This might look dulicate question and is a bit relevant to
Alphabetical sorting in treeset not working
I have a Treeset containing
TreeSet<String> ts=new TreeSet<String>(String.CASE_INSENSITIVE_ORDER);
ts.add("D1");
ts.add("D9");
ts.add("D5");
ts.add("D3");
ts.add("D8");
which gives me,
Tree set :: [D1, D3, D5, D8, D9]
but if i add double digits to my "D"
ts.add("D1");
ts.add("D9");
ts.add("D5");
ts.add("D3");
ts.add("D8");
ts.add("D11");
ts.add("D18");
ts.add("D17");
ts.add("D13");
i get,
Tree set :: [D1, D11, D13, D17, D18, D3, D5, D8, D9]
which aint correct....Please help!
Sorting is performed alphabetically - all 1 (ones) are coming before 3, then comes 5, 8, 9.
If you want to have numeric sorting you need to implement your own comparator.
Related
I had an interview yesterday and was asked to give a method to find all of the pairs of numbers from a list which add up to an integer which is given separate to the list. The list can be infinitely long, but for example:
numbers = [11,1,5,27,7,18,2,4,8]
sum = 9
pairs = [(1,8),(5,4),(7,2)]
I got as far as sorting the list and eliminating all numbers greater than the sum number and then doing two nested for loops to take each index and iterate through the other numbers to check whether they sum up to the given number, but was told that there was a more efficient way of doing it...
I've been trying to figure it out but have nothing, other than doing the nested iteration backwards but that only seems marginally more efficient.
Any ideas?
This can be done in O(n) time and O(n) auxiliary space; testing for membership of a set takes O(1) time. Since the output also takes up to O(n) space, the auxiliary space should not be a significant issue.
def pairs_sum(numbers, k):
numbers_set = set(numbers)
return [(x, y) for x in numbers if (y := k - x) in numbers_set and x < y]
Example:
>>> pairs_sum([11, 1, 5, 27, 7, 18, 2, 4, 8], 9)
[(1, 8), (2, 7), (4, 5)]
It is kind of a classic and not sure stackoverflow is the right place to ask that kind of question.
Sort the list is acsending order
Two iterators one starting from the end of the list descending i1, one starting from the beginning of the list ascending i2
Loop
while i1 > i2
if (list[i1] + list[i2] == target)
store {list[i1], list[i2]) in results pairs
i1--
i2++
else if (list[i1] + list[i2] > target)
i1--
else if (list[i1] + list[i2] < target)
i2++
This should be in O(n) with n the length of the list if you avoid the sorting algorithm which can be done with a quick sort on average in O(n log n)
Note: this algorithm doesn't take into account the case where the input list have several times the same number
I'm trying to make a LibreOffice spreadsheet formula that populates a column based on another input column, comparing each input with a series of range pairs defined in another sheet and finally outputting a symbol based on matched criteria. I have a series of ranges that specify a - output, and another series that corresponds to +, but not all inputs will fall into a category. I am using this trinary output later for another expression, which I already have in place.
My question becomes: how can I test input against each range pair without spelling out the cell coordinates for each individual cell (ie OR(AND(">= $A$1", "< $B$1"), AND(">=$A$2", "<$B$2"), ...))? Ideally I could just specify an array to compare against like $A$1:$B$4. Writing it in a python macro would work, too, since I don't plan on sharing this file.
I wrote a really quick list comp in python to illustrate what I'm after. This snippet would be one half, such as testing - qualification, and these values may be fed into a condition that outputs the symbol:
>>> def cmp(f, r):
... return r[0] <= f < r[1]
>>> f = (1, 2, 3)
>>> ranges = ((2, 5), (4, 6), (3, 8))
>>> [any([cmp(i, r) for r in ranges]) for i in f]
[False, True, True]
Here is a small test example with real input and real ranges.
Change the range pairs so that they are in two columns starting from A13. Be sure that they are in sorted order (Data -> Sort).
A B C
~~~~~~~~ ~~~~~~~~ ~
145.1000 145.5000 -
146.0000 146.4000 +
146.6000 147.0000 -
147.0000 147.4000 +
147.6000 148.0000 -
440.0000 445.0000 +
In each row, specify whether it is negative or positive. To do this, I entered the following formula in C13 and filled down. If the range pairs are not consistent enough then enter values for C13 and below manually.
=IF(ISODD(ROW());"-";"+")
Now, enter the following formula in cell C3 and fill down.
