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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)
D31 Queen's Gambit Declined (2 games)
B59 Sicilian, Boleslavsky Variation, 7.Nb3 (2 games)
E60 King's Indian Defense (2 games)
D45 Queen's Gambit Declined Semi-Slav (2 games)
C65 Ruy Lopez, Berlin Defense (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 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 Academiæ Rheno-Traiectinæ,
8) URSI websites (2016), and

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

This text by User: Tabanus. The photo was taken from

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 D 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 N 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, Nxd5, 7.O-O
23. V Soultanbeieff vs L Stumpers  ½-½461947Int BD96 Grunfeld, Russian Variation
24. L Stumpers vs A Vinken  0-1331948NED-ch sfE21 Nimzo-Indian, Three Knights
25. Prins vs L Stumpers  ½-½301948NED-ch sfD02 Queen's Pawn Game
 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  

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Premium Chessgames Member
  al wazir: <johnlspouge: @<al wazir> re. <nok>'s comments: p - 1, p, and p + 1, but p does not count as it is divisible by neither 2 nor 3.>


Well, it is arguably leaner.

I don't think I need to apologize for my proof. But I'm embarrassed that I failed to grasp the idea underlying <nok>'s.

Premium Chessgames Member
  johnlspouge: My friend's solution follows.

p prime and >3.
p^2 - 1 = (p-1)*(p+1).

p-1, p, p+1 are 3 consequtive numbers so one of them is divisible by 3 and since p is prime , 3 divides p-1 or p+1 and definitely divides their product p^2-1.

p-1 and p+1 are even numbers and so is their product .

With p-1 = 2n and p+1 = 2(n+1) it gives

p^2 - 1 = 2n*2(n+1)= 4 n(n+1) = 24*(n(n+1)/6) =24k with k=n(n+1)/6

...and I don't know whether my friend plays the piano ;>)

Premium Chessgames Member
  al wazir: Consider the following sequence: 0, 1, 3, 6, 10, 15, 21, ...

(i) Does any number in this sequence end in 2, 4, 7, or 9?

Consider the same sequence written in base-16 notation: 0, 1, 3, 6, A, F, 15, ...

(ii) Explain the difference.

Premium Chessgames Member
  al wazir: If you're puzzled by what I meant by (ii) in the post above, examine successive terms of the second sequence. Which of the 16 hexadecimal digits fail to appear in the rightmost place?
Aug-21-21  nok: I have a solution but maybe someone wants to have a go?
Premium Chessgames Member
  al wazir: <nok>: I'm afraid you'll call my solution ugly. What's yours?
Aug-22-21  nok: Nothing too pretty in fact. The general term is 1+2+...+n = n(n+1)/2 so I call f(n) the last digit of n(n+1). We look at the f(n) for n<b (the base). In base 10 the sequence is 02620... then same chunk repeats (reflected) because f(b-n)= f(n-1). No 4 or 8 in base 10 then.

Taking n, n+m < b/2, a digit repeats if f(n) = f(n+m) = f(n) + m(m+2n+1) mod b, so b must divide m(m+2n+1). But this ain't possible if b is a power of 2, because m and m+2n+1 have different parity.

Premium Chessgames Member
  al wazir: <nok>: As usual, your treatment is -- how shall I put it? -- more compact.

Here's mine:

Suppose all numbers are expressed in base m.

Then the rightmost digit R(x) of any integer x is given by

R(x) == x (mod m).

(I write it this way because <> won't let me use the regular equivalence symbol.)

Define S(n) = 1 + 2 + ... + n = n(n+1)/2.

Let a = S(m-1) = m(m-1)/2 and a' = S(m) = m(m+1)/2.

Then a' - a = m, so R(a') = R(a).

Moreover, if

b = S(m-2) = a - (m-1)


b' = S(m+1) = a' + (m+1),

then b' == b (mod m), so

R(b) = R(b').

Similarly, if c = S(m-3) and c' = S(m+2), then R(c) = R(c'), and so on.

In other words, the sequence of the first m rightmost digits repeats in reverse order in the next m rightmost digits.

Hence the rightmost digit of S(2m-2) is the same as the rightmost digit of S(1) = 1, and the rightmost digit of S(2m-1) is 0.

Of course, this follows directly from the fact that S(2m-1) = m(2m-1) is a multiple of m.

Since S(2m) = m(2m+1) is also a multiple of m, its rightmost digit is 0 also. Furthermore, S(2m+1) = S(2m) + 2m+1, so R[S(2m+1)] = 1 =R[S(1)], and S(2m+2) = S(2m+1) + 2m+2, etc., so R[S(2m+2)] = 3 = R[S(2)]. The sequence of rightmost digits keeps repeating endlessly with periodicity 2m. If a digit does not appear in the rightmost place in the first m terms, it will never appear.

Now suppose that the base is a power of 2. Let m = 2^p.

Can two numbers k = S(n) and k' = S(n') have the same rightmost digit?

