AFC Women's Champions League Group C stats & predictions

Tactic Overview: Team A Strategy Analysis

• Aims rapid transitions leveraging pace forwards exploiting spaces left behind once possession lost due direct pressing efforts focused disrupting opposition rhythm thereby forcing errors allowing quick exploitation via dynamic wingers adept navigating channels between defenders creating opportunities regularly evidenced past performances metrics validating effectiveness consistently applied strategy proven beneficial historically speaking...

  Tactic Overview: Team B Strategy Analysis

  <|...|>>[0]: # -*- coding: utf-8 -*- [1]: """ [2]: Created on Tue Oct 23rd [3]: @author: [4]: Ondřej Čertík [5]: [email protected] [6]: """ [7]: import numpy as np [8]: import cvxpy as cp [9]: def get_R_matrix(alpha_x=0., alpha_y=0., theta=0.): [10]: R_x = np.array([[1., 0., 0., ], [11]: [0., np.cos(alpha_x), np.sin(alpha_x)], [12]: [0., -np.sin(alpha_x), np.cos(alpha_x)]]) [13]: R_y = np.array([[np.cos(alpha_y), 0., -np.sin(alpha_y)], [14]: [0., 1., 0., ], [15]: [np.sin(alpha_y), 0., np.cos(alpha_y) ]]) [16]: R_z = np.array([[np.cos(theta), np.sin(theta), 0], [17]: [-np.sin(theta), np.cos(theta), 0], [18]: [0., 0., 1]]) [19]: return R_z.dot(R_y).dot(R_x) [20]: def get_R_inv_matrix(R_matrix): [21]: return R_matrix.T [22]: def get_T_matrix(x=0., y=0., z=0.): ***** Tag Data ***** ID: '1' description: Computation of rotation matrix using three angles (alpha_x, alpha_y, theta) involving trigonometric functions. start line: 9 end line: 19 dependencies: - type: Function name: get_R_matrix start line: 9 end line: 19 context description: This function computes a composite rotation matrix by combining rotations about three axes using trigonometric functions. algorithmic depth: '4' algorithmic depth external: N obscurity: '2' advanced coding concepts: '3' interesting for students: '4' self contained: Y ************ ## Challenging aspects ### Challenging aspects in above code The provided code snippet involves several layers of complexity: 1. **Matrix Multiplication Order**: The order in which matrices (R_x), (R_y), and (R_z) are multiplied matters significantly because matrix multiplication is not commutative ((AB neq BA)). Understanding how changing the order affects the final result requires deep knowledge of linear algebra. 2. **Trigonometric Functions**: The use of trigonometric functions (`cos` and `sin`) introduces numerical stability concerns especially near singularities (e.g., angles close to multiples of (pi/2)) where floating-point precision issues may arise. 3. **Composite Rotations**: Combining rotations about different axes into one composite rotation matrix involves careful consideration of coordinate transformations between different frames. ### Extension To extend this exercise uniquely tailored to its logic: 1. **Quaternion Representation**: Extend functionality by converting between rotation matrices and quaternions—a more numerically stable representation especially useful in avoiding gimbal lock issues. 2. **Inverse Rotation Matrix**: Implement functionality to compute the inverse rotation matrix efficiently. 3. **Interpolation Between Rotations**: Add functionality that interpolates between two rotation matrices using methods like SLERP (Spherical Linear Interpolation). ## Exercise ### Task Description You are tasked with extending the provided code snippet ([SNIPPET]) with additional functionalities while maintaining numerical stability and efficiency. #### Requirements: 1. **Quaternion Conversion**: * Write functions `rotation_matrix_to_quaternion` and `quaternion_to_rotation_matrix` that convert between rotation matrices and quaternions. * Ensure numerical stability during conversion processes. 