that the capabilities of LLMs have progressed dramatically in the previous couple of years, but it surely’s arduous to quantify simply how good they’ve turn into.
That received me considering again to a geometrical downside I got here throughout on a YouTube channel final yr. This was in June 2024, and I attempted to get the main massive language mannequin on the time (GPT-4o) to resolve the puzzle. It didn’t go that effectively and required rather a lot of effort to discover a resolution, and I questioned how the most recent LLMs would fare with the identical puzzle.
The puzzle
Right here’s a fast reminder of what I used to be asking the LLM to resolve again then. Assume now we have the next grid of dots/nodes. Within the x and y airplane, every node is strictly one unit away from its adjoining neighbour. It seems like this,
Now, the query I wished to reply was this,
What number of distinct squares may be drawn on this diagram?It shortly grew to become clear that GPT-4o didn’t know the reply, so I modified tack barely and as a substitute requested it this.
I would love a Python program that plots out all of the squares we will
draw on the connected diagram, assuming that the corners of any sq.
should lie on one of many spots on the diagram. Assume every adjoining spot is
1 unit aside in each the x and y instructions. Additionally print out a abstract of
the variety of squares of the identical measurement and what their facet lengths areLengthy story quick, I ultimately received GPT-4o to give you an accurate Python-based resolution. Nonetheless, it took me round two hours and effectively over 40 iterations of me going backwards and forwards with the mannequin to refine its reply till it got here up with one thing that labored.
NB, have you ever labored out the reply but? Even now, I nonetheless can’t fairly imagine that it’s 21 squares you possibly can draw on this graph.
Interplay with Sonnet 4.5
This brings us again to the current day and Anthropic’s current launch of what’s thought-about one of the best coding mannequin out there at present, specifically Claude Sonnet 4.5. It’s been round 16 months since I final checked out this, so simply how a lot have LLMs improved within the intervening interval? Let’s see how Sonnet 4.5 copes with the puzzle.
To make use of Sonnet 4.5, I visited Anthropic’s homepage.
You need to see a web page like this.
Click on the massive black “Strive Claude” button positioned close to the highest proper of the display screen. You’ll be requested to log in at this level. For those who don’t have already got an account with Anthropic, please set one up now. It’s fully free to take action.
As soon as you’re logged in, you’ll be taken to a display screen displaying a well-recognized chat-based window that means that you can work together with and ask questions of Sonnet 4.5.
Earlier than I requested the mannequin to offer a Python program to resolve the puzzle, I requested it a straight query to see if it already knew the reply.
It got here again with,
Taking a look at this grid, I have to rely all potential squares of various
sizes that may be fashioned by connecting the dots.
Let me first determine the grid construction. The dots kind:
2 dots within the prime row
2 dots within the second row
6 dots within the third row
6 dots within the fourth row
2 dots within the fifth row
2 dots within the sixth row
This seems to be a cross or plus-sign formed grid sample.
Now I will rely squares by measurement:
1×1 squares: These are fashioned by any 4 dots that make a unit sq..
Within the horizontal center part (rows 3-4): 5 squares per row × 1 = 5 squares
Within the vertical center part (columns the place the cross intersects): counting
the vertical segments offers us 5 extra squares
Complete: 10 small squares
2×2 squares: These want 4×4 preparations of dots.
Within the horizontal part: 4 squares
Within the vertical part: 4 squares
Overlap in middle: -1 (counted twice)
Complete: 7 squares of measurement 2×2
3×3 squares and bigger: The grid sample is simply 6 dots extensive at its most, however the cross form limits the place bigger squares can match. Checking the geometry:
One 3×3 sq. can match within the middle the place the arms of the cross overlap
Complete: 1 sq. of measurement 3×3
Complete: 10 + 7 + 1 = 18 squaresSo, confidently incorrect. Curious and intrigued concerning the mannequin’s reference to a 3×3 sq. (which is unimaginable on this grid), I requested it to indicate me the sq. on the diagram. It replied with this.
Wanting on the grid extra rigorously, I have to rethink whether or not a 3×3 sq.
really exists.
For a 3×3 sq. to exist, I would wish a 4×4 association of dots
(since a 3×3 sq. requires 4 dots on both sides). Let me examine the grid
construction:
The grid seems to be:
Rows 1-2: solely 2 dots every (centered)
Rows 3-4: 6 dots every (full width)
Rows 5-6: solely 2 dots every (centered)
For a 3×3 sq., I would wish 4 consecutive rows and 4 consecutive columns all with dots on the intersection factors. Nevertheless, wanting on the cross/plus sample:
The highest solely has 2 columns of dots
The center has 6 columns of dots
The underside solely has 2 columns of dots
Since rows 1-2 and rows 5-6 solely have 2 dots every (not spanning the complete width)
, there is not really a 4×4 block of dots anyplace on this grid.
