pixel] & Color.
The circle drawing one only needs you to know basic trig to get from radius and angle (0-360) to (x,y) offset from origin. You could embellish the answer with symmetry and LUTs too, but the problem statement doesn't say anything about efficiency.
I don't believe you need trig for that, it actually makes it harder if you try to iterate the angle. I believe the expected solution is to start at (R, 0) which is known belongs to the circle, and go left/top, choosing the pixel closest to the circle on each step, which does not require any floating point arithmetic.
The problem is under-specified both in terms of requirements and any implementation restrictions.
Given the lack of difficulty of the other questions (and this being pre-internet, targeted for an intern), I don't think the interviewer can have been expecting too much sophistication.
The other obvious way do do it, only requiring the same minimal realization that this is about triangles (with radius as hypotenuse) is to use r^2 = x^2 + y^2 and iterate x=0..r deriving y. You could do it without sqrt if they stipulated that.
Right, iterating through pixels is better. The tricky part about iterating the angle is that you need to choose the step size correctly or else you could skip pixels. Like if you iterate in 1-degree increments, you'll plot 360 pixels total, but the size of the circle on your canvas might be more than 360 pixels wide. I'm sure there's a way to choose the angle iteration step size to guarantee not skipping pixels, but you'd often duplicate work and re-plot the same pixel twice.
So yes, start at (R, 0), increment the y-coordinate each time and possibly decrement the x-coordinate, until x=y which will be at 45°. If the circle's center is an integer on the pixel grid, you can reflect/translate each pixel in that first octant to all eight as you go. If the center is fractionally positioned, you'd have to calculate it all the way around, iterating primarily on y or x depending on the location.
> The tricky part about iterating the angle is that you need to choose the step size correctly or else you could skip pixels
Yes, although the problem statement doesn't say if they care. In this case they are only giving a draw_pixel() primitive, but if you had draw_line() then you could use that to avoid gaps.
The other thing is that this is 90's era, with a CGA display (640x200) being mentioned in the previous question, so I'm not sure there's enough resolution to draw a real circle without gaps unless you do resort to some hack to ensure there aren't any!
As the author notes, it would certainly be a massive ramping up in difficulty if they were expecting you to reinvent the midpoint algorithm on the spot.
Assuming high colour depth, yes, but wouldn’t it have been specified as part of the question that ‘this was for four-color CGA mode’? I think 2 bits per colour for 4 colours total seem pretty sensible even in 2026. :-)
Their point, I believe, is that someone (just about any younger person in 2026) who has never seen indexed color modes, or colors taking less than one byte per pixel to encode, could reasonably be confused by the notion of "two bits per pixel".
CGA was very limited, and didn't even support full per-color indexing - instead you got to choose one of two palettes (i.e. one of two different sets of 4 predetermined colors).
CGA was followed by EGA which supported 16 individually indexed colors (with a palette of 64 colors). With dithering you could display "faded polaroid" quality photos.
How does the video whiteboard work? I presume he isn't really writing backwards or is this some sort of software that handles the video and writing surface?
I had a go at the circle one. What I tried was starting with something very straightforward (pseudocode):
for each x from -r to r
find y that minimizes abs(x^2 + y^2 - r^2)
plot(x,y)
Finding y could be done by taking sqrt(r^2 - x^2) and then taking whichever for floor or ceil of that gives less error. But I assume they probably don't want sqrt. We could find y without using sqrt by doing some kind of search--say start with 0 and increment checked x^2 + y^2 - r^2 until we find where that crosses 0 and then take whichever y made it closes to 0. But that will be even slower than sqrt.
But what if we take advantage when we are trying to find the y for x+1 that we already have found the y that minimizes the error at x? Say we wrote a next_y(y, e) function that takes the y that was used for x and the error e that would be the error if we also used y for x+1 and that function returns the y that would minimize the error at x+1.
That's pretty easy (Python):
def next_y(y, e):
if e < 0: # need to increase y
delta = 1 + 2*y
while e < 0:
if e + delta >= 0:
if abs(e+delta) < abs(e):
return y+1, e+delta
else:
return y, e
else:
e += delta
y += 1
delta += 2
elif e > 0: # need to decrease y
delta = 1 - 2*y
while e > 0:
if e + delta <= 0:
if abs(e+delta) < abs(e):
return y-1, e+delta
else:
return y, e
else:
e += delta
y -= 1
delta += 2
else:
return y, e
Here's how it is used:
def circle(r):
x = -r
y = 0
e = 0
print(f'{x},{y}')
for i in range(2r):
e += 2x + 1
x += 1
y, e = next_y(y, e)
print(f'{x},{y}')
The basic idea is that when you increase x by 1 you increase x^2 by 2x+1, and so (x+1, y) would have error 2x+1 more than (x, y). You then need to change y to compensate.
