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**AnthonyRBrown****Member**- Registered: 2024-03-09
- Posts: 64

It's all a Mystery ? Like trying to get blood out of a stone!

The Ruddest in the classroom...just might be the Genius!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,860

AnthonyRBrown wrote:

It's all a Mystery ? Like trying to get blood out of a stone!

Indeed!

phrontister wrote:

Absolute accuracy is required for all digits (right up to and including the units digit)

The rounding that occurs once QB64's double-precision limit is reached has enabled me to manufacture the following X3 + Y3 = Z3 'solution' (I adapted your FLT code):

Just BASIC (a scaled-down freeware version of Liberty BASIC) can handle HUGE numbers accurately, making 'solution' manufacturing rather difficult!

Here's an image of JB's output for the same X, Y & Z input as for the above QB64 image:

No JB solution there.

Btw, the yellow highlighted portions denote runs of similar digits in groups of cubed numbers.

*Last edited by phrontister (2024-05-21 19:28:32)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**AnthonyRBrown****Member**- Registered: 2024-03-09
- Posts: 64

It never gives up!

The Ruddest in the classroom...just might be the Genius!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,860

AnthonyRBrown wrote:

It never gives up!

But wait...there's more!

AnthonyRBrown wrote:

The most important numbers are the end numbers in brackets () the reason they are the most important is because they are the end of the process for making cube numbers! they are if you like the TRUTH numbers! its a bit like the numbers go round and round mathematically,and eventually make a whole cube number!

Ok. So, given QB64's (mis)handling of numbers that are too large for it, I thought I'd post an example demonstrating the inaccuracies that can occur in QB64 coding with the all-important last-digit TRUTH numbers in cases where the 16-digit double-precision accuracy boundary is breached:

The Program A segment of that image shows QB64's accurate output of the last digit of X & Y values, compared with the inaccurate output of the last digit of Z values, each of which overstep the double-precision boundary.

In Program B of that image, *all* X,Y,Z values overstep the double-precision boundary, hence all those inaccuracies.

There are no such inaccuracies in the Just BASIC image below, because those same X,Y,Z values that I used in QB64 (see the image above) are within JB's HUGE upper precision limit...whatever that is!

*Last edited by phrontister (2024-04-23 13:33:16)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**AnthonyRBrown****Member**- Registered: 2024-03-09
- Posts: 64

2024-04-10 11:18:46

AnthonyRBrown

Posts: 46

The FLT Demonstration program I made was made FREE! on a budget! It satisfies to show the Three Patterns A,B,C for the accuracy of the program up to certain level.

Most people read all the threads,and then walk away!

The Ruddest in the classroom...just might be the Genius!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,860

AnthonyRBrown wrote:

Most people read all the threads,and then walk away!

But I'm still here!

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**AnthonyRBrown****Member**- Registered: 2024-03-09
- Posts: 64

Truth will out.

The Ruddest in the classroom...just might be the Genius!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,860

AnthonyRBrown wrote:

Truth will out.

Nae doot, mon!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,860

phrontister (post #52) wrote:

The rounding that occurs once QB64's double-precision limit is reached has enabled me to manufacture the following X³ + Y³ = Z³ 'solution'...

I've explored that further, and found that manufacturing X3 + Y3 = Z3 'solutions' for ARB's QB64 program is fairly easily done...like so:

1. X1 = a 17-digit number.

2. Y1 = X1.

Note: {Y1 = X1+1}, {Y1 = X1+2} and {Y1 = X1+3} also work well, but yield successively fewer 'solutions' than {Y1 = X1}.

3. Z1 = the cube root of Z3: just select the first 17 digits of the cube root (the last digit may need rounding up if it's ≥5).

Using ARB's latest code (post #33), I added an option to enter custom X1,Y1,Z1 numbers instead of using the program's random numbers. Here's my adapted code:

I ran that code a few minutes ago, and here's an image from it:

In my adapted code I restricted the length of the random numbers so that their cubes display accurately, with no rounding or truncation: ie, within QB64 double-precision limits.

*Last edited by phrontister (2024-05-21 19:29:29)*

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 10,536

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,860

phrontister (post #59) wrote:

phrontister (post #52) wrote:The rounding that occurs once QB64's double-precision limit is reached has enabled me to manufacture the following X3 + Y3 = Z3 'solution'...

I've explored that further, and found that manufacturing X3 + Y3 = Z3 'solutions' for ARB's QB64 program is fairly easily done...like so:

1. X1 = a 17-digit number.

2. Y1 = X1.

Note: {Y1 = X1+1}, {Y1 = X1+2} and {Y1 = X1+3} also work well, but yield successively fewer 'solutions' than {Y1 = X1}.

3. Z1 = the cube root of Z3: just select the first 17 digits of the cube root (the last digit may need rounding up if it's ≥5).

Inspired by the second image in my post #52, my theory was that for both X1 & Y1 >17 digits, there'd be 'solutions' if their lengths are equal and their first 17 digits are equal (or nearly equal). However:

1. My first trial, with X1 & Y1 both 24 digits long & the first 18 digits equal, resulted in the 'solution' as per the image below...which yielded the coveted "WOW!".

2. My next trial, with X1 & Y1 both 30 digits long, resulted in no 'solutions' (tried equal digits from first 15 to first 30, all in vain). Hmm...

Conclusion: There are many, many 'solutions', but I haven't found the 'rule' that discovers them all.

As you can see, X1 & Y1 differ by 999,999. There would be many 'solutions' in that X1:Y1 range.

*Last edited by phrontister (2024-05-23 23:09:22)*

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**AnthonyRBrown****Member**- Registered: 2024-03-09
- Posts: 64

Proving Fermat' s Last Theorem (almost) in just 2 minutes !

https://www.youtube.com/watch?v=EymVXkPWxyc

The Ruddest in the classroom...just might be the Genius!

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