As an example question we are asked to create a variant of merge sort where it splits array in to c>2 arrays of roughly equal size (when c = 2 it will use regular merge)
This is the solution:
cmerge(a1, a2, .... ac)
if c = 2 return Merge(a1, a2)
else
b1 = cmerge(a1, a2,...a(c/2), floor(c/2))
b2 = cmerge(a(roof(c/2)),... ac, roof(c/2))
return merge(b1,b2)
I understand where they got this solution but when it asks for the reccurrnce relationi get a bit confused. in the question they state c sorted arrays can be merged in O(n log c)
which makes the first case clear
T(n) = c(T(n/c)) + nlogc if n>= c
T(n) = theta(n^2) for n<c <----------- can someone explain where they are getting this?
I just reallly don't get where the n^2 is coming from. In the second case is there not just one call to MERGE? And merge is theta(n) if I remember.
Related
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.
I have some tricky question encountered on an technical exam, hope I can get an output here
In which of the following statement(s) is(are) equivalent to the condition a AND (b OR c)
a. NOT(NOT a OR NOT b) OR NOT (NOT a OR NOT c)
b. NOT(NOT a OR (NOT b AND NOT c))
c. NOT(NOT a OR NOT b) AND (NOT a OR NOT c)
d. (NOT a AND NOT b) OR NOT c
My question, if the equation is
NOT a OR NOT b
Does it give a result b or a is acceptable since it uses OR?
The answer is to write a truth table.
OK, you might be a super-logician and break these down in your head,
but the interviewer is expecting you to draw a truth table for each of the 4 parts and say which final column matches.
So your first 3 columns are a,b,c, then some intermediate columns, then a result column.
All that matters is if the result columns match.
I am trying to write a recursive function that will return true if second number is power of first number.
For example:
find_power 3 9 will return true
find_power 2 9 will return false because the power of 2 is 8 not 9
This is what I have tried but I need a recursive solution
let rec find_power first second =
if (second mod first = 0)
return true
else
false ;;
A recursive function has the following rough form
let rec myfun a b =
if answer is obvious then
obvious_answer
else
let (a', b') = smaller_example_of_same_problem a b in
myfun a' b'
In your case, I'd say the answer is obvious if the second number is not a multiple of the first or if it's 1. That is essentially all your code is doing now, it's testing the obvious part. (Except you're not handling the 0th power, i.e., 1.)
So, you need to figure out how to make a smaller example of the same problem. You know (by hypothesis) that the second number is a multiple of the first one. And you know that x * a is a power of a if and only if x is a power of a. Since x is smaller than x * a, this is a smaller example of the same problem.
This approach doesn't work particularly well in some edge cases, like when the first number is 1 (since x is not smaller than x * 1). You can probably handle them separately.
I want to use the BLAS package. To do so, the meaning of the two first parameters of the gemm() function is not evident for me.
What do the parameters 'N' and 'T' represent?
BLAS.gemm!('N', 'T', lr, alpha, A, B, beta, C)
What is the difference between BLAS.gemm and BLAS.gemm! ?
According to the documentation
gemm!(tA, tB, alpha, A, B, beta, C)
Update C as alpha * A * B + beta*C or the other three variants according to tA (transpose A) and tB. Returns the updated C.
Note: here, alpha and beta must be float type scalars. A, B and C are all matrices. It's up to you to make sure the matrix dimensions match.
Thus, the tA and tB parameters refer to whether you want to apply the transpose operation to A or to B before multiplying. Note that this will cost you some computation time and allocations - the transpose isn't free. (thus, if you were going to apply the multiplication many times, each time with the same transpose specification, you'd be better off storing your matrix as the transposed version from the beginning). Select N for no transpose, T for transpose. You must select one or the other.
The difference between gemm!() and gemv!() is that for gemm!() you already need to have allocated the matrix C. The ! is a "modify in place" signal. Consider the following illustration of their different uses:
A = rand(5,5)
B = rand(5,5)
C = Array(Float64, 5, 5)
BLAS.gemm!('N', 'T', 1.0, A, B, 0.0, C)
D = BLAS.gemm('N', 'T', 1.0, A, B)
julia> C == D
true
Each of these, in essence, perform the calculation C = A * B'. (Technically, gemm!() performs C = (0.0)*C + (1.0)*A * B'.)
Thus, the syntax for the modify in place gemm!() is a bit unusual in some respects (unless you've already worked with a language like C in which case it seems very intuitive). You don't have the explicit = sign like you frequently do when calling functions in assigning values in a high level object oriented language like Julia.
As the illustration above shows, the outcome of gemm!() and gemm() in this case is identical, even though the syntax and procedure to achieve that outcome is a bit different. Practically speaking, however, performance differences between the two can be significant, depending on your use case. In particular, if you are going to be performing that multiplication operation many times, replacing/updating the value of C each time, then gemm!() can be a decent bit quicker because you don't need to keep re-allocating new memory each time, which does have time costs, both in the initial memory allocation and then in the garbage collection later on.
For the below quardratic formula, I have multiple a but fixed b and c.
I wish to write a partial application function, which execute efficiently, i.e., my function doesn't recompute fixed values (because of b and c).
Here is my solution
let r b c = let z = b *. b in fun a -> (-.b +. sqrt (z-.4.0*.a*.c))/.(a*.2.0);;
I guess this solution can work, but I am not sure whether it is efficient enough. I just made b^2 to be fixed as I saw other parts are all with a.
Anyone can give me a better solution?
Yeah, that's a correct way to deal with the situation at hand. The alternate form doesn't help much (as long this obtains the accuracy you require). You may want to move the 4*c out as well,
let r b c = let z = b *. b and c4 = 4.0 *. c in
fun a -> (-.b +. sqrt (z-.a*.c4))/.(a*.2.0);;