View Full Version : STEM thread part II

The Unreasoner

12-14-2015, 06:43 PM

So I will get around to PSPACE again, but first I had a physics question:

Everyone says 'spooky action at a distance' can't send information faster than light, but surely you can probabalistically? If we have 3 pairs of entangled bits separated by a light year and a prearranged time of transmission (and suitably synchronized clocks), couldn't I (the sender) measure the three bits vertically to send a one, and horizontally to send a zero? Then the receiver could measure them all vertically, and get the bit with only a one in eight chance of a false positive (and no chance of false negative). You could even scale up, and get the .125 down below an arbitrary theshhold.

The Unreasoner

12-15-2015, 12:11 AM

So I just remembered that the outcome is still randomly spin up or down. Though part of me still thinks that there must be some way to send imperfect info faster than light (when I was in HS, I didn't know gravity waves weren't instantaneous, and came up with a proof of concept project using vast chunks of antimatter to annihilate portions of distant stars, and thought I was a genius. Unfortunately, my teacher didn't know either, and couldn't correct me.

So back to Boolean formulas. Anyone who followed my last thread probably noticed how innaccesible/dull I made the tattoo flowchart after repeated 'reductions'. And some of you (Kimon) may rightly be asking yourselves, 'why should I give a shit?'

This really comes down to languages and logic. The most important question facing humanity (far greater in importance than things like stopping the environmental crises, fusion power, or how to beat ISIS (my plan? Send a token force to Dabiq, and don't turn into salt)) is: does P equal NP? The reason is this: any problem that we can check a given answer for with a Turing Machine (iow a computer, but specifically the kind we have now) quickly is part of the complexity class NP. When that same machine can actually find an answer quickly, it's in P. If they are (though it's generally believed they are not) equal (and we know how, ie no diagonalization nonconstructive bullshit), the world changes drastically. Mostly for the better/best. There are people that whine about it destroying creativity, that anyone who could recognize good art would become an artist, but it's easy enough to at least probabalistically disprove that. Though you could write 'Bach' music with a computer, in the sense that a computer couldn't tell the difference.

But what other sorts of problems can be checked quickly? Quite a few, actually. Physics, genetics, biology, chemistry, medicine would advance at (quite literally) an exponential rate, given a method to solve NP problems quickly. Cancer? Cured. Fusion? Done. Controlled mutation? Well, how many penises do you want? AI would advance to Sci-fi levels, hunan languages could be translated perfectly. Efficiency in everything. Perfect efficiency. GM yeast with DNA computers in your bloodstream that mutate as needed to produce needed meds or proteins. Maybe we'll even find a way to safely modulate our inactive genes in a way that lets us store a record of our history and knowledge, so that it could survive as long as we do. Even mathematics (real mathematics) could be partially automated.

And there is no (or almost no) exaggeration in any of that. We wouldn't need aliens or God to teach us all of the mysteries of the universe anymore. We would have the power to unlock almost any knowable one ourselves. Although I suppose God would still be technically responsible, assuming we got our mathematical abilities from him. Or maybe the aliens will be the ones to give us this Nondeterministic Oracle.

In any case, this is why I showed the reductions in the last thread. Because this means that all of these problems are essentially the same problem, and they will either stand or fall together. If just one has a chink in its armor, the first person to break through it will have all of the power of a living Oracle. This is not hyperbole.

But then, most experts feel that P is not equal to NP, so we'll need to tackle these problems one at a time, with heuristics, cunning, and a bit of luck.

Nazbaque

12-15-2015, 01:28 AM

It would still require for us to know what questions to ask and how to ask them. That is the fundamental problem. How to turn any given problem into mathematics. That's basically the difference between truly understanding art and being an advanced appreciator of it. Most artists in the whole human history didn't belong to the former. In the absolute terms this requires, most artists had simply reduced the field of their wild guess but still lacked the true understanding. Human perception is at the moment incapable of fullfilling this end of the bargain.

The Unreasoner

12-15-2015, 01:28 AM

And on that note, we'll take another look at Boolean statements. As it happens, almost any type of problem can be encoded in such a statement. You'll recall that it takes several variables, assembled in clauses, with the operators AND, OR, and NOT. The structure of the problem is embedded in the contents and format of the statement's clauses. Something like (¬x1 v x2 v x3) ^ (x1 v ¬x2 v x3). quick recap, I use 'v' as OR, '^' as AND. '¬' is NOT. Variables are either TRUE or FALSE, and the entire statement produces a single output (also TRUE or FALSE). The thing of interest isn't evaluating the function for a specific input. This is mostly straightforward, and is done quickly. As I've said, the checking the correctness of a specific answer is fast for all problems in complexity class NP. The hard part is finding them.

Some functions are easy to invert (iow, find the input(s) given an output). For instance, f(x)=x^2. If you know the output is 4, x is either 2 or -2. Even some functions with multiple variables are easily inverted. f(x,y)=x+y has a set of points as its possible inputs for any given output (in this case, the line with slope -1 and y intercept at the output). But Boolean functions are a bit trickier to work with. The variables are binary, the operators are strange, and there can be nested groups with convoluted grammar.

So how do we manipulate them? There are tools from Boolean algebra that allow us to toy with the statement; to get rid of or add nesting, to insert or remove variables from a clause. It has its own distributive property and identities. This is how I (or, more accurately, Wolfram Alpha) took the original convoluted statement with layered nesting that was derived from that tattoo flowchart and into a standardized form (specifically, Conjunctive Normal Form). While CNF may mask or distort underlying structure in a more exotic statement, it does allow for certain solving methods and reductions. One of which I will cover now.

CNF is essentially an AND list of OR lists. But Disjunctive Normal Form is practically a list of the solutions (inputs). It's an OR list of AND lists (and any AND list in it that doesn't contradict itself is a solution). So how do you get this magical DNF?

There's a proper way, but I'll show you a simpler one I usually used that may be more familiar. It also eliminates self-contradictions as you go. I call it Boolean arithmetic.

It's almost the same as the arithmetic you know, except:

Only zeroes and ones exist. One consequence of this is that any number (even if it's a variable) multiplied by itself remains unchanged. Therefore a^2*b^3*c^7 is just abc.

Addition is different. While a number plus zero remains unchanged, so does a number added to itself. This means coefficients are discarded. 5a is just a.

Every variable has a negation variable (denoted here as a'). A variable added to its negation is 1, a variable multiplied by its negation is 0.

So to recap:

0+0=0

0+1=1

1+1=1

x+0=x

x+1=1

x'+0=x'

x'+1=1

x'+x=1

x'*x=0

So, how do we use this on the example above (and repeated below)?

(¬x1 v x2 v x3) ^ (x1 v ¬x2 v x3)

Replace the negated variables with the negation variables, the AND operators with *, and the OR operators with + (I'll also switch to a b and c):

(a'+b+c)*(a+b'+c)

then distribute normally, but applying the modified rules above.

So we get:

a'a+a'b'+a'c

+

ab+b'b+bc

+

ac+b'c+cc

Then:

0+a'b'+a'c+ab+0+bc+ac+b'c+c

Since c is by itself, you can quickly group and factor terms that include it:

a'b'+ab+c*(a'+a+b'+b)

a'b'+ab+c*(1+1)

a'b'+ab+c*(1)

Finally:

a'b'+ab+c

Each term represents a suitable input: (F,F,NA; T,T,NA; NA,NA,T)

Next we'll try it on something more complicated.

The Unreasoner

12-15-2015, 01:30 AM

It would still require for us to know what questions to ask and how to ask them. That is the fundamental problem. How to turn any given problem into mathematics. That's basically the difference between truly understanding art and being an advanced appreciator of it. Most artists in the whole human history didn't belong to the former. In the absolute terms this requires, most artists had simply reduced the field of their wild guess but still lacked the true understanding. Human perception is at the moment incapable of fullfilling this end of the bargain.

Exactly. The art of asking good questions is humanity's highest. It takes a great mind to use great tools.

Nazbaque

12-15-2015, 01:52 AM

Exactly. The art of asking good questions is humanity's highest. It takes a great mind to use great tools.

Or a lucky guess.

GonzoTheGreat

12-15-2015, 04:02 AM

Some functions are easy to invert (iow, find the input(s) given an output). For instance, f(x)=x^2. If you know the output is 4, x is either 2 or -2. Even some functions with multiple variables are easily inverted. f(x,y)=x+y has a set of points as its possible inputs for any given output (in this case, the line with slope -1 and y intercept at the output). But Boolean functions are a bit trickier to work with. The variables are binary, the operators are strange, and there can be nested groups with convoluted grammar.

An additional problem (and actually a very fundamental one) is: how many variables are relevant in the function?

If you know that the answer is "4", that still doesn't tell you whether the question is the square of -2, the standard number of Beatles, or some weird Boolean combination thereof.

The Unreasoner

12-15-2015, 06:18 AM

An additional problem (and actually a very fundamental one) is: how many variables are relevant in the function?

If you know that the answer is "4", that still doesn't tell you whether the question is the square of -2, the standard number of Beatles, or some weird Boolean combination thereof.

I can't tell if you are making a joke, a subtle point, or asking a serious question.

You do, of course, need the definition of the function along with the output, or you're absolutely right. Inversion would be impossible. The relationships between variables varies wildly even among some standard functions. And if you could invert nonstandard functions easily, you could prove the Riemann Hypothesis. And the zeta function has only one variable.

Boolean sentences have many variables with often hidden or implied relationships. And they are fundamentally nonstandard (in college I found a way to convert an arbitrary sentence into a continuous, one parameter function, but to invert it you needed to evaluate a horrifying improper integral. I later found out that it was more or less just another cosine integral problem, and NPC in its own right).

