n=4
m=3
k = {}
a=1
b=0
c=0
while c< m*n:
for i in range(abs(m)):
b=b+1
c=c+1
k[(a,b)] = c
if c==m*n:
break
for i in range(abs(n-1)):
a=a+1
c=c+1
k[(a,b)] = c
if c==m*n:
break
for i in range(abs(m-1)):
b=b-1
c=c+1
k[(a,b)] = c
if c==m*n:
break
for i in range(abs(n-2)):
a= a-1
c=c+1
k[(a,b)] = c
if c==m*n:
break
print(k)
If you print this out,
{(1, 1): 1, (1, 2): 2, (1, 3): 3, (2, 3): 12, (3, 3): 13, (4, 3): 14, (4, 2): 7, (4, 1): 18, (3, 1): 19, (2, 1): 10, (2, 2): 11, (5, 3): 15, (5, 2): 16, (5, 1): 17}
It comes out like this, but it comes out as (2,3): 12
(2,3): Isn't 4 correct?
Why did you come out like this?
I think it's snail-arrangement homework.
It's not the answer, but I just created a function that allows you to take a picture of the state. While checking with this function, modify the original code to make the correct answer.
def print_k():
print('-'*15)
for i in range(1, 6):
for j in range(1, 6):
print('%4s'%(k.get((i,j), '')), end='')
print()
n=4
m=3
k = {}
a=1
b=0
c=0
while c< m*n:
for i in range(abs(m)):
b=b+1
c=c+1
k[(a,b)] = c
if c==m*n:
break
print_k()
for i in range(abs(n-1)):
a=a+1
c=c+1
k[(a,b)] = c
if c==m*n:
break
print_k()
for i in range(abs(m-1)):
b=b-1
c=c+1
k[(a,b)] = c
if c==m*n:
break
print_k()
for i in range(abs(n-2)):
a=a-1
c=c+1
k[(a,b)] = c
if c==m*n:
break
print_k()
#print(k)
print_k()
---------------
1 2 3
---------------
1 2 3
4
5
6
---------------
1 2 3
4
5
8 7 6
---------------
1 2 3
10 4
9 5
8 7 6
---------------
1 2 3
10 11 12
9 5
8 7 6
---------------
1 2 3
10 11 12
9 13
8 7 14
15
---------------
1 2 3
10 11 12
9 13
8 7 14
17 16 15
---------------
1 2 3
10 11 12
19 13
18 7 14
17 16 15
---------------
1 2 3
10 11 12
19 13
18 7 14
17 16 15
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