13
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Introduction

"Muhuhuhahahah!" The mad scientist laughs. "You're trapped in my own little game!"

In front of you is a deadly pit of snakes, while behind you is a bottomless chasm. There's no way out, you're stuck!

"Two steps in front of you is the snake pit, and two steps behind you is the chasm. But! Before you move, you MUST write down a sequence of steps, forwards and backwards, and give them to me. But! because I'm feeling a bit evil today, I can make you take, instead of every step, every nth step, where n is less than your sequence length!

Choose wisely, now."

What's the maximum number of steps you can take before your imminent death?

Task

The intro above is a twist on the Erdős discrepancy conjecture, which was recently proven true (if you want to understand more about this, go to this video, by James Grime - I "stole" the twist question off of him).

The answer to the intro is 11 steps, but I won't go too in-depth with a proof. The answer, if the distance between you and the two "dangers" were 3 steps, is 1160 steps, although that isn't validated properly yet.

Your task is to make a program that generates the longest sequence of steps you can achieve for a larger x, where x is the number of steps between you and the two "dangers". Your program must take an input for x, and output a valid sequence for that x.

For the purposes of this challenge, + represents a step forward, and - represents a step back.

So, an output for an input 2 is:

+--+-++--++

Which works, no matter what n the mad scientist chooses. For our challenge, x = 5.

NOTE: This challenge is not a dupe of this challenge or this challenge, as my challenge focuses on the output, as opposed to the code itself - in other words, it's not a code golf challenge. As well as that, these challenges are based off of x = 3, which already has an established upper bound.

Rules:

  • Your entire program should fit into your answer. However, if it doesn't fit, please provide an additional Github repository, or something similar.
  • You may update both your answer and your program, if you can get a better score via optimising your code - but by doing so, you must update everything in the list below.
  • In your answer, you must have:
    • Your program, in its entirety, or a link to a GH repository hosting your code
    • The amount of steps generated - this will be your final score.
      • You must also provide an online version of the sequence in a Pastebin, or something similar. This is so we can check your answer.
    • The time your final score was last updated, so I don't have to check your history
  • You may NOT hardcode sequences beforehand.
  • Your program must work for all x (where x is the number of steps between you and the pit & chasm), but you only need to provide the score for x = 5.

The answer with the largest score wins!

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  • \$\begingroup\$ Just to check my understanding, you need a sequence where even if you took every other, or every third element, you survive? \$\endgroup\$ – Notts90 supports Monica Jun 5 '17 at 11:23
  • 1
    \$\begingroup\$ @Notts90 Yup - it musn't just work for that, though. It must work if you took every nth step, where n is any number below your sequence size. \$\endgroup\$ – clismique Jun 5 '17 at 11:24
  • \$\begingroup\$ Very closely related. (I called this up as a potential duplicate in the sandbox; I'd have therefore expected the text of the challenge to at least discuss it.) \$\endgroup\$ – user62131 Jun 5 '17 at 16:06
  • \$\begingroup\$ I think this challenge is impossible, practically speaking. Finding the maximum discrepancy length for x=5 would require a major breakthrough that would be worthy of publication. Consider that the maximum of 1160 for x=3 was proven and published in 2014 and no further values are known.. \$\endgroup\$ – xnor Jun 5 '17 at 17:29
  • \$\begingroup\$ Is 0 a proper N? \$\endgroup\$ – tuskiomi Jun 5 '17 at 17:30
6
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Rust, score=4997216, time=2017-07-12 00:18 UTC

This improves on the best result I was able to find in the literature, which was 1148805 (Ronan Le Bras, Carla P. Gomes, Bart Selman, On the Erdős Discrepancy Problem, 2014), by a factor of 4.3.

Output sequence of length 4997216

Source code on GitHub

Running

The program accepts the maximum discrepancy as an argument (this is x − 1 in the language of the challenge, to align with the more common mathematical convention). It produces incremental outputs in a slightly compressed format that looks like this for x = 3:

$ cargo run --release 2
add +--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++-++--+-++-+
length 90
delete 12
add --++--+-++-++--+-++--+--+-++--+-
length 110
delete 4
add +-+--+-++-++--+-++--+--+-++-++--+-++-+
length 144
delete 6
add --++-++--+-++--+--++++--+--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-+-
length 214
delete 208
add --+++--+++--+-+--+++--+-+--+++--++---+-+--+-+-++-+--+++--+++--+---++-+--+-++-+++---++--+-++-++--++--+--++--+++--+-+-++-+--+-+--+++---+++-+----+++--+-++--++-+-++--+-+--+-+-++-+--+++--++--+--+--+-++-++---++-++-++-+--+-++
length 224
delete 2
add -+++--+-+--+++---++--+--
length 246
done

where add means append a sequence of signs to the end of the current sequence, delete means remove some number of signs from the end of the current sequence, and length asserts the length of the current sequence. This scheme avoids the production of many gigabytes of intermediate results as longer and longer sequences are discovered. You can extract the best result so far with the following Python program:

import sys
s = ''
for line in sys.stdin:
    cmd = line.split()
    if cmd[0] == 'delete': s = s[:-int(cmd[1])]
    elif cmd[0] == 'add': s += cmd[1]
    elif cmd[0] == 'length': assert len(s) == int(cmd[1])
print(s)

How it works

There are about a thousand lines of code here, so this is only going to be a very rough overview.

