Question

Please help with Python code.

def pyramid_blocks (n, m, h): A solid pyramid structure (although here in the ancient Mesoamerican than the more famous ancient Egyptian style) is built from layers, each layer consisting of a rectangle of identical cubic blocks. The top layer of the pyramid consists of n rows and m columns of such blocks. The layer immediately below each layer contains one more row and one more column, all the way to the bottom layer of the pyramid. If the entire pyramid consists of h such layers, how many blocks does this pyramid contain in total? Here you can solve this problem in a straightforward fashion by simply looping through the h layers and adding up all the blocks along the way in each layer. However, if you happen to know some discrete math and combinatorics, you can come up with an analytical closed form formula for the result and compute the answers much faster that way. (There is an important general principle in this for you to later ponder on your own once you have done more of these problems.) n h Expected result 2 3 1 6 2 3 10 570 10 11 12 3212 100 100 100 2318350 10**6 10**6 10**6 2333331833333500000

testing for:

def pyramid_blocks_generator(seed):
n = 300
ns = it.islice(scale_random(seed, 3, 10), n)
ms = it.islice(scale_random(seed + 1, 3, 10), n)
hs = it.islice(scale_random(seed + 2, 2, 15), n)
yield from zip(ns, ms, hs)

Answer 1

Solution : This is program to fine total blocks required in 3D pyramid. where

n = X (bredth)

m =Y (length)

h =Z (height)

any layer contains total of n*m blockes and next layer will contains one more element in X and Y direction so the next layer contain (n+1)*(m+1)

so we can say that let first layer be L(k1) elements

L(k1) = n*m

L(k2) = (n+1)*(m+1)

L(k3) = (n+2)*(m+2)

...

...

L(kh) = (n+h-1)*(m+h-1)

and the answer will be sum of all the blocks

i.e Total_bocks = L(k1)+L(k2)+L(k3)...+L(kn)

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MATHEMATICALLY

bym y (h-l) ter blocks = first layer & and layer pth layer +(1+1)(mt) + (0+2)(2012) ..(n+h) (mth-1) NXM + um ti (nem JH+ 10 4 2 (ntho)+22.- Amt fh t0) Ch-1) 4h-1 (nm 4nm 4nm.nm) 4 (1+m) (1+2+3*-- (n-1)) +1+2++2°44? (h=y? h times (6-1) times Ap wild V hxnxm +10 + m) (t+th +)) Ch-D+ (8 +) xþ ( 2XCH-1)+1) 6 Apsum formula Sume outlet Squons of fuit (n-1) Nalusal number Chxnxm + (n 4th) (k-1)8) + ht) (25+1) 8.1 totul humbur of bloody hxnm & Chimtm) + (2h)" dzo 2 (ent) 2

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FUNCTON DEFINITION IN PYTHON

############################################################

def pyramid_fast(n,m,h):
blocks = (n*h*m +(h*(h-1)//2)*(n+m)+(2*h-1)*h*(h-1)//6)
return blocks
############################################################

def pyramid(n,m,h): blocks = 0 for i in range(h): blocks+=n*m n+=1 m+=1 return blocks #function decleration take three variable #store all the sum #loop through each layer #calculate the number of blocks in layer and add to sum #increment X #increment y #after each layer completion return total def pyramid_fast(n,m,h): blocks = (n*h*m +(h* (h-1)//2)*(n+m)+(2*h-1)*h*(h-1)//6) return blocks print("n print("n print("n print("n print("n = 2,0 = 3, h = 1 : ",pyramid(2,3,1), pyramid_fast(2,3,1)) #call to the functions 2,m = 3, h = 10 : pyramid(2,3,10), pyramid_fast(2,3,10)) 10,m = 11, h = 12 : ",pyramid(10, 11, 12), pyramid_fast(10,11,12)) 100,m = 100, h = 100 : ",pyramid(100,100,100), pyramid_fast(100,100,100)) = 10**6, m = 10**6, h = 10**6 : ",pyramid (10**6,10**6,10**6), pyramid_fast(10**6,10**6,10**6))

OUT

n = RESTART: /home/deeplearningcv/Documents/number_of_pyramid_block.py n = 2,m = 3, h = 1 : 66 2,m = 3, h = 10 : 570 570 n = 10,m = 11, h 12: 3212 3212 n = 100, m = 100,h 100 : 2318350 2318350 10**6, m = 10**6, h = 10**6 : 2333331833333500000 2333331833333500000 n =