----------------------------------- cool dude Fri Jun 16, 2006 5:20 pm pythagorean triplet ----------------------------------- does anyone know of an easy algorithm (that i would understand) to get a pythagorean triplet. i know wat they are i just need an algorithm because wat i'm trying to do is find the pythagorean triplet where a + b + c = 1000. because i don't have an algorithm i'm doing guess and check. i tried making a program, but i think i'm doing something wrong so it doesn't work. ----------------------------------- cool dude Fri Jun 16, 2006 6:28 pm ----------------------------------- well i just made a program and got the numbers. :) although i did it kind of forcefully so i still want to know the algorithm. ----------------------------------- Clayton Fri Jun 16, 2006 6:33 pm ----------------------------------- well we know that the lowest pythagorean triplet is 3,4,5 right? so have a program come up with something to the effect of : (note this is in turing :() %all this is doing is taking the three sides and checking if they are in ratio with the smallest whole pythagorean triplet fcn find_triplet (a,b,c:int) :boolean if a mod 3= 0 and b mod 4=5 and c mod 5 = 0 or a mod 4=0 and b mod 3 = 0 and c mod 5 =0 then result true else result false end if end find_triplet something like that will allow you to input numbers to check and see if they are pythagorean triplets, or if you just want to find pythagorean triplets then type Pythagoreus record : a,b,c:int end record fcn give_triplet (side:Pythagoreus, max:int) :Pythagoreus for x:1..max side.a:= x*3 side.b:= x*4 side.c:= x*5 result side end for end give_triplet (note the above code is untested) basically what it does is find all pythagorean triplets up to a specified number as for real numbers im not to sure of an algorithm to solve that :( ull just have to experiment ----------------------------------- Cervantes Fri Jun 16, 2006 6:41 pm ----------------------------------- Here's what I would do: Generate the first few unique pythagorean triplets. If your search is small enough, this may be done manually. The first two are: 3, 4, 5, and 5, 12, 13. You can make a pythagorean triplet from one you already know simply by multiplying each value by a constant. a^2 + b^2 = c^2 // multiply both sides by 4 4(a^2 + b^2) = 4c^2 4a^2 + 4b^2 = 4c^2 (2a)^2 + (2b)^2 = (2c)^2 So take your first triplet: 3, 4, 5. Multiply it by 2 to get the new one: 6, 8, 10. Check if 6 + 8 + 10 == 1000. If not, move on to multiplying by 3 to get 9, 12, 15. Check. Once you get a sum that passes 1000, move on to the next triplet: 5, 12, 13. Repeat the process for this triplet. Makes sense? Now we optimize it. We don't have to iterate through all the possibilities, multiplying the base triplet by 2 then by 3, etc. until we pass our target sum. Our triplet is represented by a, b, c. We multiply each of a, b, and c by a natural number, k. Therefore, our terms are ka, kb, and kc, where k is a natural number ka + kb + kc = 1000 // For our first triplet: a, b, c = 3, 4, 5 k(3) + k(4) + k(5) = 1000 12k = 1000 k = 83.33333... But we said k is a natural number. Therefore, there is no pythagorean triplet that is a multiple of the 3, 4, 5 triplet whose sum of a, b, and c equals 1000. So, we move on to the next triplet: 5, 12, 13. This also doesn't work. k = 33.333.... Try the next triplet. Repeat until you get it. ----------------------------------- Cervantes Fri Jun 16, 2006 6:49 pm ----------------------------------- A quick search nailed it. The fourth triplet: 8, 15, 17. 1000 / (8 + 15 + 17) = 25 .: triplet is a = 25(8), b = 25(15), c = 25(17) a = 200, b = 375, c = 425 200 + 375 + 425 = 1000 ----------------------------------- Andy Fri Jun 16, 2006 7:44 pm ----------------------------------- for a small number like 1000, cervantes approach works very well, but what if it was 1351435890? then you'd need to generate a lot of pythagorean triples... let a = (n/2 - b)/(1 - b/n) for where a and b are positive integers and n is the sum given by the question. solve for b such that for 1