Performance: 80% faster Project Euler 145#10445

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cclauss merged 3 commits into from
Oct 14, 2023
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Performance: 80% faster Project Euler 145 #10445

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Performance: 80% faster Project Euler145
ManpreetXSingh Oct 14, 2023
bf77690
Added timeit benchmark
ManpreetXSingh Oct 14, 2023
c1dfb5a
>>> slow_solution() doctest
ManpreetXSingh Oct 14, 2023
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70 changes: 63 additions & 7 deletions project_euler/problem_145/sol1.py
View file
Open in desktop
Original file line numberDiff line numberDiff line change
Expand Up@@ -17,17 +17,17 @@
ODD_DIGITS = [1, 3, 5, 7, 9]


def reversible_numbers(
def slow_reversible_numbers(
remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
"""
Count the number of reversible numbers of given length.
Iterate over possible digits considering parity of current sum remainder.
>>> reversible_numbers(1, 0, [0], 1)
>>> slow_reversible_numbers(1, 0, [0], 1)
0
>>> reversible_numbers(2, 0, [0] * 2, 2)
>>> slow_reversible_numbers(2, 0, [0] * 2, 2)
20
>>> reversible_numbers(3, 0, [0] * 3, 3)
>>> slow_reversible_numbers(3, 0, [0] * 3, 3)
100
"""
if remaining_length == 0:
Expand All@@ -51,7 +51,7 @@ def reversible_numbers(
result = 0
for digit in range(10):
digits[length // 2] = digit
result += reversible_numbers(
result += slow_reversible_numbers(
0, (remainder + 2 * digit) // 10, digits, length
)
return result
Expand All@@ -67,7 +67,7 @@ def reversible_numbers(

for digit2 in other_parity_digits:
digits[(length - remaining_length) // 2] = digit2
result += reversible_numbers(
result += slow_reversible_numbers(
remaining_length - 2,
(remainder + digit1 + digit2) // 10,
digits,
Expand All@@ -76,6 +76,42 @@ def reversible_numbers(
return result


def slow_solution(max_power: int = 9) -> int:
"""
To evaluate the solution, use solution()
>>> slow_solution(3)
120
>>> slow_solution(6)
18720
>>> slow_solution(7)
68720
"""
result = 0
for length in range(1, max_power + 1):
result += slow_reversible_numbers(length, 0, [0] * length, length)
return result


def reversible_numbers(
remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
"""
Count the number of reversible numbers of given length.
Iterate over possible digits considering parity of current sum remainder.
>>> reversible_numbers(1, 0, [0], 1)
0
>>> reversible_numbers(2, 0, [0] * 2, 2)
20
>>> reversible_numbers(3, 0, [0] * 3, 3)
100
"""
# There exist no reversible 1, 5, 9, 13 (ie. 4k+1) digit numbers
if (length - 1) % 4 == 0:
return 0

return slow_reversible_numbers(length, 0, [0] * length, length)


def solution(max_power: int = 9) -> int:
"""
To evaluate the solution, use solution()
Expand All@@ -92,5 +128,25 @@ def solution(max_power: int = 9) -> int:
return result


def benchmark() -> None:
"""
Benchmarks
"""
# Running performance benchmarks...
# slow_solution : 292.9300301000003
# solution : 54.90970860000016

from timeit import timeit

print("Running performance benchmarks...")

print(f"slow_solution : {timeit('slow_solution()', globals=globals(), number=10)}")
print(f"solution : {timeit('solution()', globals=globals(), number=10)}")


if __name__ == "__main__":
print(f"{solution() = }")
print(f"Solution : {solution()}")
benchmark()

# for i in range(1, 15):
# print(f"{i}. {reversible_numbers(i, 0, [0]*i, i)}")
Comment thread
ManpreetXSingh marked this conversation as resolved.
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Performance: 80% faster Project Euler 145#10445

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Oct 14, 2023
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Performance: 80% faster Project Euler 145
#10445
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0f75804

