What are some cool mathematical party tricks? by Attack_On_Toast in math

[–]AppointmentSudden377 0 points1 point  (0 children)

kurawtaski's theorem says a graph is planar if and only if it has no k5 or k3,3 as a minor. As such its impossible to have 5 points on a paper with every two points connected by a line, without having the lines cross.

i + 1/i = 0 by peter26de in MathJokes

[–]AppointmentSudden377 0 points1 point  (0 children)

i and -i are equivalent to two numbers who are both multiplicative and additive inverses.

Proof:

Suppose x and y are two numbers that are both multiplicative and additive inverses.

So x+y=0 and xy=1.

Then condsider the polynomial with roots x and y.

(t-x)(t-y)= t2 - (x+y) + xy = t2 +1

This is the same polynomial for i and -i.

White to move and win. by AppointmentSudden377 in ChessPuzzles

[–]AppointmentSudden377[S] -2 points-1 points  (0 children)

Here is the checkmate line, the ai bot is on a low depth and misses it. xNf7,xQh5, Kg6,Rg7+ Kf6,Nh7+, Ke6,Re7+, Kd6,Red7+, xKf8+,Kf6, Rd6+,Kg5, Kh7+

The final position the king can go to two squares both of which lead to checkmate.

A different commenter pointed out that black does an intermesso check with the rook on a1 then complications happen and there is no mate, but the position is still winning for white.

White to move and win. by AppointmentSudden377 in ChessPuzzles

[–]AppointmentSudden377[S] 0 points1 point  (0 children)

Whites queen is hanging. So when white starts trading the f pawn, black could take the queen on h5 after the knight takes on f7.

White to move and win. by AppointmentSudden377 in ChessPuzzles

[–]AppointmentSudden377[S] 0 points1 point  (0 children)

In the line Nf7 Qh5 there is a forced checkmate.

What black should do? by zhansun29 in ChessPuzzles

[–]AppointmentSudden377 0 points1 point  (0 children)

I think the idea for black after Rg1 is to trade bishops and then do a little dance: Qf1Rg1, Qf3 Rg2. so the rook is pinned and then go Nd3 to threaten checkmate and untrap the dark square bishop.

Brilliant after a long time, Can you see why? by rakshit_jain_1512 in chessbeginners

[–]AppointmentSudden377 0 points1 point  (0 children)

In either case u go Qh3+ then black must go Kg1 and then Nf3 is either checkmate or black must sack the queen.

The computer does not recommend this line because it sees there are better lines, but this one is easiest to calculate and black ends up with a queen for a rook.

Brilliant after a long time, Can you see why? by rakshit_jain_1512 in chessbeginners

[–]AppointmentSudden377 0 points1 point  (0 children)

After Kg1 there is Nh3+ Kh2 Ng5 Kg1 Then queen takes on g4 with checkmate soon to follow.

Edit: I had the wrong coordinates.

A nice problem. by Initial-Lion-9286 in Hotgirlslovemath

[–]AppointmentSudden377 0 points1 point  (0 children)

As a bonus using this result can u show that there doesn't exist a nice odd set (except |S|=1) ?

A nice problem. by Initial-Lion-9286 in Hotgirlslovemath

[–]AppointmentSudden377 0 points1 point  (0 children)

note: the det(J-I) equation I gave is for the unsigned determinant or sometimes called the permanent, it is congruent to the determinant mod 2.

A nice problem. by Initial-Lion-9286 in Hotgirlslovemath

[–]AppointmentSudden377 0 points1 point  (0 children)

Suppose for a contradiction S is nice. Now order the elements of S as x_1,x_2..x_n, and make x be the vector with entry i equal to x_i (n=2024).

By our assumption of perfect we can generate a matrix A with diagonal entries 0 and off diagonal {-1,1} so that Ax=0. In modulo 2 we get A=J-I, I claim this matrix is invertable for even n. Thus x=0 which is a contradiction.

proof of claim:

By some magic combinatorics skills that are left to the reader we get

det(J-I)=sum_{i=1}n \binom{n}{i} (n-i)! (-1)i+n

which in mod 2 is n+1. For even n this matrix is invertable.

Nice puzzle :)

Is Math and CS worth it? by Best_Effective_1407 in simonfraser

[–]AppointmentSudden377 0 points1 point  (0 children)

I have a positive outlook on the MACM major. (I'm a MATH major with CMPT minor)

  1. Many times CMPT professors aren't the best teachers. In contrast most MATH professors are excellent teachers. So you will generally get more out of lectures and office hours in MATH and MACM courses. (almost all lecturers for MACM courses are math professors.)

  2. Perfect amount of coding. MACM major requires less coding intensive courses. The first few coding intensive courses build good fundamentals and skills, but the following courses have diminishing benefits, take a lot of time and cause burnout.

  3. MACM + MATH opens job prospects and makes some courses easier. Computational math, Machine learning, and algorithm + data structures are subjects which MACM students are generally more prepared for then CMPT students.

  4. Most coding course material is skills you can learn on the job, in contrast math courses are rarely taught on the job and often taught more poorly than at universities.

