> py test.py
| b = 1 + 3x-5n |
Ans: a = (x + 2 * y) / (x - 3 * y)
b = 1 + 3 * x ** (3 * n)
x = 3
y = 5
n = 2
a = (x + 2 * y) / (x - 3 * y)
b = 1 + 3 * x ** (-5 * n)
print("Output:", round(a, 4), round(b, 6))
# Output:
Output: -1.0833 1.000051
P3 5 5 255 128 128 128 128 128 128 128 128 128 128 128 128 128 128 128 128 128 128 255 128 0 255 128 0 255 128 0 128 128 128 128 128 128 255 128 0 0 128 128 255 128 0 128 128 128 128 128 128 255 128 0 255 128 0 255 128 0 128 128 128 128 128 128 128 128 128 128 128 128 128 128 128 128 128 128The header line P3 5 5 255 indicates the following:
P3: Portable Pixel Map (ppm) 5: Image is 5 pixels wide 5: Image is 5 pixels high 255: Component value range is 0 to 255.Pixels contain red, green, and blue components.
| Operator | Meaning |
|---|---|
| + | Addition Concatenation |
| - | Subtraction |
| * | Multiplication Addition |
| / | Floating Point Division |
| // | Integer Point Division |
| ** | Exponentiation |
| % | Remainder from Floating Point Division |
| = | Assignment |
n = 3 * 5This statement assigns the value on the right, which is 3 * 5 = 15 to the variable n.
< <= > >= == != and or not
a = 5
b = 7
a = 3
print(a, b)
Ans:
Variable Trace: a b
-----+-----
5 | 7
3 |
Output: 3 7
The latest value of a variable is always the one used. This is why a 3 is printed for a instead of 5# Trace1 Exampl # Perform variable trace and predict output. a = 3 b = 7 c = 5 a = 2 * a + 2 * b + c b = a + c c = 2 * c + 1 a = a + 2 b = b + 3 c = c + 4 Variable Trace: a b c -----+-----+----- 3 7 5 25 30 11 a = 2 * a + 2 * b + c 27 33 15 = 2 * 3 + 2 * 7 + 5 # Output: = 6 + 14 + 5 Output: 27 33 15 = 25