# Shortest code to generate all Pythagorean triples up to a given limit

Generate the shortest possible code in any programming language that can generate all Pythagorean triples with all values not exceeding a given integer limit. A Pythagorean triple is a set of three integers $$\(a, b, c)$/extract_tex] that satisfy the equation $$\a^2 + b^2 = c^2\$$. The program should output the triples in any format, such as a list of tuples or a newline-separated list of strings. Input: An integer limit $$\n\$$ (1 ≤ $$\n\$$$$\10^6\$$) Output: All Pythagorean triples $$\(a, b, c)\$$ such that $$\1 ≤ a, b, c ≤ n\$$ and $$\a^2 + b^2 = c^2\$$. ## Test Cases Input: 20 Output: (3, 4, 5) (5, 12, 13) (6, 8, 10) (8, 15, 17) (9, 12, 15) (12, 16, 20)  Input: 5 Output: (3, 4, 5)  Note: The output order does not matter as long as all the correct Pythagorean triples are included. Duplicate triples should not be included. But, specifying the order might help. • Can I output numbers without parentheses like 3,4,5? Feb 27 at 13:25 • Suggested test cases: a number less than 5 and a number that isn't equal to any possible \c\ value. Feb 27 at 23:28 • Suggested Test-Case: n=16, because answers iterating only over a and b might produce the wrong solution with c=17 Feb 28 at 14:14 ## 24 Answers # 05AB1E, 9 bytes L3.ÆʒnαQ  The last three bytes could alternatively be RÆ_ for the same byte-count: Explanation: L # Push a list in the range [1, (implicit) input-integer] 3.Æ # Create all triplet-combinations of this list ʒ # Filter this list of [a,b,c]-triplets by: n # Square each inner integer in the triplet  # Push all three values separated to the stack α # Get the absolute difference of the top two: |b²-c²| Q # Check if it's equal to the third one: a² == |b²-c²| # (after which the filtered list of triplets is output implicitly as result) R # Reverse the triplet to [c²,b²,a²] Æ # Reduce it by subtracting: c²-b²-a² _ # Check if this is 0  # Vyxal, 9 8 bytes ɾ3ḋ'²÷ε=  Try it Online! I'm surprised I didn't manage to make this horribly inefficient. -1 thanks to Kevin! ## Explained ɾ3ḋ'²÷ε= ɾ # Range [1, input] 3ḋ # Combinations of length 3 ' # filtered by: ² # squaring everything ÷ # dumping the triplet onto the stack in reverse order ε # absolute difference of b**2 and c**2 = # equals a**2  • ^+ can be ε for -1 byte (port of my top 05AB1E answer). Feb 27 at 13:22 • @KevinCruijssen good catch. Didn't think to think that the equation could be written in terms of equalling a squared Feb 27 at 23:10 # Factor + math.combinatorics, 57 bytes [ [1,b] 3 [ 2 v^n first3 -rot + = ] filter-combinations ]  Try it online! • [1,b] 3 [ ... ] filter-combinations select combinations of three from $$\[1..n]\$$ where... • 2 v^n square the elements of the triplet • first3 place each element onto the stack • -rot move $$\c^2\$$ from the top of the stack to the bottom • + add $$\a^2\$$ and $$\b^2\$$ • = equal? # Thunno, $$\13\log_{256}(96)\approx\$$ 10.70 bytes R1+3zQg2^Au_=  Attempt This Online! Port of Kevin Cruijssen's 05AB1E answer. #### Explanation R1+3zQg2^Au_= # Implicit input R1+ # Push range(1, input+1) 3zQ # Combinations with length 3 g # Filter by: STACK: [a, b, c] 2^ # Square each STACK: [a**2, b**2, c**2] Au # Dump onto stack STACK: a**2, b**2, c**2 _ # Subtract top two STACK: a**2, c**2 - b**2 = # Are they equal? STACK: a**2 == c**2 - b**2 # Implicit output  # Japt, 14 bytes õ à3 fÈÌ²Ñ¶Xx²  Try it õ à3 fÈÌ²Ñ¶Xx² :Implicit input of integer U õ :Range [1,U] à3 :Combinations of length 3 f :Filter by È :Passing each X through the following function Ì : Last element ² : Square Ñ : Multiply by 2 ¶ : Is equal to Xx : X reduced by addition ² : After squaring each  • 12 bytes by (roughly) porting Kevin's O5AB1E answer Feb 27 at 15:52 • Could be 11 if there were a reduce-right builtin :| Feb 27 at 16:01 • Why multiply by 2 after squaring? Feb 28 at 14:17 • @Falco the program checks if \c^2 \times 2 = a^2 + b^2 + c^2\, which simplifies to pythagorean theorem Mar 1 at 16:17 # ><> (Fish), 111 bytes i:*&43::*{:}:*+\ /*::<5[1;?)&:&:/ \:{:}(?!v~1+31. /]v?)}:{/|.!051+1~\ n$~~>1+$:@$:@= ?^50.
\' 'o}:n' 'o}:nao$~$64.


