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L Stumpers 
Louis Stumpers
Number of games in database: 55
Years covered: 1932 to 1969
Overall record: +13 -32 =10 (32.7%)*
   * Overall winning percentage = (wins+draws/2) / total games
      Based on games in the database; may be incomplete.

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)

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(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 the 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),; Digitaal Tijdschrift, 5 (255),
2) Eindhovense Schaakvereniging (2016), http://www.eindhovenseschaakverenig....
3) Noord Brabantse Schaak Bond (2016), Their main page:
4) (2016),
5) Olimpbase,
6) K. Teer, Levensbericht F. L. H. M. Stumpers, in: Levensberichten en herdenkingen, 2004, Amsterdam, pp. 90-97, Also available at
7) Catalogus Professorum Academię Rheno-Traiectinę,
8) URSI websites (2016), and

Suggested reading: Eindhovense Schaakvereniging 100 jaar 1915-2015, by Jules Welling. Stumpers' classic doctoral thesis Eenige onderzoekingen over trillingen met frequentiemodulatie (1946) can be read at

Last updated: 2016-09-30 02:49:29

 page 1 of 3; games 1-25 of 55  PGN Download
Game  ResultMoves Year Event/LocaleOpening
1. L Stumpers vs J Lehr  1-019 1932 EindhovenD18 Queen's Gambit Declined Slav, Dutch
2. Prins vs L Stumpers  1-039 1936 NED-ch prelimB20 Sicilian
3. L Stumpers vs E Spanjaard  1-055 1938 Dutch Ch prelimE02 Catalan, Open, 5.Qa4
4. E Sapira vs L Stumpers 0-125 1938 NBSB - FlandersD94 Grunfeld
5. A Wijnans vs L Stumpers  1-036 1939 NED-chB05 Alekhine's Defense, Modern
6. J van den Bosch vs L Stumpers  ½-½58 1939 NED-ch11A48 King's Indian
7. L Stumpers vs S Landau 0-141 1939 NED-ch11D33 Queen's Gambit Declined, Tarrasch
8. H van Steenis vs L Stumpers  1-025 1939 NED-ch11B02 Alekhine's Defense
9. L Stumpers vs H Kramer  0-136 1940 HilversumE25 Nimzo-Indian, Samisch
10. A J van den Hoek vs L Stumpers  1-027 1941 BondswedstrijdenB10 Caro-Kann
11. T van Scheltinga vs L Stumpers 1-035 1942 NED-ch12D94 Grunfeld
12. W Wolthuis vs L Stumpers  ½-½52 1946 NED-ch prelim IC58 Two Knights
13. L Stumpers vs J H Marwitz  1-040 1946 NED-ch prelim ID31 Queen's Gambit Declined
14. G Fontein vs L Stumpers  ½-½26 1946 NED-ch prelim ID94 Grunfeld
15. L Stumpers vs H van Steenis 0-124 1946 NED-ch prelim ID28 Queen's Gambit Accepted, Classical
16. C B van den Berg vs L Stumpers  1-058 1946 NED-ch prelim ID19 Queen's Gambit Declined Slav, Dutch
17. L Stumpers vs Euwe 0-130 1946 NED-ch prelim IE60 King's Indian Defense
18. L Stumpers vs Cortlever  ½-½50 1946 NED-ch prelim IE60 King's Indian Defense
19. L Stumpers vs Grob 1-060 1947 Int BA55 Old Indian, Main line
20. L Stumpers vs H van Steenis  0-133 1947 Int BD23 Queen's Gambit Accepted
21. Tartakower vs L Stumpers 1-024 1947 Int BD74 Neo-Grunfeld, Nxd5, 7.O-O
22. V Soultanbeieff vs L Stumpers  ½-½46 1947 Int BD96 Grunfeld, Russian Variation
23. L Stumpers vs T van Scheltinga  1-047 1948 NED-ch14C97 Ruy Lopez, Closed, Chigorin
24. Prins vs L Stumpers  ½-½30 1948 NED-chD02 Queen's Pawn Game
25. J G Baay vs L Stumpers  1-040 1948 NED-ch14E37 Nimzo-Indian, Classical
 page 1 of 3; games 1-25 of 55  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 288 OF 288 ·  Later Kibitzing>
Sep-27-16  WorstPlayerEver: 19683.
Premium Chessgames Member
  john barleycorn: 19683 is the total number of binary compositions on a set of 3 elements and 729 of them are commutative. that is the easy part.

associative compositions have probably to be tested one by one.

