51 lines
2.9 KiB
Markdown
51 lines
2.9 KiB
Markdown
|
With this problem I tried a little bit differen methology with designing the problem.
|
||
|
|
Programs with massive (16k) input are always kind of ugly in befunge because
|
||
|
|
all of the data must be in the program code, so I thought I can at least try a little bit around.
|
||
|
|
|
||
|
|
In this program I seperated the code, as much as I could, into independent subprograms.
|
||
|
|
All subprograms can use the [1,0]-[9,0] fields as temporary values, get their input
|
||
|
|
from the stack and write their output also to the stack.
|
||
|
|
Then I combined them together to build the whole program.
|
||
|
|
|
||
|
|
All the subprograms are in my [BefungePrograms git repo](https://gogs.mikescher.com/gitmirror/BefungePrograms)
|
||
|
|
|
||
|
|
This resulted in more readable code and (hopefully) snippets that I can reuse in other programs.
|
||
|
|
But the code doesn't compress as good (which nobody cares about in this problem, cause of the giant input size)
|
||
|
|
and I'm sure I could optimize it a lot by using more global state and shared variables etc.
|
||
|
|
|
||
|
|
I think for my next programs I will continue as I did before and sometimes use independent code snippets
|
||
|
|
(for example for the integer-squareroot function) but for the big main program I will write it all together.
|
||
|
|
|
||
|
|
The program works like so:
|
||
|
|
|
||
|
|
1. First we calculate a "palindromic hash value" for each input word, this is a hash algorithm that
|
||
|
|
has no collisions as long as there are max five repeated letters in a word and has the same value for
|
||
|
|
palindroms.
|
||
|
|
Practically it is a 26-digit number in base-5 where each digit denotes the amount a specific letter occurs in our word.
|
||
|
|
We can not use a larger number than 5 for our base because then we would overflow our 64bit numbers.
|
||
|
|
|
||
|
|
2. Next we go through our palindromic list and search for palindroms (words with the same hash)
|
||
|
|
With some clever sorting tricks we could do this in `log2(n)`, but I will leave this as an
|
||
|
|
exercise for ther reader and and implement in naively in `n^2`
|
||
|
|
|
||
|
|
3. For each word we iterate through all the squares with the correct digit count.
|
||
|
|
This means we start with `j`, where `j = 10^(len - 1)` and wnd with `k`, where `k = (10^len) - 1`
|
||
|
|
|
||
|
|
4. Now (for each square) we can generate the numeric value for word B with word A + square as our map.
|
||
|
|
When we generate our char->digit map (as an 26 element array) we also generate a digit->char map
|
||
|
|
to test if any digit is mapped to multiple different characters (a failure criteria)
|
||
|
|
|
||
|
|
5. Now we have square A and (possible) square B, with our optimized is-integer-squareroot function from problem 086
|
||
|
|
we test if B is a square number. And if this is the case (and B is bigger than our current candidate) we set B
|
||
|
|
as our current result candidate
|
||
|
|
|
||
|
|
6. After we have done this for all pairs we can return (= print out) our current best candidate.
|
||
|
|
|
||
|
|
|
||
|
|
Used sub programs:
|
||
|
|
- fixed_base_pow.b93
|
||
|
|
- read_single_word.b93
|
||
|
|
- get_palindromic_hash.b93
|
||
|
|
- integer-squareroot-2.b93
|
||
|
|
- is-squarenumber.b93
|
||
|
|
- length_single_word.b93
|