Performance: 80% faster Project Euler145


ManpreetXSingh Oct 14, 2023
bf77690

Added timeit benchmark


ManpreetXSingh Oct 14, 2023
c1dfb5a

>>> slow_solution() doctest


ManpreetXSingh Oct 14, 2023

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70 changes: 63 additions & 7 deletions
project_euler/problem_145/sol1.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
Expand Up@@ -17,17 +17,17 @@
ODD_DIGITS = [1, 3, 5, 7, 9]


def reversible_numbers(
def slow_reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 >>> slow_reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 >>> slow_reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 >>> slow_reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 if remaining_length == 0:
Expand All@@ -51,7 +51,7 @@ def reversible_numbers(
 result = 0
 for digit in range(10):
 digits[length // 2] = digit
 result += reversible_numbers(
 result += slow_reversible_numbers(
 0, (remainder + 2 * digit) // 10, digits, length
 )
 return result
Expand All@@ -67,7 +67,7 @@ def reversible_numbers(

 for digit2 in other_parity_digits:
 digits[(length - remaining_length) // 2] = digit2
 result += reversible_numbers(
 result += slow_reversible_numbers(
 remaining_length - 2,
 (remainder + digit1 + digit2) // 10,
 digits,
Expand All@@ -76,6 +76,42 @@ def reversible_numbers(
 return result


def slow_solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
 >>> slow_solution(3)
 120
 >>> slow_solution(6)
 18720
 >>> slow_solution(7)
 68720
 """
 result = 0
 for length in range(1, max_power + 1):
 result += slow_reversible_numbers(length, 0, [0] * length, length)
 return result


def reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 # There exist no reversible 1, 5, 9, 13 (ie. 4k+1) digit numbers
 if (length - 1) % 4 == 0:
 return 0

 return slow_reversible_numbers(length, 0, [0] * length, length)


def solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
Expand All@@ -92,5 +128,25 @@ def solution(max_power: int = 9) -> int:
 return result


def benchmark() -> None:
 """
 Benchmarks
 """
 # Running performance benchmarks...
 # slow_solution : 292.9300301000003
 # solution : 54.90970860000016

 from timeit import timeit

 print("Running performance benchmarks...")

 print(f"slow_solution : {timeit('slow_solution()', globals=globals(), number=10)}")
 print(f"solution : {timeit('solution()', globals=globals(), number=10)}")


if __name__ == "__main__":
 print(f"{solution() = }")
 print(f"Solution : {solution()}")
 benchmark()

 # for i in range(1, 15):
 # print(f"{i}. {reversible_numbers(i, 0, [0]*i, i)}")
Comment thread
ManpreetXSingh marked this conversation as resolved.


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if (__m === '*' || __re.test(location.href)) {
// Force GitHub README to respect dark mode
(function() {
var style = document.createElement('style');
style.textContent = '
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color-scheme: dark light;
}
.markdown-body pre { background: #161b22 !important; }
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.markdown-body hr { border-color: #30363d; }
';
document.head.appendChild(style);
})();
}
} catch(__e) { console.warn('[Userscript:GitHub Dark Mode README Fix]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

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Performance: 80% faster Project Euler 145#10445

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 Merged

cclauss
merged 3 commits into

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Oct 14, 2023
 Merged


Performance: 80% faster Project Euler 145
#10445
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0f75804

Performance: 80% faster Project Euler145


ManpreetXSingh Oct 14, 2023
bf77690

Added timeit benchmark


ManpreetXSingh Oct 14, 2023
c1dfb5a

>>> slow_solution() doctest


ManpreetXSingh Oct 14, 2023

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70 changes: 63 additions & 7 deletions
project_euler/problem_145/sol1.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
Expand Up@@ -17,17 +17,17 @@
ODD_DIGITS = [1, 3, 5, 7, 9]


def reversible_numbers(
def slow_reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 >>> slow_reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 >>> slow_reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 >>> slow_reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 if remaining_length == 0:
Expand All@@ -51,7 +51,7 @@ def reversible_numbers(
 result = 0
 for digit in range(10):
 digits[length // 2] = digit
 result += reversible_numbers(
 result += slow_reversible_numbers(
 0, (remainder + 2 * digit) // 10, digits, length
 )
 return result
Expand All@@ -67,7 +67,7 @@ def reversible_numbers(

 for digit2 in other_parity_digits:
 digits[(length - remaining_length) // 2] = digit2
 result += reversible_numbers(
 result += slow_reversible_numbers(
 remaining_length - 2,
 (remainder + digit1 + digit2) // 10,
 digits,
Expand All@@ -76,6 +76,42 @@ def reversible_numbers(
 return result


