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Hi bobbym.
n up to 50 already! Absolutely amazing. Beautiful work!
I've copied these 3000 digit values for x and y into a Notepad file - fortunately it is capable of word wrap. (unlike this page). So now you can restore normality to the page.
Also your conjecture I have verified to n = 33 by direct computation!
I am amazed at your power of computation! In 1989 I could only get as far as n = 13. It is great news for me to hear the conjecture has now been verified to so high a value of n.
Is your solution for n=33 printable, or would it take up the whole of the internet? Viewers might be interested to see so monstrous a solution.
Hi, bobbym... after another 6 months, I stumbled on some old notes still in my possession:
Ok, so let's go beyond n=3.
Let's do n=4 next.
Then z = 2(2^n - 1)^2 + (2^n +1)^2 ......... (just taking one of the two hypothetised values)
= 739.
Now compute the Continued Fraction for SQR(z^2 - 8). (SQR means square root of).
My notes say: "Values of x and y may be obtained by computing the proper fraction n/d from the first c denominators of the Continued Fraction for SQR(z^2 - 8).
Then x = d - n
and y = 2.d "
Explanation of c : I think you must go on with the continued fraction until the cycle of denominators reaches the critical point where it starts to repeat itself. The value of c for z=739 should be 43.
That could be a lot of fun! Want to try it?
Hi, Dave, glad to see someone taking an interest in Fractals.
Here is the essential core of my program, the process by which the points (x,y,z) are processed:
xv := xs; zv:=zs;
For xp := 0 To 799 do
begin
allzdone:=false; zv:=zs;
otd:=td; done:=0;
while allzdone=false do
begin
zv:=zv-dz; x:=xv; y:=yv; z:=zv;
xq := sqr(x); yq := sqr(y); zq:=sqr(z); r:=sqrt(xq+yq+zq);
xad:=sqrt(xq+yq);
yad:=sqrt(yq+zq);
zad:=z/2;
if zv<-3 then allzdone:=true;
cou:=0; peint:=false; sq:=2*lim;
while cou<reps do
begin
cou:=cou+1; osq:=sq;
nx := xq - 2* sqrt(zq+yq)+xad;
ny := yq - 2* sqrt(xq+zq)+yad;
nz := zq - 2* sqrt(yq+xq)+zad;
x:=nx; y:=ny; z:=nz; xq:=x*x; yq:=y*y; zq:=sqr(z); sq:=xq+yq+zq;
If sq > lim Then cou:=1000;
end;
This is part of the Delphi4 source code used to draw the picture. You can find the entire code at
http://sites.google.com/site/gurthsfiles/fractal-programs .
UPDATED RESULTS
post: #2 #3 #4 #6 #8 #9 #11 #13 #15 #16 #18 #21 #22 #23 #25
valid: 1 0 0 0 0 0 0 0 0 0 0 2 0 2 0
post: #26 #28 #29 #31 #32
valid: 3 0 4 0 1
quittyqat, well done in being the finder of the second word, after so many efforts on the part of all... but note that your Barracuda can't be considered as it appeared already in post #29. As there are 4 valid words in that list of now effectively 8 words (Aubrey having been eliminated) it would appear that Barracuda has exactly 50% chance of being valid!
You only submitted 9 words, but that is not against the rules.
This game is likely to explode soon as I am giving more valid words every day in my own lists.
Sorry about that slip. I'll give you that Aubrey was invalid.
UPDATED RESULTS
post: #2 #3 #4 #6 #8 #9 #11 #13 #15 #16 #18 #21 #22 #23 #25 #26
valid: 1 0 0 0 0 0 0 0 0 0 0 2 0 2 0 3
A warm welcome to our new participant: John E. Franklin.
UPDATED RESULTS
post: #2 #3 #4 #6 #8 #9 #11 #13 #15 #16 #18 #21 #22 #23
valid: 1 0 0 0 0 0 0 0 0 0 0 2 0 2
Interest seems to be flagging, discouraged perhaps by the string of zeros.
