that the capabilities of LLMs have progressed dramatically in the previous couple of years, however it’s arduous to quantify simply how good they’ve develop into.
That obtained me considering again to a geometrical downside I got here throughout on a YouTube channel final 12 months. This was in June 2024, and I attempted to get the main giant language mannequin on the time (GPT-4o) to unravel the puzzle. It didn’t go that properly and required lots of effort to discover a answer, and I puzzled how the newest 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 unravel again then. Assume now we have the next grid of dots/nodes. Within the x and y airplane, every node is precisely one unit away from its adjoining neighbour. It appears like this,
Now, the query I needed to reply was this,
What number of distinct squares will 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 an alternative requested it this.
I would really like a Python program that plots out all of the squares we will
draw on the hooked up 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 aspect lengths areLengthy story brief, I finally obtained GPT-4o to provide you with an accurate Python-based answer. Nonetheless, it took me round two hours and properly 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 consider that it’s 21 squares you may draw on this graph.
Interplay with Sonnet 4.5
This brings us again to the current day and Anthropic’s latest launch of what’s thought of the most effective coding mannequin accessible right this moment, 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.
It’s best to see a web page like this.
Click on the massive black “Strive Claude” button positioned close to the highest proper of the display. You’ll be requested to log in at this level. In case you 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 might be logged in, you can be taken to a display displaying a well-recognized chat-based window that permits you to work together with and ask questions of Sonnet 4.5.
Earlier than I requested the mannequin to offer a Python program to unravel 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 must rely all potential squares of various
sizes that may be fashioned by connecting the dots.
Let me first establish the grid construction. The dots type:
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 am going to 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 just 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 in regards to the mannequin’s reference to a 3×3 sq. (which is unattainable on this grid), I requested it to indicate me the sq. on the diagram. It replied with this.
Trying on the grid extra rigorously, I must 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 verify 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, trying 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 must right my reply: There's NO 3×3 sq. potential on this grid.
The full 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 essentially, it didn’t accurately say there have been 21 squares in complete.
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 really like a Python program that plots out all of the squares we will draw on
the hooked up 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 aspect 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 mixtures
from collections import defaultdict
# Outline the grid factors primarily based 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 type a sq. and return aspect 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.kind()
# For a sq.: 4 equal sides and a pair of equal diagonals
# Diagonals ought to be sqrt(2) instances the aspect 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 = aspect * 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.kind(key=angle)
return factors
# Discover all squares
squares = []
squares_by_size = defaultdict(record)
for combo in mixtures(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"Aspect 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., shade 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=shade, 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., shade 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=shade, 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 aspect size {side_length} ({len(squares_of_size)} complete)',
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 approach Sonnet may have provide you with an accurate answer 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 have been additionally output.
==================================================
SUMMARY OF SQUARES FOUND
==================================================
Aspect size 1.0: 9 sq.(s)
Aspect size 1.41: 4 sq.(s)
Aspect size 2.24: 2 sq.(s)
Aspect size 2.83: 4 sq.(s)
Aspect size 3.61: 2 sq.(s)
Complete squares: 21
==================================================
#
# Plus some graphs that I am not exhibiting 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. aspect order. Can also you will have two graphs
aspect by aspect on every "line"That is what it produced.
Stunning.
Abstract
To exhibit simply how dramatically LLMs have superior in a few 12 months, I made a decision to revisit a difficult geometric puzzle I first tried to unravel with GPT-4o again in June 2024. The puzzle was to jot down a Python program that finds and plots all potential squares on a selected cross-shaped grid of dots.
My expertise a little bit over a 12 months in the past was a wrestle; it took me roughly two hours and over 40 prompts to information GPT-4o to an accurate Python answer.
Quick ahead to right this moment, and I examined the brand new Claude Sonnet 4.5. Once I first requested the mannequin the query immediately, it didn’t calculate the proper variety of squares. Not an ideal begin, nonetheless, the actual check was giving it the very same immediate I used on GPT-4o.
To my shock, it produced a whole, right Python answer in one shot. The code it generated not solely discovered all 21 squares but additionally accurately categorised them by their distinctive aspect 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 it’s that the very act of my making an attempt to unravel this puzzle final 12 months and publishing my findings launched it to the web-o-sphere, which means Anthropic have merely crawled it and included it into their mannequin data base? Sure, I suppose that could possibly be it, however then why couldn’t the mannequin reply the primary direct query I requested it in regards to the complete 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 wrestle with the main mannequin of its time 16 months in the past is now a five-second, one-shot success with the main mannequin right this moment.