Try it

## Explanation

><> lacks a square root function so this simply tries all numbers until it finds one greater than the square of the inputs.

First, we square the input. We start searching at a,b=4,3. We check if $$\a^2\$$, (::*) + $$\b^2\$$ is more than the input squared, if so we exit.

Then we push 5, the starting third value. Square it, if it is more than a^2+b^2, go to the next iteration. Otherwise, add 1 to C.

If A^2+B^2=C^2, print A, B, and C, then go to next.

When we go to the next iteration, add 1 to B. If B>A, add 1 to A instead and reset B to 1. This is 1+$:@$:@=?~1+1. Then we jump back to the start to try another iteration.

## Scala, 71 bytes

(n:Int)=>for{c<-5 to n;b<-4 to c;a<-3 to b if a*a+b*b==c*c}yield(a,b,c)


or styled differently:

(n:Int)=>for{c<-5 to n
b<-4 to c
a<-3 to b
if a*a+b*b==c*c}yield(a,b,c)


It's not often that Scala gets a chance to do well in golfing, but the .to() method is perfect for inclusive ranges.

Try it online or on Scastie!

# Noether, 62 bytes

!aI~n(!a~bn1+({a2^!b2^+0.5^~cn<cn=|cc_=&}{aP","PbP","PcP?}b)a)


Surprisingly this is the first time I've realised the severe lack of $$\\leq\$$ and $$\\geq\$$ operators in the language.

Explanation:

!a                                                             # Initialiase variable a to 1
I~n                                                          # Store user input in n: outer loop exit condition
(                                                       ) # Outer loop
!a~b                                                     # Increment a and store value in b
n1+                                                  # Add 1 to n: inner loop exit condition
(                                             )   # Inner loop
{                         }{               }     # If ... Else
a2^                                             # Square a
!b2^                                         # Increment value of b and square
+0.5^                                    # Sum a and b and square root
~cn<                                # Store value in c and check if less than n
cn=                             # c == n
|                            # Bitwise OR
cc_=                        # c == floor(c)
&                       # Bitwise AND
aP","PbP","PcP?      # If true, output variable values, comma separated
b    # Push b to stack (inner loop exit condition)
a  # Push a to stack (outer loop exit condition)


# JavaScript, 81 bytes

n=>{for(a=0;++a<n;)for(b=a;b<n;++b)(c=(a*a+b*b)**.5)%1||c<=n&&console.log(a,b,c)}


Try it:

f=n=>{for(a=0;++a<n;)for(b=a;b<n;++b)(c=(a*a+b*b)**.5)%1||c<=n&&console.log(a,b,c)}

;[
1,
3,
5,
7,
10,
17,
20
].forEach(n=>{console.log(n+':');f(n)})

• 64 bytes Feb 27 at 15:00
• @Shaggy Thank you! But I deliberately did not write such an option because of this is the @Aranuld's answer just with for instead of while :) Feb 27 at 15:04
• @Shaggy the shorter solution errs for n=16, because then c will be 17, which is too large Feb 28 at 14:13
• @PierrePaquette There are no duplicates in either my solution or Shaggy's solution. Feb 28 at 23:38
• Indeed; I just noticed I didn’t run Shaggy’s solution properly. Apologies. Feb 28 at 23:44

# JavaScript (V8), 63 bytes

n=>{for(;p=n;n--)while(--p>(q=(n*n-p*p)**.5))q%1||print(q,p,n)}


Try it online!

