Little explorations with HP calculators (no Prime)
03-28-2017, 09:32 PM (This post was last modified: 03-28-2017 09:33 PM by pier4r.)
Post: #61
 pier4r Senior Member Posts: 1,627 Joined: Nov 2014
RE: Little explorations with the HP calculators
experiences using the EQW. Is the following a known bug or there is there a reason for the error? (three powers one over the other are refused, if split with parentheses , they are computed)

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03-29-2017, 01:58 AM (This post was last modified: 03-29-2017 01:59 AM by Han.)
Post: #62
 Han Senior Member Posts: 1,765 Joined: Dec 2013
RE: Little explorations with the HP calculators
Those are two very different numbers. The first one is equal to 27^27 (39 decimal digits) which is much smaller than the second number (3 638 334 640 025 decimal digits). Since each digit is a nibble (4 bits), there is not enough memory to hold all the digits.

The order of operations (due to the explicit parentheses) makes the difference.

Graph 3D | QPI | SolveSys
03-29-2017, 06:02 AM
Post: #63
 pier4r Senior Member Posts: 1,627 Joined: Nov 2014
RE: Little explorations with the HP calculators
True, I did not pay attention that parentheses work even in the exponential case. I'm doing too many basic errors, not good.

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03-29-2017, 07:54 AM (This post was last modified: 03-29-2017 07:55 AM by Dieter.)
Post: #64
 Dieter Senior Member Posts: 2,134 Joined: Dec 2013
RE: Little explorations with the HP calculators
(03-28-2017 07:32 PM)Han Wrote:
(03-28-2017 06:13 PM)pier4r Wrote:  Could you explain (or hint a known relationship) how did you get that d/2+x = d/2*sqrt(2) ?

The right picture can be worth a 1000 words, as they say :-)

Right you are!
The following picture should say it all.

Dieter
03-29-2017, 08:48 AM
Post: #65
 pier4r Senior Member Posts: 1,627 Joined: Nov 2014
RE: Little explorations with the HP calculators
(03-29-2017 07:54 AM)Dieter Wrote:  Right you are!
The following picture should say it all.

Neat image, which tools did you use?

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03-29-2017, 09:51 AM
Post: #66
 Simone Cerica Junior Member Posts: 46 Joined: Mar 2014
RE: Little explorations with the HP calculators
GeoGebra maybe --> https://www.geogebra.org/
03-29-2017, 11:04 AM (This post was last modified: 03-29-2017 11:07 AM by pier4r.)
Post: #67
 pier4r Senior Member Posts: 1,627 Joined: Nov 2014
RE: Little explorations with the HP calculators
Thanks Simone.

Anyway, following the recent pattern, I guess I'm missing something obvious here. I can determine almost all the angles but not the wanted ones.

Furthermore I do not even have a clue how can I employ the 50g to help me (until now I had some pseudo-working directions on the previous problems)

http://i.imgur.com/fWE9736.jpg

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03-29-2017, 11:31 AM
Post: #68
 Han Senior Member Posts: 1,765 Joined: Dec 2013
RE: Little explorations with the HP calculators
(03-29-2017 11:04 AM)pier4r Wrote:  Thanks Simone.

Anyway, following the recent pattern, I guess I'm missing something obvious here. I can determine almost all the angles but not the wanted ones.

Furthermore I do not even have a clue how can I employ the 50g to help me (until now I had some pseudo-working directions on the previous problems)

http://i.imgur.com/fWE9736.jpg

Have you considered the sum of the angles in a triangle and a quadrilateral, and properties of vertical angles?

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03-29-2017, 11:49 AM (This post was last modified: 03-29-2017 11:50 AM by pier4r.)
Post: #69
 pier4r Senior Member Posts: 1,627 Joined: Nov 2014
RE: Little explorations with the HP calculators
(03-29-2017 11:31 AM)Han Wrote:  Have you considered the sum of the angles in a triangle and a quadrilateral, and properties of vertical angles?

