Chains of 0's

Computer Science Level 3

The function

f ( x , y , z ) = x x y + x y z \displaystyle f(x, y, z) = x^{x^y} + x^{y^z}

for all positive integers. What is the length of the longest chain of consecutive 0's in the simplified form of

f ( 4 , 5 , 6 ) f(4, 5, 6)


The answer is 3.

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Sharky Kesa by Sharky Kesa , 19, United Kingdom

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6 solutions

Thaddeus Abiy
Aug 2, 2014

Python Code 

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def f(x,y,z):
    return x**(x**y) + x**(y**z)

N = str(f(4,5,6))
Size = len(N)

Longest = 0
i = 0

while i < Size:
    Count = 0
    j = i
    while j < Size:
        Char = N[j]
        if Char == "0":
            j += 1
            Count += 1
        else:
            break
    i = j
    Longest = max( Longest , Count )

print Longest

The program first converts (f(4,5,6)) into a string. It then goes through each digit one at a time until it finds a zero. If the zero is the first of a continuous sub sequence of zeros,the program activates another loop and counts the size of the sub-sequence and checks if it is the longest sub-sequence found so far.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

How many hours were used for computation?

Kenny Lau - 6 years, 8 months ago
Hunter Killer
Jan 6, 2015

Mathematica Code 

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f[x_, y_, z_] := x^x^y + x^y^z;
a = f[4, 5, 6];
b = IntegerDigits[a];
c = Split[b];
d = Cases[c, {0 ..}];
Max[Length /@ d]

It takes 0.02 sec

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Tarun Singh
Feb 2, 2015

how i solved it, 

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a=str( 4**(4**5)+4**(5**6))
if True:
    x='' 
    for i in range(20):   #assuming max consecutive zeros to be 20
        x+='0'           #x will increment to number of zeroes
        if x in a:         #check if this much consecutive zeroes exist in a
            print x
#0.07200407981872559 seconds

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Ruby Code: 

def f(x,y,z)
  return x**(x**y) + x**(y**z)
end

a = f(4,5,6).to_s.split(/[^0]/) - [""] #return array containing only strings of 0s.
z = 0
a.each { |zeros| z = zeros.size if zeros.size > z}

z = 3

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Mharfe Micaroz
Oct 30, 2014

Additional problem: What digit therefore has the longest chain?

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Answer: Digit 1 (length of longest chain is 4)

Mathematica Code 

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f[x_, y_, z_] := x^x^y + x^y^z
a = f[4, 5, 6];
b = IntegerDigits[a];
c = Split[b];
d = Table[Max[Length /@ Cases[c, {i ..}]], {i, 0, 9}]

Hunter Killer - 6 years, 5 months ago

Using Python code as follows: 

 f = x**(x**y)+x**(y**z)
 s = str(f)
 n = 0
 m = 0
 z = 'N'
 for i in range(len(s)):
      if int(s[i])==0:
            m+= 1
    z = 'Y'
elif int(s[i]) > 0 and z == 'Y':
    if m > n:
        n = m
    print m, n
    m = 0
    z = 'N'
0 Helpful 0 Interesting 0 Brilliant 0 Confused

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