=IFNA(IF(
VLOOKUP(A3;A$13:C$18;2;1) >= A3;
VLOOKUP(A3;A$13:C$18;3;1);
"None");"None")
The formula finds the closest pair and then checks if the number is inside that range or not. For better testing, I would also suggest using 145.7000 as input, which should result in no shift if I understood the question correctly.
The results in column C:
-
+
None
None
Documentation: VLOOKUP, IFNA, ROW.
EDIT:
The following formula produces correct results for the example data you gave, and it works for anything between 144.0 and 148.0.
=IFNA(VLOOKUP(A3;A$13:C$18;3;1); "None")
However, 150.0 produces - and 550.0 produces +. If that is not what you want, then use the formula above that has two VLOOKUP expressions.
I'm currently reading Programming in Lua Fourth Edition and I'm already stuck on the first exercise of "Chapter 2. Interlude: The Eight-Queen Puzzle."
The example code is as follows:
N = 8 -- board size
-- check whether position (n, c) is free from attacks
function isplaceok (a, n ,c)
for i = 1, n - 1 do -- for each queen already placed
if (a[i] == c) or -- same column?
(a[i] - i == c - n) or -- same diagonal?
(a[i] + i == c + n) then -- same diagonal?
return false -- place can be attacked
end
end
return true -- no attacks; place is OK
end
-- print a board
function printsolution (a)
for i = 1, N do -- for each row
for j = 1, N do -- and for each column
-- write "X" or "-" plus a space
io.write(a[i] == j and "X" or "-", " ")
end
io.write("\n")
end
io.write("\n")
end
-- add to board 'a' all queens from 'n' to 'N'
function addqueen (a, n)
if n > N then -- all queens have been placed?
printsolution(a)
else -- try to place n-th queen
for c = 1, N do
if isplaceok(a, n, c) then
a[n] = c -- place n-th queen at column 'c'
addqueen(a, n + 1)
end
end
end
end
-- run the program
addqueen({}, 1)
The code's quite commented and the book's quite explicit, but I can't answer the first question:
Exercise 2.1: Modify the eight-queen program so that it stops after
printing the first solution.
At the end of this program, a contains all possible solutions; I can't figure out if addqueen (n, c) should be modified so that a contains only one possible solution or if printsolution (a) should be modified so that it only prints the first possible solution?
Even though I'm not sure to fully understand backtracking, I tried to implement both hypotheses without success, so any help would be much appreciated.
At the end of this program, a contains all possible solutions
As far as I understand the solution, a never contains all possible solutions; it either includes one complete solution or one incomplete/incorrect one that the algorithm is working on. The algorithm is written in a way that simply enumerates possible solutions skipping those that generate conflicts as early as possible (for example, if first and second queens are on the same line, then the second queen will be moved without checking positions for other queens, as they wouldn't satisfy the solution anyway).
So, to stop after printing the first solution, you can simply add os.exit() after printsolution(a) line.
Listing 1 is an alternative to implement the requirement. The three lines, commented respectively with (1), (2), and (3), are the modifications to the original implementation in the book and as listed in the question. With these modifications, if the function returns true, a solution was found and a contains the solution.
-- Listing 1
function addqueen (a, n)
if n > N then -- all queens have been placed?
return true -- (1)
else -- try to place n-th queen
for c = 1, N do
if isplaceok(a, n, c) then
a[n] = c -- place n-th queen at column 'c'
if addqueen(a, n + 1) then return true end -- (2)
end
end
return false -- (3)
end
end
-- run the program
a = {1}
if not addqueen(a, 2) then print("failed") end
printsolution(a)
a = {1, 4}
if not addqueen(a, 3) then print("failed") end
printsolution(a)
Let me start from Exercise 2.2 in the book, which, based on my past experience to explain "backtracking" algorithms to other people, may help to better understand the original implementation and my modifications.
Exercise 2.2 requires to generate all possible permutations first. A straightforward and intuitive solution is in Listing 2, which uses nested for-loops to generate all permutations and validates them one by one in the inner most loop. Although it fulfills the requirement of Exercise 2.2, the code does look awkward. Also it is hard-coded to solve 8x8 board.
