RSS preview of John D. Cook

Rss preview of Blog of John D. Cook

An almost periodic function

2026-07-23 22:04:34

This post takes a more abstract view of the previous post. That post looked at the concrete question of whether a number ever has the same sine in radians as in degrees. The relation between radians and degrees is irrelevant except that π/180 is an irrational number.

Suppose α and β are two positive numbers such that α/β is irrational. In the previous post, α = 1 and β = π/180. Then the function

f(x) = sin(αx) − sin(βx)

is almost periodic: it is not periodic, but it comes close to being periodic, as close as you’d like provided you’re willing to look over a sufficiently long range of x‘s.

The identity

sin(αx) − sin(βx) = 2 cos((α + β)x/2) sin((α − β)x/2)

shows that f(x) is the product of two periodic functions but is not periodic itself. The periods of the cosine and sine above never coincide because the ratio of their frequencies is irrational.

The zeros of f are not periodic, though they can be divided into two subsequences that are periodic.

The post An almost periodic function first appeared on John D. Cook.

When sine of x degrees equals sine of x radians

2026-07-22 20:17:11

Ordinarily the sine of x radians and the sine of x degrees are very different numbers. Having your calculator in radian mode when it should be in degree mode, or vice versa, results in a major error.

But sometimes it doesn’t matter. A trivial example is when x = 0. A more interesting example is

x = 180π/(180 + π) = 3.08770208….

For that value of x,

sin(x) = sin(x°).

In this article I’ll use the common convention of using radians by default and denoting degrees with ° as above.

Note that

x = πx°/180

and so we are interested in solutions to the equation

sin(x) = sin(πx/180)

Now two angles A and B have the same sine if they differ by a multiple of 2π, or if they’re supplementary (i.e. A = π − B), or both. To put it another way, if A and B have the same sine, they are either equal mod 2π or supplementary mod 2π. This means that

sin(x) = sin(πx/180)

if and only if

x = πx/180 + 2πk

x = π − πx/180 + 2πk

for some integer k.

Therefore all solutions have the form

x = 360πk/(180 − π)

x = 180π(2k + 1)/(180 + π).

Alternative solution

The derivation above is correct, but it occurred to me later that a simpler argument would be to use the identity

sin(A) − sin(B) = 2 cos((A + B)/2) sin((A − B)/2).

Thus A and B have the same sine if

cos((A + B)/2) = 0

or if

sin((A − B)/2) = 0.

These two possibilities correspond to the two families of solutions above.

Density

When reduced modulo 2π, both families are dense in [0, 2π]. This means that for every y in [−1, 1], there is a number x such that

sin(x) = sin(x°) ≈ y

and we can make the approximation as good as we’d like.

Example 1

For example, today is July 22, so let’s set y = 0.722. We’d like to find a value of x such that the sine of x radians and the sine of x degrees both approximately equal 0.722. And let’s say our approximation tolerance is ε = 0.0001.

We can search for a value of x in the first family of solutions by looking for a value of k with

| sin(360πk/(180 − π)) − 0.722 | < 0.0001

and the smallest such k is 96343 and so

x = 360×96343 π/(180 − π) = 616093.78713621…

will do, and sin(x) = 0.72191…

Example 2

Now let’s set y = 0.2026 and look for a solution in the other family of solutions, and this time let’s set ε = 10⁻⁶. The smallest value of k such that

| sin(180π(2k + 1)/(180 + π)) − 0.2026 | < 10⁻⁶

is k = 741141. Then

sin( 4576848.310950611 ) = sin( 4576848.310950611° ) = 0.202600139…

The post When sine of x degrees equals sine of x radians first appeared on John D. Cook.

Forensic accounting in Python

2026-07-21 23:05:04

I recently had a project in which I had to reverse engineer a data analysis. There was some ambiguity regarding which of several possibilities someone chose for several of the variables, something analogous to the following example.

Suppose you have three numbers with uncertain values with a known, or at least purported, sum. The first number could be 31, 41, or 59; the second could be either 26 or 53; the last could be 58, 97, 93, or 23.

The following code enumerates all 3 × 2 × 4 = 24 possibilities and prints their sums.

from itertools import product

# Example input
possibilities = [(31, 41, 59), (26, 53), (58, 97, 93, 23)]

for combo in product(*possibilities):
    total = sum(combo) 
    print(f"Combination {combo} sums to: {total}")

In this example all the sums are unique, though of course that might not happen in practice. If, for example, you know the sum is 187, you know the three numbers were 41, 53, and 93. If the reported sum is 200, you know some assumption has been violated because none of the possible choices add up to 200.

Solutions

There are eight solutions, each being a flip or a rotation of one fundamental solution. Here’s a representative.

SWI Prolog code

:- use_module(library(clpfd)).

