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Louis Stumpers
L Stumpers 
 

Number of games in database: 63
Years covered: 1932 to 1969
Overall record: +14 -35 =14 (33.3%)*
   * Overall winning percentage = (wins+draws/2) / total games.

Repertoire Explorer
Most played openings
D94 Grunfeld (3 games)
B59 Sicilian, Boleslavsky Variation, 7.Nb3 (2 games)
D31 Queen's Gambit Declined (2 games)
D45 Queen's Gambit Declined Semi-Slav (2 games)
E60 King's Indian Defense (2 games)
E21 Nimzo-Indian, Three Knights (2 games)
C65 Ruy Lopez, Berlin Defense (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) Stumpers participated in five Dutch Chess Championships, with his high-water mark a fourth place finish 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 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 Stumpers 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, Stumpers 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) Stumpers 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: 2022-04-04 00:17:13

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 page 1 of 3; games 1-25 of 63  PGN Download
Game  ResultMoves YearEvent/LocaleOpening
1. L Stumpers vs J Lehr 1-0191932EindhovenD18 Queen's Gambit Declined Slav, Dutch
2. L 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 H Kramer  0-1361940HilversumE25 Nimzo-Indian, Samisch
10. L Stumpers vs S Landau  ½-½341940HilversumD31 Queen's Gambit Declined
11. A 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 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 N Cortlever  ½-½501946NED-ch prelim IE60 King's Indian Defense
20. L Stumpers vs H Grob 1-0601947Baarn Group BA55 Old Indian, Main line
21. L Stumpers vs H van Steenis  0-1331947Baarn Group BD23 Queen's Gambit Accepted
22. Tartakower vs L Stumpers 1-0241947Baarn Group BD74 Neo-Grunfeld, 6.cd Nxd5, 7.O-O
23. V Soultanbeieff vs L Stumpers  ½-½461947Baarn Group BD96 Grunfeld, Russian Variation
24. L Stumpers vs A Vinken  0-1331948NED-ch sfE21 Nimzo-Indian, Three Knights
25. L Prins vs L Stumpers  ½-½301948NED-ch sfD02 Queen's Pawn Game
 page 1 of 3; games 1-25 of 63  PGN Download
  REFINE SEARCH:   White wins (1-0) | Black wins (0-1) | Draws (1/2-1/2) | Stumpers wins | Stumpers loses  

Kibitzer's Corner
ARCHIVED POSTS
< Earlier Kibitzing  · PAGE 3 OF 277 ·  Later Kibitzing>
May-29-06  zarra: <themadhair> Pardon my ignorance, but what does it mean that a chord has algebraic coordinates?

<We need to compose a solution that is independant of the way the chords are generated.>

I have always thought that this isn't a problem, since we can always draw the chord first, and the triangle (and the diameter or whatever we wish to draw) only after we have drawn the chord. Thus the other drawings won't affect the chord because the chord was drawn first. But now that I thought about it a little, I'm not at all convinced that the significant points will be evenly distributed if we do it this way.

I'm not an expert on programming, but I've wondered whether it would be possible to write a program that would draw the circle and the triange and a million chords, and measure what part of the chords are longer than the triangle's side. But I suspect that this wouldn't help, since we would have to tell the computer how to draw the chord. If we told it to choose two random points from the circle and connect them, I guess that about 1/3 of the chords would be longer than the triangle's side. If we told it to pick a random midpoint, the result would probably be 1/4, and so on.

May-29-06  EmperorAtahualpa: <offramp> Please post it!
May-29-06  themadhair: <zarra>By an algebraic I mean a number that can be expressed as a solution to a polynomial equation. So fractions and surds are algebraic numbers while numbers like pi are not. The reason I choose algebraic numbers is because they are countable - and hence should not have this paradox under probability theoy.

So if we choose our circle to be the unit circle centred at (0,0) we will only consider algeraic chords. That is chords whose coordinates are given by algebraic numbers. The method <You can draw a vertical diameter to the triangle and choose a random point from the diameter. Then draw the chord through that point perpendicular to the diameter. This approach should lead to the answer 1/2.> can be adapted to consider all algebraic chords if we only consider points on the diameter that are algebraic. The probability calculated should be rotationally invariant as the concentation (bad word choice I admit) of algebraic numbers is also rotationally invariant. So I would guess that the probability of an algebraic chord being of sufficient length is ½.