Or maybe you were pointing out that PSPACE is different? In which case you are absolutely right, but I'm not quite there yet. But the teaser version: what if you ask me a (factual/logical) question, I give you the (or a) correct answer, but the answer is uncheckable for all intents and purposes? One example of such a problem comes from games: for reversi, go, checkers, or (generalized) chess on an infinite board, how do you even verify a (given) 'perfect move'? iow, I could tell you to place a stone in 3-5-9 in reversi, confident that whatever move your opponent plays next, a perfect, immutable, path to victory remains open to you. But how do you check that?

PSPACE is games where NP is puzzles (infinite sudoku is NP complete). It's SAT plus ontological operators (for all/there exists). It's a step beyond. But still an open problem. One that is closed is the one concerning hypercomputing. Hypercomputation needs an unbounded halting oracle. And considering the fact that perfect modelling of physical reality is believed to be Turing Complete, not likely physically possible.

Or maybe you think I dumbed everything down too much, and are just toying with me to point out you guys are not idiots, and I'm coming across as condescending. And if that's the case I apologize. This is already a self-selected niche thread. I don't need to teach arithmetic. I was shooting for middle ground, but I will consider the audience in the next round.

In the meantime, dors anyone understand these EM drives? I thought they were a hoax, but apparantly not.

GonzoTheGreat

12-15-2015, 06:31 AM

What I was aiming at is that being able to find such solutions would be very useful, but it doesn't help if your problem is figuring out what function is actually relevant. And that's basically what physics is all about: finding the rules that govern reality.

To show that problem for another field: consider curing cancer. You say that would be easy, with this extra mathematical method. Let's assume that this math trick had been available in the 19th century already, before anyone knew about genes, DNA and the like. Would it then really have been much use in finding a cure for cancer?

We now know a lot of reasons why it wouldn't be. But we don't know all about cancer, so there could still be unknown unknowns which would still prevent an easy solution, even if we could evaluate all the known parameters using this math trick.

The Unreasoner

12-15-2015, 03:39 PM

What I was aiming at is that being able to find such solutions would be very useful, but it doesn't help if your problem is figuring out what function is actually relevant. And that's basically what physics is all about: finding the rules that govern reality.

To show that problem for another field: consider curing cancer. You say that would be easy, with this extra mathematical method. Let's assume that this math trick had been available in the 19th century already, before anyone knew about genes, DNA and the like. Would it then really have been much use in finding a cure for cancer?

We now know a lot of reasons why it wouldn't be. But we don't know all about cancer, so there could still be unknown unknowns which would still prevent an easy solution, even if we could evaluate all the known parameters using this math trick.

Again, you are mostly right. I'm putting quite a bit of faith in the intelligence of the questioner.

But then, I think you underestimate the Oracle. While 19th century medicine couldn't exploit such an algorithm, modern medicine certainly can. For instance, cancer is already cured, from the mathematician's perspective. It's a matter of finding a protein chain that will fold into a structure that strangles cells with one genetic sequence while leaving another unharmed. Obviously you'd have a separate instance of the problem for every patient (or even the same patient with a slightly different cancer). It's not one size fits all, it's made to order. But the genetic knowledge, algorithms, and protein production capabilities we have today are enough. Only our software and hardware are lagging.

As for physics: if you tell it what you observed and which mathematical rules to obey, it will be able to find the smallest set of equations that perfectly explains all the data. And if that one doesn't work for whatever reason, it could find the second smallest. And so on.

The Unreasoner

12-15-2015, 03:56 PM

Now let's revisit tattoos using Boolean arithmetic.

From the old thread:

(PvovX)^(evfvX)^(kvlvX)^(kvmvX)^(Rvovq)^(DvXvY)^(G vXvY)^(NvXvY)^(avXvY)^(bvXvY)^(hvXvY)

A is always FALSE

B is always FALSE

C is redundant (can be either TRUE or FALSE without affecting the solution in any way)

D is always TRUE

E is TRUE iff z12 is not a member of the subset

F is TRUE iff z13 is not a member of the subset

G is always TRUE

H is always FALSE

I is redundant

J is redundant

K is TRUE iff z11 is not a member of the subset

L is TRUE iff z08 is not a member of the subset

M is TRUE iff z05 is not a member of the subset

N is always TRUE

O is TRUE iff z19 is not a member of the subset

P is TRUE iff z16 is a member of the subset

Q is TRUE iff z01 is not a member of the subset

R is TRUE iff z02 is a member of the subset

And the legend again:

A= Are you drunk?

B= Are your friends egging you on?

C= Are your friends laughing?

D= Does it have a special meaning?

E= Is it a name?

F= Is it the name of a significant other?

G= Is it unique?

H= Is it 'Steve-O' unique?

I= Are you trying to fit in?

J= Are you sure you are not trying to fit in?

K= Will it be visible when you are dressed?

L= Are you a white collar worker?

M= Will it be on your face?

N= Is it appropriate for children to see?

O= Is it going on the small of your back?

P= Do you want men to think you are easy?

Q= Are you a man?

R= Do you prefer to 'catch'?

The Unreasoner

12-15-2015, 04:21 PM

I'm on my phone, and plus signs are pain. So if it's all the same, we'll just use spaces and parentheses. Lower case letters are negation variables.

(PvovX)^(evfvX)^(kvlvX)^(kvmvX)^(Rvovq)^(DvXvY)^(G vXvY)^(NvXvY)^(avXvY)^(bvXvY)^(hvXvY)

(P o)(e f)(k l)(k m)(R o q)

That removed the known variables (if you like, you can treat them as a common factor for all the terms in the final set).

Next:

(PR Po Pq Ro o oq)(k kl km lm)(e f)

And that's actually it. Since no variable appears (or has its negation appear) in more than one clause, this is essentially DNF factored. If you choose one element from each of the three lists, it will work.

Returning the common factor:

(DGNabh)(PR Po Pq Ro o oq)(k kl km lm)(e f)

The Unreasoner

12-15-2015, 04:34 PM

You can count solutions/make the exhaustive list by reincluding the redundant variables.

(DGNabh)(PR Po Pq Ro o oq)(k kl km lm)(e f)(C c)(I i)(J j)

1*6*4*2*2*2*2=384. Which, you may note, is not 165 or whatever the other number was for the last thread. Clearly I fucked up somewhere, just don't know where.

ETA:

I forgot the list isn't exhaustive because some variables are conditionally redundant. Which may make at least one of my old answers right, though 384 is meaningless.

ETA:

Reinserting conditionally redundant variables into the ef clause raises it to three terms (everything but EF). The opqr clause goes to 11. And the klm goes to 5. So I was right about 165 after all. Now I just have to see if 8*165 equals my old guess.

The Unreasoner

12-16-2015, 05:28 PM

So let's build some functions. I think we'll look at chess pieces on a 4x4 board.

In all systems, you need some way of encoding information. We'll use 2 bits with an x tag to represent the column, and two with y to represent row.

00 (or FALSE, FALSE) is 1

01 is 2

10 is 3

11 is 4

So let's consider a function T(Rx1,Rx2,Ry1,Ry2,Hx1,Hx2,Hy1,Hy2), where T determines if a rook R threatens a piece H. We'll assume no other pieces are on the board. Essentially we need to check if H and R are in the same row or column (and if we wish, not both).

We'll do this with a function K, which checks whether two pairs of bits are the same.

Something like this:

K(a1,a2,b1,b2)=¬((a1 xor b1)v(a2 xor b2))

Now xor is just exlusive or, but in order to define everything in AND/OR/NOT gates, we'll say i xor j is Q(i,j)=(ivj)^(¬iv¬j).

And that's all we need. So T=Q(K(Rx1,Rx2,Hx1,Hx2),K(Ry1,Ry2,Hx1,Hx2)), and is TRUE iff rook R and piece H share either the same row or the same column, but not both.

Now let's turn it into CNF.

The Unreasoner

12-16-2015, 05:51 PM

First let's replace Rx1,Rx2,Ry1,Ry2,Hx1,Hx2,Hy1,Hy2 with a,b,c,d,e,f,g,h.

Then we have Q(K(a,b,e,f),K(c,d,g,h)).

Q(¬(Q(a,e)vQ(b,f)),¬(Q(c,g)vQ(d,h)))

(((¬(((ave)^(¬av¬e))v((bvf)^(¬bv¬f))))v(¬(((cvg)^( ¬cv¬g))v((dvh)^(¬dv¬h)))))^(¬(¬(((ave)^(¬av¬e))v(( bvf)^(¬bv¬f))))v¬(¬(((cvg)^(¬cv¬g))v((dvh)^(¬dv¬h) )))))

Now we just have to distribute and simplify. Though I think I'll just use Wolfram Alpha, because it's a pain to do by hand.

The Unreasoner

12-16-2015, 06:16 PM

Can anyone see what's wrong with this input? Wolfram Alpha isn't liking it.

BooleanConvert[(((! (((A || E) && (! A || ! E)) || ((B || F) && (! B || ! F)))) || (! (((C || G) && ( ! C || ! G)) || ((D || H) && (! D || ! H))))) && (! (! (((A || E) && (! A || ! E)) || (( B || F) && (! B || ! F)))) || ! (! (((C || G) && (! C || ! G)) || ((D || H) && (! D || ! H)))))),"CNF"]

GonzoTheGreat

12-17-2015, 04:09 AM

I've looked at it a bit, and the number and pairing of brackets seems all right. So that fairly obvious issue isn't it.

Then I tried feeding it to Wolfram Alpha myself, but that said the string was too long, cut of the end of it and then complained that it wasn't properly formatted. The latter was of course believable at that point, but not really helpful, since the problem was generated by WA itself.

In order to use the whole string I would need a paid version of WA, which I don't have. So I don't know what it says about the whole.

I did try to feed a (more or less randomly chosen) subset into WA, and that did not give any problems.

So I would advice you to try to chop it in half, and then feed each half separately into WA. If either half gives you the same problem back you have now, that will at least allow you to zoom in on what that problem is. Maybe you can then even solve it.

Daekyras

12-17-2015, 12:07 PM

I had a busy couple of days and I come back to find I missed this thread?

Damn!

