Let’s not beat around the bush here… I like doing my mathematics by hand.

Computers only need to be used when the paperwork becomes overwhelming. Computers should not be used when a pencil and paper will get the job done.

By hand, a high-school student should understand how to generate an enormous number of 17×17 4-colourings with FOUR monochromatic rectangles. There may be a proof that FOUR is a hard limit, so find it in this essay, and get your $289. It may also be possible that no such limit exists.

Allow me to demonstrate:

Step 1) Define 4 colour sets { A, B, C, D }, where each set contains each colour, but has nothing in common with the other sets. There are many such sets.

Example:
A = 0123
B = 1230
C = 2301
D = 3012

Step 2) Using { A, B, C, D }, construct a perfect 16×16 4-colouring, where each row has nothing in common with 3 rows, and ( 1 of each colour ) in common with the remaining 12 rows.

Example: 16×16 - A perfect 4-colouring.
0 - AAAA
1 - BBBB
2 - CCCC
3 - DDDD
4 - ABCD
5 - ACDB
6 - ADBC
7 - BADC
8 - BCAD
9 - BDCA
a - CABD
b - CBDA
c - CDAB
d - DACB
e - DBAC
f - DCBA

* I am ashamed about how long it took me to formalize these simple permutations.

Step 3) Group the rows with nothing in common together. There will be 4 groups, each containing 4 rows. This will guide Step 4.

Gp0 = { 0, 1, 2, 3 }
Gp1 = { 4, 7, c, f }
Gp2 = { 5, 9, a, e }
Gp3 = { 6, 8, b, d }

Step 4) Build a 17th column by assigning 1 colour to each Group. Simply choose the corresponding Gp#. Note that a perfect 16×20 colouring can easily be generated using this method.

Step 5) Build a 17th row by assigning 1 colour to each Column-Group with nothing in common. In the 17th row, each number represents 4 identical numbers. Notice how there will be a single position, which has not been assigned a colour, marked by X. When it gets assigned, it introduces exactly 4 mono-chromatic rectangles into an otherwise perfect 17×17 colouring. Behold the result:

0 - AAAA 0
1 - BBBB 0
2 - CCCC 0
3 - DDDD 0
4 - ABCD 1
5 - ACDB 2
6 - ADBC 3
7 - BADC 1
8 - BCAD 3
9 - BDCA 2
a - CABD 2
b - CBDA 3
c - CDAB 1
d - DACB 3
e - DBAC 2
f - DCBA 1
L - 0 1 23 X

* L = Last Row Number 17

Please do not give up hope just yet. If somebody would like to show me a 16×21 perfect 4-colouring, this might open the gates to the 17×17 perfect 4-colouring, because the 16×21 grid is the first format not easily plotted perfect by hand.

Mathematical Aside:

The “Apex Constraint Condition” - I define this condition to be a colouring where each row has exactly one position, of each colour, in common with EVERY other row.

I conjecture that an analytical proof for the (existence / non-existence) of a perfect 17×17 4-colouring will arise from a thorough investigation of colourings having the “Apex Constraint Condition”.

Here is the solution to an NP-Complete problem, known to actually have a valid solution: 10 Best 5×5 Boggle Boards - TWL06

http://www.pathcom.com/~vadco/deep.html

All the very best,

JohnPaul Adamovsky

PS - Contact me if you have questions - logarithm69@hotmail.com

Perhaps Rohan equates “unknown” with “by-hand”.

Here is the most simple explicit case:

01230123012301230
12301230123012300
23012301230123010
30123012301230120
01231230230130121
01232301301212302
01233012123023013
12300123301223011
12302301012330123
12303012230101232
23010123123030122
23011230301201233
23013012012312301
30120123230112303
30121230012323012
30122301123001231
0000111122223333X

* Think of a professional way to say: “Suck on it… Suck it long, and suck it hard.”
* The metric system, that’s right, I don’t have a job.

————————————————————————————–

Hello People,

Is the money real, or did I fall for a hoax?

Long story short. I’ve stumbled into a proof that 4 Monochromatic rectangles is the HARD-LIMIT minimum for a 17×17 4-Colouring.

Mathematics Required: High School - Basic Enumeration Techniques

STATEMENT: It is impossible to construct 5 rows in any 4-Colouring, which have nothing in common with each other.

PROOF:
r1 - 0
r2 - 1
r3 - 2
r4 - 3
r5 - X

* Filling X with any Colour will introduce a commonality in the set.

