### 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.

Pages: 1 2

### 2 Responses to “Ladder Range”

1. Jussi Piitulainen said

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

# (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)
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

# 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:")
println("expecting 3, observing ", findlower(ladder, 0.2, k * 0.2 + c))
```
2. Paul said

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