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Norma-Jean [14]
2 years ago
8

Random Walker Collisions In lecture, we saw how to model the behavior of a random walker on a 2D grid using a Monte Carlo simula

tion. In this problem, we will investigate collisions between two of these random walkers. Specifically, use your simulation to track the distances of the two walkers until they collide. 1. Start by writing a MATLAB script to simulate the path that the single random walker A takes on an 11 x 11 grid of tiles. Let this random walker use a random compass as shown in class. Thus, the probability p of moving in the North, East, South, and West directions is given by 0.2. The remaining probability is the one used to stay put. If the random walker tries to move "past" one of the boundaries on its turn, his position does not change (you can think of this as a particle bumping into a wall and staying still). Choose a random tile to let the random walker A start his walk. Lastly, write a function to update the position of the random walker A. The function must use exactly the function header shown below, i.e. the function name and number and ordering of inputs and outputs must be followed exactly. function (x,y) = Randwalk_2D (x0, y0, BC) Here, x0 and y0 are the initial positions of the random walker, BC is the array of boundary conditions, and x and y are the updated positions of the random walker. 2. Add the random walker B to your simulation who moves according to its own randomly generated values using the same compass as the random walker A in Part 1. Let the random walker B start his walk on the tile that is furthest away from the start tile of walker A. At each iteration, both walkers move simultaneously to an adjacent tile using the rules outlined in Part 1. Continue updating the position of both walkers until a collision occurs (or until the maximum number of iterations, which is set to 1000, is reached). More precisely, a collision occurs when both random walkers occupy the same tile after both have completed their move. Furthermore, a collision occurs when both random walkers are next to each other and move to the other one's tile. Note that in this instance, the collision appears on their way, but both random walkers will end up on tiles that are again next to each other. In other words, collision also appears when their paths cross. 3. Track the movement of each random walker by storing their positions in four 1D-arrays associated with their x- and y-positions in the grid. After the simulation, calculate for each iteration the distance between the random walkers A and B, and plot the results vs. the number of iterations to see when the two random walkers A and B came close to each other and when they were far away from each other.
Computers and Technology
1 answer:
vlabodo [156]2 years ago
6 0

Answer:

dang how long did i take for you to wright all this and that lecture was cool

Explanation:

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3 0
3 years ago
I need the SQL statements for these questions:
zimovet [89]

Answer:

Explanation:

/* From the information provided, For now will consider the name of table as TRIPGUIDES*/

/*In all the answers below, the syntax is based on Oracle SQL. In case of usage of other database queries, answer may vary to some extent*/

1.

Select R.Reservation_ID, R.Trip_ID , C.Customer_Num,C.Last_Name from Reservation R, Customer C where C.Customer_Num=R.Customer_Num ORDER BY C.Last_Name

/*idea is to select the join the two tables by comparing customer_id field in two tables as it is the only field which is common and then print the desired result later ordering by last name to get the results in sorted order*/

2.

Select R.Reservation_ID, R.Trip_ID , R.NUM_PERSONS from Reservation R, Customer C where C.Customer_Num=R.Customer_Num and C.LAST_NAME='Goff' and C.FIRST_NAME='Ryan'

/*Here, the explaination will be similar to the first query. Choose the desired columns from the tables, and join the two tables by equating the common field

*/

3.

Select T.TRIP_NAME from TRIP T,GUIDE G,TRIPGUIDES TG where T.TRIP_ID=TG.TRIP_ID and TG.GUIDE_NUM=G.GUIDE_NUM and G.LAST_NAME='Abrams' and G.FIRST_NAME='Miles'

/*

Here,we choose three tables TRIP,GUIDE and TRIPGUIDES. Here we selected those trips where we have guides as Miles Abrms in the GUIDES table and equated Trip_id from TRIPGUIDES to TRIP.TRIP_Name so that can have the desired results

*/

4.

Select T.TRIP_NAME

from TRIP T,TRIPGUIDES TG ,G.GUIDE

where T.TRIP_ID=TG.TRIP_ID and T.TYPE='Biking' and TG.GUIDE_NUM=G.GUIDE_NUM and G.LAST_NAME='Boyers' and G.FIRST_NAME='Rita'

/*

In the above question, we first selected the trip name from trip table. To put the condition we first make sure that all the three tables are connected properly. In order to do so, we have equated Guide_nums in guide and tripguides. and also equated trip_id in tripguides and trip. Then we equated names from guide tables and type from trip table for the desired results.

*/

5.

SELECT C.LAST_NAME , T.TRIP_NAME , T.START_LOCATION FROM CUSTOMER C, TRIP T, RESERVATION R WHERE R.TRIP_DATE='2016-07-23' AND T.TRIP_ID=R.TRIP_ID AND C.CUSTOMER_NUM=R.CUSTOMER_NUM

/*

The explaination for this one will be equivalent to the previous question where we just equated the desired columns where we equiated the desired columns in respective fields and also equated the common entities like trip ids and customer ids so that can join tables properly

*/

/*The comparison of dates in SQL depends on the format in which they are stored. In the upper case if the

dates are stored in the format as YYYY-MM-DD, then the above query mentioned will work. In case dates are stored in the form of a string then the following query will work.

SELECT C.LAST_NAME , T.TRIP_NAME , T.START_LOCATION FROM CUSTOMER C, TRIP T, RESERVATION R WHERE R.TRIP_DATE='7/23/2016' AND T.TRIP_ID=R.TRIP_ID AND C.CUSTOMER_NUM=R.CUSTOMER_NUM

*/

6.

Select R.RESERVATION_ID, R.TRIP_ID,R.TRIP_DATE FROM RESERVATION R WHERE R.TRIP_ID IN

{SELECT TRIP_ID FROM TRIP T WHERE STATE='ME'}

/*

In the above question, we firstly extracted all the trip id's which are having locations as maine. Now we have the list of all the trip_id's that have location maine. Now we just need to extract the reservation ids for the same which can be trivally done by simply using the in clause stating print all the tuples whose id's are there in the list of inner query. Remember, IN always checks in the set of values.

*/

7.

Select R.RESERVATION_ID, R.TRIP_ID,R.TRIP_DATE FROM RESERVATION WHERE

EXISTS {SELECT TRIP_ID FROM TRIP T WHERE STATE='ME' and R.TRIP_ID=T.TRIP_ID}

/*

Unlike IN, Exist returns either true or false based on existance of any tuple in the condition provided. In the question above, firstly we checked for the possibilities if there is a trip in state ME and TRIP_IDs are common. Then we selected reservation ID, trip ID and Trip dates for all queries that returns true for inner query

*/

8.

SELECT G.LAST_NAME,G.FIRST_NAME FROM GUIDE WHERE G.GUIDE_NUM IN

{

SELECT DISTINCT TG.GUIDE_NUM FROM TRIPGUIDES TG WHERE TG.TRIPID IN {

SELECT T.TRIP_ID FROM TRIP T WHERE T.TYPE='Paddling'

}

}

/*

We have used here double nested IN queries. Firstly we selected all the trips which had paddling type (from the inner most queries). Using the same, we get the list of guides,(basically got the list of guide_numbers) of all the guides eds which were on trips with trip id we got from the inner most queries. Now that we have all the guide_Nums that were on trip with type paddling, we can simply use the query select last name and first name of all the guides which are having guide nums in the list returned by middle query.

*/

4 0
2 years ago
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