I'll give it a close going over and stick in my penny's worth in a day or so. Damn!

The Unreasoner

12-17-2015, 07:53 PM

I've looked at it a bit, and the number and pairing of brackets seems all right. So that fairly obvious issue isn't it.

Then I tried feeding it to Wolfram Alpha myself, but that said the string was too long, cut of the end of it and then complained that it wasn't properly formatted. The latter was of course believable at that point, but not really helpful, since the problem was generated by WA itself.

In order to use the whole string I would need a paid version of WA, which I don't have. So I don't know what it says about the whole.

I did try to feed a (more or less randomly chosen) subset into WA, and that did not give any problems.

So I would advice you to try to chop it in half, and then feed each half separately into WA. If either half gives you the same problem back you have now, that will at least allow you to zoom in on what that problem is. Maybe you can then even solve it.

I thought it was probably brackets too, but couldn't see an issue with them. Put the first half in, didn't work. But the first and second quarters worked fine, so I'm guessing it's a length issue. Though it didn't cut it off or tell me.

On the bright side, it turns out WA does recognize xor, so I'll just feed it the Q equation version, which is much shorter.

ETA: So it could handle that version, but just gave me some Boolean Operator number or something like that. Maybe there's an index of small circuits it's referring to.

And, amusingly, it gave me the truth density of 3/8. Which works. But it's much easier to count the spaces on our chessboard and subtract the safe ones than whatever WA did.

The Unreasoner

12-18-2015, 04:40 PM

Probably more important than the conversion itself are the techniques used.

So on that note:

NOT travels inward until it attaches to a literal (variable). Functions flip:

NOT (A OR B) is (NOT A AND NOT B)

NOT (A AND B) is (NOT A OR NOT B)

Distributing across an AND:

A AND (B OR C) is (A AND B) OR (A AND C)

Distributing across an OR:

A OR (B AND C) is (A OR B) AND (A OR C)

Associative rules apply:

A AND(OR) (B AND(OR) C) is (A AND(OR) B) AND(OR) C

If you are taking it beyond CNF SAT to 3 SAT, you need a way to reduce the size of an OR clause. This is done by invoking a new variable:

(A OR B OR C OR D) is (A OR B OR X) AND (C OR D OR NOT X)

And nested clauses of the same function collapse to one level:

A AND(OR) (B AND(OR) (C AND(OR) D)) is A AND(OR) B AND(OR) C AND(OR) D

Next we'll look at circuits, then past attempts to prove P vs NP.

The Unreasoner

12-18-2015, 08:06 PM

Actually before we move on to circuits, I thought we'd cover two more things.

First: a way to test if two Boolean functions are essentially the same. This is useful if you are trying to find a more efficient formula (ie, fewer clauses or variables, less nesting, etc).

Function x is essentially equivalent (iow, equisatisfiable) to y if the function ¬(x xor y) is always true. You can confirm this by converting to CNF and checking that each OR clause has both a variable and its negation.

Another thing I thought I'd poibt out is that there are only finitely many possible Boolean functions (that are fundamentally different) for any finite set of variables. This can be seen by noting that any equation can be written in CNF. We'll look at equations in three variables for the example.

The largest possible clause contains an instance of every variable, and there are two states for each variable. So there are 2^3 clauses of size 3.

Clauses of size 2 contain all but one variable. There are 3 possibilities for the excluded variable, and 2^2 possibilities for the rest.

Finally there are 3*2 possible clauses of size one.

This gives us a total of 26 possible clauses. The number of formulas is simply the number of possible subsets aside from the empty set. This number is 2^26 -1, or 67108863.

Almost 70 million formulas in three variables.

Now a question: there are only 8 possible inputs and two possible outputs. Why aren't there only 16 formulas? (Yes, I know the answer, and I will get to it.)

The Unreasoner

12-21-2015, 02:32 PM

Okay. Slight detour from the original topic in light of some new info.

Graph Isomorphism appears to now have a far superior algorithm. It also relates to our current topic. So I'll get to unpacking some of this.

Now on YouTube: HD video of first talk

Dates: Tue Nov 10, Thu Nov 12, Tue Nov 24, Tue Dec 1

University of Chicago

Combinatorics and Theoretical Computer Science seminar

Date: Tuesday, December 1, 2015

Time: 3:00pm

Place: Ryerson 251

Speaker: László Babai (University of Chicago)

Title: Graph Isomorphism in Quasipolynomial Time III: The "Split-or-Johnson routine"

Abstract:

In this third talk of the series we present the details of the "Split-or-Johnson" routine, the second canonical partitioning algorithm required for the master algorithm. In the previous talk, the "Design Lemma," the first partitioning algorithm (reduction from k-ary to canonical binary relation) was presented. The talk can be understood without knowledge of the master algorithm outlined in the first talk. The Weisfeiler-Leman canonical refinement process discussed in the second talk will be briefly reviewed and extensively used. Basic familiarity with discrete structures such as undirected and directed graphs, bipartite graphs, hypergraphs will be assumed. For most of the talk, no group theory beyond the concept of the symmetric group will be required.

University of Chicago

Combinatorics and Theoretical Computer Science seminar

Date: Tuesday, November 24, 2015

Time: 3:00pm

Place: Ryerson 251

Speaker: László Babai (University of Chicago)

Title: Graph Isomorphism in Quasipolynomial Time II: The Design Lemma

Original subtitle: "The `Split-or-Johnson routine'"

Abstract (updated after the talk to reflect actual content and the decision to add a third part to the series):

In this second talk of the series we present the proof of the "Design Lemma," a canonical partitioning algorithm required for the master algorithm. The input to the algorithm is a k-ary relational structure with non-negligible symmetry defect; the output is either a good canonical partition or a large, canonically embedded uniprimitive coherent configuration. The the key tools are the classical and the k-dimensional Weisfeier-Leman refinement process.

This talk together with the third part of this series (Dec 1) forms a stand-alone module and can be understood without knowledge of the master algorithm outlined in the first talk. Basic familiarity with discrete structures such as undirected and directed graphs, bipartite graphs, hypergraphs will be assumed. No group theory beyond the concept of the symmetric group is required.

The University of Chicago

Group Theory seminar

Date: Thursday, November 12, 2015

Time: 4:30 pm

Place: Ryerson 251

Speaker: László Babai (University of Chicago)

Title:

A little group theory goes a long way: the group theory behind recent progress on the Graph Isomorphism problem

Abstract:

The Graph Isomorphism (GI) problem asks, given two graphs with n vertices, decide whether or not they are isomorphic. This is a classical algorithmic problem that has received considerable attention in the theory of computing because of its unsettled complexity status within the P/NP theory.

The asymptotic theory of finite permutation groups plays a central role in the study of this algorithmic problem, thanks especially to Luks's seminal 1981 paper. In 1983, Luks proved that the GI problem can be solved in exp(nlogn−−−−−√) steps. This remained the state of the art for over three decades.

The speaker's recent result brings this upper bound down to quasipolynomial, i.e., exp((logn)c). In the talk I will outline some elementary (modulo Schreier's hypothesis) group theoretic results that are at the heart of this development, and try to give some indication of their relevance to the algorithmic problem.

The University of Chicago

Combinatorics and Theoretical Computer Science seminar

Date: Tuesday, November 10, 2015

Time: 3:00pm

Place: Kent 120

Speaker: László Babai (The University of Chicago)

Title: Graph Isomorphism in Quasipolynomial Time I: The "Local Certificates algorithm"

Abstract:

In a series of two talks we outline an algorithm that solves the Graph Isomorphism (GI) problem and the related problems of String Isomorphism (SI) and Coset Intersection (CI) in quasipolynomial (exp(polylogn)) time.

The best previous bound for GI was exp(nlogn−−−−−√), where n is the number of vertices (Luks, 1983). For SI and CI the best previous bound was similar, exp(n√(logn)c), where n is the size of the permutation domain (the speaker, 1983).

In this first talk we give an overview of the algorithm and present the core group-theoretic divide-and-conquer routine, the "Local Certificates algorithm." Familiarity with undergraduate-level group theory will be assumeda

The Unreasoner

12-22-2015, 12:52 PM

So while I think I get the isomorphism algorithm, I haven't been able to verify its speed. And I'm not sure I'd be able to answer any of your questions on it (yet). He keeps bringing it back to group theory, but I think doing it with sets of functions and sets of matrices might be more intuitive (and closer to what we've been doing.

Meanwhile, something fun for the holidays.

The Kakeya Needle problem (because they look like pine needles obviously, and because of the difficulty in turning a cut Christmas tree in a dense forest):

http://m.youtube.com/watch?v=j-dce6QmVAQ

The Unreasoner

12-31-2015, 03:55 AM

So is anyone reading this thread? I'm not being needy, but if I'm just pissing people off by regularly bumping it, I'll stop. Or someone else can take over with their own areas of expertise.

Or does anyone have any questions? Without trying to be arrogant, I really am a reasonably talented and competent mathematician, and (imo) a passable teacher. I'm happy to answer questions.

Or maybe move on to another topic? To quickly wrap up Boolean stuff: the true number of unique formulas is substantially less than 70 million, because that figure contains things like (a OR b) AND (a OR NOT b) AND a. Circuits are another way of considering SAT, either real physical circuits or mathematical abstractions. It's a common way to measure a function's complexity. Either the total number of circuits needed is counted (sometimes you even restrict everything to two input NAND gates (¬a v ¬b), from which you can make AND, OR, NOT, XOR, IFF...) or you look at the depth in series circuits (levels of nesting, indicating the time complexity) and width in parallel circuits (space complexity). CNF for instance is massively parallel. So is DNF, for that matter. Some people working in optical/photonic computing believe it may be possible to use these basic facts to solve NP complete problems in linear time (but exponential space). If anyone is a fan of redstone in minecraft (or likes to build simple chips to run DIY projects), studying Circuit SAT is a profitable activity.