%% Enumerate all ways to place 5 white queens and 3 black queens
%% on a 5x5 board so that no queen attacks a queen of the opposite colour.
%% Same-colour queens are allowed to attack each other.
%%
%% Positions are numbered 1..25 in row-major order:
%%   1  2  3  4  5
%%   6  7  8  9 10
%%  11 12 13 14 15
%%  16 17 18 19 20
%%  21 22 23 24 25

solve :-
    solution(Whites, Blacks),
    print_solution(Whites, Blacks),
    fail.
solve :-
    writeln('No more solutions.').

solution(Whites, Blacks) :-
    length(Whites, 5),
    length(Blacks, 3),
    append(Whites, Blacks, All),
    All ins 1..25,
    all_distinct(All),
    % canonical order so each set of positions is generated once
    Whites = [W1,W2,W3,W4,W5],
    W1 #< W2, W2 #< W3, W3 #< W4, W4 #< W5,
    Blacks = [B1,B2,B3],
    B1 #< B2, B2 #< B3, % every black is safe from every white maplist(safe_from_all(Whites), Blacks), label(All). safe_from_all(Whites, Black) :- maplist(no_attack(Black), Whites). %% Two queens do *not* attack each other. no_attack(P, Q) :- R1 #= (P-1) div 5 + 1, C1 #= (P-1) mod 5 + 1, R2 #= (Q-1) div 5 + 1, C2 #= (Q-1) mod 5 + 1, R1 #\= R2, % different row C1 #\= C2, % different column abs(R1-R2) #\= abs(C1-C2). % different diagonal print_solution(Whites, Blacks) :- format('Whites: ~w~n', [Whites]), format('Blacks: ~w~n', [Blacks]), forall(between(1, 5, R), ( forall(between(1, 5, C), ( Pos is (R-1)*5 + C, ( memberchk(Pos, Whites) -> write('W ')
                      ; memberchk(Pos, Blacks) -> write('B ')
                      ; write('. ')
                      )
                    )),
             nl )),
    nl.

Lean 4 code

-- Lean 4 solution for the 5×5 black-and-white queens puzzle.
-- 5 white queens + 3 black queens; no opposite-colour attacks.

def absDiff (a b : Nat) : Nat :=
  if a ≥ b then a - b else b - a

def attacks (p q : Nat × Nat) : Bool :=
  p.1 == q.1 || p.2 == q.2 || absDiff p.1 q.1 == absDiff p.2 q.2

/-- All 25 squares (0-based, row-major). -/
def allPositions : List (Nat × Nat) :=
  (List.range 5).flatMap fun r =>
    (List.range 5).map fun c => (r, c)

/-- Combinations of length exactly `k` (order-preserving). -/
def combinations {α : Type} : Nat → List α → List (List α)
  | 0,     _      => [[]]
  | _,     []     => []
  | k + 1, x::ys  =>
      (combinations k ys).map (fun zs => x :: zs) ++
      combinations (k + 1) ys

def isAttackedBy (pos : Nat × Nat) (queens : List (Nat × Nat)) : Bool :=
  queens.any (fun q => attacks q pos)

/-- Every solution as a pair (white squares, black squares). -/
def solve : List (List (Nat × Nat) × List (Nat × Nat)) :=
  (combinations 3 allPositions).foldl (fun acc blacks =>
    let forbidden := allPositions.filter fun p =>
      blacks.contains p || isAttackedBy p blacks
    let available := allPositions.filter fun p =>
      !(forbidden.contains p)
    acc ++ (combinations 5 available).map fun whites => (whites, blacks)
  ) []

/-- Pretty-print one board. -/
def printBoard (whites blacks : List (Nat × Nat)) : IO Unit := do
  for r in List.range 5 do
    let mut row := ""
    for c in List.range 5 do
      let p := (r, c)
      row := row ++
        (if whites.contains p then "W "
         else if blacks.contains p then "B "
         else ". ")
    IO.println row
  IO.println ""

/-- Enumerate and display every solution. -/
def main : IO Unit := do
  let sols := solve
  IO.println s!"Found {sols.length} solutions.\n"
  for p in (List.range sols.length).zip sols do
    let i := p.1
    let (whites, blacks) := p.2
    IO.println s!"Solution {i + 1}:"
    printBoard whites blacks

#eval main

The post Solving a chess puzzle with Grok 4.5 first appeared on John D. Cook.

John D. Cook Modify

Rss preview of Blog of John D. Cook

Alternative solution

Density

Example 1

Example 2

More forensics posts

Solutions

SWI Prolog code

Lean 4 code

John D. CookModify

Rss preview of Blog of John D. Cook

Alternative solution

Density

Example 1

Example 2

More forensics posts

Solutions

SWI Prolog code

Lean 4 code

The author's social media

John D. Cook Modify