x=y=z=1; x=y=z=0.5Amber G. wrote:Okay... just for fun for Tuesday..
What are the values of (x,y,z) if:
1/x + y + z = 3
x + 1/y +z = 3
x + y + 1/z = 3
BR Maths Corner-1
Re: BR Maths Corner-1
Re: BR Maths Corner-1
Amber G. wrote:Okay... just for fun for Tuesday..
What are the values of (x,y,z) if:
1/x + y + z = 3
x + 1/y +z = 3
x + y + 1/z = 3
X = 1, Y = 1, Z = 1.
Very Nice. Are there other values? ( ) (Hint: Yes there are )x=y=z=1; x=y=z=0.5
(This problem is from an old contest between two premier institutes tournament - super easy but still fun)
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I did not know about ChatGPT. Looks interesting. Thanks.Prem Kumar wrote:Amber G: I plugged your question into ChatGPT, the latest release by OpenAI, that's got everyone super-excited. This is what it wrote back. You can see that it kind of gets into a "a = a" type of situation. But then eventually figures out at least 1 correct answer
(Just read a little about it)
(FWIW popular math software, available for years like wolfram, Maple etc solve these kind of equations within seconds without any difficulty -
But it will be interesting to know other modern AI systems available - let us know how does this AI do).
*****
FWIW:
Looks like there are some typo/errorsTo solve this system of equations, we can start by multiplying the first equation by x, the second equation by y, and the third equation by z to l.eliminate the fractions. This gives us the equations:
x^2 + y + z = 3x
xy + y^2 + z = 3y
xz + yz + z^2 = 3z
These should be:
1+yx+zx = 3x
xy +1 +yz = 3y
xz+yz+1 = 3x
So you may like to check out...
Actually - 3 equation's and 3 unknown's so equations do *not* have multiple values ... except that each equation is degree 2, so maximum number of solutions are 2*2*2 =8 (counting all permutations and double roots etc).there could be multiple sets of values that satisfy these equations. For example, one possible solution is x = 1, y = 1, and z = 1, which would give us 1/1 + 1 + 1 = 3, 1 + 1/1 + 1 = 3, and 1 + 1 + 1/1 = 3. However, there could be other solutions as well
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Is this a one off curiosity or there is deeper theory lurking behind, Amber G. ji?
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Just for fun on Saturday: Three easy problems for everyone from Stanford Math Tournament:
And a fun looking challenge (Actually this is from a national math olympiad - but not hard if you find the main reasoning).
- No calculators are needed for this one.
BRF: 202212-01
BRF: 202212-02
(Not hard but could be tricky - so be careful )
BRF: 202212-03
Here one has to find only the integer part of the expression. Again, no calculators. (If answer is 47.234.... etc, you need to find 47 only)
BRF 202212 -04 (***)
Finally the math olympiad question... very simple looking!
Give a, b , c, d are real numbers, solve:
And a fun looking challenge (Actually this is from a national math olympiad - but not hard if you find the main reasoning).
- No calculators are needed for this one.
BRF: 202212-01
BRF: 202212-02
(Not hard but could be tricky - so be careful )
BRF: 202212-03
Here one has to find only the integer part of the expression. Again, no calculators. (If answer is 47.234.... etc, you need to find 47 only)
BRF 202212 -04 (***)
Finally the math olympiad question... very simple looking!
Give a, b , c, d are real numbers, solve:
Re: BR Maths Corner-1
Wow! Excellent solutions!sudarshan wrote:
(All are correct !)
Only one comment:
For BRF: 202212-02 I said "
(Not hard but could be tricky - so be careful :) ) .. so the clear looking answer is tricky.
Yes, as you said there are four solutions to
(x^-2022)^2 -x - 2022 =0
(And some does come out to be zero - all real solutions) BUT the equation given has: *ONLY* one solution,
namely x =(1 + √8089)/2 (The correct answer, and it is not zero)
Because: f(x) = √a by definition has only one solution 9positive only). for example x^2=4 has two solutions but √4 is *only* +2.
