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Championship stats & predictions

Overview of the Ice-Hockey Championship Kazakhstan

The Ice-Hockey Championship Kazakhstan is one of the most anticipated events in the country's sports calendar. As the tournament approaches its climax, fans and enthusiasts are eagerly awaiting tomorrow's matches. With top teams battling for supremacy, the stakes are high, and the excitement is palpable. This article delves into the details of the upcoming matches, offering expert betting predictions and insights to enhance your viewing experience.

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Teams to Watch

Tomorrow's matches feature some of the strongest teams in the league. The Kazakhstani team Barys Astana, known for their strategic gameplay and skilled roster, is set to face off against Torpedo Ust-Kamenogorsk. Both teams have had stellar performances throughout the season, making this a must-watch clash.

Key Players

Several key players are expected to shine in tomorrow's matches. For Barys Astana, forward Alexei Kuznetsov has been a standout performer, consistently delivering crucial goals. On the other side, Torpedo Ust-Kamenogorsk's defenseman Mikhail Petrov is known for his defensive prowess and ability to turn defense into attack.

Match Schedule

  • Barys Astana vs Torpedo Ust-Kamenogorsk - 17:00 Local Time
  • Avtomobilist Yekaterinburg vs Nomad Astana - 19:30 Local Time
  • Saryarka Karaganda vs HK Arystan Temirtau - 21:00 Local Time

Betting Predictions and Insights

Betting enthusiasts have a lot to look forward to with tomorrow's matches. Expert analysts have provided their predictions based on team performance, player form, and historical data.

Barys Astana vs Torpedo Ust-Kamenogorsk

This match is expected to be tightly contested. Analysts predict a close game with Barys Astana having a slight edge due to their home advantage and recent form. The suggested betting odds are:

  • Barys Astana Win: 1.85
  • Torpedo Ust-Kamenogorsk Win: 2.10
  • Draw: 3.50

Avtomobilist Yekaterinburg vs Nomad Astana

Avtomobilist Yekaterinburg is favored to win this match. Their strong offensive lineup and solid defense make them a formidable opponent. Betting odds are:

  • Avtomobilist Yekaterinburg Win: 1.65
  • Nomad Astana Win: 2.40
  • Draw: 3.80

Saryarka Karaganda vs HK Arystan Temirtau

This match is expected to be an exciting encounter with both teams known for their aggressive playstyle. The odds suggest a balanced game:

  • Saryarka Karaganda Win: 2.00
  • HK Arystan Temirtau Win: 1.90
  • Draw: 3.60

Strategic Analysis of Upcoming Matches

Barys Astana's Strategy

Barys Astana will likely focus on maintaining possession and exploiting their speed advantage on the wings. Their strategy revolves around quick transitions from defense to attack, leveraging their forwards' agility.

Torpedo Ust-Kamenogorsk's Counter-Strategy

Torpedo Ust-Kamenogorsk will aim to disrupt Barys Astana's rhythm by employing a tight defensive formation and capitalizing on counter-attacks. Their key players will be crucial in breaking through Barys' defense.

Avtomobilist Yekaterinburg's Offensive Play

Avtomobilist Yekaterinburg is expected to dominate possession and apply constant pressure on Nomad Astana's defense. Their strategy includes high pressing and quick passing sequences to create scoring opportunities.

Nomad Astana's Defensive Tactics

Nomad Astana will focus on a strong defensive setup to neutralize Avtomobilist Yekaterinburg's attacks. They will look for opportunities to launch counter-attacks using their fast forwards.

Historical Context and Trends

Past Performances of Key Teams

Barys Astana has been consistent performers in previous championships, often reaching the finals. Their experience in high-pressure situations gives them an edge over less seasoned teams.

Torpedo Ust-Kamenogorsk has shown significant improvement this season, with a strong defensive record that has kept them competitive against top-tier teams.

Avtomobilist Yekaterinburg has a reputation for being a powerhouse in domestic leagues, with a history of winning multiple championships.

Trends in Betting Markets

Analyzing betting trends reveals that Barys Astana often attracts significant betting interest due to their consistent performance and star players.

Torpedo Ust-Kamenogorsk's underdog status makes them an attractive option for bettors looking for value bets.

Avtomobilist Yekaterinburg's dominance in recent matches has led to heavy betting on their victories, reflecting their strong form.

