So I have this code in python and currently it only returns the maximum value for cutting a rod. How can I modify this to also give me where the cuts were made? It takes a list of prices whose indices+1 correspond to the value of the rod at each length, and n, for length of the rod.
the problem:http://www.radford.edu/~nokie/classes/360/dp-rod-cutting.html
def cutRod(price, n):
val = [0 for x in range(n+1)]
val[0] = 0
for i in range(1, n+1):
max_val = 0
for j in range(i):
max_val = max(max_val, price[j] + val[i-j-1])
val[i] = max_val
return val[n]
If this is the question : Rod cutting
Assuming code works fine, You will have to add a condition instead of Max operation to check which of two was picked and push that one in an array :
def cutRod(price, n):
val = [0 for x in range(n+1)]
val[0] = 0
output = list()
for i in range(1, n+1):
max_val = 0
cur_max_index = -1
for j in range(i):
cur_val = price[j] + val[i-j-1]
if(cur_val>max_val):
max_val = cur_val #store current max
cur_max_index = j #and index
if cur_max_index != -1:
output.append(cur_max_index) #append in output index list
val[i] = max_val
print(output) #print array
return val[n]
I know this is old but just in case someone else has a look...I was actually just looking at this problem. I think the issue is here that these dp problems can be tricky when handling indices. The previous answer is not going to print the solution correctly simply because this line needs to be adjusted...
cur_max_index = j which should be cur_max_index = j + 1
The rest...
def cut_rod(prices, length):
values = [0] * (length + 1)
cuts = [-1] * (length + 1)
max_val = -1
for i in range(1, length + 1):
for j in range(i):
temp = prices[j] + values[i - j - 1]
if temp > max_val:
max_val = prices[j] + values[i - j - 1]
cuts[i] = j + 1
values[i] = max_val
return values[length], cuts
def print_cuts(cuts, length):
while length > 0:
print(cuts[length], end=" ")
length -= cuts[length]
max_value, cuts = cut_rod(prices, length)
print(max_value)
print_cuts(cuts, length)
Well, if you need to get the actual pieces that would be the result of this process then you'd probably need a recursion.
For example something like that:
def cutRod(price, n):
val = [0 for x in range(n + 1)]
pieces = [[0, 0]]
val[0] = 0
for i in range(1, n + 1):
max_val = 0
max_pieces = [0, 0]
for j in range(i):
curr_val = price[j] + val[i - j - 1]
if curr_val > max_val:
max_val = curr_val
max_pieces = [j + 1, i - j - 1]
pieces.append(max_pieces)
val[i] = max_val
arr = []
def f(left, right):
if right == 0:
arr.append(left)
return
f(pieces[left][0], pieces[left][1])
f(pieces[right][0], pieces[right][1])
f(pieces[n][0], pieces[n][1])
return val[n], arr
In this code, there is an additional array for pieces which represents the best way to divide our Rod with some length.
Besides, there is a function f that goes through all pieces and figures out the optimal way to divide the whole Rod.
Related
This is a program to print a matrix whose sum of each row , column or diagonal elements are equal.
I have a working code but my program gives same output each time i run it. I need a program to print different matrix output for same input.
def matrix(n):
m = [[0 for x in range(n)]for y in range(n)]
i = n // 2
j = n - 1
num = 1
while num <= (n * n):
if i == -1 and j == n:
j = n - 2
i = 0
else:
if j == n:
j = 0
if i < 0:
i = n - 1
if m[int(i)][int(j)]:
j = j - 2
i = i + 1
continue
else:
m[int(i)][int(j)] = num
num = num + 1
j = j + 1
i = i - 1
print ("Sum of eggs in each row or column and diagonal : ",int(n*(n*n+1)/2),"\n")
for i in range(0, n):
for j in range(0, n):
print('%2d ' % (m[i][j]),end = '')
if j == n - 1:
print()
n=int(input("Number of rows of the matrix : "))
matrix(n)
I am unsure whether this is what you are looking for, but one solution is to add a random number to each value of the matrix, as this doesn't break the property
a matrix whose sum of each row, column or diagonal elements are equal.
Here is how you could do it:
add = random.randint(0, 50)
m = [[v+add for v in row] for row in m]
Moreover, you can rotate and add two magic squares without loosing their property. Therefore, you can rotate the magic square you have and add it to the original. This can add some nonlinearity to the results.
def rotate(m): # 90 degrees counter clockwise
return [[m[j][i] for j in range(len(m))] for i in range(len(m[0])-1, -1, -1)]
# add the matrix with its rotated version
m = list(map(lambda e: [sum(x) for x in zip(*e)], zip(m, rotate(m))))
I hope this helps!
