7/L Segment Display

Today I was troubleshooting a sensor that was giving an error code — but Error 52 “Over Voltage” was not what I expected… regardless I checked the voltage and it looked fine. About ten minutes later I realized that the display was mounted upside down and it was Error 25, “Programming Error” — duh.

 

Which one? Choose wisely

 

While this is totally my fault it is also avoidable. So if you are a company that makes things with important codes using a seven segment display AND your display can logically be upside down then don’t use any numbers that are also valid in either orientation.

 

 

Here are the amount of numbers that satisfy that criteria:

To find this I used the criteria of a number having to be not a number upside down (because of a 7, 4 or 3) or be itself when upside down (96 -> 96).

1 Digit Numbers

2 Digit Numbers

3 Digit Numbers

4 Digit Numbers

5 Digit Numbers

 

Here is the code:

import matplotlib.pyplot as plt

def flip(num):
    flips = [
        ["0","0"],
        ["1","1"],
        ["2","2"],
        ["3","E_"],
        ["4","H_"],
        ["5","5"],
        ["6","9"],
        ["7","L_"],
        ["8","8"],
        ["9","6"]]

    new_num = ""
    for i in range(len(num)):
        new_num += flips[int(num[len(num)-1-i])][1]
    if new_num == num or len(new_num) > len(num): return num
    else: return

def find_all_digits(digits):
    valid_options = []
    for i in range(0,10**digits):
        val = str(i)
        val = "0"*(digits-len(val))+val
        if flip(val) is not None:
            valid_options.append(flip(val))
    return len(valid_options)

x,y = [],[]
for i in range(1,9):
    x.append(i)
    y.append(find_all_digits(i)/10**i)

fig = plt.figure()
ax = fig.add_subplot(111)
plt.scatter(x,y)
for i,j in zip(x,y):
    ax.annotate("("+str(i)[:4]+","+str(int(round(j*10**i)))+")",xy=(i-0.075*len(str(round(j*10**i))),j+0.01))
plt.title("Number of valid numbers that aren't confusing when upside down")
plt.xlabel("N\n Number of Digits Used")
plt.ylabel('Valid Entries / Total Possible Entries')
plt.show()

 

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