On Feb 9, 6:17 pm, "Fred." <ghrno-goo...@yahoo.com> wrote:
> Sometimes the improbable happens without intervention,
> but not often.
> Someone once gave me a thought experiment. You write
> a program to simulate random coin flips.
> Case 1: You start it up and it goes heads the 1st 6 times.
> Case 2: You start it up and it goes 1000 flips
> without 6 six heads in a row.
> Fred.
Tricky! I assume your point is that Case 2 is less likely. Indeed,
it's unbelievably less likely -- Case 1 is 1/64, Case 2 didn't happen
even once in 50000 trials (of 1000 flips each); in fact my empirical
chance of not having 7 heads in a row is under 1/10000.
> On Feb 9, 6:17 pm, "Fred." <ghrno-goo...@yahoo.com> wrote:
>> Sometimes the improbable happens without intervention,
>> but not often.
>> Someone once gave me a thought experiment. You write
>> a program to simulate random coin flips.
>> Case 1: You start it up and it goes heads the 1st 6 times.
>> Case 2: You start it up and it goes 1000 flips
>> without 6 six heads in a row.
>> Fred.
> Tricky! I assume your point is that Case 2 is less likely. Indeed,
> it's unbelievably less likely -- Case 1 is 1/64, Case 2 didn't happen
> even once in 50000 trials (of 1000 flips each); in fact my empirical
> chance of not having 7 heads in a row is under 1/10000.
> Breakeven is with about 200 flips.
> I don't see how to compute the probability.
> Charles
(63/64)^995 ~= 1 in 6,385,978? For 7 in a row: (127/128)^994 ~= 1 in
2,431. Breakeven[50%] is about 50 flips for 6 in a row, 95 flips for 7
in a row.
On Fri, 10 Feb 2012 11:34:29 -0800, Charles Brenner wrote:
> On Feb 9, 6:17 pm, "Fred." <ghrno-goo...@yahoo.com> wrote:
>> Sometimes the improbable happens without intervention,
>> but not often.
>> Someone once gave me a thought experiment. You write a program to
>> simulate random coin flips.
>> Case 1: You start it up and it goes heads the 1st 6 times.
>> Case 2: You start it up and it goes 1000 flips
>> without 6 six heads in a row.
>> Fred.
> Tricky! I assume your point is that Case 2 is less likely. Indeed, it's
> unbelievably less likely -- Case 1 is 1/64, Case 2 didn't happen even
> once in 50000 trials (of 1000 flips each); in fact my empirical chance
> of not having 7 heads in a row is under 1/10000.
> Breakeven is with about 200 flips.
> I don't see how to compute the probability.
My first run of 50,000 trials of 1000 flips got hits (6 or more in a row) 49,990 times, and misses 10 (this is using the Mersenne twister).
Out of curiosity I'm going to try it a few more times.
> > On Feb 9, 6:17 pm, "Fred." <ghrno-goo...@yahoo.com> wrote:
> >> Sometimes the improbable happens without intervention,
> >> but not often.
> >> Someone once gave me a thought experiment. You write
> >> a program to simulate random coin flips.
> >> Case 1: You start it up and it goes heads the 1st 6 times.
> >> Case 2: You start it up and it goes 1000 flips
> >> without 6 six heads in a row.
> >> Fred.
> > Tricky! I assume your point is that Case 2 is less likely. Indeed,
> > it's unbelievably less likely -- Case 1 is 1/64, Case 2 didn't happen
> > even once in 50000 trials (of 1000 flips each); in fact my empirical
> > chance of not having 7 heads in a row is under 1/10000.
> > Breakeven is with about 200 flips.
> > I don't see how to compute the probability.
> > Charles
> (63/64)^995 ~= 1 in 6,385,978?
I dismissed that calculation because it assumes that the chance of 6
heads on flips 1-6 is independent of the chance of 6 heads in flips
2-7, which is not remotely true. The vast majority of the time that
1-6 are not all heads it's because there is a tail among flips 2-6, in
which case 6 heads in flips 2-7 is precluded.
