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Including backtrace pointers with symbols in a Python output of a DP table for min edit distance algorythm

I have a working min edit distance program for two words that also iterates and outputs a DP table to the console, however I wish to add a backtrace to the programme that also prints arrow pointer symbols in the DP table to show the backtrace clearly on the output. I cannot figure out how to output these symbols correctly.
import re
import time
time1= time.time()
def printTable(table, description):
print(f'{description}\n')
current_row = current_col = 0
current_row_col = re.search("^row ([0-9]+) , col ([0-9]+)$",description)
if current_row_col:
current_row = int(current_row_col.group(1))
current_col= int(current_row_col.group(2))
row_counter=0
for row in table:
row_counter+=1
col_counter=0
for col in row:
col_counter+=1
#print(row_counter , row, current_col, col)
if (row_counter == current_row) and (col_counter == current_col):
formatting = '\033[1m'+'\033[91m' #bold + red
else: formatting = '\x1b[0m' #reset fomatting
print(formatting + str(col).rjust(10, ' '), end=' ') # rjust returns a 10-characters long, right justified version of the string
print('\n\n')
print('---------------------------------------------------------------------------------------------------------------')
# A DP-based solution for edit distance problem
def editDistDP(x,y):
leftarrow = "←"
uparrow = "↑"
diagarrow = ""
dp = [] # Create an empty table to store results of subproblems
# fill in the table with zeros
for row in range(len(x) + 1):
dp.append([0]* (len(y) + 1))
# Alternatively, you can use List Comprehension to initiate the DP table in one line of code
# dp = [[0 for column in range(len(y) + 1)] for row in range(len(x) + 1)]
# Fill in the base case (easy) subproblems, i.e. the first row and column of the DP table
# first row: base case subproblems for computing the cost of converting "" to y
for i in range(len(y) + 1):
# If x is empty then the only option is to insert all the characters of y
# Minimum number of required operations (cost) is i insertions, where i = len(y)
dp[0][i] = i
# first column: base case subproblems for computing the cost of converting x to ""
for i in range(len(x) + 1):
# If y is empty then the only option is to delete all the characters of x
# Minimum number of required operations (cost) is i deletions, where i = len(x)
dp[i][0] = i
printTable(dp,"DP table after the base case (easy) subproblems are solved");
# Fill in the rest of the DP table in a BOTTOM-UP manner
for i in range(1, len(x) + 1):
for j in range(1, len(y) + 1):
horizontal_or_insertion_cost = (dp[i][j-1] + 1)
vertical_or_deletion_cost= dp[i-1][j] + 1
# Weighted Minimum Edit Distance for sub
if x[i-1] != y[j-1] and x[i-1].isnumeric:
delta = 3
elif x[i-1] != y[j-1]:
delta = 2
else:
delta = 0
diagonal_or_substitution_cost= dp[i-1][j-1] + delta
minValue = min(horizontal_or_insertion_cost,vertical_or_deletion_cost,diagonal_or_substitution_cost)
dp[i][j] = minValue
# printTable(dp,f'row {i+1} , col {j+1}') #UNCOMMENT this line to see how the DP table is filled at each step
printTable(dp,"Completed DP table after all the subproblems are solved")
return dp[-1][-1]
str1, str2 = "intention", "execution"
print(f'edit distance between "{str1}" and "{str2}": {editDistDP(str1, str2)}')
time2 = time.time()
execTime = time2-time1
execTime = str(execTime)
print("--- Executed in: " + execTime + " seconds ---")

