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Louis Stumpers
L Stumpers 
 
Number of games in database: 56
Years covered: 1932 to 1969
Overall record: +13 -32 =11 (33.0%)*
   * Overall winning percentage = (wins+draws/2) / total games.

Repertoire Explorer
Most played openings
D94 Grunfeld (3 games)
E60 King's Indian Defense (2 games)
B59 Sicilian, Boleslavsky Variation, 7.Nb3 (2 games)
C65 Ruy Lopez, Berlin Defense (2 games)
D45 Queen's Gambit Declined Semi-Slav (2 games)
D31 Queen's Gambit Declined (2 games)


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LOUIS STUMPERS
(born Aug-30-1911, died Sep-27-2003, 92 years old) Netherlands

[what is this?]

Frans Louis Henri Marie Stumpers was born in Eindhoven, Netherlands, on 30 August 1911. (1) He was champion of the Eindhoven Chess Club in 1938, 1939, 1946, 1947, 1948, 1949, 1951, 1952, 1953, 1955, 1957, 1958, 1961 and 1963, (2) and champion of the North Brabant Chess Federation (Noord Brabantse Schaak Bond, NBSB) in 1934, 1935, 1936, 1937, 1938, 1939, 1940, 1941, 1942, 1943, 1944, 1946, 1948, 1949, 1950, 1951, 1952, 1953, 1954, 1955, 1959, 1961, 1962, 1963, 1964, 1965, 1966 and 1967. (3) He participated in five Dutch Chess Championships, with a 4th place in 1948, (4) and represented his country at the 1st European Team Championship in Vienna in 1957 (two games, vs Josef Platt and Max Dorn). (5) From 1945 and until about 1956, he was first Secretary and then Chairman of the NBSB. (3)

Stumpers was a physicist, and worked for the Philips company as an assistant from 1928. During 1934-1937, he studied at the University of Utrecht, where he took the master's degree. (6) In 1938 he was again employed at Philips, (6) and at a tournament in 1942, he supplied the hungry chess players with food from his employer. (3) After the war, he made a career in physics, with patents and awards on information ("radio") technology. He received degrees from several universities and colleges, including in Poland and Japan. (1, 3, 6) He retired from Philips in 1972, but continued teaching, (6) partly as professor at the University of Utrecht (1977-1981). (7) He was also Vice President (1975-1981) and Honorary President (1990-2003) of URSI, the International Union of Radio Science. (8)

Louis Stumpers married Mieke Driessen in 1954. They had five children, three girls and two boys. (6)

1) Online Familieberichten 1.0 (2016), http://www.online-familieberichten...., Digitaal Tijdschrift, 5 (255), http://www.geneaservice.nl/ar/2003/...
2) Eindhovense Schaakvereniging (2016), http://www.eindhovenseschaakverenig...
3) Noord Brabantse Schaak Bond (2016), http://www.nbsb.nl/pkalgemeen/pk-er... Their main page: http://www.nbsb.nl.
4) Schaaksite.nl (2016), http://www.schaaksite.nl/2016/01/01...
5) Olimpbase, http://www.olimpbase.org/1957eq/195...
6) K. Teer, Levensbericht F. L. H. M. Stumpers, in: Levensberichten en herdenkingen, 2004, Amsterdam, pp. 90-97, http://www.dwc.knaw.nl/DL/levensber... Also available at http://www.hagenbeuk.nl/wp-content/...
7) Catalogus Professorum Academiæ Rheno-Traiectinæ, https://profs.library.uu.nl/index.p...
8) URSI websites (2016), http://www.ursi.org/en/ursi_structu... and http://www.ursi.org/en/ursi_structu...

Suggested reading: Eindhovense Schaakvereniging 100 jaar 1915-2015, by Jules Welling. Stumpers' doctoral thesis Eenige onderzoekingen over trillingen met frequentiemodulatie (Studies on Vibration with Frequency Modulation) is found at http://repository.tudelft.nl/island...