### Commented

n => {                  // n = input
for(                  // outer loop:
;                   //
p = n;              //   before each iteration, initialize p to n
//   stop when n = 0
n--                 //   decrement n after each iteration
)                     //
while(              //   inner loop:
--p > (           //     decrement p and test whether it's greater
q =             //     than q defined as:
(n * n - p * p) //       the square root of the difference
** .5           //       between n² and p²
)                 //     stop as soon as the test fails
)                   //
q % 1 ||          //     if q is an integer:
print(q, p, n)  //       print the triplet (q, p, n)
}                       //

• I'm not aware of the "print" function in JavaScript. window.print() seems to print the currently open HTML document. I think you need console.log ? Feb 28 at 13:08
• @Falco This is a V8 instruction, which is unrelated to window.print(). Feb 28 at 13:26
• Oh I didn't know this. Also your program errs on n=16 (printing 17,15,8 as a valid solution, but 17 > 16) Feb 28 at 13:31
• @Falco Thanks. Now fixed. Feb 28 at 14:18

# Arturo, 50 bytes

$=>[select combine.by:3@1..&'x[=+x\0^2x\1^2x\2^2]]  Try it $=>[                      ; an anonymous function
select                ; select elements from a block
combine.by:3 @1..&    ; combinations of 3 from the range 1 to n
'x                    ; assign current triplet to x in the select
[                     ; begin select
=+x\0^2x\1^2x\2^2 ; is a^2+b^2 equal to c^2?
]                     ; end select
]                         ; end function


# Python, 91 88 bytes

lambda n,r=range:[(k,j,i)for i in r(1,n+1)for j in r(1,i)for k in r(1,j)if i*i==j*j+k*k]


Attempt This Online!

# Nekomata, 12 bytes

RS3Lᵖ{:*Ɔ$∑=  Attempt This Online! A port of @Fatalize's Brachylog answer. RS3Lᵖ{:*Ɔ$∑=
R               Range from 1 to the input
S              Subset
3L            Length should be 3
ᵖ{          Start a block; treat the block as a predicate,
and return the original value if the block does not fail
:*        Multiply by itself
Ɔ$∑= The last element should equals to the sum of the other elements  By default, the Nekomata interpreter will print all possible results. # Charcoal, 26 bytes ＩΣΣＥ⊕ＮＥιＥΦλ⁼×ιι⁺×λλ×νν⟦νλι  Attempt This Online! Link is to verbose version of code. Explanation:  Ｎ Input integer ⊕ Incremented Ｅ Map over implicit range ι Current value Ｅ Map over implicit range λ Current value Φ Filter over implicit range ×ιι Square of outer value ⁼ Equals ×λλ Square of inner value ⁺ Plus ×νν Square of innermost value Ｅ Map over matches ⟦ List of ν Innermost value λ Inner value ι Outer value Σ Flatten Σ Flatten Ｉ Cast to string Implicitly print  # MathGolf, 21 bytes ╒■■mÅ─╡gæ_▀s=gÆ²xε-┬▀  Try it online. Explanation: ╒ # Push a list in the range [1, (implicit) input-integer] ■ # Get the cartesian product with itself to create pairs ■ # Get the cartesian product with itself to create pairs of pairs m # Map over each pair of pairs, Å # using 2 characters as inner code-block: ─ # Flatten the pair to a quadruplet ╡ # Discard the last item to make it a triplet g # Filter this list of triplets by, æ # using 4 characters as inner code-block: _ # Duplicate the current list ▀ # Uniquify the values in the copy s # Sort it from lowest to highest = # Check if the two lists are still the same g # Filter this list further by, Æ # using 5 characters as inner code-block: ² # Square each integer in the triplet x # Reverse it ε # Reduce the triplet by: - # Subtracting ┬ # Check if the result of this c²-b²-a² equals 0 ▀ # After the filter: uniquify the remaining list of triplets # (after which the entire stack is output implicitly as result)  # Pyth, 14 bytes f!-F^R2_T.CSQ3  Try it online! ### Explanation  # implicitly assign Q = eval(input()) .C 3 # all sorted lists of three elements from SQ # range(1,Q+1) f # filter these lists on lambda T _T # reverse T ^R2 # map elements to their squares -F # fold on subtraction (subtract all elements from the first) ! # not (only true for zero)  # Befunge-93 (PyFunge), 125 bytes v & >35g:*25g:*+15g:*-| ,*52.g51.g52.g53<v ^p53+1_v#\g52:g53<< ^111 >$135p25g:1 5g\!#v_1+25p
^ p51+1_@#g51\g51 :p521\$<


