```
nx,ny = 0,0 #startcoordinates
for i in range(n):
result = 0
for _ in range(7):
ax = int(random.randrange(1, 16))
ay = int(random.randrange(1, 16))
distance = math.sqrt(abs(ax - nx) **2 + abs(ay - ny) **2)
Cumulative distance at result += distance #result
nx = ax #Put in the variable because the next distance must be the previous distance
ny = ay
distance = math.sqrt(abs(ax) **2 + abs(ay) **2) #sum the distance returned from the last coordinate to the origin
Result += distance # Cumulative distance in result
arr1.append(result)
a = np.array([arr1])
AVG = np.mean(a) #Mean the simulation
for i in range(n):
result = 0
for _ in range(2):
bx = int(random.randrange(16, 41))
by = int(random.randrange(16, 41))
distance = math.sqrt(abs(bx - nx) **2 + abs(by - ny) **2)
Cumulative distance at result += distance #result
nx = bx #Put in the variable because the next distance must be the previous distance
ny = by
distance = math.sqrt(abs(bx) **2 + abs(by) **2) #sum the distance returned from the last coordinate to the origin
Result += distance # Cumulative distance in result
arr2.append(result)
a = np.array([arr2])
AVG = np.mean(a) #Mean the simulation
for i in range(n):
result = 0
for _ in range(1):
cx = int(random.randrange(41, 101))
cy = int(random.randrange(41, 101))
distance = math.sqrt(abs(cx - nx) **2 + abs(cy - ny) **2)
Cumulative distance at result += distance #result
nx = cx #Put in the variable because the next distance must be the previous distance
ny = cy
distance = math.sqrt(abs(cx) **2 + abs(cy) **2) #sum the distance returned from the last coordinate to the origin
Result += distance # Cumulative distance in result
arr3.append(result)
a = np.array([arr3])
AVG = np.mean(a) #Mean the simulation
```

Hi, how are you?

We're going to randomly pick the x and y coordinates to find the distance between them I was wrong

As shown in this figure, 7

on the (15-15) coordinatesTwo on (40~40) coordinates excluding (15~15)

We tried to derive one from (100-100) coordinates excluding (40~40) but

It was a stupid way to derive this...

Can you find an expression to draw the coordinates shown in the picture above?

python

2022-09-20 14:39

```
As shown in the figure (15 to 15) 7 on coordinates
2 on [*1] (40~40) coordinates excluding (15~15)
[*2] (100-100) coordinates excluding (40~40) were trying to derive one of these
```

I think you can just code the words above.

```
import random
ax,ay = 0, 0
while(True): #16x16
ax = int(random.randrange(1, 16))
ay = int(random.randrange(1, 16))
break
while(True): #40x40
ax = int(random.randrange(1, 40))
ay = int(random.randrange(1, 40))
if not (ax < 16 and ay < 16): break # [*1]
#Exit if not in 16x16 rectangle
while(True): #100,100
ax = int(random.randrange(1, 100))
ay = int(random.randrange(1, 100))
if not ( ax < 40 and ay < 40 ): break # [*2]
#Exit if not in 40x40 rectangle
```

2022-09-20 14:39

Collecting all three from one roof.

```
import random
A, B, C = [], [], [] # A will store small square areas, middle areas in B, and random coordinates in C.
while True:
x, y = random.randrange(1, 101), random.randrange(1, 101)
# # print(x, y)
If (x <=15) and (y <=15): # The additional field is the square area
if len(A) < 15:
A.append((x, y))
elif (x <=40) and (y <=40): # Intermediate area
if len(B) < 2:
B.append((x, y))
else: # Rest of Area
if len(C) < 1:
C.append((x, y))
If (len(A) == 15) and (len(B) == 2) and (len(C) == 1): # If A, B, and C are gathered as much as you want, exit the loop.
break
for x, y in A:
# # do your math for A
for x, y in B:
# # do your math for B
for x, y in C:
# # do your math for C
```

I'll add an answer.

You can pick randomly first, and since you have gathered them, you can pick one for each area and continue to do what you want. OK?

2022-09-20 14:39

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