First Search Program - Artificial Intelligence for Robotics path printing

I'm new at Python programming and I'm doing my best to fully understand this code. Here we are printing the path for the First Search Program - Artificial Intelligence for Robotics algorithm. I know how the basic of these lines are working in general, but how they work here in this code. Could I get some clarification for this piece of code below, please?.
This is below the piece of code:
policy=[[' ' for row in range(len(grid[0]))] for col in range(len(grid))]
x=goal[0]
y=goal[1]
policy[x][y]='*'
while x !=init[0] or y !=init[1]:
x2=x-delta[action[x][y]][0]
y2=y-delta[action[x][y]][1]
policy[x2][y2]= delta_name[action[x][y]]
x=x2
y=y2
for i in range(len(policy)):
print(policy[i])
This is the code:
#grid format
# 0 = navigable space
# 1 = occupied space
grid = [[0,0,1,0,0,0],
[0,0,1,0,0,0],
[0,0,0,0,1,0],
[0,0,1,1,1,0],
[0,0,0,0,1,0]]
init = [0,0] #Start location is (0,0) which we put it in open list.
goal = [len(grid)-1,len(grid[0])-1] #Our goal in (4,5) and here are the coordinates of the cell.
#Below the four potential actions to the single field
delta = [[-1 , 0], #up by subtracting one from the first dimention, I mean the demension of (0,0)
[ 0 ,-1], #left
[ 1 , 0], #down
[ 0 , 1]] #right
delta_name = ['^','<','V','>'] #The name of above actions
cost = 1 #Each step costs you one
def search():
#open list elements are of the type [g,x,y]
#To check cells once they expanded and don't expand them again. We defined an array called closed
#and its size as our grid. It has two values 0 & 1. 0 means open and 1 means closed.
closed = [[0 for row in range(len(grid[0]))] for col in range(len(grid))]
action=[[-1 for row in range(len(grid[0]))] for col in range(len(grid))]
#We initialize the starting location as checked
closed[init[0]][init[1]] = 1
# we assigned the cordinates and g value
x = init[0]
y = init[1]
g = 0
#our open list will contain our initial value
open = [[g, x, y]]
found = False #flag that is set when search complete
resign = False #Flag set if we can't find expand
#print('initial open list:')
#for i in range(len(open)):
#print(' ', open[i])
#print('----')
while found is False and resign is False:
#Check if we still have elements in the open list
if len(open) == 0: #If our open list is empty, there is nothing to expand.
resign = True
print('Fail')
print('############# Search terminated without success')
else:
#if there is still elements on our list
#remove node from list
open.sort() #sort elements in an increasing order from the smallest g value up
open.reverse() #reverse the list
next = open.pop() #remove the element with the smallest g value from the list
#print('list item')
#print('next')
#Then we assign the three values to x,y and g. Which is our expantion.
x = next[1]
y = next[2]
g = next[0]
#Check if we are done
if x == goal[0] and y == goal[1]:
found = True
print(next) #The three elements above this "if".
print('############## Search is success')
else:
#expand winning element and add to new open list
for i in range(len(delta)): #going through all our actions the four actions
#We apply the actions to x and y with additional delta to construct x2 and y2
x2 = x + delta[i][0]
y2 = y + delta[i][1]
#if x2 and y2 falls into the grid
if x2 >= 0 and x2 < len(grid) and y2 >=0 and y2 <= len(grid[0])-1:
#if x2 and y2 not checked yet and there is not obstacles
if closed[x2][y2] == 0 and grid[x2][y2] == 0:
g2 = g + cost #we increment the cose
open.append([g2,x2,y2]) #we add them to our open list
#print('append list item')
#print([g2,x2,y2])
#Then we check them to never expand again
closed[x2][y2] = 1
action[x2][y2]=i
policy=[[' ' for row in range(len(grid[0]))] for col in range(len(grid))]
x=goal[0]
y=goal[1]
policy[x][y]='*'
while x !=init[0] or y !=init[1]:
x2=x-delta[action[x][y]][0]
y2=y-delta[action[x][y]][1]
policy[x2][y2]= delta_name[action[x][y]]
x=x2
y=y2
for i in range(len(policy)):
print(policy[i])
search()

How to add a stopping condition for Jacobian Matrix?

def jacobi(m,numiter=100):
#Number of rows determins the number of variables
numvars = m.shape[0]
#construct array for final iterations
history = np.zeros((numvars,numiter))
i = 1
while(i < numiter): #Loop for numiter
for v in range(numvars): # Loop over all variables
current = m[v,numvars] # Start with left hand side (augmented side of matrix)
for col in range(numvars): #Loop over columns
if v != col: # Don't count colume for current variable
current = current - (m[v,col]*history[col, i-1]) #subtract other guesses form previous timestep
current = current/m[v,v] #divide by current variable coefficent
history[v,i] = current #Add this answer to the rest
i = i + 1 #iterate
#plot each variable
for v in range(numvars):
plt.plot(history[v,: i]);
return history[:,i-1]
I have this code that calculates Jacobian method. How do I add a stopping condition for when the solutions converge? i.e. the values for the current iteration have changed less than some threshold e from the values for the previous iteration.
The threshold e will be an input to the function and the default value to 0.00001