As for the general case I don't have a clue.

May-29-06  ganstaman: I think this is what offramp was refering to: 3 diners go out to eat. The bill comes to $30, and being good friends they split it evenly. Each diner puts in $10 and the waiter takes it to the register.

The owner of the diner is in a jolly mood, and for whatever reason stops the waiter at the register. He says, "You know, they're good customers. Here, give them $5 off this meal." The waiter says ok and gets $5 from the register. Then he realizes that the diners split the bill, but they wouldn't be able to split $5 evenly 3 ways. Since they don't know they're getting this discount yet, the waiter keeps $2 and only gives the diners $3.

Each diner takes $1 from the 3. Each diner, then, spent $9, for a total of $27. The waiter has $2, for a sum of $2+$27=$29. $29!? We had $30 before. Somewhere, we lost a dollar. Find it.

May-29-06  zarra: <gangstaman> No, look rather this way... Let Y be the difference between the amounts of money in the envelopes. If you switch to the rich envelope, you will win Y, whereas if you switch to the poor envelope, you will lose Y. Thus, it doesn't matter whether you switch or not.
May-29-06  technical draw: You guys sure waste time on problems that were solved 50 years ago. Check out "Prisoners Dilemma"..H. W. Lewis, "Why flip a Coin"....1997...
May-29-06  ganstaman: <zarra> Obviously. I was commenting on your incorrect assertion that Sneaky's math equation was wrong. The equation is fine.

You seem to be mostly right here, though. Half the time you pick the envelope with 2x first, and half the time you pick the one with x first. Therefore, switching has an expected value of 0.5*(x)+0.5*(2x) = 1.5x. Not switching has the same equation and same expected value. Therefore, it doesn't matter which you do.

But, if someone gives you $100, and then says you can trade that in for an envelope that has either $50 or $200, you should trade it in. This is because, as I said earlier, the equation was not incorrect.

May-29-06
Premium Chessgames Member
  offramp: <ganstaman: I think this is what offramp was refering to: 3 diners go out to eat. The bill comes to $30, and being good friends they split it evenly. Each diner puts in $10 and the waiter takes it to the register. The owner of the diner is in a jolly mood, and for whatever reason stops the waiter at the register. He says, "You know, they're good customers. Here, give them $5 off this meal." The waiter says ok and gets $5 from the register. Then he realizes that the diners split the bill, but they wouldn't be able to split $5 evenly 3 ways. Since they don't know they're getting this discount yet, the waiter keeps $2 and only gives the diners $3.

Each diner takes $1 from the 3. Each diner, then, spent $9, for a total of $27. The waiter has $2, for a sum of $2+$27=$29. $29!? We had $30 before. Somewhere, we lost a dollar. Find it.>

That is indeed the problem - and very well presented!

May-29-06  ganstaman: The triangle/circle/chord problem is actually quite simple, as long as I am correct in assuming that the equilateral triangle touches the circle at each of it's 3 points (I forget the math term for this, if it exists).

The question asks for the probability that a chord will be longer than a side of the triangle. Well, it seems to me that finding the length of the sides of the triangle will be useful. I don't know if this is the cleanest method, but it's what I did (sorry if I can't explain this part well in words).

In the Cartesian plane, draw the unit circle (x^2+y^2=1) and the triangle upright, so that it touches at the point (0,1). Now look at the triangle formed by the x-axis, the y-axis, and the right side of the triangle. It's a 30-60-90 triangle, with the 60 degrees at the top, and the left leg (y-axis) has a length of 1 (radius of circle). This means that the bottom leg (x-axis) has a length of 1/sqrt(3), where sqrt(3) is the square root of 3 (this is simply obtained by sin and cos and the like).

So now we have two points along this right side of the equilateral triangle, (0,1) and (1/sqrt(3), 0). From this, we can find an equation for the line containing this side: y=1-sqrt(3)x. This is because slope=(1-0)/(0-1/sqrt(3))=-sqrt(3), and y-intercept=1. This line intercepts the circle at two points, the endpoints of this side of the triangle. We already know one point, (0,1), so let's find the other by plugging this equation into the equation for the circle.

x^2+(1-sqrt(3)x)^2=1 -> x^2+1-2sqrt(3)x+3x^2=1 -> 4x^2-2sqrt(3)x=0 -> x(4x-2sqrt(3))=0.