This text by User: Tabanus. The photo was taken from http://www.dwc.knaw.nl.

Last updated: 2018-08-17 13:29:49

 page 1 of 3; games 1-25 of 56  PGN Download
Game  ResultMoves YearEvent/LocaleOpening
1. L Stumpers vs J Lehr 1-0191932EindhovenD18 Queen's Gambit Declined Slav, Dutch
2. Prins vs L Stumpers  1-0391936NED-ch prelimB20 Sicilian
3. E Sapira vs L Stumpers 0-1251938NBSB-FlandersD94 Grunfeld
4. L Stumpers vs E Spanjaard  1-0551938NED-ch prelimE02 Catalan, Open, 5.Qa4
5. A J Wijnans vs L Stumpers  1-0361939NED-chB05 Alekhine's Defense, Modern
6. J van den Bosch vs L Stumpers  ½-½581939NED-chA48 King's Indian
7. L Stumpers vs S Landau 0-1411939NED-chD33 Queen's Gambit Declined, Tarrasch
8. H van Steenis vs L Stumpers  1-0251939NED-chB02 Alekhine's Defense
9. L Stumpers vs S Landau  ½-½341940HilversumD31 Queen's Gambit Declined
10. L Stumpers vs H Kramer  0-1361940HilversumE25 Nimzo-Indian, Samisch
11. A J van den Hoek vs L Stumpers  1-0271941BondswedstrijdenB10 Caro-Kann
12. T van Scheltinga vs L Stumpers 1-0351942NED-ch12D94 Grunfeld
13. W Wolthuis vs L Stumpers  ½-½521946NED-ch prelim IC58 Two Knights
14. L Stumpers vs J H Marwitz  1-0401946NED-ch prelim ID31 Queen's Gambit Declined
15. G Fontein vs L Stumpers  ½-½261946NED-ch prelim ID94 Grunfeld
16. L Stumpers vs H van Steenis 0-1241946NED-ch prelim ID28 Queen's Gambit Accepted, Classical
17. C B van den Berg vs L Stumpers  1-0581946NED-ch prelim ID19 Queen's Gambit Declined Slav, Dutch
18. L Stumpers vs Euwe 0-1301946NED-ch prelim IE60 King's Indian Defense
19. L Stumpers vs Cortlever  ½-½501946NED-ch prelim IE60 King's Indian Defense
20. L Stumpers vs Grob 1-0601947Int BA55 Old Indian, Main line
21. L Stumpers vs H van Steenis  0-1331947Int BD23 Queen's Gambit Accepted
22. Tartakower vs L Stumpers 1-0241947Int BD74 Neo-Grunfeld, 6.cd Nxd5, 7.O-O
23. V Soultanbeieff vs L Stumpers  ½-½461947Int BD96 Grunfeld, Russian Variation
24. L Stumpers vs C Vlagsma  0-1451948NED-ch14C65 Ruy Lopez, Berlin Defense
25. L Stumpers vs A Vinken  0-1331948NED-ch14E21 Nimzo-Indian, Three Knights
 page 1 of 3; games 1-25 of 56  PGN Download
  REFINE SEARCH:   White wins (1-0) | Black wins (0-1) | Draws (1/2-1/2) | Stumpers wins | Stumpers loses  
 

Kibitzer's Corner
< Earlier Kibitzing  · PAGE 82 OF 82 ·  Later Kibitzing>
Oct-04-19
Premium Chessgames Member
  al wazir: <johnlspouge>: You may recall my saying that I thought I might have seen this problem before. Antipodes play an important role in your argument. In my first, abortive attempt to derive a formula I mumbled something about antipodes. That might have been an instance of what psychologists call recrudescence. If so, it was very incomplete.

