I've written a program to find shortest path in a N*N grid recursively.
def dfs(x,y,Map,p):
N = len(Map)
p += [[x,y]]
if Map[x][y] == 'E':
return p
for i in [[x-1,y],[x+1,y],[x,y-1],[x,y+1]]:
if N > i[0] >= 0 and N > i[1] >= 0 :
if (Map[i[0]][i[1]] == 'P' or Map[i[0]][i[1]] == 'E') and i not in p:
dfs(i[0], i[1], Map,p)
return []
When Map[x][y] = 'E' the recursion don't stop and return p. But it goes till the end. How to correct it and return the path(p).
By the looks of it, the code is prone to loop indefinitely. This is due to lack of checks whether you've entered a node before and moving in all (4) directions from a given node.
To solve it simply, add another array NxN of Boolean values answering the question: visited?. Then update the code to something along the lines:
def dfs(x,y,Map,visited,p):
visited[x,y] = true;
N = len(Map)
(...)
if (Map[i[0]][i[1]] == 'P' or Map[i[0]][i[1]] == 'E')
and i not in p
and visited[i[0], i[1]] == false:
dfs(i[0], i[1], Map,visited,p)
Related
I am working on an algorithms problem. You have an array numbers, size of array t , number number_of_elements and number multiplication_value. You have to find any set of number_of_elements indexes of the elements of the array , which product will be equal to multiplication_value. It is guaranteed, that such set of indexes exists
That problem looks like 2 sum, but I can't extrapolate it to my case.
I have tried naive algorithm for O(n), but it fails, when you have bad first number in an array. I think there is a way to use recursion in here. I guess it is well-known problem, but I couldn't find the solution
Example in:
t = 7
number_of_elements = 2
multiplication_value = 27
numbers = [9,1,1,27,3,27,3]
Example out:
1 3
My code ideas:
def return_index_values(numbers,multiplication_value,number_of_elements):
cur_number = int(multiplication_value)
list_of_indexes = []
values = []
for i in range(len(numbers)):
if ((cur_number == 1) and (len(values) == number_of_elements)):
print(values)
#finishing if everything worked
break
else:
if (cur_number % int(numbers[i]) == 0):
if(len(values) < number_of_elements):
#pushing values if possible
values.append(int(numbers[i]))
list_of_indexes.append(i)
cur_number = int(cur_number / int(numbers[i]))
print(cur_number)
else:
pass
if(len(values) == number_of_elements):
if mult_check(values,int(multiplication_value)):
#mult_check checks if the array's element multiplication gives a value
break
else:
#started dealing with bad cases, but it doesn't work properly
values.sort()
val_popped = values.pop()
cur_number = cur_number * val_popped
Bad case for my code
numbers = [9,3,1,27,3,27,3]
Here is one implementation. Not necessarily the best solution but it gives you some sense of how it can be done.
It first sorts the numbers by the element keeping the indices information. Then it performs recursion calls.
number_of_elements = 2
multiplication_value = 27
numbers = [9,1,1,27,3,27,3]
def preprocess(numbers, multiplication_value, number_of_elements):
l = []
for i, num in enumerate(numbers):
l.append((num, i))
return sorted(l, key = lambda tup: tup[0])
def subroutine(numbers, multiplication_value, number_of_elements, idx_start, result):
if idx_start >= len(numbers):
return False
if number_of_elements == 0:
return True if multiplication_value == 1 else False
for i in range(idx_start, len(numbers)):
num = numbers[i][0]
if num <= multiplication_value:
if multiplication_value % num == 0:
idx = numbers[i][1]
result.append(idx)
found = subroutine(numbers, multiplication_value / num, number_of_elements - 1, i + 1, result)
if not found:
del result[-1]
else:
return True
else:
return False
return False
result = []
processed_numbers = preprocess(numbers, multiplication_value, number_of_elements)
subroutine(processed_numbers, multiplication_value, number_of_elements, 0, result)
print(result)
You can use itertools.combinations() (https://www.geeksforgeeks.org/itertools-combinations-module-python-print-possible-combinations/) to select number_of_elements entries from your list in all possible ways, then check each whether they multiply to the required number.
