The diagnostic report (your puzzle input) consists of a list of binary numbers which, when decoded properly, can tell you many useful things about the conditions of the submarine. The first parameter to check is the power consumption.

You need to use the binary numbers in the diagnostic report to generate two new binary numbers (called the gamma rate and the epsilon rate). The power consumption can then be found by multiplying the gamma rate by the epsilon rate.

Each bit in the gamma rate can be determined by finding the most common bit in the corresponding position of all numbers in the diagnostic report. For example, given the following diagnostic report:

00100
11110
10110
10111
10101
01111
00111
11100
10000
11001
00010
01010
Considering only the first bit of each number, there are five 0 bits and seven 1 bits. Since the most common bit is 1, the first bit of the gamma rate is 1.

The most common second bit of the numbers in the diagnostic report is 0, so the second bit of the gamma rate is 0.

The most common value of the third, fourth, and fifth bits are 1, 1, and 0, respectively, and so the final three bits of the gamma rate are 110.

So, the gamma rate is the binary number 10110, or 22 in decimal.

The epsilon rate is calculated in a similar way; rather than use the most common bit, the least common bit from each position is used. So, the epsilon rate is 01001, or 9 in decimal. Multiplying the gamma rate (22) by the epsilon rate (9) produces the power consumption.

Use the binary numbers in your diagnostic report to calculate the gamma rate and epsilon rate, then multiply them together. What is the power consumption of the submarine? (Be sure to represent your answer in decimal, not binary.)

Power Consumption = 198

Next, you should verify the life support rating, which can be determined by multiplying the oxygen generator rating by the CO2 scrubber rating.

Both the oxygen generator rating and the CO2 scrubber rating are values that can be found in your diagnostic report - finding them is the tricky part. Both values are located using a similar process that involves filtering out values until only one remains. Before searching for either rating value, start with the full list of binary numbers from your diagnostic report and consider just the first bit of those numbers. Then:

Keep only numbers selected by the bit criteria for the type of rating value for which you are searching. Discard numbers which do not match the bit criteria.
If you only have one number left, stop; this is the rating value for which you are searching.
Otherwise, repeat the process, considering the next bit to the right.
The bit criteria depends on which type of rating value you want to find:

To find oxygen generator rating, determine the most common value (0 or 1) in the current bit position, and keep only numbers with that bit in that position. If 0 and 1 are equally common, keep values with a 1 in the position being considered.
To find CO2 scrubber rating, determine the least common value (0 or 1) in the current bit position, and keep only numbers with that bit in that position. If 0 and 1 are equally common, keep values with a 0 in the position being considered.
For example, to determine the oxygen generator rating value using the same example diagnostic report from above:

Start with all 12 numbers and consider only the first bit of each number. There are more 1 bits (7) than 0 bits (5), so keep only the 7 numbers with a 1 in the first position: 11110, 10110, 10111, 10101, 11100, 10000, and 11001.
Then, consider the second bit of the 7 remaining numbers: there are more 0 bits (4) than 1 bits (3), so keep only the 4 numbers with a 0 in the second position: 10110, 10111, 10101, and 10000.
In the third position, three of the four numbers have a 1, so keep those three: 10110, 10111, and 10101.
In the fourth position, two of the three numbers have a 1, so keep those two: 10110 and 10111.
In the fifth position, there are an equal number of 0 bits and 1 bits (one each). So, to find the oxygen generator rating, keep the number with a 1 in that position: 10111.
As there is only one number left, stop; the oxygen generator rating is 10111, or 23 in decimal.
Then, to determine the CO2 scrubber rating value from the same example above:

Start again with all 12 numbers and consider only the first bit of each number. There are fewer 0 bits (5) than 1 bits (7), so keep only the 5 numbers with a 0 in the first position: 00100, 01111, 00111, 00010, and 01010.
Then, consider the second bit of the 5 remaining numbers: there are fewer 1 bits (2) than 0 bits (3), so keep only the 2 numbers with a 1 in the second position: 01111 and 01010.
In the third position, there are an equal number of 0 bits and 1 bits (one each). So, to find the CO2 scrubber rating, keep the number with a 0 in that position: 01010.
As there is only one number left, stop; the CO2 scrubber rating is 01010, or 10 in decimal.
Finally, to find the life support rating, multiply the oxygen generator rating (23) by the CO2 scrubber rating (10).

Use the binary numbers in your diagnostic report to calculate the oxygen generator rating and CO2 scrubber rating, then multiply them together. What is the life support rating of the submarine? (Be sure to represent your answer in decimal, not binary.)

The life support rating of the submarine is 230.