Im not very sure if I will translate the assignment correctly, but the bottom line is that I need to consider the function f(i) = min dist(i, S), where S is number set of length k and dist(i, S) = sum(j <= S)(a (index i) - a(index j)), where a integer array.
I wrote the following code to accomplish this task:
n, k = (map(int, input().split()))
arr = list(map(int, input().split()))
sorted_arr = arr.copy()
sorted_arr.sort()
dists = []
returned = []
ss = 0
indexed = []
pop1 = None
pop2 = None
for i in arr:
index = sorted_arr.index(i)
index += indexed.count(i)
indexed.append(i)
dists = []
if (index == 0):
ss = sorted_arr[1:k+1]
elif (index == len(arr) - 1):
sorted_arr.reverse()
ss = sorted_arr[1:k+1]
else:
if index - k < 0:
pop1 = 0
elif index + k > n - 1:
pop2 = None
else:
pop1 = index - k
pop2 = index + k + 1
ss = sorted_arr[pop1:index] + sorted_arr[index + 1: pop2]
for ind in ss:
dists.append(int(abs(i - ind)))
dists.sort()
returned.append(str(sum(dists[:k])))
print(" ".join(returned))
But I need to speed up its execution time significantly.
Here are my merge and mergeSort functions. merge merges two separate arrays and mergesSort sorts and merges them with recursion:
def merge(arrL, arrR):
arrNew = []
d = len(arrL) + len(arrR)
i = 0
j = 0
for k in range (0, d-1):
if (arrL[i] < arrR[j]) :
arrNew.append(arrL[i]) # appends to the end of the array
i = i + 1
else:
arrNew.append(arrR[j])
j = j + 1
return arrNew
def mergeSort(arr, m, n):
if (n - m == 1):
return arr[m]
else:
p = (m + n) // 2
arrL = mergeSort(arr, m, p)
arrR = mergeSort(arr, p, n)
arrNew = merge(arrL, arrR)
return arrNew
I am getting an error from lines 32, 33 and 13:
d = len(arrL) + len(arrR)
TypeError: object of type 'int' has no len()
What is causing this error? merge is taking two arrays as inputs.
What is causing this error? merge is taking two arrays as inputs.
Except when it doesn't.
if(n-m == 1):
return arr[m]
This output of mergeSort is not an array.
My guess is it's this line
if(n-m == 1):
return arr[m]
which presumably is returning the content arr[m] of the array and not an array itself.
Since your code sorts arrays, when this naked element gets recursed on, it will generate the error you're seeing.
There are multiple problems in the code:
in mergeSort, you should return arr instead of arr[m] when the length of the array is less than 2. The test if (n - m == 1) does not allow for empty arrays:
if (n - m < 2):
return arr
in merge, the main loop should run d times, ie: instead of for k in range (0, d-1): you should write:
for k in range (d):
the test in the merge loop should also check if the index value in still in range. If the second slice is exhausted, the element arrL[i] should be selected:
for k in range (d):
if i < len(arrL) and (j >= len(arrR) or arrL[i] < arrR[j]):
arrNew.append(arrL[i])
i = i + 1
else:
arrNew.append(arrR[j])
j = j + 1
Here is a modified version:
def merge(arrL, arrR):
arrNew = []
i = 0
j = 0
for k in range(len(arrL) + len(arrR)):
if i < len(arrL) and (j >= len(arrR) or arrL[i] < arrR[j]):
arrNew.append(arrL[i])
i = i + 1
else:
arrNew.append(arrR[j])
j = j + 1
return arrNew
def mergeSort(arr, m, n):
if (n - m < 2):
return arr
else:
p = (m + n) // 2
arrL = mergeSort(arr, m, p)
arrR = mergeSort(arr, p, n)
return merge(arrL, arrR)
How do you sort a list with a while loop? Having a bit of a problem, thanks very much in advance.
a = [12,0,39,50,1]
first = a[0]
i = 0
j = 1
while i < len(a):
if a[i] < first:
tmp = a[i]
a[i] = a[j]
a[j] = tmp
i += 1
print(a)
You can create an empty list that would store your sorted numbers
a = [12,0,39,50,1]
kk = len(a)
new_a = []
i = 0
while i < kk:
xx = min(a) ## This would retreive the minimum value from the list (a)
new_a.append(xx) ## You store this minimum number in your new list (new_a)
a.remove(xx) ## Now you have to delete that minimum number from the list a
i += 1 ## This starts the whole process again.
print(new_a)
Please, note that I used the original length of the list a (kk) for the while statement so as not to stop the iteration because the length of the list a decreases as we delete the minimum numbers.