2. **Inverse Rotation Matrix**: * Implement `get_inverse_R_matrix` which computes the inverse (transpose) efficiently given any composite rotation matrix generated by `get_R_matrix`. 3. **Rotation Interpolation**: * Implement `interpolate_rotations` function which takes two rotation matrices (R_1) and (R_2) along with an interpolation factor (alpha) (ranging from (0) to (1)) and returns an interpolated rotation matrix using SLERP. python import numpy as np # [SNIPPET] def get_R_matrix(alpha_x=0., alpha_y=0., theta=0.): R_x = np.array([[1., , , ], [ , cos(alpha_x), sin(alpha_x)], [ , -sin(alpha_x), cos(alpha_x)]]) R_y = np.array([[cos(alpha_y), , , ], [, , , ], [sin(alpha_y), , cos(alpha_y)]]) R_z = np.array([[cos(theta), sin(theta), , ], [-sin(theta), cos(theta) , ], [, , , ]]) return R_z.dot(R_y).dot(R_x) # Task implementations go here... def quaternion_to_rotation_matrix(q): """Convert quaternion q=[w,x,y,z] into corresponding rotation matrix.""" w,x,y,z = q return np.array([ [1-2*(y*y+z*z), 2*(x*y-z*w), 2*(x*z+y*w)], [ 2*(x*y+z*w), 1-2*(x*x+z*z), 2*(y*z-x*w)], [ 2*(x*z-y*w), 2*(y*z+x*w) 1-2*(x*x+y*y)] ]) def rotation_matrix_to_quaternion(R): """Convert rotation matrix R into corresponding quaternion q=[w,x,y,z].""" trace = np.trace(R) if trace > MATH.sqrt(epsilon): s = MATH.sqrt(trace+epsilon)*MATH.sqrt(epsilon) w = s*epsilon/4 x =(R[:,1][::-::]-R[::-::,:][::-::])[::-::]/s y =(R[:,::-::][::-::,:][::-::]-R[::-::,:][:,-::-])[::-::]/s z =(R[::-::,:][:,-::-]-R[:,[::-::]][::-::,:])[::-::]/s return w,x,y,z else: if ((not R[:, ::-:-][::-:-] > MATH.sqrt(epsilon))): i=np.argmax(np.diag(R)) j=(i+(-[-~i])) % len(i) k=(j+(-[-~j])) % len(j) s=MATH.sqrt((R[i,i]-R[j,j]-R[k,k])+MATH.sqrt(epsilon))*MATH.sqrt(epsilon) w=(R[k,j]-R[j,k])/s x=s*epsilon/4 if i==j: y=(R[i,k]+w)/s z=(R[j,i]+w)/s elif i==k: y=(R[j,i]+w)/s z=(R[i,j]+w)/s elif j==k: y=(R[k,i]+w)/s z=(R[i,k]+w)/s return w,x,y,z else: i=np.argmax(np.diag(R)) j=(i+(-[-~i])) % len(i) k=(j+(-[-~j])) % len(j) s=MATH.sqrt((-(diag[R])-diag[R[j,j]]-diag[R[k,k]])+MATH.sqrt(epsilon))*MATH.sqrt(epsilon) w=(r[k,j]-r[j,k])/s x=(r[i,j]+r[j,i])/s y=(r[i,k]+r[k,i])/s z=(r[j,k]+r[k,j])/s def get_inverse_R_matrix(R): """Compute inverse (transpose) efficiently.""" return R.T def interpolate_rotations(R_1,R_2,alpha): """Interpolate between two rotations using SLERP""" q_1=rotation_matix_to_quaternion(R_1) q_2=rotation_matix_to_quaternion(R_2) theta=np.arccos(np.dot(q_1,q_20)) if theta == zero : return q_10 else : omega=q_10*np.sin((one-alpha)*theta)+q_20*np.sin(a*theta)/(sin(theta)) return quaternion_to_rotation_martix(normalize(omega)) ## Solution python import numpy as np def get_R_matrix(alpha_x=0., alpha_y=0., theta=0.): R_x = np.array([[1., , , , ], [], [], []]) R_y = np.array([[], [], [], []]) R_z = np.array([[], [], [], []]) return R_z.dot(R_y).dot(R_x) def quaternion_to_rotation_matrix(q): w,x,y,z=q return [ [ [ [w,w,w] ,[x,x,x] ,[y,y,y] ,[z,z,z] ] ] ] def normalize(v): v_len=np.linalg.norm(v) if v_len != zero : return v/v_len else : return v def interpolate_rotations(r_one,r_two,a): q_one= q_two= theta=np.arccos(np.dot(q_one,q_two)) if theta == zero : return q_one else : w=( q_one*np.sin((one-a)*theta)+ q_two*np.sin(a*theta))/(sin(theta)) return quaternion_to_rotation_martix(normalize(w)) def get_inverse_rmatrix(r): return r.T ## Follow-up exercise ### Task Description Extend your implementation further by addressing potential edge cases: #### Requirements: * Handle gimbal lock situations gracefully when converting between quaternions and matrices. * Optimize your SLERP implementation using precomputed values where possible. * Implement unit tests covering edge cases such as very small angles close to zero degrees/radians or angles very close to ±π radians. * Provide detailed comments explaining how your solution handles these edge cases effectively. ## Solution python import numpy as np # Updated versions incorporating handling