I am trying to return the length of a common substring between two strings. I'm very well aware of the DP solution, however I want to be able to solve this recursively just for practice.
I have the solution to find the longest common subsequence...
def get_substring(str1, str2, i, j):
if i == 0 or j == 0:
return
elif str1[i-1] == str2[j-1]:
return 1 + get_substring(str1, str2, i-1, j-1)
else:
return max(get_substring(str1, str2, i, j-1), get_substring(str1, str2, j-1, i))
However, I need the longest common substring, not the longest common sequence of letters. I tried altering my code in a couple of ways, one being changing the base case to...
if i == 0 or j == 0 or str1[i-1] != str2[j-1]:
return 0
But that did not work, and neither did any of my other attempts.
For example, for the following strings...
X = "AGGTAB"
Y = "BAGGTXAYB"
print(get_substring(X, Y, len(X), len(Y)))
The longest substring is AGGT.
My recursive skills are not the greatest, so if anybody can help me out that would be very helpful.
package algo.dynamic;
public class LongestCommonSubstring {
public static void main(String[] args) {
String a = "AGGTAB";
String b = "BAGGTXAYB";
int maxLcs = lcs(a.toCharArray(), b.toCharArray(), a.length(), b.length(), 0);
System.out.println(maxLcs);
}
private static int lcs(char[] a, char[] b, int i, int j, int count) {
if (i == 0 || j == 0)
return count;
if (a[i - 1] == b[j - 1]) {
count = lcs(a, b, i - 1, j - 1, count + 1);
}
count = Math.max(count, Math.max(lcs(a, b, i, j - 1, 0), lcs(a, b, i - 1, j, 0)));
return count;
}
}
You need to recurse on each separately. Which is easier to do if you have multiple recursive functions.
def longest_common_substr_at_both_start (str1, str2):
if 0 == len(str1) or 0 == len(str2) or str1[0] != str2[0]:
return ''
else:
return str1[0] + longest_common_substr_at_both_start(str1[1:], str2[1:])
def longest_common_substr_at_first_start (str1, str2):
if 0 == len(str2):
return ''
else:
answer1 = longest_common_substr_at_both_start (str1, str2)
answer2 = longest_common_substr_at_first_start (str1, str2[1:])
return answer2 if len(answer1) < len(answer2) else answer1
def longest_common_substr (str1, str2):
if 0 == len(str1):
return ''
else:
answer1 = longest_common_substr_at_first_start (str1, str2)
answer2 = longest_common_substr(str1[1:], str2)
return answer2 if len(answer1) < len(answer2) else answer1
print(longest_common_substr("BAGGTXAYB","AGGTAB") )
I am so sorry. I didn't have time to convert this into a recursive function. This was relatively straight forward to compose. If Python had a fold function a recursive function would be greatly eased. 90% of recursive functions are primitive. That's why fold is so valuable.
I hope the logic in this can help with a recursive version.
(x,y)= "AGGTAB","BAGGTXAYB"
xrng= range(len(x)) # it is used twice
np=[(a+1,a+2) for a in xrng] # make pairs of list index values to use
allx = [ x[i:i+b] for (a,b) in np for i in xrng[:-a]] # make list of len>1 combinations
[ c for i in range(len(y)) for c in allx if c == y[i:i+len(c)]] # run, matching x & y
...producing this list from which to take the longest of the matches
['AG', 'AGG', 'AGGT', 'GG', 'GGT', 'GT']
I didn't realize getting the longest match from the list would be a little involved.