We can restrict consideration to cases 0 < n < n' < m because of the way the sequence repeats.

The condition that k and k' have the same rightmost digit is k' == k (mod m), or

k' - k = n'(n'+1)/2 - n(n+1)/2 = qm

for some number q. If we set n' = n+d, this becomes

(n+d)(n+d+1) - n(n+1) = 2qm.

This equation can be rewritten as

d(2n+d+1) = 2^(p+1) q.

The left hand side of the equation must be divisible by 2^(p+1) = 2m.

If d is even, then 2n+d+1 is odd, so d must be divisible by 2^(p+1). But d < m by definition, so this is impossible.

If d is odd, then 2n+d+1 = n+n'+1 must be divisible by 2^(p+1). But n+n'+1 < 2m, so this too is impossible. Hence R(k) and R(k') are different.

Therefore all m digits must occur in the rightmost position of the first m terms of the sequence.

But I don't have a simple way of predicting which digits will fail to appear in the rightmost place when m is *not* a power of 2 .

Premium Chessgames Member
  al wazir: A long slow introduction, followed by some short questions.

Suppose you have a chessboard made of some flexible, stretchable material. If you fasten the left edge and the right edge of the board together so that the a-file and h-file are adjacent, a king, for example, could move directly between the two files. (You don't have to spoil a perfectly good vinyl chessboard for this; just imagine that they are adjacent, i.e., hold these relationships in your mind.) Now the chessboard has the topology of a cylinder. A second way to turn the chessboard into a cylinder is to identify the top and bottom edges, so that a king can move directly between the first and eighth ranks.

Now imagine *both* pairs of opposite sides joined, left to right and top to bottom. This turns the chessboard into a torus. It can be done either by identifying the left and right edges first and then the top and bottom, or by identifying the top and bottom edges first and then the left and right. I'll call these two kinds of torus “horizontal” and “vertical,” respectively. In the first kind the files become major circumferences and the ranks become minor circumferences. In the second kind it's the other way around. The two are distinct but completely equivalent topologically.

Now a king on a1 is able to move directly to a8, b8, h1, h2, and h8, as well as the usual moves to a2, b1, and b2. A knight on a1 can move directly to b7, c8, g2, h3, g8, and h7, as well as the usual moves to b3 and c2. But there is nothing special about a1. No square has any property that isn't shared by all the others. None is in a privileged position.

Depending on which way the board is curved in the first step, the playing surface can end up either on the inside or the outside of the torus. So altogether there are four ways to make the topology of the chessboard toroidal: “horizontal inside,” “horizontal outside,” “vertical inside,” and “vertical outside,” or HI, HO, VI, and VO, respectively.

Imagine that some kind of intelligent two-dimensional life form lives on the playing surface of the torus. It can distinguish left and right, forward and backward, but it has no way of perceiving up and down, the third dimension. It understands topology. It therefore knows that it is living on a toroidal surface, but it can no more visualize a three-dimensional torus than I can visualize a hypersphere. Assume that it has thoroughly explored its 2D world. It has traversed all the major and minor circumferences (ranks and files, and the diagonals as well). It has even labeled them so that it can tell where it is (but of course it doesn't use algebraic notation, because that would presuppose a way of telling which are ranks and which are files).

(a) Show, or at least convince yourself, that it can't determine by its own devices which kind of torus it is living on (HI, HO, VI, or VO).

(b) Suppose it somehow learns (let's say, from an alien life form) that it is living on the outside surface. How can it then determine whether it is an HO or VO torus?

(c) Suppose it somehow learns that it is living on a horizontal torus. How can it then determine whether it is on an inside or outside surface?

Premium Chessgames Member
  al wazir: I'll rephrase the questions.

There are two ways to make a circuit that comes back to the starting point, one that encircles the hole in a doughnut (a major circumference) and one that doesn't (a minor circumference). The torus has an inner surface and an outer surface.

(a) Does a 2D being have any way to tell which is the major circumference and which is the minor? Does it have any way to tell whether it is living on the inner surface or the outer?

(b) If it somehow learns which surface it is on, can it then tell which is the major circumference and which is the minor?

(c) If it somehow learns which circumference is major and which is minor, can it then tell which surface it is on?

Premium Chessgames Member
  thegoodanarchist: <al wazir> Here is a little quiz for you. It is in reference to this music video by Weird Al Yankovic:

A couple of times in the video, Weird Al and some other dude I don't recognize are dancing in front of a white screen. The screen has a mathematical equation on it.

Q1) Fill in the blanks:

What is shown is the __________ equation for a __________ .

Q2) (A two-part question) Is there one or more errors in the equation? If so, what is/are the errors?

Posted at 1:34 PM Pacific time.

Premium Chessgames Member
  al wazir: <thegoodanarchist:>: It's the Schroedinger equation for the spatial dependence of the wave function of an electron in the field of a singly charged ion (for hydrogen, a proton), in the center-of-mass coordinate system. (The time-dependence has been factored out. So has the factor describing the motion of the center of mass, which no one ever bothers to write down.)