So, x=0 and sqrt(3)/2. We find the two points to be (0,1) and (sqrt(3)/2, -1/2). The length of this side of the triangle (and all sides of it) is sqrt[(1+1/2)^2+(sqrt(3)/2)^2] = sqrt(9/4+3/4) = sqrt(12/4) = sqrt(3).

The length of a side of the triangle is sqrt(3). How long is a chord, then? well, anywhere from 0 to the diameter of the circle (2 in this case). The probability that a chord will be longer than the side of the triangle should be (2-sqrt(3))/2 = 13.397%, which wasn't one of the choices.

Did I go wrong somewhere?

May-29-06  themadhair: <ganstaman><Did I go wrong somewhere?>No. I've been scouring the internet for the past while and managed to find this (http://www.cut-the-knot.org/bertran...). This is apparently called Bertand's paradox. And apparently all these answers are valid!? If that doesn't make your head spin then what will?

It does appear though that the paradox arises because the set of chords is uncountable, so my previous posts aren't complete crap.

May-29-06  EmperorAtahualpa: <I think this is what offramp was refering to: 3 diners go out to eat. The bill comes to $30, and being good friends they split it evenly. Each diner puts in $10 and the waiter takes it to the register.

The owner of the diner is in a jolly mood, and for whatever reason stops the waiter at the register. He says, "You know, they're good customers. Here, give them $5 off this meal." The waiter says ok and gets $5 from the register. Then he realizes that the diners split the bill, but they wouldn't be able to split $5 evenly 3 ways. Since they don't know they're getting this discount yet, the waiter keeps $2 and only gives the diners $3.

Each diner takes $1 from the 3. Each diner, then, spent $9, for a total of $27. The waiter has $2, for a sum of $2+$27=$29. $29!? We had $30 before. Somewhere, we lost a dollar. Find it.>

LOL, nice problem. I think instead of ADDING 2 to 27, you have to subtract it so you get 25, right? 25 is the cost of the meal, 3 euros goes back to the customers, and 2 to the waiter.

May-29-06
Premium Chessgames Member
  OhioChessFan: <One of the rounds involves three doors - behind one of them is a pot of cash and behind the other two is nothing. You choose a door but you choice is not yet fixed. To spice things up on this gameshow one of the remaining two doors are shown to be empty and you are given the choice to switch. Which is better - to stick or switch? Can you see why they do not offer the same probability of success?>

<themadhair>, I've been thinking about this, and it seems to me you should always switch. If an empty door is always shown, then you are de facto making a 50% choice by changing. If you don't switch, you are remaining with your original 33% chance.

May-29-06
Premium Chessgames Member
  OhioChessFan: When a woman says, "We need to talk", why doesn't she ever mean about football?
May-29-06  blingice: <EmperorAtahualpa: 25 is the cost of the meal, 3 euros goes back to the customers, and 2 to the waiter.>

Comon, <EA>, get on the ball. We're talkin dollars here.

Anyway, my contribution to this solution is this: I used to be quite interested in magic. There was this one trick I picked up (but never used, as it was so lame n' cheesy) where you say "I can mathematically prove that I have 11 fingers." So, you count off your fingers, one at a time: "1, 2, 3, 4, 5, 6, 7, 8, 9, 10...?" You look confused, and say, "I swear I had eleven..." So, you decide to count backwards: "10, 9, 8, 7, 6..." So here, you've gone through your first five fingers, and have five remaining. You then look at the other five and say, "Plus five, equals 11!" [and hold for stunned looks and applause...]

May-29-06  notsodeepthought: <arifattar: <madhair> I think it would 142857 x 5> 142857 is a cute number for another reason - since some kibitzers may not know this, here's why:

142857 * 2 = 285714
142857 * 3 = 428571
142857 * 4 = 571428
142857 * 5 = 714285
142857 * 6 = 857142

In other words, multiplying 142857 by 2, 3, 4, 5 and 6 gives a number that, in each case, contains the same digits as 142857, and in the same order, except (of course) for starting with a different digit in each case.

And, to end with a flourish:

142857 * 7 = 999999

A little trick to impress your friends (or at least those friends of yours that happen to be geeks).

