Fractal Maze

Logic Level 2

Above is the fractal maze , where region A , B , C A,B,C are exact copies of the maze itself. Is it possible to go from + + sign to − - sign (of the same depth as the starting point)?

Note: You can't go from one path to another by the intersections. These are actually high-way overpasses.

No Yes

Alice Smith by Alice Smith , 20, China

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2 solutions

David Vreken
May 20, 2020

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Wow, easy approach! But I think there is something wrong here. I meant the negative sign of the same depth as the positive sign.

Alice Smith - 1 year ago

The - sign that you enter is not at the same depth as the + sign you leave. You are still in the C square when you enter the -sign.

Joshua Lowrance - 1 year ago

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Hmmm ... that stipulation was not in the original problem. I'll have to look at this again.

David Vreken - 1 year ago

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Okay, I updated my solution.

David Vreken - 1 year ago

@David Vreken , Similar reasoning. By the way, @Alice Smith Fantastic problem. The most unique one I've seen in a while. A whole new concept like a fractal in a maze is just awesome!!

Mahdi Raza - 1 year ago

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Yeah, I just found it on the Internet and decided to post it here. Actually, it's a very good problem!

Alice Smith - 1 year ago

Ohh, can you please share the original link. Thanks!

Mahdi Raza - 1 year ago

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http://www.mathpuzzle.com/18Nov2003.html

Alice Smith - 1 year ago

Nice thank you for sharing another resource! The article itself has another hard fractal puzzle. I'll try to solve it!

Mahdi Raza - 1 year ago

I wrote this Python computer program to help me find the solution: 

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# Fractal Maze
# Python

# set up variables
D = []
NA = []
NB = []
NC = []
for a in range(0, 17):
    D.append([])
    NA.append([])
    NB.append([])
    NC.append([])

# set up pathway choices
# Key:
#    12 11 10 09
# 13             08
#
# 14             07
#
# 15             06
#
# 16             05
#    01 02 03 04

D[0] = []
D[1] = ["A08", "10", "C12", "B06"]
D[2] = ["15", "A15"]
D[3] = ["16", "05", "A02"]
D[4] = ["B10"]
D[5] = ["03", "16", "A02"]
D[6] = ["C07", "A10"]
D[7] = ["B05"]
D[8] = ["B09"]
D[9] = ["B12"]
D[10] = ["A08", "01", "C12", "B06"]
D[11] = ["A09"]
D[12] = ["A12", "13", "14"]
D[13] = ["A12", "12", "14"]
D[14] = ["A12", "12", "13"]
D[15] = ["A15", "02"]
D[16] = ["03", "05", "A02"]

NA[0] = []
NA[1] = ["-"]
NA[2] = ["16", "03", "05"]
NA[3] = ["A13"]
NA[4] = ["C16"]
NA[5] = ["B13"]
NA[6] = []
NA[7] = []
NA[8] = ["10", "01", "C12", "B06"]
NA[9] = ["11"]
NA[10] = ["06", "C07"]
NA[11] = []
NA[12] = ["12", "13", "14"]
NA[13] = ["A03"]
NA[14] = []
NA[15] = ["02"]
NA[16] = []

NB[0] = []
NB[1] = []
NB[2] = ["C09"]
NB[3] = []
NB[4] = []
NB[5] = ["07"]
NB[6] = ["01", "10", "A08", "C12"]
NB[7] = []
NB[8] = []
NB[9] = ["08"]
NB[10] = ["04"]
NB[11] = []
NB[12] = ["09"]
NB[13] = ["A05"]
NB[14] = []
NB[15] = ["C14"]
NB[16] = []

NC[0] = []
NC[1] = []
NC[2] = []
NC[3] = ["+"]
NC[4] = []
NC[5] = ["C06"]
NC[6] = ["C05"]
NC[7] = ["06", "A10"]
NC[8] = []
NC[9] = ["B02"]
NC[10] = []
NC[11] = []
NC[12] = ["B06", "01", "A08", "10"]
NC[13] = []
NC[14] = ["B15"]
NC[15] = []
NC[16] = ["A04"]

# function that takes in a path sequence and returns a list of
#   possible next path sequences ("~" means "leaving")
def g(s):
    tilde = (s[-4:-3] == "~")
    letter = s[-3:-2]
    number = int(s[-2:])
    if tilde:
        unaff = s[:-5]
        pletter = s[-5:-4]
    else:
        unaff = s[:-4]
        pletter = s[-4:-3]
    nstr = []
    rstr = []
    if tilde:
        if letter == "A":
            for a in range(0, len(NA[number])):
                nstr.append(NA[number][a])
        elif letter == "B":
            for a in range(0, len(NB[number])):
                nstr.append(NB[number][a])
        elif letter == "C":
            for a in range(0, len(NC[number])):
                nstr.append(NC[number][a])
        for a in range(0, len(nstr)):
            if len(nstr[a]) == 1:
                rstr.append("")
            elif len(nstr[a]) == 2:
                rstr.append(unaff + "~" + pletter + nstr[a])
            elif len(nstr[a]) == 3:
                rstr.append(unaff + pletter + nstr[a])
    else:
        for a in range(0, len(D[number])):
            nstr.append(D[number][a])
        for a in range(0, len(nstr)):
            if len(nstr[a]) == 1:
                rstr.append("")
            elif len(nstr[a]) == 2:
                rstr.append(unaff + pletter + "~" + letter + nstr[a])
            elif len(nstr[a]) == 3:
                rstr.append(unaff + pletter + letter + nstr[a])
    for a in range(0, len(nstr)):
        if len(rstr[a]) > 0:
            if rstr[a][0] == "~" and not (
                rstr[a][1] == "A" or rstr[a][1] == "B" or rstr[a][1] == "C"):
                rstr[a] = ""
    return rstr

# Go through each path sequence starting from "C03" until "~A01" is found.
#   The p list keeps track of each possible path sequence by string,
#   index sequence, and number of moves: p[moves][(path string, path seq)]
m = 0
p = []
p.append([])
p[0].append(("C03", ""))
fs = ""
while fs == "":
    m += 1
    p.append([])
    for a in range(0, len(p[m - 1])):
        np = g(p[m - 1][a][0])
        for b in range(0, len(np)):
            if np[b] <> "":
                p[m].append((np[b], p[m - 1][a][1] + str(b)))
    for a in range(0, len(p[m])):
        if p[m][a][0] == "~A01":
            fs = p[m][a][1]

# Using the finish sequence (fs), display the steps
for a in range(0, len(fs) + 1):
    for b in range(0, len(p[a])):
        if p[a][b][1] == fs[:a]:
            print(p[a][b][0])

David Vreken - 1 year ago

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Brilliant solution！

Alice Smith - 1 year ago

I just found the shortest path: + to A then go to the first one on the bottom and that goes to the - sign - that gave me the answer Yes.

0 Helpful 0 Interesting 0 Brilliant 1 Confused

Oh, I forget to say that you can't go from one path to another by the intersections. These are actually high-way streets:)

Alice Smith - 1 year ago

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Well, I didn't know at the time...

A Former Brilliant Member - 1 year ago

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Don't worry. You will eventually find the solution:)

Alice Smith - 1 year ago

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