Premium Chessgames Member
  john barleycorn: <diceman: ...

Besides Abdel?>

Maybe <Abdel> finds time on his <10k Jihad> to find the number. However, someone should tell him that it has nothing to do with fractions.

Premium Chessgames Member
  diceman: <john barleycorn:

<diceman: Besides Abdel?>

However, someone should tell him that it has nothing to do with fractions.>

His specialty area is "factions."

Premium Chessgames Member
  john barleycorn: <diceman> there are 8 translations for "faction" from English to German and I think each is a fit for the great autodidact.
Premium Chessgames Member
  Sneaky: Wow Mr. Stumpers looks very stern in that photo.
Premium Chessgames Member
  john barleycorn: hi <Sneaky>, would you have time to hack out the possible associative binary compositions on a 3-set with your computer?
Premium Chessgames Member
  beatgiant: <john barleycorn> I wrote a little script to generate them by brute force and got 1673. I checked associativity on a small sample and it looked correct. I can post one or two of them for you if you want.

Still, I'd be happier if we could come up with a formula.

If this is for academic purposes, you can site "beatgiant,, Oct. 2016" (I'm kidding)

Premium Chessgames Member
  john barleycorn: <beatgiant> thanks a lot for your work. 1673 is at least some good guideline number. I got a dozen or so and no clue how find a formula.
Premium Chessgames Member
  beatgiant: <john barleycorn> To sanity check, I added code to also output the breakdown by number of distinct result values per composition. For example, if the composition is { aa=c, ab=c, ac=c, ba=c, bb=c, bc=c, ca=c, cb=c, cc=c }, it has one distinct result value.

It gave the following breakdown:
2 with 1 distinct result value
210 with 2 distinct result values
1461 with 3 distinct result values

That shows the script was wrong, because there should be 3 with one distinct result value (the three "constant" compositions). I quick fixed that and ran again and now got 1674. But given the above, I wouldn't trust this answer too much, except as a lower bound.

But there are definitely a lot more than just a dozen or so. I can post a random selection of them if it would help.

Premium Chessgames Member
  john barleycorn: <beatgiant> yes, the 2 with 1 distict result value surprised me immediately before I read further.

Still, I do not have a clue about a way to determine the number. Probably not many have an idea that why being "associative" is the last property one would give up. It gets messy either by finding a substitute for it or doing all cases one by one. A horror dream.

Premium Chessgames Member
  beatgiant: <john barleycorn> I printed out some more compositions and spot checked them and unfortunately found some violations of associativity. I debugged that and ran again, this time also printing out all the compositions, and got the following:

3 with 1 distinct result value
42 with 2 distinct result values
68 with 3 distinct result values
total: 113

I spot checked a few and did not find any more violations.

I won't have time to debug this any further. But I could post a reasonable subset of the compositions if you want to spot check them yourself.

Premium Chessgames Member
  john barleycorn: <beatgiant> that is great news. I will open my forum where you can post them.
Premium Chessgames Member
  beatgiant: <john barleycorn> Posted them all in your forum.

I can post the python script there too, but the indentation will get mangled and the increment operator += will get converted to . (I used html entities as workaround in the previous sentence.) If you don't mind repairing those yourself, I'll post the script.

Premium Chessgames Member
  john barleycorn: thanks <beatgiant>. however, I do not have the slightest idea what all the computer stuff is about. :-) I grew up with a crayon in my hands.

Please, give me some time to digest your work.

I had some look at the commutative compositions and because of their low number (729) and simple structure (symmetric multiplication table) it should be advantageous to start with them. To determine the amount of commutative plus associative compositions should be a piece of cake compared to the original task

Premium Chessgames Member
  beatgiant: <commutative plus associative compositions> I can try modifying the script to output those too, if it would help.