Note:
x≈-45.458 solution is not valid ... (Note that x, if real, has to be positive by definition)
x≈44.458 solution not valid check it out, when you substitute in x≈44.458 you get √(2022 +√(2022 + 44.458)) is NOT 44.458
x≈-44.469 Not valid.
x≈45.469 is the only valid solution.
In any case excellent solution and wonderful write up.
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Sudashanji - I like the way your writing and the way it was posted in brf... Just curious do you use LaTeX.. (would be nice if Brf had interface with LaTeX ....
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^^ Thanks!
With the first problem - I used to have a very hard time working with "a^2 + ab + b^2" or "a^2 - ab + b^2" situations. It's not quite (a+b)^2, nor (a-b)^2. Then I realized that it works much better if you go with the cubes: a^3 - b^3 or: a^3 + b^3.
The second problem took the longest, and I thought the trick was that you have to be aware that there are negative solutions. I guess it's a convention in math that the sqrt sign implies only the positive square root? All four roots were also very easy to identify using Excel, so I could verify that the sum was zero, but I'm not sure how to actually solve a quartic equation analytically.
The third problem was a standard bounds estimation, which occurs rather frequently in engg. scenarios.
The fourth one - I just fortuitously noticed that the cube terms could be canceled out, if one considered all the equations as an aggregate.
I've used LaTeX, but the material above was actually generated using MS Word equation editor. You can see the red underlines, where Word over-enthusiastically corrects the "typos" in the text. Then I just did a screen grab and posted the resulting image. Otherwise, it's a mega pain, trying to put all that math stuff in a text message.
With the first problem - I used to have a very hard time working with "a^2 + ab + b^2" or "a^2 - ab + b^2" situations. It's not quite (a+b)^2, nor (a-b)^2. Then I realized that it works much better if you go with the cubes: a^3 - b^3 or: a^3 + b^3.
The second problem took the longest, and I thought the trick was that you have to be aware that there are negative solutions. I guess it's a convention in math that the sqrt sign implies only the positive square root? All four roots were also very easy to identify using Excel, so I could verify that the sum was zero, but I'm not sure how to actually solve a quartic equation analytically.
The third problem was a standard bounds estimation, which occurs rather frequently in engg. scenarios.
The fourth one - I just fortuitously noticed that the cube terms could be canceled out, if one considered all the equations as an aggregate.
I've used LaTeX, but the material above was actually generated using MS Word equation editor. You can see the red underlines, where Word over-enthusiastically corrects the "typos" in the text. Then I just did a screen grab and posted the resulting image. Otherwise, it's a mega pain, trying to put all that math stuff in a text message.
Re: BR Maths Corner-1
Okay one more for this Sunday...
(Question is from a well known contest - It is not hard (if one hits upon the right idea, kind of simple) and kind of cute)
One can assume all x's are real numbers.
(Question is from a well known contest - It is not hard (if one hits upon the right idea, kind of simple) and kind of cute)
One can assume all x's are real numbers.
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sire my shAshTaanga pranaamsudarshan wrote:[img]...https://live.staticflickr.com/65535/525 ... 4e41_o.png[/img]
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^ Mine too, in response .
The problem is non-linear because of the 1/x2, 1/x3... terms, but if it were linear, then it would lead to a "sparse matrix," which is much easier to solve (diagonalize) than a regular matrix. But even in the non-linear case, the same principle applies - eliminate one variable at a time (having only two variables at a time in any equation makes this pretty easy). So after eliminating x2, x3, x4,... etc. in succession, with the final equation, it ends up relating the first variable to itself, which is pretty easy to deal with.
The problem is non-linear because of the 1/x2, 1/x3... terms, but if it were linear, then it would lead to a "sparse matrix," which is much easier to solve (diagonalize) than a regular matrix. But even in the non-linear case, the same principle applies - eliminate one variable at a time (having only two variables at a time in any equation makes this pretty easy). So after eliminating x2, x3, x4,... etc. in succession, with the final equation, it ends up relating the first variable to itself, which is pretty easy to deal with.
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Sudashanji - Excellent. Aryabhatta (also Brahmgupta) would be proud.
(The problem, this or similar, IIRC appeared once in Putnam Competition !)