-- Listing 2
local function allsolutions (a)
-- generate all possible permutations
for c1 = 1, N do
a[1] = c1
for c2 = 1, N do
a[2] = c2
for c3 = 1, N do
a[3] = c3
for c4 = 1, N do
a[4] = c4
for c5 = 1, N do
a[5] = c5
for c6 = 1, N do
a[6] = c6
for c7 = 1, N do
a[7] = c7
for c8 = 1, N do
a[8] = c8
-- validate the permutation
local valid
for r = 2, N do -- start from 2nd row
valid = isplaceok(a, r, a[r])
if not valid then break end
end
if valid then printsolution(a) end
end
end
end
end
end
end
end
end
end
-- run the program
allsolutions({})
Listing 3 is equivalent to List 2, when N = 8. The for-loop in the else-end block does what the whole nested for-loops in Listing 2 do. Using recursive call makes the code not only compact, but also flexible, i.e., it is capable of solving NxN board and board with pre-set rows. However, recursive calls sometimes do cause confusions. Hope the code in List 2 helps.
-- Listing 3
local function addqueen (a, n)
n = n or 1
if n > N then
-- verify the permutation
local valid
for r = 2, N do -- start from 2nd row
valid = isplaceok(a, r, a[r])
if not valid then break end
end
if valid then printsolution(a) end
else
-- generate all possible permutations
for c = 1, N do
a[n] = c
addqueen(a, n + 1)
end
end
end
-- run the program
addqueen({}) -- empty board, equivalent allsolutions({})
addqueen({1}, 2) -- a queen in 1st row and 1st column
Compare the code in Listing 3 with the original implementation, the difference is that it does validation after all eight queens are placed on the board, while the original implementation validates every time when a queen is added and will not go further to next row if the newly-added queen causes conflicts. This is all what "backtracking" is about, i.e. it does "brute-force" search, it abandons the search branch once it finds a node that will not lead to a solution, and it has to reach a leaf of the search tree to determine it is a valid solution.
Back to the modifications in Listing 1.
(1) When the function hits this point, it reaches a leaf of the search tree and a valid solution is found, so let it return true representing success.
(2) This is the point to stop the function from further searching. In original implementation, the for-loop continues regardless of what happened to the recursive call. With modification (1) in place, the recursive call returns true if a solution was found, the function needs to stop and to propagate the successful signal back; otherwise, it continues the for-loop, searching for other possible solutions.
(3) This is the point the function returns after finishing the for-loop. With modification (1) and (2) in place, it means that it failed to find a solution when the function hits this point, so let it explicitly return false representing failure.
Just as a side note, this isn't asking someone to help with homework it's just a example question in my lecture notes that i'm not grasping and would greatly appreciate if someone could explain to me how exactly to work it out.
It says simply this:
Let G = {3,5,7}. Write down some examples of 4-tuples.
Thank you to anyone who tries to help, this is mathematics to understand a systems unit :)
Your collection G is a set, which is unordered and non-repeating, containing three elements. You want some 4-tuples, which is an ordered collection of possibly-repeating elements, and there must be 4 elements.
We show tuples by using parentheses around the collection, while a set like G is written using braces (curly brackets). Some examples of 4-tuples using the elements of G are
(3, 3, 3, 3)
(3, 3, 3, 5)
(3, 3, 3, 7)
(3, 3, 5, 3)
...
(7, 7, 7, 5)
(7, 7, 7, 7)
That list of mine was in a particular order, called the lexigraphical order. Since there are 4 elements and each element has 3 choices regardless of the other choices, the total number of 4-tuples is
3x3x3x3 = 81
As another answer implied, your question is somewhat ambiguous. I assumed that each 4-tuple was to have elements taken from your set G, but your question did not actually say that. It does seem to be implied, however.
A 4-tuple, in math, is any vector populated with 4 objects. An example of a 4-tuple is {1, 5, 3, 7}.
Without more context, I can't use the fact that G = {3, 5, 7}.
I am designing an 8-bit comparator in Xilinx ISE Project Navigator. My goal is to add four 2-bit comparators, as shown at the picture. The input is a 16-bit literal, of which the first 8 bits are number A, the second are number B (SW(15:8) -> A; SW(7:0) -> B). There are two inputs BTN0 and BTN1, I use BTN0 to give the first comparator the EQ input value 1.
In ISim, the comparison works fine if the two numbers are equal, but gets weird when I try with two different numbers. I am working from several sources and I'm a beginner at all this, so there could be easily a bug/error I didn't think about.
http://25.media.tumblr.com/4e443e33d84b43e80e4f595b0044ab86/tumblr_mjd7vttpuc1r65yueo1_1280.png
I am afraid the 2-bit comparator is not correct. For example, if A1 = 1, A0 = 0, B1 = 0, and B0 = 0, the output of the AND3B1 is 0, and the output of the AND4B1 will also be 0, so AG = 0.