That said, I think that in my previous message I must have
misinterpreted my table of results. Trying again and comparing with
your formula:
Let n=200 flips, checking for 6 heads in a row:
By your formula, approximate chance to not have a run of 6
heads=(63/64)*195=0.046.
By experiment, max run of heads < 6 occurred 20% (among 10000
sequences of n=200)
Formula 5x too low.
Let n=500 flips:
Your formula: 1/2400
Experiment: 1.5% (close to the chance of 6 conseq heads)
formula 37x too low
Let n=1000 flips.
Your formula=1/6 million
Experiment: 6/10000
formula 3800x too low.
As a sort of confirmation of my revised "experiment" figures, they are
psychologically consistent with Fred's trivia (whomever invented his
examples probably knew that the number 1000 couldn't have been
dramatically lowered).
> On Fri, 10 Feb 2012 11:34:29 -0800, Charles Brenner wrote:
> > On Feb 9, 6:17 pm, "Fred." <ghrno-goo...@yahoo.com> wrote:
> >> Sometimes the improbable happens without intervention,
> >> but not often.
> >> Someone once gave me a thought experiment. You write a program to
> >> simulate random coin flips.
> >> Case 1: You start it up and it goes heads the 1st 6 times.
> >> Case 2: You start it up and it goes 1000 flips
> >> without 6 six heads in a row.
> >> Fred.
> > Tricky! I assume your point is that Case 2 is less likely. Indeed, it's
> > unbelievably less likely -- Case 1 is 1/64, Case 2 didn't happen even
> > once in 50000 trials (of 1000 flips each); in fact my empirical chance
> > of not having 7 heads in a row is under 1/10000.
> > Breakeven is with about 200 flips.
> > I don't see how to compute the probability.
> My first run of 50,000 trials of 1000 flips got hits (6 or more in a row)
> 49,990 times, and misses 10 (this is using the Mersenne twister).
> Out of curiosity I'm going to try it a few more times.
Your result agrees with what I'm now claiming. I think my mistake
wasn't so subtle as an inferior random number generator; more like
being cross-eyed.
On Friday, February 10, 2012 2:34:29 PM UTC-5, Charles Brenner wrote:
> On Feb 9, 6:17 pm, "Fred." <ghrno-goo...@yahoo.com> wrote:
> > Sometimes the improbable happens without intervention,
> > but not often.
> > Someone once gave me a thought experiment. You write
> > a program to simulate random coin flips.
> > Case 1: You start it up and it goes heads the 1st 6 times.
> > Case 2: You start it up and it goes 1000 flips
> > without 6 six heads in a row.
> > Fred.
> Tricky! I assume your point is that Case 2 is less likely. Indeed,
> it's unbelievably less likely -- Case 1 is 1/64, Case 2 didn't happen
> even once in 50000 trials (of 1000 flips each); in fact my empirical
> chance of not having 7 heads in a row is under 1/10000.
> Breakeven is with about 200 flips.
> I don't see how to compute the probability.
> Charles
The real point is that your conclusion from the first
event is that something is "fixed", i.e. your algorithm
for the random number generator has a problem. And, you
have a fairly high probability of being right.
However, your conclusion from the second event is the
same with much more likelihood.
When I get dealt a 0-loser hand, I look for the deck
stacker, and you will not convincing me that that person doesn't exist, but there is some chance, however
small, that I am wrong, just as I could be wrong when
the starting 6 heads sends me looking for the error in
the geneator.
> On Feb 9, 6:17 pm, "Fred."<ghrno-goo...@yahoo.com> wrote:
>> Sometimes the improbable happens without intervention,
>> but not often.
>> Someone once gave me a thought experiment. You write
>> a program to simulate random coin flips.
>> Case 1: You start it up and it goes heads the 1st 6 times.
>> Case 2: You start it up and it goes 1000 flips
>> without 6 six heads in a row.
>> Fred.