You could add another condition to your while loop, so when it reaches your error threshold it stops.
def jacobi(m,numiter=100, error_threshold = 1e-4):
#Number of rows determins the number of variables
numvars = m.shape[0]
#construct array for final iterations
history = np.zeros((numvars,numiter))
i = 1
err = 10*error_threshold
while(i < numiter and err > error_threshold): #Loop for numiter and error threshold
for v in range(numvars): # Loop over all variables
current = m[v,numvars] # Start with left hand side (augmented side of matrix)
for col in range(numvars): #Loop over columns
if v != col: # Don't count colume for current variable
current = current - (m[v,col]*history[col, i-1]) #subtract other guesses form previous timestep
current = current/m[v,v] #divide by current variable coefficent
history[v,i] = current #Add this answer to the rest
#check error here. In this case the maximum error
if i > 1:
err = max((history[:,i] - history[:,i-1])/history[:,i-1])
i = i + 1 #iterate
#plot each variable
for v in range(numvars):
plt.plot(history[v,: i]);
return history[:,i-1]

Spawning objects in groups when the first object of the group was spawned randomly Python

I'm currently doing a project, and in the code I have, I'm trying to get trees .*. and mountains .^. to spawn in groups around the first tree or mountain which is spawned randomly, however, I can't figure out how to get the trees and mountains to spawn in groups around a single randomly generated point. Any help?
grid = []
def draw_board():
row = 0
for i in range(0,625):
if grid[i] == 1:
print("..."),
elif grid[i] == 2:
print("..."),
elif grid[i] == 3:
print(".*."),
elif grid[i] == 4:
print(".^."),
elif grid[i] == 5:
print("[T]"),
else:
print("ERR"),
row = row + 1
if row == 25:
print ("\n")
row = 0
return

There's a number of ways you can do it.
Firstly, you can just simulate the groups directly, i.e. pick a range on the grid and fill it with a specific figure.
def generate_grid(size):
grid = [0] * size
right = 0
while right < size:
left = right
repeat = min(random.randint(1, 5), size - right) # *
right = left + repeat
grid[left:right] = [random.choice(figures)] * repeat
return grid
Note that the group size need not to be uniformly distributed, you can use any convenient distribution, e.g. Poisson.
Secondly, you can use a Markov Chain. In this case group lengths will implicitly follow a Geometric distribution. Here's the code:
def transition_matrix(A):
"""Ensures that each row of transition matrix sums to 1."""
copy = []
for i, row in enumerate(A):
total = sum(row)
copy.append([item / total for item in row])
return copy
def generate_grid(size):
# Transition matrix ``A`` defines the probability of
# changing from figure i to figure j for each pair
# of figures i and j. The grouping effect can be
# obtained by setting diagonal entries A[i][i] to
# larger values.
#
# You need to specify this manually.
A = transition_matrix([[5, 1],
[1, 5]]) # Assuming 2 figures.
grid = [random.choice(figures)]
for i in range(1, size):
current = grid[-1]
next = choice(figures, A[current])
grid.append(next)
return grid
Where the choice function is explained in this StackOverflow answer.