Yes I considered the angles in a triangle (180) and quadrilateral (360), I considered that an angle is the same when opposite or the complementary that have a sum of 180. Unless I missed some internal shape, I got all the possibilities of which I'm aware of.

What I did see after your question is the concave pentagon. And maybe this will help.

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03-29-2017, 01:07 PM
Post: #70
 Dieter Senior Member Posts: 2,134 Joined: Dec 2013
RE: Little explorations with the HP calculators
(03-29-2017 08:48 AM)pier4r Wrote:  Neat image, which tools did you use?

This was a quick job in OpenOffice Draw, exported as PNG.

Dieter
03-29-2017, 06:15 PM
Post: #71
 Han Senior Member Posts: 1,765 Joined: Dec 2013
RE: Little explorations with the HP calculators
(03-29-2017 11:49 AM)pier4r Wrote:
(03-29-2017 11:31 AM)Han Wrote:  Have you considered the sum of the angles in a triangle and a quadrilateral, and properties of vertical angles?

Yes I considered the angles in a triangle (180) and quadrilateral (360), I considered that an angle is the same when opposite or the complementary that have a sum of 180. Unless I missed some internal shape, I got all the possibilities of which I'm aware of.

What I did see after your question is the concave pentagon. And maybe this will help.

So $$\angle BED$$ , $$\angle BDE$$, $$\angle CDE$$ and $$\angle AED$$ should be the only angles whose measure you cannot determine directly from the properties mentioned. What if you set up an appropriate system of equations involving these angles?

Graph 3D | QPI | SolveSys
03-29-2017, 06:35 PM
Post: #72
 pier4r Senior Member Posts: 1,627 Joined: Nov 2014
RE: Little explorations with the HP calculators
Interesting idea, again I missed that, I'll try as soon as I finish the other stuff (I just discovered DIR ... END on another thread)

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03-29-2017, 06:58 PM
Post: #73
 Isaac S. Friedman Junior Member Posts: 3 Joined: Feb 2017
RE: Little explorations with the HP calculators
(03-16-2017 11:06 AM)Logan Wrote:
(03-15-2017 11:56 PM)Thomas Okken Wrote:  in fact, many such problems are notoriously difficult, for example:

find a, b, c, n such that

a^n + b^n = c^n

It's trivial to find solutions to this, but if you add these conditions:

a, b, c, n ∈ N
a, b, c > 0
n > 2

It's suddenly a lot harder. :-)

That's not hard, I could write down my marvelous little proof but this text box is too narrow.

K Fermat
03-30-2017, 08:30 AM (This post was last modified: 03-30-2017 08:39 AM by pier4r.)
Post: #74
 pier4r Senior Member Posts: 1,627 Joined: Nov 2014
RE: Little explorations with the HP calculators
My Hp 50g is actually computing since 10+ hours (and I dread that it will take days to finish) so I cannot readily try the suggestions of Han.

The code, that maybe is extra unoptimized (note: I do not like to just optimize the code, it has to be also readable if I want to expand it in the future, without remembering much about it. So also readability is a factor) is the following - me trying the DIR END structure but before applying the idea of compiled local variables :

(It still has to go through some modifications, I decided to collect the results for the round robin format but, after testing for 2 repetitions, I could not imagine such lengthy process and I cannot estimate when it will finish too. So it may finish in the next minute. This is a clear example why my low quality batteries last few hours. Holy usb chargers directly from the power outlet!

If you have any tip by chance [the file is long and ugly, but it works], I'm listening

Further versions, if any, will be stored here )
Code:

%%HP: T(0)A(D)F(.);
@ You may edit the T(0)A(D)F(.) parts.
@ The earlier parts of the line are used by Debug4x.

@ in npp, just select language "normal" to get rid of the highlight.