I have to appropriate my reply: There's NO 3×3 sq. potential on this grid.
The overall ought to be: 10 (1×1 squares) + 7 (2×2 squares) = 17 squares
I apologize for the error in my preliminary rely. The cross-shaped sample
prevents any sq. bigger than 2×2 from being fashioned.Alright, it semi-redeemed itself with that reply, however basically, it didn’t accurately say there have been 21 squares in whole.
The ultimate reply
To maintain issues truthful with the mannequin comparability over the 16 months, I requested Sonnet 4.5 the very same query I had requested GPT-4o, which, as a reminder, was this.
I would love a Python program that plots out all of the squares we will draw on
the connected diagram, assuming that the corners of any sq. should lie on one
of the spots on the diagram. Assume every adjoining spot is 1 unit aside in
each the x and y instructions. Additionally print out a abstract of the variety of
squares of the identical measurement and what their facet lengths areSonnet 4.5 thought for all of 5 seconds earlier than producing its Python code.
import matplotlib.pyplot as plt
import numpy as np
from itertools import combos
from collections import defaultdict
# Outline the grid factors based mostly on the picture
factors = [
# Top two rows (2 dots each, centered)
(3, 6), (4, 6),
(3, 5), (4, 5),
# Middle two rows (6 dots each, full width)
(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4),
(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3),
# Bottom two rows (2 dots each, centered)
(3, 2), (4, 2),
(3, 1), (4, 1),
]
def is_square(p1, p2, p3, p4):
"""Verify if 4 factors kind a sq. and return facet size if true."""
factors = [p1, p2, p3, p4]
# Calculate all pairwise distances
distances = []
for i in vary(len(factors)):
for j in vary(i + 1, len(factors)):
dist = np.sqrt((factors[i][0] - factors[j][0])**2 +
(factors[i][1] - factors[j][1])**2)
distances.append(dist)
distances.type()
# For a sq.: 4 equal sides and a pair of equal diagonals
# Diagonals ought to be sqrt(2) occasions the facet size
if len(distances) == 6:
sides = distances[:4]
diagonals = distances[4:]
# Verify if all sides are equal
if np.allclose(sides, [sides[0]] * 4, atol=1e-6):
# Verify if diagonals are equal
if np.allclose(diagonals, [diagonals[0]] * 2, atol=1e-6):
# Verify if diagonal = facet * sqrt(2)
if np.allclose(diagonals[0], sides[0] * np.sqrt(2), atol=1e-6):
return True, sides[0]
return False, 0
def order_square_points(p1, p2, p3, p4):
"""Order factors of a sq. in counterclockwise order ranging from bottom-left."""
factors = [p1, p2, p3, p4]
# Discover centroid
cx = sum(p[0] for p in factors) / 4
cy = sum(p[1] for p in factors) / 4
# Kind by angle from centroid
def angle(p):
return np.arctan2(p[1] - cy, p[0] - cx)
factors.type(key=angle)
return factors
# Discover all squares
squares = []
squares_by_size = defaultdict(listing)
for combo in combos(factors, 4):
is_sq, side_length = is_square(*combo)
if is_sq:
ordered = order_square_points(*combo)
squares.append(ordered)
squares_by_size[round(side_length, 2)].append(ordered)
# Print abstract
print("=" * 50)
print("SUMMARY OF SQUARES FOUND")
print("=" * 50)
total_squares = 0
for side_length in sorted(squares_by_size.keys()):
rely = len(squares_by_size[side_length])
total_squares += rely
print(f"Facet size {side_length}: {rely} sq.(s)")
print(f"nTotal squares: {total_squares}")
print("=" * 50)
# Create visualization
fig, axes = plt.subplots(2, 2, figsize=(14, 14))
axes = axes.flatten()
# Plot 1: All dots
ax = axes[0]
x_coords = [p[0] for p in factors]
y_coords = [p[1] for p in factors]
ax.scatter(x_coords, y_coords, c='black', s=100, zorder=3)
ax.set_xlim(0, 7)
ax.set_ylim(0, 7)
ax.set_aspect('equal')
ax.grid(True, alpha=0.3)
ax.set_title('Authentic Grid', fontsize=14, fontweight='daring')
ax.set_xlabel('X')
ax.set_ylabel('Y')
# Plot 2: All squares overlaid
ax = axes[1]
ax.scatter(x_coords, y_coords, c='black', s=100, zorder=3)
colours = plt.cm.rainbow(np.linspace(0, 1, len(squares)))
for sq., coloration in zip(squares, colours):
square_closed = sq. + [square[0]]
xs = [p[0] for p in square_closed]
ys = [p[1] for p in square_closed]
