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retroforth/example/Gott.retro
crc d8d1b0b759 updated Gott.retro from Kiyoshi 
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2019-12-26 21:26:29 +00:00

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(~~~
:rl (-)_ReLoad reset d:wipe 'Gott file:load.package ;
(~~~
# Gott
This program takes
- $g$ in $]0, infty[$, a random time gone since something began.
- $p$ in $]0, 1[$, the portion of time that you consider to have elapsed.
and produces the interval in which that something should end.
The method is taken from
William Poundstone; The Doomsday Calculation; Little, Brown Spark, 2019
The relevant part of this book can be read as
Amazon preview
https://www.amazon.com/Doomsday-Calculation-Equation-Transforming-Everything-ebook/dp/B07J4WCSMR
If you prefer an audio introduction, watch on YouTube
The Doomsday Calculation: Book Trailer
https://youtu.be/jr693Q6M8OY
Also, an interview with the author is available as
Michael Shermer with William Poundstone
— The Doomsday Calculation (SCIENCE SALON # 76)
https://youtu.be/M0tHz4BdrvA
In pp.14-18 of Poundstone's book there is a description of a prediction
J. Richard Gott III made regarding the fall of the Berlin wall.
The input data are
- In 1961 the wall was built.
- In 1969 Gott visited the wall, which he considers that
to have been a random moment in time.
- He proposes to make a prediction with a 50% level of confidence,
which is equivalent to saying that at least 25% of the total duration
is assumed to have elapsed.
so that
- `gone` is 1969 - 1961 = 7 years.
- `portion` of the time gone is a half the 50% or 1/4.
which produces the interval $[3, 24]$, which translates to
- There is a 50% chance that the wall falls sometime between
$1969 + 3 = 1972$ and $1969 + 24 = 1993$ .
The wall actually fell in 1989.
Since the method is controversial, read the book before applying
it to your personal problems.
Pay attention to the randomness of `gone` and time scale invariance.
# Program
Consider something of interest that began at time `start`.
Take the origin 0 of the time axis to be a random time `now`.
Assuming that the end does come, it has to be within finite time from `now`.
Let that maximum time be the unit $1$.
Since the maximum time is unknown,
the time unit 1 is unknown in any time scale.
```
now maximum time
| |
...--+---------+---------+---------+---------+--...--> time
0 1
```
Consider the $100 \, q$% confidence interval within which the period ends.
```
100 q %
now |<--- confidence interval --->| max
| | | |
...--+----v****+*********+*********+****v----+--...--> time
0 (1-q)/2 1-(1-q)/2 1
```
Let $p := (1 - q)/2$.
The `end` falls somewhere within the confidence interval $[p, 1-p]$
with probability $q$.
```
now |<------------ q ------------>| max
| | | |
...--+----v****+*********+*********+****v----+--...--> time
0 p 1-p 1
```
The `end` will come with probability 1 sometime after 0,
but with probability $q$ only after $p$ and before $1-p$.
This means that with probability $q$ the time elapsed between `start` and
`now`
equals $p$.
Let the time `gone` be $g :=$ `now` $-$ `start`.
This quantity comes with a time scale since `now` and `start` are measured in
some unit such as days or years.
Now let $\ell$ be the time corresponding to $p$ in the scale of $g$.
Then $\ell/g = p/(1-p)$ so that
$$
\ell = \frac{p}{1-p} \, g
$$
Similarly, the time corresponding to $1-p$ in the scale of $g$ is
$$
u = \frac{1-p}{p} \, g
$$
Now the end will come neither until the next moment of `now`, which is 0,
nor after the earliest end within the confidence interval which is at $p$.
To sum up, with $100 \, q$% confidence
$l := g \, r$ and $u := g / r$, with $r := p/(1-p)$.
```
start l u
| |<------------ q ------------>|
V | |
--+--...--+----v****+*********+*********+****v----+--...--> time
0 p 1-p 1
```
Implementing this method, I make a function called `gott` which takes
`start`, `now`, `gone` and returns `l` and `u`, where
- `l`: shortest time from `now` that the event may take place
- `u`: longest time from `now` that the event may take place
both with $100 \, q$% confidence.
Helpers.
~~~
:f:rot- (-_abc-cab) f:rot f:rot ;
:n:to-f (n-_-n) n:to-float ;
:. (-) dump-stack
#0 f:depth lt? [ nl 'f_ s:put f:dump-stack ] if
#0 f:adepth lt? [ nl 'fa_ s:put f:dump-astack ] if ;
:f:reset (-__..-) f:depth [ f:drop ] times ;
:f:areset (-__-__..-) f:adepth [ f:pop f:drop ] times ;
:r (..-__..-__..-) reset f:reset f:areset ;
:f:complement (-__n-n)_1-f .1. f:swap f:- ;
~~~
Here is the program.
~~~
{{
:f:short (-__p-r)_r=p/(1-p)_where_p=portion,_p=<1/2
f:dup f:complement f:/ ;
:f:lo (-_gp-s)_g=gone f:short f:* ;
:f:hi (-_gp-l) f:short f:/ ;
---reveal---
:f:future.lo (-__ngp-l)_n=now f:lo f:+ ;
:f:future.hi (-__ngp-h) f:hi f:+ ;
:f:future (-__ngp-lh)
#3 [ f:dup f:push f:rot- ] times f:future.lo
#3 &f:pop times f:future.hi ;
}}
~~~
## Berlin wall
This problem has been taken from Poundstone's book.
