GSV_Empiricist
@gsv-empiricist.bsky.social
Love math and programming (paid for C# but Python is my guilty pleasure, still trying to grok Prolog).
Smacked a time or two with the autism stick, don't get offended if I insist on calling your foot-powered digging implement a spade.
Smacked a time or two with the autism stick, don't get offended if I insist on calling your foot-powered digging implement a spade.
Doing a four-way (set-inclusion) test, we can solve this by only considering eight possibilities:
This gives only one solution,
a = 6
b = 9
c = -3
d = -25
e = 5
f = -1
g = -7
h = -2
i = 8 ✅
This gives only one solution,
a = 6
b = 9
c = -3
d = -25
e = 5
f = -1
g = -7
h = -2
i = 8 ✅
May 29, 2025 at 6:47 PM
Doing a four-way (set-inclusion) test, we can solve this by only considering eight possibilities:
This gives only one solution,
a = 6
b = 9
c = -3
d = -25
e = 5
f = -1
g = -7
h = -2
i = 8 ✅
This gives only one solution,
a = 6
b = 9
c = -3
d = -25
e = 5
f = -1
g = -7
h = -2
i = 8 ✅
Ignoring negatives for the moment, factorization and comparison gets us two base solutions:
1/
1/
May 29, 2025 at 5:57 PM
Ignoring negatives for the moment, factorization and comparison gets us two base solutions:
1/
1/
Initial analysis,
February 12, 2025 at 12:34 PM
Initial analysis,
There is already a 1,
so 2^d != 1
so d > 0
so d = 1 and c = 0
then f > 1
so f = 2 and e = 0
then a > 2
so a = 3 and b = 0
which gives us a single solution,
so 2^d != 1
so d > 0
so d = 1 and c = 0
then f > 1
so f = 2 and e = 0
then a > 2
so a = 3 and b = 0
which gives us a single solution,
February 5, 2025 at 5:41 PM
There is already a 1,
so 2^d != 1
so d > 0
so d = 1 and c = 0
then f > 1
so f = 2 and e = 0
then a > 2
so a = 3 and b = 0
which gives us a single solution,
so 2^d != 1
so d > 0
so d = 1 and c = 0
then f > 1
so f = 2 and e = 0
then a > 2
so a = 3 and b = 0
which gives us a single solution,
Factoring and distributing factors get us this far by forced moves.
Now we have a constraints problem where
a..f in Z >= 0
a + b = 3
c + d = 1
e + f = 2
a + c + e = 3
b + d + f = 3
1/
Now we have a constraints problem where
a..f in Z >= 0
a + b = 3
c + d = 1
e + f = 2
a + c + e = 3
b + d + f = 3
1/
February 5, 2025 at 5:37 PM
Factoring and distributing factors get us this far by forced moves.
Now we have a constraints problem where
a..f in Z >= 0
a + b = 3
c + d = 1
e + f = 2
a + c + e = 3
b + d + f = 3
1/
Now we have a constraints problem where
a..f in Z >= 0
a + b = 3
c + d = 1
e + f = 2
a + c + e = 3
b + d + f = 3
1/
Adding those constraints collapses the problem to a single solution,
February 1, 2025 at 3:28 PM
Adding those constraints collapses the problem to a single solution,
Which gets us to
February 1, 2025 at 3:11 PM
Which gets us to
Propagating these new constraints we get
February 1, 2025 at 2:58 PM
Propagating these new constraints we get
Of note:
column 30 must contain 1 and 9 as its square values
column 475 must contain 25 and 225 as its square values and 75 and 150 as its non-squares
row 130 must contain 9
row 258 must contain 225
That gets us to
column 30 must contain 1 and 9 as its square values
column 475 must contain 25 and 225 as its square values and 75 and 150 as its non-squares
row 130 must contain 9
row 258 must contain 225
That gets us to
February 1, 2025 at 2:30 PM
Of note:
column 30 must contain 1 and 9 as its square values
column 475 must contain 25 and 225 as its square values and 75 and 150 as its non-squares
row 130 must contain 9
row 258 must contain 225
That gets us to
column 30 must contain 1 and 9 as its square values
column 475 must contain 25 and 225 as its square values and 75 and 150 as its non-squares
row 130 must contain 9
row 258 must contain 225
That gets us to
Using that basis to generate combinations, and requiring that grey cells be square, gets us to
February 1, 2025 at 2:06 PM
Using that basis to generate combinations, and requiring that grey cells be square, gets us to
Quick analysis:
January 28, 2025 at 2:16 PM
Quick analysis:
I wrote a Python script which finds all left-truncatable primes in about 3 seconds on my cellphone.
January 26, 2025 at 3:27 AM
I wrote a Python script which finds all left-truncatable primes in about 3 seconds on my cellphone.
And I'll leave the final solution as an exercise to the reader. 🫡
January 24, 2025 at 2:10 PM
And I'll leave the final solution as an exercise to the reader. 🫡
Adding constraints for those gets us to:
January 24, 2025 at 1:55 PM
Adding constraints for those gets us to:
We now have single combinations for row 232 and column 32,
January 24, 2025 at 1:17 PM
We now have single combinations for row 232 and column 32,
Propagating constraints from that greatly reduces the number of possibilities:
January 24, 2025 at 1:09 PM
Propagating constraints from that greatly reduces the number of possibilities:
Looking at column 178:
The second combination is not possible, because there are no values which can intersect with row 28 (56 and 80 are both too large).
