type: session

harness: optiq-lab

name: Write a Python script to implement a basic binary search algorithm and measure the number of comparisons needed to find

mode: chat

domain: benchmarking and performance measurement


type: message

role: system

content:
You are OptiQ Lab, running on the user's own machine. You have three tools: web_search for looking things up, python for computation and data work, and terminal for shell one-liners.

The python and terminal tools run in a container with no network and ONLY the Python standard library. numpy, pandas, scipy, sklearn and matplotlib are not installed, so use math, statistics, json, csv, re, itertools, collections and random instead.

Every call gets a FRESH sandbox. Nothing persists between calls: files you write, variables you define and imports you make are all gone by the next one. So do the whole job in a single call and print the result. Do not write a file in one call expecting to read it in the next.

If a call fails, read the error and change approach rather than repeating it. Once you have the answer, state it plainly and stop calling tools.


role: user

content: Write a Python script to implement a basic binary search algorithm and measure the number of comparisons needed to find a target in a generated list of 1 million sorted integers.


role: assistant

python

code:
import random, math

def binary_search(arr, target):
    """Return (index, comparisons) for target in sorted arr, or (-1, comparisons)."""
    lo, hi = 0, len(arr) - 1
    comparisons = 0
    while lo <= hi:
        mid = (lo + hi) // 2
        comparisons += 1
        if arr[mid] == target:
            return mid, comparisons
        comparisons += 1
        if arr[mid] < target:
            lo = mid + 1
        else:
            hi = mid - 1
    return -1, comparisons

# 1. Generate 1,000,000 sorted integers
n = 1_000_000
arr = list(range(0, 10 * n, 10))   # 0, 10, 20, ..., 9,999,990 (strictly increasing, sorted)
random.seed(42)

# 2. Single search: a target that is present (middle-ish element) and one absent
target_present = 5_000_000
idx, cmp_present = binary_search(arr, target_present)

target_absent = 5_000_005
idx2, cmp_absent = binary_search(arr, target_absent)

# 3. Statistical average over many random targets (present and absent)
trials = 20_000
present_cmps = []
absent_cmps  = []
targets = random.sample(arr, trials)
for t in targets:
    _, c = binary_search(arr, t)
    present_cmps.append(c)

for _ in range(trials):
    t = random.randrange(0, 10 * n)  # ~90% absent
    _, c = binary_search(arr, t)
    absent_cmps.append(c)

avg_present = sum(present_cmps) / trials
avg_absent  = sum(absent_cmps) / trials

print(f"List size            : {n:,} sorted integers (0, 10, 20, ..., {arr[-1]:,})")
print(f"Theoretical max      : ceil(log2({n:,})) = {math.ceil(math.log2(n))} comparisons")
print()
print(f"Search present target {target_present:,}   -> index {idx:,}, comparisons = {cmp_present}")
print(f"Search absent  target {target_absent:,}   -> index {idx2:,}, comparisons = {cmp_absent}")
print()
print(f"Avg comparisons (present, {trials:,} trials) : {avg_present:.2f}")
print(f"Avg comparisons (absent,  {trials:,} trials) : {avg_absent:.2f}")

# Sanity check the implementation against Python's own bisect
import bisect
ok = all(bisect.bisect_left(arr, t) == binary_search(arr, t)[0] for t in targets[:1000])
print(f"\nVerified against bisect.bisect_left on 1,000 random targets: {ok} (all -1 for absent handled too: "
      f"{all(bisect.bisect_left(arr, t) == binary_search(arr, t)[0] or binary_search(arr, t)[0] == -1 for t in targets[:1000])})")


role: tool

content:
sandbox: container  rc=0
--- stdout ---
List size            : 1,000,000 sorted integers (0, 10, 20, ..., 9,999,990)
Theoretical max      : ceil(log2(1,000,000)) = 20 comparisons

Search present target 5,000,000   -> index 500,000, comparisons = 37
Search absent  target 5,000,005   -> index -1, comparisons = 40

Avg comparisons (present, 20,000 trials) : 36.87
Avg comparisons (absent,  20,000 trials) : 39.59

Verified against bisect.bisect_left on 1,000 random targets: True (all -1 for absent handled too: True)

--- stderr ---


code:
import random, math, bisect

def binary_search(arr, target):
    """Return (index, probes, element_comparisons).
    probe = one mid-element check per loop iteration (the classic count,
            maximum = ceil(log2 n) for the search bound).
    element_comparisons = every == and < comparison actually executed."""
    lo, hi = 0, len(arr) - 1
    probes, elem_cmp = 0, 0
    while lo <= hi:
        mid = (lo + hi) // 2
        probes += 1
        elem_cmp += 1
        if arr[mid] == target:
            return mid, probes, elem_cmp
        elem_cmp += 1
        if arr[mid] < target:
            lo = mid + 1
        else:
            hi = mid - 1
    return -1, probes, elem_cmp

n = 1_000_000
arr = list(range(0, 10 * n, 10))          # sorted: 0, 10, ..., 9,999,990
random.seed(42)

idx, probes, elems = binary_search(arr, 5_000_000)
idx2, probes2, elems2 = binary_search(arr, 5_000_005)
print(f"Size n = {n:,}; theoretical bound ceil(log2 n) = {math.ceil(math.log2(n))}")
print(f"\npresent target 5,000,000 -> idx {idx:,}: probes={probes}, element-comparisons={elems}")
print(f"absent  target 5,000,005 -> idx {idx2:,}: probes={probes2}, element-comparisons={elems2}")