In pp.14-18 there is a description of a prediction J. Richard Gott III
made regarding the fall of the Berlin wall.
### 50% confidence
The input items are:
- .1961. `start` : floating time when it began
- .1969. `now` : floating present time
- .0.25 `portion` : floating portion `gone`, between 0 and 1
It is crucial that `now` may be considered a random moment
in time line after `start`.
This is a special case of the Copernican principle.
1961 is when the wall was built.
1969 is the time in which Gott visited the wall.
He considers this point to be a random time in
the period of existence of the wall.
0.5 is the confidence level; the prediction is made so that there is
a 50% chance that the end of the wall will fall
within the time interval to be produced below.
~~~
r 'Berlin_wall_1/2 s:put nl
#1969 n:to-f (now) f:dup #1961 n:to-f (now_now_start)
f:- (now_gone) .1. .4. f:/ (now_gone_portion)
f:future . nl nl
~~~
gives
f 1971.666667 1993.000000
meaning that there is a 50% chance that the wall will fall
sometime between 1972 and 1993.
"His 1967 prediction was that there was a 50 percent chance that
the wall would stand at least 2.67 years after his visit but no
more than 24 years." (p.16)
### 95% confidence
To recalculate with a confidence level of 95% as in p.19 of the book,
~~~
r 'Berlin_wall_95% s:put nl
#1969 n:to-f (now) f:dup #1961 n:to-f (now_now_start)
f:- (now_gone) .0.025 (now_gone_portion)
f:future . nl nl
~~~
which gives
f 1969.205128 2281.000000
or between 1969 and 2281; reasonable but uninteresting.
## Diana and Charles
Now try the relationship duration between Diana and Charles that
comes as the first example in the book, p.3.
~~~
r 'Di_&_Charles s:put nl
.1993 f:dup (now_now)
.1981 .7 .12 f:/ f:+ (now_now_start) f:- (now_gone)
.0.1 (now_gone_portion)
f:future.lo . nl nl
~~~
giving
f 1994.268519
or after 1994.
`future.hi` is irrelevant, considering the couple's life span.
Note that the value of `Gone` is set to 0.1 rather than 0.05,
even though the level of confidence is stated to be 90%.
Regarding this result the book says
"Gott's formula predicted a 90 percent chance that the royal marriage
would end in as little as 1.3 more years."
which is misleading: if a 90% confidence level is assumed
the split will not take place at least until 1.3 years from "now."
"The split was formalized on August 28, 1996."
## The Third Reich
Poundstone book p.82.
- 1934-09 The Third Reich proclaimed.
- 20 months before, Hitler rised into power.
- 95% confidence.
~~~
r '3rd_Reich s:put nl
.1934 .9 .12 f:/ f:+ f:dup f:dup (now_now_now)
.20 .12 f:/ f:- (now_now_start) f:- (now_gone)
.0.05 .2. f:/ (now_gone_portion)
f:future . nl nl
~~~
f 1934.792735 1999.750000
"A Copernican would have predicted the Nazi state to survive somewhere
between another two weeks and another sixty-five years (at 95 percent
conficence). The Third Reich lasted another eleven years."
# Personal applications
Dates are in YYMMDD.
## 190816--191126 UL project
190726 Brasilia Time (BRT)
Project proposal from a software company FS to
a chemical company UL, both in Brazil, upon a request from UL.
190815 Received a message from FS that UL has
not yet made a decision.
19 days have passed since the proposal.
~~~
r 'UL_BR s:put nl
.0. .19. .0.2 (now_gone_portion}
f:future . nl nl
~~~
f 4.750000 76.000000
So there is an 80% chance that UL will not give its decision
within 5 days.
190822 No notice yet, as predicted.
Since 76 days are about 2.5 months, the chance that a notification
will arrive by the end of November is 80%.
191112 Project cancelled due to a change in market prices.
This is also as predicted.
## 190816 HE worker evaluation
190722 HE, a Japanese company, asked me to comment on
its employee's work, so they can make a decision on her.
190724 HE sent me related data.
190730 I turned in my report to HE.
190801 HE asked me to wait for their decision.
190816 Today.
~~~
r 'HE_on_an_employee s:put nl
.16. f:dup (now_now) .-1. (now_now_start)
f:- (now_gone) .0.2 (now_gone_portion)
f:future . nl nl
~~~
f 20.250000
So there is an 80% chance that HE will not send me
their decision before 190820.
190822 Received HE's decision on its employe.
This is as predicted.
## 190922 War Japan
~~~
r 'War_JP s:put nl
.2019. f:dup .1945. (now_now_start) f:- (now_gone)
#2 [ f:dup-pair f:push f:push ] times (fa:gone_now_gone_now)
.0.05 (now_gone_portion)
f:future.lo
f:pop f:pop .0.1 (now_gone_portion)
f:future.lo
f:pop f:pop .0.2 (now_gone_portion)
f:future . nl
~~~
f 2022.894737 2027.222222 2037.500000 2315.000000
There are 90, 80, and 70% chances that Japan will not go into a war
until 2023, 2027, and 2038, respectively.
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