Lets say we have a car rental company and that we have multiple cars of the same type. When a user books a car he doesn’t book a specific car but one of its type.
Which algorithm is the best to display/arrange/sort the bookings on a timeline table, looking like this:
/*
* | 09 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 |
* | A A A A A |
* | D D D B B B |
* | D D D B B B |
* | C C C C |
* | C C C C |
* | C C C C |
* */
Where each line represents a car of the same type, and each letter represents 1 booking (1 user can book multiple cars). The system has built in detection so, you cannot book a car if there are no cars left (like from 11. to 14. there are no cars available).
There are multiple ways to sort the bookings, but an algorithm (or some pointers on how to) that always returns the same order and works for all (or close) different cases would be great.
Related
I have the below question:
In the first part of the question, is says the probability that the selected person will be a male is 0.44, it means the number of males is 25*0.44 = 11. That's ok
In the second part, the probability of the selected person will be a male who was born before 1960 is 0.28, Does that mean 0.28 out of the total number which is 25 or out of the number of males?
I mean should the number of male who was born before 1960 equals into 250.28 OR 110.28
I find it easiest to think of these sorts of problems as contingency tables.
You use a maxtrix layout to express the distributions in terms of two or more factors or characteristics, each having two or more categories. The table can be constructed either with probabilities (proportions) or with counts, and switching back and forth is easy based on the total count in the table. Entries in the table are the intersections of the categories, corresponding to and in a verbal description. The numbers to the right or at the bottom of the table are called marginals, because they're found in the margins of the tables, and are always the sum of the table row or column entries in which they occur. The total probability (or count) in the table is found by summing across all the rows and columns. The marginal distribution of gender would be found by summing across rows, and the marginal distribution of birthdays would be found by summing across the columns.
Based on this, you can inferentially determine other values as indicated by the entries in parentheses below. With one more entry, either for gender or in the marginal row for birthdays, you'd be able to fill in the whole table inferentially. (This is related to the concept of degrees of freedom - how many pieces of info can you fill in independently before the others are determined by the known constraint that the totals are fixed or that probability adds to 1.)
Probabilities
Birthday
< 1960 | >= 1960
_______________________
G | | |
e F | | | (0.56)
n __|_________|__________|
d | | |
e M | 0.28 | (0.16) | 0.44
r __|_________|__________|______
? ? | 1.00
Counts
Birthday
< 1960 | >= 1960
_______________________
G | | |
e F | | | (14)
n __|_________|__________|
d | | |
e M | 7 | (4) | 11
r __|_________|__________|_____
? ? | 25
Conditional probability corresponds to limiting yourself to the subset of rows or columns specified in the condition. If you had been asked what is the probability of a birthday < 1960 given the gender is male, i.e., P{birthday < 1960 | M} in relatively standard notation, you'd be restricting your focus to just the M row, so the answer would be 7/11 = 0.28/0.44. Computationally, you take the probabilities or counts in the qualifying table entries and express them as a proportion of the probabilities or counts of the specified (given) marginal entries. This is often written in prob & stats texts as P(A|B) = P(AB)/P(B), where AB is a set shorthand for A and B (intersection).
0,44 = 11 / 25 people are male.
0,28 = 7 / 25 people are male & born before 1960.
I have a system where people rate different items on a scale from 0-5. The issue is, not everyone rates the same items, and the scoring is not objective. The goal is to achieve a fair comparison between items so that an item's score is not affected too much if one of the scorers is very "lenient" or "harsh." In actuality, there may be 100 items, each one scored twice, but here is an example dataset where 4 people scored 12 items, each one being scored twice:
| Item | Score 1 | Score 2 |
_____
1 | 5 | | 4 |
2 | 5 | | 3 | C
3 | 4 | |_2_|
4 A | 5 | | 5 |
5 | 3 | | 0 |
6 |_5_| | 3 |
7 | 3 | | 1 | D
8 | 4 | | 1 |
9 B | 4 | |_2_|
10 | 4 | | 3 |
11 | 4 | | 3 | C
12 |_5_| | 4 |
In this table, the boxes represent a single person's set of scores. We can label the person who gave score 1 to items 1-6 person A, the one who gave score 1 to 7-12 person B, the one who gave score 2 to 1-3 and 10-12 person C, and the one who gave score 2 to to 4-9 person D.
Informally, if we assume person C was the closest to each item's objective score, we might reason as follows:
Person A generally gave higher scores than C on items 1-3 so he is "lenient."
D gave low scores to all of them except for item 4 which then must have been truly good. He gave scores generally lower than A, so his scores should be adjusted slightly upwards perhaps.
B gave higher scores than D, and a bit higher than C, so a bit "lenient".