> Tricky! I assume your point is that Case 2 is less likely. Indeed,
> it's unbelievably less likely -- Case 1 is 1/64, Case 2 didn't happen
> even once in 50000 trials (of 1000 flips each); in fact my empirical
> chance of not having 7 heads in a row is under 1/10000.
> Breakeven is with about 200 flips.
> I don't see how to compute the probability.
mmmh, I'll give it a try.
Chance of six heads in a row on the first six flips: 1/64.
I.e., chance of "surviving" the first six flips: 63/64,
where 32/64 have a tail on flip 6, and 31/64 a head.
On any of the flips from 7 to 1000:
Let p(n) be the probability that after n flips the condition
has not occurred. I.e., p(6)=63/64.
Let p_i(n) be the probability that after n flips the
condition has not occurred, and the exactly the previous
i flips were head.
I.e., p_0(6)=32/64, p_1(6)=16/64, p_2(6)=8/64, p_3(6)=4/64,
p_4(6)=2/64, and p_5(6)=1/64. And, for all n,
p(n)=p_0(n) + ... p_5(n).
On Fri, 10 Feb 2012 13:32:37 -0800, Charles Brenner wrote:
> On Feb 10, 1:22 pm, HoneyMonster <some...@someplace.invalid> wrote:
>> On Fri, 10 Feb 2012 11:34:29 -0800, Charles Brenner wrote:
>> > On Feb 9, 6:17 pm, "Fred." <ghrno-goo...@yahoo.com> wrote:
>> >> Sometimes the improbable happens without intervention,
>> >> but not often.
>> >> Someone once gave me a thought experiment. You write a program to
>> >> simulate random coin flips.
>> >> Case 1: You start it up and it goes heads the 1st 6 times.
>> >> Case 2: You start it up and it goes 1000 flips
>> >> without 6 six heads in a row.
>> >> Fred.
>> > Tricky! I assume your point is that Case 2 is less likely. Indeed,
>> > it's unbelievably less likely -- Case 1 is 1/64, Case 2 didn't happen
>> > even once in 50000 trials (of 1000 flips each); in fact my empirical
>> > chance of not having 7 heads in a row is under 1/10000.
>> > Breakeven is with about 200 flips.
>> > I don't see how to compute the probability.
>> My first run of 50,000 trials of 1000 flips got hits (6 or more in a
>> row)
>> 49,990 times, and misses 10 (this is using the Mersenne twister).
>> Out of curiosity I'm going to try it a few more times.
> Your result agrees with what I'm now claiming. I think my mistake wasn't
> so subtle as an inferior random number generator; more like being
> cross-eyed.
Results for the next few runs:
49983 17
49981 19
49989 11
49990 10
I think the moral of the story is this:
"One of these days in your travels, a guy is going to come up to you and show you a nice brand-new deck of cards on which the seal is not yet broken, and this guy is going to offer to bet you that he can make the Jack of Spades jump out of the deck and squirt cider in your ear. But, son, do not bet this man, for as sure as you are standing there, you are going to end up with an earful of cider."
> > On Feb 9, 6:17 pm, "Fred."<ghrno-goo...@yahoo.com> wrote:
> >> Sometimes the improbable happens without intervention,
> >> but not often.
> >> Someone once gave me a thought experiment. You write
> >> a program to simulate random coin flips.
> >> Case 1: You start it up and it goes heads the 1st 6 times.
> >> Case 2: You start it up and it goes 1000 flips
> >> without 6 six heads in a row.
> >> Fred.
> > Tricky! I assume your point is that Case 2 is less likely. Indeed,
> > it's unbelievably less likely -- Case 1 is 1/64, Case 2 didn't happen
> > even once in 50000 trials (of 1000 flips each); in fact my empirical
> > chance of not having 7 heads in a row is under 1/10000.
> > Breakeven is with about 200 flips.
> > I don't see how to compute the probability.
> mmmh, I'll give it a try.
> Chance of six heads in a row on the first six flips: 1/64.
> I.e., chance of "surviving" the first six flips: 63/64,
> where 32/64 have a tail on flip 6, and 31/64 a head.