def slow_solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
 >>> slow_solution(3)
 120
 >>> slow_solution(6)
 18720
 >>> slow_solution(7)
 68720
 """
 result = 0
 for length in range(1, max_power + 1):
 result += slow_reversible_numbers(length, 0, [0] * length, length)
 return result


def reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 # There exist no reversible 1, 5, 9, 13 (ie. 4k+1) digit numbers
 if (length - 1) % 4 == 0:
 return 0

 return slow_reversible_numbers(length, 0, [0] * length, length)


def solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
Expand All@@ -92,5 +128,25 @@ def solution(max_power: int = 9) -> int:
 return result


def benchmark() -> None:
 """
 Benchmarks
 """
 # Running performance benchmarks...
 # slow_solution : 292.9300301000003
 # solution : 54.90970860000016

 from timeit import timeit

 print("Running performance benchmarks...")

 print(f"slow_solution : {timeit('slow_solution()', globals=globals(), number=10)}")
 print(f"solution : {timeit('solution()', globals=globals(), number=10)}")


if __name__ == "__main__":
 print(f"{solution() = }")
 print(f"Solution : {solution()}")
 benchmark()

 # for i in range(1, 15):
 # print(f"{i}. {reversible_numbers(i, 0, [0]*i, i)}")
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ManpreetXSingh marked this conversation as resolved.


, 'i');
if (__m === '*' || __re.test(location.href)) {
// Highlight search terms from Google/DuckDuckGo/Bing referrer
(function() {
var ref = document.referrer;
var terms = [];
if (ref.includes('google.com') || ref.includes('duckduckgo.com') || ref.includes('bing.com')) {
var url = new URL(ref);
var q = url.searchParams.get('q') || url.searchParams.get('p');
if (q) {
terms = q.split(/\s+/).filter(function(t) { return t.length > 2; });
}
}
if (terms.length === 0) return;
var style = document.createElement('style');
style.textContent = '.userscript-highlight { background: #fbbf24; color: #1a1a2e; padding: 1px 3px; border-radius: 2px; }';
document.head.appendChild(style);
function highlight(node) {
if (node.nodeType === 3) { // text node
var text = node.textContent;
var found = false;
terms.forEach(function(term) {
var regex = new RegExp('(' + term.replace(/[.*+?^${}()|[\]\\]/g, '\\') + ')', 'gi');
if (regex.test(text)) {
found = true;
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parts.forEach(function(part, i) {
if (i % 2 === 0) {
frag.appendChild(document.createTextNode(part));
} else {
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frag.appendChild(span);
}
});
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}
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if (!skipTags.includes(node.tagName)) {
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}
}
}
highlight(document.body);
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mutations.forEach(function(m) {
m.addedNodes.forEach(function(node) {
if (node.nodeType === 1 || node.nodeType === 3) highlight(node);
});
});
});
observer.observe(document.body, { childList: true, subtree: true });
})();
}
} catch(__e) { console.warn('[Userscript:Highlight Search Terms]', __e); }
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Performance: 80% faster Project Euler 145#10445

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cclauss
merged 3 commits into

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Oct 14, 2023
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Performance: 80% faster Project Euler 145
#10445
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0f75804

Performance: 80% faster Project Euler145


ManpreetXSingh Oct 14, 2023
bf77690

Added timeit benchmark


ManpreetXSingh Oct 14, 2023
c1dfb5a

>>> slow_solution() doctest


ManpreetXSingh Oct 14, 2023

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70 changes: 63 additions & 7 deletions
project_euler/problem_145/sol1.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
Expand Up@@ -17,17 +17,17 @@
ODD_DIGITS = [1, 3, 5, 7, 9]


def reversible_numbers(
def slow_reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 >>> slow_reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 >>> slow_reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 >>> slow_reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 if remaining_length == 0:
Expand All@@ -51,7 +51,7 @@ def reversible_numbers(
 result = 0
 for digit in range(10):
 digits[length // 2] = digit
 result += reversible_numbers(
 result += slow_reversible_numbers(
 0, (remainder + 2 * digit) // 10, digits, length
 )
 return result
Expand All@@ -67,7 +67,7 @@ def reversible_numbers(

 for digit2 in other_parity_digits:
 digits[(length - remaining_length) // 2] = digit2
 result += reversible_numbers(
 result += slow_reversible_numbers(
 remaining_length - 2,
 (remainder + digit1 + digit2) // 10,
 digits,
Expand All@@ -76,6 +76,42 @@ def reversible_numbers(
 return result