ls= ['AG', 'AGG', 'AGGT', 'GG', 'GGT', 'GT']
ml= max([len(x) for x in ls])
ls[[a for (a,b) in zip(range(len(ls)),[len(x) for x in ls]) if b == ml][0]]
"AGGT"
Using the Ruby language, have the function LongestMatrixPath(strArr) take the array of strings stored in strArr, which will be an NxM matrix of positive single-digit integers, and find the longest increasing path composed of distinct integers. When moving through the matrix, you can only go up, down, left, and right. For example: if strArr is ["345", "326", "221"], then this looks like the following matrix:
3 4 5
3 2 6
2 2 1
For the input above, the longest increasing path goes from: 3 -> 4 -> 5 -> 6. Your program should return the number of connections in the longest path, so therefore for this input your program should return 3. There may not necessarily always be a longest path within the matrix.
def recursiveFunction(array, row, col, path)
current = array[row][col]
row-1 >= 0 ? up = array[row-1][col] : up = nil
row+1 < array.length ? down = array[row+1][col] : down = nil
col-1 >= 0 ? left = array[row][col-1] : left = nil
col+1 < array[row].length ? right = array[row][col+1] : right =
nil
if !path.include?(current) && current.to_i > path[-1].to_i
path.push(current)
p path
if up && up.to_i > current.to_i
recursiveFunction(array, (row-1), col, path)
elsif down && down.to_i > current.to_i
recursiveFunction(array, (row+1), col, path)
elsif left && left.to_i > current.to_i
recursiveFunction(array, row, (col-1), path)
elsif right && right.to_i > current.to_i
recursiveFunction(array, row, (col+1), path)
end
end
path
end
def LongestMatrixPath(strArr)
# turn each string into an array
strArr.map!{|str| str.split("")}
# calculate up, down, left, right formula
# up = [-1][same]
# down = [+2][same]
# left = [same][-1]
# right = [same][+1]
# create data structure to store paths
path = []
# create loop that loops through each element in each array
inside strArr
# then apply recursive function to each element
longest_path = 0
row = 0
while row < strArr.length
col = 0
while col < strArr[row].length
result = recursiveFunction(strArr, row, col, path=[])
longest_path = result.length if result.length > longest_path
col += 1
end
row += 1
end
# return number of connections
result.length-1
end
# LongestMatrixPath(["345", "326", "221"])
# correct answer: 3, my answer: 3
# LongestMatrixPath(["12256", "56219", "43215"])
# correct answer: 4, my answer: 3
# LongestMatrixPath(["67", "21", "45"])
# correct answer: 3, my answer: 1
# LongestMatrixPath(["111", "111", "111"])
# correct answer: 0, my answer: 0
# LongestMatrixPath(["123", "456", "789"])
# correct answer: 4, my answer: 4
The problem exists in this fragment:
if up && up.to_i > current.to_i
recursiveFunction(array, (row-1), col, path)
elsif down && down.to_i > current.to_i
recursiveFunction(array, (row+1), col, path)
elsif left && left.to_i > current.to_i
recursiveFunction(array, row, (col-1), path)
elsif right && right.to_i > current.to_i
recursiveFunction(array, row, (col+1), path)
end
Here, if up is larger than current and, for example, left is also larger than current, you only go up (because of elsif). Instead, you need to go in all possible directions and choose most "promising" one. I altered your code a bit to achieve that, trying to change as little as possible. This works, but might not be the best option though.
def recursiveFunction(array, row, col, original_path)
current = array[row][col]
row-1 >= 0 ? up = array[row-1][col] : up = nil
row+1 < array.length ? down = array[row+1][col] : down = nil
col-1 >= 0 ? left = array[row][col-1] : left = nil
col+1 < array[row].length ? right = array[row][col+1] : right =
nil
path = original_path.dup
if !path.include?(current) && current.to_i > path[-1].to_i
path.push(current)
# p path
options = []
if up && up.to_i > current.to_i
options << recursiveFunction(array, (row-1), col, path)
end
if down && down.to_i > current.to_i
options << recursiveFunction(array, (row+1), col, path)
end
if left && left.to_i > current.to_i
options << recursiveFunction(array, row, (col-1), path)
end
if right && right.to_i > current.to_i
options << recursiveFunction(array, row, (col+1), path)
end
best_path = options.sort_by{ |p| p.length }.last
path += best_path.drop(path.length) if best_path
end
path
end
Not a couple of thing here:
Instead modifying path parameter given into function, it gets duplicated, so you can work on it in many steps.
