Step by Step solution about How to resolve the algorithm Taxicab numbers step by step in the Python programming language

1. Identifying Cube Roots

cube2n = {x**3:x for x in range(1, 1201)}
sums2cubes = defaultdict(set)

• cube2n is a dictionary that maps cube values (from 1 to 1,200,000) to their corresponding roots (from 1 to 1200).
• sums2cubes is a defaultdict that maps sums of two cubes to sets of tuples representing the cube roots that produce those sums.

2. Computing Sums of Two Cubes

for c1, c2 in product(cube2n, cube2n):
   if c1 >= c2: sum2cubes[c1 + c2].add((cube2n[c1], cube2n[c2]))

• product(cube2n, cube2n) generates all possible pairs of cube values.
• For each pair (c1, c2) where c1 >= c2, the sum c1 + c2 is calculated and used as a key in the sums2cubes dictionary.
• A tuple (cube2n[c1], cube2n[c2]) is then added to the set associated with that key, representing the pair of cube roots that produce the given sum.

3. Identifying Taxicab Numbers

taxied = sorted((k, v) for k,v in sum2cubes.items() if len(v) >= 2)

• sorted(...) sorts the sum2cubes dictionary based on the keys (sums of two cubes).
• The (k, v) tuples represent pairs of sum and their associated set of tuples.
• The len(v) >= 2 filter ensures that only sums with multiple representations as sums of two cubes are considered.

4. Printing the First 25 and Last 6 Taxicab Numbers

#pp(len(taxied))  # 2068
for t in enumerate(taxied[:25], 1):
   pp(t)
print('...')    
for t in enumerate(taxied[2000-1:2000+6], 2000):
   pp(t)

• pp(len(taxied)) prints the total number of taxicab numbers found (2068).
• The code then prints the first 25 and the last 6 taxicab numbers in a tuple format showing the sum and its representations as sums of two cubes.

5. Computing Taxicab Numbers with Cube Roots

cubes, crev = [x**3 for x in range(1,1200)], {}
# for cube root lookup
for x,x3 in enumerate(cubes): crev[x3] = x + 1

sums = sorted(x+y for x in cubes for y in cubes if y < x)

idx = 0
for i in range(1, len(sums)-1):
   if sums[i-1] != sums[i] and sums[i] == sums[i+1]:
       idx += 1
       if idx > 25 and idx < 2000 or idx > 2006: continue

       n,p = sums[i],[]
       for x in cubes:
           if n-x < x: break
           if n-x in crev:
               p.append((crev[x], crev[n-x]))
       print "%4d: %10d"%(idx,n),
       for x in p: print " = %4d^3 + %4d^3"%x,
       print

• This section uses a more efficient way to compute taxicab numbers by leveraging precomputed cube values and cube root lookups.
• It prints the first 25 and the last 6 taxicab numbers in a different representation, where the sum is followed by a list of tuples representing the cube roots that produce that sum.

6. Generating Taxicab Numbers with a Priority Queue

from heapq import heappush, heappop

def cubesum():
   h,n = [],1
   while True:
       while not h or h[0][0] > n**3: # could also pre-calculate cubes
           heappush(h, (n**3 + 1, n, 1))
           n += 1

       (s, x, y) = heappop(h)
       yield((s, x, y))
       y += 1
       if y < x:    # should be y <= x?
           heappush(h, (x**3 + y**3, x, y))

def taxis():
   out = [(0,0,0)]
   for s in cubesum():
       if s[0] == out[-1][0]:
           out.append(s)
       else:
           if len(out) > 1: yield(out)
           out = [s]

• This section demonstrates a different way of generating taxicab numbers using a priority queue (heapq).
• The cubesum() generator yields tuples representing sums of two cubes, starting with the smallest possible sum (1^3 + 1^3).
• The taxis() function then groups these sums into taxicab numbers and yields them as lists of tuples.

7. Printing the First 2006 Taxicab Numbers

n = 0
for x in taxis():
   n += 1
   if n >= 2006: break
   if n <= 25 or n >= 2000:
       print(n, x)

• The code iterates over the taxicab numbers generated by taxis(), printing the first 2006 numbers.
• It filters out numbers that are already printed in the previous section (those in the first 25 or last 6 positions).

from collections import defaultdict
from itertools import product
from pprint import pprint as pp

cube2n = {x**3:x for x in range(1, 1201)}
sum2cubes = defaultdict(set)
for c1, c2 in product(cube2n, cube2n):
	if c1 >= c2: sum2cubes[c1 + c2].add((cube2n[c1], cube2n[c2]))
	
taxied = sorted((k, v) for k,v in sum2cubes.items() if len(v) >= 2)

#pp(len(taxied))  # 2068
for t in enumerate(taxied[:25], 1):
    pp(t)
print('...')    
for t in enumerate(taxied[2000-1:2000+6], 2000):
    pp(t)


cubes, crev = [x**3 for x in range(1,1200)], {}
# for cube root lookup
for x,x3 in enumerate(cubes): crev[x3] = x + 1

sums = sorted(x+y for x in cubes for y in cubes if y < x)

idx = 0
for i in range(1, len(sums)-1):
    if sums[i-1] != sums[i] and sums[i] == sums[i+1]:
        idx += 1
        if idx > 25 and idx < 2000 or idx > 2006: continue

        n,p = sums[i],[]
        for x in cubes:
            if n-x < x: break
            if n-x in crev:
                p.append((crev[x], crev[n-x]))
        print "%4d: %10d"%(idx,n),
        for x in p: print " = %4d^3 + %4d^3"%x,
        print


from heapq import heappush, heappop

def cubesum():
    h,n = [],1
    while True:
        while not h or h[0][0] > n**3: # could also pre-calculate cubes
            heappush(h, (n**3 + 1, n, 1))
            n += 1

        (s, x, y) = heappop(h)
        yield((s, x, y))
        y += 1
        if y < x:    # should be y <= x?
            heappush(h, (x**3 + y**3, x, y))

def taxis():
    out = [(0,0,0)]
    for s in cubesum():
        if s[0] == out[-1][0]:
            out.append(s)
        else:
            if len(out) > 1: yield(out)
            out = [s]

n = 0
for x in taxis():
    n += 1
    if n >= 2006: break
    if n <= 25 or n >= 2000:
        print(n, x)