def slow_solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
 >>> slow_solution(3)
 120
 >>> slow_solution(6)
 18720
 >>> slow_solution(7)
 68720
 """
 result = 0
 for length in range(1, max_power + 1):
 result += slow_reversible_numbers(length, 0, [0] * length, length)
 return result


def reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 # There exist no reversible 1, 5, 9, 13 (ie. 4k+1) digit numbers
 if (length - 1) % 4 == 0:
 return 0

 return slow_reversible_numbers(length, 0, [0] * length, length)


def solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
Expand All@@ -92,5 +128,25 @@ def solution(max_power: int = 9) -> int:
 return result


def benchmark() -> None:
 """
 Benchmarks
 """
 # Running performance benchmarks...
 # slow_solution : 292.9300301000003
 # solution : 54.90970860000016

 from timeit import timeit

 print("Running performance benchmarks...")

 print(f"slow_solution : {timeit('slow_solution()', globals=globals(), number=10)}")
 print(f"solution : {timeit('solution()', globals=globals(), number=10)}")


if __name__ == "__main__":
 print(f"{solution() = }")
 print(f"Solution : {solution()}")
 benchmark()

 # for i in range(1, 15):
 # print(f"{i}. {reversible_numbers(i, 0, [0]*i, i)}")
Comment thread
ManpreetXSingh marked this conversation as resolved.


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// Strip utm_, fbclid, gclid, etc. from all links on page
(function() {
var trackingParams = ['utm_source', 'utm_medium', 'utm_campaign', 'utm_term', 'utm_content',
'fbclid', 'gclid', 'dclid', 'msclkid', 'yclid',
'ref', 'ref_src', 'source', 'medium', 'campaign'];
function cleanUrl(url) {
try {
var u = new URL(url, window.location.origin);
var changed = false;
trackingParams.forEach(function(p) {
if (u.searchParams.has(p)) {
u.searchParams.delete(p);
changed = true;
}
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} catch (e) {
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}
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if (clean !== a.href) a.href = clean;
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mutations.forEach(function(m) {
m.addedNodes.forEach(function(node) {
if (node.nodeType === 1) {
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node.querySelectorAll('a[href]').forEach(function(a) {
var clean = cleanUrl(a.href);
if (clean !== a.href) a.href = clean;
});
}
});
});
});
observer.observe(document.body, { childList: true, subtree: true });
})();
}
} catch(__e) { console.warn('[Userscript:Remove Tracking Parameters from Links]', __e); }
})();
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try {
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var __re = new RegExp('^' + "youtube\\.com" + '

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Oct 14, 2023
 Merged


Performance: 80% faster Project Euler 145
#10445
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0f75804

Performance: 80% faster Project Euler145


ManpreetXSingh Oct 14, 2023
bf77690

Added timeit benchmark


ManpreetXSingh Oct 14, 2023
c1dfb5a

>>> slow_solution() doctest


ManpreetXSingh Oct 14, 2023

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70 changes: 63 additions & 7 deletions
project_euler/problem_145/sol1.py

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Open in desktop

Original file line numberDiff line numberDiff line change
Expand Up@@ -17,17 +17,17 @@
ODD_DIGITS = [1, 3, 5, 7, 9]


def reversible_numbers(
def slow_reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 >>> slow_reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 >>> slow_reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 >>> slow_reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 if remaining_length == 0:
Expand All@@ -51,7 +51,7 @@ def reversible_numbers(
 result = 0
 for digit in range(10):
 digits[length // 2] = digit
 result += reversible_numbers(
 result += slow_reversible_numbers(
 0, (remainder + 2 * digit) // 10, digits, length
 )
 return result
Expand All@@ -67,7 +67,7 @@ def reversible_numbers(

 for digit2 in other_parity_digits:
 digits[(length - remaining_length) // 2] = digit2
 result += reversible_numbers(
 result += slow_reversible_numbers(
 remaining_length - 2,
 (remainder + digit1 + digit2) // 10,
 digits,
Expand All@@ -76,6 +76,42 @@ def reversible_numbers(
 return result


