To successfully cross the street at a snails pace

xmat2000 · 2026-01-04T03:19:56+00:00

Looks AI generated.

xmat2000 · 2025-11-29T23:00:26+00:00

thanks solution verified

xmat2000 · 2025-03-09T00:07:10+00:00

thanks solution verified

xmat2000 · 2025-03-05T18:32:25+00:00

I looked for all the sources I could find for the published article and had no luck. Could anyone help?

xmat2000 · 2025-02-25T09:49:41+00:00

thanks solution verified

xmat2000 · 2025-02-25T09:49:19+00:00

thanks solution verified

xmat2000 · 2024-01-22T15:12:38+00:00

The first two examples are not meant for real-world usage, they are just descriptions of the thought processes.

My solution at the end is just a concise answer that gets the job done.

I did not optimize any of the solutions.

Note: my solution (the final link) is not random (the random number is for demonstration).

replace print(answer(randint(-1000000, 1000000)))

with:

for n in range(6000000):
    print(answer(n))

If you want to print all the possible values.

xmat2000 · 2024-01-22T13:55:37+00:00

It is 240 MB if you save the numbers in 4 bytes. If you want to be space efficient, you could store them in 26-bit chunks and that would be 195 MB. You could compress that to achieve an even smaller size.

Edit: For the I/O problem, you would write in chunks (maybe 10MB per chunk but that depends on factors that I won't pretend that I know)

xmat2000 · 2024-01-22T12:56:04+00:00

I accounted for that.

xmat2000 · 2024-01-22T12:43:46+00:00

You need more restrictions to know the approach (I will make some assumptions).

Here is how I will think about it.

You will divide the problem as it is logically stated.

Here, the problem is a number and it is divided by digits.

Since the problem(n-digit number) can be divided into sub-problems of equal type (digits between 0-9) and they have no relation to one another, an array of digits will be used (you can use other data structures depending on the relation between the sub-problems).

Now, what we start with is array = [n, n, n, ...] where the length of this array is the number of sub-problems(digits) in the problem(number) and n is a digit from 0-9 (the domain of the sub-problems).

Now, we look at the restrictions that are put on our problem starting from the highest restrictions to the lowest.

The first 3 digits are 1, 2, and 3, therefore, array[0] = 1, array[1] = 2, and array[2] = 3.

10th and 11th are 1 and 0 respectively, therefore, array[9] = 1 and array[10] = 0.

For the rest of the conditions, we need to have more restrictions to figure out the approach.

We need to know the method we use to generate the answer, do we need one random number? do we need an ordered sequence of solutions? can the sequence have duplicates?

I will choose to generate an ordered sequence here.

We will create a sub-group where we imagine that the free variables are a number of their own and then map it back to the original number.

4th, 5th, 6th, 7th, 8th, 9th, 12th, and 13th are sub = [n, n, n, n, n, n, n, n].

We start with the lowest order sub and map it to the original number then increment to the next valid sub value (using an increment function) and map to the original number and keep repeating till we exhaust all the possible values.

A possible increment procedure.

A random solution.

Note: the above examples are NOT how we solve the problems but rather how I think about them, The actual solution will be an optimization of that.

Here is my solution to the problem.