The current state of the proof of P vs NP is essentially:

If they are equal, the field is mostly open, except SAT cannot be solved in linear time. Some problems (like the Traveling Salesman) have powerful heuristics, while others (like Clique) cannot have powerful heuristics, or even mediocre ones.

Proving that they are not equal, while it is generally assumed to be true, seems to be far off. We now know that any such proof requires mathematics not yet conceived (iow, we know no tools currently available are up to the task). The last attempt to be given a serious chance was based on circuit complexity. Specifically, someone showed that any algorithm for the Clique problem cannot have a small circuit if it only uses AND and OR gates. This got a lot of traction, and people assumed (or maybe 'hoped' is more accurate) that it was only a matter of time before he extended it to NOT gates as well (because, unlike the NAND gate, AND and OR by themselves are not functionally complete: there are functions you cannot construct with just the two). This would have (more or less) sealed the deal.

So, what now? Another specific problem (Traveling Salesman, Clique)? Another topic altogether? I couldn't cover any topic in mathematics, but I think you (and I) would be surprised to see how many I actually can.

Here's a few that I'd be happy to go into, are example heavy, and be able to answer (almost) any questions you might have:

The Four Color Theorem (with basic topology and graph theory)

The Goldbach Conjecture/Riemann Hypothesis (to teach number theory)

Various topics in statistics/probability/combinatorics (illustrated with gambling games and tge bond markets and stock option pricing)

Classical cryptography (with Kryptos as the example)

Modern cryptography (with Bitcoin and some of the Snowden revelations (Dual EC) for the examples)

Game theory (with the 2012 GOP primary as the example)

Lattices and Fourier analysis (not really related to each other, but both related to the example that is Terezian analysis)

Algorithms and heuristics (we'll use some finite automata as examples)

There are probably more I could pit together a good syllabus for on short notice, so feel free to ask.

Or one of you take over the thread. That was the original plan. I like to learn too. Or maybe a joint effort course?

Frenzy

12-31-2015, 10:44 AM

i'm reading this, but it's all going over my head. i haven't taken a pure math class since high school Calculus in the early 90s, and the closest i come to pure math these days is helping my son with his Geometry homework.

The Unreasoner

12-31-2015, 11:15 AM

If you have any questions, feel free to ask. I also take requests, and could probably find specific examples (charts, youtube videos, articles) to clarify things left unclear. I probably could have picked a better topic to start with, but this is close to what I do for a living anyway, and I've been enjoying myself. Part of me always wanted tobe a math professor. But I realized I might be the only one having fun, and thought I'd test the water.

That being said, the original plan was to have several posters cover topics of their choice. And I think I closed it off nicely this first round, so I think I'll wait a few days, answer any final questions, and then let someone else have the floor (if they want it). If they don't, but still want the thread updated, I'll cover the Four Color theorem next, barring a request for something else. Much less jargon or technical complexity, far more intuitive. It may not even seem recognizable as mathematics.

The Unreasoner

01-07-2016, 03:46 PM

So it looks like EM drives aren't a real thing, despite what recent articles led me to believe. Apparently the measured thrust is within the margin of measurement errors.

That being said, unless some has questions about complexity theory or Boolean algebra, or wants to take over the thread with their own area of expertise, I will start the next topic tomorrow (Four color, unless someone wants to request something else).

GonzoTheGreat

01-08-2016, 03:50 AM

So it looks like EM drives aren't a real thing, despite what recent articles led me to believe. Apparently the measured thrust is within the margin of measurement errors.

Well, duh!

From what I've read of it*, their propulsion was to be generated by assuming that Newton's "action is minus reaction" didn't apply if you simply left it out of your calculations. I'm not particularly surprised that this works better on paper than in reality.

* Note: I base this assessment on a description and the schematics of the EM drive that I've seen on the Net. If that description was incomplete, then it may be that there is another problem with the concept, rather than the one that was apparent to me from the sales presentation.

The Unreasoner

01-08-2016, 10:53 AM

I'm not a physicist. And the first I heard about this thing was that it got positive test results from NASA. I vaguely thought it might work like those tinfoil glider things.

I didn't know until I looked into it that it's almost certainly a hoax. Why NASA even bothered to test it, and why several normally reliable news sources failed to convey the information that this thing is about as credible as that cold fusion thing by that Italian guy, I do not know.

GonzoTheGreat

01-08-2016, 12:26 PM

Correction to what I wrote in my previous post: I think now that I saw the description of the EM drive in a New Scientist. Which is more trustworthy than a random web page, but not on a par with a peer reviewed article.

The Unreasoner

01-12-2016, 03:49 PM

The Four Color Theorem was proposed as conjecture by a math student (I forget his name) around 100 years ago. The student noticed that any map he had to color never needed more than four colors to ensure that each region was colored differently than all regions it shared a border with. He brought this to his instructor, who was intrigued, but neither of them could prove it. It was one of the most falsely proved theories for quite some time (both for and against).

The actual proof is far too complicated to go into here (at least in its current form), mainly because it was one of the first proofs to be largely aided by a computer. Mathematicians broke down the infinite set of possible maps into thousands of 'families', then had a computer check each one. The proof is valid because the reasoning was sound, and the algorithm was correct (we won't go into metaproofs).

The theorem has almost no practical use. While it is true that any planar region broken up into contiguous pieces (like a puzzle) can be colored with four colors, it is possible that those valid colorings can be slightly hard to find. Also, real world maps have three issues (or rather, potential issues) that may render the theorem inapplicable. But first, some examples:

Four coloring of the continental US:http://web.stonehill.edu/compsci/lc/four-color/usa.gif

A false counterexample:

http://tdboui.com/blog/wp-content/uploads/2012/06/morons-on-math.jpg

Another false counterexample, and the corrected version:

https://upload.wikimedia.org/wikipedia/commons/thumb/a/a6/4CT_Non-Counterexample_1.svg/2000px-4CT_Non-Counterexample_1.svg.png

http://upload.wikimedia.org/wikipedia/commons/7/7a/4CT_Non-Counterexample_2.svg

The Unreasoner

01-12-2016, 04:27 PM

Sorry. Didn't realize that one was so damn big. And I screwed up the spacing on the US one, but for some reason the edit button is missing, so we'll all have to deal.

Anyway, in order to understand the topics at hand, two ideas worth keeping in mind:

What matters is shared borders, not common corners. Also, the actual shape of the map is mostly irrelevant. The only relevant fact is shared borders. This means we can smoothly deform (or really, continuously deform, the derivative need not be continuous. If two figures, surfaces, or n-dimensional shapes can be described by a continuous function, and the difference between the functions is continuous, they are topologically more or less the same. Homotopic is the technical word. This has to do with the Poincare Conjecture as well, proved by Perelman. It says any finite n-dimensional shape wothout holes is homotopic to an n-dimensional sphere). So consider this:

https://www.mathsisfun.com/activity/images/coloring-9.gif

Sometimes it helps to work with more rigorous tools, for that we change the map to a graph (mathematical graph, not the plot/picture of a function):

http://world.mathigon.org/resources/Graph_Theory/fourcolour.png

Those who were around for part one may think that this is familiar. Specifically, deciding whether or not a particular graph is colorable in three colors is NP complete. One important point is that the graphs for 3-colorability need not be planar (so lines/edges can intersect) while for the four-color theorem, they must be. This can be seen intuitively if you try to draw a map corresponding to a graph with such an intersection: at that point, you would need to have two countries in the same place. Consider the Four Corners in the US: if Colorado and Arizona shared a small border at the point of intersection, Utah and New Mexico could not.

Graphs are useful in understanding the deformations, however, because you can feel free to move everything around as you wish, so long as you do not cross any lines.

The other point is that any region must be considered as an independent node. This means one region cannot be broken up into multiple parts and keep the same color (or at least this cannot be guaranteed. In general, the maximum number of colors needed is four plus the number of regions with distinct parts, though in practice you rarely need so many).

So the three possible issues with four-coloring maps in practice are:

Disjointed regions, like the US with Alaska, or Michigan's separated piece.

Bodies of water, which tend to require the same color. It can be seen, however, that this requirement is exactly the same as the previous one: the bodies of water can be treated as disjointed regions of a single entity.

Two examples:

http://world.mathigon.org/resources/Graph_Theory/colouring.png

As you can see, Michigan is colored differently here (though there is a valid way to color the US with four colors that does not have this problem, see above).

And here, there appear to be quite a few countries underwater:

https://upload.wikimedia.org/wikipedia/commons/5/5d/Five_colors_world_map_%28Malawi_and_African_Great_ Lakes_problem%29.svg

The third issue is the fact that real maps may not be manifolds (projections) of a flat surface.

Now a question: is a map on a globe always four-colorable (ignoring the disjointed region issue)?

Also, a somewhat tricky example you can try yourselves:

http://www.vb-helper.com/FourColorMap1.gif

One Answer (http://www.alternatievewiskunde.nl/fcp/FourColorMap1.jpg)

The Unreasoner

01-12-2016, 04:36 PM

It looks like that world map has a few issues (Japan, for instance), but it gets the point across.

In any case, for the sphere question, the answer is yes. This can be seen if you imagine the Earth stretched and skewed and molded in such a way that a single country (probably easiest to visualize with Antarctica, but possible for even Vatican City) takes up an entire hemisphere, than flatten it into a disc. You will have two maps: one of just Antarctica (or whatever), and another with everything else surrounded by a border of the one country. These maps are both flat, and deforming the globe (or, by the Poincare conjecture, any finite 3d shape without any holes in it) was continuous (so it didn't add or remove any borders).

Now what about maps on a donut?

Also: would anyone like me to walk them through the invalid proofs? Even if invalid, they can help us gain insight into the problem, especially if we look at why. They are also very intuitive.

The Unreasoner

01-12-2016, 05:57 PM

I think I'll give a rough sketch of the proofs anyway.

