Finding Collatz-like rules with many loops

Question

This code-challenge is related to the code-golf question Analyzing Collatz-like sequences but the goal is quite different here.

If you are familiar with Collatz-like sequences you can skip to the section "The task".

We define a Collatz-like rule with 3 positive integers:

• d > 1 divisor
• m > 1 multiplier
• i > 0 increment

(In the original Collatz sequence d = 2, m = 3 and i = 1.)

For a Collatz-like rule and a positive integer n starting value we define a sequence s in the following manner:

• s(0) = n
• if k > 0 and s(k-1) mod d = 0 then s(k) = s(k-1) / d
• if k > 0 and s(k-1) mod d != 0 then s(k) = s(k-1) * m + i

An example sequence with d = 2, m = 3, i = 5 and n = 80 will be s = 80, 40, 20, 10, 5, 20, 10, 5, 20, ....

Every sequence will either reach higher values than any given bound (i.e. the sequence is divergent) or get into an infinite loop (for some t and u (t!=u) the s(t) = s(u) equality will be true).

Two loops are said to be different if they don't have any common elements.

The task

You should write a program which finds a Collatz-like rule with many different loops. Your score will be the number of found loops. (You don't have to find all of the possible loops and higher score is better.)

The program could be specialized in any manner to produce the best score.

For proof the program should output the values of d, m, i and the smallest values from every found loop.

Example

For d=3 m=7 i=8 the loops are 
LOOP 15 5 43 309 103 729 243 81 27 9 3 1
LOOP 22 162 54 18 6 2
LOOP 36 12 4
So a valid proof with a score of 3 could be 
d=3 m=7 i=8 min_values=[1 4 2]

You can validate your result with this Python3 code. The program expects a list of space-separated integers d m i min_value_1 min_value_2 ... as input.

• orlp: It's not immediately obvious to me that there exist no combination of (d, m, i), all positive, that has infinite amount of loops. Even harder to disprove, who says there exist no algorithm that can produce a combination of (d, m, i) that has a finite but arbitrarily large amount of loops.
• orlp: @PeterTaylor I thought of that, but i has to be positive, so it can't be 0.
• flawr: I do not see the challenge, you can just bruteforce up to an arbitrary number...
• Martin Ender: @flawr Until Peter's answer there was no proof that the number of loops is actually unbounded.
• randomra: @flawr Finding big primes is still interesting despite the fact that there are simple algorithms that can yield arbitrarily large ones in very long time. I think this is a similar situation. (If you could suggest a modification of the question to make it clearer that would be great.)

Answer 1

GolfScript, arbitrary score

~:l 2:m.@?):d1円 l,{m\)?d-*}/abs.{1 m 3$)?-*m@)?d-m(*/}+l,/]' '*

This takes an integer on stdin and outputs a sequence with that many loops, in the input format of the Python tester. The online demo runs it for l = 4 and produces output 17 2 1755 117 405 1365 26325; this fork of the tester verifies it.

Explanation

In order to keep it simple, I restricted the analysis to primitive loops: ones with only one division.

(x -> mx + i)^n = (x -> m^n x + (m^n - 1)/(m-1) i)

so this means that I'm looking for solutions to

m^n x + (m^n - 1)/(m - 1)i = dx

or

(m^n - d) x + (m^n - 1)/(m - 1)i = 0

When n > 0, m > 1, m^n - d != 0 this has exactly one solution for x. m - 1 > 0 and m^n - 1 > 0, so in order for the solution to be positive we require m^n - d < 0, and in order for the solution to be an integer we require d - m^n to be a factor of (m^n - 1)/(m - 1)i.

The rest may already be obvious. We pick the desired score, l. We're going to find primitive loops of length 1 to l, so we want d - m^l > 0. To get a smallish output we pick m = 2, d = 2^l + 1. Then to get an easy guarantee of the divisibility constraints we pick i to be the absolute value of (m - d)(m^2 - d)...(m^l - d).

• orlp: This was exactly my conjecture why this golf's scoring is flawed - it can be arbitrary. Your answer can however be beat by a solution that provides infinite score, if it exists.
• Peter Taylor: @orlp, what's the distinction you're drawing between arbitrarily large and infinite?
• orlp: An infinite solution would have one m, d, i combination that would contain an infinite amount of cycles. Your solution can generate a m, d, i for an arbitrary, but finite score. I do not know if an infinite solution exists, but I do not find it obvious that it does not.
• Peter Taylor: It's not obvious that there isn't an infinite solution in that sense, but it would surprise me if there is one. If you fix m, d, and the loop structure (in terms of the sequences of multiply or divide steps) then there are 0 or 1 loops of that structure, and one of the conditions required for it to exist is a divisibility constraint on i. Unless there are an infinite number of loop structures which induce the same divisibility constraint, which would be surprising, any finite i can only satisfy a finite number of them.