Since the kinetic energy term is written using the Laplacian, it's three-dimensional. (The angular dependence of the wave function hasn't been factored out.) The argument r of psi, the wave function, should therefore be bold-face to show this.

It's correct as written, but there is a trivial ambiguity. The Laplacian is written in terms of the vector displacement between the proton and the electron, not the laboratory position of the electron, and the r in the denominator is the scalar distance between them.

Premium Chessgames Member
  thegoodanarchist: <al wazir: <thegoodanarchist:>: It's the Schroedinger equation for the spatial dependence of the wave function of an electron in the field of a singly charged ion (for hydrogen, a proton), in the center-of-mass coordinate system... etc. etc.>

I typed a long reply. Then, before posting, I realized although your answer has multiple merits going for it, there are a couple of things that still leave me puzzled and unsure about <aw> the person.

Oh, and unless my eyes or memory deceived me, you missed one error.

I apologize for bothering you this day, especially since my selfish purpose was thwarted.

I wish you all the best.

Premium Chessgames Member
  al wazir: <thegoodanarchist>: What's the one error?

Before posting I did a little googling and discovered that there is widespread confusion about the definition of the center-of-mass coordinate system and the reduced mass mu. That discovery was reflected in my response.

Unfortunately, <> doesn't allow me to use the Greek alphabet or common mathematical symbols. This put a crimp in my style.

Premium Chessgames Member
  al wazir: Enough time has elapsed since I posted that topology problem for anyone to answer who wants to.

The answers to part (a) (as rephrased) are No and No,

At the time I posted I thought that the answers to parts (b) and (c) are Yes and Yes.

Since then I have thought some more and I now think I was wrong. There is no way for 2D individuals living on a toroidal surface to discover which is the minor and which is the major circumference, even if they know whether they are on the inner or outer side, or to know which side they are on even if they know which circumference is which.

But if someone has different answers, and arguments supporting them, I'd like to see them.

Premium Chessgames Member
  thegoodanarchist: <al wazir: ... Before posting I did a little googling and discovered that there is widespread confusion about the definition of the center-of-mass coordinate system and the reduced mass mu. That discovery was reflected in my response.>

I am happy to agree that you were "spot-on" (as the British Soccer commentators say) in regards to the reduced mass. In fact, I trust you on that particular point more than I trust myself.

In any event, I wasn't much interested in that portion of your answer. And I did no math or googling whatsoever to confirm my impression that you're spot-on. Color me lazy or unconcerned. Guilty of both.

<Unfortunately, <> doesn't allow me to use the Greek alphabet or common mathematical symbols. This put a crimp in my style.>

You and me both.

<What's the one error?>

In the kinetic energy term of the Hamiltonian, Planck's constant should be the reduced Planck's constant, aka h-bar. I knew I could be wrong, since I am having to remember 18 years ago (the last time I solved any quantum mechanics problems). So I checked two trusted sources, and they validated my recollection.

Or maybe one of us just needs a new prescription for eye glasses? Doubtful, because I am almost certain the equation contains 'h' and not 'h-bar'.

Premium Chessgames Member
  thegoodanarchist: <aw> If you're interested, here's one on-line source that confirms it:

And I am sure my grad school QM textbook would confirm it as well, but unfortunately I donated it to a library in 2018, and then moved 2 thousand miles away from the library.

Premium Chessgames Member
  al wazir: <thegoodanarchist: <What's the one error?>

In the kinetic energy term of the Hamiltonian, Planck's constant should be the reduced Planck's constant, aka h-bar.>

You're right, of course.

(I'll have to take your word for it that the equation shown had h in place of h-bar because I'm not going to watch that stupid video a second time, though my handwritten copy has h-bar. I scribbled the equation on a piece of paper rather than use a screen shot. My bad.)

Premium Chessgames Member
  al wazir: There is a word in English that is pronounced the same way with four letters removed: queue.

Now find another one.

(Here's a helpful hint: You don't have to look far.)

Premium Chessgames Member
  al wazir: No, not "queues." It's not that close.
Premium Chessgames Member
  johnlspouge: α β γ δ ε
Premium Chessgames Member
  johnlspouge: HTML also contains math symbols.

[ ]

Premium Chessgames Member
  johnlspouge: aitch
Premium Chessgames Member
  al wazir: <johnlspouge: aitch> Excellent!

Did you make use of my clue? Not only is 'h' fairly close to 'q' in the alphabet, the preceding thread in this forum (just a few posts up on the page) dealt with various forms of 'h' in quantum mechanics.

Premium Chessgames Member
  johnlspouge: Use all the data ;-)

Your discussion about the radial Schrodinger equation may have suggested h to my unconscious, particularly after I started thinking about letters of the alphabet because of your hint.

I was impressed by your ability to recognize the radial equation so specifically, so h or h-bar may have stuck in my mind.

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