def slow_solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
 >>> slow_solution(3)
 120
 >>> slow_solution(6)
 18720
 >>> slow_solution(7)
 68720
 """
 result = 0
 for length in range(1, max_power + 1):
 result += slow_reversible_numbers(length, 0, [0] * length, length)
 return result


def reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 # There exist no reversible 1, 5, 9, 13 (ie. 4k+1) digit numbers
 if (length - 1) % 4 == 0:
 return 0

 return slow_reversible_numbers(length, 0, [0] * length, length)


def solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
Expand All@@ -92,5 +128,25 @@ def solution(max_power: int = 9) -> int:
 return result


def benchmark() -> None:
 """
 Benchmarks
 """
 # Running performance benchmarks...
 # slow_solution : 292.9300301000003
 # solution : 54.90970860000016

 from timeit import timeit

 print("Running performance benchmarks...")

 print(f"slow_solution : {timeit('slow_solution()', globals=globals(), number=10)}")
 print(f"solution : {timeit('solution()', globals=globals(), number=10)}")


if __name__ == "__main__":
 print(f"{solution() = }")
 print(f"Solution : {solution()}")
 benchmark()

 # for i in range(1, 15):
 # print(f"{i}. {reversible_numbers(i, 0, [0]*i, i)}")
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Performance: 80% faster Project Euler 145
#10445
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0f75804

Performance: 80% faster Project Euler145


ManpreetXSingh Oct 14, 2023
bf77690

Added timeit benchmark


ManpreetXSingh Oct 14, 2023
c1dfb5a

>>> slow_solution() doctest


ManpreetXSingh Oct 14, 2023

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70 changes: 63 additions & 7 deletions
project_euler/problem_145/sol1.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
Expand Up@@ -17,17 +17,17 @@
ODD_DIGITS = [1, 3, 5, 7, 9]


def reversible_numbers(
def slow_reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 >>> slow_reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 >>> slow_reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 >>> slow_reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 if remaining_length == 0:
Expand All@@ -51,7 +51,7 @@ def reversible_numbers(
 result = 0
 for digit in range(10):
 digits[length // 2] = digit
 result += reversible_numbers(
 result += slow_reversible_numbers(
 0, (remainder + 2 * digit) // 10, digits, length
 )
 return result
Expand All@@ -67,7 +67,7 @@ def reversible_numbers(

 for digit2 in other_parity_digits:
 digits[(length - remaining_length) // 2] = digit2
 result += reversible_numbers(
 result += slow_reversible_numbers(
 remaining_length - 2,
 (remainder + digit1 + digit2) // 10,
 digits,
Expand All@@ -76,6 +76,42 @@ def reversible_numbers(
 return result


def slow_solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
 >>> slow_solution(3)
 120
 >>> slow_solution(6)
 18720
 >>> slow_solution(7)
 68720
 """
 result = 0
 for length in range(1, max_power + 1):
 result += slow_reversible_numbers(length, 0, [0] * length, length)
 return result


def reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 # There exist no reversible 1, 5, 9, 13 (ie. 4k+1) digit numbers
 if (length - 1) % 4 == 0:
 return 0

 return slow_reversible_numbers(length, 0, [0] * length, length)


def solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
Expand All@@ -92,5 +128,25 @@ def solution(max_power: int = 9) -> int:
 return result


def benchmark() -> None:
 """
 Benchmarks
 """
 # Running performance benchmarks...
 # slow_solution : 292.9300301000003
 # solution : 54.90970860000016

 from timeit import timeit

 print("Running performance benchmarks...")

 print(f"slow_solution : {timeit('slow_solution()', globals=globals(), number=10)}")
 print(f"solution : {timeit('solution()', globals=globals(), number=10)}")


if __name__ == "__main__":
 print(f"{solution() = }")
 print(f"Solution : {solution()}")
 benchmark()

 # for i in range(1, 15):
 # print(f"{i}. {reversible_numbers(i, 0, [0]*i, i)}")
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// Remove or un-stick sticky/fixed headers that block content
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function unstick() {
document.querySelectorAll('header, nav, [role="banner"], .header, .navbar, .sticky, .fixed-top, [style*="position: fixed"], [style*="position:sticky"]').forEach(function(el) {
if (el.style.position === 'fixed' || el.style.position === 'sticky' || getComputedStyle(el).position === 'fixed' || getComputedStyle(el).position === 'sticky') {
el.style.position = 'static';
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}
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observer.observe(document.body, { childList: true, subtree: true, attributes: true, attributeFilter: ['style', 'class'] });
})();
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} catch(__e) { console.warn('[Userscript:Kill Sticky Headers]', __e); }
})();
(function(){
try {
var __m = "*";
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Performance: 80% faster Project Euler 145#10445