That is quite a remarkable achievement; looking through a shorter list of scrabble words, I found almost 3% of the words were valid. (But please keep off unusual words!).
You know one of the words in list 1 is valid. Stare at them long and hard, and one of them is sure to blush and you'll be able to say: "Your slip is showing!". It's too easy, leave your imagination at rest, what you want to see is in plain sight. If no more valid words are discovered soon, I'll submit a list with a couple in it.
UPDATED RESULTS
post: #2 #3 #4 #6 #8 #9 #11 #13 #15 #16 #18
valid: 1 0 0 0 0 0 0 0 0 0 0
#9
Enraptured, impish, demonic solver Zheroically tries similar triangles, favourite ------ theorems (6).
It's to late, bobby. I've been antisocial ever since I was a kid. I just don't like people. That's the point I don't like mingling. Once I get certain people close to me I never let go, and everyone else I ignore.
You sound like my ideal person. If everybody were like you, the world would be perfect.
As it is, I think it is already perfect. If "God" (supposing for a moment there is something like that) were to say to me: "I'll improve the world in any way you want", I'd have to reply: "I honestly can't think of any possible improvement."
The world will end at the right time. The only time.
UPDATED RESULTS
post: #2 #3 #4 #6 #8 #9 #11 #13 #15 #16
valid: 1 0 0 0 0 0 0 0 0 0
quittyqat, it's great to have you join in... these games of mine have a few quirks, as the "old hands" have discovered... the rules are mostly hidden, and get found out as you go along!
One rule is: words that have been valid on first appearance, will no longer be treated as valid in further lists... so your sensible aim of trying to find out which word in the first list is valid, will require other tactics.
Another tip which old players will probably remember: I don't favour complex rules involving "and" or "or" (e.g. "a valid word must contain p or z), so my main rule is always essentially a very simple one. In this particular puzzle, a Grade 1 pass in English (junior school) would be ample qualification to solve this puzzle.
UPDATED RESULTS
post: #2 #3 #4 #6 #8 #9 #11 #13
valid: 1 0 0 0 0 0 0 0
Let P, Q be the points where the tangent touches the bigger and smaller arcs.
Let AM intersect XY at Z.
Compute length AM.
AZ = 2(MZ)...... (ratio of the 2 similar rt-angled triangles); so AZ = 2/3 AM.
Solve the triangles APZ and MQZ.
Compute angles BAZ and AZX: gives the slope of XY.
We know XY goes through the point P which we can compute.
So we can get the equation of line XY.
Solve against the equations for FE and CD to get the coordinates of X and Y.
From these compute length XY.
UPDATED RESULTS
post: #2 #3 #4 #6 #8 #9 #11
valid: 1 0 0 0 0 0 0
#7
Get big warning (5).
#8
Excellent, or terrible (7)
ZHero, as you are working on #2: If you want another hint, the first letter is .
Well done, ZHero, you finally join the champs. Nine out of ten is totally convincing, "able" is incorrect and I can't fathom why you included it, but that's your prerogative. Maybe as the zbumi necessary to art!
I think phrontister's and ZHero's final lists also define the rule.
ZHero: #2: no.
#6: plausible, because people do say "o" for zero; but not what I was thinking of.
ZHero: a very nice inspiration of yours, you've got the WITHOUT correct... and if the clue had been (8, 7, 7) your answer would have been "Colander without caldron" which would have made sense and forced acceptance.
What makes this clue perhaps inadmissibly hard is that I have used an additional skin of disguise, for instance if colander/caldron hade been my IDEA, I might have given the clue as "e (5, 7, 3)", the answer to that being "sieve without pot". That answer would be obviously correct to me, but not to anyone else. (I like the idea of cryptic answers to cryptic clues, answers that do not give the game away even when labelled as correct).
In case anyone wants to take poison at this stage, let me assure them that only #1 is as foxy as this.
ZHero, you put me onto it! I don't know how this could have foxed us so long!