CPMagical CP Library

Problem solution · Turing

CCC 2006 J3 - Cell-Phone Messaging

CCC 2006 J3 - Cell-Phone Messaging: a Turing solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Direct simulation
Source
CCCSolutions
Length
49 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For CCC 2006 J3 - Cell-Phone Messaging, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 49 lines of Turing from the credited upstream file ccc06j3.t.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2006 J3 - Cell-Phone Messaging · TuringTuring
Use this to learn the idea, then write your own version.
% CCC 2006 Junior problem 3% Cell-Phone Messaging%% determine the time to type a message on a numeric pad% one sec per keystroke and two sec per pause between% consecutive letters on the same key.%% simple string processing and if's var s : stringvar t : int loop    get s    exit when s = "halt"    t := 0    for i : 1 .. length (s)        if index("adgjmptw", s (i)) > 0 then            t := t + 1        elsif index ("behknqux", s (i)) > 0 then            t := t + 2        elsif index ("cfilorvy", s (i)) > 0 then            t := t + 3        elsif index ("sz", s (i)) > 0 then            t := t + 4        end if        if i > 1 and ( (index ("abc", s (i - 1)) > 0 and                index ("abc", s (i)) > 0) or                (index ("def", s (i - 1)) > 0 and                index ("def", s (i)) > 0) or                (index ("ghi", s (i - 1)) > 0 and                index ("ghi", s (i)) > 0) or                (index ("jkl", s (i - 1)) > 0 and                index ("jkl", s (i)) > 0) or                (index ("mno", s (i - 1)) > 0 and                index ("mno", s (i)) > 0) or                (index ("pqrs", s (i - 1)) > 0 and                index ("pqrs", s (i)) > 0) or                (index ("tuv", s (i - 1)) > 0 and                index ("tuv", s (i)) > 0) or                (index ("wxyz", s (i - 1)) > 0 and                index ("wxyz", s (i)) > 0)) then            t := t + 2        end if    end for    put tend loop

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