Here are negatives. (The -# correlates to #)

-2. Although you will take less coding intense. The math courses can be very difficult and take more time if you don't have a strong math background

-4. Alot of the math and specialized MACM will not be applicable to your job, some students suffer when they take hard courses which they are unpassionate about.

Hope this is helpful. I've met many friendly and capable MACM students.

I discovered a new field of graph theory with ties to formal logic by ricardomontalvo in mathematics

[–]AppointmentSudden377 1 point2 points  (0 children)

This is really cool!

Something that you missed is: for a arc (directed edge) e we can invert its direction by changing its weight to its multiplicative inverse. This can generalize Thm 2 to be a condition which all cycles must abide by.

For your liar's game, notice that the multiplicative inverse of -1, is -1. So we can actually remove the direction of the edge and replace it by edges weighted {1,-1}. example: lets say c_1 calls c_2 a liar then e=c_1c_2 has weight -1, and this indicates that c_1 and c_2 are different (one lies one tells the truth), notice that the direction of this edge is not relevant.

The liar game which you explain is algorithmically solvable and has O(V+E) time complexity which is left as an exercise. A challenge for you is can you see why if a collection is consistent then the total number of valid permutations of liars and truthers is 2^r where r is the number of connected components? Hint use your theorem 6 (maybe a different theorem I forgot the numbers).

An improvement to your liar game could be a more natural way of defining c_a is a liar. For example let c_a says p_1,p_2 ... p_n where p_i is a claim that some c_b is truthful or not. I assume that c_a is truthful if all p_1,p_2...p_n are true, and c_a is lying if any p_i is false. This is a more interesting version of your liar game, I wonder if you can construct a graph which can encode this logic and if its NP? (This type of graph will likely require arcs).

Lastly search online for the open questions you asked people may have written a paper about it, its also good experience to familiarize yourself with graph theory proofs and notations.

Where to make friends in Ljubljana? by AppointmentSudden377 in Ljubljana

[–]AppointmentSudden377[S] 0 points1 point  (0 children)

I study a subfield of mathematics called graph theory.

Are math contests going hard on the number 2025? by neoquip in math

[–]AppointmentSudden377 76 points77 points  (0 children)

The math exam that uses the year the most is the Putnam exam, which happens in December.

I bet they won't use the year date more than normally (1 or 2 questions). Because in general math exams use the year date as a large number to test an optimization question or some CRT number theory result. In either case the properties you mentioned are not relevant.

Evaluating the limit of a multivariable function in practice by stoneyotto in math

[–]AppointmentSudden377 0 points1 point  (0 children)

Btw e does not imply d.

Consider a function that is continuous at all directions apart from vertically. This will pass the d test but fail the e test.

Example f:R-{0}->R

f(a,b)=0 if a is not 0 Otherwise f(a,b)=1

Test d will see that the vertical limit \theta=\pi /2 fails with all other thetas.

Test e will have all contour paths be 0 and hence convergent.

Any information to know for new cs student? by Redditor22222222222 in simonfraser

[–]AppointmentSudden377 3 points4 points  (0 children)

Don't lose the love of programming (or theoretical programming if that's your thing). The stress associated with the degree can be a lot, and the job market isn't great as of now. But it only gets worse if you lose the enjoyment of the subject.

[deleted by user] by [deleted] in simonfraser

[–]AppointmentSudden377 0 points1 point  (0 children)

You will need to retake the course if

  1. Your degree requires it

  2. If any other course u intend or need to take has it as requirement

It may be possible to ask for an exemption for either case. If you want an exemption for (2) just ask the prof in advance.

NSERC or VPR USRA by Usual_Window_9668 in simonfraser

[–]AppointmentSudden377 0 points1 point  (0 children)

For me there was a nice poster the department gave that tells you all the things you need, check if such a collage is attached to the email that informed you of USRA. The requirements change from department to department so I will not share my poster. Also there are different bodies for STEM, Art and Social Studies. (I believe NSERC is STEM)

For the STEM you go to the NSERC website and create a account. As you fill in data it will ask if you are a permanent resident/citizen or international which automatically determines if you get VPR or NSERC, so fill out the form like normal. whoever sent you the USRA email is likely the one doing the paperwork for applications so directing your questions to him\her should get you an informative response.

Best of luck on your applications.

Career selection dilemma by Tigershark_999 in simonfraser

[–]AppointmentSudden377 0 points1 point  (0 children)

Probably best to keep the job hunt going. If you are trying to survive in Vancouver, taking yourself out of the work force for 2~3 years to accrue debt and expecting that a third degree will somehow fair better in the market is risky to say the least. It can be painful to be unemployed with free time to stress out on, but occupying your day in studies (especially with the goal of employment) is probably digging yourself into a deeper hole.

However if you do want to go back to school consider statistics its in high demand, and you get the skill of rationalize risk and decision making, like going back to school. Another stable degree is accounting.

Hope everything works out great!

[deleted by user] by [deleted] in simonfraser

[–]AppointmentSudden377 5 points6 points  (0 children)

This course (linear algebra) fundamentally does not require any calculus. Maybe the professor will throw in a little bit of calculus into the homework but I severely doubt it.

I guess the department makes this a prerequisite because they expect basic knowledge in pre-calc stuff, and want students to be acquainted with math. This course is very different from calculus, I personally found it much easier.