How I microwave food

September 20, 2024 (1y ago)

To optimize my microwave process and get back to coding as quickly as possible, I need to find the most efficient way to enter cooking times.

When I microwave, I don’t press 1, 0, and 0 to set a one-minute timer. Instead, I press 6 and 0. This way, I can save a third of the energy used every time I microwave food for some time of less than or equal to 1 minute and 39 seconds. But the calculation doesn’t end here. There must be a way to determine the best sequence for entering a time on the microwave keypad.

Now, to determine the best sequence we have to calculate the distance between each button. First, we can convert the microwave keypad into a coordinate system:

The button 1 is positioned at (-1, 3).

The button 2 is positioned at (0, 3).

The button 3 is positioned at (1, 3).

The button 4 is positioned at (-1, 2).

The button 5 is positioned at (0, 2).

The button 6 is positioned at (1, 2).

The button 7 is positioned at (-1, 1).

The button 8 is positioned at (0, 1).

The button 9 is positioned at (1, 1).

The button 0 is positioned at (0, 0).

Next, to calculate the minimum distance, we can use the Euclidean distance formula:

import math
 
def dist(x1: int, y1: int, x2: int, y2: int):
    return math.sqrt((x2 - x1)**2 + (y2 - y1)**2)

The total distance for a sequence would be calculated as the sum of individual differences.

Next, when determining the optimized sequence for entering a specific time, there might be multiple ways to input the sequence. If the seconds component is less or equal to 39, there are potentially two options to represent the time:

  1. Entering the time as MMSS, where M represents the minutes and S represents the seconds.
  2. Adjusting the sequence so that the total seconds fit within a single minute, which can be written as (MM-1)(SS+60).

For example, when computing 2 minutes and 39 seconds, the valid sequences are “239” and “199”.

If we calculate the distance travelled:

This effectively saves about 5.72% of the original energy required.

Another example would include computing 5 minutes and 31 seconds, which has valid sequences “531” and “491”.

If we calculate the distance travelled:

That’s about 32.62% of the energy saved!

The algorithm has spoken!

# some python code I wrote in <5 minutes
# single digit inputs do not work, because why would you compute them anyways?
 
import math
 
keypad_coords = {
    '1': (-1, 3), '2': (0, 3), '3': (1, 3),
    '4': (-1, 2), '5': (0, 2), '6': (1, 2),
    '7': (-1, 1), '8': (0, 1), '9': (1, 1),
    '0': (0, 0)
}
 
def calculate_distance(b1, b2):
    x1, y1 = keypad_coords[b1]
    x2, y2 = keypad_coords[b2]
    return dist(x1, y1, x2, y2)
 
def total_distance(sequence):
    distance = 0
    for i in range(len(sequence) - 1):
        distance += calculate_distance(sequence[i], sequence[i + 1])
    return distance
 
def optimized_sequence(minutes, seconds):
    seq1_minutes = minutes
    seq1_seconds = seconds
 
    seq1 = f"{seq1_minutes if seq1_minutes > 0 else ''}{seq1_seconds:02}"
 
    seq2 = None
    if minutes > 0 and seconds <= 39:
        seq2_minutes = minutes - 1
        seq2_seconds = seconds + 60
        # Option 2: (MM-1)(SS+60) format
        seq2 = f"{seq2_minutes if seq2_minutes > 0 else ''}{seq2_seconds:02}"
 
    dist1 = total_distance(seq1)
    dist2 = total_distance(seq2) if seq2 else float('inf')
 
    if dist1 <= dist2:
        return seq1, dist1, seq2, dist2
    else:
        return seq2, dist2, seq1, dist1
 
time_minutes = 1
time_seconds = 39
 
best_sequence, min_distance, worst_sequence, max_distance = optimized_sequence(time_minutes, time_seconds)
 
print(f"Best sequence for {time_minutes} minutes and {time_seconds} seconds: {best_sequence}")
print(f"Minimum distance: {min_distance:.4f}")
print(f"Worst sequence: {worst_sequence}")
print(f"Maximum distance: {max_distance:.4f}")
print(f"Percent saved: {((1 - min_distance / max_distance) * 100):.2f}%")
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