One that was reasonably well-regarded considered the colorability of a map with only triangular regions. Any polygon can be broken up into triangles, and any region on the map can be continuously deformed into some kind of polygon (not necessarily regular). This can be done in such a way that each triangle's edge is only shared by one other triangle. Obviously, such a map could be four-colored. The (flawed) proof claimed that the triangular map was equally colorable as the arbitrary map it was derived from. While this is technically true, it isn't actually a case of implication (or at least whoever it was couldn't show it was actually an implication). Both statements are true, but the (at least known) direction of the implication went the opposite way.

The other proof is a bit stronger (and I actually personally worked on a version of it, and independently came up with a very strong version that's inductive) but I'll need to draw some diagrams for it, so stay tuned.

Also, any questions? Comments? Some feedback would be appreciated. (Constructive) Criticisms on style are particularly helpful.

The Unreasoner

01-12-2016, 08:25 PM

So for the other proof, consider this diagram:

http://theoryland.com/vbulletin/picture.php?albumid=7&pictureid=133

In a sense, adjacency is transitive. The entire pseudoproof is way more complicated than I remember, and relies on algorithms and finite groups and modularity (I made the thing to demonstrate an algorithm), but here's a rough description (again, not a real proof, not valid, but it gets the idea across, more or less):

Any region or node that is forced to become a fifth color necessarily touches four regions that are forced to be different colors. Since any forcing is always relative, there will always be an affine mapping (possibly more than one) from one coloring on one 'sub map' to another, after adding some adjacency from one to the other. But without that adjacency, any permutation of the color set can be applied to both, simultaneously. And the amount of adjacency needed between two sub-maps is at minimum three shared borders (or edges). Now the 'unproven' part:

Since adjacency is in some way transitive, and since a unique affine transformation is only obtained when there is a sufficient number of shared borders, you can treat any sub-map as a single region. Therefore, the simplest case for a (necessarily) five-colored map is demonstrated above, if one exists. Specifically, it would need to have a region (or node) at one of the pink diamonds, and it must share a border with (at least) four other regions/nodes. That means lines drawn from one to four circles, with no intersections. But, every diamond can only reach three. Therefore, the four-color theorem is true.

Again, the proof is an invalid one. I modified an old one, (and made it substantially stronger) to show some people the potential strength of metaproofs. Basically I showed that if you had an algorithm that colors as the map is converted to a graph, step by step, there would only ever be three regions on the outermost border of the map (if you treat regions of the same color as a single region). The main problem with my version of the proof is that it required an abstract version of the map where two regions could exist in the same space, and I never bothered to show (nor did I really see how to show) that this didn't fundamentally change the problem.

GonzoTheGreat

01-13-2016, 03:34 AM

What matters is shared borders, not common corners. Also, the actual shape of the map is mostly irrelevant. The only relevant fact is shared borders. This means we can smoothly deform (or really, continuously deform, the derivative need not be continuous.

https://www.mathsisfun.com/activity/images/coloring-9.gif

Note: some changes in weather patterns may result if you do this in reality. Still, it would be fascinating to hear the politicians try to come to grips with it, if they one day wake up to such a new situation.

The Unreasoner

01-13-2016, 11:01 PM

First off, I'm not a programmer. I can write a bit in C and Fortran (and java, but I don't think that will work here), but for the most part, when I need some specialized process done (iow, that can't be done by excell (side note, you can build a SAT circuit in excell. Or even something more conplicated using operators other than and/or/not. It's actually kind of fun, very visual) or the modelling software I work with), I just map it out and someone else builds the tool.

But I'm on my own here. I need to build some kind of tool to process a certain type of large file (like maybe around 1 gigabyte). The specific operations it would need to do I can code. But do any of you know how to bundle it together? Excell can't handle the data. The modelling software can, but it's slow and inefficient. Is there a way to build a simple program (with some kind of command line interface) that processes a few million entries then pauses, displays stats, and resumes after getting revised intructions? How would the data be stored? I was thinking a basic text file with each entry in a specific line, that was treated like a tape in a Turing macine might work, only keeping what it needs atm in memory. But is that the best way, in terms of efficiency? I have 8 gigs of RAM available atm, does this mean I should/can store the entire file in memory? While I can translate algorithms to code, I honestly have no idea how to turn a handful of algorithms or routines into a cohesive program. Certainly not efficiently. Can I overclock my cpu within the program? Will overclocking normally let those extra cycles be used by a program I made without programming in the possibility?

jarno87

01-14-2016, 02:28 AM

I do quite some programming, mostly scientific programs at work. And usually they involve a lot of data, easily several to tens of GB. So some tips.

First off, I'm not a programmer. I can write a bit in C and Fortran (and java, but I don't think that will work here), but for the most part, when I need some specialized process done (iow, that can't be done by excell (side note, you can build a SAT circuit in excell. Or even something more conplicated using operators other than and/or/not. It's actually kind of fun, very visual) or the modelling software I work with), I just map it out and someone else builds the tool.

But I'm on my own here. I need to build some kind of tool to process a certain type of large file (like maybe around 1 gigabyte). The specific operations it would need to do I can code. But do any of you know how to bundle it together? Excell can't handle the data. The modelling software can, but it's slow and inefficient. Is there a way to build a simple program (with some kind of command line interface) that processes a few million entries then pauses, displays stats, and resumes after getting revised intructions?

Most of the tools we write at my work are command line programs, as they are the easiest. One of the arguments would be the filename of the file containing all your data. My usual approach would be to break things down in three parts:

- First read in/preprocess your data, and store it in memory if it fits, just in a large array (in C/C++, use fopen, fread,fclose and new[] for creating the array)

- 2nd step do your calculation

- Print your statistics to the standard output using printf(). Or write them to some text file.

If the processing is such that you would only need to read each part of your data once, you can integrate the firs and second step, and process the data in small blocks, just keeping aggregated results.

I wouldn't recommend programming pausing and allowing renewed instructions. This probably requires parsing command by hand and that is a big hastle. Just exit after showing results. If needed the user can just rerun the program/command with adjusted parameters

How would the data be stored? I was thinking a basic text file with each entry in a specific line, that was treated like a tape in a Turing macine might work, only keeping what it needs atm in memory. But is that the best way, in terms of efficiency?

Text representations of data take usually more space than the raw once. A normal integer is 32 bits, i.e. 4 bytes but written out might take many characters. Furthermore, this requires converting between numbers and strings and v.v. all the time which is very inefficient. If needed I would recommend storing information in some binary format. (Using fwrite/fread)

I have 8 gigs of RAM available atm, does this mean I should/can store the entire file in memory?

Preferably yes, make sure to compile your program for 64 bits. On linux that would be the default, on windows check your compiler settings. Nowadays it is very easy to use large arrays, if your computer has sufficient RAM. Makes the programming much easier. I have processes using 2 to 10 GB all the time, with exceptions going as high as 100GB on a 128 GB RAM machine.

While I can translate algorithms to code, I honestly have no idea how to turn a handful of algorithms or routines into a cohesive program. Certainly not efficiently. Can I overclock my cpu within the program? Will overclocking normally let those extra cycles be used by a program I made without programming in the possibility?

Overclocking is not something you do in a program. Usually this is done in the BIOS/UEFI settings. It then starts your computer with the processor always a certain % faster. So yes any program would use the extra cycles, they don't have to/ cannot be aware of this. A normal user program is not able to overclock your processor. However, overclocking comes with risks to your processor and system stability. If you don't know what you are doing ,stay away from it. In most cases it would not give you more than 10 to say 30 % extra anyway, so just let it run a bit longer.

Hope this helps.

GonzoTheGreat

01-14-2016, 03:30 AM

Most of the tools we write at my work are command line programs, as they are the easiest. One of the arguments would be the filename of the file containing all your data. My usual approach would be to break things down in three parts:

- First read in/preprocess your data, and store it in memory if it fits, just in a large array (in C/C++, use fopen, fread,fclose and new[] for creating the array)

- 2nd step do your calculation

- Print your statistics to the standard output using printf(). Or write them to some text file.

If the processing is such that you would only need to read each part of your data once, you can integrate the firs and second step, and process the data in small blocks, just keeping aggregated results.

I could be wrong, but I have the impression that it isn't really known in advance what precise settings would be needed, but that this should become clearer when the program runs with somewhat flawed settings. So then restarting with an updated settings file would make sense.

In such a case, it may be more efficient to store the actual data (not the settings) in some "raw" format, basically as a memory dump.

If true, then you should first work out how you want to store your bulk data in your program, then write a program to read it in from whatever format it is now (probably a text file of some kind) into an array of the type you would you in your actual program, and dump this into a file. That would hopefully be smaller than a full text file, since text isn't really an efficient storage format (though it is a very versatile one). However, this depends on what kind of data you're dealing with. If it is lots of numbers, then this procedure would be useful. If it is a large bulk of Swahili text, then keeping it as plain text would probably be best.

Next, you would almost certainly want to have at least two separate files: one with the data, and one with the settings that you can change to do whatever fiddling you can't leave to the program. I think that a third file, with the results, would be advisable as well.

So, you would have a program with (at least) three command line options: the data file, the settings file and the output file.

Depending on the type of task, I would use either C (probably C++ light, for this) or Fortran, with a greater likelihood of me using C++ nowadays. But if what you have to do is mostly lots of calculation, then Fortran might be the best option.

The Unreasoner

01-14-2016, 03:21 PM

Thanks you guys. Working on the preprocessor now. The data is technically a directed graph, where each node has an associated 2d table and string (other than its number/name). Right now it's in a csv text file for the nodes and edges, and another csv for the associated data. And I'll have to find out how that one was written, because most of the tables are 64x64, while the rest are 4x4. The entries are mostly functions and variables, though some variables are integers, and some are strings. And the 4x4 are floating point. Is an array of arrays efficient? Or should I call on several distinct arrays (assuming I go with Gonzo's suggestion on the memory dump, and jarno's on compiling for 64 bits)?

ETA: I repped you both, but am on my phone so I couldn't comment. But thanks for your help.

