I am studying data structures right now and in specific Hash Tables. I came across the follow question:
Imagine that we have placed the following keys
in an initial empty hash table with a length of 7
with linear probing, using the following table of hash-values:
key: A B C D E F G
hash: 3 1 4 1 5 2 5
Which of the following arrays could be the linear-probing array?
1.
0 1 2 3 4 5 6
G B D F A C E
2.
0 1 2 3 4 5 6
B G D F A C E
3.
0 1 2 3 4 5 6
E G F A B C D
When I create the linear-probing array I get this:
0 1 2 3 4 5 6
G B D A C E F
Could somebody please tell me why I am wrong and whats the right answer?
Notice how the question doesn't specify the order in which the keys are inserted, so your answer is only correct assuming that the keys are actually inserted in the order A-B-C-D-E-F-G, but since the question doesn't explicitly state the order, you need to dig deeper.
What you do know, however, is that one of those keys will be inserted first and it will go to its designated slot as shown in the Key-to-Hash diagram, since the hash table is initially empty. This immediately discards option choice 2 because none of the keys are in their designated array entry, leaving you with choice 1 and 3.
For table 1, B is in slot 1, which corresponds to its hash value and for table 3, keys F and A are in their initial hash-value spots.
It's simple to prove that no sequence of key inserts on table 3 after inserting F and A will yield table 3 as a result. And its likewise easy to prove that the sequence of key inserts B-D-F-A-C-E-G will result in table 1.
Although this is a question based on hash tables, I honestly don't consider it a good way to assess your knowledge on linear probing, this is more of a puzzle, as #gnasher729 mentioned.
Related
probably very simple question but yeah, dont know how of course.
I have 3 columns:
A B C
-------
5 1 A
5 2 B
5 3 C
and want to multiply column A and B, and just copy C column. All with one expresion.
So Result of multiplication A and B should be in D and result of copying should be in F:
Final result should be:
D F
---
5 A
10 B
15 C
Is there any simple method to do something like this?
this should do:
=INDEX({A1:A3*B1:B3, C1:C3})
or if you are on non-english sheet:
=INDEX({A1:A3*B1:B3\ C1:C3})
I have a problem with pd.Series(a).unique()
I made a Series, and I used .unique().
However, this deletes the pd.Series index.
How can I made unique Array with original index?
Instead of using .unique() you can use .drop_duplicates():
x = pd.Series([1,2,3,1,1,2,4,5,6], index=list("abcdefghi"))
print(x)
a 1
b 2
c 3
d 1
e 1
f 2
g 4
h 5
i 6
dtype: int64
.drop_duplicates() will remove all duplicates from the Series while maintaining reference to the index. You can choose whether you want to keep the index location of the "first" or the "last" duplicated item via the keep argument:
# Keep the first entry of each duplicated value
x.drop_duplicates(keep="first")
a 1
b 2
c 3
g 4
h 5
i 6
dtype: int64
# Keep the last entry of each duplicated item
x.drop_duplicates(keep="last")
c 3
e 1
f 2
g 4
h 5
i 6
dtype: int64
Can anyone explain to me how to go about figuring out which dependencies the following instance satisfies?
A B C
1 0 1
1 1 1
I know it satisfies B-> A, B->C, A->C, C->A (and other implied dependencies)
but I haven't been able to grasp the concept of how to just view that from this chart. Can anyone explain how I am supposed to read it and go about determining what it satisfies with only the 0's and 1's ?
Adding another example to help better understand:
A B C
1 0 1
1 1 1
2 2 1
Since there is only one row where B = 0 and B = 2, can you base the B -> A off of only one row with one unique value. Like since there is only one place where B = 0 and A = 1 does that mean it automatically holds since there is no other value of B with a 0 ?
A way of answering to a question like this is to look at all the possible proper subsets of columns, let's call them X1, X2, ... , starting first with single columns (so in this case we start with X1=A, X2=B, X3=C, and trying to see, for identical value in Xi, which other columns have identical values.
For instance, starting with A, we discover that for A=1, B has two different values: this means that B cannot depend on A, (that is has not the same value for the value of A, which is the definition of functional dependency), while C has the same value (1), so that we know that this instance of relation satisfies A → C.
Looking at B, we discover that all the values are different, so we can say that all the other columns are dependent on it, and we add B → A, B → C. Finally, in analysing C, we discover that only the values of A are equal when the values of C are equal, so that C → A.
We can stop here, without considering the pairs of attributes AB, AC, and BC, since in this simple case every attribute is the determinant of some dependency, so that dependencies with set of attributes as determinant are implied by the dependencies already found.
In summary
In a certain instance, to know if a dependency X -> Y hold or not, we check:
if all the values of X are different, then the dependency hold; if there are rows with repeat values, then, if for each row with the same value of X the value of Y is always the same, then the dependency holds, otherwise no.
Here is another example:
A B C
1 2 2
0 3 3
1 2 4
2 2 4
In this instance A → B ? Yes, since the are two rows (the first and the fourth) with the same value of A (1), and in both rows the value of B is equal (2). Is A → C ? No, since C has two different values in the first and fourth row.
B → A ? No, since B has three rows with the same value (2) and A has different values in the same rows (1 and 2).
I can't seem to find something quite like this problem...
I have an array table where each row contains a random assortment of numbers 1-N
On another sheet, I have a table with column and row headers numbered 1-N
I want to count how many rows in the array contain both the column and row headers for a given cell in the table. Since countifs only reference the current cell in the specified array, they don't seem to be working in this scenario.
Example array:
A B C D
1 3 5 7
1 2 3 4
2 3 4 5
2 4 6 8
...
Table results (symmetrical about the diagonal):
A B C D E F
. 1 2 3 4 5 ...
1 - 1 2 1 1
2 1 - 2 2 1
3 2 2 - 2 2
4 1 2 2 - 1
5 1 1 2 1 -
Would using nested countifs work?
I don't agree with your results corresponding to 4/2, which surely should be 3, not 2, but this formula, based on the array table being in Sheet1 A1:D4 and the results table being in Sheet2 A1:F6, placed in cell B2 of the latter, should work:
=IF($A2=B$1,"-",SUMPRODUCT(N(MMULT(N(COUNTIF(OFFSET(Sheet1!$A$1:$D$1,ROW(Sheet1!$A$1:$D$4)-MIN(ROW(Sheet1!$A$1:$D$4)),),CHOOSE({1,2},B$1,$A2))>0),{1;1})=2)))
Copy across and down as required.
Note: If your actual table is in fact much larger than that given, it will probably be worth adding a simple clause into the above to the effect that the results for approximately half of the cells are obtained from their symmetrical counterparts, rather than via calculation of this construction, thus saving resource.
Regards
I have two Matlab arrays A(containing groups of numbers) and B(containing values that belong to the groups in A), there are repeats in array A
A = [1 1 1 2 2 3 4 4 4 4 4]
B = [1 2 3 3 5 4 4 1 6 7 8]
Now i would like to get the following two maps:
C = ['1': {1,2,3}, '2': {3,5}, '3':{4}, '4':{1,6,7,8}]
where C gives a map with the group number as index and related values in that particular group.
D = ['1':{2,4},'2':{1},'3':{4},'4':{1}]
Where D gives a map with the group number from A as index. The values are the group numbers from A for which there are repeated values in B for that particular sub group.
What is the most efficient way of dealing with this problem? Are maps a good data structure to store this kind of data. I know the first one can be dealt with a for loop which i would like to avoid.
I don't understand how you get to D.
For C, you can use accumarray:
C = accumarray(A,B,[],#(x){x})
C{1} is [1 2 3]