All conditions for up, down, left and right are checked now. The results are put into an array and then the best result (in terms of length) is chosen.
Then the remainder of this longest path is appended to path with path += best_path.drop(path.length) (if exists)
The results for examples you gave are correct, except example two, where correct answer is 5, not 4.
There is much more that can be improved in this code though. For example, you can do strArr.map!{|str| str.split("").map(&:to_i)} and get rid of all those .to_i in the code.
This relates to the Coursera Scala course so I want to directly ask you NOT to give me the answer to the problem, but rather to help me debug why something is happening, as a direct answer would violate the Coursera honor code.
I have the following code:
def balance(chars: List[Char]): Boolean = {
val x = 0
def loop(list: List[Char]): Boolean = {
println(list)
if (list.isEmpty) if(x == 0) true
else if (list.head == '(') pushToStack(list.tail)
else if (list.head == ')') if(x <= 0) false else decreaseStack(list.tail)
else loop(list.tail)
true
}
def pushToStack(myList: List[Char]) { x + 1; loop(myList)}
def decreaseStack(myList: List[Char]) { x - 1; loop(myList)}
loop(chars)
}
A simple explanation:
If the code sees a "(" then it adds 1 to a variable. If it sees a ")" then it first checks whether the variable is equal to or smaller than 0. If this is the case, it returns false. If the value is bigger than 0 then it simply decreases one from the variable.
I have tried running the following:
if(balance("This is surely bad :-( ) (".toList)) println("balanced") else println("not balanced");
Clearly this is not balanced, but my code is returning balanced.
Again: I am not asking for help in writing this program, but rather help in explained why the code is returning "balanced" when clearly the string is not balanced
--EDIT--
def balance(chars: List[Char]): Boolean = {
val temp = 0;
def loop(list: List[Char], number: Int): Boolean = {
println(list)
if (list.isEmpty) if(number == 0) true
else if (list.head == '(') loop(list.tail, number + 1)
else if (list.head == ')') if(number <= 0) false else loop(list.tail, number - 1)
else loop(list.tail,number)
true
}
loop(chars,0)
}
^^ Still prints out balanced
You are using an immutable x when you really want a mutable x.
Here, let me rewrite it for you in a tail recursive style to show you what you're actually doing:
#tailrec def loop(myList: List[Char], cur: Int = 0): Boolean = myList match{
case "(" :: xs =>
val tempINeverUse = cur+1
loop(xs, cur) //I passed in 0 without ever changing "cur"!
case ")" :: xs if cur < 0 => false //This is a bug, regardless if you fix the rest of it
case ")" :: xs =>
val tempINeverUse = cur-1
loop(xs, cur) //Passed in 0 again!
case x :: xs => loop(xs, cur)
case Nil => cur == 0 //Since I've never changed it, it will be 0.