GonzoTheGreat

01-15-2016, 03:37 AM

I think that an array of arrays would be slightly more efficient, but I doubt the difference would be noticeable unless you paid really close attention. Speed improvements would depend far more on choosing the right overall algorithm, and on figuring out which things can be most efficiently done once instead of a hundred million* times. In general, computers are very good at array manipulation, because it is such a common problem. All processors that I know the machine code of have been optimised for this to a great extend. On the other hand, using a number of arrays would at most add a small bit of overhead, if you set it up correctly. So even if you don't use this optimised approach here, the difference wouldn't be anywhere near as big as the gains you could get from doing it in a way that is easy for you. Unless you really botch your algorithm, in which case you're screwed even with full optimisation anyway.

So this shouldn't be much of a worry. Of course, I would think about it anyway; but I like excessive optimisation.

* Suppose you have to multiply three numbers from different data sets; let's designate those sets A, B and C, with individual numbers from each a, b and c. Then you would have to get the result a*b*c for each combination. Now, if set A has a hundred million members, and sets B and C have four members each, then it would be a lot more efficient to calculate d=b*c outside the inner loop, and multiply that with each individual a only. But how to achieve such improvements I'll leave up to you; I don't know your actual problem after all.

The Unreasoner

01-20-2016, 04:50 PM

So I'm kinda busy right now, but quickly on map coloring:

Coloring a torus requires 7 colors. I'll look for an example. But this can be seen intuitively (if you take FCT as given) by embedding the torus in the plane and forcing one region in the intersection of the boundaries. You can try cyllinders and mobius bands yourselves if you want further practice. There is no 3d analogue with shared faces. You can think about jenga towers and their extensions to see why.

And just so I don't forget, I want to do Terezian analysis next:

https://upload.wikimedia.org/wikipedia/commons/thumb/2/27/Lattice-reduction.svg/450px-Lattice-reduction.svg.png

The Unreasoner

01-22-2016, 09:52 AM

For those that haven't seen Pixels:

9944

Σ f(x)=f(25872248)

x=1

What is f(x)?

GonzoTheGreat

01-23-2016, 03:27 AM

f(x)=2601.794851166532582461786001609

Admittedly, that's a bit trite, but it does work. :p

Nazbaque

01-23-2016, 03:40 AM

f(x)=2601.794851166532582461786001609

Admittedly, that's a bit trite, but it does work. :p

No it doesn't. f(x)=0 would be the trite answer.

The Unreasoner

01-23-2016, 03:40 AM

Lol. I didn't have that one. There are infinite answers though.

It's the question I submitted for the employee screening test.

The most common correct answer is f(x)=0. The one I haven't seen yet is f(x)=xth prime. I'll have to take a look at yours when I'm less tired.

The Unreasoner

01-24-2016, 01:33 PM

f(x)=2601.794851166532582461786001609

Admittedly, that's a bit trite, but it does work. :p

lol. I think you missed the fact that the right side is also within the function. The simplest linear answer (other than f(x)=0) is f(x)=x-(23574292/9943).

ETA:

I love the question because they know it has infinite answers, while some of them are more interesting than others. It almost functions like a personality test. Someone who gives f(x)=0 is a smartass. Someone who uses Gonzo's answer plus the floor function (to make it work. And I have seen this one) seems to be adaptable. Someone who gives the general form of the linear solutions is meticulous. Someone who does the same for the geometric series is a show-off.

Nazbaque

01-24-2016, 02:46 PM

f(x)=sin(x*pi)? Smart ass squared?

The Unreasoner

01-24-2016, 06:25 PM

f(x)=sin(x*pi)? Smart ass squared?

f(x)=(x+1)! mod 2

Smartass factorial.

Nazbaque

01-25-2016, 04:45 AM

f(x)=(x+1)! mod 2

Smartass factorial.

f(x)=cos((x+1)*pi/2)?

The Unreasoner

01-25-2016, 10:31 AM

f(x)=cos((x+1)*pi/2)?

That's pretty good. But you can do better. This repeats every four terms. But 8 is the largest common factor. I'm looking at a model for elliptic curves over finite fields right now to take advantage of the fact that the right side is 1 plus a prime.

GonzoTheGreat

01-25-2016, 11:01 AM

f(x) = tan(x!)

Proving that one wrong wouldn't be a trivial matter, I think. Possible, I suspect, but it would require very high precision calculations.

I admit that in my previous answer, I'd overlooked that the function appeared on both sides.

The Unreasoner

01-25-2016, 11:38 AM

Did you mean to put a pi in there?

And either way: do we just need to prove it wrong, or do we need to actually evaluate either side?

Nazbaque

01-25-2016, 11:48 AM

That's pretty good. But you can do better. This repeats every four terms. But 8 is the largest common factor.

But that would have been showing off. With pi/2 you just get 0+1+0-1... which eventually cancels out unless the number of terms is 4n+2 (4n cancels out by default 4n+1 and 4n+3 cancel out if you start at the right spot). With pi/4 you get those messy ∓1/√2 terms. It's crossing the line to gaudy when a bit more modesty would have had elegance.

Now I'm wondering if you could use f(x)=(1-q)q^(x-1) so that the sum becomes 1-q+q-q^2+q^2....-q^9944 which becomes 1-q^9944 and is in itself relatively neat but then the final equation becomes 1-q^9944=q^25872247-q^25872248 and solving the q in that is something of a pain.

The Unreasoner

01-25-2016, 12:19 PM

But that would have been showing off.

What do you think we're doing here?

Watch:

f(x) = tan(x!)

Proving that one wrong wouldn't be a trivial matter, I think. Possible, I suspect, but it would require very high precision calculations.

(infinity+)

Sigma ((z^25872248)*(e^(1/z))-((z*e^(1/z))*(1-z^9944)/(1-z)))dz

(0)

is not equal to zero.

ETA:

hmm. I forgot about the tan. I'll get back to you.

Nazbaque

01-25-2016, 01:00 PM

What do you think we're doing here?

Showing off, but the goal is not to let others realize it. The goal is to be elegant. If you over do it, you're just gaudy.

The Unreasoner

01-25-2016, 06:44 PM

f(x) = tan(x!)

Proving that one wrong wouldn't be a trivial matter, I think. Possible, I suspect, but it would require very high precision calculations.

Do you know the answer? I can show the infinite series that describes the difference converges to a nonzero value...'probably'. And that took a significant proportion of computer power away from the model builder (turns out after preprocessing and computing common values ahead of time it blew up to 11 gigs).

But doing it by hand requires some identities no one has bothered to work out (and are quite probably analytic).

The Unreasoner

01-25-2016, 07:34 PM

Oh and if anyone is still interested in the Chess CNF, I got it with some circuit script from work:

(AvCv!Ev!G)^(AvDv!Ev!H)^(AvGv!Cv!E)^(AvHv!Dv!E)^(B vCv!Fv!G)^(BvDv!Fv!H)^(BvGv!Cv!F)^(BvHv!Dv!F)^(CvE v!Av!G)^(CvFv!Bv!G)^(DvEv!Av!H)^(DvFv!Bv!H)^(EvGv! Av!C)^(EvHv!Av!D)^(FvGv!Bv!C)^(FvHv!Bv!D)^(AvBvCvD vEvFvGvH)^(AvBvCvEvFvGv!Dv!H)^(AvBvDvEvFvHv!Cv!G)^ (AvBvEvFv!Cv!Dv!Gv!H)^(AvCvDvEvGvHv!Bv!F)^(AvCvEvG v!Bv!Dv!Fv!H)^(AvDvEvHv!Bv!Cv!Fv!G)^(AvEv!Bv!Cv!Dv !Fv!Gv!H)^(BvCvDvFvGvHv!Av!E)^(BvCvFvGv!Av!Dv!Ev!H )^(BvDvFvHv!Av!Cv!Ev!G)^(BvFv!Av!Cv!Dv!Ev!Gv!H)^(C vDvGvHv!Av!Bv!Ev!F)^(CvGv!Av!Bv!Dv!Ev!Fv!H)^(DvHv! Av!Bv!Cv!Ev!Fv!G)^(!Av!Bv!Cv!Dv!Ev!Fv!Gv!H)

ETA:

if any of you have miniSAT (free):

c

c THATFUCKINGCHESSTHING

c

c

c DIMACS

c

c

p cnf 8 32

1 3 -5 -7 0

1 4 -5 -8 0

1 7 -3 -5 0

1 8 -4 -5 0

2 3 -6 -7 0

2 4 -6 -8 0

2 7 -3 -6 0

2 8 -4 -6 0

3 5 -1 -7 0

3 6 -2 -7 0

4 5 -1 -8 0

4 6 -2 -8 0

5 7 -1 -3 0

5 8 -1 -4 0

6 7 -2 -3 0

6 8 -2 -4 0

1 2 3 4 5 6 7 8 0

1 2 3 5 6 7 -4 -8 0

1 2 4 5 6 8 -3 -7 0

1 2 5 6 -3 -4 -7 -8 0

1 3 4 5 7 8 -2 -6 0

1 3 5 7 -2 -4 -6 -8 0

1 4 5 8 -2 -3 -6 -7 0

1 5 -2 -3 -4 -6 -7 -8 0

2 3 4 6 7 8 -1 -5 0

2 3 6 7 -1 -4 -5 -8 0

2 4 6 8 -1 -3 -5 -7 0

2 6 -1 -3 -4 -5 -7 -8 0

3 4 7 8 -1 -2 -5 -6 0

3 7 -1 -2 -4 -5 -6 -8 0

4 8 -1 -2 -3 -5 -6 -7 0

-1 -2 -3 -4 -5 -6 -7 -8 0

ETA2:

Does anyone remember where RJ talked about using Boolean logic to work out the cultures in WoT? Terez's thread (and this one) made me remember it.