The only thing I managed to prove about the relationship is the obvious fact that, in associative compositions, every power of an element commutes with every other power of an element: a(aa) = (aa)a (so elements commute with their squares) and (aa)(aaa) = (aaa)(aa) (so squares of elements commute with cubes), etc.

Oct-12-16  nok: I see people are having "fun" here.
Premium Chessgames Member
  beatgiant: <john barleycorn> Better sharpen your crayon again; I posted some more data to your forum.
Oct-14-16  sea otter: <Circle packing in a square>

This might be a fun visual challenge. The aim is to pack n unit circles into the smallest possible square. For example, packing five circles in a square would look like:

Just change the number in the link for more examples. It only goes to 20. Here's a link to all 20 solutions:

I've never tried it myself, but it looks fun. Try to get all the way to 25.

Premium Chessgames Member
  al wazir: <sea otter>: Thanks for that link. The solution for 13 is really cool. It reminds me of some Islamic tiling patterns.

It's hard to believe that the solution given for 7 is really optimum.You can use coins of the same denomination or pawns from a chess set to test different arrangements. The obvious first try is to surround one circle with six others. (They fit snugly.) This can be enclosed in a square of size 6.0, which is certainly bigger than 5.732... But is that the smallest square that will enclose them? Suppose you rotate the square slightly. Does that allow it to shrink?

I'm having trouble calculating the dimensions of the square as a function of the rotation angle. Can someone help me?

Premium Chessgames Member
  al wazir: Here's a somewhat related topic.

The flag of the United States has always had 13 stripes alternating red and white, with a blue rectangular field in an upper corner (conventionally portrayed on the left) on which n stars are placed, where n is the number of states in the union at a given time. The 39 historical flags are shown here:

In most cases the stars were arranged in highly symmetrical patterns. For example, the current 50-star pattern is symmetric with respect to 180-degree rotation and inversion about either the vertical or horizontal axis, but not with respect to 90-degree rotations.

Some of the earlier star patterns were more symmetric, but some (e.g., those with 21 and 23 stars) were less symmetric. Can you think of improved designs? Of course those flags were not in use very long, and your designs will never be adopted in the future unless secession takes place, which seems unlikely, so this is not a practical exercise.

But the most symmetric arrangement for 51 stars, in case Puerto Rico or the District of Columbia becomes a state does have a potential application. That could actually happen. (Both have more inhabitants than Vermont and Wyoming.) What about 52, in case both join? Or 53, 54, etc., if some of our other island possessions do.

The idea is to arrange n identical objects in a planar pattern with the highest possible symmetry, for arbitrary n.

Premium Chessgames Member
  al wazir: <sea otter>: I've thought some more about packing seven identical circles in a square box and found the answer to the question I asked.

Start with the arrangement I suggested (one circle in contact with six others surrounding it), enclosed by a square of size 6.0 (three times the diameter of a unit circle), with three of the circles aligned either horizontally or vertically. The square *can* be shrunk by rotating it. The minimum size is reached with a 15-degree rotation. The dimension of the square then is 6.0 cos(15) = 5.79555.... That, however, is bigger than the solution given at the website you linked to, 5.73205...

The arrangement I suggested is indeed the most compact one. It minimizes the space between circles. But the wasted space *outside the circles*, i.e., between the circles and the walls of the box, increases.

Oct-17-16  sea otter: <al wazir>, you may find the solution to N=23 on this link interesting:

It's pretty fun to observe the different grain boundaries when N is large, for example around N=100. The site also has circle packing in a Circle, when hexagonal lattices come into play pretty early sometimes.

Oct-17-16  sea otter: The 51 star problem is especially appealing to me, since 17 is a very nice number imo, and three way symmetry is often very pretty. I'm going to try some of those out later today.
Premium Chessgames Member
  al wazir: <sea otter: you may find the solution to N=23 on this link interesting>. Yes! That's the pattern that should have been used for the flag! But flag manufacturers of the 19th century would have been unhappy with it.

I noticed that the circle-packing solution for N=13 given at the Magdeburg site is *not* isomorphic to the one given at the Wiki site; but in both the size of the square is 2[2+sqrt(3)] = 7.4641... .

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