(Continued fractions is one of my favorite techniques which is not that much taught in formal courses, but I like to teach ..).
The technique - if x = f(f(x) or (x = f(f(f....(f(x) for that matter then f(x) = x is a solution. Quite intuitive but deep. Sometimes, it is the only solution.
The same theme was in my Q BRF: 202212-02.
One can reduce the equation x = √(2022+√(2022+x)) by the same logic to x = √(2022+x) resulting to simple x = (1+√(4*2022+1))/2 ..(Taking *only* one of the (positive) root of the resulting quadrating equation).
(may be in some next year SMT IMO or JEE one would see the problem like x = √(2023+√(2023+(√2023+x))).. (This will return in 6 degree equation but still only one valid solution)
****
(The problem, this or similar, IIRC appeared once in Putnam Competition !)
(Continued fractions is one of my favorite techniques which is not that much taught in formal courses, but I like to teach ..).
The technique - if x = f(f(x) or (x = f(f(f....(f(x) for that matter then f(x) = x is a solution. Quite intuitive but deep. Sometimes, it is the only solution.
The same theme was in my Q BRF: 202212-02.
One can reduce the equation x = √(2022+√(2022+x)) by the same logic to x = √(2022+x) resulting to simple x = (1+√(4*2022+1))/2 ..(Taking *only* one of the (positive) root of the resulting quadrating equation).
(may be in some next year SMT IMO or JEE one would see the problem like x = √(2023+√(2023+(√2023+x))).. (This will return in 6 degree equation but still only one valid solution)
****
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Okay... here is one solution to problem posted above
Along with Sudarshan's solution above, let me post it here... (Since BRF does not have LaTeX interface, I too am posting screenshots :
QED!
Along with Sudarshan's solution above, let me post it here... (Since BRF does not have LaTeX interface, I too am posting screenshots :
QED!
Re: BR Maths Corner-1
Amber ji, this is a great solution, but doesn't it assume that x1/x2, x2/x3, etc. are all positive? Otherwise the sqrt in the geometric mean will not yield real numbers. It's true that the sqrt terms all cancel out, but are complex GMs defined? If so, is there a good definition of >= for complex numbers?Amber G. wrote:Okay... here is one solution to problem posted above
Along with Sudarshan's solution above, let me post it here... (Since BRF does not have LaTeX interface, I too am posting screenshots :
[img]...[/img]
QED!
Re: BR Maths Corner-1
^^^^ Very good point..(When I gave this problem to some students, I made it a little easier and explicitly said x1,x2.. all are positive real numbers.. In BRF I was careful to point it out that assume all *are* *real*)
--- But a moments thought, one can 'enhance' the proof to include *all* real numbers. For complex numbers, a little more work is needed.
(but can be done). (For complex numbers, one can use the polar form (r*e^(i theta) with r as real and >0)
The trick is to introduce |x_1| which is always positive and then √(|x1|/|x2|) can be well defined. and AM-GM can be used. One then introduce that |x| >= x (does not matter x is positive or not. Yes it requires some more work (and needs to be shown if problem solution is to be graded ...
--- But a moments thought, one can 'enhance' the proof to include *all* real numbers. For complex numbers, a little more work is needed.
(but can be done). (For complex numbers, one can use the polar form (r*e^(i theta) with r as real and >0)
The trick is to introduce |x_1| which is always positive and then √(|x1|/|x2|) can be well defined. and AM-GM can be used. One then introduce that |x| >= x (does not matter x is positive or not. Yes it requires some more work (and needs to be shown if problem solution is to be graded ...
Last edited by Amber G. on 13 Dec 2022 07:04, edited 1 time in total.
Re: BR Maths Corner-1
This is very interesting, I'll have to try this out.The technique - if x = f(f(x) or (x = f(f(f....(f(x) for that matter then f(x) = x is a solution. Quite intuitive but deep. Sometimes, it is the only solution.
The same theme was in my Q BRF: 202212-02.
One can reduce the equation x = √(2022+√(2022+x)) by the same logic to x = √(2022+x) resulting to simple x = (1+√(4*2022+1))/2 ..(Taking *only* one of the (positive) root of the resulting quadrating equation).