CONCLUSION:

Four sets of 4 rows with nothing in common is an absolute hard limit for a 16×16 perfect 4-Colouring. This represents a “Minimal Constraint Condition” for 16×16 Colourings. It can also be shown that the “Apex Constraint Condition” where every row has 1 position of each Colour in common with each other row is impossible to construct for a 16×16 grid, and trivial for a 16×20 grid.

T1 - Any 4-Colouring can only contain a maximum of 4 rows with nothing in common.

PROOF BY CONTRADICTION:

Proposition: A Perfect 17×17 4-Colourings Does Exist

Thus, this Colouring must contain a perfect 16×16 Colouring.
By T1, and demonstrated in above post, the Optimal 16×16 Perfect Colouring has 4 sets of 4 rows with nothing in common.

Even with this optimal arrangement, there will be 4 sets of columns with nothing in common, and there will be 4 row sets with nothing in common.

Each row and column set can therefore be assigned a unique Colour, which will be added to the 17th row and column respectively. Adding the 16-element-row and 16-element-column will then introduce ZERO monochromatic rectangles.

The final element in the Bottom-Right corner is marked with an “X”.

Leaving the “X” blank, even in the most optimal case, the 17th row, and 17th column have exactly 1 Colour in common with each other corresponding row or column.

Filling any Colour into position X will thus introduce exactly 4 monochromatic rectangles. Simply inspect the Enumeration examples in the post directly above this one…

The proposition is a logical flaw, because assuming it to be true, PROVES THAT IT IS FALSE!

Therefore, A Perfect 17×17 4-Colourings Does NOT Exist

PS - I prefer cash, so please find a colleague of yours that lives in Toronto, who can hand it to me, and then you can wire them the money.

PPS - I prefer not to deal with banks until I get a sincere-written-apology from CIBC for claiming that they saved my life by emptying the $5000 in my account, while I was in a 25 day-long coma. I saved up that money working construction to pay for a trip to find a merit-based Aerospace Engineering job. I’m pretty sure that bankers aren’t in the business of saving lives, and I don’t give up, so I am still unemployed.

PPPS - I’m thinking this was a hoax, because I doubt that a tenured computer scientist would have any trouble putting together this bush-league high-school proof after a thorough and systematic investigation. I’ve recovered from massive head trauma, so even if it was a hoax, I’d at least like some peer review, and then my money in cash.

PPPPS - I also wrote an extremely powerful search algorithm in C with meticulous and space-saving record keeping, so as to completely eliminate cyclic redundant analysis, while being extremely greedy. Row isolated deviations allowed my quad core to analyse 2,733,499,642,000 of the best colourings in 9 hours and change. The program tanks out at 4 monochromatic rectangles, moves around the 6 to 10 space, and finds the next closest 4-mono-Colourings (typically 20 of them).

5 Monochromatic Rectangles are never found.

Maybe put out a bounty to prove that they don’t exist.

Either your intuition was way off, and-or you’re a stooge.

All the very best,

JohnPaul Adamovsky

Thank you!

Are the odds for two years simply 1/15th of 60%, that is 4%? And are the odds for the next week simply 1/52nd of 1/30th of 60%– or is this not the way to calculate the answer? Or is there no way to calculate the probability of a single event occurring in a fraction of the 30 year period?

I recognize that this is not the best forum for a question like this, but I am a regular reader of you blog, and I hope that someone here can assistt a simpleton with such a simple, trivial problem.

I’ve been having a debate about this question with my girlfriend, neither of us can resolve it, and I can’t seem to find an answer to it on google (see: http://www.google.com/search?hl=en&q=%22probability+over+time%22&sourceid=navclient-ff&rlz=1B3GGGL_enUS353US354&ie=UTF-8).

Would anyone on this blog please help us with this?

{1, 1, 2, 1, 2, 2, 3, 3, 4, 3, 4, 4, 1, 3, 2, 4},
{4, 1, 1, 2, 1, 2, 2, 3, 3, 4, 3, 4, 4, 1, 3, 2},
{2, 4, 1, 1, 2, 1, 2, 2, 3, 3, 4, 3, 4, 4, 1, 3},
{3, 2, 4, 1, 1, 2, 1, 2, 2, 3, 3, 4, 3, 4, 4, 1},
{1, 3, 2, 4, 1, 1, 2, 1, 2, 2, 3, 3, 4, 3, 4, 4},
{4, 1, 3, 2, 4, 1, 1, 2, 1, 2, 2, 3, 3, 4, 3, 4},
{4, 4, 1, 3, 2, 4, 1, 1, 2, 1, 2, 2, 3, 3, 4, 3},
{3, 4, 4, 1, 3, 2, 4, 1, 1, 2, 1, 2, 2, 3, 3, 4},
{4, 3, 4, 4, 1, 3, 2, 4, 1, 1, 2, 1, 2, 2, 3, 3},
{3, 4, 3, 4, 4, 1, 3, 2, 4, 1, 1, 2, 1, 2, 2, 3},
{3, 3, 4, 3, 4, 4, 1, 3, 2, 4, 1, 1, 2, 1, 2, 2},
{2, 3, 3, 4, 3, 4, 4, 1, 3, 2, 4, 1, 1, 2, 1, 2},
{2, 2, 3, 3, 4, 3, 4, 4, 1, 3, 2, 4, 1, 1, 2, 1},
{1, 2, 2, 3, 3, 4, 3, 4, 4, 1, 3, 2, 4, 1, 1, 2},
{2, 1, 2, 2, 3, 3, 4, 3, 4, 4, 1, 3, 2, 4, 1, 1},
{1, 2, 1, 2, 2, 3, 3, 4, 3, 4, 4, 1, 3, 2, 4, 1}