@ Objective:
@ - We want to compare the expected performances of teams of a certain
@   average strength in some types of tournaments to determine
@   what is the one that determines the best team.
@   Assuming that the best team is the one to which is given higher strength
@   a priori. Of course the simulation of the tournaments
@   should be done some times to determine the percentage of winning for each team.
@   1. round robin. Every team meets everyone else, at least twice.
@      (it could be extended with four times, six and so on)
@      Often used in long tournaments.
@      This is assumed as the best format and will be used as reference.
@   2. swiss tournament
@      Random pairings, and then teams with high score meet other teams
@      with high score, if possible avoiding pairings already done.
@      The minimum is that teams meet once.
@      Often used in chess, short but maybe effective.
@   3. knockout tournaments. Random pairing at the start and then
@      the ones that wins proceed.
@      Often used in "show tournaments", short but maybe not so effective.
@   4. only one match against all the others (like half of a round robin)
@      Often used in show tournaments. Short but maybe not so effective.

@ Remarks: the comments try to give an idea, the values in the comments may
@ be different from the actual code though.

@TODO: to pass variables between programs, try to use the compiled local variables.
@ so one does not have persistent variables after the execution.
@ Although it is not so bad to have global variables and modifying them for the next
@ trial. It is like using inputs, without having to type them on the stack.

DIR
main
\<<
0 @tmpV to use for short time, value can be changed in other functions!
0 @tmpV2
0 @rrResArr
0 @rrResList (not sure what I'll use)
\->
@external inputs
@local var
tmpV
tmpV2

rrResArr
rrResList
\<<
@creating/updating global variables
@to use among programs without too much passing
@of values
'n*25+1200' 'n' 1 16 1 SEQ 'listTeams' STO
@ list of teams with a priori strength score
listTeams SIZE 'numTeams' STO
'0' 'n' 1 numTeams 1 SEQ 'finalTournResList' STO

creMatrVarProb
@Eval not needed when calling globals

1 maxTournRepV
FOR counter
rrFunc
DUP @leaving a copy of the final list of rrFunc

@using the list returned by rrFunc
updWinList
NEXT

finalTournResList
\>>
\>>

@####################
@creating/updating global variables
@to use among programs without too much passing
@of values

@remark: to recall and put in local var only when speed is needed.

@flag for team A winning
uFteamAwon
10

@temp flag, not possible to expect to persist
@in different block executions
uFtmp
20

@flag for finding a probability
uFmatchingProbFound
30

listTeams
{ 0 }

numTeams
0

@max cumulative probability
maxSumProbV
0

@number of possible choices or probabilties
numProbV
0

@column for cumulative probability values
colSumProb
1

@column for single probabilities
colProb
2

@column for variance result
colVar
3

@final matrix with cumulative probability, probability, variance results.
mVarProb
0

@points assigned for a win
pointsForWin
1

@repeating the same tournament type to get the winners.
maxTournRepV
1000

finalTournResList
0

@end globals
@####################

creMatrVarProb
\<<
@createMatrixWithVarianceAndProbability

0 @tmpV
0 @tmpV2
\->
@external inputs
@local var
tmpV
tmpV2
\<<
@ creating the matrix containing the probability of the points. Variance
@ between -400 points and +400 points

@ every drop or increment of the base strength is made in 25 points.
@ those points have a probability. So I assign "probability tokens" at every
@ change. For example.
@ -400  has 25 prob tokens. -375 has 50, -350 has 75 and so on, until the
@ base value then goes back. +25 has 400 tokens, +50 has 375 and so on.
@ so I want to model those tokens.
'n' 'n' 25 400 25 SEQ
425 +
@ expanding the previous list with 425
'n' 'n' 400 25 -25 SEQ
@ Or REVLIST could have been used.
+ 'tmpV' STO
@ storing the list
@ the complete list now goes from 25 to 425 and back to 25.