Also thanks for the clarification on the complex number thing.
Re: BR Maths Corner-1
Just got from Prof Sachdeva ( DR Subir Sachdev - Renowned prof of SYK fame.Prem Kumar wrote:Amber G: I plugged your question into ChatGPT, the latest release by OpenAI, that's got everyone super-excited. This is what it wrote back. You can see that it kind of gets into a "a = a" type of situation. But then eventually figures out at least 1 correct answer
Just impressive in one sense..Asked ChatGPT: write a sonnet on the SYK model:
In a world of chaotic chaos,
There lies a model, simple yet profound,
The SYK model, with its symmetry unbound,
Describes a world in which all is lost.
Its quenched randomness doth make it so,
That chaos reigns in all its glory,
But still this model doth hold the key,
To unlocking secrets of the unknown.
Its beauty lies in its simplicity,
And yet within its depths so much is found,
From quantum mechanics to black holes,
The SYK model doth truly astound.
In this world of chaos and confusion,
The SYK model doth provide a solution.
Re: BR Maths Corner-1
Thanks. I looked up ChatGPT and got some very interesting feedback from friends...We have come a long way from early AI of my era (Eliza of 1970's, 80's)...Prem Kumar wrote:Amber G: I plugged your question into ChatGPT, the latest release by OpenAI, that's got everyone super-excited. This is what it wrote back. You can see that it kind of gets into a "a = a" type of situation. But then eventually figures out at least 1 correct answer
Wrt ChatGPT, IMO, - Large language models like this can spout nonsense in a confident, authoritative tone. This overconfidence - which reflects the data they’re trained on - makes them more likely to mislead.
While real experts know when to sound confident, and when to let others know they’re at the boundaries of their knowledge. Experts know, and can describe, the boundaries of what they know.
Key is to build a large language models that can accurately decide when to be confident and when not to will reduce their risk of misinformation and build trust.
--- Pretty obvious, the output 'solution' put by chatGPT reeks of impressive 'math language' - even one correct solution(which most people will get in 15 seconds anyway) - but quite non-sense mumbo-jumbo and full of math-errors, as described.
*
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Along with x= y= z = 1ricky_v wrote:x=y=z=1; x=y=z=0.5Amber G. wrote:Okay... just for fun for Tuesday..
What are the values of (x,y,z) if:
1/x + y + z = 3
x + 1/y +z = 3
x + y + 1/z = 3
and x = y = z = 1/2
we also have [x= -1/3, y = z = 3] and symmetric [y= -1/3, z=x=3], [z= -1/3, x=y=3]
-
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Re: BR Maths Corner-1
Agree Amber G. Unlike say Google, which gives you references and assumes no authority itself, ChatGPT, because its a text generator/summarizer, gives an authoritative answer, which could be very wrong. Its still in Beta. They cannot release it in the wild and open themselves to lawsuits without adequate legal protection.
But we can imagine where this tech will be, 5 years from now. We are in the exponential part of the hockey stick
But we can imagine where this tech will be, 5 years from now. We are in the exponential part of the hockey stick
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^^^ Thanks for mentioning ChatCPT. Some of the AI tools for Math does fairly good job for these easy type problems - even some IMO level problems writing proofs. Of course, solving the problem posted here are quite easy for any standard popular tools (like maple, mathcad etc).
(Unlike chatCPT, from what I saw they do not write poems about STK model )
Anyway what would be nice - what I try to teach these bright students are concept like: given
1/x + y + z = 3 (Eq 1)
x + 1/y +z = 3 (Eq 2)
x + y + 1/z = 3 (Eq 3)
-- Notice symmetry (any permutation of solution will also be a solution)
-- Case A - if all variables are equal
==> 1/x+x+x = 3 ==> 1+2x^2 -3x = (2x-1)(x-1)=0 ==> x= 1 or 1/2 (Thus x=y=z =1 or x=y=z = 1/2)]
Case B - at least one pair is not equal (say x≠y) (Remember symmetry so we can use x and y WLOG)
==> Eq 1 - Eq2 ==> (y-x) + (1/x-1/y) = 0 or (y-x)+(y-x)/xy = (y-x)(1+xy)=0 since (x-y) ≠ 0 ==> 1+xy = 0 or y = -1/x
===> Eq 2 now gives 1/x-1/x +z= 3 ==> z=3
Eq 3 Now gives x - 1/x + 1/3 = 3 or x-1/x = 3-1/3 ==> x = 3 or x = -1/3 ==> Solution (3,-1/3,3) or any of the permutation.
