Laver Cup stats & predictions

Rafael Nadal

• Vittorie nei tornei dello Slam: 20 (record)
• Punti totalizzati nelle ultime sei settimane: +5000 (classifica ATP)
• Punti percentuali nei duelli contro Medvedev: Vinte il 60% delle partite disputate

Daniil Medvedev

• Vittorie nei tornei dello Slam: Australian Open (2021)
• Punti totalizzati nelle ultime sei settimane: +4500 (classifica ATP)
• Punti percentuali nei duelli contro Nadal: Perde il 60% delle partite disputate

Roger Federer

• Vittorie nei tornei dello Slam: Record conosciuto (20 titoli)
• Punti totalizzati nelle ultime sei settimane: +3000 (classifica ATP)
• Punti percentuali nei duelli contro Zverev: Vinte l'80% delle partite disputate

Alexander Zverev

• Vittorie nei tornei dello Slam: US Open (2020)
• Punti totalizzati nelle ultime sei settimane: +4000 (classifica ATP)
• Punti percentuali nei duelli contro Federer: Perde l'80% delle partite disputate

Dominic Thiem

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• Vittorie nei tornei dello Slam: US Open (2020)Punti totalizzati nelle ultime sei settimane: +3500 (classifica ATP)
• Punti percentuali nei duelli contro Isner: Vinte l'70% delle partite disputate /li>

John Isner

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• Vittorie nei tornei dello Slam: N/D
• Punti totalizzati nelle ultime sei settimane:+3000 (classifica ATP)
• Punti percentuali nei duelli contro Thiem : Perde l'70% delle partite disputate /li>

Marcel Granollers & Horacio Zeballos

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• Titoli vinti insieme: ATP Finals (2019), Monte Carlo Masters (2018)
• Punteggio doppio nelle ultime sei settimane:+2500 (classifica ATP)
• Punti percentuali nei duelli contro Ram & Salisbury : Vinte l'65% delle partite disputate /li>

Rajeev Ram & Joe Salisbury

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• Titoli vinti insieme:N/D
• Punteggio doppio nelle ultime sei settimane:+2000 (classifica ATP)
• Punti percentuali nei duelli contro Granollers & Zeballos : Perde l'65% delle partite disputate /li>

S<|end_of_document|>assistantTo solve the problem of finding the number of distinct ways to color a (2 times n) grid using (k) colors such that no two adjacent cells have the same color and no color is repeated in any row or column more than once consecutively across both rows simultaneously, we can use a dynamic programming approach. ### Problem Breakdown 1. **Grid Structure**: We have a (2 times n) grid. 2. **Coloring Constraints**: - No two adjacent cells can have the same color. - No color can be repeated consecutively in any row or column across both rows. ### Dynamic Programming Approach Define (dp[i][c1][c2]) as the number of ways to color the first (i) columns of the grid where: - (c1) is the color of the cell at position ((1, i)). - (c2) is the color of the cell at position ((2, i)). ### Transition For each column (i), we need to determine valid colors for (c1) and (c2) based on the previous column's colors (c1') and (c2'): - (c1 neq c1') (no two adjacent cells in the same row can have the same color). - (c2 neq c2') (no two adjacent cells in the same row can have the same color). - (c1 neq c2) (no two adjacent cells in the same column can have the same color). - (c1 neq c2') and (c2 neq c1') (no color is repeated consecutively across both rows). ### Base Case For the first column ((i = 1)), any pair of different colors ((c1, c2)) is valid: [ dp[1][c1][c2] = 1 quad text{for } c1 neq c2 ] ### Recurrence Relation For each subsequent column (i), update the DP table: [ dp[i][c1][c2] = sum_{substack{c1' neq c1 \ c2' neq c2 \ c1' neq c2 \ c1 neq c2'}} dp[i-1][c1'][c2'] ] ### Final Calculation The total number of ways to color the entire grid is obtained by summing over all valid configurations for the last column: [ text{Total Ways} = sum_{substack{c1 \ c2 \ c1 neq c2}} dp[n][c1][c2] ] ### Complexity The time complexity is (O(n times k^3)) because for each column and each pair of colors for that column, we consider up to (k^2) previous states. ### Implementation Here's a Python function implementing this approach: python def count_colorings(n, k): # Initialize DP table dp = [[[0] * k for _ in range(k)] for _ in range(n + 1)] # Base case for the first column for c1 in range(k): for c2 in range(k): if c1 != c2: dp[1][c1][c2] = 1 # Fill DP table for i in range(2, n + 1): for c1 in range(k): for c2 in range(k): if c1 != c2: dp[i][c1][c2] = sum( dp[i-1][c1_prime][c2_prime] for c1_prime in range(k) for c2_prime in range(k) if c1_prime != c1 and c2_prime != c2 and c1_prime != c2_prime and c1 != c2_prime and c2 != c1_prime ) # Calculate total number of ways total_ways = sum(dp[n][c1][c2] for c1 in range(k) for c2 in range(k) if c1 != c2) return total_ways # Example usage: n = 3 k = 3 print(count_colorings(n, k)) # Output the number of ways to color a 2x3 grid with 3 colors This function calculates the number of valid ways to color a (2 times n) grid with (k) colors under the given constraints.