Following is the implementation of basic sorting using two while loops.
In every iteration, the minimum element (considering ascending order) from the unsorted subarray is picked and moved to the sorted subarray.
:
a=[12,0,39,50,1]
i=0
while i<len(a):
key=i
j=i+1
while j<len(a):
if a[key]>a[j]:
key=j
j+=1
a[i],a[key]=a[key],a[i]
i+=1
print(a)
# By using For loop
def ascending_array(arr):
print(f"Original array is {arr}")
arr_length = len(arr)
if arr_length <= 1:
return arr
for i in range(len(arr)):
for j in range(i+1, len(arr)):
if arr[i] >= arr[j]:
arr[i], arr[j] = arr[j], arr[i]
print(f"The result array is {arr}") # [0,0,0,1,10,20,59,63,88]
arr = [1,10,20,0,59,63,0,88,0]
ascending_array(arr)
# By using While loop
def ascending_array(arr):
print(f"Original array is {arr}")
arr_length = len(arr)
if arr_length <= 1:
return arr
i = 0
length_arr = len(arr)
while (i<length_arr):
j = i+1
while (j<length_arr):
if arr[i] > arr[j]:
arr[i], arr[j] = arr[j], arr[i]
j+=1
i+=1
print(f"The result array is {arr}") # [0,0,0,1,10,20,59,63,88]
arr = [1,10,20,0,59,63,0,88,0]
ascending_array(arr)
For-loop is best in terms of performance. while-loop is checking condition every iteration.
You can also concatenate two lists and sort them in decreasing/increasing order using this example:
x = [2,9,4,6]
y = [7,8,3,5]
z = []
maxi = x[0]
pos = 0
print('x: '+str(x))
print('y: '+str(y))
for i in range(len(y)):
x.append(y[i])
for j in range(len(x)-1):
maxi = x[0]
for i in range(len(x)):
if maxi < x[i]:
maxi = x[i]
pos = i
z.append(maxi)
del x[pos]
z.append(x[0])
print('z: '+str(z))
def Checker(n):
p = [[7,3,5,6,1],[2,6,7,0,2],[3,5,7,8,2],[7,6,1,1,4],[6,7,4,7,8]] #profit of each cell
cost = [[0 for j in range(n)] for i in range(n)]
w = [[0 for j in range(n)] for i in range(n)] #w[i, j] store the column number (j) of the previous square from which we moved to the current square at [i,j]
for j in range(1,n):
cost[1][j] = 0
for i in range(2,n):
for j in range(1,n):
max = cost[i-1][j] + p[i-1][j]
w[i][j] = j
if (j > 1 and cost[i-1][j-1] + p[i-1][j-1] > max):
max = cost[i-1][j-1] + p[i-1][j-1]
w[i][j] = j-1
if (j < n and cost[i-1][j+1] + p[i-1][j+1] > max):
max = cost[i-1][j+1] + p[i-1][j+1]
w[i][j] = j+1
cost[i][j] = max
print cost[i][j]
maxd = cost[1][1]
maxj = 1
for j in range(2,n):
if cost[1][j] >maxd:
maxd = cost[1][j]
maxj = j
print "Maximum profit is: ",maxd
printsquares(w,n,maxj)
def printsquares(w,i,j):
if i == -1:
return
print "Square at row %d and column %d"%(i,j)
printsquares(w,i-1,w[i][j])
if __name__ == '__main__':
print "5*5 checker board problem"
n = 5
Checker(n)
The above program is implementation of checker board algorithm in python.
when i run the above code the following error is shown:
if (j < n and cost[i-1][j+1] + p[i-1][j+1] > max): IndexError: list
index out of range
what am i doing wrong and any one would propose solution for it?
To avoid the IndexError exceptions,
if (j < n and cost[i-1][j+1] + p[i-1][j+1] > max):
max = cost[i-1][j+1] + p[i-1][j+1]
should be written:
if (j < n-1 and cost[i-1][j+1] + p[i-1][j+1] > max):
max = cost[i-1][j+1] + p[i-1][j+1]
and
printsquares(w,i-1,w[i][j])
becomes
printsquares(w,i-1,w[i-1][j])
But as said by other fellows, I'm not certain that the algorithm is correctly implemented.
You're pretty much trying to do this:
L = range(5)
print L[5+1]
At cost[i-1][j+1].
There is no 6th element. There's only five. Hence the IndexError.
As for a solution, you probably only want n and not n+1 if you want the last element. However, I'm not 100% sure.
It looks like you are trying to adapt an algorithm from a language where lists start from 1 whereas they start from 0 in Python. As far I as checked, you never access cost[i][0].