Solutions (x,y,z) = (1,1,1), ot (1/2, 1/2 , 1/2) or (-1/3,3,3) or (3,-1/3,3) or (3,3,-1/3)
(Unlike chatCPT, from what I saw they do not write poems about STK model )
Anyway what would be nice - what I try to teach these bright students are concept like: given
1/x + y + z = 3 (Eq 1)
x + 1/y +z = 3 (Eq 2)
x + y + 1/z = 3 (Eq 3)
-- Notice symmetry (any permutation of solution will also be a solution)
-- Case A - if all variables are equal
==> 1/x+x+x = 3 ==> 1+2x^2 -3x = (2x-1)(x-1)=0 ==> x= 1 or 1/2 (Thus x=y=z =1 or x=y=z = 1/2)]
Case B - at least one pair is not equal (say x≠y) (Remember symmetry so we can use x and y WLOG)
==> Eq 1 - Eq2 ==> (y-x) + (1/x-1/y) = 0 or (y-x)+(y-x)/xy = (y-x)(1+xy)=0 since (x-y) ≠ 0 ==> 1+xy = 0 or y = -1/x
===> Eq 2 now gives 1/x-1/x +z= 3 ==> z=3
Eq 3 Now gives x - 1/x + 1/3 = 3 or x-1/x = 3-1/3 ==> x = 3 or x = -1/3 ==> Solution (3,-1/3,3) or any of the permutation.
Solutions (x,y,z) = (1,1,1), ot (1/2, 1/2 , 1/2) or (-1/3,3,3) or (3,-1/3,3) or (3,3,-1/3)
Re: BR Maths Corner-1
On the 135th anniversary.... 0<a<b+1/2
Happy Birthday Ramanujan!
Happy Birthday Ramanujan!
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They say.. this ChatGPT is incredible ..
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Doh. It is still two hours.
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Interesting q & a with ChatGPT. While the response from ChatGPT is a howler, some things to consider, as below (just out of academic interest ):
1. Many people also reason this way, and have to learn by experience that such reasoning is incorrect (one can surmise that maybe, the AI picked up on this kind of faulty *human* reasoning, from trawling through the internet)
2. The AI is an evolving entity, and can also learn from "experience" (in this case - the experience comes from the programmers initially, beyond that - it's anybody's guess, how well the AI is going to evolve on its own)
3. The answer to the problem: that 8 cars will also take the same 2 hours as 4 cars - does this imply that the answer is always 2 hours, irrespective of the number of cars?
Not quite. The answer is only valid, so long as one is not constrained by the finiteness of the road itself.
For example - how about 8 million cars? Can they all fit on the road? If not, the number of cars which can fit, depends on the width and length of the road itself. Beyond that limit, and going to more and more cars (eventually - infinite cars) - the travel time indeed becomes linear with the number of cars! Because only so many cars can fit on the road, traveling at a finite speed, and one set of cars has to get off the road before the next set can even get on (assuming infinite parking lots on either side, with no delay in getting on and off).
So the solution to the problem starts off with a constant time, regardless of the number of cars, then goes through an intermediate stage, before hitting a linear asymptote.
1. Many people also reason this way, and have to learn by experience that such reasoning is incorrect (one can surmise that maybe, the AI picked up on this kind of faulty *human* reasoning, from trawling through the internet)
2. The AI is an evolving entity, and can also learn from "experience" (in this case - the experience comes from the programmers initially, beyond that - it's anybody's guess, how well the AI is going to evolve on its own)
3. The answer to the problem: that 8 cars will also take the same 2 hours as 4 cars - does this imply that the answer is always 2 hours, irrespective of the number of cars?