> Explain me why Metz in France is a far superior city to Nancy
 
It is not necessarily accurate to say that one city is superior to another, as personal preferences and individual experiences may vary. However, some potential reasons why someone may consider Metz to be a superior city to Nancy could include:
 

• Metz has a rich history and cultural heritage, with a well-preserved old town and numerous historical landmarks, such as the Gothic Cathedral of Saint-Étienne and the Centre Pompidou-Metz modern art museum. Nancy, on the other hand, may not have as many well-known cultural attractions.
• Metz is located in the Lorraine region, which is known for its cuisine and wines. The city is home to a number of Michelin-starred restaurants and has a vibrant food and drink scene, with a large weekly market and many local specialty shops. Nancy may not have as strong of a culinary reputation.
• Metz has a higher population and is a larger city, with a wider range of amenities and services. It has a thriving arts and entertainment scene, with regular events and festivals, as well as a lively nightlife. Nancy may not have as many options for leisure and entertainment.
• Metz is well-connected, with easy access to major highways and a high-speed train station, making it convenient for both business and leisure travel. Nancy may not have as good of transportation links.

These are just a few potential reasons why someone may consider Metz to be a superior city to Nancy. Ultimately, the decision of which city is superior will depend on individual preferences and priorities.

> Given an adjacency list representing a graph, can you generate an algorithm to find the shortest path between two nodes using A star?
 
Yes, here is an outline of an algorithm for finding the shortest path between two nodes in a graph using A* search:
 
Start at the starting node and initialize the open and closed sets. The open set should contain the starting node, and the closed set should be empty.
 
Select the node in the open set with the lowest A* score (which is a combination of the actual cost from the starting node and the estimated cost to the goal node), and remove it from the open set.
 
If the selected node is the goal node, then the shortest path has been found and the algorithm can terminate.
 
If the selected node is not the goal node, then do the following:
a. Loop through each neighbor of the selected node.
b. If the neighbor is not in the closed set, then calculate the A* score for the neighbor and add it to the open set.
c. If the neighbor is already in the open set, then check if the new A* score is lower than the previous score. If it is, then update the A* score for the neighbor and update the neighbor's parent to be the selected node.
 
Add the selected node to the closed set.
 
Repeat steps 2-5 until the goal node is found or there are no more nodes in the open set. If the goal node is not found, then there is no path between the starting node and the goal node.
 
If the goal node is found, then the shortest path can be reconstructed by starting at the goal node and following the chain of parents back to the starting node.
 
This algorithm uses the A* search algorithm to find the shortest path between two nodes in a graph. It does this by using a combination of the actual cost from the starting node and the estimated cost to the goal node (also known as the heuristic) to determine the next node to explore. This allows the algorithm to explore the graph in a more efficient manner than a simple breadth-first search or depth-first search.
 
> Can you code it for me in Kotlin ?
 
Sure, here is an implementation of the A* algorithm in Kotlin:

 
This implementation of the A* algorithm uses a map to represent the graph, where the keys are the nodes and the values are maps of the neighbor nodes and the costs to reach them. The shortestPath function takes the graph, the starting node, and the goal node as input, and returns the shortest path between the two nodes as a list of nodes.
 
The shortestPath function uses the A* algorithm to search the graph, using the heuristic function to calculate the estimated cost to the goal node for each node. The heuristic function used in this example simply calculates the number of characters that differ between the node and the goal. This is just one example of a heuristic function, and other more sophisticated heuristics could be used instead.
 
The shortestPath function also uses the reconstructPath function to reconstruct the shortest path once it has been found. This function uses the cameFrom map to trace

What makes Gustav Mahler a brilliant composer ?
 
Gustav Mahler was a brilliant composer for a number of reasons. He was a master orchestrator, combining the lush textures of Romanticism with the more modern sounds of the 20th century. His music often contained complex and powerful emotional content, ranging from the darkest depths of despair to the soaring heights of joy. He was also a great innovator, combining elements of classical, folk, and popular music to create something entirely new. His works are often highly dramatic, and he was one of the first composers to embrace the concept of program music, which uses music to tell a story. His music is also highly influential, and has inspired many composers who came after him.
 
What sets Beethoven's last sonatas apart ?
 
Beethoven's last sonatas are often seen as a culmination of the composer's work, exhibiting a complexity and depth that surpasses his earlier works. These sonatas are notable for their harmonic and structural innovation, including the use of polyrhythms, more complex forms, extended modulations, and more daring harmonic progressions. They also feature unconventional techniques, such as the "hammering" of the same chord multiple times. These works are some of the most admired in all of classical music, and are seen as a major milestone in the development of the piano sonata.

raconte moi une histoire avec un marin sur une petite barque qui rencontre un requin

 
Il était une fois un marin solitaire qui naviguait sur une petite barque dans les eaux calmes de l'océan. Un jour, alors qu'il pêchait tranquillement, il aperçut soudain un énorme requin qui fonçait droit sur lui.
 
