Challenges

The first student enrolled in Discrete Mathematics 2024 who sends a sophisticated proof to submit@discmath.ch will receive chocolate in the next exercise session (indicate which exercise session you will attend, you can obsviously try any challenge you like :)) - be honest and don't use external tools :). Before sending a mail, please check that this challenge is unsolved.

Week 5

deployment 21.10.2024 18:32

πŸ”“

SOLVED by nwessbecher, 00:20h after deployment (solution file):

Equivalence Relation on Vector SpacesSource:Β  Tobias Steinbrecher

Let VV be a real vector space, and define a relation ∼\sim on VV as follows: For u,v∈Vu, v \in V, we say u∼vu \sim v if and only if uβˆ’v∈Wu - v \in W, where WW is a subspace of VV. Prove that this relation is an equivalence relation.

Week 6

deployment 28.10.2024 18:17

πŸ”“

SOLVED by leoschmidt, 3d 05:41h after deployment (solution file):

Two be or not two beDifficulty: 5/5Relevance: πŸͺ©Source:Β  Yannick Funke, Max Obreiter

Prove the following statement:

∣N/ ⁣≑2∣=2 \left| \mathbb N / \! \equiv_2 \right| = 2

You are NOT allowed to use:

  • Proof by Induction
  • Results from Chapter 4 and 5 unless explicitely allowed

Assume the following:

  • The universe is U=ZU = \mathbb Z
  • The binary functions β‹…,+,βˆ’\cdot, +, - and the unary function βˆ’- are defined as usual. You can also use the properties of these functions (e.g. distributivity, etc.).
  • The predicates ≀\le and == are defined as usual. You can also use the properties of the combination of these predicates and the functions (e.g. the order axioms).
  • (N;≀)(\mathbb N; \le) is well-ordered
  • a≑2b⟺defβˆƒk(2k=aβˆ’b)a \equiv_2 b \overset{\small def}{\Longleftrightarrow} \exists k (2k = a - b)

This challenge is very difficult, so don't worry if you need to look at the hints!

There are multiple ways to prove this statement, the hint/answer show the intended way.

πŸ”“

SOLVED by leoschmidt, 3d 05:41h after deployment (solution file):

Counting the UncountableDifficulty: 3/5Relevance: πŸŽ“πŸŽ“πŸŽ“Source:Β  Yannick Funke

Let BB be an arbitrary but fixed set. We then define QQ as:

Q={(A,f)∈(P(B)Γ—BB)β€…β€Šβˆ£β€…β€Šf∈AA∧Aβͺ―N∧A≁N} Q = \left\{ (A, f) \in \left( \mathcal P(B) \times B^B \right) \; \vert \; f \in A^A \land A \preceq \mathbb N \land A \not\sim \mathbb N \right\}

(If this "arbitrary" BB looks scary, you can assume that QQ contains all elements of the form (A,f∈AA)(A, f \in A^A) where AA is finite. Note that this set is not defined though.)

We define ∼\sim on QQ as:

(A1,f1)∼(A2,f2)β€…β€ŠβŸΊβ€…β€ŠdefΒ thereΒ existsΒ aΒ bijectionΒ g:A1β†’A2,Β suchΒ that:Β f1=gβˆ’1∘f2∘g (A_1, f_1) \sim (A_2, f_2) \overset{\text{\small def}}{\iff}\text{ there exists a bijection $g: A_1 \to A_2$, such that: }f_1 = g^{-1} \circ f_2 \circ g

Prove that QQ is an equivalence relation (for a fixed BB). Then prove that (Q/∼)(Q/\sim) is countable (for a fixed BB).

πŸ”“

SOLVED by leoschmidt, 3d 04:33h after deployment (solution file):

Linear CombinationsDifficulty: 2/5Relevance: πŸŽ“πŸŽ“πŸŽ“Source:Β  Tobias Steinbrecher

Prove that the set of all finite rational linear combinations of elements in a countably infinite set S={s1,s2,s3,… }S = \{s_1, s_2, s_3, \dots\} is countable.