Changing y by k changes y^2 by 2yk + k^2. Changing k by 1 changes the amount that y^2 changes by 2y + 2k + 1. Note that 2y + 2k + 1 goes up by 2 every time k goes up. In other words if you look at the sequence y^2, (y+1)^2, (y+2)^2, ... the third order diffs are 2. So to see how consecutive changes to y affect y, we can just keep track of the adjusted e and the current second order diff. Then for each consecutive y we can add the second order diff in to update the adjusted e to for the next y, and add the third order diff (the constant 2) into the second order diff.
I split next_y into two cases depending on whether we need to increase y or decrease y. There is probably some way to combine them but it is late and I'm half asleep. Oh, and I realize half of the abs() calls can be removed. I thought it looked clearer with them.
For a radius 50 half circle here is how times it took N tries to find the right y:
3 0
68 1
18 2
6 3
2 4
1 5
2 10
i.e., for 68 values of x, the first y it looked at was the right one. Here are the numbers for a radios 300 half circle:
That was a fun little exercise. It has two multiplies, but they are both by 2 so could easily be replaced by add or shift.
An improvement would be to make more use of symmetry. The curve looks best when x is between -r/2 and r/2. It would be better to draw just those parts and then use symmetry to fill the rest.
I interviewed at Microsoft in Redmond in 1997 and got zero programming questions. They were all knowledge-based or brain-teaser questions so I don't know if I believe that they gave 4 programming questions in 1994.
It's been a long time since I did any interviewing, but for a whiteboard task I'd ask for something very simple like reversing a list. It's basically a lying test more than a coding test - and sad how many people claiming multiple years of C++ could not do it.
Pitch is the offset between the start of consecutive rows of pixels in the image, used to convert y coordinates into the start of any given row, so you access a pixel as buffer[y*pitch+x]. Often this is the image width, but can be greater depending on required alignment.
So at the end of each run increment the relevant pointer by (pitch - w) not pitch which I'm sure it's one of the bugs they saw all the time in this interview :)
Well in an interview I guess something like "Of course we shouldn't allow C-strings in general outside of syscalls and argv, but for the purpose of the exercise...." And now you've shown that you know what you're talking about and that you won't be difficult to work with.
1) https://www.computerenhance.com/p/microsoft-intern-interview...
2) https://www.computerenhance.com/p/microsoft-intern-interview...
3) https://www.computerenhance.com/p/microsoft-intern-interview...
4) https://www.computerenhance.com/p/efficient-dda-circle-outli...
You could iterate over the degrees 0 to 360, use trig functions to figure out where the point at that angle is, and plot the closest point. Might need to be a bit tricky about the step size, but I bet you could compute a decent guess from the radius using more trig functions. Of course that probably doesn't work if you don't have floating point and trig functions.
You could also split it into four quadrants. Plot each of the top, bottom, right, and left points by direct computation of adding/subtracting the radius, then for each quadrant, you start from the computed point, check 3 specific points in the appropriate direction for which way you're going in that quadrant, plot whichever one is closest to the radius away from the center, then repeat from that plotted point until you reach the computed point for the next quadrant. It would be tedious and repetitive, but should be doable. You could probably also avoid computing square roots by comparing the squares instead. So I guess you'd want something based on that idea to do it fast on 90s hardware.
In fact, coming up with the CS-perfect solution immediately may be a bit of an anti-signal. I want the person who can think their way through to a solution to a problem that's new to them reasonably well. The fact that you happened to have memorized the best algorithm for this and can recite it on command doesn't tell me much useful, because nobody has the perfect algorithm for everything memorized.
https://en.wikipedia.org/wiki/Midpoint_circle_algorithm
Programming has changed over time, but the change has been so gradual I hadn't even realized this until this article. These days I'm pondering how the profession has changed in the last 2 years due to AI. Feels a lot more like a step change. And yet I'm having more fun than I've had in a long time, both at work and at home, throwing Claude at problems. I still don't fully understand why.
Writing that oracle after 20+ years would have been left as an exercise to the reader.
It was an awesome package - I shipped many products which used it in that era - and it was always a bit amusing to me that it seemed like Borland had almost everything they needed to write an operating system. But, they didn't.
Additionally, I was allowed to store Color however I wanted — so if I needed some precomputation, I was allowed to bake it in there.
I believe it can be done in three operations, not including the precomputation.
The naive approach’s assumes you can iterate over the first string until it terminates.
It’s a bit trickier if you do not assume the memory regions cannot overlap.
See memcpy vs memmove: https://man7.org/linux/man-pages/man3/memmove.3.html
y -= x >> 4;
x += y >> 4;
Certainly works, and seems to require 100 iterations to get a full circle.
Are there other approximations, taking smaller angular steps, to get a better circle?
For the circle drawing exercise, it just seems that the interviewer did not do a good job hinting the author. Fun fact: a person I know got this question on their Microsoft interview in around 2016. I guess, it the question works, why bother changing it!
this is probably a rhetorical question, but lemme answer anyways: By packing the colour channels into a single byte. So, for example, you'd have RRGGBBAA within a single byte, for each pixel. Giving you 64 possible colours, with 4 steps of alpha.