How to deal with very big Bitboards

I'm working on a 2-player board game (e.g. connect 4), with parametric board size h, w. I want to check for winning condition using hw-sized bitboards.
In game like chess, where board size is fixed, bitboards are usually represented with some sort of 64-bit integer. When h and w are not constant and maybe very big (let's suppose 30*30) are bitboards a good idea? If so, are the any data types in C/C++ to deal with big bitboards keeping their performances?
Since I'm currently working on python a solution in this language is appreciated too! :)
Thanks in advance

I wrote this code while ago just to play around with the game concept. There is no intelligence behaviour involve. just random moves to demonstrate the game. I guess this is not important for you since you are only looking for a fast check of winning conditions. This implementation is fast since I did my best to avoid for loops and use only built-in python/numpy functions (with some tricks).
import numpy as np
row_size = 6
col_size = 7
symbols = {1:'A', -1:'B', 0:' '}
def was_winning_move(S, P, current_row_idx,current_col_idx):
#****** Column Win ******
current_col = S[:,current_col_idx]
P_idx= np.where(current_col== P)[0]
#if the difference between indexes are one, that means they are consecutive.
#we need at least 4 consecutive index. So 3 Ture value
is_idx_consecutive = sum(np.diff(P_idx)==1)>=3
if is_idx_consecutive:
return True
#****** Column Win ******
current_row = S[current_row_idx,:]
P_idx= np.where(current_row== P)[0]
is_idx_consecutive = sum(np.diff(P_idx)==1)>=3
if is_idx_consecutive:
return True
#****** Diag Win ******
offeset_from_diag = current_col_idx - current_row_idx
current_diag = S.diagonal(offeset_from_diag)
P_idx= np.where(current_diag== P)[0]
is_idx_consecutive = sum(np.diff(P_idx)==1)>=3
if is_idx_consecutive:
return True
#****** off-Diag Win ******
#here 1) reverse rows, 2)find new index, 3)find offest and proceed as diag
reversed_rows = S[::-1,:] #1
new_row_idx = row_size - 1 - current_row_idx #2
offeset_from_diag = current_col_idx - new_row_idx #3
current_off_diag = reversed_rows.diagonal(offeset_from_diag)
P_idx= np.where(current_off_diag== P)[0]
is_idx_consecutive = sum(np.diff(P_idx)==1)>=3
if is_idx_consecutive:
return True
return False
def move_at_random(S,P):
selected_col_idx = np.random.permutation(range(col_size))[0]
#print selected_col_idx
#we should fill in matrix from bottom to top. So find the last filled row in col and fill the upper row
last_filled_row = np.where(S[:,selected_col_idx] != 0)[0]
#it is possible that there is no filled array. like the begining of the game
#in this case we start with last row e.g row : -1
if last_filled_row.size != 0:
current_row_idx = last_filled_row[0] - 1
else:
current_row_idx = -1
#print 'col[{0}], row[{1}]'.format(selected_col,current_row)
S[current_row_idx, selected_col_idx] = P
return (S,current_row_idx,selected_col_idx)
def move_still_possible(S):
return not (S[S==0].size == 0)
def print_game_state(S):
B = np.copy(S).astype(object)
for n in [-1, 0, 1]:
B[B==n] = symbols[n]
print B
def play_game():
#initiate game state
game_state = np.zeros((6,7),dtype=int)
player = 1
mvcntr = 1
no_winner_yet = True
while no_winner_yet and move_still_possible(game_state):
#get player symbol
name = symbols[player]
game_state, current_row, current_col = move_at_random(game_state, player)
#print '******',player,(current_row, current_col)
#print current game state
print_game_state(game_state)
#check if the move was a winning move
if was_winning_move(game_state,player,current_row, current_col):
print 'player %s wins after %d moves' % (name, mvcntr)
no_winner_yet = False
# switch player and increase move counter
player *= -1
mvcntr += 1
if no_winner_yet:
print 'game ended in a draw'
player = 0
return game_state,player,mvcntr
if __name__ == '__main__':
S, P, mvcntr = play_game()
let me know if you have any question
UPDATE: Explanation:
At each move, look at column, row, diagonal and secondary diagonal that goes through the current cell and find consecutive cells with the current symbol. avoid scanning the whole board.
extracting cells in each direction:
column:
current_col = S[:,current_col_idx]
row:
current_row = S[current_row_idx,:]
Diagonal:
Find the offset of the desired diagonal from the
main diagonal:
diag_offset = current_col_idx - current_row_idx
current_diag = S.diagonal(offset)
off-diagonal:
Reverse the rows of matrix:
S_reversed_rows = S[::-1,:]
Find the row index in the new matrix
new_row_idx = row_size - 1 - current_row_idx
current_offdiag = S.diagonal(offset)