> On any of the flips from 7 to 1000:
> Let p(n) be the probability that after n flips the condition
> has not occurred. I.e., p(6)=63/64.
> Let p_i(n) be the probability that after n flips the
> condition has not occurred, and the exactly the previous
> i flips were head.
> I.e., p_0(6)=32/64, p_1(6)=16/64, p_2(6)=8/64, p_3(6)=4/64,
> p_4(6)=2/64, and p_5(6)=1/64. And, for all n,
> p(n)=p_0(n) + ... p_5(n).
> Getting too late now for me to try to get
> this into one concise formula for p(n).
> Thomas
Nice -- that's how to do it. Assuming that
p(n+1)=p(n)-1/128*p(n-6)
is correct, there's a mechanistic way to write an equation for p(n)
from this. The process is tedious -- it requires writing down the
roots of the characteristic polynomial
x^7 - x^6 + 1/128, then p(n) is a linear combination of the n-th
powers of those roots (finding the coefficients more tedium)
notwithstanding that (I guess) six of the seven roots are complex.
Maybe Mathematica. Anyway my point is that conversely the formula that
this procedure produces, with all it's irrational complex numbers, is
as concise an answer as there is.
> On Feb 10, 1:40 pm, Thomas Dehn<thomas-use...@arcor.de> wrote:
>> On 02/10/2012 08:34 PM, Charles Brenner wrote:
>>> On Feb 9, 6:17 pm, "Fred."<ghrno-goo...@yahoo.com> wrote:
>>>> Sometimes the improbable happens without intervention,
>>>> but not often.
>>>> Someone once gave me a thought experiment. You write
>>>> a program to simulate random coin flips.
>>>> Case 1: You start it up and it goes heads the 1st 6 times.
>>>> Case 2: You start it up and it goes 1000 flips
>>>> without 6 six heads in a row.
>>>> Fred.
>>> Tricky! I assume your point is that Case 2 is less likely. Indeed,
>>> it's unbelievably less likely -- Case 1 is 1/64, Case 2 didn't happen
>>> even once in 50000 trials (of 1000 flips each); in fact my empirical
>>> chance of not having 7 heads in a row is under 1/10000.
>>> Breakeven is with about 200 flips.
>>> I don't see how to compute the probability.
>> mmmh, I'll give it a try.
>> Chance of six heads in a row on the first six flips: 1/64.
>> I.e., chance of "surviving" the first six flips: 63/64,
>> where 32/64 have a tail on flip 6, and 31/64 a head.
>> On any of the flips from 7 to 1000:
>> Let p(n) be the probability that after n flips the condition
>> has not occurred. I.e., p(6)=63/64.
>> Let p_i(n) be the probability that after n flips the
>> condition has not occurred, and the exactly the previous
>> i flips were head.
>> I.e., p_0(6)=32/64, p_1(6)=16/64, p_2(6)=8/64, p_3(6)=4/64,
>> p_4(6)=2/64, and p_5(6)=1/64. And, for all n,
>> p(n)=p_0(n) + ... p_5(n).
>> Getting too late now for me to try to get
>> this into one concise formula for p(n).
>> Thomas
> Nice -- that's how to do it. Assuming that
> p(n+1)=p(n)-1/128*p(n-6)
> is correct, there's a mechanistic way to write an equation for p(n)
> from this. The process is tedious -- it requires writing down the
> roots of the characteristic polynomial
> x^7 - x^6 + 1/128, then p(n) is a linear combination of the n-th
> powers of those roots (finding the coefficients more tedium)
> notwithstanding that (I guess) six of the seven roots are complex.
> Maybe Mathematica. Anyway my point is that conversely the formula that
> this procedure produces, with all it's irrational complex numbers, is
> as concise an answer as there is.
Real solutions:
Root 1: -0.420154549170266
Root 2: 0.5
Root 3: 0.991791421712163
Complex roots:
Root 4: -0.23096445325876+0.359572188979042 * i
Root 5: -0.23096445325876-0.359572188979042 * i
Root 6: 0.195146016987811-0.408930828152264 * i
Root 7: 0.195146016987811+0.408930828152264 * i
Yes, 1/2 is one of the roots.