I followed your derivation of the formula f(N) = N^2 - 2N + 2. But when I got to the paragraph beginning "Third," I had some difficulty, because the terminology is confusing. You have introduced a plethora of poles. In ordinary parlance every sphere has but one equator and two poles at a time. (It is possible to carry out a coordinate transformation, but in the new coordinate system there will still be only one pair of poles.) I don't understand what is meant by a "successful choice." A choice of N points that lie within a given hemisphere? A choice of hemispheres that contain all of N given random points? I see that there is a one-to-one correspondence between the f(N) spherical areas and ... something.

I glanced at Wendel's paper, but it seemed to add a whole new set of complexities, so I quit reading after that glance. While I was pondering this, a glimmering -- I won't say a light, that would be immodest -- a faint adumbration of an idea struck me.

As I said, I have been unable to find anything wrong in my code since the last bug I fixed in the counting routine.

And I am too respectful of your mathematical acumen to suppose that you would post a mistaken demonstration of a theorem -- not to mention the countless other mathematicians and Putnam contestants who have solved the problem, and the examiners who checked their work.

That leaves one logical possibility: We are both right.

We are calculating two *different* probabilities!

Oct-04-19
Premium Chessgames Member
  al wazir: Gastropod is singular. Pseudopod is singular. Therefore the singular of antipodes must be antipod.
Oct-04-19
Premium Chessgames Member
  johnlspouge: < <al wazir> wrote: Gastropod is singular. Pseudopod is singular. Therefore the singular of antipodes must be antipod. >

In English, the singular is "antipode", derived by back-formation from "antipodes" in the early 17th century. You are have etymology on your side, but the first page of a Google search on "antipod" has no hits in English, and just one in a Serbo-Croation dictionary.

[ https://en.wiktionary.org/wiki/anti... ]

Oct-04-19
Premium Chessgames Member
  beatgiant: The singular of <kibitzers> is <pedant>.
Oct-04-19
Premium Chessgames Member
  johnlspouge: < <beatgiant> wrote: The singular of <kibitzers> is <pedant>. >

ROFL

Oct-04-19
Premium Chessgames Member
  johnlspouge: < <al wazir> wrote: That leaves one logical possibility: We are both right.

We are calculating two *different* probabilities! >

I would not be surprised. After all, I entered the thread by pointing everyone to Bertrand's Paradox.

[ https://en.wikipedia.org/wiki/Bertr... ]

My screed abstracted the problem. The probability distribution you are calculating lacks one of the following properties: (1) it chooses points independently from a continuous distribution; or (2) if it does, it does not choose antipodes with equal probability densities.

Oct-04-19
Premium Chessgames Member
  johnlspouge: The case N=4 is simpler than the general case and may aid in understanding. To start, choose any point on the sphere. Now, choose any three pairs of antipodes in general position (so no two pairs lie on the same great circle with the point). Now from each pair of antipodes, choose one with probability 1/2. The antipodes form 8 spherical triangles, all triangles have equal probability when one from each pair is chosen, so the probability that the initial point lies inside any one of the triangles is 1/8.

The case N=4 is unusually simple, because the great circles through pairs of antipodes form spherical polygons that are triangles with the antipodes for vertices.

Oct-04-19
Premium Chessgames Member
  al wazir: <johnlspouge: The probability distribution you are calculating lacks one of the following properties: (1) it chooses points independently from a continuous distribution; or (2) if it does, if it does, it does not choose antipodes with equal probability densities.>

The distribution is as "continuous" as a discrete numerical calculation performed by a digital computer using 32-bit arithmetic can be. The choice is as "independent" as an operation carried out deterministically (i.e., repeatably) can be.

My random number generator is ran2, copied from Numerical Recipes. The authors say this: <We think that, within the limits of its floating-point precision, ran2 provides perfect random numbers; a practical definition of “perfect” is that we will pay $1000 to the first reader who convinces us otherwise (by finding a statistical test that ran2 fails in a nontrivial way, excluding the ordinary limitations of a machine’s floating-point representation).> https://pdfs.semanticscholar.org/cc...