Answer 2

Python 3, (d = 2, m = 3, i = 96521425) → Score = 2369

The program below writes the output to collatz.txt, and also prints summaries to the shell.

I used the observation that, if (d, m, i) had a lot of loops, then in general (d, m, i*k) also tended to have a lot of loops for different values of k. I started with (2, 3, 13), and eventually worked i up like so:

13 -> 65 -> 1235 -> 35815 -> 250705 -> 2757755 -> 13788775 -> 96521425

before ending up with the lengthy output that can be found here.

The respective scores of the above chain are (using a bound of 10000):

10 -> 16 -> 28 -> 146 -> 304 -> 535 -> 766 -> 1094

2369 was obtained by bumping the bound up to 100000.

This raises a question similar to the one that @orlp has brought up — if I keep doing this multiplication process, will I always be able to get more and more loops? If anyone can prove that, then they'd probably win :)

from collections import defaultdict
import itertools
def f(d, m, i, bound=1000):
 D = defaultdict(set)
 S = set()
 for j in range(1, bound):
 A = [j]
 B = {j}
 for _ in range(bound):
 if j > bound**3:
 break
 if j % d == 0:
 j //= d
 else:
 j = j*m + i
 if j in B:
 S.add(min(A[A.index(j):]))
 break
 A.append(j)
 B.add(j)
 return S
N = 1000
THRESHOLD = 0
BOUND = 10000
MULTIPLIER = 96521425
with open("collatz.txt", "w") as outfile:
 for d, m, i in itertools.product([2], [3], range(1, N)):
 result = f(d, m, MULTIPLIER*i, bound=BOUND)
 if len(result) > THRESHOLD:
 print(d, m, MULTIPLIER*i, " ".join(map(str, sorted(result))), file=outfile, flush=True)
 print(d, m, MULTIPLIER*i, len(result))

---

(Proof in progress for a possibly different arbitrary score method)

Pick d, m, i such that d and i are coprime and suppose we have a loop for (d, m, i):

a_0 -> a_1 -> a_2 -> ... -> a_(n-1) -> a_n = a_0

Now pick k coprime to i and d, and we want to look at (d, m, i*k). Consider the number k*a for some a.

• If d divides a then d divides k*a, and in this case we map k*a -> k*a/d = k*(a/d).
• Otherwise, d can't divide k*a since d and k are coprime, and we map k*a -> m*(k*a) + i*k = k*(m*a + i).

Hence one step for a in (d, m, i) is equivalent to one step for k*a in (d, m, i*k). This forms a bijection between the loop in (d, m, i) with the loop

k*a_0 -> k*a_1 -> k*a_2 -> ... k*a_(n-1) -> k*a_n = k*a_0

in (d, m, i*k).

This means that (d, m, i*k) has at least as many loops as (d, m, i). But since i is coprime to d, we can apply the same reasoning to any loops in (d, m, k).

(This explains why the multiplication method managed to give at least as many loops each time, but the proof is incomplete until I can constrain i, k such it always gives strictly more loops each time).

Answer 3

Java, arbitrary score with relatively small numbers

In my previous answer, I considered only primitive loops. It turns out that there's a massive benefit to be gained by considering more complex loops.

Let the loop structure n_0 ... n_{k-1} denote a loop which has n_0 multiply operations, then a division, then n_1 multiplications, then a division, etc. If we start with x_0 and loop back to x_0 then we get the constraint

-(d^k - m^n) x + (d^{b_0}m^0 + d^{b_1}m^1 + ... + d^{b_{n-1}}m^{n-1}) i

where n = n_0 + ... + n_{k-1} and k > b_0 >= b_1 >= ... >= b_{n-1} with the transitions at points corresponding to division steps.

The big bonus is the divisibility constraint: (d^k - m^n) | i is sufficient for x to exist and doesn't depend on the breakdown of the loop structure. However, because we have small d we have to worry about a new problem.

The map x -> mx + i (mod d) is a permutation of 0 .. d-1, and the length of the cycle which contains 0 places an upper bound on the length of a step in the loop sequence. So we have to pick d to satisfy two constraints: d^k - m^n > 0 and the cycle length of 0 under the permutation x -> mx - m^n is greater than n.

