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								---
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								id: 5900f4021000cf542c50ff13
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								title: 'Problem 149: Searching for a maximum-sum subsequence'
							 
						 
					
						
							
								
									
										
										
										
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								challengeType: 5
							 
						 
					
						
							
								
									
										
										
										
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								forumTopicId: 301778
							 
						 
					
						
							
								
									
										
										
										
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								dashedName: problem-149-searching-for-a-maximum-sum-subsequence
							 
						 
					
						
							
								
									
										
										
										
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								---
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								
							 
						 
					
						
							
								
									
										
										
										
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								# --description--
  
						 
					
						
							
								
									
										
										
										
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								Looking at the table below, it is easy to verify that the maximum possible sum of adjacent numbers in any direction (horizontal, vertical, diagonal or anti-diagonal) is $16 (= 8 + 7 + 1)$.
							 
						 
					
						
							
								
									
										
										
										
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								$$\begin{array}{|r|r|r|r|}
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								  \hline
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
										 
							
							
								  − &   5 &   3 &  2 \\\\ \hline
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
										 
							
							
								   9 &  − &   5 &  1 \\\\ \hline
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								   3 &   2 &   7 &  3 \\\\ \hline
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
										 
							
							
								  − &   8 &  − &  8 \\\\ \hline
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								\end{array}$$
							 
						 
					
						
							
								
									
										
										
										
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								Now, let us repeat the search, but on a much larger scale:
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								First, generate four million pseudo-random numbers using a specific form of what is known as a "Lagged Fibonacci Generator":
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								
							 
						 
					
						
							
								
									
										
										
										
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								For $1 ≤ k ≤ 55$, $s_k = (100003 − − 
							 
						 
					
						
							
								
									
										
										
										
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								For $56 ≤ k ≤ 4000000$, $s_k = (s_{k − − − 
							 
						 
					
						
							
								
									
										
										
										
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								Thus, $s_{10} = − 
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
										 
							
							
								The terms of $s$ are then arranged in a 2000× 
							 
						 
					
						
							
								
									
										
										
										
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								Finally, find the greatest sum of (any number of) adjacent entries in any direction (horizontal, vertical, diagonal or anti-diagonal).
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								
							 
						 
					
						
							
								
									
										
										
										
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								# --hints--
  
						 
					
						
							
								
									
										
										
										
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								`maximumSubSequence()`  should return `52852124` . 
						 
					
						
							
								
									
										
										
										
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								```js
							 
						 
					
						
							
								
									
										
										
										
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								assert.strictEqual(maximumSubSequence(), 52852124);
							 
						 
					
						
							
								
									
										
										
										
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								```
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								
							 
						 
					
						
							
								
									
										
										
										
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								# --seed--
  
						 
					
						
							
								
									
										
										
										
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								## --seed-contents--
  
						 
					
						
							
								
									
										
										
										
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								```js
							 
						 
					
						
							
								
									
										
										
										
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								function maximumSubSequence() {
							 
						 
					
						
							
								
									
										
										
										
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								  return true;
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								}
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								
							 
						 
					
						
							
								
									
										
										
										
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								maximumSubSequence();
							 
						 
					
						
							
								
									
										
										
										
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								```
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								
							 
						 
					
						
							
								
									
										
										
										
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								# --solutions--
  
						 
					
						
							
								
									
										
										
										
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								```js
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								// solution required
							 
						 
					
						
							
								
							 
							
								
							 
							
								 
							
								
									
								 
							
							
								```