Recall the definition of the set of rational linear combinations of some vectors SS:

lin(S)={βˆ‘i=1nΞ»isi∣n∈N∧[βˆ€β€…β€Š1≀i≀n:β€…β€Šsi∈S∧λi∈Q]} \text{lin}(S) = \left\{\left. \sum_{i = 1}^n \lambda_i s_i \right| n \in \mathbb N \land \left[ \forall \; 1 \le i \le n: \; s_i \in S \land \lambda_i \in \mathbb Q \right] \right\}

Week 8

deployment 11.11.2024 19:50

πŸ”“

SOLVED by leoschmidt, 21:05h after deployment (solution file):

Isoproductphism?!

Prove or disprove the following assertion. Let G,HG, H and KK be groups. If GΓ—K≃HΓ—KG \times K \simeq H \times K, then G≃HG \simeq H.

Week 10

deployment 26.11.2024 20:39

πŸ”“

SOLVED by habdel, 1d 19:19h after deployment (solution file)
(Bonus) SOLVED by okovacs, 41d 11:28h after deployment (using sage math)

Ueli's Security AgencySource:Β  chopingu, Tobias Steinbrecher

Alice and Bob are top operatives of rival intelligence agencies. Despite their rivalry, a crisis has forced them to collaborate. They want to securely communicate using their only shared secret mm, a critical code that unlocks a ton of sensitive information essential to global catastrophe. You know the two public keys of the two organizations aa and bb. Because they are rivals, aa and bb are obviously coprime. Both Alice and Bob already know mm, but they need to ensure its secrecy during the exchange. Unfortunately, years of rivalry and agency-specific training have left Alice and Bob with incompatible cryptographic practices. Their instructions mistakenly consist of elements of RSA and Diffie-Hellman protocols, leading to a flawed scheme, which makes it possible for you to intercept A=Rp(ma)A = R_p(m^a) and B=Rp(mb)B = R_p(m^b) for some big prime pp also known to you (such that m<p<min⁑(ma,mb)m < p < \min(m^a, m^b)). A third party, USA (Ueli's Security Agency), hires you. How could you recover mm?

Bonus:

For training purposes, your boss provided you with the following values to practice before working with the actual data:

  • A=0x185ea966031f220259b35d26f7f3aa059427f91ae95fd51d4fcce086645ebd0f64cf0bedb51a5c2118b7620c7725cdcc4d7bbeb36362c546956ef7ccdee566aff98fdae92791ab7d5b1dfd7633fded925c586db9363d615ed0304ff93d688e9f1e40260b62ef32e0e220c7f20cf12d481663965a67d8d03f99ca66a158f3c311

  • B=0xafd19f1f20b823ca16cbea684b892240289feaac141615c859698fe5b22a6653a48acf68675e130466d1ba3faeaf8d5d57c03bd3ece06f7ab803aaafe1e30062ac9ebd532589e590345cf22bf564d539d092ffb5cdfde4149118796b4517186aac3b8e6868c49903b14bc7021927f963b784c2bdf2b7f7f674a9bb7b75bba3cd

  • a=0xab75c71b2577e217c7ab5d04a7342e4539ca07bd33c0dd55639fd3df593a38ef612114799b1da377fada3748f47960aead03edd0c6c9aaa936c5b35bbff9b76d

  • b=0xea186540a997c53185faeebf9a2861432a1e03445b772ae1a554a556fb8304ca06260874d792ff2ce5474670ce670478ac91174092f88ddc51e2464865a04279

  • p=0xbabb7eb5c93c2c584f69efd5755aaea9b602c16e8fa469c67182dc0b873bb53a287f00bf970873acc43caa2260782c4a5caa60be176e88601c7dbc3fe467c10af91524f9ef0d6db540c0c6f8d87c272f9a90fa2c4668694105ec02216d730c78f832bf2d1ccbe4162036c04deb99b7f2aface73cf80fc9051bff4ae875c2a377

Is there any information in here?

πŸ”“

SOLVED by leoschmidt, 4d 20:40h after deployment (solution file):

Chinese Lagrange InterpolationSource:Β  Tobias Steinbrecher

There is a deep connection between Lagrange Interpolation and the Chinese Remainder Theorem. Find this connection, i.e. formulate a theorem on a more general level (list all the necessary conditions for a Ring to satisfy, for this Theorem to hold) and show how it gives rise to Lagrange Interpolation and the Chinese Remainder Theorem as special cases.

The idea of challenges is inspired by aellison (opens in a new tab).