In-depth Player Analysis

Alexei Kuznetsov (Barys Astana)

Alexei Kuznetsov has been instrumental in Barys Astana's success this season. His ability to read the game and position himself effectively makes him a constant threat in the attacking third.

<|repo_name|>jasonmeng/CS-124<|file_sep|>/hw1/hw1.md # CS124 HW1 This assignment can be found [here](https://docs.google.com/document/d/1Jb8XGONW7ZdIi7qRzb8u6EoZcMk7x5r9wN8eWnI4DfY/edit). ## Question A For this question we were given three vectors `u`, `v`, `w` as well as a matrix `A` as follows: python u = [1., -1., -1., -1., -1., -1., -1., -1., -1., -1.] v = [0., -4., -4., -4., -4., -4., -4., -4., -4., -4.] w = [-5., -4., -7., -5., -7., -6., -6., -6., -7., -6.] A = [ [0,-9,-8,-8,-9,-9,-8,-7,-9,-9], [-9,0,-5,-5,-5,-4,-5,-4,-5,-5], [-8,-5,0,-7,-8,-6,-8,-7,-6,-8], [-8,-5,-7,0,-7,-6,-6,-6,-6,-6], [-9,-5,-8,-7,0,-7,-10, -8, -9, -9], [-9,-4,-6,-6,-7,0, -8, -7, -8, -8], [-8,-5,-8,-6,-10, -8,0, -6, -7, -10], [-7,-4,-7,-6, -8 , -7 ,-6 ,0 ,-5 ,-5], [-9 ,-5 ,-6 ,-6 ,-9 ,-8 ,-7 ,-5 ,0 ,-4], [-9 ,-5 ,-8 ,-6 ,-9 ,-8 ,-10 ,-5 ,-4 ,0] ] ### Part A In this part we were asked to implement `matrix_vector_mult(A,v)` function that takes as input matrix `A` and vector `v` (both are Python lists) and returns `Av` (also as Python list). We were also asked to verify that our implementation was correct by showing that it gives correct results on two test cases. python def matrix_vector_mult(A,v): """ Computes Av where A is an n x n matrix stored as an n x n Python list, v is an n-dimensional vector (represented as Python list), and returns a new vector of dimension n (also represented as Python list). Args: A: An n x n matrix represented as Python list. v: An n-dimensional vector represented as Python list. Returns: The product Av represented as Python list. """ Av = [] for i in range(len(A)): sum = float(0) for j in range(len(A[0])): sum += A[i][j]*v[j] Av.append(sum) return Av We can verify that our implementation works correctly by running it on two test cases: python >>> print(matrix_vector_mult([[2],[2]],[[2],[2]])) [4] [4] >>> print(matrix_vector_mult([[2],[2]],[[2],[2]])) [[4], [4]] ### Part B In this part we were asked to implement `matrix_matrix_mult(A,B)` function that takes two matrices `A`, `B` (both are Python lists) and returns `AB` (also as Python list). We were also asked to verify that our implementation was correct by showing that it gives correct results on two test cases. python def matrix_matrix_mult(A,B): """ Computes AB where A is an m x n matrix stored as an m x n Python list, B is an n x p matrix (represented as Python list), and returns an m x p matrix representing AB (also represented as Python list). Args: A: An m x n matrix represented as Python list. B: An n x p matrix represented as Python list. Returns: The product AB represented as Python list. """ AB = [] for i in range(len(A)): row = [] for j in range(len(B[0])): sum = float(0) for k in range(len(B)): sum += A[i][k]*B[k][j] row.append(sum) AB.append(row) return AB We can verify that our implementation works correctly by running it on two test cases: python >>> print(matrix_matrix_mult([[2],[2]],[[2],[2]])) [[4]] >>> print(matrix_matrix_mult([[2],[2]],[[2],[2]])) [[[4]]] ### Part C In this part we were asked to implement `scalar_mult(c,v)` function that takes scalar `c` and vector `v` (Python list) and returns `cv` (as Python list). We were also asked to verify that our implementation was correct by showing that it gives correct results on two test cases. python def scalar_mult(c,v): """ Computes cv where c is a scalar number stored as floating point number, v is an n-dimensional vector (represented as Python list), and returns cv represented as Python list. Args: c: A scalar number stored as floating point number. v: An n-dimensional vector represented as Python list. Returns: The product cv represented as Python list. """ cv = [] for i in range(len(v)): cv.append(c*v[i]) return cv We can verify that our implementation works correctly by running it on two test cases: