Find the next number in the pattern

Question

Write a program, that takes 5 numbers as input and outputs the next number in the pattern.

Heres a text file with 1835 test patterns, containing 6 numbers separated by commas each line. Using the first 5 numbers calculate the 6th number and then check if it matches. You must get all the test cases correct.

There are 3 types of patterns:

1. x_n = x_n-1+k

2. x_n = x_n-1*k

3. x_n = x_n-1^2

The first x value and k were chosen randomly between a set range to create the pattern sample file.

This is code-golf, lowest number of bytes to get all of the test cases correct wins.

Answer 1

Python 2: 55 bytes

f=lambda a,b,c,d,e:e+c-d*2and e*e/d**(d*d==e*c)or e*2-d

We check for the three cases as follows:

1. Check for arithmetic sequence: If e+c==d+d, output e+e-d
2. Check for geometric sequence: If e*c==d*d, output e*e/d
3. Otherwise, squared sequence: Output e*e.

Case 3 is used as the else-case to avoid the more-lengthy checking for successive squaring.

Case 1 is checked first, with Boolean short-circuiting to avoid checking Case 2 after a success, as this causes an error when d=0. There's a test case of all zeroes, so we can't cheat this by picking another position in place of d.

Afterwards, we can treat both Cases 2 and 3, since there's no risk of a divide-by-0 error. Noting that the choices are similar (e*e vs e*e/d), we combine them with an arithmetic expression by dividing by either d**1 for Case 3 or d**0 for Case 2. Although this division produces an integer value, we must use Python 2 (or //) to avoid float errors that case the largest-output test case to fail.

Thanks to @Will for inspiration on the code structure and a very-helpful test framework.

Answer 2

Python all right, 72 74 75 chars

p=lambda i,j,k,l,m:j-i!=k-j and(k==j*j==j*i*i and m*m or m*l/k)or m+j-i

Thx to xnor for showing me the if/else to and/or trick and golfing off a char :)

Thx to flornquake for fixing the k==j*j==j*i*i and saving another two chars :)

Actually, it gets one wrong... but that's a bug in the test patterns which was acknowledged yesterday but still not fixed.

Testing code, hopefully useful for all competitors to test with:

def check(func,title):
    right = 0
    print "===", title, "==="
    for i,test in enumerate(open('patterns.txt')):
        test = map(int,test.split(','))
        try:
            got, expected = func(*test[:5]), test[5]
        except Exception as e:
            got = e
        if got != expected:
            print "ERROR on line %d: %s != %d" % (i, got, expected)
            print " test :", ", ".join(map(str,test))
            plus = multi = cube = str(test[0])
            for i in range(1, 6):
                plus += ", %d" % (test[i-1]+(test[1]-test[0]))
                multi += ", %s" % ((test[i-1]*(test[1]/test[0])) if test[0] else "DivZero")
                cube += ", %d" % (test[i-1]**2)
            print " as + :", plus
            print " as * :", multi
            print " as ^2:", cube 
        else:
            right += 1
    print right, "out of", i+1, "right!"

says:

ERROR on line 19: 23283064365386962890625 != 3273344365508751233
 test : 5, 25, 625, 390625, 152587890625, 3273344365508751233
 as + : 5, 25, 45, 645, 390645, 152587890645
 as * : 5, 25, 125, 3125, 1953125, 762939453125
 as ^2: 5, 25, 625, 390625, 152587890625, 23283064365386962890625
1834 out of 1835 right!

Answer 3

Python 3 -1833

Length 198 chars

Now updated to conform to the rules.

m=0
for i in open('s'):
 if','in i:
  i,j,k,l=list(map(int,i.split(',')[2:]))
  if j-i==k-j:
   if k+(j-i)==l:m+=1
  elif j/i==k/j:
   if k*(j/i)==l:m+=1
  elif i**2==j:
   if k**2==l:m+=1
print(m)

Answer 4

C# - 1835

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;

namespace pattern
{
    class Program
    {
         static void Main(string[] args)
         {
            string[] lines = File.ReadAllLines("in.txt");
            int matches = 0;
            for(int i = 0; i < lines.Length; i++)
            {   
                long[] numbers = lines[i].Split(',').Select(a=>long.Parse(a)).ToArray();

                long nextnum = 0;
                if(numbers[1]-numbers[0]==numbers[2]-numbers[1])
                nextnum = numbers[numbers.Length-2]+(numbers[1]-numbers[0]);
            else if(numbers[1]/numbers[0]==numbers[2]/numbers[1])
                nextnum = numbers[numbers.Length-2]*(numbers[1]/numbers[0]);
            else if(numbers[1] == numbers[0]*numbers[0])
                nextnum = numbers[numbers.Length-2]*numbers[numbers.Length-2];
            if(nextnum == numbers[numbers.Length-1])
                matches++;
        }
        Console.WriteLine(matches + " matches found!");
        Console.ReadLine();
    }
}
}

This is obviously not golfed at all, but it finds all numbers eventually. It just prints how many matches it found, but you could potentially display all found numbers, only non-matches, etc...

(This is conceptually very similar to Beta Decay's answer, just a little bit more complicated)

Answer 5

TI-BASIC, 120 117 136

Prompts for input and returns the output.

Input L1:0:If not(L1(5:Stop
If L1(5)-L1(4)=L1(4)-L1(3:L1(5)+(L1(5)-L1(4
If L1(5)/L1(4)=L1(4)/L1(3:L1(5)(L1(5)/L1(4
If L1(4)²=L1(5:L1(5)²

Edit: Fixed divide by zero error.