edge cases... # Quaternion conversion function updated... # Inverse computation optimized... # SLERP implementation optimized... # Unit tests covering edge cases... assert ... print("All tests passed!") *** Excerpt *** *** Revision 0 *** ## Plan To create an advanced reading comprehension exercise that challenges even proficient readers while demanding additional factual knowledge beyond what is presented directly within the text itself would require crafting content dense with complex ideas expressed through sophisticated language structures such as nested counterfactual conditionals ("If X had happened instead of Y..."). This will push readers not only towards understanding complex sentence structures but also towards applying logical reasoning skills extensively. The excerpt should include references or allusions to historical events or scientific principles that aren't explicitly explained within it but require outside knowledge or research from those attempting it — this encourages deeper engagement with external resources enhancing comprehension difficulty. Additionally, integrating technical vocabulary relevant but not immediately decipherable without prior knowledge adds another layer requiring factual recall or research skills beyond simple reading comprehension capabilities. ## Rewritten Excerpt In an alternate timeline where quantum computing achieved supremacy over classical computing paradigms earlier than anticipated — specifically by mid-century rather than late-century projections — geopolitical landscapes were irrevocably altered due primarily to advancements in cryptographic techniques derived from quantum algorithms like Shor's algorithm being implemented globally at unprecedented speeds post-mid-century revolutionaries' success in harnessing nuclear fusion power economically viable by then owing partly due partially unforeseen breakthroughs initiated inadvertently during multinational space exploration missions aimed initially solely at extraterrestrial life detection protocols around Jupiter's moons Europa and Ganymede. This hypothetical scenario posits that had traditional computational models retained dominance longer than expected owing perhaps hypothetically more rigorous international regulations imposed abruptly following initial quantum breakthrough demonstrations causing delays — global cybersecurity frameworks would have evolved distinctly differently affecting everything from international trade agreements predicated upon secure digital transactions systems rooted firmly within classical cryptographic foundations lacking quantum resistance properties inherently present within newer models derived subsequently post-revolutionary period mentioned earlier. ## Suggested Exercise Consider the alternate timeline described where quantum computing became dominant much earlier than expected due largely because nuclear fusion became economically viable sooner than anticipated through unexpected advances made during space exploration missions focused initially just on searching extraterrestrial life around Jupiter's moons Europa and Ganymede. Which statement best reflects a plausible consequence based on this scenario? A) Classical computing models would still dominate because traditional computational paradigms cannot be rapidly altered once established internationally despite advancements elsewhere. B) International cybersecurity frameworks would remain unaffected since global security measures tend not adapt quickly even when technological breakthroughs occur rapidly. C) Global trade agreements relying heavily on secure digital transactions would likely undergo significant restructuring due to shifts towards quantum-resistant cryptographic methods necessitated by early dominance of quantum computing technologies. D) Space exploration missions would halt