As for your second point, I tried inserting a branch (pseudo)randomly replacing the calculated point with its antipod in the loop where I calculate n (pseudo)random points inside the unit sphere. I didn't think it would change the results, aside from statistical error. It didn't.

Also, as <Tiggler> correctly noted, it doesn't make any difference whether or not I project the points onto the surface of the sphere. (I tried it both ways.) A plane drawn through the center of a sphere and two other points is independent of where those points are on their respective radii.

Let me try to pin down exactly what you are doing in your derivation.

Suppose the sphere is the Earth. I choose a set of n points on the surface and find their n antipodal antipoints. Together the points and antipoints comprise an augmented set of 2n distinct elements. Each point and its antipoint are associated with the great circle equidistant from the two, their equatorial great circle (EGC). All the EGCs are different.

Let's suppose that n=5, and five elements of the ten in the augmented set are at Chicago, Detroit, New York City, Pittsburgh, and Washington. I don't know which of these five are points and which are antipoints; I just know they are all in the augmented set.

It's clear that any hemisphere must contain five points of the augmented set. This applies to a hemisphere centered on one of those five cities (equivalently, a hemisphere for which one of those cities is a point or antipoint, or equally equivalently, a hemisphere bounded by the EGC of one of those cities). So what?

Since f(5) = 22, the five EGCs demarcate 22 "spherical areas" on the Earth's surface. But the five cities are so close together that all five lie in the *same* spherical area. I can travel between any two of them along a geodesic without crossing an EGC.

What is the one-to-one correspondence of which you wrote?

Oct-04-19
Premium Chessgames Member
  al wazir: <johnlspouge: In English, the singular is "antipode", derived by back-formation from "antipodes" in the early 17th century.>

Bah! I suppose you think that "media" is a singular noun. Wrong. It is the plural of "medium."

But Media is a city in southeastern Pennsylvania.

Oct-04-19
Premium Chessgames Member
  johnlspouge: @<al wazir>: You have put the 10 = 2*5 antipodes into 2 clusters, but there are still 5-subsets of the 10 antipodes that lie in a single hemisphere and contain points from both clusters. Despite your clustering, the ECGs of antipodes still partition the sphere, but some spherical areas become very thin and short. In other words, the thinness of such an area corresponds to the precision required to position any hemisphere so it contains 5 fixed points from both of your clusters. If an arc joins pairs of points in such an area, _that_ arc is not bisected by a ECG.
Oct-04-19
Premium Chessgames Member
  johnlspouge: < <al wazir> wrote: Bah! I suppose you think that "media" is a singular noun. >

No. I do not agree with the majority usage and use "media" as a plural. At my biological work daily, I use "data" as a plural. Kindly do not lecture me on the plural of 2nd declension neuter Latin nouns. We pedants pride ourselves on the speed with which we can recite the declension.

If you are fond enough of "antipod" to use it, I can only suggest you learn Serbo-Croatian. Perhaps regrettably, "antipod" is not any standard of English.

Oct-04-19
Premium Chessgames Member
  johnlspouge: < <johnlspouge> wrote: We pedants pride ourselves on the speed with which we can recite the declension. >

Still under 3 seconds on the first try.

Not bad :)

Oct-04-19
Premium Chessgames Member
  al wazir: <johnlspouge: You have put the 10 = 2*5 antipodes into 2 clusters, but there are still 5-subsets of the 10 antipodes that lie in a single hemisphere and contain points from both clusters.> I'm afraid that doesn't answer the question. In my example, where is the one-to-one correspondence? What corresponds to what?
Oct-04-19
Premium Chessgames Member
  al wazir: <johnlspouge>: "Gastropod, "pseudopod," and "tripod" are English words. "Gastropode" and "pseudopode" are not. They are French. I don't know if "tripode" is a word in any language.

"Antipod" may be Croatian, but it is certainly not Serbian. Антипод might be.

