EE2026 (misc)

Hardware challenge requiring vivado and some understanding of circuit schematics.

The challenge author provides us with a folder containing a bunch of files. This seems to be his submission for some EE (Electrical Engineering) assignment. The most important file given is the pdf file, which describes the assignment.

The assignment starts off by telling us the format of a student identification number. It starts with a one-digit value A, followed the personal password which is 5 characters containing digits padded with the letter X, followed by an alphabet B and ending with the digit C. For example, if A=1, personal password = 1234X, A='B' and C=0, then the ID no. would be 11234XB0. Note that the digits in B must be sorted and it cannot have duplicate digits.

The assignment requires the student to program a basys3 board with several requirements.

Firstly, when the 16 active-high switches SW0-SW15 are off during initialisation, all active-high LEDs LD0-LD15 should be off. Based on the value A, we desire the following display on the 7 segment display of the board:

Secondly, whenever any of the switches SW0-SW9 are on, their corresponding LEDs LD0-LD9 should also turn on. LD10-LD14 should not have any constraints.

Thirdly, the switches SW0-SW9 represent a user's "guesses" of the 5-character password B. For example if the password is 1234X, then turning SW1-4 on and leaving SW5-9 off is correct. When the guess is correct, LD15 should turn on. LD15 should be off if the combination of on switches is wrong.

Lastly, when the correct password is entered, the 7 segment display should show the following depending on the alphabet B:

When the correct password is entered, the anodes on the 7 segment display which light up should be the following depending on the digit C. What are anodes? We will get to this later.

So we have to somehow reverse engineer the given files to find out what student number is the correct input for a particular basys3 board design.

Looking through the files, some of them are graded_post_lab_assignment_1.xpr, vivado.jou, vivado.log, basys3_constraints.xdc and we also have .dcp files.

Upon further research, this appears to be a design project done using vivado. So we need to install vivado in order to analyse the author's project.

To open it in vivado, simply go to Open Project and choose the .graded_post_lab_assignment_1.xprfile.

The initial project which vivado shows us isn't helpful for solving the challenge (I think).

In order to view the circuit design we need to go to Flow > Open Implemented Design > impl_1... first which shows us the following:

This still doesn't seem to be very meaningful, so after poking around a bit I eventually went to Tools > Schematic which shows a very nice schematic of the circuit board design. I broke it into 3 pictures since it's too big:

We already know that SW stands for switches and LD stands for the LEDs on the board. But what are the alphabets A-G, OBUF, IBUF, AN, LUT5 and LUT6? After researching abit on this board, I came to realise the following:

LUT5 and LUT6 are simply programmable Look-Up Tables which we can change the logic of.
OBUF and IBUF are output and input buffers and we don't need to care much about them.
The different ANx values and the alphabets A-G represent different components of the 7 segment display according to the figure below:

-> The anodes AN0-AN3 each represent a different digit in the seven-segment display, so each account for a quarter of the display.

-> The letters A-G in the schematic represent the 7 different parts of a digit in the seven-segment display.

After more research, I found that:

To get a certain part of the display to light up, for example anode 1 segment G, the schematic's input to AN1 needs to be 0 and the schematic's input to the cathode G needs to be 0 (both active low).
As mentioned above, the LEDs and switches are active-high (on switches output a 1, which turns an LED on).

It appears that for this LUT5, the only way for the output to be 0 is for I4=I3=1 and I2=I1=I0=0.

For the LUT5 below we have this truth table excerpt:

So having I4=I3=I2=0 and I0=I1=1 is the only way to output a 1.

For LUT6 we have only one row outputting a 1:

Now that we have understood the logic, we can find the digit A by imagining what the 7 segment display would show if all switches SW0-SW9 were turned off. Looking at the bottom-right of the schematic, we know that AN3=A=E=0 forever as they are all grounded. We also know that B=F=1 forever as they are always powered.

Tracing the rest of the 7 segment display components, the output of LUT6 would be 0, the output of the top LUT5 would be 1 and the output of the bottom LUT5 would be 0. Therefore AN0=C=1 and AN1=AN2=D=G=0. Hence out of the four anodes, AN1, AN2, AN3 would light up, and the segments lit up would be A, D, E, G. Referring to the 7 segment display again, this configuration corresponds to 2. So the first digit of the ID number is 2.

Now let's find out the 5 characters in personal password. To do this, we have to figure out which combination of SW0-SW9 would turn LD15 on. This means that the output of the bottom LUT5 must be 1. The corresponding inputs should be I4=I3=I2=0 and I0=I1=1 (as mentioned above). This means that we must have SW0=SW7=SW9=0 so they should be off. SW8=1 so switch 8 should be on, and the output from LUT6 should be 1. If the output from LUT6 should be 1 then LUT6's I2=I4=I5=1 and I0=I1=I3=0. This means that SW1=SW2=SW4=1 and SW6=SW5=SW3=0.

-> Thus to produce the correct output we should only turn on switches 1, 2, 4 and 8. Hence the 5-char password is 1248X.

Now let's solve for alphabet B. To figure this out, we need to know which of the 7 segments display's segments are on (the segment's input is 0) when the correct password is entered. We know that when the correct password is entered, the top LUT5 outputs 0. So C=0. The bottom LUT5 would output 1, so D=G=1. And we know that B and F are always 1, while A and E are always 0. So we need to see which alphabet B has the segments A, E, C on. This turns out to be the alpabet G.

Lastly, let's solve for the value C. To find this we need to find out which anodes are on (ANx=0) when the correct password is entered. Since the top LUT5 outputs 0, AN0=0. Since the bottom LUT5 outputs 1, AN1=AN2=1. And AN3=0 as it is always grounded. So AN0 and AN3 would be on. Hence the last digit is 8.

Putting everything together, the 8-character ID would be 21248XG8. The flag is therefore grey{21248xG8}.

Last updated 3 months ago