Thus, we might produce adjusted scores for each item. For example, even though item 2 has a higher average score than item 9, they are probably on par considering A is generally lenient and D is generally harsh. The question is, how do we do this programmatically. I thought we might make several transformation functions which transform a raw score into an adjusted score, say A, B, C, and D. For example, we might have A(5)=3.7 because when A rates an item as 5, it is really in the 3-4 range. Then, we want to minimize
|A(x_0a)-C(x_0c)|^2 + |D(x_1d)-A(x_1a)|^2 + |B(x_2b)-D(x_2d)|^2 + |C(x_3c)-B(x_3b)|^2
where x_ip is a vector which consists of person p's ratings for items 3i+1, 3i+2, and 3i+3. We might make A, B, C, and D linear transformations, for example. How then do you optimize it? And is this the best way to eliminate one the harshness or leniency of scorers without throwing away their ratings?
I have a timetable like this:
+-----------+-------------+------------+------------+------------+------------+-------+----+
| transport | trainnumber | departcity | arrivecity | departtime | arrivetime | price | id |
+-----------+-------------+------------+------------+------------+------------+-------+----+
| Q | Q00 | BJ | TJ | 13:00:00 | 15:00:00 | 10 | 1 |
| Q | Q01 | BJ | TJ | 18:00:00 | 20:00:00 | 10 | 2 |
| Q | Q02 | TJ | BJ | 16:00:00 | 18:00:00 | 10 | 3 |
| Q | Q03 | TJ | BJ | 21:00:00 | 23:00:00 | 10 | 4 |
| Q | Q04 | HA | DL | 06:00:00 | 11:00:00 | 50 | 5 |
| Q | Q05 | HA | DL | 14:00:00 | 19:00:00 | 50 | 6 |
| Q | Q06 | HA | DL | 18:00:00 | 23:00:00 | 50 | 7 |
| Q | Q07 | DL | HA | 07:00:00 | 12:00:00 | 50 | 8 |
| Q | Q08 | DL | HA | 15:00:00 | 20:00:00 | 50 | 9 |
| ... | ... | ... | ... | ... | ... | ... | ...|
+-----------+-------------+------------+------------+------------+------------+-------+----+
In this table, there 13 cities and 116 routes altogether and the smallest unit of time is half an hour.
There are difference transports, which doesn't matter. As you can see, there can be multiple edges with same departcity and arrivecity but difference time and difference price. The time is constant everyday.
Now, here arises a problem.
A user wonder how he can travel from city A to city B (A and B may be one city), with passing zero or some cities C, D...(whether they should be in order depends on whether the user wants it to be, that is, there are two problems), within X hours and also least costs under above conditions.
Before this problem, I have solved another simpler problem.
A user wonder how he can travel from city A to city B (A and B may be one city), with passing zero or some cities C, D...(whether they should be in order depends), with least costs under above conditions.
Here is how I solve it (just take not in order as an example):
Sort the must-pass cities:C1, C2, C3...Cn. Let C0 = A, C(n+1) = B, minCost.cost = INFINITE;
i = 0, j = 1, W = {};
Find a least cost way S from Ci to Cj using Dijkstra Algorithm with price as the weight of edges. W=W∪S;
i = i + 1, j = j + 1;
If j <= n + 1, goto 3;
if W.cost < minCost.cost, minCost = W;
If next permutation for C1...Cn exists, rearrange list C1...Cn in order of the next permutation for C1...Cn and goto 2;
Return minCost;
However, I cannot come up with a efficient solution to the first problem, Please help me, thanks.
I'll be appreciated if anyone can solve another problem:
A user wonder how he can travel from city A to city B (A and B may be one city), with passing zero or some cities C, D...(whether they should be in order depends), within least time under above conditions.
It's quite a big problem, so I will just sketch a solution.
First, remodel your graph as follows. Instead of each vertex representing a city, let a vertex represent a tuple of (city, time). This is feasible as there are only 13 cities and only (time_limit - current_time) * 2 possible time slots as the smallest unit of time is half an hour. Now connect vertices according to the given timetable with prices as their weights as before. Don't forget that the user can stay at any city for any amount of time for free. All nodes with city A are start nodes, all nodes with city B are target nodes. Take the minimum value of all (B, time) vertices to get the solution with least cost. If there are multiple, take the one with the smallest time.
Now on towards forcing the user to pass through certain cities in order. If there are n cities to pass through (plus start and target city), you need n+2 copies of the same graph which act as different levels. The level represents how many cities of your list you have already passed. So you start in level 0 on vertex A. Once you get to C1 in level 0 you move to the vertex C1 in level 1 of the graph (connect the vertices by 0-weight edges). This means that when you are in level k, you have already passed cities C1 to Ck and you can get to the next level only by going through C(k+1). The vertices of city B in the last level are your target nodes.