GonzoTheGreat

01-26-2016, 02:57 AM

Do you know the answer? I can show the infinite series that describes the difference converges to a nonzero value...'probably'. And that took a significant proportion of computer power away from the model builder (turns out after preprocessing and computing common values ahead of time it blew up to 11 gigs).

But doing it by hand requires some identities no one has bothered to work out (and are quite probably analytic).

I would say that it is almost certainly false. Both sides of the equation are as near to random values as makes no difference, but if you want to actually be certain (rather than guessing, which isn't how mathematics should work) then you have a problem. You would need to calculate extremely big factorials, then figure out precisely how often Pi fits into those results, calculate the resulting tangents and finally start adding (or subtracting) them all. The tangent thing and the additions are trivial (apart from also having an infinite number of decimals involved), but the "modulo Pi" bit is where things get really non-trivial. It is a trick I learned from a book by Richard Feynman, who himself learned it from a fellow who was even cleverer than he was.

The Unreasoner

01-27-2016, 05:11 PM

I would say that it is almost certainly false.

Yeah. On my second pass I thought 'that can't possibly be true'. But as you note, in mathematics, that sentence only works if you mean it literally.

It is a trick I learned from a book by Richard Feynman, who himself learned it from a fellow who was even cleverer than he was.

It looks like some stuff I've seen from von Neumann. Probably the greatest mathematician since Euler, along with Godel.

The Unreasoner

01-27-2016, 08:48 PM

On the model-builder:

I'm using a quad-core processor with two threads per core. But the program never seems to use more than one thread (even when nothing else is running, it just jumps between 12 and 13 percent of processor time). Did I fuck up compiling? Or does each thread need to be called on explicitly?

jarno87

01-28-2016, 01:43 AM

On the model-builder:

I'm using a quad-core processor with two threads per core. But the program never seems to use more than one thread (even when nothing else is running, it just jumps between 12 and 13 percent of processor time). Did I fuck up compiling? Or does each thread need to be called on explicitly?

In general, programs don't use more than 1 core automatically. It has to be programmed in specifically. One of the reasons is that the code must be Thread-safe (google it), and so satisfy some requirements (to avoid race-conditions, for one). There exists frameworks to help you do this for big computations, like MPI, but they have their own requirements.

So the short answer: yes you have to create each thread explicitly. Depending on the program this doesn't have to be a lot of work. I have made an existing program multi-threaded in an hour, once. However, it was quite suitable and well written when I started.

The Unreasoner

01-28-2016, 01:37 PM

(google it)

lol.

So the short answer: yes you have to create each thread explicitly. Depending on the program this doesn't have to be a lot of work. I have made an existing program multi-threaded in an hour, once. However, it was quite suitable and well written when I started.

Thanks. When I finished the preprocessing, it was made so that nothing was used twice until the end. So I just split the data into different smaller arrays and ran multiple instances. Still only hovering around 50%, but it's a significant improvement. I'll have to look and see if I can make the last step 'thread safe'.

The Unreasoner

01-28-2016, 01:45 PM

but the "modulo Pi" bit is where things get really non-trivial.

I meant to say this earlier:

Computationally difficult, but not the hardest mathematically. The factorial is a pain, but you can work with it. And over any finite set, you can create a lattice basis that includes both integer multiples of Pi and the multiples of the integers. When I was working on it the hardest part seemed to be getting the product of the sin values. And then you have to hope the function never zeros out over the positive real numbers, because then you need to take a cosine integral (which, as we've stated, is NP-Complete).

The Unreasoner

01-28-2016, 02:43 PM

ETA2:

Does anyone remember where RJ talked about using Boolean logic to work out the cultures in WoT? Terez's thread (and this one) made me remember it.

Took awhile to find, but here:

Therese Littleton

Are there particular historical eras that influence your stories?

Robert Jordan

Well, to give you an example of the way these things work... the Aiel. They have some bits of Japanese in them. Also some bits of the Zulu, the Berbers, the Bedouin, the Northern Cheyenne, the Apache, and some things that I added in myself. They are in no way a copy of any of these cultures, because what I do is say, "If A is true, what else has to be true about this culture? If B is true, what else has to be true?" And so forth.

In this way I begin to construct a logic tree, and I begin to get out of this first set of maybe 10, maybe 30 things that I want to be true about this culture. I begin to get around an image of this culture, out of just this set of things, because these other things have to be true. Then you reach the interesting part, because this thing right here has to be true, because of these things up here. But, this thing right here has to be false, because of those things up there. Now, which way does it go, and why? You've just gotten one of the interesting things about the culture, one of the really interesting little quirks.

To me, that in itself is a fascinating thing—the design of a culture. So that's how the Aiel came about. There are no cultures that are a simple lift of Renaissance Italy or 9th-century Persia or anything else. All of them are constructs.

The Unreasoner

02-02-2016, 01:31 PM

The program is taking a ridiculously long time to run. And apparently web surfing cuts efficiency dramatically. One block completed in 28000 seconds while I was sleeping. Another is running now, and we're up to 123000.

Was the guy Von Neumann?

The Unreasoner

02-14-2016, 01:52 AM

So it takes over an hour to read in the data after every iteration...should I just learn how to program in pauses and prompts? Or would the speedup of not restarting each iteration not be worth it?

GonzoTheGreat

02-14-2016, 03:16 AM

It might be worthwhile to try to figure out why it takes so long to read. Maybe it will help you come up with a more efficient reading mechanism. If you're really lucky, it could even be something that would be useful in some other cases too.

The Unreasoner

02-15-2016, 02:26 PM

The problem may be that it's in a text format (predicting the type and location of the outputs seemed like a pain in the ass). Or that I didn't actually 'compile' anything in the end, I decided instead to use some tool a friend had that let me use either Fortran or Maple's language (which I am most comfortable with). I'll run it by some people who have some specific knowledge of the problem I'm working on. Though he hates being bothered, so it would be nice to have something to show him.

Anyway, on Terezian analysis:

So apparently my lattice idea is, and I quote, 'fucking stupid', mostly for reasons I was aware of. The person I ran it by said while it might be the way to go in theory (though we had a debate on the efficiency of regression techniques I would rely on to sort elements into spines), in practice it is too slow. He did like my original idea (encoding music into an image, then using various techniques for finding edges, eigenfaces, and stegonographic information to build the metrics and operators), but mostly because it draws on some fields that are well-understood and documented. I'll walk you guys through both if you care, or I can give it another pass first. Also, something crawled across my feeds today that may be worth looking into for guidance:

http://blog.leapmotion.com/wp-content/uploads/2016/02/Lyra.gif

GonzoTheGreat

02-16-2016, 02:25 AM

The problem may be that it's in a text format (predicting the type and location of the outputs seemed like a pain in the ass). Or that I didn't actually 'compile' anything in the end, I decided instead to use some tool a friend had that let me use either Fortran or Maple's language (which I am most comfortable with). I'll run it by some people who have some specific knowledge of the problem I'm working on. Though he hates being bothered, so it would be nice to have something to show him.

I could be wrong, but I don't think the problem is one of interpretation instead of compilation here. Reading in text files is a bit of a pain anyway, so for that the disadvantage of using an interpreter is lower than it would be for the actual computations.

What might be the problem (though I don't know your code, so I'm guessing based on ignorance*) could be the way you add each newly read item to memory. If that involves rearranging the whole rest of what was already read, then you've created an N-squared mess right there. And if it also means making memory assignments, then it would be a very inefficient N-squared mess to boot. If this is the case, then the solution would be: first read, then sort in place.

* Which, of course, is the best kind of guess to make. It gives one oodles of plausible deniability, and if one is right then that's amazing.

The Unreasoner

03-04-2016, 03:45 PM

Okay, I think the problem is the fact that I'm treating some very large lists as sets. A few thousand sets with a few thousand elements in each. So, since I'm using the Maple language for those, I'm guessing that Maple doesn't check for duplicate elements very efficiently.

The Unreasoner

03-04-2016, 03:47 PM

As for this:

http://blog.leapmotion.com/wp-content/uploads/2016/02/Lyra.gif

They don't have anything really novel in terms of audio encoding.

The Unreasoner

03-04-2016, 10:43 PM

When I compile it's too slow, and when I use the interpreter it can't read in efficiently. I can't even begin to describe how pissed I am. I'm ready to fucking crucify someone for this. And if it's me, so be it. As long as someone bleeds.

Anyway, on a lighter note, something I've worked on before is reversible computing as SAT. It's closer to what we've done before in this thread, and pretty interesting in its own right. Unless someone wants to take over, or wants to see the Fourier+lattice techniques for music theory (so we'll ignore computability issues), or wants this thread to die.

Nazbaque

03-04-2016, 10:52 PM

How can you crucify yourself? Suppose you manage to nail down one hand, what will you nail the other hand with? The first hand being full at that point. Of bloody nails that is. Not the kind at the ends of your fingers. Man there are a lot of puns in this line of thought.

GonzoTheGreat

03-05-2016, 03:15 AM

How can you crucify yourself? Suppose you manage to nail down one hand, what will you nail the other hand with? The first hand being full at that point. Of bloody nails that is. Not the kind at the ends of your fingers. Man there are a lot of puns in this line of thought.

Design a Rube Goldberg Contraption to fire off the right number of nail guns at the same time, then stand in front of it and start the thing. If necessary, rebuild it so that you can actually reach the starting mechanism from where you're standing waiting to be crucified.

No problem, apart perhaps from some engineering stuff which a proper scientist dismisses as "trivial".

The Unreasoner

03-16-2016, 10:14 PM

As Gonzo notes, designing a crucifixion machine is basically just an engineering problem, and well within the scope of this thread. Though my Catholic background compels me to note that the nails don't actually go through the hand (you can ripthe hand away). They gobetween the major bones in the forearm.

Why can interpreters run in multiple threads without special coding?

Also, I might get to reversible SAT later, but anyone's interested in Terezian analysis...

