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Life to the Max - 2012 The Birth of Max 1-1 was released on:
USA: 7 October 2006
1 answer
Use the median-of-three algorithm:
int min (int a, int b) { return a<b?a:b; }
int max (int a, int b) { return a<b?b:a; }
int median_of_three (int a, int b, int c) { return max (min (a, b), min (max (a, b), c)); }
Note that the algorithm does not cater for equal values which creates a problem when any two values are equal, because there are only two values to play with, neither of which can be regarded as being the middle value. If the equal value is the lower of the two values, the largest value is returned if and only if it is the last of the three values, otherwise the lowest value is returned. But when the equal value is the larger of the two values, the largest value is always returned.
Lowest value is equal:
Input: 0, 0, 1 = max (min (0, 0), min (max (0, 0), 1)) = max (0, min (0, 1)) = max (0, 1) = 1
Input: 0, 1, 0 = max (min (0, 1), min (max (0, 1), 0)) = max (0, min (1, 0)) = max (0, 0) = 0
Input: 1, 0, 0 = max (min (1, 0), min (max (1, 0), 0)) = max (0, min (1, 0)) = max (0, 0) = 0
Highest value is equal:
Input: 0, 1, 1 = max (min (0, 1), min (max (0, 1), 1)) = max (0, min (1, 1)) = max (0, 1) = 1
Input: 1, 0, 1 = max (min (1, 0), min (max (1, 0), 1)) = max (0, min (1, 1)) = max (0, 1) = 1
Input: 1, 1, 0 = max (min (1, 1), min (max (1, 1), 0)) = max (1, min (1, 0)) = max (1, 0) = 1
The only way to resolve this problem and produce a consistent result is to sum all three inputs then subtract the minimum and maximum values:
int median_of_three (int a, int b, int c) { return a + b + c - min (min (a, b), c) - max (max (a, b), c)); }
Lowest value is equal:
Input: 0, 0, 1 = 0 + 0 + 1 - min (min (0, 0), 1) - max (max (0, 0), 1) = 1 - 0 - 1 = 0
Input: 0, 1, 0 = 0 + 1 + 0 - min (min (0, 1), 0) - max (max (0, 1), 0) = 1 - 0 - 1 = 0
Input: 1, 0, 0 = 1 + 0 + 0 - min (min (1, 0), 0) - max (max (1, 0), 0) = 1 - 0 - 1 = 0
Highest value is equal:
Input: 0, 1, 1 = 0 + 1 + 1 - min (min (0, 1), 1) - max (max (0, 1), 1) = 2 - 0 - 1 = 1
Input: 1, 0, 1 = 1 + 0 + 1 - min (min (1, 0), 1) - max (max (1, 0), 1) = 2 - 0 - 1 = 1
Input: 1, 1, 0 = 1 + 1 + 0 - min (min (1, 1), 0) - max (max (1, 1), 0) = 2 - 0 - 1 = 1
This makes sense because when we sort 0, 0, 1 in ascending order, 0 is in the middle, while 0, 1, 1 puts 1 in the middle.
1 answer
Roswell - 1999 Max to the Max 1-20 was released on:
USA: 1 May 2000
Japan: 20 October 2001
1 answer
Max 1. DIMONT has written:
'The indestructable Jew'
1 answer
Max and Ruby - 2002 Max's Music 1-47 was released on:
UK: January 2003
USA: 1 February 2003
Australia: 7 November 2008
1 answer
The Max It Show - 2013 1-1 was released on:
USA: 23 March 2013
1 answer
Max Headroom - 1987 Blipverts 1-1 is rated/received certificates of:
Iceland:L
1 answer
Mighty Max - 1993 Bring Me the Head of Mighty Max 1-10 was released on:
USA: 10 November 1993
1 answer
Max and Ruby - 2002 Max Cleans Up Max's Cuckoo Clock Ruby's Jewelry Box 1-7 was released on:
USA: October 2002
1 answer
Max Headroom - 1987 Blipverts 1-1 was released on:
USA: 31 March 1987
1 answer
Max Monroe - 1990 Legacy 1-1 was released on:
USA: 5 January 1990
1 answer
Assuming the numbers are integers, and they are stored in an array called myArray. The following has not been tested, but gives the general idea: int max; max = myArray[0]; for (int i = 1; i < myArray.length(); i++) } if (myArray[i] > max) max = myArray[i]; System.out.println(max); }
Assuming the numbers are integers, and they are stored in an array called myArray. The following has not been tested, but gives the general idea: int max; max = myArray[0]; for (int i = 1; i < myArray.length(); i++) } if (myArray[i] > max) max = myArray[i]; System.out.println(max); }
Assuming the numbers are integers, and they are stored in an array called myArray. The following has not been tested, but gives the general idea: int max; max = myArray[0]; for (int i = 1; i < myArray.length(); i++) } if (myArray[i] > max) max = myArray[i]; System.out.println(max); }
Assuming the numbers are integers, and they are stored in an array called myArray. The following has not been tested, but gives the general idea: int max; max = myArray[0]; for (int i = 1; i < myArray.length(); i++) } if (myArray[i] > max) max = myArray[i]; System.out.println(max); }
2 answers
This is best done with an array; that way, you can easily extend to more than three numbers. In Java, it would be something like this:
int myNumbers[] = {1, 5, 3};
int max = myNumbers[1];
for (int i = 1; i < myNumbers.length(); i++)
if (myNumbers[i] > max) max = myNumbers[i];
This is best done with an array; that way, you can easily extend to more than three numbers. In Java, it would be something like this:
int myNumbers[] = {1, 5, 3};
int max = myNumbers[1];
for (int i = 1; i < myNumbers.length(); i++)
if (myNumbers[i] > max) max = myNumbers[i];
This is best done with an array; that way, you can easily extend to more than three numbers. In Java, it would be something like this:
int myNumbers[] = {1, 5, 3};
int max = myNumbers[1];
for (int i = 1; i < myNumbers.length(); i++)
if (myNumbers[i] > max) max = myNumbers[i];
This is best done with an array; that way, you can easily extend to more than three numbers. In Java, it would be something like this:
int myNumbers[] = {1, 5, 3};
int max = myNumbers[1];
for (int i = 1; i < myNumbers.length(); i++)
if (myNumbers[i] > max) max = myNumbers[i];
2 answers