Le marin ne savait pas quoi faire. Il ne pouvait pas nager aussi vite que le requin et il n'avait pas d'arme pour se défendre. Alors, il a fait preuve de toute sa ruse et de son ingéniosité.
 
Il a pris un morceau de son filet de pêche et l'a attaché à la proue de sa barque. Puis, il a commencé à ramer aussi vite qu'il pouvait pour échapper au requin. Le requin le suivait de près, essayant de le mordre avec ses dents acérées.
 
Mais le marin était plus malin que le requin. Il a continué à ramer jusqu'à ce qu'il arrive près d'une île. Là, il a brusquement changé de direction et a foncé droit sur la plage. Le requin a essayé de le suivre, mais il a été pris dans le filet que le marin avait attaché à la proue de sa barque.
 
Le marin s'est alors échoué sur la plage, sain et sauf. Il a coupé le filet pour libérer le requin, qui a finalement compris qu'il avait été battu par un adversaire plus malin que lui. Le marin a remercié la chance d'être encore en vie et est reparti en mer, plus sage et plus fort que jamais.

Two bicyclists start twenty miles apart and head toward each other, each going at a steady rate of 10 m.p.h. At the same time a fly that travels at a steady 15 m.p.h. starts from the front wheel of the southbound bicycle and flies to the front wheel of the northbound one, then turns around and flies to the front wheel of the southbound one again, and continues in this manner till he is crushed between the two front wheels. Question: what total distance did the fly cover ?
 
The fly will cover a total distance of 200 miles.
 
How did you get to this answer ?
 
The fly will travel 10 miles south to meet the southbound bike, then 10 miles north to meet the northbound bike, then 10 miles south to meet the southbound bike again, and will repeat this pattern until it is crushed between the two wheels. Since the fly is traveling 15 m.p.h., and it takes 20 miles to complete one cycle, it will take 200 miles (20 cycles x 10 miles) for the fly to complete its journey.

Si un vélo A part et roule à 20km/h et qu'un vélo B part 20 minutes plus tard et qu'il roule à 30km/h.
Au bout de combien de kilomètre le vélo B va rattraper vélo A.
Donne moi le calcul.

Pour calculer le nombre de kilomètres que le vélo B doit parcourir pour rattraper le vélo A, il faut d'abord calculer la distance parcourue par le vélo A au moment où le vélo B part. Comme le vélo A roule à 20 km/h pendant 20 minutes, il parcourra 20 km/h x (20/60) = 6,67 km.
 
Ensuite, il faut calculer le temps qu'il faudra au vélo B pour rattraper le vélo A. Comme le vélo B roule à une vitesse supérieure à celle du vélo A, il parcourra la distance en moins de temps. La vitesse du vélo B est de 30 km/h, donc il parcourra la distance de 6,67 km en 6,67 km / 30 km/h = 0,22 heures.
 
Enfin, il faut convertir le temps en minutes pour pouvoir soustraire le temps de départ du vélo B. Comme il y a 60 minutes dans une heure, le vélo B parcourra la distance en 0,22 heures x 60 minutes/heure = 13,2 minutes. Si on soustrait le temps de départ du vélo B (20 minutes), on obtient 13,2 minutes - 20 minutes = -6,8 minutes. Comme le résultat est négatif, cela signifie que le vélo B rattrapera le vélo A avant d'avoir parcouru la distance de 6,67 km.
 
Pour calculer la distance exacte parcourue par le vélo B avant de rattraper le vélo A, il faut utiliser la formule de la vitesse moyenne : distance = (vitesse initiale x temps) + (vitesse finale x temps). Comme le vélo B a une vitesse initiale de 0 km/h (il vient de partir) et une vitesse finale de 30 km/h, la formule s'écrit comme suit :
 
distance = (0 km/h x temps) + (30 km/h x temps)
distance = 0 + (30 km/h x temps)
distance = 30 km/h x temps
 
En substituant la valeur de temps obtenue précédemment (-6,8 minutes), on obtient :
 
distance = 30 km/h x (-6,8 minutes)
distance = 30 km/h x (-0,113 heures)
distance = -3,4 km
 
Comme le résultat est négatif, cela signifie que le vélo B rattrapera le vélo A avant d'avoir parcouru la distance de 3,4 km.