If we do that, though, the net result is that for much smaller increments than in the primitive case we can get a number of loops whose growth is related to necklaces. We have two free variables: in the interests of keeping it simple, we'll pick n = 2k. Then the score is OEIS A082936 (k+1).

This program allows you to specify k via the command line. If k isn't specified it defaults to 15 (and will take a while to run).

import java.math.BigInteger;
import java.util.*;
public class PPCG47835 {
 public static void main(String[] args) {
 int k = args.length > 0 ? Integer.parseInt(args[0]) : 15;
 BigInteger TWO = BigInteger.valueOf(2);
 int n = 2 * k;
 // We want d to allow steps of up to n without getting a division "too early".
 BigInteger d, i;
 nd: for (d = BigInteger.valueOf((k + 1) | 1); true; d = d.add(TWO)) {
 i = d.pow(k).subtract(BigInteger.ONE.shiftLeft(n));
 if (i.signum() < 0) continue;
 // Find multiplicative inverse of m and additive inverse of i.
 BigInteger half = d.add(BigInteger.ONE).shiftRight(1);
 BigInteger negI = d.subtract(i.mod(d));
 BigInteger inv = BigInteger.ZERO;
 for (int j = 0; j < n; j++) {
 inv = inv.add(negI).multiply(half).mod(d);
 // If we get a division too early, reject this value of d.
 if (inv.signum() == 0) continue nd;
 }
 break;
 }
 System.out.print(d); System.out.print(" ");
 System.out.print("2 "); // m is hard-coded for optimisation reasons
 System.out.print(i); System.out.print(" ");
 visit(d, i, new int[k], 0, n);
 }
 private static long visit(BigInteger d, BigInteger i, int[] a, int off, int unused) {
 long count = 0;
 if (off == a.length - 1) {
 a[off] = unused;
 if (canonical(a)) {
 BigInteger x0 = BigInteger.ZERO, dj = BigInteger.ONE;
 int tail = 0;
 for (int step : a) tail += step;
 for (int step : a) {
 x0 = x0.add(dj.shiftLeft(tail));
 tail -= step;
 x0 = x0.subtract(dj.shiftLeft(tail));
 dj = dj.multiply(d);
 }
 // Find min element of loop
 BigInteger x = x0, min = x0;
 for (int step : a) {
 x = x.add(i).shiftLeft(step).subtract(i).divide(d);
 if (x.compareTo(min) < 0) min = x;
 }
 System.out.print(min); System.out.print(" ");
 count++;
 }
 }
 else {
 for (a[off] = 0; a[off] <= unused; a[off]++) {
 count += visit(d, i, a, off + 1, unused - a[off]);
 }
 }
 return count;
 }
 private static boolean canonical(int[] a) {
 for (int i = 1; i < a.length; i++) {
 // If (a[i], a[i+1], ..., a[n], a[0], ...) < (a[0], a[1], ..., a[n]) then not canonical
 int cmp = 0;
 for (int j = 0; j < a.length; j++) {
 int l = a[(i + j) % a.length];
 if (l < a[j]) return false;
 if (l > a[j]) { cmp = 1; break; }
 }
 }
 return true;
 }
}

A table of values for small k is

k d m i count time*
1 5 2 1 1 < 10 ms
2 9 2 65 3 < 10 ms
3 11 2 1267 10 < 10 ms
4 11 2 14385 43 < 10 ms
5 13 2 370269 201 ~ 20 ms
6 19 2 47041785 1038 ~ 100 ms
7 19 2 893855355 5538 ~ 200 ms
8 19 2 16983497505 30667 ~ 200 ms
9 25 2 3814697003481 173593 ~ 1 s
10 29 2 420707232251625 1001603 ~ 6 s
11 29 2 12200509761511525 5864750 ~ 40 s
12 29 2 353814783188691825 34769374 ~ 5 mins
13 29 2 10260628712891493325 208267320 ~ 30 mins
14 37 2 9012061295994739864233 1258579654 Est. 3 hours
15 37 2 333446267951814233346669 7663720710 Est. 20 hours
16 37 2 12337511914217162067306945 46976034379 Est. 6 days
17 37 2 456487940826035138224277733 289628805623 Est. 40 days
18 53 2 10888439761782913818653903872953 1794932468571 Est. 200 days
19 53 2 577087307374494432392024159626573 11175157356522 Est. 4 years
20 53 2 30585627290848204916790749477648625 69864075597643 Est. 30 years
21 53 2 1621038246414954860589963598385138149 438403736549145 Est. 200 years
22 53 2 85915027059992607611268286218691365993 2760351032959050 Est. 1 millennium
23 53 2 4553496434179608203397220031607758574013 17433869214973754 Est. 7 millennia
24 53 2 241335311011519234780052665123279669128225 110420300879752990 Est. 50 millennia

* Time estimates are for my desktop and do not take into account Moore's law.