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 Merged

cclauss
merged 3 commits into

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Oct 14, 2023
 Merged


Performance: 80% faster Project Euler 145
#10445
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0f75804

Performance: 80% faster Project Euler145


ManpreetXSingh Oct 14, 2023
bf77690

Added timeit benchmark


ManpreetXSingh Oct 14, 2023
c1dfb5a

>>> slow_solution() doctest


ManpreetXSingh Oct 14, 2023

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70 changes: 63 additions & 7 deletions
project_euler/problem_145/sol1.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
Expand Up@@ -17,17 +17,17 @@
ODD_DIGITS = [1, 3, 5, 7, 9]


def reversible_numbers(
def slow_reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 >>> slow_reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 >>> slow_reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 >>> slow_reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 if remaining_length == 0:
Expand All@@ -51,7 +51,7 @@ def reversible_numbers(
 result = 0
 for digit in range(10):
 digits[length // 2] = digit
 result += reversible_numbers(
 result += slow_reversible_numbers(
 0, (remainder + 2 * digit) // 10, digits, length
 )
 return result
Expand All@@ -67,7 +67,7 @@ def reversible_numbers(

 for digit2 in other_parity_digits:
 digits[(length - remaining_length) // 2] = digit2
 result += reversible_numbers(
 result += slow_reversible_numbers(
 remaining_length - 2,
 (remainder + digit1 + digit2) // 10,
 digits,
Expand All@@ -76,6 +76,42 @@ def reversible_numbers(
 return result


def slow_solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
 >>> slow_solution(3)
 120
 >>> slow_solution(6)
 18720
 >>> slow_solution(7)
 68720
 """
 result = 0
 for length in range(1, max_power + 1):
 result += slow_reversible_numbers(length, 0, [0] * length, length)
 return result


def reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 # There exist no reversible 1, 5, 9, 13 (ie. 4k+1) digit numbers
 if (length - 1) % 4 == 0:
 return 0

 return slow_reversible_numbers(length, 0, [0] * length, length)


def solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
Expand All@@ -92,5 +128,25 @@ def solution(max_power: int = 9) -> int:
 return result


def benchmark() -> None:
 """
 Benchmarks
 """
 # Running performance benchmarks...
 # slow_solution : 292.9300301000003
 # solution : 54.90970860000016

 from timeit import timeit

 print("Running performance benchmarks...")

 print(f"slow_solution : {timeit('slow_solution()', globals=globals(), number=10)}")
 print(f"solution : {timeit('solution()', globals=globals(), number=10)}")


if __name__ == "__main__":
 print(f"{solution() = }")
 print(f"Solution : {solution()}")
 benchmark()

 # for i in range(1, 15):
 # print(f"{i}. {reversible_numbers(i, 0, [0]*i, i)}")
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ManpreetXSingh marked this conversation as resolved.


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if (__m === '*' || __re.test(location.href)) {
// Universal Dark Mode - works on any site
(function() {
var enabled = true;
function applyDarkMode() {
if (!enabled) return;
// Create style element if it doesn't exist
var style = document.getElementById('universal-dark-mode-style');
if (!style) {
style = document.createElement('style');
style.id = 'universal-dark-mode-style';
document.head.appendChild(style);
}
// Dark mode CSS - inverts colors but preserves images/video
style.textContent = '
/* Invert everything except media */
html {
filter: invert(1) hue-rotate(180deg) !important;
background: #1a1a2e !important;
}
/* Restore images, videos, iframes, canvas */
img, video, iframe, canvas, svg, picture, [style*="background-image"] {
filter: invert(1) hue-rotate(180deg) !important;
}
/* Preserve specific elements that should not be inverted */
.no-dark-mode, .no-dark-mode *,
[data-theme="light"], [data-theme="light"],
.ace_editor, .ace_editor *,
.CodeMirror, .CodeMirror *,
.monaco-editor, .monaco-editor *,
.markdown-body pre, .markdown-body pre *,
.highlight, .highlight *,
pre code, pre code * {
filter: none !important;
}
/* Fix common UI elements */
.modal, .popup, .dropdown-menu, .tooltip, .popover {
filter: invert(1) hue-rotate(180deg) !important;
background: #2d2d44 !important;
border-color: #444 !important;
}
/* Scrollbars */
::-webkit-scrollbar { background: #1a1a2e !important; }
::-webkit-scrollbar-thumb { background: #444 !important; }
::-webkit-scrollbar-thumb:hover { background: #555 !important; }
/* Selection */
::selection { background: #4ecdc4 !important; color: #1a1a2e !important; }
::-moz-selection { background: #4ecdc4 !important; color: #1a1a2e !important; }
';
}
function removeDarkMode() {
var style = document.getElementById('universal-dark-mode-style');
if (style) style.remove();
}
// Toggle with Alt+Shift+D
document.addEventListener('keydown', function(e) {
if (e.altKey && e.shiftKey && e.key === 'D') {
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Performance: 80% faster Project Euler 145#10445