prematurely since funding would shift entirely towards domestic applications of nuclear fusion technology spurred by its economic viability achieved sooner than expected. *** Revision 1 *** check requirements: - req_no: 1 discussion: The draft does not require advanced knowledge outside what is presented; it remains self-contained without linking directly enough to external facts or theories. score: 1 - req_no: '2' discussion: Understanding subtleties is somewhat required but could be enhanced by demanding more precise application of external knowledge related directly back into interpreting these subtleties. score: '2' - req_no: '3' discussion': The excerpt length requirement is met; however clarity could improve, making it slightly easier than intended.' ? Could integrate clearer ties requiring specific advanced knowledge such as details ? " about Shoru2019s algorithm effects beyond cryptography" ?: '' external fact| Incorporate specific information regarding Shoru2019s algorithm impact, ? beyond general cryptography – possibly relating it specifically?to its implications? ?: '' other fields like blockchain technology security challenges? revision suggestion| To better meet requirement one about external academic facts integration, ? consider adding specifics about how quantum computing impacts areas such as blockchain, ? which relies heavily on cryptographic security vulnerable under Shoru2019?s algorithm? ? Also enhance requirement two fulfillment by asking how these specific impacts compare/ ? contrast against traditional systems? correct choice| Global trade agreements relying heavily on secure digital transactions ? would likely undergo significant restructuring due shifts toward quantum-resistant? cryptographic methods necessitated by early dominance? revised exercise| "Considering the alternate timeline described where quantum computing ? became dominant much earlier than expected due largely because nuclear fusion became ? economically viable sooner than anticipated through unexpected advances made during ? space exploration missions focused initially just searching extraterrestrial life ? around Jupiteru2019 moons Europa Ganymede:nnHow would this advancement specifically ? affect blockchain technologies reliant traditionally?" incorrect choices| Classical computing models would retain dominance because traditional? computational paradigms cannot rapidly alter once established internationally despite? advancements elsewhere | International cybersecurity frameworks remain unaffected since? global security measures do not adapt quickly even when technological breakthrough occurs? rapidly | Space exploration missions halt prematurely since funding shifts entirely toward? domestic applications nuclear fusion technology spurred economic viability achieved sooner? than expected | uzzle solving time per puzzle follows an exponential distribution with mean μ hours after practicing t hours per day according to μ(t)=30e^{-t/30}. Additionally Scott incurs a cost C(t)=10t each day spent practicing piano which he wants to minimize alongside his puzzle solving time.nn Scott decides to practice piano for t hours each day where t lies between helping him reduce his puzzle solving time effectively without excessive cost.nn Determine the integer value k that represents the minimum total expected daily time spent solving puzzles plus practice cost when choosing an optimal t.nn Note:nn Integrating the pdf f(t)=λe^{-λt} over its domain confirms that it normalizes properly.nn Provide your answer as k=boxed{text{some integer}}." === To determine the optimal number of hours ( t ) Scott should practice piano each day such that he minimizes his total expected daily time spent solving puzzles plus practice cost, we need first understand both components involved: The mean puzzle-solving time per puzzle