9
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Background

In this challenge, a base-b representation of an integer n is an expression of n as a sum of powers of b, where each term occurs at most b-1 times. For example, the base-4 representation of 2015 is

4^5 + 3*4^4 + 3*4^3 + 4^2 + 3*4 + 3

Now, the hereditary base-b representation of n is obtained by converting the exponents into their base-b representations, then converting their exponents, and so on recursively. Thus the hereditary base-4 representation of 2015 is

4^(4 + 1) + 3*4^4 + 3*4^3 + 4^2 + 3*4 + 3

As a more complex example, the hereditary base-3 representation of

7981676788374679859068493351144698070458

is

2*3^(3^(3 + 1) + 2) + 3 + 1

The hereditary base change of n from b to c, denoted H(b, c, n), is the number obtained by taking the hereditary base-b representation of n, replacing every b by c, and evaluating the resulting expression. For example, the value of

H(3, 2, 7981676788374679859068493351144698070458)

is

2*2^(2^(2 + 1) + 2) + 2 + 1 = 2051

The Challenge

You are given as input three integers b, c, n, for which you may assume n >= 0 and b, c > 1. Your output is H(b, c, n). The shortest byte count wins, and standard loopholes are disallowed. You can write either a function or a full program. You must be able to handle arbitrarily large inputs and outputs (bignums).

Test Cases

4 2 3 -> 3
2 4 3 -> 5
2 4 10 -> 1028
4 4 40000 -> 40000
4 5 40000 -> 906375
5 4 40000 -> 3584
3 2 7981676788374679859068493351144698070458 -> 56761
2 3 2051 -> 35917545547686059365808220080151141317047

Fun Fact

For any integer n, the sequence obtained by

n1 = n
n2 = H(2, 3, n1) - 1
n3 = H(3, 4, n2) - 1
n4 = H(4, 5, n3) - 1
....

eventually reaches 0. This is known as Goodstein's theorem.

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6
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CJam, 60 58 45 43 41 38 36 bytes

Thanks to Optimizer for saving two bytes.

l~:C;:B;{Bb)1$,,@f{1$~=C@)F#*+}~}:F~

Test it here.

Takes input in order n b c.

You can use this to test run all test cases:

"3 4 2 
3 2 4 
10 2 4 
40000 4 4 
40000 4 5 
40000 5 4 
7981676788374679859068493351144698070458 3 2 
2051 2 3 "N/
{
~:C;:B;{Bb)1$,,@f{1$~=C@)F#*+}~}:F~
p}/

Explanation

This is a fairly direct implementation of the process explained in the challenge, except that I interleave recursive base expansion, base substitution and computation of the final result:

l~:C;:B;{Bb)1$,,@f{1$~=C@)F#*+}~}:F~
l~:C;:B;                             "Read and evaluate input, store b and c in B and C.";
        {                       }:F  "Define a block F. This performs the required conversion.";
         Bb                          "Get digits of input number in base B.";
           )                         "Split off 0-power digit.";
            1$,                      "Copy remaining digits. Get their length n.";
               ,                     "Make array [0 1 ... n-1].";
                @                    "Pull up remaining digits.";
                 f{           }      "Map this block onto the range, passing in the digits
                                      as a second argument each time.";
                   1$~=              "Copy current i, bitwise complement, access digit array.
                                      This accesses the digits in reverse order.";
                       C             "Push the new base C.";
                        @)           "Pull up current i and increment to get power.";
                          F          "Apply F recursively.":
                           ~         "Raise C to the resulting power.";
                            *        "Multiply by digit.";
                             +       "Add to running total.";
                               ~     "The result will be in an array. Unwrap it.";
                                   ~ "Execute F on the input n.";
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8
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Python 2, 55

H=lambda b,c,n,s=0:n and n%b*c**H(b,c,s)+H(b,c,n/b,s+1)

A recursive solution. Like the recursive algorithm to convert between bases, except it recurses on the exponent as well.

We split n into two parts, the current digit n%b, and all other digits n/b. The current place value is stored in the optional parameter s. The current digit is converted to base c with c** and the exponent s is converted recursively. The remainder is then converted the same way, as +H(b,c,n/b,s+1) but the place value s is one higher.

Unlike base conversion, hereditary base conversion required remembering the current place value in the recursion for it to converted.

For ease of reading, here's what it looks like when b and c are fixed global constants.

H=lambda n,s=0:n and n%b*c**H(s)+H(n/b,s+1)
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  • \$\begingroup\$ I've posted this mostly because I didn't realize you could use named arguments in pyth: D(GHY=Z0)R&Y+*%YG^H(GHZ)(GH/YGhZ. Feel free to add it if you want (I'm off to tips for golfing in pyth :D ) \$\endgroup\$ – FryAmTheEggman Jan 12 '15 at 3:56

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