Choose one decisive position—such as the first run, a peak location, or the first forced tile—so every arrangement enters exactly one case.
Runs can overlap, a page pattern can begin partway through a block, and one arrangement may contain several interesting positions. Classify by a canonical feature—usually the first decisive position.
The string 11111000 contains a four-1 block beginning at positions 1 and 2. Counting it in both start cases would duplicate it.
Assign the same string to the case “the first block 1111 begins at position 1.” It now has one label.
Give an outcome one decisive position or feature, so it cannot appear in two rows.
Build a valid pattern before checking. A run beginning at either end leaves four possible places for the separate hit. Each of the four interior starts leaves three: adjacent positions would extend the run to four. Total: 4 + 3 + 3 + 3 + 3 + 4 = 20.
Click a shot to switch between miss and hit.
At the two ends, the fourth hit has four legal places. In each interior case, it has three.
After placing the triple, reject either adjacent position for the separate hit. At an end, only one adjacent position exists.
The worked example uses eight digits made from 1 and 2. Here we rename every 2 as 0. This is a reversible relabeling, so the count stays the same. These are strings, not eight-digit numbers: a leading 0 is allowed. Assign each string to its first four-1 window, even when a longer run creates more windows.
Assign a string to its first matching run. A longer run still has only one first position.
The largest number, 5, must be at a peak: position 2 or position 4. If 5 is at position 2, choose the number at position 1 in four ways. Of the three remaining numbers, put the largest at position 4; the other two can occupy positions 3 and 5 in two orders. This gives 4 × 2 = 8. Reversing the row gives eight more with 5 at position 4. The two cases cannot overlap, so 8 + 8 = 16.
The page checks all 5!=120 permutations and keeps only those with both required peaks.
The largest number 5 must occupy position 2 or position 4. These are your two main cases.
For 1–6 with peaks exactly at 2 and 5, split the positions into triples (1,2,3) and (4,5,6). Choose the first triple’s three values in 20 ways. Each triple must put its largest value in the middle, with its other two values in either order. This gives 20 × 2 × 2 = 80. Positions 3 and 4 cannot be peaks because each has a larger neighbor at 2 or 5.
Load “1–6, peaks 2 and 5.”
A peak cannot be an endpoint. Adjacent positions cannot both exceed each other.
Use complete groups of four pages first, then examine the remaining pages for each starting phase.
Draw a horizontal dividing line above the bottom 2 × 2 block. Either no domino crosses it, or both columns have a crossing domino. One crossing alone would leave an odd number of cells in the bottom block.
Let T(n) count a 2 × n rectangle. At its left edge, use one vertical domino, leaving T(n−1), or two horizontal dominoes, leaving T(n−2). Thus T(n)=T(n−1)+T(n−2). Start with T(0)=1 (nothing left to fill) and T(1)=1.
| n | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| T(n) | 1 | 1 | 2 | 3 | 5 | 8 | 13 |
No crossing: 13 × 2 = 26. Two crossings: the bottom row is forced horizontal, and the two top-right cells are also forced horizontal. Only a 2 × 4 rectangle remains: 5 ways. Total: 26 + 5 = 31.
For the first-cell split, horizontal first gives 2 × T(4) + T(2) = 12; vertical first gives 2 × T(5) + T(3) = 19. In each expression, the first term has no crossing and the second has two crossings. These disjoint cases again give 31.
Cover the fixed L-shaped board with 1×2 dominoes, each covering two edge-adjacent cells with no overlap or gap. Count placements on this fixed board; do not merge placements by rotating or reflecting them. Matching numbers in the drawing label the two cells of one domino. The upper-left cell must pair either horizontally or vertically—two disjoint, exhaustive first cases.
tilings
tilings
The enumeration always fills the first uncovered cell, so no tiling is repeated or omitted.
At the joint, zero or two dominoes cross. Count the remaining rectangles in each case.
Eight objects have different positive integer weights, each at most 15 g. First split all eight objects into two groups of four and keep the group with greater total weight. Split those four into two pairs and keep the heavier pair. Compare that pair and keep the heavier object. No comparison is tied. The final object is actually the fifth-heaviest.
Label the weights a₁ > a₂ > … > a₈. Let x=a₅ be the final winner. Its winning group of four must contain exactly one heavier object H: with none, it loses to the four largest; with two or more, x and its lighter final opponent cannot beat the other pair. Call the two lighter group members y > z. The best pair containing x is x+y, so survival requires x+y > H+z. Let t be the remaining lighter object and S the sum of the other three heavier objects.
The weights are 15, 14, 13, 12, 11, 10, 5, 1. Check: 15+11+10+5=41 > 14+13+12+1=40; 11+10=21 > 15+5=20; 11>10. These inequalities prove both necessity and possibility.
Call the fifth-heaviest weight x. Compare the best pair containing x with the other pair, then combine this with the first comparison.
Correct all eight answers to complete the workshop: 20; 48; 16; 80; 72; 2; 31; 11. Count first run positions, split by the largest peak, and include each permitted starting phase. The two illustration totals at 99 pages are 74 and 75.
Answers: 6; 20; 16; 31; 5. Run starts are positions 1–6. Shooting cases total 4+3+3+3+3+4. The peak cases give 8+8. The fixed-board tilings give 12+19. The weight proof forces the eight weights 15,14,13,12,11,10,5,1, so the second-lightest is 5.
This certifies that the learner built complete, non-overlapping classifications for runs, peaks, fragments, tilings, and comparison trees.