@ Now we want to create the sum of all those tokens,
@ one value for each token, We can use STREAM but we need to recall
@ objects from the tack. I do not like stack operations because
@ they are unreadable but when they are a few it is ok.
@ otherwise one should comment the actions that they do.
1 'tmpV2' STO
@ counter, it needs to start from 1.
tmpV
\<<
@ having 25 50 75 100
@ STREAM  25  2 Pick 25 + 25 STREAM 25 Pick 2 25 + 25  STREAM 25  Pick 2 25   + 25
@         50         50   75        75        75   75         75         75     75
@                    25             75        75   150        150        150    150
@                                             75              100        100    250
@                                                                        150
2 PICK
+
1 'tmpV2' STO+
@we need to count the objects that we leave on the stack. Plus 1 added at the start.
\>>
STREAM
@now we have the list with all the sums
@before making a vector out of it we save the last value that is the maximum value.
DUP 'maxSumProbV' STO
@we continue
tmpV2
@ the number of objects.
\->ARRY
@ transformed in a vector

tmpV
@ the first list
OBJ\->
@ exploded
@before saving it as vector I save the number of elements or
@probility tokens (equals to variances)
DUP 'numProbV' STO
\->ARRY
@ transformed in a vector

'n' 'n' -400 -25 25 SEQ
0 +
'n' 'n' 25 400 +25 SEQ
+ OBJ\->
\->ARRY
@ transformed in a vector

3 COL\-> 'mVarProb' STO
@final matrix with columns: sum of probability tokens, probability tokens for variance, variance in points.

@#########
@output
\>>
\>>

getVarFunc
\<<
@program to extract the variance out of the built matrix

@output, the variance of the strength points out of the matrix

0 @curProbV
0 @tmpV
\->
@external inputs
@local var
curProbV
tmpV
\<<
maxSumProbV RAND * 1 + IP 'curProbV' STO
@ we get a random vnumber that has to be compared with the probability sums

1 'tmpV' STO
@with tmpV we will use it as a counter
uFmatchingProbFound CF
WHILE
uFmatchingProbFound FC?
tmpV numProbV \<=

AND
REPEAT
IF
curProbV 'mVarProb(tmpV, colSumProb)' EVAL \<=
@accessing the matrix in algebraic mode, slower but more readable.
@ than the messy RPN (RPN is not always less readable, but in some cases)
@maintenance and debug use resources as well that often are more valuable.
THEN
@ found a probability sum bigger than the actual random value.
uFmatchingProbFound SF
ELSE
1 'tmpV' STO+
END
END

'mVarProb(tmpV, colVar)' EVAL
\>>
\>>

matchResFunc
\<<
@program to compute the result of a match based on the variance and probability of it.
@ should be used by all the tournaments.

0 @teamAtmpV
0 @teamBtmpV
\->
@external inputs
teamAstrV
teamBstrV
@local var
teamAtmpV
teamBtmpV
\<<
@In input are expected the strength of the two teams
@ and a user flag for the result
@as for of variables already existing.
@teamA
@teamB
@userFlagNumber

@in output a flag is set
@if teamA won, otherwise teamB

uFteamAwon CF

@ teamA streangth modified
teamAstrV getVarFunc
+ 'teamAtmpV' STO

@ teamB streangth modified
teamBstrV getVarFunc
+ 'teamBtmpV' STO

@there is no draw possible, at least for now.
IF
teamAtmpV teamBtmpV >
THEN
@teamA won.
uFteamAwon SF
ELSE
@ teamA with lower or equal strength
IF
teamAtmpV teamBtmpV <
THEN
@teamA lost
uFteamAwon CF
ELSE
@ same strength
@ coin toss
IF
RAND 10 * IP
5 <
THEN
@teamA won
uFteamAwon SF
ELSE
@teamA lost
uFteamAwon CF
END
END
END
\>>
\>>

IncListElFunc
\<<
@ program to increase the value of element in a list

@ input on the stack see variables
@ see the following, that if I keep adding variables on the main
@ program it gets too unreadable.