So column 178 must be {1, 35} in the grey cells, 2 in the bottom cell, 140 in the top cell.
The second combination is not possible, because there are no values which can intersect with row 28 (56 and 80 are both too large).
So column 178 must be {1, 35} in the grey cells, 2 in the bottom cell, 140 in the top cell.
January 24, 2025 at 12:57 PM
Looking at column 178:
The second combination is not possible, because there are no values which can intersect with row 28 (56 and 80 are both too large).
So column 178 must be {1, 35} in the grey cells, 2 in the bottom cell, 140 in the top cell.
The second combination is not possible, because there are no values which can intersect with row 28 (56 and 80 are both too large).
So column 178 must be {1, 35} in the grey cells, 2 in the bottom cell, 140 in the top cell.
So here's my initial analysis of possible combinations for each row and column.
The only constraints so far are
1. combinations of 4 unique factors of 560
2. sum to required values
3. grey cells are odd
1/
The only constraints so far are
1. combinations of 4 unique factors of 560
2. sum to required values
3. grey cells are odd
1/
January 24, 2025 at 12:43 PM
So here's my initial analysis of possible combinations for each row and column.
The only constraints so far are
1. combinations of 4 unique factors of 560
2. sum to required values
3. grey cells are odd
1/
The only constraints so far are
1. combinations of 4 unique factors of 560
2. sum to required values
3. grey cells are odd
1/
A non-spoiler-y head start. 👍
January 22, 2025 at 3:54 PM
A non-spoiler-y head start. 👍
In the interests of not spoiling it, I won't show the final solution; but this gets you 90% of the way there. 👍
January 20, 2025 at 1:46 PM
In the interests of not spoiling it, I won't show the final solution; but this gets you 90% of the way there. 👍
Plan:
I will represent the board as an 8×7 grid where each cell is a boolean "occupied" flag.
That gives a total of 56 bits, so can map to a single uint64 value.
Using an extra column along the right edge lets me pre-mark it as already-occupied,
1/
I will represent the board as an 8×7 grid where each cell is a boolean "occupied" flag.
That gives a total of 56 bits, so can map to a single uint64 value.
Using an extra column along the right edge lets me pre-mark it as already-occupied,
1/
January 16, 2025 at 5:48 AM
Plan:
I will represent the board as an 8×7 grid where each cell is a boolean "occupied" flag.
That gives a total of 56 bits, so can map to a single uint64 value.
Using an extra column along the right edge lets me pre-mark it as already-occupied,
1/
I will represent the board as an 8×7 grid where each cell is a boolean "occupied" flag.
That gives a total of 56 bits, so can map to a single uint64 value.
Using an extra column along the right edge lets me pre-mark it as already-occupied,
1/
3/5 (780 - v) + 2/3 v = 780 - 300
9 (780 - v) + 10v = 15 • 480
7020 + v = 7200
v = 180
(1 - 2/3) 180 = 60 👈
Erika had $600 and spent $360
Veronica had $180 and spent $120
Veronica has $60 remaining. 👍
9 (780 - v) + 10v = 15 • 480
7020 + v = 7200
v = 180
(1 - 2/3) 180 = 60 👈
Erika had $600 and spent $360
Veronica had $180 and spent $120
Veronica has $60 remaining. 👍
January 16, 2025 at 5:04 AM
3/5 (780 - v) + 2/3 v = 780 - 300
9 (780 - v) + 10v = 15 • 480
7020 + v = 7200
v = 180
(1 - 2/3) 180 = 60 👈
Erika had $600 and spent $360
Veronica had $180 and spent $120
Veronica has $60 remaining. 👍
9 (780 - v) + 10v = 15 • 480
7020 + v = 7200
v = 180
(1 - 2/3) 180 = 60 👈
Erika had $600 and spent $360
Veronica had $180 and spent $120
Veronica has $60 remaining. 👍
Let g be the number of girls
where g in Z | 0 <= g <= 130
then 130 - g is the number of boys
(130 - g)/3 + 3g/7 = 50
7(130 - g) + 9g = 21 • 50
910 + 2g = 1050
g = (1050 - 910)/2 = 70
3/7 (70) = 30 👈
So 30 girls (out of 70) prefer fiction. 👍
where g in Z | 0 <= g <= 130
then 130 - g is the number of boys
(130 - g)/3 + 3g/7 = 50
7(130 - g) + 9g = 21 • 50
910 + 2g = 1050
g = (1050 - 910)/2 = 70
3/7 (70) = 30 👈
So 30 girls (out of 70) prefer fiction. 👍
January 16, 2025 at 4:54 AM
Let g be the number of girls
where g in Z | 0 <= g <= 130
then 130 - g is the number of boys
(130 - g)/3 + 3g/7 = 50
7(130 - g) + 9g = 21 • 50
910 + 2g = 1050
g = (1050 - 910)/2 = 70
3/7 (70) = 30 👈
So 30 girls (out of 70) prefer fiction. 👍
where g in Z | 0 <= g <= 130
then 130 - g is the number of boys
(130 - g)/3 + 3g/7 = 50
7(130 - g) + 9g = 21 • 50
910 + 2g = 1050
g = (1050 - 910)/2 = 70
3/7 (70) = 30 👈
So 30 girls (out of 70) prefer fiction. 👍