Rise, sequence, rise

Question

We have a strictly increasing sequence of non-negative integers, like:

12 11 10

Wait! This sequence isn't strictly increasing, is it? Well, the numbers are written in different bases. The least possible base is 2, the biggest is 10.

The task is to guess bases each number is written, so that:

• the sequence is strictly increasing,
• the sum of the bases is maximised.

For instance, the solution for the sample will be:

6 8 10

because under those bases the sequence becomes 8 9 10 decimal - a strictly increasing sequence, and we are not capable of finding bases for which the sequence remains strictly increasing and whose sum is bigger than 6+8+10 .

Due to the second limitation a solution 3 5 7 is not satisfactory: in spite of fact that the sequence becomes 5 6 7 under those bases - we need to maximise the bases sum, and 3+5+7 < 6+8+10.

If under no bases 2<=b<=10 is it possible for the series to be strictly increasing, for instance:

102 10000 10

single

0

should be output.

The input sequence can be passed in the way it's most convenient for your solution (standard input / command line parameters / function arguments...).

Answer 1

Pyth, 31 30 29 bytes

e+0f.x!sgM.:iVczdT2ZosN^STlcz

1 byte thanks to @Jakube.

Demonstration. Test Harness.

Input is given on STDIN, space separated. If newline separated input is allowed, I can shorten the program by 2 bytes.

Explanation:

e+0f.x!sgM.:iVczdT2ZosN^STlcz
                                  Implicit: z = input(), T = 10, Z = 0, d = ' '
                        ST        [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
                          lcz     len(z.split())
                       ^          All combinations w/replacement of that length.
                    osN           Order by increasing sum.
   f                              Filter on
              czd                 z.split(' ')
            iV   T                Vectorize the "Convert to base" operation over 
                                  the integers as strings and the base sequence.
          .:      2               Take length 2 subsequences.
        gM                        Map the >= operation over them.
      !s                          Sum and logically negate.
    .x             Z              If that throws an error, returns 0 (e.g. reject)
 +0                               Prepend a 0, in case no sequences are found.
e                                 Take the end of the list.

Including 1 in the list of possible bases is safe because i, which uses Python's int builtin, doesn't allow 1 as a base, and therefore always throws an error, which is caught and filtered out.

Answer 2

CJam, 43 bytes

0B,2>ea,m*{:+~}${ea::~_2$.b__Q|$=*@.b=}=p];

Reads command-line arguments and prints an array.

Try it online in the CJam interpreter.

Examples

$ cjam rise.cjam 12 11 10
[6 8 10]
$ cjam rise.cjam 19 18 17
0

How it works

0       e# Push a 0 (default return value).
B,2>    e# Push [0 ... 10] and remove the first two elements.
ea,     e# Push the number of command-line arguments (n).
m*      e# Cartesian power. Pushes all vectors of {2 ... 10}^n.
{:+~}$  e# Sort by the negated sums.
{       e# Find; for each vector V in {2 ... 10}^n:
  ea::~ e#   Evaluate each character of each command-line argument.
  _2$   e#   Copy the results and V.
  .b    e#   Vectorized base conversion (list to integer).
  __    e#   Push two copies.
  Q|$   e#   Deduplicate and sort the last copy.
  =     e#   Compare it to the first. Pushes 1/0 if equal/unequal.
  *     e#   Repeat the original result of .b that many times.
  @.b   e#   Vectorized base conversion (integer to list).
  =     e#   Compare the result to the modified command-line arguments.
        e#   Equality makes sure that the base was greater than all digits.
}=      e# If pushed 1, push V and break.
p       e# Print. Either prints the last V or 0 if none matched.
];      e# Clear the stack to avoid implicitly printing the 0 (if still present).

Answer 3

Julia, 176 156 145 118 109 99 97 bytes

A->try p=NaN;flipud(map(i->(k=11;t=p;while t<=(p=parseint("$i",k-=1))end;k),flipud(A)))catch;0end

Ungolfed:

function anonfunc(i)
  # Start with k=11 so that it evaluates to 10 on first while iteration
  k=11
  # set t to the previous value of p
  # Note: p here gets held over between iterations within the map
  t=p
  # Iterate through, dropping k by 1 and evaluating the integer in
  # base k and stopping if the value drops below t
  # Note: "p=" expression inside conditional to ensure k-=1 is evaluated
  # at least once (to make NaN work as desired)
  while t<=(p=parseint("$i",k-=1))
  end
  # if it dropped below t, return the base, k to be the corresponding
  # element in the map
  return k
end

function f(A)
  # Using try/catch to return 0 if no acceptable base found
  try
    # This is a trick to make sure the comparison in the while loop
    # evaluates to false on the first use of it (last value in A)
    p=NaN
    # Apply anonfunc to each element of A, starting with the last element
    # and store the result in S
    S=map(anonfunc,flipud(A))
    # S is backwards, so flip it and return it
    return flipud(S)
  catch
    # Will throw to here if parseint fails with the base due to having
    # a digit not acceptable in the base
    return 0
  end
end

Used with a 1d array input. If the function is assigned to c, then you would call c([12,11,10]) and it would output [6,8,10].

Note: I had used dec(i) inside the parseint command, but because i is a single-character variable name, and I don't need to access a component, I used "$i" to get the same result.

Answer 4

Julia, 259 204 183 bytes

Saved a bunch with help from Glen O.

A->(M(x)=maxabs(digits(x))+1:10;S=[];X={};for i=M(A[1]),j=M(A[2]),k=M(A[3]) s=map(parseint,map(dec,A),[i,j,k]);all(diff(s).>0)&&(S=[S,sum(s)];X=[X,{[i,j,k]}])end;X==[]?0:X[indmax(S)])

Ungolfed + explanation:

function f(A)
    # Define a function to obtain the smallest possible base range
    M(x) = (maxabs(digits(x)) + 1):10

    # Define container arrays for the sums and bases
    S = []
    X = {}

    # Loop over all possible bases for each of the elements
    for i = M(A[1]), j = M(A[2]), k = M(A[3])
        # Parse each element of the input as a string
        # in the given base
        s = map(parseint, map(dec, A), [i,j,k])

        # Push the sum and bases if s is rising
        if all(diff(s) .> 0)
            S = [S, sum(s)]
            X = [X, {[i,j,k]}]
        end
    end

    # If X is empty, return 0, otherwise return the bases
    isempty(X) ? 0 : X[indmax(S)]
end

Answer 5

CJam (39 bytes)

{Afb:X,9,2f+m*{X\.b__$_&=*},{:+}$0\+W=}

This is an anonymous function which takes the input as an array of decimal integers on the stack and leaves the output as an array or the integer 0 on the stack. Online demo.

Answer 6

Python 2 (147 bytes)

def x(s):
 q=int;c=10;o=q(s[-1])+1;l=[]
 for i in map(str,s)[::-1]:
    n=q(i,c)
    while o<=n:
        c-=1;n=q(i,c)
        if 3>c:return 0
    l=[c]+l;o=n
 return l

Call the function x with a list of the ints.

Example:

print x([12,11,10])

prints

[6, 8, 10]