I'm a little surprised that someone who knows the Latin second declension -- and maybe the third, fourth, and fifth as well -- has fallen under the sway of a 17th-century solecism.

Oct-04-19
Premium Chessgames Member
  beatgiant: <al wazir> English got "gastropods", "pseudopods" and "tripods" via Latin, but "antipodes" directly from Greek, or so say a couple of dictionaries I checked. And the singular in Greek would be "antipus," like "platypus" (although the plural of "platypus" is not "platypodes").

English is notoriously full of all kinds of non-uniformity.

Oct-04-19
Premium Chessgames Member
  al wazir: <beatgiant: English got "gastropods", "pseudopods" and "tripods" via Latin>. I doubt if the Romans had any use for the first two, though educated Romans spoke Greek. More likely medieval or Renaissance savants coined them. "Pseudo" is definitely Greek, not Latin.
Oct-05-19
Premium Chessgames Member
  beatgiant: <al wazir>
The sources say "Latin", but not necessarily "ancient Latin". As you suggested, it's not unusual for much later scholars to coin new words based on Latin. And it's also not unusual for Latin to be based on Greek.

But my main point is, we don't need to speculate. There are good quality specialized works called etymological dictionaries, where a person like you has already done the job of finding out and publishing the answer.

Would you believe... etymonline.com?

Oct-05-19
Premium Chessgames Member
  al wazir: <johnlspouge> and <Tiggler>: I found a typo in my code -- this despite my having gone over and over it so many times previously.

My numbers are now consistent, to ~1%, with the ones your formula gives:

All 4 pts in 1 half in 0.8810 of trials.
All 5 pts in 1 half in 0.6859 of trials.
All 6 pts in 1 half in 0.5042 of trials.
All 7 pts in 1 half in 0.3440 of trials.
All 8 pts in 1 half in 0.2308 of trials.
All 9 pts in 1 half in 0.1408 of trials.
All 10 pts in 1 half in 0.0924 of trials.
All 11 pts in 1 half in 0.0537 of trials.
All 12 pts in 1 half in 0.0318 of trials.
All 13 pts in 1 half in 0.0180 of trials.
All 14 pts in 1 half in 0.0116 of trials.
All 15 pts in 1 half in 0.0066 of trials.
All 16 pts in 1 half in 0.0030 of trials.
All 17 pts in 1 half in 0.0011 of trials.
All 18 pts in 1 half in 0.0012 of trials.
All 19 pts in 1 half in 0.0011 of trials.
All 20 pts in 1 half in 0.0004 of trials.

Pleas accept my crestfallen apology. All I can say in my defense is that I never swore that there could not possibly be any mistakes.

Oct-05-19
Premium Chessgames Member
  johnlspouge: Cool.

Apparently, all we have left to argue about is the singular of “antipodes”.

And where you are concerned, I am more than happy being right 1/2 the time :)

Oct-05-19  Tiggler: <al wazir> Apology accepted. And I'll stop calling you <Al>, but please note: though my name is not <tiggler> you can continue to call me that anyway.
Oct-07-19
Premium Chessgames Member
  johnlspouge: < <Tiggler> wrote: And I'll stop calling you <Al> >

Hmm. I don't know why, but this scene came to mind...

[ https://www.youtube.com/watch?v=rO1... ]

Oct-09-19  Tiggler: I was expecting this one: https://www.youtube.com/watch?v=uq-...
Oct-09-19  Tiggler: <johnl> works near Chevy Chase, I believe.
Oct-10-19
Premium Chessgames Member
  johnlspouge: < <Tiggler> wrote: I was expecting this one >

Yours is more on the mark for the nickname; but less, for the message ;>)

Oct-10-19
Premium Chessgames Member
  johnlspouge: < <Tiggler> wrote: <johnl> works near Chevy Chase, I believe. >

Nahh, but that other <johnl> is one hell of a guy ;>)

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