August 2020 - mathfun@mailman.xmission.com

Re: [math-fun] 'in-band' signaling in voting systems
by Keith F. Lynch 16 Aug '20

16 Aug '20

Henry Baker <hbaker1(a)pipeline.com> wrote: > I.e., in order to properly test the voting network, it may be > necessary to inject a *known* number of 'false' votes into the > network, and subsequently subtract these false votes from the > totals. How would that help confirm that the votes are getting through, unless the total number of votes for candidate X were less than the number of test votes for candidate X? Perhaps test votes should be for a non-existent candidate. Hypothetical hacked software can't accept those votes while rejecting votes for a real candidate, unless the name of one or both candidates appears somewhere in the code, which it shouldn't. Of course an encrypted version of one or both names could be hidden in the code. A better approach is to use the live votes as tracers. Every voter is encouraged to make a note of their ballot serial number. The whole database of who everyone voted for is made publicly available online, except for the link between the voter's name and the ballot serial number. Each voter could look up his own ballot and see that it's correct. There are two downsides to that. One downside is shared with absentee voting -- the voter could have been coerced or rewarded to vote in a specific way. The advantage of in-person voting is that that's not possible, at least not unless the coercer or rewarder hooks the voter up to a reliable lie detector. The other downside is that the voter could falsely claim that his vote was missing or wrong. Also, this doesn't address the opposite problem, ballot-box stuffing. > Clearly, there are great dangers in a poorly implemented voting > system with 'in-band' signaling. However, there are also great > dangers in *not testing* a voting system for link failures and > active hacking attacks. Re hacking, who says voting has to rely on computers? See https://xkcd.com/2030/ Of course non-computerized voting systems are also subject to hacking. But it's harder to automate, and harder to hide, just as before computers spies could steal secret information, but couldn't steal millions of pages of it at once. > Has anyone else considered these kinds of problems? Of course. For instance it's a perennial topic on the Risks Digest, which has been running for 35 years, making it perhaps the oldest still extant email list. Note that even if there's an easy and foolproof solution, that doesn't mean that it will be implemented if those in charge prefer the flaws in the current system. For instance they've been simultaneously making government-issued ID harder to get, and required for voting. Here in Virginia, the ID requirement was recently abolished, but the online portal for requesting an absentee ballot won't let you submit the request unless you provide the serial number on your current ID. Analogously, there's an obvious solution to the many forensic lab scandals: Double-blind testing. Never tell the techs what samples the prosecutors want to match what other samples. But since prosecutors run the system, that will never happen. Some labs are actually paid per conviction, which is an enormous conflict of interest. And some are overworked to the point where it's physically impossible to actually run all the tests.

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[math-fun] Geometry puzzle
by Dan Asimov 16 Aug '20

16 Aug '20

Consider the cubic 3-dimensional torus T^3 = R^3/Z^3, a unit cube with opposite faces identified. Now connect each of 4 alternate corners of the cube, say {(1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1)}, to the center (1/2,1/2,1/2) of the cube by a line segment. Now we want to shorten this union of 4 segments with these moves: 1) Any vertex (endpoint of a segment) may be moved continuously, while keeping the same connectivity (segments that connect 2 vertices will still connect them even if they are moved); and 2) Any vertex v may split in two to become 2 new vertices v', v''. v' and v'' are assumed to be connected by a line segment of length 0. When that happens, one or more segments that had v as an endpoint (but not all of them) are assigned to have v' as endpoint, and the remaining segments are assigned to have v'' as endpoint. 3) The reverse of 2): any 2 vertices may (by 1) coalesce to form a new vertex. Puzzle: Using only these moves, what is the shortest possible network that the original network of 4 segments can be transformed to? —Dan

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[math-fun] Sage on iMac ?
by Fred Lunnon 16 Aug '20

16 Aug '20

I'm currently attempting to install SageMath on my iMac, and would appreciate hearing (directly please) from anyone who has done so: what version of OSX, what version of Sage, and what outcome? WFL

