Ladder Range
July 25, 2017
Once a year, or thereabouts, I pick up Jon Bentley’s book Programming Pearls and read a random chapter; even though I’ve read it before, re-reading always makes the material seem fresh. Today’s exercise is Exercise 7 from Chapter 4 about binary search:
A colleague faced the following problem in a program to draw lines on a bit-mapped display. An array of n pairs of reals (ai, bi) defined the n lines y = mi x +
bi. The lines were ordered in the x-interval [0, 1] in the sense that yi < yi+1 for all values of i between 0 and n − 2 and all values of x in [0, 1].[ Here Bentley has a picture of a ladder with rungs at various angles to the horizontal. We won’t reproduce it here; get the book if you want to see it. ]
Less formally, the lines don’t touch in the vertical slab. Given a point (x, y), where 0 ≤ x ≤ 1, he wanted to determine the two lines that bracket the point. How could he solve the problem quickly?
Your task is to write a program to solve Bentley’s exercise. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
Binary search is where I tend to make the relevant assumptions explicit as comments, as below (in Julia). Otherwise I keep getting something off by one or some condition accidentally reversed.
I had two typos in my test code on the first run. I had also written the test invocations to expect zero-based indexes, so the correct one-based indexes were reported as off-by one on that first run :)
My function returns only the lower-rung index. The corresponding upper-rung index is the next integer.
module Ladder export findlower function findlower(ladder, x, y) n = length(ladder) # (x, y) is above (or on) rung 1 # (x, y) is below rung n b, e = 1, n while b + 1 < e # (x, y) is above (or on) rung b # (x, y) is below rung e m = b + div(e - b, 2) # b < m < e # (m is half-way between b and e) k, c = ladder[m] if y < k * x + c # (x, y) is below rung m e = m else # (x, y) is above (or on) rung m b = m end end # b + 1 == e # (x, y) is above (or on) rung b # (x, y) is below rung e b end end using Ladder # Praxis test slopes and intercepts but as Float64 ladder = collect(zip((0., 2., 10., 6., 7., -1., -11., 0., -7., 11., 0.), (0., 1., 4., 9., 16., 25., 36., 49., 64., 81., 100.))) println("Testing Praxis points (but Julia indexing is 1-based):") println("expecting 8, observing ", findlower(ladder, 0.5, 50.)) println("expecting 6, observing ", findlower(ladder, 0.25, 25.)) println() println("Testing a point on the third rung:") k, c = ladder[3] println("expecting 3, observing ", findlower(ladder, 0.2, k * 0.2 + c))In Python using the bisect module for the search. As bisect needs a list, the class LazyRungs is used, that behaves like a list and only calculates the y-value when needed.
from bisect import bisect_left as bisect class LazyRungs(object): 'rungs is a list of tuples (left, right)' def __init__(self, rungs, x): self.rungs = rungs self.x = x def __len__(self): return len(self.rungs) def __getitem__(self, i): (left, right), x = rungs[i], self.x return left * (1-x) + right * x def ladder(x, y, rungs): 'return 2 bracketing rungs (a, b), such that a < y <= b' i = bisect(LazyRungs(rungs, x), y) # i is the index such that rungs[i] >= y if not 0 < i < len(rungs): raise ValueError("(x,y) outside range") return i-1, i left = map(int, '0 1 4 9 16 25 36 49 64 81 100'.split()) right = map(int, '0 3 14 15 23 24 25 49 57 92 100'.split()) rungs = list(zip(left, right)) print(ladder(0.5, 50, rungs)) # (7, 8) print(ladder(0.25, 25, rungs)) # (5, 6)