Not quite. The answer is only valid, so long as one is not constrained by the finiteness of the road itself.
For example - how about 8 million cars? Can they all fit on the road? If not, the number of cars which can fit, depends on the width and length of the road itself. Beyond that limit, and going to more and more cars (eventually - infinite cars) - the travel time indeed becomes linear with the number of cars! Because only so many cars can fit on the road, traveling at a finite speed, and one set of cars has to get off the road before the next set can even get on (assuming infinite parking lots on either side, with no delay in getting on and off).
So the solution to the problem starts off with a constant time, regardless of the number of cars, then goes through an intermediate stage, before hitting a linear asymptote.
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Happy New Year!
2023 = (2+0+2+3)(2²+0²+2²+3²)²
How cool is this!
2023 is a Harshad Number. In mathematics, a harshad number in a given number base is an integer that is divisible by the sum of its digits when written in that base. Harshad numbers in base n are also known as n-harshad (or n-Niven) numbers. Harshad (Happy) numbers were defined by Prof D. R. Kaprekar.
Interestringly Ramanujan's Taxi Cab number 1729 is a famous harshad number too.
BTW, Very rare but 2022,2023,2024, and 2025 are all harshad numbers --- something not going to happen in near future again!
Happy New Year.
2023 = (2+0+2+3)(2²+0²+2²+3²)²
How cool is this!
2023 is a Harshad Number. In mathematics, a harshad number in a given number base is an integer that is divisible by the sum of its digits when written in that base. Harshad numbers in base n are also known as n-harshad (or n-Niven) numbers. Harshad (Happy) numbers were defined by Prof D. R. Kaprekar.
Interestringly Ramanujan's Taxi Cab number 1729 is a famous harshad number too.
BTW, Very rare but 2022,2023,2024, and 2025 are all harshad numbers --- something not going to happen in near future again!
Happy New Year.
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Meanwhile .. my kind of PM: "This is personal for me", Rishi Sunak
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Dear Guru log,
One of my nephews published this as part of his Phd.
https://eccc.weizmann.ac.il/report/2022/038/
Thank you.
One of my nephews published this as part of his Phd.
https://eccc.weizmann.ac.il/report/2022/038/
While I'm proud of him, this is way above anything I can understand. Can anyone interested take a look and tell me what such work can be applied on to solve what kind of real life problems?Abstract:
For every prime p > 0, every n > 0 and \kappa = O(logn), we show the existence of an unsatisfiable system of polynomial equations over O(n log n) variables of degree O(log n) such that any Polynomial Calculus refutation over F_p with M extension variables, each depending on at most \kappa original variables requires size exp(\Omega(n2/(\kappa^2*2^\kappa(M+nlog(n))))) .
Thank you.
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Is his advisor Russel Impagliazzo? Wow. That is as high as it can get in Computational Complexity Theory and Theoretical CS in general. Impagliazzo's advisor Manuel Blum was Turing Award winner and several of his students won Turing awards (including Shaffi Godlwaaser - in cryptography). I had the good fortune to hear Prof. Blum talk when he was invited to address our graduate seminar class.
Your nephew must be crypto researcher then. F_p is a Prime field (a finite field over a prime p) (or notated as or a Galois field GF(p) (usually p is used for prime in Number Theory).
Another wow. Pitassi is Stephen Cook's student. Stephen Cook showed that 3SAT is NP-complete. That is the first problem to be shown to be NP-complete. Then Richard Karp showed the method of reductions to show more problems to be NP-complete.
Your nephew must be crypto researcher then. F_p is a Prime field (a finite field over a prime p) (or notated as or a Galois field GF(p) (usually p is used for prime in Number Theory).
Another wow. Pitassi is Stephen Cook's student. Stephen Cook showed that 3SAT is NP-complete. That is the first problem to be shown to be NP-complete. Then Richard Karp showed the method of reductions to show more problems to be NP-complete.
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As for real-life problems, new (and possibly faster) encryption methods as well as cryptanalysis, malliable/non-malliable cryptography, digital signatures, ... But why bother about applications when they are actually pushing the envelope of computational complexity theory?!
