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Comment by vjerancrnjak | Hacker News Reader

Yes, it's a question with a silly trick. Completely useless if one knows the answer, hard to spot if the person feigns struggle.

Although, n•(n+1)/2 formula is not necessary. One can start with an xor sum and find the duplicate by xor adding elements again. This is another silly trick.

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That’s a very elegant solution. Hard looking problems that had simple, trick solutions were called Jewish Problems at the entrance of some universities.

This would let the examiner fail a student for subjective reasons instead of academic success. The examiner would then be able to say « I failed them because of their skills, not because of their religions, look how easy the solution is ».

Perhaps this is what we are reproducing as an industry with all our convoluted interview processes, and it may be a decorum to choose the candidates we want instead of using objective criteria.

The explanation of the name somehow is a lot better than what I assumed it was referring to. The phrasing on the second sentence is... unfortunate.

I’m referring to the easy looking, tricky to solve problems. I’ve heard them referred to by that term because they have been used at Moscow University to discriminate against Jewish students but I don’t know if they have other names.

There is some discussions and examples of such problems at https://arxiv.org/pdf/1110.1556.pdf and http://3038.org/press/shen.pdf.

It's interesting that doing competitive math makes it really easy to spot these kinds of problems and guess at their "true" difficulty. Honestly, I'm kind of surprised that "this problem is short" or "the solution is short" would ever fly as a measure of how easy a problem is, but I assume that a system that is intended to encode racism doesn't need a whole lot of logic behind it…

Of course, just the phrasing made me think the "trick solution" was eugenics, not a university admission trick. Unfortunately, historically "Jewish question" or "Jewish problem" has referred to these concerns [0]. Of course, following your first link it seems like the native phrase for it is in no way better, but that seems predictable as it's stemming from a racist premise in the first place.

[0] https://en.wikipedia.org/wiki/Jewish_question

I think you missed the point of the second sentence.

It's like saying the phrasing on the Chinese Exclusion Act is unfortunate. It's not the phrasing that's unfortunate, but the history of excluding Chinese. These questions were designed to allow screeners to discriminate, and were named after the group designed to be kept out.

edit: If you edit a comment in response to a response, it's polite to say [edited]. Otherwise, the conversation is a non-sequitur, which seems to be happening here a bit.

I didn't edit my comment, at least not from what I remember. May have been a typo fix in the first few seconds though, which I do a bit too often.

Not sure if my comment was clear, but I was referring to "hard problems with trick solutions were called Jewish Problems", which sounds like it's referring to the Final Solution.

Reminds me of a funny anecdote, I was staying with a friend of mine in Berlin and asked her what the second most common religion was in Germany. She casually said "used to be Jewish, but not sure now..." She was of course referring to the influx of refugees from Africa, but for a very brief moment it looked like her life flashed in front of her eyes.

Could also just use a hashtable, the keys are the array elements and the items are the occurrence of each element, then return the one with an occurrence of 2.

Or just a hashset and return when about to insert an existing element.

What is the closed form solution of the xor-sum though?

s(n) =

• n if n mod 4 = 0,

• 1 if n mod 4 = 1,

• n + 1 if n mod 4 = 2,

• 0 if n mod 4 = 3.

Or alternatively:

• s(4n) = 4n

• s(4n + 1) = 1

• s(4n + 2) = 4n + 3

• s(4n + 3) = 0

Proof by induction:

• s(4n + 1) = s(4n) ⊕ (4n + 1) = 4n ⊕ (4n + 1) = 1

• s(4n + 2) = s(4n + 1) ⊕ (4n + 2) = 1 ⊕ (4n + 2) = 4n + 3

• s(4n + 3) = s(4n + 2) ⊕ (4n + 3) = (4n + 3) ⊕ (4n + 3) = 0

• s(4n + 4) = s(4n + 3) ⊕ (4n + 4) = 0 ⊕ (4n + 4) = 4n + 4

You already have to spend the time to go through the list. Just add up the xor as you iterate through the list.

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