python >>> print(scalar_mult(2,[2])) [4] >>> print(scalar_mult(2,[2])) [4] ### Part D In this part we were asked to implement `vector_sum(v,w)` function that takes two vectors `v`, `w` (both are Python lists) of equal length and returns `v + w` (as Python list). We were also asked to verify that our implementation was correct by showing that it gives correct results on two test cases. python def vector_sum(v,w): """ Computes v + w where v,w are both vectors of dimension n (represented as Python lists) of equal length. Args: v : An n-dimensional vector represented as Python list. w : An n-dimensional vector represented as Python list. Returns: The sum v + w represented as Python list. Raises: ValueError if v,w have different lengths. """ if len(v) != len(w): raise ValueError('Different Length') vw_sum = [] for i in range(len(v)): vw_sum.append(v[i] + w[i]) return vw_sum We can verify that our implementation works correctly by running it on two test cases: python >>> print(vector_sum([1], [1])) [2] >>> print(vector_sum([1], [1])) [2] ## Question B For this question we were given a function called `f(x,y)` defined below: python def f(x,y): return ((x+5)**(y+1)) / ((y+5)**(x+1)) and we were asked to evaluate it at all pairs `(x,y)` such that `-100 <= x <=100` and `-100 <= y <=100`. ### Part A To solve this problem I wrote down a nested loop from `-100` up until `100`, then I used the built-in function called **map** which applies function **f** over each pair `(x,y)`. python for i in range(-100 ,101): for j in range(-100 ,101): print(map(f,(i,j))) The output looks like below: ![output](output.png) ### Part B To solve this problem I wrote down another nested loop from `-100` up until `100`, then I used **if** statement inside inner loop so whenever `(x,y)` satisfies equation `(y-x)^{(y+x)} >= f(x,y)` we add it into a **list** called **result**. python result = [] for i in range(-100 ,101): for j in range(-100 ,101): if ((j-i)**(i+j)) >= f(i,j): result.append((i,j)) print(result) The output looks like below: ![output](outputb.png) ### Part C To solve this problem I wrote down another nested loop from `-100` up until `100`, then I used **if** statement inside inner loop so whenever `(x,y)` satisfies equation `(y-x)^{(y+x)} >= f(x,y)` we add it into a **list** called **result**. Then after nested loops finished I calculated sum of all first elements inside each tuple from **result** using built-in function called **sum**. python result = [] for i in range(-100 ,101): for j in range(-100 ,101): if ((j-i)**(i+j)) >= f(i,j): result.append((i,j)) sum_result = sum(list(map(lambda t : t[0], result))) print(sum_result) The output looks like below: ![output](outputc.png) ## Question C In this question we were given some vectors: python u = [1.,-1.,-1.,-1.,-1.,-1.,-1.,-1.,-1.,-1.] v = [0.-4.-4.-4.-4.-4.-4.-4.-4.-4.] w = [-5.-4.-7.-5.-7.-6.-6.-6.-7.-6.] A = [ [0.-9.-8.-8.-9.-9.-8.-7.-9.-9], [-9.0.-5.-5.-5.-4.-5.-4.-5.-5], [-8.-5.0.-7.-8.-6.-8.-7.-6.-8], [-8.-5.-7.0.-7.-6.-6.-6-.06-.06], [-9..-5..-8..-7..0..-7..-10..-8..-9..-9], [-9..-4..-6..-6..-7..0...-8...-7...-8...-8], [-8..-5..-8..-6...-10...-8...0...-6...-7...-10], [-7..-4..-7..-.06....08....07....06....00....05....05], [-9..-.05....06....-.06....-.09....-.08....00....-.04], [-9..-.05....-.08....-.06....-.09....-.08....-.10....00] ] b = [14,.33,.16,.15,.16,.19,.20,.19,.20,.22] c = [-60,.35,.12,.05,.17,.15,.15,.21,.18,.21] and we were asked to solve linear equations using Gaussian elimination method. To do so first I created helper functions called **swap_rows**, **scale_row**, **add_multiple_times_row**, **print_matrix**, **create_identity_matrix** which are self-explanatory from their names. Then I created another helper function called **gaussian_elimination** which performs Gaussian elimination algorithm over matrix passed into it. Afterwards I created another helper function called **back_substitution** which performs back substitution algorithm over matrix passed into it. Lastly I created main function called **solve_linear_equ