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@Cyrano gaaru (and anyone else following this thread), proof complexity has applications in program proving.
A program (== any program) calculates and outputs a function value on its input(s). But how does one know whether the program calculated the function correctly?
Can the program self-check? If it can, can it output a certificate of its own correctness? The program can always output itself as a certificate of its correctness. Then we have to check the program for correctness.
Question: Can the program output a proof that is shorter than the program itself?
If so, we need to check a short(er) certificate (shorter than the program) of the program correctness.
That is the importance of proof complexity.
A program (== any program) calculates and outputs a function value on its input(s). But how does one know whether the program calculated the function correctly?
Can the program self-check? If it can, can it output a certificate of its own correctness? The program can always output itself as a certificate of its correctness. Then we have to check the program for correctness.
Question: Can the program output a proof that is shorter than the program itself?
If so, we need to check a short(er) certificate (shorter than the program) of the program correctness.
That is the importance of proof complexity.
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Thank you for the lumière Vayutuvan garu. Yes, my nephew is also the son of the man who wowed the world in March 2019 and a lot more for the triad, nuff said.
He wants to pursue a year or two of post doc at SUPSI Lugano now, and perhaps come back to a teaching/research position at one of the IITs after that.
He wants to pursue a year or two of post doc at SUPSI Lugano now, and perhaps come back to a teaching/research position at one of the IITs after that.
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What is the full name of Mouli Senior?
Re: BR Maths Corner-1
Small world in some sense. Few tidbits...Cyrano wrote:Thank you for the lumière Vayutuvan garu. Yes, my nephew is also the son of the man who wowed the world in March 2019 and a lot more for the triad, nuff said.
He wants to pursue a year or two of post doc at SUPSI Lugano now, and perhaps come back to a teaching/research position at one of the IITs after that.
- I am a physicist - but am familiar with the work and interest in the field (for last 20 years or so) - mainly because of the work of Manandra Agrawal (along with Neeraj Kayal, and Nitin Saxena) who became a well-known people in the world with there Primes in P work. This was, among other things, a front page story in NYTimes.. and we hosted Prof. Agrawal in US tour.. Of course there are *MANY* items in this math dhaga for example <here> (do a search - there are probably 20+ posts over the last 15-20 years)... you (and your nephew will enjoy this).
- Russell Impagliazzo & Toniann Pitassi - I have seen a few papers (they have worked with Prof Agrawal .
- Beauty of some of this work is - it is understandable (and thus enjoyable) by even non-experts - lot of this could be understood by even who are familiar with few courses in Number Theory.. (This is why I have discussed some of interesting points in brf dhaga--
- Both Prof Agrawal and Prof Vidyasagar (Who are in IIT's as engineering departments but are first-rate mathematicians too) are involved (leaders) of SUTRA model - about which I have talked quite often in other dhagas.
- Your nephew's paper talks about "Algebraic system Polynomial Calculus(PC),
... a dynamic generalization of Nullstellensatz. Interestingly Prof Vidyasagar's handle in Twitter is nullstellensatz... How cool is that..
- Any relation with Prof Satapathy?
Last edited by Amber G. on 07 Jan 2023 06:15, edited 1 time in total.
Re: BR Maths Corner-1
Mouli Sr (lol reminds me of a poem from Sumathi shatakamu) - heads the labs now, he is my BiL
Well since it's anyway in public domain
https://www.drdo.gov.in/labs-establishm ... atory-drdl
Well since it's anyway in public domain
https://www.drdo.gov.in/labs-establishm ... atory-drdl
Re: BR Maths Corner-1
Lo ! And this nugget pops up in my YT feed !
Can't keep pi out of unusual places!
Can't keep pi out of unusual places!
Re: BR Maths Corner-1
^All the best to you and your family...
You may already know this, but IIT Kanpur has very nice group ..(in computational complexity theory as well as pure mathematics) which your nephew may like to look at. https://iitk.ac.in/new/manindra-agrawal
You may already know this, but IIT Kanpur has very nice group ..(in computational complexity theory as well as pure mathematics) which your nephew may like to look at. https://iitk.ac.in/new/manindra-agrawal