@output, the list in input, modified, on the stack
\->
@external inputs
lList
@list
lPosV
@position of the element
lincV
@increase
@local var
\<<
lList lPosV
@placing list and position on the stack
@for storing, later.
lList lPosV GET
lincV +
@getting the value and increasing it
PUT
@putting the increasing value back
\>>
\>>

updWinList
\<<
@program that updates the winner list given the
@last tournament result list

0 @maxV
0 @tmpV
0 @tmpList
\->
lastTresList
maxV
tmpV
tmpList
\<<
uFtmp CF
@we use it to determine if we are done

lastTresList SORT REVLIST 'tmpList' STO
@to find the maximum quickly

WHILE
uFtmp FC?
REPEAT

IF
tmpV maxV \>=
THEN
@valid new max
tmpV 'maxV' STO

@make tmpList smaller.
tmpList TAIL 'tmpList' STO

@increase the value in the position of the team with highest value
@ and saving.
finalTournResList
lastTresList maxV POS
1
IncListElFunc
'finalTournResList' STO
ELSE
@we are done
uFtmp SF
END
END
\>>
\>>

rrFunc
\<<
@program to compute a round robin tournament

0 @rrResList
0 @teamAstrV
0 @teamBstrV
\->
@external inputs
@numTeams
@listTeams
@pointsForWin
@uFteamAwon
@local var
rrResList
teamAstrV
teamBstrV
\<<
@program to compute a round robin tournament

@ given a list of teams with their strength and other

@output, the list of teams with final points.

'0' 'n' 1 numTeams 1
SEQ 'rrResList' STO
@saving the result list that at the start is just zeroes

1
numTeams 1 -
FOR teamPosA
listTeams teamPosA GET 'teamAstrV' STO
@get first team

teamPosA 1 +
numTeams
FOR teamPosB
listTeams teamPosB GET 'teamBstrV' STO
@get second team team

@compute the result of a match. twice
@as round robin requirement.
1 2
FOR counter
teamAstrV teamBstrV
matchResFunc

IF
uFteamAwon FS?
THEN
@update results
rrResList teamPosA pointsForWin
IncListElFunc
'rrResList' STO
ELSE
@update results
rrResList teamPosB pointsForWin
IncListElFunc
'rrResList' STO
END
NEXT
NEXT
NEXT

rrResList
\>>
\>>

@directory end
@###############################################################################​
END

@ log {
@   20:59 29.03.2017 {
@     After the discovery of the DIR END structure, I'm going to refactor the program.
@   }
@   12:05 29.03.2017{
@     Generating list of 16 teams between 1200 and 1600 strength points
@     'n*25+1200' 'n' 1 16 1 SEQ
@   }
@ }

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03-30-2017, 12:13 PM
Post: #75
 Dieter Senior Member Posts: 2,134 Joined: Dec 2013
RE: Little explorations with the HP calculators
(03-30-2017 08:30 AM)pier4r Wrote:  My Hp 50g is actually computing since 10+ hours (and I dread that it will take days to finish) so I cannot readily try the suggestions of Han.

?!? – *what* is it computing?
Obviously this in not related to the triangle problem we are discussing.

Dieter
03-30-2017, 12:26 PM
Post: #76
 Dieter Senior Member Posts: 2,134 Joined: Dec 2013
RE: Little explorations with the HP calculators
(03-29-2017 06:15 PM)Han Wrote:  So $$\angle BED$$ , $$\angle BDE$$, $$\angle CDE$$ and $$\angle AED$$ should be the only angles whose measure you cannot determine directly from the properties mentioned. What if you set up an appropriate system of equations involving these angles?

I have tried this as well. Let <BDE = u, <BED = v, <CDE = x and <AED = y and it is possible to set up several linear equations, for instance...

u + v = 162
x + y = 111
u + x = 132
v + y = 141

...but I cannot get four independent ones, so there is no solution for a linear equation system. What do you get?