We limit the search to completely multiplicative sequences (those with f(a·b) = f(af(b)), because that means we need only concern ourselves with bounding the partial sums for n = 1, and the partial sums for n ≥ 2 will satisfy the same bound automatically.

For any substring f(i + 1), …, f(j) of a partially assigned sign sequence (so each element is ‘+’, ‘−’, or unknown), define danger+(i, j) to be twice the number of ‘+’s minus the length ji (for convenience, we allow i to be as small as −x + 2 and assume f(−x + 3) = ⋯ = f(0) = ‘+’ for this purpose). Define danger(i, j) similarly. Then the bound on partial sums for n = 1 is equivalent to the condition that whenever ijx (mod 2), danger+(i, j) ≤ x − 2 and danger(i, j) ≤ x − 2.

We build an incremental data structure that supports constant time queries for the substring with largest danger, with logarithmic time updates. It works by associating four values:

  • danger(i, j),
  • maxikj danger(i, k),
  • maxikj danger(k, j), and
  • maxiklj danger(k, l),

with every string of length 2, every other string of length 4, every fourth string of length 8, and so on. The values associated with a longer string can be computed in constant time from the values associated with its two halves.

This structure, augmented with some auxiliary information, lets us perform constraint propagation and conflict detection on partial sequences very quickly. We use it to do a CDCL-like search, with unit propagation, decision levels, and non-chronological backtracking (but without clause learning, for now), for full sequences of longer and longer lengths.

At each search step, we make a guess for the earliest unassigned sign. The heuristic used to make this guess is very important for avoiding a lot of backtracking; we use f(3·k) = −f(k), f(3·k + 1) = ‘+’, f(3·k + 2) = ‘−’.

Results

The discrepancy 0, 1, and 2 searches immediately find the optimal completely multiplicative sequences of length 0, 9, and 246.

The discrepancy 3 search gets stuck within seconds at 41319, which is pretty far from the known optimal completely multiplicative sequence of length 127645 found by Le Bras et al., 2014 (and a very slightly longer non-multiplicative extension of length 130000 found shortly thereafter), but much better than the best known sequence before that of length 17000.

The discrepancy 4 search improves the longest sequence more or less continually for about five minutes until it gets stuck at 4997216. The best previously known sequence of length 1148805 = 9·127645 was expanded from the discrepancy 3 sequence by replacing every sign s with +−−+−++−s. As far as I can tell, sequences this long are too hard for a general SAT solver to reasonably improve directly (but maybe you, dear reader, can prove me wrong!).

I expect I’ll need to add some kind of clause learning to my program to get past these barriers.

The sequence as a 2187×2285 bitmap

(Click to view at full resolution.)

the sequence as a 2187×2285 bitmap

| improve this answer | |
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  • \$\begingroup\$ 127465 is for completely multiplicative sequences, right? \$\endgroup\$ – CalculatorFeline Jul 12 '17 at 1:29
  • \$\begingroup\$ "Clause learning"? \$\endgroup\$ – CalculatorFeline Jul 12 '17 at 1:38
  • \$\begingroup\$ See conflict-driven clause learning—it’s how modern SAT solvers work. \$\endgroup\$ – Anders Kaseorg Jul 12 '17 at 1:39
  • \$\begingroup\$ What's all the fuzz in the bitmap? \$\endgroup\$ – S.S. Anne Feb 29 at 17:45
  • \$\begingroup\$ @S.S.Anne Places where the program discovered that a deviation from the heuristic leads to a longer sequence. \$\endgroup\$ – Anders Kaseorg Feb 29 at 21:29
3
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Haskell, score=9020, time=2017-06-09 00:52 UTC

f 1=""
f n="+-"++do c<-f(n-1)++"-";"-+-++-"++c:"+-"

Try it online!

This scores (9n − 1 − 1)⋅11/8 for all n. Here are the first few outputs:

n=1, length=0: 
n=2, length=11: +--+-++--+-
n=3, length=110: +--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++--+-
n=4, length=1001: +--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++--+-
n=5, length=9020: 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