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Re: [math-fun] nth digit * nth term
by Éric Angelini 16 Aug '20

16 Aug '20

Many thanks Hans — and beautiful illustration! I will submit this sequence together with Kris Katterjohn in a couple of hours and link it to your blue spikes, if you don’t mind. Best, É. Catapulté de mon aPhone > Le 16 août 2020 à 14:39, Hans Havermann <gladhobo(a)bell.net> a écrit : > > EA: "We start S with a(1) = 2 and always extend S with the product « nth digit x nth term of S ». (When the product is = 0, we don’t extend S with 0 but with the smallest integer not yet present in S.)" > > See my graph of S to 10^5 here: > > http://chesswanks.com/num/nthDigit*nthTerm.png > > Large values in S are created by lengthy non-zero stretches in the digit list, appearing in the graph as spikes that culminate in a peak when the digit is again zero. Based only on the digit list for my calculated S, the peak at a(28385) can be predicted to be superseded by a slightly larger peak at a(350165). > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] nth digit * nth term
by Éric Angelini 16 Aug '20

16 Aug '20

Hello Math-fun, We start S with a(1) = 2 and always extend S with the product « nth digit x nth term of S ». (When the product is = 0, we don’t extend S with 0 but with the smallest integer not yet present in S.) S = 2, 4, 16, 16, 96, 96, 576, 5184, 31104, 279 936,... We see that 2 x 2 = 4 = a(2) 4 x 4 = 16 = a(3) 1 x 16 = 16 = a(4) 6 x 16 = 96 = a(5) 1 x 96 = 96 = a(6) 6 x 96 = 576 = a(7) 9 x 576 = 5184 = a(8) 6 x 5184 = 31104 = a(9) 9 x 31104 = 279 936 = a(10) 6 x 279 936 = 1 679 616 = a(11) 5 x 1679 616 = ... etc. The first column of the above array is the succession of the digits of S; the second column is the succession of the terms of S. Question: what is the biggest of the first 100 terms of S? Best, É. à+ É. Catapulté de mon aPhone

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[math-fun] 'in-band' signaling in voting systems
by Henry Baker 15 Aug '20

15 Aug '20

A voting system is a communications system with some extremely weird characteristics. Nevertheless, this communication system still shares problems with more traditional communications systems. The simplest problem is ensuring that the system is 'live' -- i.e., that it actually is carrying information. Also, voting systems are highly targeted by hackers, so standard problems like 'replay attacks' must be solved. When Akamai (or any other large-scale communications company) wants to map out the internet, it sends packets, sometimes in pairs, in order to detect connectivity, latency, throughput, error rates. These packets aren't 'payload', in the sense that Akamai doesn't get *paid* for transmitting these packets, but it needs to ascertain the characteristics of the various communications links in order to better plan for actual paying traffic. The voting system also desperately needs to map out its network, and make sure that it is working, even in the presence of 'noise', out of order delivery, random delays and/or active *jamming*. Due to the weird nature of the voting network, however, these 'signaling' packets need to be distinguished from 'regular' packets (payloads). But due to the nature of detecting hacking, jamming and other 'active attacks', it may be necessary to make the signaling packets indistinguishable from payload packets as they are crossing the network, and the only knowledge of which packets are which is found in the testing logic itself. I.e., in order to properly test the voting network, it may be necessary to inject a *known* number of 'false' votes into the network, and subsequently subtract these false votes from the totals. Such 'false' votes can also be used to better protect voters' privacy in precincts where there are only a handful of voters. In these precincts, the 'false' votes can temporarily drown out the real votes, but these 'false' votes will later be subtracted from the totals for the 'real' vote. Clearly, there are great dangers in a poorly implemented voting system with 'in-band' signaling. However, there are also great dangers in *not testing* a voting system for link failures and active hacking attacks. Has anyone else considered these kinds of problems?