I agree but the symmetry question is interesting in and of itself even for smaller grids and fewer colors. What is interesting, is how many equivalence classes are there for a given grid size and number of colors. The first thing to notice is that there are more than the (17!)^2 symmetries from row permutations and column permutations as has been pointed out by a number of commentators. Jim Ward (Jan 8) mentions rotation, reflection, and color permutations. But now there is some overlap because one can obtain 1) reflection around a horizontal axis by permuting rows, 2) reflection about a vertical axis by permuting columns, 3) rotations by using reflections, and maybe even more. Thus multiplying (17!)^2 by no more than 96 new symmetries, making a negligible dent in the original problem.

The situation is even worse because for certain, maybe even most, equivalence classes the symmetries are not distinct. The size of each equivalence class is equal to the number of distinct symmetries it has.
So the number of equivalence classes will be greater than 4^289
divided by 96x(17!)^2.

For example in the smallest case, there are 16 2×2 grids of 2 colors and thus no more than 16 symmetries. But in fact there are 4 equivalence classes, with the following representatives.
00
00 has 2 members,

00
01 has 8 members,

00
11 has 4 members, and

01
10 has 2 members.

I guess the interest is because the problem seems so approachable. But proving the nonexistence of a solution is not very approachable.

This idea of interpolating between the maximum element and a random element reminds me of the Lp norms,
http://en.wikipedia.org/wiki/Lp_space

Last try. For real. My previous explanation only works for m=0.5…

Let the indices start at 0. We want the probability of choosing index n to be m*(1-m)^n. (That is, skipping n indices and then not skipping the next.)

Choose x at random between 0 and 1. Find an integer n that satisfies:

(1-m)^(n+1) < x ≤ (1-m)^n

Observe that these intervals are disjoint for distinct n.

What is the probability of choosing a particular n? Since x is uniformly distributed between 0 and 1, that is just the difference between the left-hand and right-hand side of the above inequality… Which works out to m*(1-m)^n, as desired.

So now just solve the inequality for n. Take the log base 1-m — which is a decreasing function because 1-m is less than one, meaning the inequality gets reversed.

Result is mccoyn’s floor function, as desired.

Sorry for the multi-posting.

From:

(n+1)*log(m) < log(x) ≤ n*log(m)

When m < 1, log(m) is negative, so dividing through by it reverses the inequalities:

n+1 > log(x)/log(m) ≥ n

…and from there the “floor” formula follows immediately.

Wish we could edit these replies. Guess I need to be more patient and use the preview more…

I meant:

When m is-less-than 1,

n+1 > log(x)/log(m) >= n

…from which the floor-log solution follows.

If you have a random number generator that returns an infinite-precision uniformly chosen number between 0 and 1, you only need to call it once… Because it provides an infinite number of random bits. :-)

Let m be your parameter.

Then choose x at random between 0 and 1.

You seek an n satisfying:

m^(n+1) < x <= m^n

(This is obvious if you consider where the powers of m fall on the segment between 0 and 1.)

Logarithm is an increasing function, so take the log of all three, and recall that log(a^b) = b*log(a). Result:

(n+1)*log(m) < log(x) <= n*log(m)

Now divide through by log(m). m=1 is a special case that blows up — so treat it as a special case. When m < 1, log(m) log(x)/log(m) >= n

Therefore n = floor(log(x)/log(m))

Recall that log(x)/log(m) is log-base-m of x, and you get mmcoyn’s function. When m is a power of 2 — or more clearly, a power of 1/2 — this floor-of-log is just a fraction of the exponent on the floating-point representation. See frexp().

You have to start worrying about your double (or “long double”) precision when m gets too close to 0 or to 1. Details left as exercise for the reader…