May-29-06  notsodeepthought: <EmperorAtahualpa: LOL, nice problem. I think instead of ADDING 2 to 27, you have to subtract it so you get 25, right? 25 is the cost of the meal, 3 euros goes back to the customers, and 2 to the waiter.> Yep. The question is deliberately misleading - the 2 are part of the 27 that have already been paid by the customers, so they cannot be "re-added" to the 27. As <EA> noted, the correct breakdown is 25+2 (= 27, the total "net" paid by the customers) +3 (change) = 30.

Incidentally, the 2 that the waiter pockets would be a perfectly reasonable tip in many European countries, but in the US it would be considered almost insultingly low (the "standard" is typically 15%, so for a $25 meal, one "should" leave $3.75 or so - assuming decent service of course). European friends of mine on a visit to the US have told me of one experience where, at the end of a meal, they left 10% or so - which they felt was generous - and left the restaurant, only to be chased down in the street by a waitress who apparently was on the verge of tears because of the "meager" tip.

May-30-06  ganstaman: 1/7 = 0.142857142857142857.... which is the repeating pattern 142857.

We are using the digits 1, 2, 4, 5, 7, and 8. 1/7 starts with the 1, 2/7 starts with the 2, 3/7 starts with the 4, etc. So 1/7=0.142857....., and 2/7=0.285714...., 3/7=0.428571...

While this is hardly an explaination, it's in many ways another version of <notsodeepthought>'s math. It took me a little bit to realize this, but I knew that 142857 looked familiar.

May-30-06  zarra: <themadhair> <algebraic numbers etc.> I can understand what you mean, but my knowledge of math isn't good enough to say whether that might be right or not. All I can say is that I don't by any means disagree. Or more simply said, I believe you.

<gangstaman> That's a nice solution. I had known that the side of the triangle is sqrt 3, but it never occured to me that the probability might simply be <(2-sqrt(3))/2 = 13.397%>.

<OhioChessFan> <Three doors problem> In fact you don't make a 50 % choice, but the probability for getting the pot of cash is as much as 67 % if you switch. When you pick a door in the beginning, you pick the pot with a probability of 1/3 and an empty door with a probability of 2/3. After the host opens an empty door, you have two doors remaining. Behind one of them is the cash pot and the other one is empty. But since the door you first chose is empty by a probability of 2/3, then the probability that the cash is behind the other door is the same, 2/3. Therefore you should definitely switch.

<All> Does anyone know what Newcomb's paradox is?

May-30-06  themadhair: <Newcomb's paradox>I have never got the reason that it is a paradox. If the being is who they say they are then you take one box. End of. The arguement that taking both boxes will always give X amount more doesn't hold up if this superior being is really physic. All the second guessing must be irrelevant - if it's physic then it knows what you are going to do. End of. That should open up a can of worms.
May-30-06  technical draw: <zarra> <et al>...The three door problem is a "problem" only because there exists a good probability of guessing which door has the prize at first guess. Change from three doors to a million doors and you will have no chance of guessing at first try. Now keep on opening doors until you only have two left and ask the question..."want to switch"? Unless you are a fool and thought you hit the million to one odds you will switch. The same principle applys with x doors. I learned this when I was 13, (I'm now 57). Nuff said....By the way I'm not psychic but I think you mean "psychic"...
May-30-06  themadhair: <By the way I'm not psychic but I think you mean "psychic">I did. Cheers.
May-30-06  apawnandafool: The paradox of the missing USD reminds me of some of the Tangram paradoxes, where the same set of shapes are rearranged to form different areas. A famous one is the Chessboard paradox, where a chessboard is cut into 4 pieces and rearranged differently to form 65 squares.

Another one is illustrated in the photo link below. A triangle is placed on a square grid, then cut into 4 pieces and rearranged to form a new triangle with a different area.

Your job is to account for the missing "square".

http://www.mathematik.uni-bielefeld...

May-30-06  notsodeepthought: http://www.mathematik.uni-bielefeld... Hint: consider the properties of/requirements for similar triangles...
May-30-06  themadhair: Hint number 2 - <apawnandafool> has told a porky...
May-30-06  apawnandafool: last one.

tangram truth, or fiction:
http://www.hokus-pokus.no/images/im...

tangram printout for above:
http://dimacs.rutgers.edu/nj_math_c...

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