ax.plot(xs, ys, c=coloration, linewidth=2, alpha=0.6)
ax.set_xlim(0, 7)
ax.set_ylim(0, 7)
ax.set_aspect('equal')
ax.grid(True, alpha=0.3)
ax.set_title(f'All {total_squares} Squares Overlaid', fontsize=14, fontweight='daring')
ax.set_xlabel('X')
ax.set_ylabel('Y')
# Plot 3 & 4: Squares by measurement
plot_idx = 2
for side_length in sorted(squares_by_size.keys()):
if plot_idx >= len(axes):
break
ax = axes[plot_idx]
ax.scatter(x_coords, y_coords, c='lightgray', s=100, zorder=2)
squares_of_size = squares_by_size[side_length]
colours = plt.cm.viridis(np.linspace(0, 1, len(squares_of_size)))
for sq., coloration in zip(squares_of_size, colours):
square_closed = sq. + [square[0]]
xs = [p[0] for p in square_closed]
ys = [p[1] for p in square_closed]
ax.plot(xs, ys, c=coloration, linewidth=2.5, marker='o', markersize=8)
ax.set_xlim(0, 7)
ax.set_ylim(0, 7)
ax.set_aspect('equal')
ax.grid(True, alpha=0.3)
ax.set_title(f'Squares with facet size {side_length} ({len(squares_of_size)} whole)',
fontsize=14, fontweight='daring')
ax.set_xlabel('X')
ax.set_ylabel('Y')
plot_idx += 1
plt.tight_layout()
plt.present()At first sight, I assumed there was no method Sonnet may have give you an accurate resolution so shortly. GPT-4o had taken hours. Nevertheless, I put in all of the required libraries the code wanted and copied and pasted it right into a Jupyter Pocket book cell, operating it and acquiring the next output. I’ve omitted the graphs for now, which had been additionally output.
==================================================
SUMMARY OF SQUARES FOUND
==================================================
Facet size 1.0: 9 sq.(s)
Facet size 1.41: 4 sq.(s)
Facet size 2.24: 2 sq.(s)
Facet size 2.83: 4 sq.(s)
Facet size 3.61: 2 sq.(s)
Complete squares: 21
==================================================
#
# Plus some graphs that I am not displaying right here
#That shocked me. The reply was completely spot on.
The one slight factor the mannequin didn’t fairly get proper was that it didn’t output a plot of every set of in another way sized squares. It simply did the 9 1x1s and the 4 √2x√2 ones. I solved that by asking Sonnet to incorporate these, too.
Are you able to print the graphs in sq. facet order. Can also you've two graphs
facet by facet on every "line"That is what it produced.
Lovely.
Abstract
To show simply how dramatically LLMs have superior in a couple of yr, I made a decision to revisit a difficult geometric puzzle I first tried to resolve with GPT-4o again in June 2024. The puzzle was to put in writing a Python program that finds and plots all potential squares on a particular cross-shaped grid of dots.
My expertise just a little over a yr in the past was a battle; it took me roughly two hours and over 40 prompts to information GPT-4o to an accurate Python resolution.
Quick ahead to at present, and I examined the brand new Claude Sonnet 4.5. After I first requested the mannequin the query straight, it did not calculate the right variety of squares. Not an awesome begin, nonetheless, the actual take a look at was giving it the very same immediate I used on GPT-4o.
To my shock, it produced a whole, appropriate Python resolution in one shot. The code it generated not solely discovered all 21 squares but in addition accurately categorised them by their distinctive facet lengths and generated detailed plots to visualise them. Whereas I wanted one fast follow-up immediate to good the plots, the core downside was solved immediately.
May or not it’s that the very act of my attempting to resolve this puzzle final yr and publishing my findings launched it to the web-o-sphere, that means Anthropic have merely crawled it and included it into their mannequin information base? Sure, I suppose that could possibly be it, however then why couldn’t the mannequin reply the primary direct query I requested it concerning the whole variety of squares accurately?
To me, this experiment starkly illustrates the unbelievable leap in LLM functionality. What was as soon as a two-hour iterative battle with the main mannequin of its time 16 months in the past is now a five-second, one-shot success with the main mannequin at present.