By way of comparison, the primitive loop approach gets a count of 256 with d, m, i

115792089237316195423570985008687907853269984665640564039457584007913129639937 2 49968490581887083932095804336538857276757417490402562927136208951271487539689095350518315671301404278005363619443112350923124996893730693441853802988787649829173834378882653691501262688830470259060532631533427430770702136299497849324717113003384169491948585144662309301555418354928369764175214625528498298281514394917957697007610895507068653221093095188675091865216617272197943427128945145535282933335835999158551191526601683314776122600532692429865604302347050773307867947871242247467417554911259346113701538927646726497472509852234411078217429031179784822107433104045508215699906272669550200824625122676996125246083825817358721488729539820382911790387362505222144036190526712328090026898643367818652593793545401098423008361620007066091731759647650767785112637095783967446745061562940336964274874939952845810757383710907696922802160547621885891275181382881315205841975743181786600619939726953558998318001503462956122792894679599223590901815297738260321839835509661476390130530211756244463205868100449183431705062206670445120771257613033170733226704746855964393508142206414546777215253150392801313925328908299101701300676970320804094742804393381647592715119303325138677025707665442285411323759289471830684060945833475938328005642804427622906358274525832409978699538436928881922260679479595091188536546688424867586685739453609590901458114420828646341675639858393765027850268225431713269426496582339871392757218592043290353835198068238612124250576139380049311650436985118635646174627087745565825616719407169728941806489390066578961154288629594552384172309467136790675954898168626916443491474862068947798499576390075637734891858257154829186292418746469533939685899739837260557581405096709651311307842577526587467702995228758369679164175586245800565166450258185997461077868111951176109069620072117908733988857052934321247935362102983225520842987661982680693860628808054915992220832993354643182332600510561245789234769915565946271051413257200145979042629750575885631601279833943463371495594511849212232740461662165099231724532972763088660059706401609719744472536401854830499389962909139679904172189684566122922880928798535805265433824541033533239322217610437526501009585105145468363346653018444546327067325798081560766434573441063566637173199520662054221368155845574982742874163068099483335163380595124002858360317680377387288538666550856978356289617459609365791500140305940510997366117308287849520902479251497170884801775860919730775159201637834268735863579303973551177393818232325693935844429212250562928164989461805619412895259696170039604225916775322325816954944228946728265014348350107111445486932327217655248588835015845641316141062327352382735682719421143498298798125490111306274989321402199637873036663920472953471301618796885083908114564570523686410878842444327176956928617406201155686422090096852189709409958234070215084032130616015198336189199149497499325607299448636944086048856415648877223135692598854638245608116295668807663626020012518782077948092918132181223394468212203077651412354720824430117663039804754199346922916494265207869382103831596948188859072724868680345961830969467839486417736592350598653322073697531917019530742717205093541634143665305091102887782953647153044594814197262695911111130338314764675973362782025256381353873801968881081209289299122065927588730368417431740592835618492705895469735186709335746090178621014542113218144295159656057829559385548357639666536568773781147012141790455178325125187259679604746885488456909904032525706852966347353838250308514907578360708484872177271431506301226325823267135246712164313678979671857319011509789782361328586963988787707090857146113396544593817634214541406752347105460060137273110421632802623585765555747292820406973666424354869115637362289788365228626731735661914673151628370860744114408929983716179450500036905009051518961140247520336301488975330963759879464842309451607155817395430589784169646238858612671370723308241834173243124718366744466140108760081336397903798238645892196315508514934051795715214144939062117857161014899379444763186820552545701005775676740900729995672865635386747452728586547395967811407500887267019830634675686889697525413526627246932217730474894533557035929176177780588870599088435994931881789678787008746864521432447971488195130058872464308945520952427755129904453498616226930053479082160817407727976458424404567950270305767639906044730208010036318849170958344743043316016925926437610321306545206851214315391175153156932682252008424063692741427320559722711075163517767077183715332126127321591070135206303549158060859578478201446106801614436432651357690911157352401724754883365300369356210889875297667847042733104820589217338267235040676953036551240577457779986956523394013134172376005453783592699837828264331875024571732285180560046353235675806883400159739847741656614770907697513467366386953103469683977580305011615663038524262172652557223985911910190983514737219037843441651343635897414058881810120659426751473593438711874591030928662440412291081099932233743426995478153867452872028431879622861888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• randomra: Fantastic answer! I will digest the details in the next few days.