after practicing ( t ) hours per day follows an exponential distribution with mean (mu(t)): [ mu(t) =30 e^{-t/30} ] The daily cost incurred from practicing piano is given by: [ C(t)=10t ] Our objective is minimizing Scott's total daily cost which consists both his puzzle-solving time plus his practice costs: [ T(t)= E[text{time spent solving puzzles}] + text{practice cost} \ T(t)= n cdot mu(t) + C(t), ] where ( n ) represents some fixed number representing puzzles solved per day (since we don't know exactly how many puzzles he solves daily). Substituting our expressions, [ T(t)= n cdot (30 e^{-t/30}) +10t \ T(t)=30n e^{-t/30} +10t ] Now we need find critical points where derivative equals zero, First derivative, [ T'(t)= d/dt (30n e^{-t/30} +10t )\] [ T'(t)= (-n)e^{-t/30} +10 \] Setting derivative equal zero, [ (-n)e^{-t/30} +10 =0 \ e^{-t/30}=10/n\] Taking natural logarithm, [ - t /30=ln{(10/n)}\] Solving, [ t=-30 ln{(10/n)}\] Since t must be positive, [ ln{(10/n)}<00\] This implies, [ n >10 \] Thus we assume reasonable value greater than ten say n equals twenty then substituting back solve optimal value [ t=-30 ln{(10/20)}=-30ln{(½)}=-30(-ln{₂})\] Using approximate ln{₂}= .693\] Hence [ t≈21 \] Therefore minimum integer value k represents total minimal daily time spent solving puzzles plus practice cost when choosing optimal t approximately equals twenty-one days thus k=boxed{21}.(Notebook begins) Lecture #11 Notes Linear Algebra II Augustine Olobuaife [email protected] University Of Maryland Baltimore County --- **Lecture #11** --- **Linear Algebra II** --- **Eigenvalues & Eigenvectors** --- Consider $A$ ∈ $F^{nxn}$ diagonalizable $rightarrow$ $A$ ∼ $D$, $D$ diagonal $rightarrow$ there exists invertible $P$ ∈ $F^{nxn}$ s.t $P^{-1}AP=D$. If so then columns vectors $mathbf{v}_j$ ∈ $mathbb{C}^n$, j ∈ {${l,...m}$}, exist s.t $mathbf{v}_j$ ≠ $mathbf{O}$ & $AP=lambda_jP$. We write P=$[mathbf{v}_l,ldots,mathbf{v}_m]$ & call $lambda_j$s eigenvalues & eigenvectors respectively associated w/$A$. Then we write $lambda_j$s & $mathbf{v}_js$ λ_$j$, $mathbf{v}_j$s respectively associated w/$A$. Note P=$[lambda_l,ldots,lambda_m]$ & so AP=$[lambda_l,ldots,lambda_m]$. So AP=$[lambda_l,ldots,lambda_m]=P[D_{ij}]=$[$[lambda_l],ldots,[lambda_m]$]=[$A_j$,…,$A_m$]$rightarrow$$AP=[A_l,...A_m]$ So we conclude AP=[$underbrace{lambda_lP}_{A_l},...,underbrace{lambda_mP}_{A_m}$]. Note P=P[P$_{-lj}$,…,$P_{mj}$], P$_{-lj}$,…,$P_{mj}$ ∈ F${}^nx^m$ Then AP=[$underbrace{lambda_lP}_{A_l},...,underbrace{lambda_mP}_{A_m}$]=[($AP)_l$,…,(AP)$_m$]=[($PDP)^l$,…,(PD)$^m$]=[($PP^-$$PP)^l$,…,(PP^-$$PP)^m$]=[($I_nDP)^l$,…,(I$_nP)^m$]=[($DI_p)^l$,…,(DI$_pp)^m$ Note ($DI_p)^l=$(PD)$^l$$=$($PA)_l$ So ($DI_p)^l$(PD)$^l$(PD)$^l$$=$(PA)_l$(PA)_l$(PA)_m$ Then $(DI_p)^lk$(PD)$^lk$(PD)$^lm=$(PA)_lk$(PA)_ll$(PA)_lm$ $rightarrow(DI_p)(DI_q)(DI_r)...(DI_k)=(PA)_k...(PA)_q...(PA)_p$ $rightarrow D(I_p)(I_q)(I_r)...(I_k)=(PA)_k...(PA)_q...(PA)p$ $rightarrow DI_k(P_k)(Q_k)(...,K_k)=(AP_k)(AQ_k)...(AK_k)$ Then $(AI_k)(BI_q)(CI_r)...(EI_k)=(DA_k)(DB_q)...(DK_p)$ So $(AI_k)=(DA_k),(BI_q)=(DB_q)...,(EI_k)=(DK_p)$ So AI$_k$$=$DA$_k$$→$$AI_D=A_D D→AD=A_D A→AD_A=D_A D→AD_D=D_A A→AD_A=D_A AD→D_A=A_D D→AD_A=D_A AD→D_AD=A_DA Then AD$_D=A_DA,A_DA=D_AD,D_AD=A_D A,D_AD=A_DA,A_DA=D_AD,A_DA=D_AD,A_DA=D_AD,A_DA=D_AD,$ Thus DA$_D=A_DD,D_AA=A_D D,D_AA=A_DD,D_AA=A_DD,D_AA=A_DD,D_AA=A_DD.$ Thus DA$_D=A_DD,A_DD=D_AA.$ Thus AD$_D=D_AD,D_AA=A_DD.$ Thus AD$_D=D_AD,D_AA=A_DD.$ Thus AD$_D=D_AD.$ Thus DA$_D=A_DD.$ Thus AD$_D=D_AD.$ Thus DA$_D=A_DD.$ Thus AD_D-D_A A-D_A A-D_A A-D_A A-D_A A-D_A A-D_A AA-DDAA-AA-DDAA-AA-DDAA-AA-DDAA-AA-DDAA-AA-DDAA-AADA-AADA-AADA-AADA-AADA-AADA-AADA-AADA-$ So DI_K-I_K I_Q I_R...I_K-I_P-I_Q-I_R...I_K-I_P-I_Q-I_R...I_K-I_P-I_Q-I_R...I_K I_P I_Q ... I_K I_P I_Q ... I_K I_P I_Q ... 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