Dieter
03-30-2017, 04:14 PM
Post: #77
 Han Senior Member Posts: 1,765 Joined: Dec 2013
RE: Little explorations with the HP calculators
(03-30-2017 12:26 PM)Dieter Wrote:
(03-29-2017 06:15 PM)Han Wrote:  So $$\angle BED$$ , $$\angle BDE$$, $$\angle CDE$$ and $$\angle AED$$ should be the only angles whose measure you cannot determine directly from the properties mentioned. What if you set up an appropriate system of equations involving these angles?

I have tried this as well. Let <BDE = u, <BED = v, <CDE = x and <AED = y and it is possible to set up several linear equations, for instance...

u + v = 162
x + y = 111
u + x = 132
v + y = 141

...but I cannot get four independent ones, so there is no solution for a linear equation system. What do you get?

Dieter

I had not actually attempted this problem when I posed that quest. My question about considering a system of linear equations was merely that -- a question (i.e. was it tried, and does it work) -- and was not meant as an indication that I had solved it through linear systems.

You are correct, though, that the resulting system is not sufficient for solving the problem (the system does have a solution; however it appears it has an infinite number of solutions assuming we are not restricted to whole degrees for the angles).

I did eventually make some progress on this problem. I think I have a solution but I wonder if it is correct since it is not the cleanest. It is an elementary solution, but nowhere near what Paul Erdos would consider as being from "The Book." But what I ended up doing making use of the law of sines and cosines. Here's an outline of my solution.

Let F be the point along CD such that AF is perpendicular to CD. G be the intersection of the blue lines. Observe that AC and AD are equal in length. Call this length L. Use basic trigonometry determine CF (in terms of L), which then gives you CD (since CD is twice the length of CF). The law of sines applied to triangle CAE will give you the length of CE in terms of L. Then use the law of cosines in triangle CDE to determine ED (again in terms of L). Lastly, apply the law of sines to triangle CGE to determine EG in terms of L. Now you can apply the law of sines to EDG to obtain the angle GDE (which is the same as angle CDE).

I got approximately 76.32 degrees for the angle through a bit of scribbling (very likely I made a arithmetic mistake, I would imagine) and a basic scientific calculator (HP 32S).

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03-30-2017, 05:47 PM (This post was last modified: 03-30-2017 07:55 PM by Gerson W. Barbosa.)
Post: #78
 Gerson W. Barbosa Senior Member Posts: 1,048 Joined: Dec 2013
RE: Little explorations with the HP calculators
(03-30-2017 12:26 PM)Dieter Wrote:
(03-29-2017 06:15 PM)Han Wrote:  So $$\angle BED$$ , $$\angle BDE$$, $$\angle CDE$$ and $$\angle AED$$ should be the only angles whose measure you cannot determine directly from the properties mentioned. What if you set up an appropriate system of equations involving these angles?

I have tried this as well. Let <BDE = u, <BED = v, <CDE = x and <AED = y and it is possible to set up several linear equations, for instance...

u + v = 162
x + y = 111
u + x = 132
v + y = 141

...but I cannot get four independent ones, so there is no solution for a linear equation system. What do you get?

Dieter

That's what I'd done too, but the solution is very simple, I realize now. No need to solve any linear system.

u = 72
v = 90
x = 60
y = 51

Explanation later.

Gerson.

PS: From E draw a perpendicular line to line AB, which intercepts it at F. Now we have two similar right-triangles: BEF and EFD. Angle DEF = angle EBF = 18 degrees, as we already know. Angle BDE, which you have named 'u' is its complement, 72 degrees. Finally, angle EDC, or your 'x', can be easily determined as 180 - 72 - 48 = 60 degrees.

PPS: Please do not consider this. Without any grounds I had assumed angle BED was a right angle. Fooled by my own drawing...
03-30-2017, 06:02 PM (This post was last modified: 03-30-2017 06:12 PM by pier4r.)
Post: #79
 pier4r Senior Member Posts: 1,627 Joined: Nov 2014
RE: Little explorations with the HP calculators
(03-30-2017 12:13 PM)Dieter Wrote:  ?!? – *what* is it computing?
Obviously this in not related to the triangle problem we are discussing.