I confirmed that x=0.991791421712163 is a good approximation
of a root. As its absolute value is much larger than
for any of the other roots, for practical purposes
this is the speed of the decay (assuming I did not make
a mistake somewhere).
> On 02/11/2012 02:41 AM, Charles Brenner wrote:
> > On Feb 10, 1:40 pm, Thomas Dehn<thomas-use...@arcor.de> wrote:
> >> On 02/10/2012 08:34 PM, Charles Brenner wrote:
> >>> On Feb 9, 6:17 pm, "Fred."<ghrno-goo...@yahoo.com> wrote:
> >>>> Sometimes the improbable happens without intervention,
> >>>> but not often.
> >>>> Someone once gave me a thought experiment. You write
> >>>> a program to simulate random coin flips.
> >>>> Case 1: You start it up and it goes heads the 1st 6 times.
> >>>> Case 2: You start it up and it goes 1000 flips
> >>>> without 6 six heads in a row.
> >> p(n+1)=p(n)-1/128*p(n-6).
> > Nice -- that's how to do it. Assuming that
> > p(n+1)=p(n)-1/128*p(n-6)
> > is correct, there's a mechanistic way to write an equation for p(n)
> > from this. The process is tedious -- it requires writing down the
> > roots of the characteristic polynomial
> > x^7 - x^6 + 1/128, then p(n) is a linear combination of the n-th
> > powers of those roots (finding the coefficients more tedium)
> > notwithstanding that (I guess) six of the seven roots are complex.
> > Maybe Mathematica. Anyway my point is that conversely the formula that
> > this procedure produces, with all it's irrational complex numbers, is
> > as concise an answer as there is.
> As its absolute value is much larger than
> for any of the other roots, for practical purposes
> this is the speed of the decay (assuming I did not make
> a mistake somewhere).
> On Feb 10, 9:24 pm, Thomas Dehn<thomas-use...@arcor.de> wrote:
>> On 02/11/2012 02:41 AM, Charles Brenner wrote:
>>> On Feb 10, 1:40 pm, Thomas Dehn<thomas-use...@arcor.de> wrote:
>>>> On 02/10/2012 08:34 PM, Charles Brenner wrote:
>>>>> On Feb 9, 6:17 pm, "Fred."<ghrno-goo...@yahoo.com> wrote:
>>>>>> Sometimes the improbable happens without intervention,
>>>>>> but not often.
>>>>>> Someone once gave me a thought experiment. You write
>>>>>> a program to simulate random coin flips.
>>>>>> Case 1: You start it up and it goes heads the 1st 6 times.
>>>>>> Case 2: You start it up and it goes 1000 flips
>>>>>> without 6 six heads in a row.
>>>> p(n+1)=p(n)-1/128*p(n-6).
>>> Nice -- that's how to do it. Assuming that
>>> p(n+1)=p(n)-1/128*p(n-6)
>>> is correct, there's a mechanistic way to write an equation for p(n)
>>> from this. The process is tedious -- it requires writing down the
>>> roots of the characteristic polynomial
>>> x^7 - x^6 + 1/128, then p(n) is a linear combination of the n-th
>>> powers of those roots (finding the coefficients more tedium)
>>> notwithstanding that (I guess) six of the seven roots are complex.
>>> Maybe Mathematica. Anyway my point is that conversely the formula that
>>> this procedure produces, with all it's irrational complex numbers, is
>>> as concise an answer as there is.
This is a polynomial where all coefficients are integer,
and furthermore the coefficient at the highest power of
y is 1 (or -1).
It is pretty easy to see that any rational solution is
integer (*), and must furthermore be a divisor of 128.
So all you have to do is try all divisors of 128:
1, -1, 2, -2, 4, -4, 8, -8, 16, -16, ..., 128, -128.