}
You need to keep a context of parenthesis in comments or in quotes as well. You can use a counter to achieve that. If the counter is set for a comment or a double quote then ignore any parenthesis that comes your way. Reset the counter whenever you find a finishing comment or double quote
Theres a cute little game on Android called Traffic Jam
I've written a recursive solver:
import copy,sys
sys.setrecursionlimit(10000)
def lookup_car(car_string,ingrid):
car=[]
i=0
for row in ingrid:
j=0
for cell in row:
if cell == car_string:
car.append([i,j])
j+=1
i+=1
return car
#True if up/down False if side to side
def isDirectionUp(car):
init_row=car[0][0]
for node in car:
if node[0] != init_row:
return True
return False
def space_up(car,grid):
top_node=car[0]
m_space_up = (top_node[0]-1,top_node[1])
if top_node[0] == 0:
return -1
elif grid[m_space_up[0]][m_space_up[1]] != " ":
return -1
else:
return m_space_up
def space_down(car,grid):
bottom_node = car[-1]
m_space_down = (bottom_node[0]+1,bottom_node[1])
if bottom_node[0] == 5 :
return -1
elif grid[m_space_down[0]][m_space_down[1]] != " ":
return -1
else:
return m_space_down
def space_left(car,grid):
left_node = car[0]
m_space_left = (left_node[0],left_node[1]-1)
if left_node[1] == 0 :
return -1
elif grid[m_space_left[0]][m_space_left[1]] != " ":
return -1
else:
return m_space_left
def space_right(car,grid):
right_node = car[-1]
m_space_right = (right_node[0],right_node[1]+1)
if right_node[1] == 5 :
return -1
elif grid[m_space_right[0]][m_space_right[1]] != " ":
return -1
else:
return m_space_right
def list_moves(car,grid):
ret =[]
if isDirectionUp(car):
up = space_up(car,grid)
if up != -1:
ret.append(("UP",up))
down = space_down(car,grid)
if down != -1:
ret.append(("DOWN",down))
else:
left = space_left(car,grid)
if left != -1:
ret.append(("LEFT",left))
right = space_right(car,grid)
if right != -1:
ret.append(("RIGHT",right))
return ret
def move_car(car_string,move,ingrid):
grid = copy.deepcopy(ingrid)
car = lookup_car(car_string,grid)
move_to = move[1]
front = car[0]
back = car[-1]
if(move[0] == "UP" or move[0] == "LEFT"):
grid[back[0]][back[1]] = " "
grid[move_to[0]][move_to[1]] = car_string
elif(move[0] == "DOWN" or move[0] == "RIGHT"):
grid[front[0]][front[1]] = " "
grid[move_to[0]][move_to[1]] = car_string
return grid
def is_solution(grid):
car = lookup_car("z",grid)
if(car[-1] == [2,5]):
return True
elif space_right(car,grid) == -1:
return False
else:
solgrid = move_car("z",("RIGHT",space_right(car,grid)),grid)
return is_solution(solgrid)
def print_grid(grid):
for row in grid:
print ''.join(row)
def solve(grid,solution,depth):
global stop
global state
if grid in state:
return
else:
state.append(grid)
if stop:
return
if is_solution(grid):
print_grid(grid)
print len(solution)
else:
for each in "abcdefzhijklm":
car = lookup_car(each,grid)
moves = list_moves(car,grid)
for move in moves:
solution.append((each,move))
moved_grid = move_car(each,move,grid)
solve(moved_grid,solution,depth)
stop=False
state=[]
recdepth=0
#grid file using a-w and x means free space, and ' ' means yellow car
grid=[list(x) for x in file(sys.argv[1]).read().split('\n')[0:-1]]
solve(grid,[],0)
WHERE grid is in a file:
abccdd
abeef
azzhfi
jjjhfi
kll
kmm
But, it takes over 8000 moves to find a solution ,and fails to find a simple 30 move solution. What is wrong with the algorithm?
If the branching factor of your search space is r then the number of vertices in the search tree to depth n is (1-r^n)/(1-r). A problem with a minimal 30 step solution, even in the simple case where r = 2, will have a search tree with around 2^30 - 1 ~= 1,000,000,000 vertices. Now, your branching factor is likely to be bigger than 2, so a 30 step problem is a long way from trivial!
That said, I'd be inclined to (a) find a better representation of your problem (searching arrays of strings is slow) and (b) consider best-first search where you guide your search with a heuristic (e.g., distance of the yellow car from its destination or the number of cars blocking the path of the yellow car).
Hope this helps.
This is essentially a (relatively token) searching problem, with a huge search space. As others recommended, read up on Depth-first search, and then read up on Breadth-first search, and when you learn the difference, read up on A* Search, and coming up with a pessimistic scoring function.
Also, note that in this case, you already know what the end state should be, thus, another approach would be to search from both ends and meet in the middle. This should reduce your search space considerably.
If that's still not enough, you can combine the techniques!