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 Merged

cclauss
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Oct 14, 2023
 Merged


Performance: 80% faster Project Euler 145
#10445
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0f75804

Performance: 80% faster Project Euler145


ManpreetXSingh Oct 14, 2023
bf77690

Added timeit benchmark


ManpreetXSingh Oct 14, 2023
c1dfb5a

>>> slow_solution() doctest


ManpreetXSingh Oct 14, 2023

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70 changes: 63 additions & 7 deletions
project_euler/problem_145/sol1.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
Expand Up@@ -17,17 +17,17 @@
ODD_DIGITS = [1, 3, 5, 7, 9]


def reversible_numbers(
def slow_reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 >>> slow_reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 >>> slow_reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 >>> slow_reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 if remaining_length == 0:
Expand All@@ -51,7 +51,7 @@ def reversible_numbers(
 result = 0
 for digit in range(10):
 digits[length // 2] = digit
 result += reversible_numbers(
 result += slow_reversible_numbers(
 0, (remainder + 2 * digit) // 10, digits, length
 )
 return result
Expand All@@ -67,7 +67,7 @@ def reversible_numbers(

 for digit2 in other_parity_digits:
 digits[(length - remaining_length) // 2] = digit2
 result += reversible_numbers(
 result += slow_reversible_numbers(
 remaining_length - 2,
 (remainder + digit1 + digit2) // 10,
 digits,
Expand All@@ -76,6 +76,42 @@ def reversible_numbers(
 return result


def slow_solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
 >>> slow_solution(3)
 120
 >>> slow_solution(6)
 18720
 >>> slow_solution(7)
 68720
 """
 result = 0
 for length in range(1, max_power + 1):
 result += slow_reversible_numbers(length, 0, [0] * length, length)
 return result


def reversible_numbers(
 remaining_length: int, remainder: int, digits: list[int], length: int
) -> int:
 """
 Count the number of reversible numbers of given length.
 Iterate over possible digits considering parity of current sum remainder.
 >>> reversible_numbers(1, 0, [0], 1)
 0
 >>> reversible_numbers(2, 0, [0] * 2, 2)
 20
 >>> reversible_numbers(3, 0, [0] * 3, 3)
 100
 """
 # There exist no reversible 1, 5, 9, 13 (ie. 4k+1) digit numbers
 if (length - 1) % 4 == 0:
 return 0

 return slow_reversible_numbers(length, 0, [0] * length, length)


def solution(max_power: int = 9) -> int:
 """
 To evaluate the solution, use solution()
Expand All@@ -92,5 +128,25 @@ def solution(max_power: int = 9) -> int:
 return result


def benchmark() -> None:
 """
 Benchmarks
 """
 # Running performance benchmarks...
 # slow_solution : 292.9300301000003
 # solution : 54.90970860000016

 from timeit import timeit

 print("Running performance benchmarks...")

 print(f"slow_solution : {timeit('slow_solution()', globals=globals(), number=10)}")
 print(f"solution : {timeit('solution()', globals=globals(), number=10)}")


if __name__ == "__main__":
 print(f"{solution() = }")
 print(f"Solution : {solution()}")
 benchmark()

 # for i in range(1, 15):
 # print(f"{i}. {reversible_numbers(i, 0, [0]*i, i)}")
Comment thread
ManpreetXSingh marked this conversation as resolved.