Exploring NATURAL JOIN

The NATURAL JOIN operation in SQL automatically joins two tables on their common columns. This saves us a step, as we don’t have to explicitly specify the keys for the join. When performing a NATURAL JOIN, the operation:

Matches rows in both tables based on all columns with the same name.
Combines rows where the values in the common columns match.
Consider the Following Tables:

Table R:

A B
a1 b1
a2 b1
a3 b2
a4 b4

Table S:

B C
b1 c1
b1 c2
b2 c3
b3 c4

You can load these tables in R using the following code:

rm(list = ls())
library(sqldf)
R

Question

Exploring NATURAL JOIN

The NATURAL JOIN operation in SQL automatically joins two tables on their common columns. This saves us a step, as we don’t have to explicitly specify the keys for the join. When performing a NATURAL JOIN, the operation:

Matches rows in both tables based on all columns with the same name.
Combines rows where the values in the common columns match.
Consider the Following Tables:

Table R:

A	B
    a1   	    b1    
    a2    	    b1    
    a3    	    b2    
    a4    	    b4

Table S:

B	C
    b1    	    c1    
    b1    	    c2    
    b2    	    c3    
    b3    	    c4

You can load these tables in R using the following code:

rm(list = ls())
library(sqldf)
R <- read.csv("https://www.dropbox.com/scl/fi/empw9y5394bdatvwf119o/R.csv?rlkey=m1uqky67r6zp7mec7a3cgw5hq&st=vhazojgh&dl=1")
S <- read.csv("https://www.dropbox.com/scl/fi/vrr73ncmuyk3xah748ng6/S.csv?rlkey=ap8fzt4dyttfdab194yfm4e72&st=nro3921a&dl=1")
Question:

What is the cardinality (number of rows) of the result of the following query?

SELECT *  FROM R  NATURAL JOIN S;

Key Considerations:

The NATURAL JOIN will automatically join tables R and S on their common column(s).

Only rows with matching values in column B will appear in the result.

Each matching pair will produce a new row in the output, including columns from both tables.

BlackTom AI · Accepted Answer

First, identify the common column(s) used by NATURAL JOIN between tables R and S. In this scenario, the shared column is B, so the join is performed on R.B = S.B.
Next, list the rows in each table by their B values to see where matches occur:
- R has rows with B values: b1, b1, b2, b4
- S has rows with B values: b1, b1, b2, b3
This means potential matches exist for B = b1 and B = b2, while B = b4 in R has no corresponding B value in S, and B = b3 in S has no corresponding B value in R.
Now, count the matches for each matching B value:
- For B = b1: R provides 2 rows (a1,b1) and (a2,b1), and S provides 2 rows (b1,c1) and (b1,c2). Each R row combines with each S row sharing B = b1, yielding 2 × 2 = 4 output rows.
- For B = b2: R provides 1 row (a3,b2) and S provides 1 row (b2,c3). This yields 1 × 1 = 1 output row.
- For B = b4: there are no matching rows in S, so no output rows from this B value.
- For B = b3 in S: there are no matching rows in R, so no output rows from this B value either.
Summing the contributions from all matching B values gives 4 + 1 = 5 output rows in the NATURAL JOIN result.
Therefore, the cardinality of SELECT * FROM R NATURAL JOIN S is 5, since the join combines each matching pair of rows across the common column(s) and includes all columns from both tables for those joined rows.

ECON3310001.1251 Lab Five

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