Or if you don't need to have alpha, you could pack it even further down to RRGGBB in a byte, which leaves 2 bits left over for the next pixel. Via that, you can pack four pixels worth of colour data into 3 bytes: RRGGBBRR|GGBBRRGG|BBRRGGBB (italics for delineting pixels, vertical bars for delineting bytes)
The latter is a tradeoff between compression and a more complex accessing pattern.
A bit of a tangent, some system used RRRGGGBB for colours, because the eyes are the least sensitive to differentiating the amount of blue, so that's another way to use up a full byte per pixel.
https://en.wikipedia.org/wiki/Color_depth
But the original post actually talks about CGA [1] with just four colors. Encoding a color needs two bits then, so each byte encodes four pixels.
[1]: https://en.wikipedia.org/wiki/Color_Graphics_Adapter
The flood fill color detection one would be fastest with a look up table returning a bitmask of colors contained in the byte, with the Color param also as a bitmask, then the result is just lut[Pixel] & Color.
The circle drawing one only needs you to know basic trig to get from radius and angle (0-360) to (x,y) offset from origin. You could embellish the answer with symmetry and LUTs too, but the problem statement doesn't say anything about efficiency.
Given the lack of difficulty of the other questions (and this being pre-internet, targeted for an intern), I don't think the interviewer can have been expecting too much sophistication.
The other obvious way do do it, only requiring the same minimal realization that this is about triangles (with radius as hypotenuse) is to use r^2 = x^2 + y^2 and iterate x=0..r deriving y. You could do it without sqrt if they stipulated that.
So yes, start at (R, 0), increment the y-coordinate each time and possibly decrement the x-coordinate, until x=y which will be at 45°. If the circle's center is an integer on the pixel grid, you can reflect/translate each pixel in that first octant to all eight as you go. If the center is fractionally positioned, you'd have to calculate it all the way around, iterating primarily on y or x depending on the location.
Yes, although the problem statement doesn't say if they care. In this case they are only giving a draw_pixel() primitive, but if you had draw_line() then you could use that to avoid gaps.
The other thing is that this is 90's era, with a CGA display (640x200) being mentioned in the previous question, so I'm not sure there's enough resolution to draw a real circle without gaps unless you do resort to some hack to ensure there aren't any!
As the author notes, it would certainly be a massive ramping up in difficulty if they were expecting you to reinvent the midpoint algorithm on the spot.
Assuming high colour depth, yes, but wouldn’t it have been specified as part of the question that ‘this was for four-color CGA mode’? I think 2 bits per colour for 4 colours total seem pretty sensible even in 2026. :-)
CGA was followed by EGA which supported 16 individually indexed colors (with a palette of 64 colors). With dithering you could display "faded polaroid" quality photos.
I think he wears a shirt with mirrored writing so it looks correct to the viewers.
But what if we take advantage when we are trying to find the y for x+1 that we already have found the y that minimizes the error at x? Say we wrote a next_y(y, e) function that takes the y that was used for x and the error e that would be the error if we also used y for x+1 and that function returns the y that would minimize the error at x+1.
That's pretty easy (Python):
Here's how it is used:def circle(r): x = -r y = 0 e = 0 print(f'{x},{y}') for i in range(2r): e += 2x + 1 x += 1 y, e = next_y(y, e) print(f'{x},{y}')
Here are some half circles drawn with it: https://imgur.com/a/5aYzPqS
The basic idea is that when you increase x by 1 you increase x^2 by 2x+1, and so (x+1, y) would have error 2x+1 more than (x, y). You then need to change y to compensate.
Changing y by k changes y^2 by 2yk + k^2. Changing k by 1 changes the amount that y^2 changes by 2y + 2k + 1. Note that 2y + 2k + 1 goes up by 2 every time k goes up. In other words if you look at the sequence y^2, (y+1)^2, (y+2)^2, ... the third order diffs are 2. So to see how consecutive changes to y affect y, we can just keep track of the adjusted e and the current second order diff. Then for each consecutive y we can add the second order diff in to update the adjusted e to for the next y, and add the third order diff (the constant 2) into the second order diff.
I split next_y into two cases depending on whether we need to increase y or decrease y. There is probably some way to combine them but it is late and I'm half asleep. Oh, and I realize half of the abs() calls can be removed. I thought it looked clearer with them.
For a radius 50 half circle here is how times it took N tries to find the right y:
i.e., for 68 values of x, the first y it looked at was the right one. Here are the numbers for a radios 300 half circle: That was a fun little exercise. It has two multiplies, but they are both by 2 so could easily be replaced by add or shift.An improvement would be to make more use of symmetry. The curve looks best when x is between -r/2 and r/2. It would be better to draw just those parts and then use symmetry to fill the rest.
while (*to++ = *from++) ;