Geometry of Music

Sunday, February 14, 2016: 10:00 AM-11:30 AM

Marshall Ballroom West (Marriott Wardman Park)

Dmitri Tymoczko, Princeton University, Princeton, NJ

In my talk I am going to build on my earlier work into the non-Euclidean geometry of Western harmony (work which led to two papers published in Science, the book "A Geometry of Music," published by Oxford University Press, and an interactive sculpture in the National Museum of Mathematics in Manhattan). I will show how the mathematical notion of a "vector" corresponds to what musicians refer to as a "voice leading." Once we understand this correspondence, we can use mathematical and computational tools to study the vectors found in musical structure -- leading to new insights about musical style and also particular pieces. I will show how these vectors are crucial to understanding the geometrical structure of musical harmonies, and how that geometrical structure in turn helps us expand our conception of a musical vector. And we will develop new insights into the Renaissance composer Luca Marenzio's musical representations of the afterlife.

This happened. Apparently he was light on details, but perhaps worth looking into if I can find any specifics on his work. Or I could just ask him...

Anyway: one of the above, or another pass at Kryptos?

Nazbaque

03-16-2016, 10:39 PM

As Gonzo notes, designing a crucifixion machine is basically just an engineering problem, and well within the scope of this thread. Though my Catholic background compels me to note that the nails don't actually go through the hand (you can ripthe hand away). They gobetween the major bones in the forearm.

I actually am aware of this, but treating these practicalities as obvious would have implied the trial and error kind of experience which led Romans to those practical methods, so I went with the established myth. This was merely to prevent people from seeing me as some sort of nutjob and had nothing to do with my desire to follow the pun chain.

The Unreasoner

03-18-2016, 05:04 PM

Okay, a quick rundown on some mathematics of nonstandard computing. True quantum computers (iow, based on entanglement, not the Canadian adiabatic bullshit) rely on quantum bits that each take the form of a probability distribution. You can set each one independently, and when entangled they are understood by the model of a 2^n dimensional circle. From here you perform a number of flips and rotations to the system until it encodes the problem under consideration. At this point, you perform a quantum operation (like the quantum fourier transformation) that collapses the state. A quantum computer is 'complete' if it has a functionally complete set of logical operators and at least one quantum operator. It's important to note that results are only probabalistic, the qubits are not entirely independent, the functional operators need to be reversible, and the quantum operator is irreversible (you can't uncollapse a state).

There are other nonstandard computing models, like the analog computer the Dutch use to manage their unique water issues. I believe it uses voltage and current to directly model the water. Other models use photonic processes, pneumatic processes, mechanical processes (some of these are really cool. I met a local artist who builds some amazing clockwork computers for collectors), or even purely hypothetical ones.

And the model I am currently trying to design is another. As we've seen, any operation computable by a Turing machine can be reduced to a circuit of AND/OR/NOT gates. But we can intuitively see that more information per 'unit' (bit, qubit, obit) means greater efficiency. As it happens, my model uses sets of elements in a finite field instead of bits. But, with just a little tweaking, we can see that we stil have gates that function like AND, OR, and NOT; and De Morgan's laws still apply. In addition, you can develop special operators that act directly on the elements of the sets themselves (for instance, if the finite field was integers modulo 7, you could have an affine transformation operator). So: AND becomes the intersection of the sets, OR becomes the union, and NOT becomes the complement (sticking with integers modulo 7, the complement of {0,1,2,3} would be {4,5,6}). The empty/null set is analogous to FALSE, and the full set corresponds to TRUE. Partial sets are used to rigorously manipulate discrete probability distributions of nondetermistic variables.

I have to say, I am really fucking proud of the work I've done in the field, and my model will work with certain hardware better than anything else currently out there. Unfortunately the guy in charge of the project just wants to throw computing power at a primitive genetic algorithm to build the model. He has no understanding of what it will look like, what the extra operators can be, and how they behave. So like I said before, I'm on my own.

The Unreasoner

03-27-2016, 09:45 PM

Okay, I was bored and already testing some things in Maple when I decided to take another look at the problem I got hung up on last time. Turns out I just misinterpreted a line (and I confused the technique with another I'm more familiar with. Though the authors could have been a bit clearer in their notation). Turns out s is just twice the target minus the summation of the multiset. And since lambda is just there to ensure what we might call 'technical compliance', we'll ignore it, and just look at the straight residue (as opposed to the quadratic diophantine version). But, to show the relationship:

Quadratic Diophantine:

A*x^2+B*y=C with A, B, C, x, and y all in N (natural numbers).

Since you can assume A, B, and C are all relatively prime (no common factors. And you can assume it because the Euclidean algorithm makes it easy to find gcd(gcd(A,B),C), and you can then just factor it out), you can rewrite it as:

A*x^2=C-B*y

A*x^2 = C mod B

x^2 = C*A^(-1) mod B

Since C*A^(-1) mod B is easily worked out, you can show that if the square of a natural number x is congruent to some number Q modulo N (where Q=C*A^(-1), and N=B), then there exists a solution to the subset-sum problem used to generate Q and N, provided x is less than some positive number t (also generated from the subset-sum problem. It's essentially a holdout of the restriction of the domain on the Quadratic Diophantine prolem).

You find Q using the Chinese Remainder Theorem on the following congruences:

Q=t^2 mod P

and

Q=s^2 mod 2^(m+1)

while N is equal to P*2^(m+1)

P is the product of a set of distinct prime numbers (other than 2) equal in size to the multiset raised to the power m (the number of bits n the summation of all the elements in the multiset), while s is the summation of a set of values obtained with the Chinese Remainder Theorem on the following congruences:

value[i]=multiset[i] mod 2^m

and

value[i]=0 mod P*primeset[i]^(-m)

You must also ensure that the values are not congruent to zero modulo the corresponding prime.

Now, this problem is NP-Complete, which means that it is a deciscion problem. Finding a suitable x only means that a solution to the subset-sum problem exists. To deconstruct the x value into the members of the subsets, you need to test each member (appropriately tweaking the values) and see if the congruence still holds (but these are much easier to do than finding the initial satisfiable x). That being said, here are the Q, N, and t values for the tattoo problem, along with a satisfying x:

Q=

13336001699742420686934233547272257409195346594156 82724534116136000988597221417465786711259086166981 18171681770816177404214012655963100828749652830208 54284290255887671935141179833791741577165394255213 75283889443416823485393064744585773608773611552790 75397713471851802327593114552596382780605290453652 07339740598748555180385668281023749958339598750179 20885085371761835725344335119084050637212831886860 15401275935321702864399786025

N=

17637633085770622730167837158713586993955291176605 22975805775356565230649360756421710721188160488822 39650926420928748408886434689636325996966686327607 33996956386638348176583805095951413458974055718811 51510682141417609797054161585105219795825853625704 22912499358432081648005992562325172825869068050796 05252968582528272876272691000467076037340989243503 05805319436099832720320261381112451194704105295364 60897537747220000000000000000

t=

14617619954027659287490340612037252463102826513514 76973197696361287780797932664617294740806305686793 85859046671490428808078940755393128988066259149277 37188540029303521452816955002512704789065694036647 13578929639092785029041463685283867334672994721157 83464940159141940503120547586825551777893350315713 84559579256550517603272266403906272756186227781821 64874364346618291695662594231636567551201082790345 54834031231243103645

x=

14617619633525147492129890288700729114868914534751 46277820262346268891743548951014757483385733388587 36365592106335149370810012590102613115015950247250 45588814771855724069959090134347241495837143808517 32054100734545378712219504178935025638417080045028 60889944097138095731539434077244907991703641901644 39733968042396973272497368511048003935827847781320 03469245351589859856719536806406258223874158777088 79853608867961853645

As you can see, even a relatively small problem blows up pretty quickly under repeated reductions.

The Unreasoner

03-28-2016, 12:21 AM

Just for fun, I took a factoring problem as 3 SAT and converted to a Quadratic Residue problem. The number being factored was 6. And look at what I got: 51997253222067379089319540632601559669005042061175 56637289779477435766558348655888120797686376205713 66817852368272041101568515731129459426988680734924 97121123141893672238060887148367921867243470949721 63079195543006011995195942318767451628999722625526 87071685899663428729235205797440768502726473537355 54157638572402310610922543578821807037948401319654 24273783987609525212172854590568202235016380500900 42407964475693355830792419784983592726945270918830 46770756831347192641393664992273321695322733653800 03361167250057574362677508458793019349885303902967 87626528110029131351614600873167912415715149704702 43233437945756842088798003526542646604937257199352 10160049895759890263228607678442225684149482171696 10051048785356980445389901308640159184497630976102 88996010089824079698876214018392650897852234340233 34300955347230657552579292067550390047470887926085 02526366632624033848409555466442603478616726682030 14053722734026306902094104970363760272775110200373 11849782384032944339284846395486410498419090648028 33881533304855686685307809488808927769396228229195 53235033118102247438146849539314772135017055407785 78206387382731276237647539797627942075567378923332 04992471884470732735089484712495313883635424538003 01966523463740104394477564020953790336579957665395 88657254664943914910129366025586001717400425304085 07856217734002436762355936824509443645999112110102 14037942743618506434185787437735043384233822710353 84334936861382942238034695460630193055822868366830 27629401550412605642534807856676367418011018289341 31217918493840474211990621332141467654871399007546 95663668381985876701916386505480438785605410495930 86580782147354481745785233916987938863322422319915 89666666492456743607703253513608238168647116858964 60107306963118407273139670134578632282611769526223 97409831999688284057026248035184181632063802304484 15186127893134727458996161415676313692089013489501 63354646709731320143477067263468330837624231132571 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The Unreasoner

03-29-2016, 07:46 PM

So I don't know if this really belongs here, but I have a question...

You know those pop rock things that have some kind of firework stuff wrapped in tissue paper? Well there's another one that I got once in Hawaii that's similar, but its actually a solid blueish-purple pellet sold in a kind of pill container (like a plastic tray with little dips in it sealed with paper).

Does anyone know what's that's called?

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