Note: I said copies of the same graph, but that is not exactly true. You can't allow the user to reach C(k+2), ..., B in level k, that would violate the required order.
To enforce passing cities in any order, a different scheme of connecting the levels (and modifying them during runtime) is required. I'll leave this to you.
i'm trying to do two calculations in one procedure in pl/sql. The simple version would look like
select a1+a2
into a
from table c;
select b1+b2
into b
from table c;
...
in the end i would like to further calculate the results of a and b, so let's say
select (a*b)
into c
from x
insert
into x (A,B,C)
values (a,b,c)
would that be possible? In case I split the two into two procedures (to first calculate a, b and then a second procedure to calculate c) i got results like
A | B | C
a | b | null
null | null | c
what i want to get is
A | B | C
a | b | c
I've heard about the Apriori algorithm several times before but never got the time or the opportunity to dig into it, can anyone explain to me in a simple way the workings of this algorithm? Also, a basic example would make it a lot easier for me to understand.
Apriori Algorithm
It is a candidate-generation-and-test approach for frequent pattern mining in datasets. There are two things you have to remember.
Apriori Pruning Principle - If any itemset is infrequent, then its superset should not be generated/tested.
Apriori Property - A given (k+1)-itemset is a candidate (k+1)-itemset only if everyone of its k-itemset subsets are frequent.
Now, here is the apriori algorithm in 4 steps.
Initially, scan the database/dataset once to get the frequent 1-itemset.
Generate length k+1 candidate itemsets from length k frequent itemsets.
Test the candidates against the database/dataset.
Terminate when no frequent or candidate set can be generated.
Solved Example
Suppose there is a transaction database as follows with 4 transactions including their transaction IDs and items bought with them. Assume the minimum support - min_sup is 2. The term support is the number of transactions in which a certain itemset is present/included.
Transaction DB
tid | items
-------------
10 | A,C,D
20 | B,C,E
30 | A,B,C,E
40 | B,E
Now, let's create the candidate 1-itemsets by the 1st scan of the DB. It is simply called as the set of C_1 as follows.
itemset | sup
-------------
{A} | 2
{B} | 3
{C} | 3
{D} | 1
{E} | 3
If we test this with min_sup, we can see {D} does not satisfy the min_sup of 2. So, it will not be included in the frequent 1-itemset, which we simply call as the set of L_1 as follows.
itemset | sup
-------------
{A} | 2
{B} | 3
{C} | 3
{E} | 3
Now, let's scan the DB for the 2nd time, and generate candidate 2-itemsets, which we simply call as the set of C_2 as follows.
itemset | sup
-------------
{A,B} | 1
{A,C} | 2
{A,E} | 1
{B,C} | 2
{B,E} | 3
{C,E} | 2
As you can see, {A,B} and {A,E} itemsets do not satisfy the min_sup of 2 and hence they will not be included in the frequent 2-itemset, L_2
itemset | sup
-------------
{A,C} | 2
{B,C} | 2
{B,E} | 3
{C,E} | 2
Now let's do a 3rd scan of the DB and get candidate 3-itemsets, C_3 as follows.
itemset | sup
-------------
{A,B,C} | 1
{A,B,E} | 1
{A,C,E} | 1
{B,C,E} | 2
You can see that, {A,B,C}, {A,B,E} and {A,C,E} does not satisfy min_sup of 2. So they will not be included in frequent 3-itemset, L_3 as follows.
itemset | sup
-------------
{B,C,E} | 2
Now, finally, we can calculate the support (supp), confidence (conf) and lift (interestingness value) values of the Association/Correlation Rules that can be generated by the itemset {B,C,E} as follows.
Rule | supp | conf | lift
-------------------------------------------
B -> C & E | 50% | 66.67% | 1.33
E -> B & C | 50% | 66.67% | 1.33
C -> E & B | 50% | 66.67% | 1.77
B & C -> E | 50% | 100% | 1.33
E & B -> C | 50% | 66.67% | 1.77
C & E -> B | 50% | 100% | 1.33
See Top 10 algorithms in data mining (free access) or The Top Ten Algorithms in Data Mining. The latter gives a detailed description of the algorithm, together with details on how to get optimized implementations.
Well, I would assume you've read the wikipedia entry but you said "a basic example would make it a lot easier for me to understand". Wikipedia has just that so I'll assume you haven't read it and suggest that you do.
Read the wikipedia article.
The best introduction to Apriori can be downloaded from this book:
http://www-users.cs.umn.edu/~kumar/dmbook/index.php
you can download the chapter 6 for free which explain Apriori very clearly.
Moreover, if you want to download a Java version of Apriori and other algorithms for frequent itemset mining, you can check my website:
http://www.philippe-fournier-viger.com/spmf/