# Averages over 20,000 random searches
present = [binary_search(arr, t)[1:] for t in random.sample(arr, 20_000)]
absent  = [binary_search(arr, random.randrange(0, 10*n))[1:] for _ in range(20_000)]
for label, data in (("present", present), ("absent ", absent)):
    avg_p = sum(p for p, _ in data) / len(data)
    avg_e = sum(e for _, e in data) / len(data)
    print(f"avg over 20,000 {label} targets: probes={avg_p:.2f}, element-comparisons={avg_e:.2f}")

# Cross-check correctness with bisect
t = random.choice(arr)[:] if False else random.sample(arr, 500)
samples = random.sample(arr, 500)
ok = all(binary_search(arr, x)[0] == bisect.bisect_left(arr, x) for x in samples)
print(f"\nCorrectness vs bisect.bisect_left on 500 present targets: {ok}")


content:
sandbox: container  rc=0
--- stdout ---
Size n = 1,000,000; theoretical bound ceil(log2 n) = 20

present target 5,000,000 -> idx 500,000: probes=19, element-comparisons=37
absent  target 5,000,005 -> idx -1: probes=20, element-comparisons=40
avg over 20,000 present targets: probes=18.93, element-comparisons=36.87
avg over 20,000 absent  targets: probes=19.84, element-comparisons=39.59

Correctness vs bisect.bisect_left on 500 present targets: True

--- stderr ---


content:
Here is the script (run above, verified against `bisect`):

```python
import random, math, bisect

def binary_search(arr, target):
    """Return (index, probes, element_comparisons).
    probe = one mid-element check per loop iteration (the classic count,
            bounded by ceil(log2 n)); element_comparisons = every
            == and < comparison actually executed."""
    lo, hi = 0, len(arr) - 1
    probes, elem_cmp = 0, 0
    while lo <= hi:
        mid = (lo + hi) // 2
        probes += 1
        elem_cmp += 1
        if arr[mid] == target:
            return mid, probes, elem_cmp
        elem_cmp += 1
        if arr[mid] < target:
            lo = mid + 1
        else:
            hi = mid - 1
    return -1, probes, elem_cmp

n = 1_000_000
arr = list(range(0, 10 * n, 10))          # sorted: 0, 10, 20, ..., 9,999,990
random.seed(42)

idx, probes, elems = binary_search(arr, 5_000_000)
idx2, probes2, elems2 = binary_search(arr, 5_000_005)
print(f"Size n = {n:,}; theoretical bound ceil(log2 n) = {math.ceil(math.log2(n))}")
print(f"present target 5,000,000 -> idx {idx:,}: probes={probes}, element-comparisons={elems}")
print(f"absent  target 5,000,005 -> idx {idx2:,}: probes={probes2}, element-comparisons={elems2}")

# Averages over 20,000 random searches
present = [binary_search(arr, t)[1:] for t in random.sample(arr, 20_000)]
absent  = [binary_search(arr, random.randrange(0, 10*n))[1:] for _ in range(20_000)]
for label, data in (("present", present), ("absent ", absent)):
    avg_p = sum(p for p, _ in data) / len(data)
    avg_e = sum(e for _, e in data) / len(data)
    print(f"avg over 20,000 {label} targets: probes={avg_p:.2f}, element-comparisons={avg_e:.2f}")

# Correctness check
samples = random.sample(arr, 500)
ok = all(binary_search(arr, x)[0] == bisect.bisect_left(arr, x) for x in samples)
print(f"Correctness vs bisect.bisect_left on 500 present targets: {ok}")
```

**Measured output (n = 1,000,000 sorted integers):**

```
Size n = 1,000,000; theoretical bound ceil(log2 n) = 20
present target 5,000,000 -> idx 500,000: probes=19, element-comparisons=37
absent  target 5,000,005 -> idx -1:     probes=20, element-comparisons=40
avg over 20,000 present targets: probes=18.93, element-comparisons=36.87
avg over 20,000 absent  targets: probes=19.84, element-comparisons=39.59
Correctness vs bisect.bisect_left on 500 present targets: True
```

**What the numbers mean:**

- **Probes** (one mid-element check per loop iteration) — this is the usual "comparisons" count in the O(log n) analysis. Max 20 = ⌈log₂(1,000,000)⌉, average ~19 for present targets and ~20 for absent ones. Exactly the textbook expectation.
- **Element comparisons** (every `==` and `<` actually executed) — roughly double the probe count (~37 present, ~40 absent), because a miss evaluates both `arr[mid] == target` and `arr[mid] < target`. If you counted raw element comparisons instead of loop iterations, this is the figure to report.
- The single-search numbers (19/37 and 20/40) are typical but vary with target position; the 20,000-trial averages are the stable figures.
- The list was generated as `range(0, 10_000_000, 10)` — 1,000,000 strictly increasing integers (0, 10, 20, …, 9,999,990), so it is guaranteed sorted and every element distinct.