(*) To see that any rational solution is integer, consider
y^n + a_(n-1)*y^(n-1) + ... + a_0 = 0,
where all a_i are integer.
Let y=p/q be any non-integer rational solution, where p and
q are integers, have no common divisors, and q>1.
Insert y=y/q, multiply the equation by q^n:
p^n + a_(n-1)*p^(n-1)*q + ... + a_0*q^n = 0.
If you look at this modulo q, p^n mod q is not 0, as
p and q have no common divisors, and all other parts
are zero modulo q. This is a contradiction, and thus
such a non-integer rational solution y=p/q cannot exist.
> Warning: off-topic mathematics below.
> No bridge content in this posting.
> On 02/11/2012 10:27 AM, Charles Brenner wrote:
> > On Feb 10, 9:24 pm, Thomas Dehn<thomas-use...@arcor.de> wrote:
> >> On 02/11/2012 02:41 AM, Charles Brenner wrote:
> >>> On Feb 10, 1:40 pm, Thomas Dehn<thomas-use...@arcor.de> wrote:
> >>>> On 02/10/2012 08:34 PM, Charles Brenner wrote:
> >>>>> On Feb 9, 6:17 pm, "Fred."<ghrno-goo...@yahoo.com> wrote:
> >>>>>> Sometimes the improbable happens without intervention,
> >>>>>> but not often.
> >>>>>> Someone once gave me a thought experiment. You write
> >>>>>> a program to simulate random coin flips.
> >>>>>> Case 1: You start it up and it goes heads the 1st 6 times.
> >>>>>> Case 2: You start it up and it goes 1000 flips
> >>>>>> without 6 six heads in a row.
> >>>> p(n+1)=p(n)-1/128*p(n-6).
> >>> Nice -- that's how to do it. Assuming that
> >>> p(n+1)=p(n)-1/128*p(n-6)
> >>> is correct, there's a mechanistic way to write an equation for p(n)
> >>> from this. The process is tedious -- it requires writing down the
> >>> roots of the characteristic polynomial
> >>> x^7 - x^6 + 1/128, then p(n) is a linear combination of the n-th
> >>> powers of those roots (finding the coefficients more tedium)
> >>> notwithstanding that (I guess) six of the seven roots are complex.
> >>> Maybe Mathematica. Anyway my point is that conversely the formula that
> >>> this procedure produces, with all it's irrational complex numbers, is
> >>> as concise an answer as there is.
> > Euler and Ramanujan would have noticed that. I didn't.
> It is pretty easy to find the 1/2 root
> because of the simple structure
> of the characteristic polynomial.
I had in mind the famous conversation between Ramanujan and Hardy, in
which Hardy happened to mention the number 1729 and Ramanujan
instantly noted that it was the sum of cubes in two different ways.
That fact was obvious to Hardy as well, once alerted to it, for when
Hardy ("naturally") asked for a corresponding number for 4th powers
Ramanujan, knowing his friend, replied that there was no similarly
*obvious* answer. (Euler also explored the question.)
With this polynomial, maybe no manipulation is necessary to guess 1/2
as a root once you think to look. I was just admitting that I didn't
look. I only blindly attacked the polynomial with numeric tools.
> This is a polynomial where all coefficients are integer,
> and furthermore the coefficient at the highest power of
> y is 1 (or -1).
> It is pretty easy to see that any rational solution is
> integer (*), and must furthermore be a divisor of 128.
> So all you have to do is try all divisors of 128:
> 1, -1, 2, -2, 4, -4, 8, -8, 16, -16, ..., 128, -128.
> (*) To see that any rational solution is integer, consider
> y^n + a_(n-1)*y^(n-1) + ... + a_0 = 0,
> where all a_i are integer.
> Let y=p/q be any non-integer rational solution, where p and
> q are integers, have no common divisors, and q>1.
> Insert y=y/q, multiply the equation by q^n:
> p^n + a_(n-1)*p^(n-1)*q + ... + a_0*q^n = 0.