Dieter

No, not to that. So during an international tournament that used the knockout format (you know, teams meet and the winner proceeds) I realized that I could compare tournaments formats somehow. (in short: pick one format as reference, assign team strength, assign variance, do the math, see who wins in the reference tournament and in the other formats)

I did a little search and I found out the swiss tournament format that is pretty neat, but I would have liked to know the statistical properties of this format compared to other formats.

I started a comparison model on paper, using the sharp el 506w as random generator (and also for multiplications). I tested two or three cases and the comparison procedure looked promising.

So I said, ok let's automate the process. But then something else happened and I put the idea aside. The other day I decided to resume the idea from my todos collection (note 1) and since I am also in a period where I'm trying to refresh and expand my user RPL / math knowledge, I decided to implement it on the 50g.

First I did it in a program with nested programs as functions (see also this topic), then I refactored it to use the newly discovered (for me) DIR ... END structure.(Pretty neat)
So I was all excited (and I still am excited) that finally I could manage better a large structure in a single file without clumsy workarounds or multiple tabs.

After some hours of coding and debugging (debugging on the hp 50g, I mean the physical one, takes time. Although I discovered a neat solution thanks to nested programs. See topic mentioned above) finally the code worked for the first tournament format. The round robin. So I did two iterations as tests and I did not detect major flaws (could still be that there are some, I hope not). So I launched 1000 iterations to get the statistics back.

I launched them at 0:56 30.03.2017 , local time. The calculator is still working (it is 20:09 local time).
My fault was that when I launched the two iterations I did not take the execution time, I was eating. So I estimated some hours of work (at the end I still usel userRPL that is heavily emulated) but now I wonder if it will be finished before the weekend. I have to rethink about the role of the hp 50g, maybe it is not so suited for such tasks unless I go for hpgcc and hpgcc-based languages (I will likely skip sysrpl).

I posted the code of the current working version of the tournament simulator in a previous post. I do not know if I missed obvious optimizations but I really thought that a tournament with 16 teams, so 240 pairings (120 pairings twice), repeated 1000 times would not take such amount of time.

It is not the first time that I end up waiting hours/day on the hp 50 though.

(note 1) About todos I do not know what the others do, but I try to collect my interesting todos. I always end up with more ideas than I'm able to process. I'm also pretty sure that there are plenty of people that have the same problem if not bigger (more todos than todos processing ability). Anyway at least I try to collect my todos, could be that someone will do it in the future.

Wikis are great, Contribute :)
03-30-2017, 06:36 PM (This post was last modified: 03-30-2017 07:04 PM by Han.)
Post: #80
 Han Senior Member Posts: 1,765 Joined: Dec 2013
RE: Little explorations with the HP calculators
(03-30-2017 05:47 PM)Gerson W. Barbosa Wrote:  That's what I'd done too, but the solution is very simple, I realize now. No need to solve any linear system.

I suspected there had to be a simple approach. I'll try your hints and also rework my solution to see where I may have made mistakes.

EDIT (after some initial thoughts):

Quote:PS: From E draw a perpendicular line to line AB, which intercepts it at F. Now we have two similar right-triangles: BEF and EFD. Angle DEF = angle EBF = 18 degrees, as we already know. Angle BDE, which you have named 'u' is its complement, 72 degrees. Finally, angle EDC, or your 'x', can be easily determined as 180 - 72 - 48 = 60 degrees.

Are you saying draw EF so that $$\angle BFE$$ is 90 degrees? If so, how does it follow that triangle BEF is similar to triangle EFD without also knowing that $$\angle BED$$ is 90 degrees (which does not seem obvious to me).

Or are you suggesting we make $$\angle BEF$$ equal to 90 degrees? This would make it even less obvious how triangle EFD is even a right triangle.

EDIT 2: Your solution seems to suggest that we only need angles A and C. Am I missing another simple observation ?

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