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[math-fun] More spirals and 2 x 2 square squares
by Éric Angelini 15 Aug '20

15 Aug '20

Hello Math-fun, We've started a spiral of distinct integers with 0 in the middle of the plane, then 1, then 2. At this point the next integer must be 6 as we want the content of EVERY 2 x 2 SQUARE of the plane to sum up to a square (0+1+2+6=9). When we have a choice, we always extend the spiral with the smallest unused term. Thus here, after 6, the term 3. This 3 forces us to fill the next cell with 7 (indeed 0+6+3+7=16) and not 8 (the sum 17 is not a square) or 40 (though 0+6+3+40=49 is also a square) -- remember, we must always extend the spiral with the smallest unused integer. This means that the only constraint on the next integer (after 7) is to be the smallest unused one, thus 4. Now to complete the next 2 x 2 square we must use 5 (as 0+7+4+5=16), etc. +---+---+---+---+---+---+---+---+---+---+ | | | | | | | | | | | +---+---+---+---+---+---+---+---+---+---+ | | | | | | | | | | | +---+---+---+---+---+---+---+---+---+---+ | | |13 |38 |34 |15 |31 | A | | | +---+---+---+---+---+---+---+---+---+---+ | | |26 | 4 | 5 |10 | 8 | B | | | +---+---+---+---+---+---+---+---+---+---+ | | |12 | 7 | 0 | 1 |17 | C | | | +---+---+---+---+---+---+---+---+---+---+ | | |14 | 3 | 6 | 2 |16 | . | | | +---+---+---+---+---+---+---+---+---+---+ | | |11 |21 |19 |22 | 9 | . | | | +---+---+---+---+---+---+---+---+---+---+ | | | | | | | . | . | | | +---+---+---+---+---+---+---+---+---+---+ | | | | | | | | | | | +---+---+---+---+---+---+---+---+---+---+ This spiral leads to the lexicographically earliest sequence of distinct positive terms 0, 1, 2, 6, 3, 7, 4, 5, 10, 8, 17, 16, 9, 22, 19, 21, 11, 14, 12, 26, 13, 38, 34, 15, 31, A, B, C, ... which is not in the OEIS and perhaps of general interest (caveat: handmade, as usual). (Our next move after dinner, soccer on TV, sleep, breakfast? Each 2 x 2 square sums up to a prime, of course). Best, É.

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[math-fun] Rearranging digits
by James Propp 15 Aug '20

15 Aug '20

This tweet from @AlgebraFact gave me a good way to get my brain working this morning: “Suppose n is a power of 2. Then you cannot rearrange the base 10 digits of n to form a different power of 2.” What about Fibonacci numbers? Jim Propp

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[math-fun] Sum of consecutive terms = shortcut
by Éric Angelini 15 Aug '20

15 Aug '20

Hello Math-fun, could I recommend to our members who like erratic behaviors to have a look at http://oeis.org/A336897 Best, É. Catapulté de mon aPhone

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[math-fun] intriguing partial result
by Bill Gosper 15 Aug '20

15 Aug '20

4 (Re[(1-I) hilbert[EllipticTheta[4, 1/2]]]) | (2 + √2) EllipticTheta[4,1/√2] - (√2 - 2) EllipticTheta[3,1/√2] /.R : EllipticTheta[__] :> Round[R, 2^-691] // N[#, 69] & During evaluation of In[679]:= EllipticTheta::argtu: EllipticTheta called with 1 argument; 2 or 3 arguments are expected. {1.78030371851856017082096982647118260720864907614581765107300156501703} | 1.78030371851856017082096982647118260720864907614581765107300156501703 (The extra braces are to allow hilbert to be multivalued.) Kids: The pattern EllipticTheta[___] means EllipticTheta of any number of arguments. So why is it complaining? And how do I shut it up? (I won't say if nobody asks!) Kids: True or false: Im[x]==Re[I*x] —rwg

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