Try it online!

Note: The three "1"s shown here on line 6 ((1,5),(2,5) and (3,5) in Funge-Space) are actually be U+0001 in the source code. This is reflected correctly in TIO.

I may be able to save some bytes by moving those storage characters to line 1 or 2, and send line 6 left and wrap it, and if I can find a way to reduce the number of "g"s required to get the incremented values.

I'll do a nice writeup for this when I have time.

# Python 3, 112 104 bytes

lambda n,r=range:{(*sorted((i,j,k)),)for i in r(1,n+1)for j in r(1,n+1)for k in r(1,n+1)if i*i+j*j==k*k}


Literally the same thing as the other answers.

# Python, 98 bytes

Two solutions already exist for Python, but this one uses a slightly different approach that I thought might be informative:

from itertools import *
lambda n:[(k,j,i)for(k,j,i)in combinations(range(0,n+1),3)if i*i==j*j+k*k]


# Brachylog, 13 bytes

⟦₁⊇Ṫ.^₂ᵐṅᵗ+0∧


Try it online!

### Explanation

This is a generator, which will unify with each triplet.

⟦₁                 Range [1, …, N]
⊇Ṫ.              Subset of 3 elements
.^₂ᵐ          Square each element
ṅᵗ        Negate the last one (which is the biggest)
+0∧    The sum must be 0


### Much faster, 19 bytes

≥~hṪ≥₁ℕ₁ᵐ^₂ᵐṅʰ+0∧Ṫ≜


Try it online!

Takes about 1.5s for N = 200 on TIO. This uses integer constraint programming mechanisms which is way more efficient than brute forcing combinations in a range, but is longer to express in this case.

≥~hṪ                 A triplet of elements whose head is smaller than N
≥₁               The triplet is non-increasing
ℕ₁ᵐ            Each element is in [1, +inf)
^₂ᵐ         Map square
ṅʰ       Negate the first element
+0∧    The sum must be 0
Ṫ≜  Assign values to satisfy these constraints


# C (gcc), 113 + 7 bytes

Compiled with -lm -DM=<input>

r,s,t;main(){for(;s<M;)for(t=++s;(r=sqrt(2*s*++t))<M;)r-sqrt(2*s*t)||(r+=s+t)>M||printf("%d %d %d\n",r-t,r-s,r);}


Try it online!

# JavaScript 72 bytes

c=>{for(a=b=c;c;--a||(a=--b)||(a=b=--c))c*c-b*b-a*a||console.log(c,b,a)}


Formatted:

c => {
for( a=b=c; c; /* until c is 0 */
/* count down a, when a==0 reduce b and reset a=b, the same with c */
--a || (a = --b) || (a = b = --c)
)
/* if c²-b²-a² == 0 log the tuple */
c*c - b*b - a*a || console.log(c,b,a)
}


Try it online

q l=[(a,b,c)|c<-[1..l],a<-[1..c],b<-[a..l],a^2+b^2==c^2]

[range(3;.)]|combinations(3)|select(map(pow(.;2))|.[0]==.[1]+.[2]and.[1]>.[2])
`