> If you look at this modulo q, p^n mod q is not 0, as
> p and q have no common divisors, and all other parts
> are zero modulo q. This is a contradiction, and thus
> such a non-integer rational solution y=p/q cannot exist.
> I had in mind the famous conversation between Ramanujan and Hardy, in
> which Hardy happened to mention the number 1729 and Ramanujan
> instantly noted that it was the sum of cubes in two different ways.
> That fact was obvious to Hardy as well, once alerted to it, for when
> Hardy ("naturally") asked for a corresponding number for 4th powers
> Ramanujan, knowing his friend, replied that there was no similarly
> *obvious* answer. (Euler also explored the question.)
> With this polynomial, maybe no manipulation is necessary to guess 1/2
> as a root once you think to look. I was just admitting that I didn't
> look. I only blindly attacked the polynomial with numeric tools.
> Charles
Although correct, apparently Euler discovered that:
> > I had in mind the famous conversation between Ramanujan and Hardy, in
> > which Hardy happened to mention the number 1729 and Ramanujan
> > instantly noted that it was the sum of cubes in two different ways.
> > That fact was obvious to Hardy as well, once alerted to it, for when
> > Hardy ("naturally") asked for a corresponding number for 4th powers
> > Ramanujan, knowing his friend, replied that there was no similarly
> > *obvious* answer. (Euler also explored the question.)
> > With this polynomial, maybe no manipulation is necessary to guess 1/2
> > as a root once you think to look. I was just admitting that I didn't
> > look. I only blindly attacked the polynomial with numeric tools.
> > Charles
> Although correct, apparently Euler discovered that:
> 158^4 + 59^4 = 134^4 + 133^4 = 635,318,657
> Gerben
Yes, and needless to say he discovered it by brainpower and not by
sheer calculating. In fact he may (my memory is vague) have had an
algebraic identity of the form, hence an infinite family of solutions.
He also (of course) explored the corresponding question for fifth
powers as did I with a 1960's computer (the 7090, "-ty" standing for
Transistorized), and as have others with much more powerful computers
and more ingenious programs. Discuss: Which is the stronger evidence
that there is no solution for fifth powers, that computers haven't
found one, or that Euler didn't?
In article <804945be-ec83-47a0-b1bd-fbe199ac9...@l14g2000vbe.googlegroups.com>, Gerben Dirksen <gerbe...@hotmail.com> writes:
>> With this polynomial, maybe no manipulation is necessary to guess 1/2
>> as a root once you think to look. I was just admitting that I didn't
>> look. I only blindly attacked the polynomial with numeric tools.
When I learned of his method to factor numbers that can be expressed as
the sum of two positive squares in two ways, I coded it up. Unfortunately,
the only way that I knew to find decompositions into the sum of two squares
was brute force, so the program was painfully slow.
Are there any good ways to perform this decomposition? (I wouldn't
consider "factor it and reverse Euler's method" to be very good.)
-- Michael F. Stemper
#include <Standard_Disclaimer>
This message contains at least 95% recycled bytes.
> In article<804945be-ec83-47a0-b1bd-fbe199ac9...@l14g2000vbe.googlegroups.com>, Gerben Dirksen<gerbe...@hotmail.com> writes:
>>> With this polynomial, maybe no manipulation is necessary to guess 1/2
>>> as a root once you think to look. I was just admitting that I didn't
>>> look. I only blindly attacked the polynomial with numeric tools.
>> Although correct, apparently Euler discovered that:
>> 158^4 + 59^4 = 134^4 + 133^4 = 635,318,657
> Presumably, he used this to factor 635,318,657.
> When I learned of his method to factor numbers that can be expressed as
> the sum of two positive squares in two ways, I coded it up. Unfortunately,
> the only way that I knew to find decompositions into the sum of two squares
> was brute force, so the program was painfully slow.
> Are there any good ways to perform this decomposition? (I wouldn't
> consider "factor it and reverse Euler's method" to be very good.)
Depends on the size of numbers you are thinking about.
I'd guess that computing a rainbow table once isn't
that expensive for, say, all numbers up to 10^12.
